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653 symbol spellings in 2141 mathematical passages across 19 selected source files. Subscripts and superscripts have their own entries alongside their base symbols.
The same symbol can have several meanings here. A passage classified as a definition may define another part of its expression. Read the linked context before assigning a meaning.
All 653 entries are available below.
+Introduction passage · 201 passages
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Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
=Introduction passage · 626 passages
Search aliases: =
Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
AIntroduction passage · 35 passages
Search aliases: A
Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
Exact mathematical passage
A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
aIntroduction passage · 126 passages
Search aliases: a
Introduction passage · appendix_operator.tex, line 176
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
a_j
a^2Introduction passage · 14 passages
Search aliases: a^2
Introduction passage · chapters15_16.tex, line 153
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
Exact mathematical passage
U(a^2)=\frac{T_{2m+1}-cT_{2m-1}}a
a_jIntroduction passage · 10 passages
Search aliases: a_j
Introduction passage · appendix_operator.tex, line 176
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
a_j
a_{j+1}Introduction passage · 1 passage
Search aliases: a_j+1
Introduction passage · appendix_operator.tex, line 177
ogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its differen…
Exact mathematical passage
a_j+4(j+1)(M-j)a_{j+1}=0.
BIntroduction passage · 34 passages
Search aliases: B
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
B_0Introduction passage · 2 passages
Search aliases: B_0
Introduction passage · chapters06_07.tex, line 104
ial at each fixed shell Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime index p in a p-adic or real variable.
Exact mathematical passage
\label{eq:pn5} P_n(5)=288\tau(n),\qquad B_0(n)=288|\Vn|.
B_jIntroduction passage · 10 passages
Search aliases: B_j
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
cIntroduction passage · 49 passages
Search aliases: c
Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
Exact mathematical passage
A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
c_iIntroduction passage · 10 passages
Search aliases: c_i
Introduction passage · chapters06_07.tex, line 98
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
v_i=i+5c_i
c_jIntroduction passage · 7 passages
Search aliases: c_j
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
DIntroduction passage · 68 passages
Search aliases: D
Introduction passage · appendix_operator.tex, line 45
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
D\ne0
EIntroduction passage · 35 passages
Search aliases: E
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
E_4Introduction passage · 12 passages
Search aliases: E_4
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
E_4^2Introduction passage · 3 passages
Search aliases: E_4^2
Introduction passage · chapters01_03.tex, line 100
E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_8=E_4^2
E_4^3Introduction passage · 4 passages
Search aliases: E_4^3
Introduction passage · chapters01_03.tex, line 97
E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
\label{eq:deltaeisen} \Delta=\frac{E_4^3-E_6^2}{1728}.
E_6Introduction passage · 7 passages
Search aliases: E_6
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
E_6^2Introduction passage · 6 passages
Search aliases: E_6^2
Introduction passage · chapters01_03.tex, line 97
E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
\label{eq:deltaeisen} \Delta=\frac{E_4^3-E_6^2}{1728}.
E_8Introduction passage · 3 passages
Search aliases: E_8
Introduction passage · chapters01_03.tex, line 100
E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_8=E_4^2
fIntroduction passage · 50 passages
Search aliases: f
Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
Exact mathematical passage
A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
HIntroduction passage · 67 passages
Search aliases: H
Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
\mathcal HIntroduction passage · 2 passages
Search aliases: H mathcal
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
\mathcal H_MIntroduction passage · 1 passage
Search aliases: H_M mathcal
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
iIntroduction passage · 83 passages
Search aliases: i
Introduction passage · chapters06_07.tex, line 98
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
v_i=i+5c_i
jIntroduction passage · 117 passages
Search aliases: j
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
kIntroduction passage · 87 passages
Search aliases: k
Introduction passage · appendices.tex, line 15
The symbol T_k(p) in the trace-formula chapter denotes a Hecke operator on the weight-k cusp-form space. It is an operator, whereas T_r in a scalar recurrence is an integer or an integer polynomial. The weight k and the exponent r serve different roles.
Exact mathematical passage
T_k(p)
\lfloorIntroduction passage · 6 passages
Search aliases: lfloor
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
\logIntroduction passage · 5 passages
Search aliases: log
Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
MIntroduction passage · 83 passages
Search aliases: M
Introduction passage · appendix_operator.tex, line 171
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
M
mIntroduction passage · 112 passages
Search aliases: m
Introduction passage · chapters15_16.tex, line 142
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
Exact mathematical passage
m\ge0
M_cIntroduction passage · 6 passages
Search aliases: M_c
Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
Exact mathematical passage
A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
M_fIntroduction passage · 6 passages
Search aliases: M_f
Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
Exact mathematical passage
A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
nIntroduction passage · 155 passages
Search aliases: n
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
PIntroduction passage · 29 passages
Search aliases: P
Introduction passage · appendix_operator.tex, line 171
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
P
pIntroduction passage · 324 passages
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Introduction passage · appendices.tex, line 15
The symbol T_k(p) in the trace-formula chapter denotes a Hecke operator on the weight-k cusp-form space. It is an operator, whereas T_r in a scalar recurrence is an integer or an integer polynomial. The weight k and the exponent r serve different roles.
Exact mathematical passage
T_k(p)
p^{2m+1}Introduction passage · 3 passages
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Introduction passage · chapters15_16.tex, line 143
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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R(p^{2m+1})=\tau(p)U_{p,m}.
P_nIntroduction passage · 6 passages
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Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
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\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
QIntroduction passage · 77 passages
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Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
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\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
qIntroduction passage · 68 passages
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Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
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E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
q^nIntroduction passage · 11 passages
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Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
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E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
RIntroduction passage · 65 passages
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Introduction passage · chapters15_16.tex, line 143
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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R(p^{2m+1})=\tau(p)U_{p,m}.
rIntroduction passage · 77 passages
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Introduction passage · appendices.tex, line 15
The symbol T_k(p) in the trace-formula chapter denotes a Hecke operator on the weight-k cusp-form space. It is an operator, whereas T_r in a scalar recurrence is an integer or an integer polynomial. The weight k and the exponent r serve different roles.
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T_r
\rfloorIntroduction passage · 6 passages
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Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
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\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
sIntroduction passage · 127 passages
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Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
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\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
s_2Introduction passage · 43 passages
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Introduction passage · appendix_operator.tex, line 180
H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its difference from P is divisible by s_2.
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s_2
s_2^jIntroduction passage · 7 passages
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Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
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\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
s_{2j}Introduction passage · 3 passages
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Introduction passage · chapters04_05.tex, line 91
Trace moments and a unit triangular inversion. The Hurwitz-moment description. Another exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displ…
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s_{2j}(p)=\frac12\sum_{|t|<2\sqrt p}t^{2j}H(4p-t^2).
s_fIntroduction passage · 3 passages
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Introduction passage · appendix_workbook.tex, line 292
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
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s_f
s_jIntroduction passage · 1 passage
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Introduction passage · chapters04_05.tex, line 94
exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displayed range. The normalization, including the factor 1/2, must be kept fixed when a trac…
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s_j(v)
TIntroduction passage · 74 passages
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Introduction passage · appendices.tex, line 15
The symbol T_k(p) in the trace-formula chapter denotes a Hecke operator on the weight-k cusp-form space. It is an operator, whereas T_r in a scalar recurrence is an integer or an integer polynomial. The weight k and the exponent r serve different roles.
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T_k(p)
tIntroduction passage · 44 passages
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Introduction passage · chapters04_05.tex, line 91
Trace moments and a unit triangular inversion. The Hurwitz-moment description. Another exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displ…
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s_{2j}(p)=\frac12\sum_{|t|<2\sqrt p}t^{2j}H(4p-t^2).
t^2Introduction passage · 2 passages
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Introduction passage · chapters04_05.tex, line 91
Trace moments and a unit triangular inversion. The Hurwitz-moment description. Another exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displ…
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s_{2j}(p)=\frac12\sum_{|t|<2\sqrt p}t^{2j}H(4p-t^2).
t^{2j}Introduction passage · 1 passage
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Introduction passage · chapters04_05.tex, line 91
Trace moments and a unit triangular inversion. The Hurwitz-moment description. Another exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displ…
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s_{2j}(p)=\frac12\sum_{|t|<2\sqrt p}t^{2j}H(4p-t^2).
T^jIntroduction passage · 2 passages
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Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
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\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
T_{2m+1}Introduction passage · 4 passages
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Introduction passage · chapters15_16.tex, line 153
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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U(a^2)=\frac{T_{2m+1}-cT_{2m-1}}a
T_{2m-1}Introduction passage · 2 passages
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Introduction passage · chapters15_16.tex, line 153
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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U(a^2)=\frac{T_{2m+1}-cT_{2m-1}}a
T_kIntroduction passage · 4 passages
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Introduction passage · appendices.tex, line 15
The symbol T_k(p) in the trace-formula chapter denotes a Hecke operator on the weight-k cusp-form space. It is an operator, whereas T_r in a scalar recurrence is an integer or an integer polynomial. The weight k and the exponent r serve different roles.
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T_k(p)
T_rIntroduction passage · 16 passages
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Introduction passage · appendices.tex, line 15
The symbol T_k(p) in the trace-formula chapter denotes a Hecke operator on the weight-k cusp-form space. It is an operator, whereas T_r in a scalar recurrence is an integer or an integer polynomial. The weight k and the exponent r serve different roles.
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T_r
UIntroduction passage · 28 passages
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Introduction passage · chapters15_16.tex, line 142
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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U_{p,m}
U_{p,0}Introduction passage · 1 passage
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Introduction passage · chapters15_16.tex, line 152
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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U_{p,0}=1
U_{p,m}Introduction passage · 2 passages
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Introduction passage · chapters15_16.tex, line 142
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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U_{p,m}
vIntroduction passage · 180 passages
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Introduction passage · chapters04_05.tex, line 94
exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displayed range. The normalization, including the factor 1/2, must be kept fixed when a trac…
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s_j(v)
v_2Introduction passage · 6 passages
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Introduction passage · chapters15_16.tex, line 147
For every odd prime p≠5 and every m≥0, an odd integer U_p,m satisfies R(p^2m+1)=τ(p)U_p,m. The multiplier is defined by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are…
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v_2(R(p^{2m+1}))=v_2(\tau(p)).
v_iIntroduction passage · 35 passages
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Introduction passage · chapters06_07.tex, line 98
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
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v_i=i+5c_i
\VnIntroduction passage · 23 passages
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Introduction passage · chapters06_07.tex, line 98
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
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v\in\Vn
\ZIntroduction passage · 19 passages
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Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
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A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
\Z^{16\times18}Introduction passage · 1 passage
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Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
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A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
\Z^{6\times16}Introduction passage · 1 passage
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Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
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A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
\Z^{6\times18}Introduction passage · 1 passage
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Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
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A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
|Introduction passage · 26 passages
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Introduction passage · chapters04_05.tex, line 91
Trace moments and a unit triangular inversion. The Hurwitz-moment description. Another exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displ…
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s_{2j}(p)=\frac12\sum_{|t|<2\sqrt p}t^{2j}H(4p-t^2).
\cdotIntroduction passage · 14 passages
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Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
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\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
\timesIntroduction passage · 6 passages
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Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
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A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
\DeltaIntroduction passage · 71 passages
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Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
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\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
\Delta_HIntroduction passage · 23 passages
Search aliases: Δ_H Delta
Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
\Delta_H^jIntroduction passage · 1 passage
Search aliases: Δ_H^j Delta
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
\sigmaIntroduction passage · 18 passages
Search aliases: σ sigma
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
\sigma_3Introduction passage · 5 passages
Search aliases: σ_3 sigma
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
\sigma_5Introduction passage · 2 passages
Search aliases: σ_5 sigma
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
\tauIntroduction passage · 168 passages
Search aliases: τ tau
Introduction passage · chapters06_07.tex, line 104
ial at each fixed shell Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime index p in a p-adic or real variable.
Exact mathematical passage
\label{eq:pn5} P_n(5)=288\tau(n),\qquad B_0(n)=288|\Vn|.
\nablaIntroduction passage · 10 passages
Search aliases: ∇ nabla
Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
\nabla_HIntroduction passage · 7 passages
Search aliases: ∇_H nabla
Introduction passage · appendix_operator.tex, line 46
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
\label{appop:conjugation} \frac{\Delta_H(DQ)}{D} =\Delta_HQ+2\nabla_H\log D\cdot\nabla_HQ.
\inIntroduction passage · 46 passages
Search aliases: ∈ in
Introduction passage · appendix_workbook.tex, line 288
The degree-at-most-22 certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let A∈Z^6×16, M_c∈Z^6×18, M_f∈Z^16×18. The fine-to-coarse parent map completely specifies A: its column j has a 1 in the parent row and zero elsewhere. The original fine masses are given below. Let s_f denote their sixteen-entry column vector. Indices start at 0.
Exact mathematical passage
A\in\Z^{6\times16},\quad M_c\in\Z^{6\times18},\quad M_f\in\Z^{16\times18}.
\prodIntroduction passage · 24 passages
Search aliases: ∏ prod
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
\prod_{i<j}Introduction passage · 4 passages
Search aliases: ∏_i<j prod
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
\sumIntroduction passage · 106 passages
Search aliases: ∑ sum
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
\sum_{j=0}^{10}Introduction passage · 3 passages
Search aliases: ∑_j=0^10 sum
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
\sum_{j=0}^{\lfloor M/2\rfloor}Introduction passage · 2 passages
Search aliases: ∑_j=0^lfloor M/2rfloor sum lfloor rfloor
Introduction passage · appendix_operator.tex, line 172
For a homogeneous polynomial P of degree M, a concrete top-degree harmonic projection in dimension four is H_M(P)=∑_j=0^lfloor M/2rfloor frac(-1)^j(M-j)!4^j j!M! s_2^jΔ_H^jP. For M=0 or 1 the sum contains only j=0. In general, if a_j denotes its coefficient, the product rule gives the cancellation condition a_j+4(j+1)(M-j)a_j+1=0. The factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its…
Exact mathematical passage
\label{appop:hprojection} \mathcal H_M(P)=\sum_{j=0}^{\lfloor M/2\rfloor} \frac{(-1)^j(M-j)!}{4^j j!M!}\,s_2^j\Delta_H^jP.
\sum_{n\ge1}Introduction passage · 7 passages
Search aliases: ∑_n≥1 sum ge
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
\sum_{v\in\Vn}Introduction passage · 13 passages
Search aliases: ∑_v∈Vn sum in Vn
Introduction passage · chapters06_07.tex, line 99
A genuine auxiliary polynomial and its local roots. A polynomial at each fixed shell. Write v_i=i+5c_i for v∈Vn. Define P_n(T)=∑_v∈Vn∏_i<j[(i-j)+T(c_i-c_j)] =∑_j=0^10B_j(n)T^j. For a fixed n, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at T=5 returns the Vandermonde numerators: P_n(5)=288τ(n), B_0(n)=288|Vn|. The variable T is auxiliary. Nothing in the construction says that τ(p) is one analytic function of the prime …
Exact mathematical passage
\label{eq:pn} P_n(T)=\sum_{v\in\Vn}\prod_{i<j}\bigl[(i-j)+T(c_i-c_j)\bigr] =\sum_{j=0}^{10}B_j(n)T^j.
\sum_{|t|<2\sqrt p}Introduction passage · 1 passage
Search aliases: ∑_|t|<2sqrt p sum sqrt
Introduction passage · chapters04_05.tex, line 91
Trace moments and a unit triangular inversion. The Hurwitz-moment description. Another exact route uses Hurwitz class numbers. Let H(D) denote the conventional weighted Hurwitz class number for a positive discriminant parameter D, with the standard exceptional automorphism weights. At a prime p define s_2j(p)=frac12∑_|t|<2sqrt pt^2jH(4p-t^2). Here the symbol s_2j(p) is a trace moment and is unrelated to the coordinate power sums s_j(v) introduced later. The argument of H is positive in the displ…
Exact mathematical passage
s_{2j}(p)=\frac12\sum_{|t|<2\sqrt p}t^{2j}H(4p-t^2).
\neIntroduction passage · 43 passages
Search aliases: ≠ ne
Introduction passage · appendix_operator.tex, line 45
On the open set where D≠0, the product rule gives fracΔ_H(DQ)D =Δ_HQ+2∇_Hlog D·∇_HQ. Here ∇log D denotes the rational logarithmic derivative ∇ D/D. It does not require choosing a complex logarithm or assuming that D is positive. For symmetric Q, the apparent poles cancel and the right side extends to a polynomial on all of H.
Exact mathematical passage
D\ne0
\geIntroduction passage · 77 passages
Search aliases: ≥ ge
Introduction passage · chapters01_03.tex, line 92
Eisenstein coefficients. With the standard constant-one normalization, E_4(q) · =1+240∑_n≥1σ_3(n)q^n, E_6(q) · =1-504∑_n≥1σ_5(n)q^n, and Δ=fracE_4^3-E_6^21728. Since E_8=E_4^2, one may equivalently write (E_4E_8-E_6^2)/1728. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator 1728 is exact, and an integer implementation can test the required divisibility before extracting a coefficient.
Exact mathematical passage
E_4(q)&=1+240\sum_{n\ge1}\sigma_3(n)q^n,\\ E_6(q)&=1-504\sum_{n\ge1}\sigma_5(n)q^n,
*Use in source · 2 passages
Search aliases: *
Use in source · chapters08_10.tex, line 36
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
L^*/L\cong A_4^*/5A_4, \qquad |L^*/L|=5^5=3125.
\mathbf1Use in source · 5 passages
Search aliases: 1 mathbf
Use in source · appendix_operator.tex, line 9
Two descriptions of the same metric. The orthogonal projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:projection} P=I_5-\frac15\mathbf1\mathbf1^T,
\mathbf1^TUse in source · 3 passages
Search aliases: 1^T mathbf
Use in source · appendix_operator.tex, line 9
Two descriptions of the same metric. The orthogonal projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:projection} P=I_5-\frac15\mathbf1\mathbf1^T,
A'Use in source · 1 passage
Search aliases: A'
Use in source · appendix_workbook.tex, line 51
Jacobi gives a third engine. Put T_j=j(j+1)/2 and a_j=(-1)^j(2j+1). The series A(q)=∏_k≥1(1-q^k)^3=1+∑_j≥1a_jq^T_j satisfies F=A^8. The identity AqF'=8FqA' yields m f_m=∑_substackj≥1
Exact mathematical passage
AqF'=8FqA'
a^{2m}Use in source · 1 passage
Search aliases: a^2m
Use in source · chapters15_16.tex, line 63
First suppose a≠0, and write t=v_p(a). In T_2m, the term indexed by j has valuation at least 2mt+j(11-2t). If t≤4, the j=0 term a^2m has strictly least valuation 2mt. Likewise T_2m-2 has valuation (2m-2)t. The correction p^10T_2m-2 has valuation 2mt+(10-2t)>2mt. Thus v_p(R(p^2m))=2mt, t≤4.
Exact mathematical passage
a^{2m}
a^3Use in source · 1 passage
Search aliases: a^3
Use in source · chapters01_03.tex, line 144
The first local blocks are T_2=a^2-b, T_3=a^3-2ab, T_4=a^4-3a^2b+b^2. Even if a≠0, it is not formally obvious from these expressions that none vanishes. The reduction to primes needs a short arithmetic argument.
Exact mathematical passage
T_2=a^2-b,\qquad T_3=a^3-2ab,\qquad T_4=a^4-3a^2b+b^2.
a^4Use in source · 1 passage
Search aliases: a^4
Use in source · chapters01_03.tex, line 144
The first local blocks are T_2=a^2-b, T_3=a^3-2ab, T_4=a^4-3a^2b+b^2. Even if a≠0, it is not formally obvious from these expressions that none vanishes. The reduction to primes needs a short arithmetic argument.
Exact mathematical passage
T_2=a^2-b,\qquad T_3=a^3-2ab,\qquad T_4=a^4-3a^2b+b^2.
A^8Use in source · 1 passage
Search aliases: A^8
Use in source · appendix_workbook.tex, line 51
Jacobi gives a third engine. Put T_j=j(j+1)/2 and a_j=(-1)^j(2j+1). The series A(q)=∏_k≥1(1-q^k)^3=1+∑_j≥1a_jq^T_j satisfies F=A^8. The identity AqF'=8FqA' yields m f_m=∑_substackj≥1
Exact mathematical passage
F=A^8
a^{r-2j}Use in source · 1 passage
Search aliases: a^r-2j
Use in source · chapters15_16.tex, line 53
The recurrence polynomial and untied valuations. For p≡4pmod5, χ=1 and a≡0pmod5. The exact recurrence expansion is T_r=∑_j=0^lfloor r/2rfloor (-1)^jbinomr-jjb^ja^r-2j. It follows by induction: multiplying the previous row by a and subtracting b times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are 1, so a strictly least valuation at either end cannot be raised by a coefficient divisible by p.
Exact mathematical passage
\label{eq:trexpansion} T_r=\sum_{j=0}^{\lfloor r/2\rfloor} (-1)^j\binom{r-j}{j}b^ja^{r-2j}.
a_1Use in source · 1 passage
Search aliases: a_1
Use in source · chapters17_18.tex, line 120
For that t, set L_i(v)=∏_j≠ i fracE_t(v)-e_j(s_2(v)/C)^5e_i-e_j. Each factor is homogeneous of degree 10, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree 10(N-1). On the shell, s_2/C=1, so L_i(v_j)=δ_ij. Consequently, any prescribed rational orbit values a_1,…,a_N are interpolated by Q=∑_ia_iL_i. If N=1, an ordinary constant suffices.
Exact mathematical passage
a_1,\ldots,a_N
a_{16}Use in source · 2 passages
Search aliases: a_16
Use in source · chapters08_10.tex, line 147
Moment identities that introduce other forms. The harmonic polynomial D(130s_4-47s_2^2) has weight-16 theta form. Its first coefficient is -280. Write E_4Δ=∑ a_16(n)q^n. Then 13∑_v∈VnW(v)s_4(v) =470n^2τ(n)-28a_16(n). Likewise D(90s_2s_4-27s_2^3-320s_3^2) gives 900n∑ W(v)s_4(v)-27000n^3τ(n) -320∑ W(v)s_3(v)^2 =3600a_18(n), where E_6Δ=∑ a_18(n)q^n and the packet sums are on Vn. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer non…
Exact mathematical passage
E_4\Delta=\sum a_{16}(n)q^n
a_{18}Use in source · 2 passages
Search aliases: a_18
Use in source · chapters08_10.tex, line 153
Moment identities that introduce other forms. The harmonic polynomial D(130s_4-47s_2^2) has weight-16 theta form. Its first coefficient is -280. Write E_4Δ=∑ a_16(n)q^n. Then 13∑_v∈VnW(v)s_4(v) =470n^2τ(n)-28a_16(n). Likewise D(90s_2s_4-27s_2^3-320s_3^2) gives 900n∑ W(v)s_4(v)-27000n^3τ(n) -320∑ W(v)s_3(v)^2 =3600a_18(n), where E_6Δ=∑ a_18(n)q^n and the packet sums are on Vn. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer non…
Exact mathematical passage
\label{eq:moment18} 900n\sum W(v)s_4(v)-27000n^3\tau(n) -320\sum W(v)s_3(v)^2 =3600a_{18}(n),
A_4Use in source · 4 passages
Search aliases: A_4
Use in source · chapters08_10.tex, line 35
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
A_4=H\cap\Z^5
A_4^*Use in source · 1 passage
Search aliases: A_4^*
Use in source · chapters08_10.tex, line 36
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
L^*/L\cong A_4^*/5A_4, \qquad |L^*/L|=5^5=3125.
a_5Use in source · 1 passage
Search aliases: a_5
Use in source · appendices.tex, line 124
S13: packet cardinality. Garvan, Kim and Stanton's Theorem~4 and equation~(5.3) give the five-core count [source reference]. In the present conventions this reads a_5(n-1)=∑_dmid nchifive(d)frac nd. For a prime different from 5, the shell cardinality is p+chifive(p). This is the number of packets, not their weighted sum.
Exact mathematical passage
a_5(n-1)=\sum_{d\mid n}\chifive(d)\frac nd.
a_iUse in source · 6 passages
Search aliases: a_i
Use in source · chapters08_10.tex, line 41
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
x=\frac1{\sqrt5}(a_i+t/5)_{i=0}^4, \qquad0\le t<5,\qquad\sum_i a_i=-t.
a_i^2Use in source · 3 passages
Search aliases: a_i^2
Use in source · chapters08_10.tex, line 49
For a regular representative, ∑_i a_i^2 is divisible by 5, since the squared residues sum to 30. It is even, since a_i^2≡ a_ipmod2 and ∑_i a_i=0. The quadratic exponent ∑_i a_i^2/10 is integral. The Weil operator T therefore acts as 1 on the sign line. The Weil action commutes with S_5, so this line is invariant. The rank is even, and the representation descends to SL_2(Z). On the line the relation (ST)^3=S^2 reduces to S^3=S^2, giving S=1.
Exact mathematical passage
\sum_i a_i^2
a_NUse in source · 1 passage
Search aliases: a_N
Use in source · chapters17_18.tex, line 120
For that t, set L_i(v)=∏_j≠ i fracE_t(v)-e_j(s_2(v)/C)^5e_i-e_j. Each factor is homogeneous of degree 10, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree 10(N-1). On the shell, s_2/C=1, so L_i(v_j)=δ_ij. Consequently, any prescribed rational orbit values a_1,…,a_N are interpolated by Q=∑_ia_iL_i. If N=1, an ordinary constant suffices.
Exact mathematical passage
a_1,\ldots,a_N
a_pUse in source · 3 passages
Search aliases: a_p
Use in source · chapters19_21.tex, line 197
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
a_p=\tau(p)\quad(p\ne q),\qquad a_q=0.
a_{p,j}Use in source · 2 passages
Search aliases: a_p,j
Use in source · chapter22.tex, line 82
A sufficient sign input. A second precise route would construct permitted corrections with an independently proved sign. At each remaining prime, choose finitely many rational universal null polynomials Q_p,j and rational coefficients a_p,j, and put F_p(v)=1+∑_j a_p,jQ_p,j(v). The choice may depend on p. Universal nullness must hold after the choice, so its weighted sum is still the actual τ(p).
Exact mathematical passage
a_{p,j}
a_qUse in source · 2 passages
Search aliases: a_q
Use in source · chapters19_21.tex, line 197
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
a_p=\tau(p)\quad(p\ne q),\qquad a_q=0.
bUse in source · 26 passages
Search aliases: b
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
b=p^{11}
b^2Use in source · 1 passage
Search aliases: b^2
Use in source · chapters01_03.tex, line 144
The first local blocks are T_2=a^2-b, T_3=a^3-2ab, T_4=a^4-3a^2b+b^2. Even if a≠0, it is not formally obvious from these expressions that none vanishes. The reduction to primes needs a short arithmetic argument.
Exact mathematical passage
T_2=a^2-b,\qquad T_3=a^3-2ab,\qquad T_4=a^4-3a^2b+b^2.
b^jUse in source · 1 passage
Search aliases: b^j
Use in source · chapters15_16.tex, line 53
The recurrence polynomial and untied valuations. For p≡4pmod5, χ=1 and a≡0pmod5. The exact recurrence expansion is T_r=∑_j=0^lfloor r/2rfloor (-1)^jbinomr-jjb^ja^r-2j. It follows by induction: multiplying the previous row by a and subtracting b times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are 1, so a strictly least valuation at either end cannot be raised by a coefficient divisible by p.
Exact mathematical passage
\label{eq:trexpansion} T_r=\sum_{j=0}^{\lfloor r/2\rfloor} (-1)^j\binom{r-j}{j}b^ja^{r-2j}.
b^{m-1}Use in source · 1 passage
Search aliases: b^m-1
Use in source · chapters15_16.tex, line 157
by the recurrence polynomial, including when τ(p)=0. If τ(p)≠0, then v_2(R(p^2m+1))=v_2(τ(p)). For m=0, take U_p,0=1. For m≥1, use the quotient U(a^2)=fracT_2m+1-cT_2m-1a as polynomial division. The constant term in a^2 comes from the linear-in-a terms in the two blocks. By [source reference], it is (-1)^mb^m-1((m+1)b+mc). For an odd prime p≠5, both b=p^11 and c=chifive(p)p^10 are odd. The bracket in [source reference] is congruent to (m+1)+m=2m+1 modulo 2. Thus the constant term is odd.
Exact mathematical passage
\label{eq:oddconstant} (-1)^mb^{m-1}\bigl((m+1)b+mc\bigr).
B_{10}Use in source · 1 passage
Search aliases: B_10
Use in source · chapters06_07.tex, line 123
A simple root at texorpdfstringp=499p=499. The exact coefficients B_0,…,B_10 at 499 are ( · 144000,-360000,1063040,28418240,-148872808,
Exact mathematical passage
B_0,\ldots,B_{10}
b_aUse in source · 1 passage
Search aliases: b_a
Use in source · appendix_operator.tex, line 24
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
b_a=e_a-e_4
B_{\mathrm{cris}}Use in source · 1 passage
Search aliases: B_cris mathrm
Use in source · chapters19_21.tex, line 145
The realization theorem and conventions. Colmez--Fontaine's theorem makes a weakly admissible filtered module admissible [source reference]. In the present setting K=K_0=Q_p and the base Frobenius is the identity. The theorem supplies a two-dimensional semistable Q_p-representation realizing the filtered module. Since N=0, the comparison is crystalline, using B_cris=B_st^N=0.
Exact mathematical passage
B_{\mathrm{cris}}=B_{\mathrm{st}}^{N=0}
B_kUse in source · 2 passages
Search aliases: B_k
Use in source · chapters13_14.tex, line 160
Localization by cancellation, rather than indicators. Let the columns of a rational matrix N form a basis of ker M_k. Then B_k=M_k+1N, AB_k=0. The columns of B_k transfer mass only within the parents: their effects on every parent cancel. They are still combinations of universal null polynomials. This is the valid repair of the failed sector-indicator proposal.
Exact mathematical passage
\label{eq:detailmatrix} B_k=M_{k+1}N,\qquad AB_k=0.
B_{\mathrm{st}}^{N=0}Use in source · 1 passage
Search aliases: B_st^N=0 mathrm
Use in source · chapters19_21.tex, line 145
The realization theorem and conventions. Colmez--Fontaine's theorem makes a weakly admissible filtered module admissible [source reference]. In the present setting K=K_0=Q_p and the base Frobenius is the identity. The theorem supplies a two-dimensional semistable Q_p-representation realizing the filtered module. Since N=0, the comparison is crystalline, using B_cris=B_st^N=0.
Exact mathematical passage
B_{\mathrm{cris}}=B_{\mathrm{st}}^{N=0}
\bmodUse in source · 10 passages
Search aliases: bmod
Use in source · appendices.tex, line 42
The corresponding five-core partition has size n-1. In particular, the coefficient at n=1 corresponds to the empty partition. The involution is J(v)_i=-v_-ibmod5. A J-orbit may contain one or two packets, and its mass is the sum of their weights.
Exact mathematical passage
J(v)_i=-v_{-i\bmod5}
\bottomruleUse in source · 1 passage
Search aliases: bottomrule
Use in source · chapters11_12.tex, line 87
[3pt] 2, k≥2 · (-25M-5,10M+2,-15M+4,15M-4,15M+3) · (10M-5,10M+2,20M+4,-20M-4,-20M+3) [3pt] 0, k≥3 · (-20M-15,10M-3,10M+4,20M+1,-20M+13) · (15M-15,-25M-3,10M+4,-15M+1,15M+13) The appropriate powers of 2 make each entry divisible by 7. After division, each vector has the standard residues, sum zero and norm squared 10N_k. The norm identities before division are ∑_i(7v_i)^2=∑_i(7w_i)^2 =1400M^2+140M+70γ_k=490N_k. In the first two rows, (w-v)/(5M)=(1,0,1,-1,-1). In the third it is (1,-1,0,-1,1). Thu…
Exact mathematical passage
3pt] 2, $k\ge2$&$(-25M-5,10M+2,-15M+4,15M-4,15M+3)$&$(10M-5,10M+2,20M+4,-20M-4,-20M+3)$\\[3pt] 0, $k\ge3$&$(-20M-15,10M-3,10M+4,20M+1,-20M+13)$&$(15M-15,-25M-3,10M+4,-15M+1,15M+13)$\\ \bottomrule \end{tabular}\end{center} The appropriate powers of $2$ make each entry divisible by $7$. After division, each vector has the standard residues, sum zero and norm squared $10N_k$. The norm identities before division are \[ \sum_i(7v_i)^2=\sum_i(7w_i)^2 =1400M^2+140M+70\gamma_k=490N_k.
CUse in source · 23 passages
Search aliases: C
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
\CUse in source · 1 passage
Search aliases: C
Use in source · chapters17_18.tex, line 6
Let Q be any finite symmetric polynomial over C, restricted to ∑_i v_i=0. Suppose ∑_v∈VnW(v)Q(v)=0 for every positive integer n. Write Q=∑_dQ_d in homogeneous degrees. Then every Q_d is separately universally null. Moreover, the weighted Q-sum is zero in each exact-content sector of every shell.
Exact mathematical passage
\C
C^2Use in source · 1 passage
Search aliases: C^2
Use in source · chapters17_18.tex, line 83
Interpolation is not universal nullness. Coordinates that identify exactly the orbits. Fix a nonempty shell and put C=10n. Consider the normalized even symmetric invariants I(v)=(fracs_4C^2,fracs_3^2C^3, fracs_3s_5C^4,fracs_5^2C^5). They are unchanged by J.
Exact mathematical passage
\label{eq:invariantmap} I(v)=\left(\frac{s_4}{C^2},\frac{s_3^2}{C^3}, \frac{s_3s_5}{C^4},\frac{s_5^2}{C^5}\right).
c^2Use in source · 1 passage
Search aliases: c^2
Use in source · chapters19_21.tex, line 38
Let c be complex conjugation. Then c^2=1 and the cyclotomic character sends c to -1. Consequently detρ_ℓ(c)=-1. In characteristic zero an involution is diagonalizable, with eigenvalues among 1,-1. The determinant forces one of each, so Trρ_ℓ(c)=0 as an exact Z_ℓ identity. This reasoning also covers ℓ=2: the trace is computed before reduction. Diagonalizing an involution in characteristic 2 would not justify the same step.
Exact mathematical passage
c^2=1
C^3Use in source · 1 passage
Search aliases: C^3
Use in source · chapters17_18.tex, line 83
Interpolation is not universal nullness. Coordinates that identify exactly the orbits. Fix a nonempty shell and put C=10n. Consider the normalized even symmetric invariants I(v)=(fracs_4C^2,fracs_3^2C^3, fracs_3s_5C^4,fracs_5^2C^5). They are unchanged by J.
Exact mathematical passage
\label{eq:invariantmap} I(v)=\left(\frac{s_4}{C^2},\frac{s_3^2}{C^3}, \frac{s_3s_5}{C^4},\frac{s_5^2}{C^5}\right).
C^4Use in source · 1 passage
Search aliases: C^4
Use in source · chapters17_18.tex, line 83
Interpolation is not universal nullness. Coordinates that identify exactly the orbits. Fix a nonempty shell and put C=10n. Consider the normalized even symmetric invariants I(v)=(fracs_4C^2,fracs_3^2C^3, fracs_3s_5C^4,fracs_5^2C^5). They are unchanged by J.
Exact mathematical passage
\label{eq:invariantmap} I(v)=\left(\frac{s_4}{C^2},\frac{s_3^2}{C^3}, \frac{s_3s_5}{C^4},\frac{s_5^2}{C^5}\right).
C^5Use in source · 1 passage
Search aliases: C^5
Use in source · chapters17_18.tex, line 83
Interpolation is not universal nullness. Coordinates that identify exactly the orbits. Fix a nonempty shell and put C=10n. Consider the normalized even symmetric invariants I(v)=(fracs_4C^2,fracs_3^2C^3, fracs_3s_5C^4,fracs_5^2C^5). They are unchanged by J.
Exact mathematical passage
\label{eq:invariantmap} I(v)=\left(\frac{s_4}{C^2},\frac{s_3^2}{C^3}, \frac{s_3s_5}{C^4},\frac{s_5^2}{C^5}\right).
c^rUse in source · 2 passages
Search aliases: c^r
Use in source · chapters17_18.tex, line 54
Separating the quasimodular weights. We need only an elementary weight-separation statement. Suppose finitely many quasimodular functions F_k of distinct nonnegative integer weights satisfy F_k(γ z)=∑_r=0^p_kf_k,r(z)c^r(cz+d)^k-r, f_k,0=F_k, p_k≤ k, for γ=[a · b
Exact mathematical passage
F_k(\gamma z)=\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r}, \qquad f_{k,0}=F_k,\qquad p_k\le k,
C_jUse in source · 4 passages
Search aliases: C_j
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
C_rUse in source · 2 passages
Search aliases: C_r
Use in source · chapters15_16.tex, line 41
A residue eliminates three prime classes. Suppose p≠5 and put χ=chifive(p). Modulo 5, χ≡ p^2, so the two recurrence roots may be taken as p and χ: p+χ≡ a, pχ≡ p^11pmod5. The complete homogeneous recurrence coefficient is T_r≡ C_r=∑_i=0^r p^r-iχ^ipmod5. Also χ p^10≡1pmod5. Since χ^2=1, for r≥2 the sum telescopes: R(p^r)≡ C_r-C_r-2 =p^r-1(p+χ)pmod5. The same expression holds at r=1. It is nonzero for p≡1,2,3pmod5. It proves nonvanishing for all positive exponents in those prime classes. Only p=5 a…
Exact mathematical passage
T_r\equiv C_r=\sum_{i=0}^r p^{r-i}\chi^i\pmod5.
C_{r-2}Use in source · 1 passage
Search aliases: C_r-2
Use in source · chapters15_16.tex, line 45
lasses Suppose p≠5 and put χ=chifive(p). Modulo 5, χ≡ p^2, so the two recurrence roots may be taken as p and χ: p+χ≡ a, pχ≡ p^11pmod5. The complete homogeneous recurrence coefficient is T_r≡ C_r=∑_i=0^r p^r-iχ^ipmod5. Also χ p^10≡1pmod5. Since χ^2=1, for r≥2 the sum telescopes: R(p^r)≡ C_r-C_r-2 =p^r-1(p+χ)pmod5. The same expression holds at r=1. It is nonzero for p≡1,2,3pmod5. It proves nonvanishing for all positive exponents in those prime classes. Only p=5 and p≡4pmod5 remain.
Exact mathematical passage
\label{eq:rmod5} R(p^r)\equiv C_r-C_{r-2} =p^{r-1}(p+\chi)\pmod5.
\chifiveUse in source · 29 passages
Search aliases: chifive
Use in source · appendices.tex, line 18
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d, 1 · d≡1,4pmod5, -1 · d≡2,3pmod5. The character value and the tenth power in the scaling formula are both essential.
Exact mathematical passage
\chifive(d)= \begin{cases} 0&5\mid d,\\ 1&d\equiv1,4\pmod5,\\ -1&d\equiv2,3\pmod5. \end{cases}
\mathrm{cris}Use in source · 1 passage
Search aliases: cris mathrm
Use in source · chapters19_21.tex, line 145
The realization theorem and conventions. Colmez--Fontaine's theorem makes a weakly admissible filtered module admissible [source reference]. In the present setting K=K_0=Q_p and the base Frobenius is the identity. The theorem supplies a two-dimensional semistable Q_p-representation realizing the filtered module. Since N=0, the comparison is crystalline, using B_cris=B_st^N=0.
Exact mathematical passage
B_{\mathrm{cris}}=B_{\mathrm{st}}^{N=0}
dUse in source · 74 passages
Search aliases: d
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
D'Use in source · 2 passages
Search aliases: D'
Use in source · chapters19_21.tex, line 131
Checking weak admissibility. For the chosen cohomological convention, define t_H(D)=∑_i idimgr^iD, t_N(D)=v_p(detφ). Weak admissibility requires t_H(D)=t_N(D) and t_H(D')≤ t_N(D') for every φ-stable subobject D', with the induced filtration. The total equality is t_H(D)=11=t_N(D).
Exact mathematical passage
t_H(D')\le t_N(D')
d^{10}Use in source · 5 passages
Search aliases: d^10
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
d^2Use in source · 5 passages
Search aliases: d^2
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
d^jUse in source · 1 passage
Search aliases: d^j
Use in source · chapters01_03.tex, line 54
Exact coefficient engines. The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
\sigma_j(n)=\sum_{d\mid n}d^j
d^rUse in source · 1 passage
Search aliases: d^r
Use in source · appendix_workbook.tex, line 274
These four coefficients were checked independently by the Eisenstein convolution. If σ_r(n)=∑_dmid nd^r, the coefficient of (E_4E_8-E_6^2)/1728 is obtained from 1728τ(n)= · 240σ_3(n)+480σ_7(n)+1008σ_5(n)
Exact mathematical passage
\sigma_r(n)=\sum_{d\mid n}d^r
d^\perpUse in source · 1 passage
Search aliases: d^⊥ perp
Use in source · chapters11_12.tex, line 60
Project e_i-e_j orthogonally to d^⊥. Its squared norm is 2-fracδ^24. Because m⊥ d, Cauchy's inequality gives m ^2≥frac(a|δ|+1)^22-δ^2/4. For |δ|=1 this is 4(a+1)^2/7. For |δ|=2 it is (2a+1)^2, a larger quantity. Therefore 10n≥4a^2+frac47(a+1)^2, which simplifies to the lower bound inside the ceiling in [source reference]. Taking the ceiling proves that no earlier shell can be mixed. Since powers of 2 modulo 7 cycle as 2,4,1, direct division gives the displayed γ_k values.
Exact mathematical passage
d^\perp
d_iUse in source · 1 passage
Search aliases: d_i
Use in source · chapters11_12.tex, line 51
If d ^2≥6, [source reference] gives n≥frac154M^2>frac100M^2+10M+235, M≥2. The remaining case is d ^2=4. Then d has two coordinates 1, two -1 and one 0. Of the ten coordinate pairs, two have equal d entries and eight have unequal entries. For a pair with δ=d_i-d_j, (v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2δ^2. The two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of D(v) and D(w). N…
Exact mathematical passage
\delta=d_i-d_j
d_jUse in source · 1 passage
Search aliases: d_j
Use in source · chapters11_12.tex, line 51
If d ^2≥6, [source reference] gives n≥frac154M^2>frac100M^2+10M+235, M≥2. The remaining case is d ^2=4. Then d has two coordinates 1, two -1 and one 0. Of the ten coordinate pairs, two have equal d entries and eight have unequal entries. For a pair with δ=d_i-d_j, (v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2δ^2. The two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of D(v) and D(w). N…
Exact mathematical passage
\delta=d_i-d_j
\detUse in source · 5 passages
Search aliases: det
Use in source · chapters06_07.tex, line 34
Integrality follows from an integer-valued basis. For every integer x, including negative integers, the generalized binomial coefficient binomxj is an integer. Its leading coefficient as a polynomial in x is 1/j!. Thus det[binomv_ij]_0≤ i,j≤4 =frac∏_i<j(v_j-v_i)∏_j=0^4j!. There are ten pairs, so reversing every difference changes the sign by (-1)^10=1. The determinant is W(v) and is integral. This proof is independent of the fact that the total in [source reference] is an integer.
Exact mathematical passage
\det\left[\binom{v_i}{j}\right]_{0\le i,j\le4} =\frac{\prod_{i<j}(v_j-v_i)}{\prod_{j=0}^4j!}.
\dimUse in source · 4 passages
Search aliases: dim
Use in source · appendix_workbook.tex, line 340
The corresponding upper bounds have direct structural explanations. Each correction matrix has column sum zero, giving bounds 5 and 15. Every row of A has a nonempty, disjoint support, so rank A=6 and dimker A=10. Since AB=0, the detail rank is at most 10. The upper and lower bounds establish the displayed ranks exactly.
Exact mathematical passage
\dim\ker A=10
\DQUse in source · 10 passages
Search aliases: DQ
Use in source · chapters01_03.tex, line 55
Exact coefficient engines. The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
\label{eq:logdelta} \DQ\Delta=\left(1-24\sum_{m\ge1}\sigma(m)q^m\right)\Delta, \qquad \DQ=q\frac{d}{dq}.
\DQ^2Use in source · 1 passage
Search aliases: DQ^2 DQ
Use in source · chapters04_05.tex, line 40
DQ E_6 · =(E_2E_6-E_4^2)/2. These normalized relations are recorded in [source reference], and the constant-one formula for Δ is checked in [source reference]. They define a derivation on the polynomial ring Q[E_2,E_4,E_6]. Repeated differentiation and collection of terms yield E_2DQ^2E_2 · =4DQ^3E_2+frac310DQ^2E_4, E_2DQ^4E_2 · =frac125DQ^5E_2+frac17DQ^4E_4-frac2435Δ. These are polynomial identities in the differential ring. They can be verified by expanding each derivative using the three rela…
Exact mathematical passage
E_2\DQ^2E_2&=4\DQ^3E_2+\frac3{10}\DQ^2E_4,\label{eq:d2}\\ E_2\DQ^4E_2&=\frac{12}{5}\DQ^5E_2+\frac17\DQ^4E_4-\frac{24}{35}\Delta.\label{eq:d4}
\DQ^3Use in source · 1 passage
Search aliases: DQ^3 DQ
Use in source · chapters04_05.tex, line 40
DQ E_6 · =(E_2E_6-E_4^2)/2. These normalized relations are recorded in [source reference], and the constant-one formula for Δ is checked in [source reference]. They define a derivation on the polynomial ring Q[E_2,E_4,E_6]. Repeated differentiation and collection of terms yield E_2DQ^2E_2 · =4DQ^3E_2+frac310DQ^2E_4, E_2DQ^4E_2 · =frac125DQ^5E_2+frac17DQ^4E_4-frac2435Δ. These are polynomial identities in the differential ring. They can be verified by expanding each derivative using the three rela…
Exact mathematical passage
E_2\DQ^2E_2&=4\DQ^3E_2+\frac3{10}\DQ^2E_4,\label{eq:d2}\\ E_2\DQ^4E_2&=\frac{12}{5}\DQ^5E_2+\frac17\DQ^4E_4-\frac{24}{35}\Delta.\label{eq:d4}
\DQ^4Use in source · 4 passages
Search aliases: DQ^4 DQ
Use in source · chapters04_05.tex, line 40
DQ E_6 · =(E_2E_6-E_4^2)/2. These normalized relations are recorded in [source reference], and the constant-one formula for Δ is checked in [source reference]. They define a derivation on the polynomial ring Q[E_2,E_4,E_6]. Repeated differentiation and collection of terms yield E_2DQ^2E_2 · =4DQ^3E_2+frac310DQ^2E_4, E_2DQ^4E_2 · =frac125DQ^5E_2+frac17DQ^4E_4-frac2435Δ. These are polynomial identities in the differential ring. They can be verified by expanding each derivative using the three rela…
Exact mathematical passage
E_2\DQ^2E_2&=4\DQ^3E_2+\frac3{10}\DQ^2E_4,\label{eq:d2}\\ E_2\DQ^4E_2&=\frac{12}{5}\DQ^5E_2+\frac17\DQ^4E_4-\frac{24}{35}\Delta.\label{eq:d4}
\DQ^5Use in source · 3 passages
Search aliases: DQ^5 DQ
Use in source · chapters04_05.tex, line 40
DQ E_6 · =(E_2E_6-E_4^2)/2. These normalized relations are recorded in [source reference], and the constant-one formula for Δ is checked in [source reference]. They define a derivation on the polynomial ring Q[E_2,E_4,E_6]. Repeated differentiation and collection of terms yield E_2DQ^2E_2 · =4DQ^3E_2+frac310DQ^2E_4, E_2DQ^4E_2 · =frac125DQ^5E_2+frac17DQ^4E_4-frac2435Δ. These are polynomial identities in the differential ring. They can be verified by expanding each derivative using the three rela…
Exact mathematical passage
E_2\DQ^2E_2&=4\DQ^3E_2+\frac3{10}\DQ^2E_4,\label{eq:d2}\\ E_2\DQ^4E_2&=\frac{12}{5}\DQ^5E_2+\frac17\DQ^4E_4-\frac{24}{35}\Delta.\label{eq:d4}
\DQ^jUse in source · 2 passages
Search aliases: DQ^j DQ
Use in source · chapters17_18.tex, line 37
Each application of DQ=q d/dq multiplies the shell coefficient by v ^2/10. Therefore θ_s_2^jh=10^jDQ^jθ_h. Differentiating a modular transformation law shows that DQ^jf, for modular weight w, is quasimodular of weight w+2j and depth at most j. Combining [source reference] and [source reference], the homogeneous shell series F_d(τ)=∑_n≥1(∑_v∈VnW(v)Q_d(v))q^n is quasimodular of weight (M-2j+2)+2j=12+d, independent of the radial index j. Its depth is at most d/2. Polynomial growth and Gaussian deca…
Exact mathematical passage
\label{eq:radialderivative} \theta_{s_2^jh}=10^j\DQ^j\theta_h.
eUse in source · 21 passages
Search aliases: e
Use in source · appendix_operator.tex, line 24
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
b_a=e_a-e_4
e_1Use in source · 3 passages
Search aliases: e_1
Use in source · appendix_operator.tex, line 136
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
e_1=s_1=0
E_2Use in source · 6 passages
Search aliases: E_2
Use in source · chapters04_05.tex, line 33
Deriving the identity in the differential ring. Set A(q)=∑_n≥1σ(n)q^n, B(q)=∑_n≥1σ_3(n)q^n. Then E_2=1-24A and E_4=1+240B. The Ramanujan differential relations are DQ E_2 · =(E_2^2-E_4)/12,
Exact mathematical passage
E_2=1-24A
e_2Use in source · 6 passages
Search aliases: e_2
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
E_2^2Use in source · 1 passage
Search aliases: E_2^2
Use in source · chapters04_05.tex, line 34
Deriving the identity in the differential ring. Set A(q)=∑_n≥1σ(n)q^n, B(q)=∑_n≥1σ_3(n)q^n. Then E_2=1-24A and E_4=1+240B. The Ramanujan differential relations are DQ E_2 · =(E_2^2-E_4)/12, DQ E_4 · =(E_2E_4-E_6)/3, DQ E_6 · =(E_2E_6-E_4^2)/2. These normalized relations are recorded in [source reference], and the constant-one formula for Δ is checked in [source reference]. They define a derivation on the polynomial ring Q[E_2,E_4,E_6]. Repeated differentiation and collection of terms yield E_2DQ…
Exact mathematical passage
\DQ E_2&=(E_2^2-E_4)/12,\\ \DQ E_4&=(E_2E_4-E_6)/3,\\ \DQ E_6&=(E_2E_6-E_4^2)/2.
e_3Use in source · 4 passages
Search aliases: e_3
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
e_4Use in source · 4 passages
Search aliases: e_4
Use in source · appendix_operator.tex, line 24
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
b_a=e_a-e_4
e_5Use in source · 5 passages
Search aliases: e_5
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
e_aUse in source · 1 passage
Search aliases: e_a
Use in source · appendix_operator.tex, line 24
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
b_a=e_a-e_4
e_iUse in source · 4 passages
Search aliases: e_i
Use in source · chapters11_12.tex, line 45
in one sector. Replace w by Jw if needed so that w-v=5Md, d∈ A_4∖{0}. Set a=5M/2 and m=(v+w)/2. Since M is even, both are integral. Equal norms give m· d=0,qquad10n= m ^2+a^2 d ^2. The squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be 2 here. Such a vector is ±(e_i-e_j), and m· d=0 would force v_i-v_j=±5M. This is divisible by 5, contrary to the distinct coordinate residues.
Exact mathematical passage
\pm(e_i-e_j)
e_jUse in source · 4 passages
Search aliases: e_j
Use in source · appendix_operator.tex, line 132
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
e_j
E_kUse in source · 6 passages
Search aliases: E_k
Use in source · chapters13_14.tex, line 126
Put H_k=ker(1^T:Q^s_k→Q), T_k=im M_k, E_k=H_k/T_k. The quotient E_k is the vector space of total-preserving changes that are unavailable from the chosen family. It is not a count of failed sign certificates. A sign certificate may exist even when E_k is nonzero.
Exact mathematical passage
H_k=\ker(\mathbf1^T:\Q^{s_k}\to\Q),\qquad T_k=\im M_k,\qquad E_k=H_k/T_k.
E_{k+1}Use in source · 2 passages
Search aliases: E_k+1
Use in source · chapters13_14.tex, line 135
The exact refinement sequence. There is a short exact sequence 0⟶ fracker AM_k+1(ker M_k) ⟶ E_k+1⟶ E_k⟶0. Consequently, if r_k=rank M_k, dim E_k+1-dim E_k =(s_k+1-s_k)-(r_k+1-r_k). The aggregation map sends H_k+1 onto H_k. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.
Exact mathematical passage
\label{eq:refinement} 0\longrightarrow \frac{\ker A}{M_{k+1}(\ker M_k)} \longrightarrow E_{k+1}\longrightarrow E_k\longrightarrow0.
E_tUse in source · 4 passages
Search aliases: E_t
Use in source · chapters17_18.tex, line 102
An explicit homogeneous interpolation basis. Let v_1,…,v_N represent the distinct J-orbits. For an integer t≥0, define the degree-10 symmetric polynomial E_t=s_4s_2^3+t s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2. For two distinct orbits, E_t(v_i)-E_t(v_j) is a nonzero polynomial in t of degree at most 3, by Proposition~[source reference]. It has at most three roots. Among all pairs there are at most 3N(N-1)/2 forbidden values. Hence some integer 0≤ t≤frac3N(N-1)2 separates all e_i=E_t(v_i).
Exact mathematical passage
\label{eq:et} E_t=s_4s_2^3+t\,s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2.
\expUse in source · 1 passage
Search aliases: exp
Use in source · chapters17_18.tex, line 30
Radial factors become derivatives. For an alternating harmonic h_ℓ, write θ_h(τ)=∑_substackv_i≡ i (5) ∑ v_i=0 h(v)exp(2π iτfrac v ^210). Chapter~[source reference] establishes that it is a scalar cusp form of weight ℓ+2. The normalization includes only a fixed scalar if one divides by 288.
Exact mathematical passage
\theta_h(\tau)=\sum_{\substack{v_i\equiv i\ (5)\\\sum v_i=0}} h(v)\exp\left(2\pi i\tau\frac{\|v\|^2}{10}\right).
FUse in source · 41 passages
Search aliases: F
Use in source · appendix_operator.tex, line 12
Two descriptions of the same metric. The orthogonal projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
F
\mathbb FUse in source · 2 passages
Search aliases: F mathbb
Use in source · appendix_workbook.tex, line 338
The rational row-reduction results are rank M_c=5, rank M_f=15, rank B=10. An independent check obtains the same lower bounds over both prime fields F_65537 and F_1000003. Every denominator used in a reduction is invertible there. A nonzero minor in either field proves a nonzero rational minor after clearing denominators.
Exact mathematical passage
\mathbb F_{65537}
\mathbf FUse in source · 1 passage
Search aliases: F mathbf
Use in source · chapters06_07.tex, line 134
· -11893316304,-1160632800). Their weighted valuations are (3,5,3,4,4,5,7,11,8,9,12). Hence μ_499=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)…
Exact mathematical passage
\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\quad\text{in }\mathbf F_5[U].
F'Use in source · 2 passages
Search aliases: F'
Use in source · appendix_workbook.tex, line 32
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
q\frac{F'(q)}{F(q)}=-24\sum_{j\ge1}\sigma_1(j)q^j
f_0Use in source · 3 passages
Search aliases: f_0
Use in source · appendices.tex, line 11
oefficient and its local parameters The normalized form is Δ(q)=q∏_m≥1(1-q^m)^24 =∑_n≥1τ(n)q^n, τ(1)=1. All coefficient indices in this book are positive integers. A stored value at index zero is a programming sentinel unless a displayed formula explicitly introduces the coefficients of Δ/q. For the latter coefficients, f_m=τ(m+1) and f_0=1.
Exact mathematical passage
f_0=1
F_1Use in source · 1 passage
Search aliases: F_1
Use in source · chapters15_16.tex, line 114
For m=1, the polynomial is simply F_1(x)=x-(p+1). For larger m, its degree grows, but the same constant-term and residue argument controls any integral square root. That is why no bound on m is needed.
Exact mathematical passage
F_1(x)=x-(p+1)
f_1Use in source · 1 passage
Search aliases: f_1
Use in source · appendix_workbook.tex, line 23
For direct product expansion, terms of degree at most m depend only on factors with k≤ m. The exact binomial expansion (1-q^k)^24=∑_j=0^24(-1)^jbinom24jq^kj therefore supplies a finite algorithm for each desired coefficient. Three short calculations are f_1 · =-24, f_2 · =binom242-24=276-24=252, f_3 · =-binom243+(-24)(-24)-24 · =-2024+576-24=-1472. The last line comes respectively from the q^3 term in the first factor, the product of its q term with the second factor's q^2 term, and the third fa…
Exact mathematical passage
f_1&=-24,\\ f_2&=\binom{24}{2}-24=276-24=252,\\ f_3&=-\binom{24}{3}+(-24)(-24)-24\\ &=-2024+576-24=-1472.
\mathbb F_{1000003}Use in source · 1 passage
Search aliases: F_1000003 mathbb
Use in source · appendix_workbook.tex, line 338
The rational row-reduction results are rank M_c=5, rank M_f=15, rank B=10. An independent check obtains the same lower bounds over both prime fields F_65537 and F_1000003. Every denominator used in a reduction is invertible there. A nonzero minor in either field proves a nonzero rational minor after clearing denominators.
Exact mathematical passage
\mathbb F_{1000003}
f_2Use in source · 1 passage
Search aliases: f_2
Use in source · appendix_workbook.tex, line 23
For direct product expansion, terms of degree at most m depend only on factors with k≤ m. The exact binomial expansion (1-q^k)^24=∑_j=0^24(-1)^jbinom24jq^kj therefore supplies a finite algorithm for each desired coefficient. Three short calculations are f_1 · =-24, f_2 · =binom242-24=276-24=252, f_3 · =-binom243+(-24)(-24)-24 · =-2024+576-24=-1472. The last line comes respectively from the q^3 term in the first factor, the product of its q term with the second factor's q^2 term, and the third fa…
Exact mathematical passage
f_1&=-24,\\ f_2&=\binom{24}{2}-24=276-24=252,\\ f_3&=-\binom{24}{3}+(-24)(-24)-24\\ &=-2024+576-24=-1472.
f_3Use in source · 2 passages
Search aliases: f_3
Use in source · appendix_workbook.tex, line 23
For direct product expansion, terms of degree at most m depend only on factors with k≤ m. The exact binomial expansion (1-q^k)^24=∑_j=0^24(-1)^jbinom24jq^kj therefore supplies a finite algorithm for each desired coefficient. Three short calculations are f_1 · =-24, f_2 · =binom242-24=276-24=252, f_3 · =-binom243+(-24)(-24)-24 · =-2024+576-24=-1472. The last line comes respectively from the q^3 term in the first factor, the product of its q term with the second factor's q^2 term, and the third fa…
Exact mathematical passage
f_1&=-24,\\ f_2&=\binom{24}{2}-24=276-24=252,\\ f_3&=-\binom{24}{3}+(-24)(-24)-24\\ &=-2024+576-24=-1472.
\mathbf F_5Use in source · 1 passage
Search aliases: F_5 mathbf
Use in source · chapters06_07.tex, line 134
· -11893316304,-1160632800). Their weighted valuations are (3,5,3,4,4,5,7,11,8,9,12). Hence μ_499=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)…
Exact mathematical passage
\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\quad\text{in }\mathbf F_5[U].
\mathbb F_{65537}Use in source · 1 passage
Search aliases: F_65537 mathbb
Use in source · appendix_workbook.tex, line 338
The rational row-reduction results are rank M_c=5, rank M_f=15, rank B=10. An independent check obtains the same lower bounds over both prime fields F_65537 and F_1000003. Every denominator used in a reduction is invertible there. A nonzero minor in either field proves a nonzero rational minor after clearing denominators.
Exact mathematical passage
\mathbb F_{65537}
F_dUse in source · 2 passages
Search aliases: F_d
Use in source · chapters17_18.tex, line 41
dq multiplies the shell coefficient by v ^2/10. Therefore θ_s_2^jh=10^jDQ^jθ_h. Differentiating a modular transformation law shows that DQ^jf, for modular weight w, is quasimodular of weight w+2j and depth at most j. Combining [source reference] and [source reference], the homogeneous shell series F_d(τ)=∑_n≥1(∑_v∈VnW(v)Q_d(v))q^n is quasimodular of weight (M-2j+2)+2j=12+d, independent of the radial index j. Its depth is at most d/2. Polynomial growth and Gaussian decay justify the termwise diff…
Exact mathematical passage
F_d(\tau)=\sum_{n\ge1}\left(\sum_{v\in\Vn}W(v)Q_d(v)\right)q^n
F_kUse in source · 5 passages
Search aliases: F_k
Use in source · chapters17_18.tex, line 53
Separating the quasimodular weights. We need only an elementary weight-separation statement. Suppose finitely many quasimodular functions F_k of distinct nonnegative integer weights satisfy F_k(γ z)=∑_r=0^p_kf_k,r(z)c^r(cz+d)^k-r, f_k,0=F_k, p_k≤ k, for γ=[a · b
Exact mathematical passage
F_k
f_{k,0}Use in source · 1 passage
Search aliases: f_k,0
Use in source · chapters17_18.tex, line 54
Separating the quasimodular weights. We need only an elementary weight-separation statement. Suppose finitely many quasimodular functions F_k of distinct nonnegative integer weights satisfy F_k(γ z)=∑_r=0^p_kf_k,r(z)c^r(cz+d)^k-r, f_k,0=F_k, p_k≤ k, for γ=[a · b
Exact mathematical passage
F_k(\gamma z)=\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r}, \qquad f_{k,0}=F_k,\qquad p_k\le k,
f_{k,r}Use in source · 2 passages
Search aliases: f_k,r
Use in source · chapters17_18.tex, line 54
Separating the quasimodular weights. We need only an elementary weight-separation statement. Suppose finitely many quasimodular functions F_k of distinct nonnegative integer weights satisfy F_k(γ z)=∑_r=0^p_kf_k,r(z)c^r(cz+d)^k-r, f_k,0=F_k, p_k≤ k, for γ=[a · b
Exact mathematical passage
F_k(\gamma z)=\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r}, \qquad f_{k,0}=F_k,\qquad p_k\le k,
F_mUse in source · 3 passages
Search aliases: F_m
Use in source · chapters15_16.tex, line 97
t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading term comes from u^2m. The constant term comes from H_2m(0)=(-p)^m and H_2m-2(0)=(-p…
Exact mathematical passage
F_m(x)
f_mUse in source · 6 passages
Search aliases: f_m
Use in source · appendices.tex, line 11
The coefficient and its local parameters. The normalized form is Δ(q)=q∏_m≥1(1-q^m)^24 =∑_n≥1τ(n)q^n, τ(1)=1. All coefficient indices in this book are positive integers. A stored value at index zero is a programming sentinel unless a displayed formula explicitly introduces the coefficients of Δ/q. For the latter coefficients, f_m=τ(m+1) and f_0=1.
Exact mathematical passage
f_m=\tau(m+1)
f_{m-j}Use in source · 1 passage
Search aliases: f_m-j
Use in source · appendix_workbook.tex, line 36
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
m f_m=-24\sum_{j=1}^{m}\sigma_1(j)f_{m-j} \qquad(m\ge1). \label{eq:workbook-divisor-recurrence}
f_{m-T_j}Use in source · 1 passage
Search aliases: f_m-T_j
Use in source · appendix_workbook.tex, line 52
Jacobi gives a third engine. Put T_j=j(j+1)/2 and a_j=(-1)^j(2j+1). The series A(q)=∏_k≥1(1-q^k)^3=1+∑_j≥1a_jq^T_j satisfies F=A^8. The identity AqF'=8FqA' yields m f_m=∑_substackj≥1 T_j≤ m a_j(9T_j-m)f_m-T_j. At m=3, only T_1=1, T_2=3 contribute. Its numerator is (-3)(9-3)(252)+5(27-3)(1)=-4536+120=-4416. Division by 3 again gives -1472. The direct product, divisor recurrence and triangular recurrence reach the same integer through different algebraic operations.
Exact mathematical passage
m f_m=\sum_{\substack{j\ge1\\T_j\le m}} a_j(9T_j-m)f_{m-T_j}. \label{eq:workbook-triangular-recurrence}
f_{n-1}Use in source · 1 passage
Search aliases: f_n-1
Use in source · appendix_workbook.tex, line 16
Write F(q)=fracΔ(q)q=∏_k≥1(1-q^k)^24 =∑_m≥0f_mq^m. Then f_0=1 and τ(n)=f_n-1. This shift is essential. A program that stores τ(0)=0 as a sentinel must not confuse that entry with the constant coefficient f_0.
Exact mathematical passage
\tau(n)=f_{n-1}
F_pUse in source · 2 passages
Search aliases: F_p
Use in source · chapter22.tex, line 83
A sufficient sign input. A second precise route would construct permitted corrections with an independently proved sign. At each remaining prime, choose finitely many rational universal null polynomials Q_p,j and rational coefficients a_p,j, and put F_p(v)=1+∑_j a_p,jQ_p,j(v). The choice may depend on p. Universal nullness must hold after the choice, so its weighted sum is still the actual τ(p).
Exact mathematical passage
F_p(v)=1+\sum_j a_{p,j}Q_{p,j}(v).
F_{p,0}Use in source · 1 passage
Search aliases: F_p,0
Use in source · chapters19_21.tex, line 201
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
F_{p,0}=1,\qquad F_{p,1}=a_p,\qquad F_{p,r}=a_pF_{p,r-1}-p^{11}F_{p,r-2}.
F_{p,1}Use in source · 1 passage
Search aliases: F_p,1
Use in source · chapters19_21.tex, line 201
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
F_{p,0}=1,\qquad F_{p,1}=a_p,\qquad F_{p,r}=a_pF_{p,r-1}-p^{11}F_{p,r-2}.
F_{p,r}Use in source · 2 passages
Search aliases: F_p,r
Use in source · chapters19_21.tex, line 201
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
F_{p,0}=1,\qquad F_{p,1}=a_p,\qquad F_{p,r}=a_pF_{p,r-1}-p^{11}F_{p,r-2}.
F_{p,r-1}Use in source · 1 passage
Search aliases: F_p,r-1
Use in source · chapters19_21.tex, line 201
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
F_{p,0}=1,\qquad F_{p,1}=a_p,\qquad F_{p,r}=a_pF_{p,r-1}-p^{11}F_{p,r-2}.
F_{p,r-2}Use in source · 2 passages
Search aliases: F_p,r-2
Use in source · chapters19_21.tex, line 201
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
F_{p,0}=1,\qquad F_{p,1}=a_p,\qquad F_{p,r}=a_pF_{p,r-1}-p^{11}F_{p,r-2}.
F_{p,r_p}Use in source · 1 passage
Search aliases: F_p,r_p
Use in source · chapters19_21.tex, line 205
Set prime parameters a_p=τ(p) (p≠ q), a_q=0. For every prime define integral local blocks F_p,0=1, F_p,1=a_p, F_p,r=a_pF_p,r-1-p^11F_p,r-2. For n=∏ p^r_p, define f(n)=∏_pF_p,r_p. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.
Exact mathematical passage
f(n)=\prod_pF_{p,r_p}
F_{q,2m}Use in source · 1 passage
Search aliases: F_q,2m
Use in source · chapters19_21.tex, line 224
The full size bound survives. At primes other than q, the Hecke roots have absolute value p^11/2 by Deligne. Formula [source reference] therefore gives |τ(p^r)|≤(r+1)p^11r/2. At the modified prime, F_q,2m=(-q^11)^m, F_q,2m+1=0. These obey the same local bound at every exponent, including zero. Multiplying the local estimates gives |f(n)|≤∏_p^r_pVert n(r_p+1)p^11r_p/2 =d(n)n^11/2. A zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified paramet…
Exact mathematical passage
F_{q,2m}=(-q^{11})^m,\qquad F_{q,2m+1}=0.
F_{q,2m+1}Use in source · 1 passage
Search aliases: F_q,2m+1
Use in source · chapters19_21.tex, line 224
The full size bound survives. At primes other than q, the Hecke roots have absolute value p^11/2 by Deligne. Formula [source reference] therefore gives |τ(p^r)|≤(r+1)p^11r/2. At the modified prime, F_q,2m=(-q^11)^m, F_q,2m+1=0. These obey the same local bound at every exponent, including zero. Multiplying the local estimates gives |f(n)|≤∏_p^r_pVert n(r_p+1)p^11r_p/2 =d(n)n^11/2. A zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified paramet…
Exact mathematical passage
F_{q,2m}=(-q^{11})^m,\qquad F_{q,2m+1}=0.
F_{q,r}Use in source · 1 passage
Search aliases: F_q,r
Use in source · chapters19_21.tex, line 209
Agreement at every fixed modulus and finite prefix. Only the parameter at q has changed. Every n≤ B is coprime to q, so f(n)=τ(n) there. At q, the old and new first parameters agree modulo 10M. Induction in the integral recurrence gives F_q,r≡τ(q^r)pmod10M for every exponent. All other local blocks agree exactly. Multiplying local congruences gives f(n)≡τ(n)pmod10M for every n, not only through a finite range.
Exact mathematical passage
F_{q,r}\equiv\tau(q^r)\pmod{10M}
\FilUse in source · 1 passage
Search aliases: Fil
Use in source · chapters19_21.tex, line 115
1 · 0, Trφ=0, detφ=p^11, φ^2=-p^11I. Choose any Q_p-line L⊂ D and put Fil^iD= D · i≤0, L · 1≤ i≤11, 0 · i≥12. The two nonzero associated graded pieces are gr^0D=D/L and gr^11D=L, each of dimension 1.
Exact mathematical passage
\Fil^iD= \begin{cases} D&i\le0,\\ L&1\le i\le11,\\ 0&i\ge12. \end{cases}
\Fil^iUse in source · 1 passage
Search aliases: Fil^i Fil
Use in source · chapters19_21.tex, line 115
1 · 0, Trφ=0, detφ=p^11, φ^2=-p^11I. Choose any Q_p-line L⊂ D and put Fil^iD= D · i≤0, L · 1≤ i≤11, 0 · i≥12. The two nonzero associated graded pieces are gr^0D=D/L and gr^11D=L, each of dimension 1.
Exact mathematical passage
\Fil^iD= \begin{cases} D&i\le0,\\ L&1\le i\le11,\\ 0&i\ge12. \end{cases}
\FrobUse in source · 1 passage
Search aliases: Frob
Use in source · chapters19_21.tex, line 26
Integral representations and complex conjugation. For each prime ℓ, the representation attached to Δ has the arithmetic-Frobenius normalization ρ_ℓ:G_Q⟶ GL_2(Q_ℓ), is continuous and unramified outside ℓ, and satisfies det(X-ρ_ℓ(Frob_p))=X^2-τ(p)X+p^11, p≠ℓ. Its determinant is the eleventh power of the ℓ-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the prima…
Exact mathematical passage
\label{eq:frobpoly} \det(X-\rho_\ell(\Frob_p))=X^2-\tau(p)X+p^{11}, \qquad p\ne\ell.
\Frob_pUse in source · 1 passage
Search aliases: Frob_p Frob
Use in source · chapters19_21.tex, line 26
Integral representations and complex conjugation. For each prime ℓ, the representation attached to Δ has the arithmetic-Frobenius normalization ρ_ℓ:G_Q⟶ GL_2(Q_ℓ), is continuous and unramified outside ℓ, and satisfies det(X-ρ_ℓ(Frob_p))=X^2-τ(p)X+p^11, p≠ℓ. Its determinant is the eleventh power of the ℓ-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the prima…
Exact mathematical passage
\label{eq:frobpoly} \det(X-\rho_\ell(\Frob_p))=X^2-\tau(p)X+p^{11}, \qquad p\ne\ell.
GUse in source · 38 passages
Search aliases: G
Use in source · appendix_operator.tex, line 25
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
\label{appop:chartmetric} G=I_4+\mathbf1\mathbf1^T, \qquad G^{-1}=I_4-\frac15\mathbf1\mathbf1^T,
gUse in source · 33 passages
Search aliases: g
Use in source · appendices.tex, line 45
The content of a packet is the positive gcd of its five coordinates. Content-one packets are primitive. Multiplication followed by the prescribed residue reindexing gives τ(n)=∑_g^2mid nchifive(g)g^10R(n/g^2), R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). The boundary values are R(1)=1 and R(p)=τ(p). At a prime shell the only possible content is one, so the extraction has not created a new prime coefficient.
Exact mathematical passage
\tau(n)=\sum_{g^2\mid n}\chifive(g)g^{10}R(n/g^2), \qquad R(n)=\sum_{d^2\mid n}\mu(d)\chifive(d)d^{10}\tau(n/d^2).
G'Use in source · 1 passage
Search aliases: G'
Use in source · chapters06_07.tex, line 145
evant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)≠0. Thus v_5(G_499(U))=v_5(U-ρ) on that residue class. At U=1 the difference from ρ has valuation 1, giving v_5(τ(499))=3+1=4. The root has been located in the auxiliary variable…
Exact mathematical passage
G'(1)\ne0
G'_{16249}Use in source · 2 passages
Search aliases: G'_16249
Use in source · appendix_workbook.tex, line 229
G_16249(1) · =61339951998700724370240. The first equals 288τ(16249), and the second equals the first divided by 625. Modulo 5, the normalized polynomial is 3+U^2+U^3=(U-1)^2(U-2). Modulo 25, the table gives G_16249(U)≡13+15U+11U^2+11U^3+15U^5. Thus G_16249(1)≡15pmod25 while G'_16249(1)≡0pmod5. For every integer or 5-adic integer h, polynomial Taylor expansion gives G_16249(1+5h)≡ G_16249(1)+5hG'_16249(1) ≡15pmod25. Terms of order at least two have a factor 25. This is a uniform obstruction throu…
Exact mathematical passage
G'_{16249}(1)\equiv0\pmod5
G'_{499}Use in source · 1 passage
Search aliases: G'_499
Use in source · chapters06_07.tex, line 138
· -11893316304,-1160632800). Their weighted valuations are (3,5,3,4,4,5,7,11,8,9,12). Hence μ_499=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)…
Exact mathematical passage
G_{499}(1)\equiv10\pmod{25},\qquad G'_{499}(1)\equiv1\pmod5.
G'_pUse in source · 2 passages
Search aliases: G'_p
Use in source · chapters06_07.tex, line 165
A double residual root with no lift. At p=16249, the shell has 16250 packets. The weighted coefficient valuations of P_p(5U) are (4,5,4,4,6,5,7,8,9,9,11), so μ_p=4. The reduction is overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)pmod5. The residual root at 1 is double. Exact coefficient reduction gives G_p(1)≡15pmod25, G'_p(1)≡0pmod5. For every y∈Z_5, polynomial Taylor expansion has an integral remainder: G_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y), H∈Z_5[y]. Consequently G_p(1+5y)≡15pmod25 for every y∈Z_5. No …
Exact mathematical passage
G_p(1)\equiv15\pmod{25},\qquad G'_p(1)\equiv0\pmod5.
G^{-1}Use in source · 2 passages
Search aliases: G^-1
Use in source · appendix_operator.tex, line 25
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
\label{appop:chartmetric} G=I_4+\mathbf1\mathbf1^T, \qquad G^{-1}=I_4-\frac15\mathbf1\mathbf1^T,
g^{10}Use in source · 3 passages
Search aliases: g^10
Use in source · appendices.tex, line 45
The content of a packet is the positive gcd of its five coordinates. Content-one packets are primitive. Multiplication followed by the prescribed residue reindexing gives τ(n)=∑_g^2mid nchifive(g)g^10R(n/g^2), R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). The boundary values are R(1)=1 and R(p)=τ(p). At a prime shell the only possible content is one, so the extraction has not created a new prime coefficient.
Exact mathematical passage
\tau(n)=\sum_{g^2\mid n}\chifive(g)g^{10}R(n/g^2), \qquad R(n)=\sum_{d^2\mid n}\mu(d)\chifive(d)d^{10}\tau(n/d^2).
g^{10+d}Use in source · 1 passage
Search aliases: g^10+d
Use in source · chapters13_14.tex, line 62
g(v)=1W(v)Q(v). Symmetry removes the residue permutation from Q, and homogeneity gives Q(v)=g^dQ(u). The scaling bijection yields ∑_v∈VnW(v)Q(v) =∑_g^2mid nchifive(g)g^10+dR_Q(n/g^2). The left side is zero. At n=1, only g=1 contributes, so R_Q(1)=0. Inductively, the g=1 term is R_Q(n) and every nonzero term with g>1 has smaller argument. Thus R_Q(n)=0 for every n. The contribution of exact content g is its scaling factor times R_Q(n/g^2), so it is zero as well.
Exact mathematical passage
\label{eq:thetacontent} \sum_{v\in\Vn}W(v)Q(v) =\sum_{g^2\mid n}\chifive(g)g^{10+d}R_Q(n/g^2).
g^2Use in source · 8 passages
Search aliases: g^2
Use in source · appendices.tex, line 45
The content of a packet is the positive gcd of its five coordinates. Content-one packets are primitive. Multiplication followed by the prescribed residue reindexing gives τ(n)=∑_g^2mid nchifive(g)g^10R(n/g^2), R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). The boundary values are R(1)=1 and R(p)=τ(p). At a prime shell the only possible content is one, so the extraction has not created a new prime coefficient.
Exact mathematical passage
\tau(n)=\sum_{g^2\mid n}\chifive(g)g^{10}R(n/g^2), \qquad R(n)=\sum_{d^2\mid n}\mu(d)\chifive(d)d^{10}\tau(n/d^2).
g^dUse in source · 2 passages
Search aliases: g^d
Use in source · chapters13_14.tex, line 61
g(v)=1W(v)Q(v). Symmetry removes the residue permutation from Q, and homogeneity gives Q(v)=g^dQ(u). The scaling bijection yields ∑_v∈VnW(v)Q(v) =∑_g^2mid nchifive(g)g^10+dR_Q(n/g^2). The left side is zero. At n=1, only g=1 contributes, so R_Q(1)=0. Inductively, the g=1 term is R_Q(n) and every nonzero term with g>1 has smaller argument. Thus R_Q(n)=0 for every n. The contribution of exact content g is its scaling factor times R_Q(n/g^2), so it is zero as well.
Exact mathematical passage
Q(v)=g^dQ(u)
G_{16249}Use in source · 5 passages
Search aliases: G_16249
Use in source · appendix_workbook.tex, line 198
begingroupsmall setlengthtabcolsep4pt r r r r j · B_j in P_16249 · C_j in G_16249 · j+v_5(B_j)
Exact mathematical passage
G_{16249}
G_{499}Use in source · 8 passages
Search aliases: G_499
Use in source · appendix_workbook.tex, line 121
begingroupsmall setlengthtabcolsep5pt r r r r j · B_j in P_499 · C_j in G_499 · j+v_5(B_j)
Exact mathematical passage
G_{499}
G_nUse in source · 6 passages
Search aliases: G_n
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
G_pUse in source · 4 passages
Search aliases: G_p
Use in source · chapters06_07.tex, line 165
A double residual root with no lift. At p=16249, the shell has 16250 packets. The weighted coefficient valuations of P_p(5U) are (4,5,4,4,6,5,7,8,9,9,11), so μ_p=4. The reduction is overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)pmod5. The residual root at 1 is double. Exact coefficient reduction gives G_p(1)≡15pmod25, G'_p(1)≡0pmod5. For every y∈Z_5, polynomial Taylor expansion has an integral remainder: G_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y), H∈Z_5[y]. Consequently G_p(1+5y)≡15pmod25 for every y∈Z_5. No …
Exact mathematical passage
G_p(1)\equiv15\pmod{25},\qquad G'_p(1)\equiv0\pmod5.
G_{\Q}Use in source · 4 passages
Search aliases: G_Q Q
Use in source · chapters19_21.tex, line 22
Integral representations and complex conjugation. For each prime ℓ, the representation attached to Δ has the arithmetic-Frobenius normalization ρ_ℓ:G_Q⟶ GL_2(Q_ℓ), is continuous and unramified outside ℓ, and satisfies det(X-ρ_ℓ(Frob_p))=X^2-τ(p)X+p^11, p≠ℓ. Its determinant is the eleventh power of the ℓ-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the prima…
Exact mathematical passage
\rho_\ell:G_{\Q}\longrightarrow GL_2(\Q_\ell),
\gcdUse in source · 1 passage
Search aliases: gcd
Use in source · chapters01_03.tex, line 131
Why the problem lives at primes. The Hecke contracts. The normalized form Δ is a Hecke eigenform. Its coefficients satisfy τ(mn) · =τ(m)τ(n), · · gcd(m,n)=1, τ(p^r+1) · =τ(p)τ(p^r)-p^11τ(p^r-1), · · r≥1. The exponent 11 again records weight 12. The constant term for the local recurrence is τ(1)=1, and the first parameter is a=τ(p). These classical statements occur in Lehmer's original paper and in the modern source [source reference].
Exact mathematical passage
\tau(mn)&=\tau(m)\tau(n),&&\gcd(m,n)=1,\label{eq:mult}\\ \tau(p^{r+1})&=\tau(p)\tau(p^r)-p^{11}\tau(p^{r-1}),&&r\ge1.\label{eq:hecke}
\operatorname{gr}Use in source · 3 passages
Search aliases: gr operatorname
Use in source · chapters19_21.tex, line 123
0 · i≥12. The two nonzero associated graded pieces are gr^0D=D/L and gr^11D=L, each of dimension 1.
Exact mathematical passage
\operatorname{gr}^0D=D/L
\operatorname{gr}^0Use in source · 1 passage
Search aliases: gr^0 operatorname
Use in source · chapters19_21.tex, line 123
0 · i≥12. The two nonzero associated graded pieces are gr^0D=D/L and gr^11D=L, each of dimension 1.
Exact mathematical passage
\operatorname{gr}^0D=D/L
\operatorname{gr}^{11}Use in source · 1 passage
Search aliases: gr^11 operatorname
Use in source · chapters19_21.tex, line 123
0 · i≥12. The two nonzero associated graded pieces are gr^0D=D/L and gr^11D=L, each of dimension 1.
Exact mathematical passage
\operatorname{gr}^{11}D=L
\operatorname{gr}^iUse in source · 1 passage
Search aliases: gr^i operatorname
Use in source · chapters19_21.tex, line 127
Checking weak admissibility. For the chosen cohomological convention, define t_H(D)=∑_i idimgr^iD, t_N(D)=v_p(detφ). Weak admissibility requires t_H(D)=t_N(D) and t_H(D')≤ t_N(D') for every φ-stable subobject D', with the induced filtration. The total equality is t_H(D)=11=t_N(D).
Exact mathematical passage
t_H(D)=\sum_i i\dim\operatorname{gr}^iD,\qquad t_N(D)=v_p(\det\varphi).
hUse in source · 21 passages
Search aliases: h
Use in source · appendix_operator.tex, line 164
Radial factors and harmonic projection. Let h_ℓ be homogeneous harmonic of degree ℓ on the rank-four space H. Euler's identity is v·∇_Hh_ℓ=ℓ h_ℓ. The product rule for the radial factor s_2^j gives Δ_H(s_2^jh_ℓ) =4j(ℓ+j+1)s_2^j-1h_ℓ, j≥1. At j=0 the left side is zero. At j=1,ℓ=0 the coefficient is 8, agreeing with Δ_Hs_2. At j=1,ℓ=10 it is 48, agreeing with Δ_H(s_2D)=48D.
Exact mathematical passage
h_\ell
h^2Use in source · 1 passage
Search aliases: h^2
Use in source · chapters06_07.tex, line 6
The Nekrasov--Okounkov identity in Han's formulation is ∏_m≥1(1-q^m)^z-1 =∑_λ q^|λ|∏_h∈ H(λ)(1-frac zh^2). At z=25, a partition having a hook of length 5 contributes zero. The surviving terms are the 5-cores. Multiplication by q shows that the coefficient τ(n) is the sum of the hook products over 5-cores of size n-1.
Exact mathematical passage
\label{eq:hookidentity} \prod_{m\ge1}(1-q^m)^{z-1} =\sum_\lambda q^{|\lambda|}\prod_{h\in H(\lambda)}\left(1-\frac z{h^2}\right).
H_0Use in source · 1 passage
Search aliases: H_0
Use in source · chapters15_16.tex, line 90
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
H_0=1,\qquad H_1=u,\qquad H_r=uH_{r-1}-pH_{r-2}.
H_1Use in source · 1 passage
Search aliases: H_1
Use in source · chapters15_16.tex, line 90
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
H_0=1,\qquad H_1=u,\qquad H_r=uH_{r-1}-pH_{r-2}.
\mathcal H_{12}Use in source · 1 passage
Search aliases: H_12 mathcal
Use in source · appendix_operator.tex, line 191
Harmonic projection is not preservation of shell values. The elementary counterexample is P=s_2D, of degree 12. Since Δ_HP=48D and the next Laplacian is zero, H_12(s_2D)=s_2D-frac148s_2(48D)=0. Yet on Vn the original polynomial equals 10nD, which is nonzero at every packet. The operation removes radial information. It cannot silently turn a shell interpolant into a universally null polynomial with the same values.
Exact mathematical passage
\mathcal H_{12}(s_2D)=s_2D-\frac1{48}s_2(48D)=0.
H_{2m}Use in source · 2 passages
Search aliases: H_2m
Use in source · chapters15_16.tex, line 94
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
\label{eq:tied} R(p^{2m})=p^{10m}\bigl(H_{2m}(u)-H_{2m-2}(u)\bigr).
H_{2m-2}Use in source · 2 passages
Search aliases: H_2m-2
Use in source · chapters15_16.tex, line 94
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
\label{eq:tied} R(p^{2m})=p^{10m}\bigl(H_{2m}(u)-H_{2m-2}(u)\bigr).
H_kUse in source · 2 passages
Search aliases: H_k
Use in source · chapters13_14.tex, line 126
Put H_k=ker(1^T:Q^s_k→Q), T_k=im M_k, E_k=H_k/T_k. The quotient E_k is the vector space of total-preserving changes that are unavailable from the chosen family. It is not a count of failed sign certificates. A sign certificate may exist even when E_k is nonzero.
Exact mathematical passage
H_k=\ker(\mathbf1^T:\Q^{s_k}\to\Q),\qquad T_k=\im M_k,\qquad E_k=H_k/T_k.
H_{k+1}Use in source · 1 passage
Search aliases: H_k+1
Use in source · chapters13_14.tex, line 147
inement There is a short exact sequence 0⟶ fracker AM_k+1(ker M_k) ⟶ E_k+1⟶ E_k⟶0. Consequently, if r_k=rank M_k, dim E_k+1-dim E_k =(s_k+1-s_k)-(r_k+1-r_k). The aggregation map sends H_k+1 onto H_k. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.
Exact mathematical passage
H_{k+1}
h_{M-2j}Use in source · 2 passages
Search aliases: h_M-2j
Use in source · appendix_operator.tex, line 183
Repeating the construction on that lower-degree quotient produces the harmonic decomposition P=∑_j=0^lfloor M/2rfloors_2^jh_M-2j. It is unique. If such a sum were zero, apply the Laplacian as many times as the largest occupied radial exponent. All smaller-exponent terms vanish. Formula [source reference] leaves a nonzero scalar times the last harmonic polynomial, so that polynomial must be zero. Descending removes every term.
Exact mathematical passage
\label{appop:decomposition} P=\sum_{j=0}^{\lfloor M/2\rfloor}s_2^jh_{M-2j}.
H_rUse in source · 3 passages
Search aliases: H_r
Use in source · chapters15_16.tex, line 86
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
H_r(u)=p^{-5r}T_r.
H_{r-1}Use in source · 1 passage
Search aliases: H_r-1
Use in source · chapters15_16.tex, line 90
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
H_0=1,\qquad H_1=u,\qquad H_r=uH_{r-1}-pH_{r-2}.
H_{r-2}Use in source · 1 passage
Search aliases: H_r-2
Use in source · chapters15_16.tex, line 90
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
H_0=1,\qquad H_1=u,\qquad H_r=uH_{r-1}-pH_{r-2}.
h_\ellUse in source · 9 passages
Search aliases: h_ℓ ell
Use in source · appendix_operator.tex, line 164
Radial factors and harmonic projection. Let h_ℓ be homogeneous harmonic of degree ℓ on the rank-four space H. Euler's identity is v·∇_Hh_ℓ=ℓ h_ℓ. The product rule for the radial factor s_2^j gives Δ_H(s_2^jh_ℓ) =4j(ℓ+j+1)s_2^j-1h_ℓ, j≥1. At j=0 the left side is zero. At j=1,ℓ=0 the coefficient is 8, agreeing with Δ_Hs_2. At j=1,ℓ=10 it is 48, agreeing with Δ_H(s_2D)=48D.
Exact mathematical passage
h_\ell
IUse in source · 6 passages
Search aliases: I
Use in source · appendix_operator.tex, line 9
Two descriptions of the same metric. The orthogonal projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:projection} P=I_5-\frac15\mathbf1\mathbf1^T,
I_4Use in source · 1 passage
Search aliases: I_4
Use in source · appendix_operator.tex, line 25
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
\label{appop:chartmetric} G=I_4+\mathbf1\mathbf1^T, \qquad G^{-1}=I_4-\frac15\mathbf1\mathbf1^T,
I_5Use in source · 1 passage
Search aliases: I_5
Use in source · appendix_operator.tex, line 9
Two descriptions of the same metric. The orthogonal projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:projection} P=I_5-\frac15\mathbf1\mathbf1^T,
\iffUse in source · 2 passages
Search aliases: iff
Use in source · chapter22.tex, line 34
R(n)=0iffτ(n)=0 · Primitive extraction encodes the same zero problem, with R(p)=τ(p).
Exact mathematical passage
R(n)=0\iff\tau(n)=0
\imUse in source · 1 passage
Search aliases: im
Use in source · chapters13_14.tex, line 126
Put H_k=ker(1^T:Q^s_k→Q), T_k=im M_k, E_k=H_k/T_k. The quotient E_k is the vector space of total-preserving changes that are unavailable from the chosen family. It is not a count of failed sign certificates. A sign certificate may exist even when E_k is nonzero.
Exact mathematical passage
H_k=\ker(\mathbf1^T:\Q^{s_k}\to\Q),\qquad T_k=\im M_k,\qquad E_k=H_k/T_k.
JUse in source · 32 passages
Search aliases: J
Use in source · appendices.tex, line 42
The corresponding five-core partition has size n-1. In particular, the coefficient at n=1 corresponds to the empty partition. The involution is J(v)_i=-v_-ibmod5. A J-orbit may contain one or two packets, and its mass is the sum of their weights.
Exact mathematical passage
J(v)_i=-v_{-i\bmod5}
KUse in source · 7 passages
Search aliases: K
Use in source · appendix_operator.tex, line 198
Two explicit harmonic moment checks. The degree-four correction K_4=130s_4-47s_2^2 has LK_4 =130frac1885s_2-47(104s_2)=0. Thus DK_4 is harmonic. At the first packet, W=1 and K_4=130·34-47·100=-280. Its theta form has weight 16, whose cusp space is spanned by Δ E_4 with first coefficient one. Consequently the theta form is -280Δ E_4, giving the moment identity already stated in Chapter~[source reference].
Exact mathematical passage
K_4=130s_4-47s_2^2
k^4Use in source · 2 passages
Search aliases: k^4
Use in source · chapters04_05.tex, line 4
The differential residual that survives extraction. A divisor convolution with the original cusp term. Define the fourth divisor moment S_4(n)=∑_k=1^n-1k^4σ(k)σ(n-k). It is a positive integer when n>1. The weights k^4 and the two divisor sums suggest a route through elementary arithmetic. The exact identity, however, retains the cusp coefficient: 840S_4(n)=n^4[50σ_3(n)+(35-84n)σ(n)]-τ(n). Thus the difference between two explicit divisor expressions is exactly τ(n). Positivity of S_4(n) does not …
Exact mathematical passage
S_4(n)=\sum_{k=1}^{n-1}k^4\sigma(k)\sigma(n-k).
K_0Use in source · 1 passage
Search aliases: K_0
Use in source · chapters19_21.tex, line 145
The realization theorem and conventions. Colmez--Fontaine's theorem makes a weakly admissible filtered module admissible [source reference]. In the present setting K=K_0=Q_p and the base Frobenius is the identity. The theorem supplies a two-dimensional semistable Q_p-representation realizing the filtered module. Since N=0, the comparison is crystalline, using B_cris=B_st^N=0.
Exact mathematical passage
K=K_0=\Q_p
K_4Use in source · 4 passages
Search aliases: K_4
Use in source · appendix_operator.tex, line 198
Two explicit harmonic moment checks. The degree-four correction K_4=130s_4-47s_2^2 has LK_4 =130frac1885s_2-47(104s_2)=0. Thus DK_4 is harmonic. At the first packet, W=1 and K_4=130·34-47·100=-280. Its theta form has weight 16, whose cusp space is spanned by Δ E_4 with first coefficient one. Consequently the theta form is -280Δ E_4, giving the moment identity already stated in Chapter~[source reference].
Exact mathematical passage
K_4=130s_4-47s_2^2
K_6Use in source · 2 passages
Search aliases: K_6
Use in source · appendix_operator.tex, line 209
Likewise, set K_6=90s_2s_4-27s_2^3-320s_3^2. Using [source reference]--[source reference], its operator image is LK_6= · (5760-5760)s_4
Exact mathematical passage
K_6=90s_2s_4-27s_2^3-320s_3^2.
\kerUse in source · 11 passages
Search aliases: ker
Use in source · appendix_workbook.tex, line 327
The full machine certificate is small pathevidence/computation/replays/transport/p31_transport_certificate.json. It supplies an 18×13 rational matrix N whose columns form a basis of ker M_c, and the detail matrix B=M_fN of size 16×13. The basic exact checks are AM_f=M_c, As_f=s_c, M_cN=0, B=M_fN, AB=0. The last equation shows that a detail correction changes no coarse mass.
Exact mathematical passage
\ker M_c
LUse in source · 27 passages
Search aliases: L
Use in source · appendices.tex, line 180
The CRT/NTT computation returns residues modulo 2^19. It is not a reconstruction of those large integer tau coefficients. Within each convolution, however, the integer coefficient is reconstructed before reduction. With L=1032191, M=524288, the nonnegative representative bound is L(M-1)^2=283725419316756479 <998244353·1004535809 =1002772198720536577. The common transform length 2^21 divides both finite-field multiplicative periods. This bound, primitive-root checks and integer-type limits are pa…
Exact mathematical passage
L=1032191,\qquad M=524288,
\mathcal LUse in source · 23 passages
Search aliases: L mathcal
Use in source · appendices.tex, line 53
Three polynomial operations. The ordinary hyperplane Laplacian is Δ_H. The conjugated operator L is characterized by Δ_H(DQ)=D LQ. Harmonicity of DQ means LQ=0. It does not mean that Q itself is harmonic.
Exact mathematical passage
\mathcal L
lUse in source · 1 passage
Search aliases: l
Use in source · chapters11_12.tex, line 87
[3pt] 2, k≥2 · (-25M-5,10M+2,-15M+4,15M-4,15M+3) · (10M-5,10M+2,20M+4,-20M-4,-20M+3) [3pt] 0, k≥3 · (-20M-15,10M-3,10M+4,20M+1,-20M+13) · (15M-15,-25M-3,10M+4,-15M+1,15M+13) The appropriate powers of 2 make each entry divisible by 7. After division, each vector has the standard residues, sum zero and norm squared 10N_k. The norm identities before division are ∑_i(7v_i)^2=∑_i(7w_i)^2 =1400M^2+140M+70γ_k=490N_k. In the first two rows, (w-v)/(5M)=(1,0,1,-1,-1). In the third it is (1,-1,0,-1,1). Thu…
Exact mathematical passage
3pt] 2, $k\ge2$&$(-25M-5,10M+2,-15M+4,15M-4,15M+3)$&$(10M-5,10M+2,20M+4,-20M-4,-20M+3)$\\[3pt] 0, $k\ge3$&$(-20M-15,10M-3,10M+4,20M+1,-20M+13)$&$(15M-15,-25M-3,10M+4,-15M+1,15M+13)$\\ \bottomrule \end{tabular}\end{center} The appropriate powers of $2$ make each entry divisible by $7$. After division, each vector has the standard residues, sum zero and norm squared $10N_k$. The norm identities before division are \[ \sum_i(7v_i)^2=\sum_i(7w_i)^2 =1400M^2+140M+70\gamma_k=490N_k.
L^*Use in source · 2 passages
Search aliases: L^*
Use in source · chapters08_10.tex, line 36
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
L^*/L\cong A_4^*/5A_4, \qquad |L^*/L|=5^5=3125.
L_0Use in source · 5 passages
Search aliases: L_0
Use in source · chapters19_21.tex, line 32
A continuous representation of the compact group G_Q has a stable Z_ℓ lattice. To see this directly, take an initial lattice L_0. Compactness bounds all its translates inside ℓ^-aL_0 for some a. The module L=∑_g∈ G_Qρ_ℓ(g)L_0 is contained in ℓ^-aL_0 and contains L_0. It is finitely generated, since a submodule of a finite free Z_ℓ module is finitely generated, and it is invariant by construction. Reductions modulo ℓ^e are therefore defined. No image-surjectivity theorem is required.
Exact mathematical passage
L_0
L_2Use in source · 6 passages
Search aliases: L_2
Use in source · chapters04_05.tex, line 101
The Eichler--Selberg trace identities give, for r≥1, ∑_j=0^r(-1)^r-jbinomr+jr-jp^r-js_2j(p) =-1-Tr T_2r+2(p), together with s_0(p)=p. Here T_k(p) is the Hecke operator at p on S_k(SL_2(Z)). It is distinct from the scalar recurrence notation T_r=τ(p^r) at a fixed prime. The trace normalization and Catalan inversion are checked against [source reference], Sections 4--6. The cusp spaces in weights 4,6,8,10 are zero. The weight-12 cusp space is spanned by Δ, so its Hecke trace is τ(p).
Exact mathematical passage
S_k(SL_2(\Z))
L_iUse in source · 3 passages
Search aliases: L_i
Use in source · chapters17_18.tex, line 112
For that t, set L_i(v)=∏_j≠ i fracE_t(v)-e_j(s_2(v)/C)^5e_i-e_j. Each factor is homogeneous of degree 10, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree 10(N-1). On the shell, s_2/C=1, so L_i(v_j)=δ_ij. Consequently, any prescribed rational orbit values a_1,…,a_N are interpolated by Q=∑_ia_iL_i. If N=1, an ordinary constant suffices.
Exact mathematical passage
\label{eq:li} L_i(v)=\prod_{j\ne i} \frac{E_t(v)-e_j(s_2(v)/C)^5}{e_i-e_j}.
\lceilUse in source · 1 passage
Search aliases: lceil
Use in source · chapters11_12.tex, line 20
The sharp threshold. Let k≥1, M=2^k, and set N_k=lceilfrac100M^2+10M+235rceil =frac20M^2+2M+γ_k7, where γ_k= 7 · k≡1pmod3,
Exact mathematical passage
\label{eq:nk} N_k=\left\lceil\frac{100M^2+10M+2}{35}\right\rceil =\frac{20M^2+2M+\gamma_k}{7},
\log_{10}Use in source · 1 passage
Search aliases: log_10 log
Use in source · chapters01_03.tex, line 121
figplaceholdercoefficients512The exact coefficients through 512 on a signed logarithmic scale. The plotted ordinate is sgn(τ(n))log_10(1+|τ(n)|). Sign changes and large values are visible, while every plotted coefficient is nonzero. This finite plot does not establish the conjecture.
Exact mathematical passage
\sgn(\tau(n))\log_{10}(1+|\tau(n)|)
\LongleftrightarrowUse in source · 8 passages
Search aliases: Longleftrightarrow
Use in source · chapters01_03.tex, line 34
Integrality makes a residue argument useful. If x∈Z, |x|<L and Lmid x, then x=0. If instead a nonzero residue modulo L is established, then x≠0 without any size estimate. At primes p≥5, [source reference] yields 2p^11/2<p^6. Therefore τ(p)=0 Longleftrightarrow p^6midτ(p), p≥5. This is an exact criterion. It replaces equality to zero by a divisibility condition with a modulus depending on p. It does not show that the divisibility condition fails.
Exact mathematical passage
\label{eq:p6} \tau(p)=0\quad\Longleftrightarrow\quad p^6\mid\tau(p),\qquad p\ge5.
\LongrightarrowUse in source · 1 passage
Search aliases: Longrightarrow
Use in source · chapters19_21.tex, line 253
The logical conclusion and the remaining information. The theorem refutes the listed arithmetic properties Longrightarrow every coefficient is nonzero. The construction does not impose a modular-form realization, the packet-shell identity or the compatible global Galois representations of Δ. Those are potential sources of additional information, rather than properties already contained in the weaker premises.
Exact mathematical passage
\text{the listed arithmetic properties}\quad\Longrightarrow\quad \text{every coefficient is nonzero}.
m^{10}Use in source · 1 passage
Search aliases: m^10
Use in source · chapters13_14.tex, line 37
The square-supported arithmetic kernel in [source reference] has inverse d^2⟼μ(d)chifive(d)d^10. Indeed, at a square index m^2, the convolution of the two kernels is chifive(m)m^10∑_dmid mμ(d), which is 1 at m=1 and 0 otherwise. At nonsquare indices both kernels have no support. Consequently R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). Both formulas are exact for every positive integer. They are not fitted from the first few shells.
Exact mathematical passage
\chifive(m)m^{10}\sum_{d\mid m}\mu(d),
M^2Use in source · 3 passages
Search aliases: M^2
Use in source · chapters11_12.tex, line 20
The sharp threshold. Let k≥1, M=2^k, and set N_k=lceilfrac100M^2+10M+235rceil =frac20M^2+2M+γ_k7, where γ_k= 7 · k≡1pmod3,
Exact mathematical passage
\label{eq:nk} N_k=\left\lceil\frac{100M^2+10M+2}{35}\right\rceil =\frac{20M^2+2M+\gamma_k}{7},
m^2Use in source · 1 passage
Search aliases: m^2
Use in source · chapters13_14.tex, line 36
The square-supported arithmetic kernel in [source reference] has inverse d^2⟼μ(d)chifive(d)d^10. Indeed, at a square index m^2, the convolution of the two kernels is chifive(m)m^10∑_dmid mμ(d), which is 1 at m=1 and 0 otherwise. At nonsquare indices both kernels have no support. Consequently R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). Both formulas are exact for every positive integer. They are not fitted from the first few shells.
Exact mathematical passage
m^2
M_1Use in source · 1 passage
Search aliases: M_1
Use in source · chapters13_14.tex, line 182
The stored two-stage certificate first eliminates five coarse masses, using only five elimination equations. It then uses coefficient vectors in ker M_1 to eliminate all but one fine mass while preserving every coarse total. The survivor is -52843168. The product engine independently identifies this with τ(31). The solve does not require a preassigned tau value in its elimination equations. The final survivor remains an exact finite calculation, not a formula that independently forces a nonzero …
Exact mathematical passage
\ker M_1
m_1Use in source · 1 passage
Search aliases: m_1
Use in source · chapters19_21.tex, line 75
The quantifiers cannot be reversed. Given a fixed finite list of moduli m_1,…,m_s, apply Theorem~[source reference] to their least common multiple. Infinitely many actual primes defeat every test in that list whose decisive step is finding a nonzero residue of τ(p) or any R(p^r). This remains true even if the test is allowed all positive local exponents.
Exact mathematical passage
m_1,\ldots,m_s
m_iUse in source · 4 passages
Search aliases: m_i
Use in source · chapters11_12.tex, line 52
If d ^2≥6, [source reference] gives n≥frac154M^2>frac100M^2+10M+235, M≥2. The remaining case is d ^2=4. Then d has two coordinates 1, two -1 and one 0. Of the ten coordinate pairs, two have equal d entries and eight have unequal entries. For a pair with δ=d_i-d_j, (v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2δ^2. The two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of D(v) and D(w). N…
Exact mathematical passage
(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\delta^2.
m_jUse in source · 2 passages
Search aliases: m_j
Use in source · chapters11_12.tex, line 52
If d ^2≥6, [source reference] gives n≥frac154M^2>frac100M^2+10M+235, M≥2. The remaining case is d ^2=4. Then d has two coordinates 1, two -1 and one 0. Of the ten coordinate pairs, two have equal d entries and eight have unequal entries. For a pair with δ=d_i-d_j, (v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2δ^2. The two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of D(v) and D(w). N…
Exact mathematical passage
(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\delta^2.
M_kUse in source · 10 passages
Search aliases: M_k
Use in source · chapters13_14.tex, line 113
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
t_{k,r}=\sum_{v\in S_{k,r}}W(v),\qquad (M_k)_{rj}=\sum_{v\in S_{k,r}}W(v)Q_j(v).
M_{k+1}Use in source · 6 passages
Search aliases: M_k+1
Use in source · chapters13_14.tex, line 120
Every depth-k+1 sector belongs to exactly one parent at depth k. Let A be the corresponding 0--1 aggregation matrix. Summing child contributions gives the exact identities t_k=At_k+1, M_k=AM_k+1. Every occupied parent has a child. Each correction column has global total zero.
Exact mathematical passage
\label{eq:aggregate} t_k=At_{k+1},\qquad M_k=AM_{k+1}.
m_pUse in source · 2 passages
Search aliases: m_p
Use in source · chapters15_16.tex, line 119
All prime cases have now been covered. If s=∏ p^m_p, multiplicativity gives R(s^2)=∏_pR(p^2m_p). Every factor is nonzero, and R(1)=1. This proves Theorem~[source reference].
Exact mathematical passage
s=\prod p^{m_p}
m_sUse in source · 1 passage
Search aliases: m_s
Use in source · chapters19_21.tex, line 75
The quantifiers cannot be reversed. Given a fixed finite list of moduli m_1,…,m_s, apply Theorem~[source reference] to their least common multiple. Infinitely many actual primes defeat every test in that list whose decisive step is finding a nonzero residue of τ(p) or any R(p^r). This remains true even if the test is allowed all positive local exponents.
Exact mathematical passage
m_1,\ldots,m_s
\midUse in source · 24 passages
Search aliases: mid
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
\minUse in source · 2 passages
Search aliases: min
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
\min_{B_j(n)\ne0}Use in source · 1 passage
Search aliases: min_B_j(n)≠0 min ne
Use in source · chapters06_07.tex, line 110
Normalize the coefficient valuations by μ_n=min_B_j(n)≠0(j+v_5(B_j(n))), G_n(U)=5^-μ_nP_n(5U). Every coefficient of G_n is integral, and at least one is a unit. Because 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). This formula remains meaningful if τ(n)=0, with valuation ∞. The normalization does not assume nonvanishing: its constant coefficient comes from a positive packet count on a nonempty shell.
Exact mathematical passage
\mu_n=\min_{B_j(n)\ne0}\bigl(j+v_5(B_j(n))\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U).
\min_jUse in source · 1 passage
Search aliases: min_j min
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
NUse in source · 37 passages
Search aliases: N
Use in source · appendices.tex, line 143
S09: weak admissibility and admissibility. The Colmez--Fontaine author manuscript supplies the implication from weak admissibility to admissibility and the N=0 crystalline realization in the specified setting [source reference]. The constructed rank-two module has equal Hodge and Newton totals and no proper Frobenius-stable rational line. Those checks concern a local object. They do not identify it with the global representation of Δ.
Exact mathematical passage
N=0
n^{11/2}Use in source · 2 passages
Search aliases: n^11/2
Use in source · chapters19_21.tex, line 180
every real B≥1 and integer M≥1, there exist a prime q>B and a multiplicative integer sequence f with the following properties: item f(1)=1 and f(n)=τ(n) for every n≤ B. item f(n)≡τ(n)pmod10M for every positive integer n. item Every weight-12 prime-power recurrence holds: f(p^r)=f(p)f(p^r-1)-p^11f(p^r-2), r≥2. item |f(n)|≤ d(n)n^11/2, where d(n) counts the positive divisors of n. item f(n) is odd exactly when n is an odd square, and f(n)≡ nσ(n)pmod5. item The arithmetic primitive transform R_f is…
Exact mathematical passage
|f(n)|\le d(n)n^{11/2}
n^2Use in source · 1 passage
Search aliases: n^2
Use in source · chapters08_10.tex, line 148
Moment identities that introduce other forms. The harmonic polynomial D(130s_4-47s_2^2) has weight-16 theta form. Its first coefficient is -280. Write E_4Δ=∑ a_16(n)q^n. Then 13∑_v∈VnW(v)s_4(v) =470n^2τ(n)-28a_16(n). Likewise D(90s_2s_4-27s_2^3-320s_3^2) gives 900n∑ W(v)s_4(v)-27000n^3τ(n) -320∑ W(v)s_3(v)^2 =3600a_18(n), where E_6Δ=∑ a_18(n)q^n and the packet sums are on Vn. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer non…
Exact mathematical passage
\label{eq:moment16} 13\sum_{v\in\Vn}W(v)s_4(v) =470n^2\tau(n)-28a_{16}(n).
n^3Use in source · 1 passage
Search aliases: n^3
Use in source · chapters08_10.tex, line 153
Moment identities that introduce other forms. The harmonic polynomial D(130s_4-47s_2^2) has weight-16 theta form. Its first coefficient is -280. Write E_4Δ=∑ a_16(n)q^n. Then 13∑_v∈VnW(v)s_4(v) =470n^2τ(n)-28a_16(n). Likewise D(90s_2s_4-27s_2^3-320s_3^2) gives 900n∑ W(v)s_4(v)-27000n^3τ(n) -320∑ W(v)s_3(v)^2 =3600a_18(n), where E_6Δ=∑ a_18(n)q^n and the packet sums are on Vn. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer non…
Exact mathematical passage
\label{eq:moment18} 900n\sum W(v)s_4(v)-27000n^3\tau(n) -320\sum W(v)s_3(v)^2 =3600a_{18}(n),
n^4Use in source · 1 passage
Search aliases: n^4
Use in source · chapters04_05.tex, line 8
l that survives extraction A divisor convolution with the original cusp term. Define the fourth divisor moment S_4(n)=∑_k=1^n-1k^4σ(k)σ(n-k). It is a positive integer when n>1. The weights k^4 and the two divisor sums suggest a route through elementary arithmetic. The exact identity, however, retains the cusp coefficient: 840S_4(n)=n^4[50σ_3(n)+(35-84n)σ(n)]-τ(n). Thus the difference between two explicit divisor expressions is exactly τ(n). Positivity of S_4(n) does not imply that this differenc…
Exact mathematical passage
\label{eq:s4general} 840S_4(n)=n^4\bigl[50\sigma_3(n)+(35-84n)\sigma(n)\bigr]-\tau(n).
N_3Use in source · 1 passage
Search aliases: N_3
Use in source · chapters11_12.tex, line 34
6 · k≡0pmod3. Every depth-k sector in every shell n<N_k is sign-pure. The shell n=N_k has a mixed sector. This threshold ranges over all positive integer shell indices. It does not assert that N_k is prime. For example, N_3=186 is composite.
Exact mathematical passage
N_3=186
N_kUse in source · 6 passages
Search aliases: N_k
Use in source · chapters11_12.tex, line 20
The sharp threshold. Let k≥1, M=2^k, and set N_k=lceilfrac100M^2+10M+235rceil =frac20M^2+2M+γ_k7, where γ_k= 7 · k≡1pmod3,
Exact mathematical passage
\label{eq:nk} N_k=\left\lceil\frac{100M^2+10M+2}{35}\right\rceil =\frac{20M^2+2M+\gamma_k}{7},
\nmidUse in source · 3 passages
Search aliases: nmid
Use in source · chapters13_14.tex, line 3
Primitive content and conserved arithmetic sectors. The coordinate gcd is a real invariant. For v∈Vn, let g(v) be the positive gcd of its five coordinates. Since the coordinates occupy all residues modulo 5, one has 5nmid g(v). Call g(v) the content of the packet. A primitive packet has content 1.
Exact mathematical passage
5\nmid g(v)
\notUse in source · 3 passages
Search aliases: not
Use in source · chapter22.tex, line 62
A sufficient adaptive residue input. One precise route would be a uniform construction of residue certificates. Suppose there are explicitly defined functions on prime indices, with integer values m(p)≥2 and b(p), for which one proves τ(p)≡ b(p)pmodm(p), b(p)not≡0pmodm(p) for every prime in [source reference]. These two assertions would settle that family and hence Lehmer. The modulus may depend on p, so the fixed-modulus survivor theorem does not refute this proposed form of input.
Exact mathematical passage
\label{eq:adaptiveresidueinput} \tau(p)\equiv b(p)\pmod{m(p)}, \qquad b(p)\not\equiv0\pmod{m(p)}
OUse in source · 1 passage
Search aliases: O
Use in source · bibliography.tex, line 77
bibitemIW Tomoyoshi Ibukiyama and Brandon Williams, Hermitian modular forms and algebraic modular forms on SO(6), https://arxiv.org/pdf/2505.23497arXiv:2505.23497v1, header 29 May 2025, internal date 12 June 2025. S3.4, printed pp.~18--19, gives vector-valued harmonic theta modularity and weight ν+n/2. The scalar full-level sign-line descent is a project argument. The paper's conjectural Hermitian correspondences are not assumed.
Exact mathematical passage
SO(6)
oUse in source · 1 passage
Search aliases: o
Use in source · chapters11_12.tex, line 87
[3pt] 2, k≥2 · (-25M-5,10M+2,-15M+4,15M-4,15M+3) · (10M-5,10M+2,20M+4,-20M-4,-20M+3) [3pt] 0, k≥3 · (-20M-15,10M-3,10M+4,20M+1,-20M+13) · (15M-15,-25M-3,10M+4,-15M+1,15M+13) The appropriate powers of 2 make each entry divisible by 7. After division, each vector has the standard residues, sum zero and norm squared 10N_k. The norm identities before division are ∑_i(7v_i)^2=∑_i(7w_i)^2 =1400M^2+140M+70γ_k=490N_k. In the first two rows, (w-v)/(5M)=(1,0,1,-1,-1). In the third it is (1,-1,0,-1,1). Thu…
Exact mathematical passage
3pt] 2, $k\ge2$&$(-25M-5,10M+2,-15M+4,15M-4,15M+3)$&$(10M-5,10M+2,20M+4,-20M-4,-20M+3)$\\[3pt] 0, $k\ge3$&$(-20M-15,10M-3,10M+4,20M+1,-20M+13)$&$(15M-15,-25M-3,10M+4,-15M+1,15M+13)$\\ \bottomrule \end{tabular}\end{center} The appropriate powers of $2$ make each entry divisible by $7$. After division, each vector has the standard residues, sum zero and norm squared $10N_k$. The norm identities before division are \[ \sum_i(7v_i)^2=\sum_i(7w_i)^2 =1400M^2+140M+70\gamma_k=490N_k.
\overline GUse in source · 2 passages
Search aliases: overline G overline
Use in source · chapters06_07.tex, line 134
· -11893316304,-1160632800). Their weighted valuations are (3,5,3,4,4,5,7,11,8,9,12). Hence μ_499=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)…
Exact mathematical passage
\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\quad\text{in }\mathbf F_5[U].
\overline G_{499}Use in source · 1 passage
Search aliases: overline G_499 overline
Use in source · chapters06_07.tex, line 134
· -11893316304,-1160632800). Their weighted valuations are (3,5,3,4,4,5,7,11,8,9,12). Hence μ_499=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)…
Exact mathematical passage
\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\quad\text{in }\mathbf F_5[U].
\overline G_pUse in source · 1 passage
Search aliases: overline G_p overline
Use in source · chapters06_07.tex, line 161
A double residual root with no lift. At p=16249, the shell has 16250 packets. The weighted coefficient valuations of P_p(5U) are (4,5,4,4,6,5,7,8,9,9,11), so μ_p=4. The reduction is overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)pmod5. The residual root at 1 is double. Exact coefficient reduction gives G_p(1)≡15pmod25, G'_p(1)≡0pmod5. For every y∈Z_5, polynomial Taylor expansion has an integral remainder: G_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y), H∈Z_5[y]. Consequently G_p(1+5y)≡15pmod25 for every y∈Z_5. No …
Exact mathematical passage
\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\pmod5.
p^{-5r}Use in source · 1 passage
Search aliases: p^-5r
Use in source · chapters15_16.tex, line 86
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
H_r(u)=p^{-5r}T_r.
p^{10}Use in source · 9 passages
Search aliases: p^10
Use in source · chapters15_16.tex, line 4
Every primitive square survives. The local numerator introduced by content. The transform in [source reference] is a Dirichlet convolution of multiplicative functions, so R is multiplicative. At a prime p, write a=τ(p), b=p^11, c=chifive(p)p^10. The local generating factor is ∑_r≥0R(p^r)X^r =frac1-cX^21-aX+bX^2. It follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In par…
Exact mathematical passage
a=\tau(p),\qquad b=p^{11},\qquad c=\chifive(p)p^{10}.
p^{10m}Use in source · 1 passage
Search aliases: p^10m
Use in source · chapters15_16.tex, line 94
The tied valuation texorpdfstringt=5t=5. The preceding endpoint arguments leave t=5. Write a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading ter…
Exact mathematical passage
\label{eq:tied} R(p^{2m})=p^{10m}\bigl(H_{2m}(u)-H_{2m-2}(u)\bigr).
p^{11}Use in source · 28 passages
Search aliases: p^11
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
b=p^{11}
p^{11/2}Use in source · 7 passages
Search aliases: p^11/2
Use in source · appendices.tex, line 90
S06: Deligne's bound and normalization. The modular representation and purity inputs give |τ(p)|≤2p^11/2 [source reference]. The modular paper's representation construction and the Weil theorem supply different parts of this implication. Their Frobenius conventions require translation. This upper bound includes zero and is never used as a positive lower bound.
Exact mathematical passage
|\tau(p)|\le2p^{11/2}
p^{11m-1}Use in source · 2 passages
Search aliases: p^11m-1
Use in source · chapters15_16.tex, line 79
If a=0, do not assign it a finite t. Direct substitution gives R(p^2m)=(-1)^mp^11m-1(p+1), which is nonzero. This branch is indispensable: a theorem purporting to survive a hypothetical prime zero must actually evaluate that case.
Exact mathematical passage
\label{eq:evenzeroa} R(p^{2m})=(-1)^mp^{11m-1}(p+1),
p^{11r/2}Use in source · 1 passage
Search aliases: p^11r/2
Use in source · chapters19_21.tex, line 220
The full size bound survives. At primes other than q, the Hecke roots have absolute value p^11/2 by Deligne. Formula [source reference] therefore gives |τ(p^r)|≤(r+1)p^11r/2. At the modified prime, F_q,2m=(-q^11)^m, F_q,2m+1=0. These obey the same local bound at every exponent, including zero. Multiplying the local estimates gives |f(n)|≤∏_p^r_pVert n(r_p+1)p^11r_p/2 =d(n)n^11/2. A zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified paramet…
Exact mathematical passage
|\tau(p^r)|\le(r+1)p^{11r/2}.
p^{11r_p/2}Use in source · 1 passage
Search aliases: p^11r_p/2
Use in source · chapters19_21.tex, line 228
The full size bound survives. At primes other than q, the Hecke roots have absolute value p^11/2 by Deligne. Formula [source reference] therefore gives |τ(p^r)|≤(r+1)p^11r/2. At the modified prime, F_q,2m=(-q^11)^m, F_q,2m+1=0. These obey the same local bound at every exponent, including zero. Multiplying the local estimates gives |f(n)|≤∏_p^r_pVert n(r_p+1)p^11r_p/2 =d(n)n^11/2. A zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified paramet…
Exact mathematical passage
|f(n)|\le\prod_{p^{r_p}\Vert n}(r_p+1)p^{11r_p/2} =d(n)n^{11/2}.
P^2Use in source · 1 passage
Search aliases: P^2
Use in source · appendix_operator.tex, line 12
Two descriptions of the same metric. The orthogonal projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
P^2=P
p^2Use in source · 11 passages
Search aliases: p^2
Use in source · chapters04_05.tex, line 14
At a prime p, σ(p)=1+p and σ_3(p)=1+p^3. Put C(p)=85-49p-84p^2+50p^3. Equation [source reference] becomes τ(p)=p^4C(p)-840S_4(p). For p>7, the coefficient 840 is a p-adic unit. Reducing modulo p^4 gives p^4mid S_4(p) Longleftrightarrow p^4midτ(p). The proposed assertion that p^4 never divides S_4(p) would therefore assert a uniform bound v_p(τ(p))≤3. It is a sufficient strengthening of Lehmer, not a consequence of the displayed convolution.
Exact mathematical passage
C(p)=85-49p-84p^2+50p^3.
p^{2m}Use in source · 8 passages
Search aliases: p^2m
Use in source · chapters15_16.tex, line 23
The even primitive blocks are not the same as the even tau blocks. For example, R(p^2)=a^2-p^10(p+chifive(p)). An argument that τ(p^2m)≠0 does not alone imply the corresponding difference is nonzero. We now prove the primitive statement directly.
Exact mathematical passage
\tau(p^{2m})\ne0
p^{2m_p}Use in source · 1 passage
Search aliases: p^2m_p
Use in source · chapters15_16.tex, line 120
All prime cases have now been covered. If s=∏ p^m_p, multiplicativity gives R(s^2)=∏_pR(p^2m_p). Every factor is nonzero, and R(1)=1. This proves Theorem~[source reference].
Exact mathematical passage
R(s^2)=\prod_pR(p^{2m_p}).
p^3Use in source · 5 passages
Search aliases: p^3
Use in source · chapters04_05.tex, line 13
At a prime p, σ(p)=1+p and σ_3(p)=1+p^3. Put C(p)=85-49p-84p^2+50p^3. Equation [source reference] becomes τ(p)=p^4C(p)-840S_4(p). For p>7, the coefficient 840 is a p-adic unit. Reducing modulo p^4 gives p^4mid S_4(p) Longleftrightarrow p^4midτ(p). The proposed assertion that p^4 never divides S_4(p) would therefore assert a uniform bound v_p(τ(p))≤3. It is a sufficient strengthening of Lehmer, not a consequence of the displayed convolution.
Exact mathematical passage
\sigma_3(p)=1+p^3
p^4Use in source · 11 passages
Search aliases: p^4
Use in source · chapter22.tex, line 75
The index-dependent criterion already available from Deligne's bound is τ(p)=0iff p^6midτ(p), p≥5. Combined with the divisor convolution, it becomes 840S_4(p)not≡ p^4C(p)pmodp^6, p>7, as the exact desired exclusion. This gives a correctly typed target for a convolution route. No uniform arithmetic property excluding that residue has been proved here. A stronger valuation bound may suffice, but its proof must permit the zero case as a possibility until the bound itself excludes it.
Exact mathematical passage
\label{eq:closeoutconvolution} 840S_4(p)\not\equiv p^4C(p)\pmod{p^6}, \qquad p>7,
p^5Use in source · 2 passages
Search aliases: p^5
Use in source · chapters04_05.tex, line 114
The resulting first moments are s_2(p) · =p^2-1, s_4(p) · =2p^3-3p-1, s_6(p) · =5p^4-9p^2-5p-1, s_8(p) · =14p^5-28p^3-20p^2-7p-1, s_10(p) · =42p^6-90p^4-75p^3-35p^2-9p-1-τ(p). The integer 42 is the fifth Catalan number. It records the contribution from s_0(p)=p after inversion. The lower-degree terms come from the constant -1 in the trace formula. The coefficient of τ(p) is -1 because the last diagonal entry is 1.
Exact mathematical passage
s_2(p)&=p^2-1,\\ s_4(p)&=2p^3-3p-1,\\ s_6(p)&=5p^4-9p^2-5p-1,\\ s_8(p)&=14p^5-28p^3-20p^2-7p-1,\\ s_{10}(p)&=42p^6-90p^4-75p^3-35p^2-9p-1-\tau(p). \label{eq:s10}
p^6Use in source · 7 passages
Search aliases: p^6
Use in source · chapter22.tex, line 71
The index-dependent criterion already available from Deligne's bound is τ(p)=0iff p^6midτ(p), p≥5. Combined with the divisor convolution, it becomes 840S_4(p)not≡ p^4C(p)pmodp^6, p>7, as the exact desired exclusion. This gives a correctly typed target for a convolution route. No uniform arithmetic property excluding that residue has been proved here. A stronger valuation bound may suffice, but its proof must permit the zero case as a possibility until the bound itself excludes it.
Exact mathematical passage
\tau(p)=0\iff p^6\mid\tau(p),\qquad p\ge5.
p^{m-1}Use in source · 2 passages
Search aliases: p^m-1
Use in source · chapters15_16.tex, line 98
refeq:trexpansion shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading term comes from u^2m. The constant term comes from H_2m(0)=(-p)^m and H_2m-2(0)=(-p)^m-1.
Exact mathematical passage
\label{eq:fm} F_m(0)=(-1)^mp^{m-1}(p+1),\qquad F_m(x)\equiv x^{m-1}(x-1)\pmod p.
p^{m_p}Use in source · 1 passage
Search aliases: p^m_p
Use in source · chapters15_16.tex, line 119
All prime cases have now been covered. If s=∏ p^m_p, multiplicativity gives R(s^2)=∏_pR(p^2m_p). Every factor is nonzero, and R(1)=1. This proves Theorem~[source reference].
Exact mathematical passage
s=\prod p^{m_p}
p^rUse in source · 15 passages
Search aliases: p^r
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
T_r=\tau(p^r)
p^{r+1}Use in source · 1 passage
Search aliases: p^r+1
Use in source · chapters01_03.tex, line 131
Why the problem lives at primes. The Hecke contracts. The normalized form Δ is a Hecke eigenform. Its coefficients satisfy τ(mn) · =τ(m)τ(n), · · gcd(m,n)=1, τ(p^r+1) · =τ(p)τ(p^r)-p^11τ(p^r-1), · · r≥1. The exponent 11 again records weight 12. The constant term for the local recurrence is τ(1)=1, and the first parameter is a=τ(p). These classical statements occur in Lehmer's original paper and in the modern source [source reference].
Exact mathematical passage
\tau(mn)&=\tau(m)\tau(n),&&\gcd(m,n)=1,\label{eq:mult}\\ \tau(p^{r+1})&=\tau(p)\tau(p^r)-p^{11}\tau(p^{r-1}),&&r\ge1.\label{eq:hecke}
p^{r-1}Use in source · 3 passages
Search aliases: p^r-1
Use in source · chapters01_03.tex, line 131
Why the problem lives at primes. The Hecke contracts. The normalized form Δ is a Hecke eigenform. Its coefficients satisfy τ(mn) · =τ(m)τ(n), · · gcd(m,n)=1, τ(p^r+1) · =τ(p)τ(p^r)-p^11τ(p^r-1), · · r≥1. The exponent 11 again records weight 12. The constant term for the local recurrence is τ(1)=1, and the first parameter is a=τ(p). These classical statements occur in Lehmer's original paper and in the modern source [source reference].
Exact mathematical passage
\tau(mn)&=\tau(m)\tau(n),&&\gcd(m,n)=1,\label{eq:mult}\\ \tau(p^{r+1})&=\tau(p)\tau(p^r)-p^{11}\tau(p^{r-1}),&&r\ge1.\label{eq:hecke}
p^{r-2}Use in source · 1 passage
Search aliases: p^r-2
Use in source · chapters19_21.tex, line 177
For every real B≥1 and integer M≥1, there exist a prime q>B and a multiplicative integer sequence f with the following properties: item f(1)=1 and f(n)=τ(n) for every n≤ B. item f(n)≡τ(n)pmod10M for every positive integer n. item Every weight-12 prime-power recurrence holds: f(p^r)=f(p)f(p^r-1)-p^11f(p^r-2), r≥2. item |f(n)|≤ d(n)n^11/2, where d(n) counts the positive divisors of n. item f(n) is odd exactly when n is an odd square, and f(n)≡ nσ(n)pmod5. item The arithmetic primitive transform R_…
Exact mathematical passage
f(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\qquad r\ge2.
p^{r-i}Use in source · 1 passage
Search aliases: p^r-i
Use in source · chapters15_16.tex, line 41
A residue eliminates three prime classes. Suppose p≠5 and put χ=chifive(p). Modulo 5, χ≡ p^2, so the two recurrence roots may be taken as p and χ: p+χ≡ a, pχ≡ p^11pmod5. The complete homogeneous recurrence coefficient is T_r≡ C_r=∑_i=0^r p^r-iχ^ipmod5. Also χ p^10≡1pmod5. Since χ^2=1, for r≥2 the sum telescopes: R(p^r)≡ C_r-C_r-2 =p^r-1(p+χ)pmod5. The same expression holds at r=1. It is nonzero for p≡1,2,3pmod5. It proves nonvanishing for all positive exponents in those prime classes. Only p=5 a…
Exact mathematical passage
T_r\equiv C_r=\sum_{i=0}^r p^{r-i}\chi^i\pmod5.
p^{r-j}Use in source · 2 passages
Search aliases: p^r-j
Use in source · chapters04_05.tex, line 97
The Eichler--Selberg trace identities give, for r≥1, ∑_j=0^r(-1)^r-jbinomr+jr-jp^r-js_2j(p) =-1-Tr T_2r+2(p), together with s_0(p)=p. Here T_k(p) is the Hecke operator at p on S_k(SL_2(Z)). It is distinct from the scalar recurrence notation T_r=τ(p^r) at a fixed prime. The trace normalization and Catalan inversion are checked against [source reference], Sections 4--6. The cusp spaces in weights 4,6,8,10 are zero. The weight-12 cusp space is spanned by Δ, so its Hecke trace is τ(p).
Exact mathematical passage
\label{eq:traceidentity} \sum_{j=0}^r(-1)^{r-j}\binom{r+j}{r-j}p^{r-j}s_{2j}(p) =-1-\Tr T_{2r+2}(p),
p^{r_p}Use in source · 3 passages
Search aliases: p^r_p
Use in source · chapters01_03.tex, line 203
· and τ(p)=0. In particular, Lehmer's conjecture is equivalent to nonvanishing at every prime. Factor n=∏_p p^r_p. Multiplicativity expresses τ(n) as a finite product of local blocks. An integral product is zero exactly when one factor is zero. Apply [source reference] to each factor.
Exact mathematical passage
n=\prod_p p^{r_p}
p_1Use in source · 1 passage
Search aliases: p_1
Use in source · chapters11_12.tex, line 176
Take the prime indices (u_1,p_1)=(15,245759), (u_2,p_2)=(47,770047). The exact values are τ(245759) · =77900595298037436891536424960,
Exact mathematical passage
(u_1,p_1)=(15,245759),\qquad(u_2,p_2)=(47,770047).
P_{16249}Use in source · 2 passages
Search aliases: P_16249
Use in source · appendix_workbook.tex, line 198
begingroupsmall setlengthtabcolsep4pt r r r r j · B_j in P_16249 · C_j in G_16249 · j+v_5(B_j)
Exact mathematical passage
P_{16249}
p_2Use in source · 1 passage
Search aliases: p_2
Use in source · chapters11_12.tex, line 176
Take the prime indices (u_1,p_1)=(15,245759), (u_2,p_2)=(47,770047). The exact values are τ(245759) · =77900595298037436891536424960,
Exact mathematical passage
(u_1,p_1)=(15,245759),\qquad(u_2,p_2)=(47,770047).
P_{31}Use in source · 1 passage
Search aliases: P_31
Use in source · chapters06_07.tex, line 203
Individual packet polynomials have rational finite roots in [-4,4], since each root is -(i-j)/(c_i-c_j) with a nonzero integer denominator. The aggregate polynomial need not retain that root region. Exact values at p=31 are P_31(10)=-56767541184<0, P_31(11)=4645170845280>0. It has a real root between 10 and 11. Summing signed real-rooted factors does not preserve their root bounds.
Exact mathematical passage
P_{31}(10)=-56767541184<0,\qquad P_{31}(11)=4645170845280>0.
P_{499}Use in source · 2 passages
Search aliases: P_499
Use in source · appendix_workbook.tex, line 121
begingroupsmall setlengthtabcolsep5pt r r r r j · B_j in P_499 · C_j in G_499 · j+v_5(B_j)
Exact mathematical passage
P_{499}
p_kUse in source · 1 passage
Search aliases: p_k
Use in source · chapters17_18.tex, line 54
Separating the quasimodular weights. We need only an elementary weight-separation statement. Suppose finitely many quasimodular functions F_k of distinct nonnegative integer weights satisfy F_k(γ z)=∑_r=0^p_kf_k,r(z)c^r(cz+d)^k-r, f_k,0=F_k, p_k≤ k, for γ=[a · b
Exact mathematical passage
F_k(\gamma z)=\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r}, \qquad f_{k,0}=F_k,\qquad p_k\le k,
P_pUse in source · 1 passage
Search aliases: P_p
Use in source · chapters06_07.tex, line 156
A double residual root with no lift. At p=16249, the shell has 16250 packets. The weighted coefficient valuations of P_p(5U) are (4,5,4,4,6,5,7,8,9,9,11), so μ_p=4. The reduction is overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)pmod5. The residual root at 1 is double. Exact coefficient reduction gives G_p(1)≡15pmod25, G'_p(1)≡0pmod5. For every y∈Z_5, polynomial Taylor expansion has an integral remainder: G_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y), H∈Z_5[y]. Consequently G_p(1+5y)≡15pmod25 for every y∈Z_5. No …
Exact mathematical passage
P_p(5U)
P_vUse in source · 3 passages
Search aliases: P_v
Use in source · appendix_workbook.tex, line 84
For a packet v, set c_i=(v_i-i)/5 and form P_v(T)=∏_0≤ i<j≤4((i-j)+T(c_i-c_j)). Its physical value is P_v(5)=288W(v). Summing the full shell gives P_n(T)=∑_v∈VnP_v(T)=∑_j=0^10B_jT^j, P_n(5)=288τ(n). The polynomial has a different variable from the prime index. A root in T or in a rescaled variable does not represent a nearby prime coefficient.
Exact mathematical passage
P_v(T)=\prod_{0\le i<j\le4}\bigl((i-j)+T(c_i-c_j)\bigr).
P_zUse in source · 2 passages
Search aliases: P_z
Use in source · chapters17_18.tex, line 61
Fix z in the upper half-plane. The polynomial P_z(c,d)=∑_k∑_r f_k,r(z)c^r(cz+d)^k-r vanishes at every relatively prime integer pair (c,d), because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime c. Infinitely many integers d are coprime to it, so the polynomial in d vanishes identically. Each of its coefficient polynomials in c then vanishes at infinitely many primes, so is identically zero. Hen…
Exact mathematical passage
\label{eq:weightpolynomial} P_z(c,d)=\sum_k\sum_r f_{k,r}(z)c^r(cz+d)^{k-r}
\pmodUse in source · 63 passages
Search aliases: pmod
Use in source · appendices.tex, line 18
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d, 1 · d≡1,4pmod5, -1 · d≡2,3pmod5. The character value and the tenth power in the scaling formula are both essential.
Exact mathematical passage
\chifive(d)= \begin{cases} 0&5\mid d,\\ 1&d\equiv1,4\pmod5,\\ -1&d\equiv2,3\pmod5. \end{cases}
\QUse in source · 20 passages
Search aliases: Q
Use in source · appendix_workbook.tex, line 342
The coarse coefficient vector a∈Q^18 makes x=s_f+M_fa, Ax=(0,0,0,0,0,-52843168)^t. The certificate then chooses a coefficient vector d∈ker M_c. Its final fine vector y=x+M_fd is zero except at index 6, where it equals -52843168.
Exact mathematical passage
a\in\Q^{18}
q^{(2j+1)^2}Use in source · 1 passage
Search aliases: q^(2j+1)^2
Use in source · chapters01_03.tex, line 111
Parity from the actual product. Modulo 2, the Frobenius identity gives ∏_m≥1(1-q^m)^24 ≡∏_m≥1(1-q^8m)^3. Using [source reference], whose coefficients 2j+1 are odd, gives Δ(q)≡∑_j≥0q^1+4j(j+1) =∑_j≥0q^(2j+1)^2pmod2. Distinct nonnegative j give distinct exponents. Therefore τ(n) is odd Longleftrightarrow n is an odd square. In particular, τ(p) is even at every odd prime. This elementary global property will later make the primitive odd-block multiplier odd. It is useful arithmetic, but is consiste…
Exact mathematical passage
\Delta(q)\equiv\sum_{j\ge0}q^{1+4j(j+1)} =\sum_{j\ge0}q^{(2j+1)^2}\pmod2.
q^1Use in source · 1 passage
Search aliases: q^1
Use in source · chapters08_10.tex, line 112
Why a finite initial segment suffices. The level-one valence formula implies that a nonzero holomorphic modular form of weight k has order at the cusp at most k/12 [source reference]. A cusp form whose coefficients of q^1,…,q^lfloor k/12rfloor all vanish has order at least lfloor k/12rfloor+1>k/12. It must be zero.
Exact mathematical passage
q^1,\ldots,q^{\lfloor k/12\rfloor}
q^{1+4j(j+1)}Use in source · 1 passage
Search aliases: q^1+4j(j+1)
Use in source · chapters01_03.tex, line 111
Parity from the actual product. Modulo 2, the Frobenius identity gives ∏_m≥1(1-q^m)^24 ≡∏_m≥1(1-q^8m)^3. Using [source reference], whose coefficients 2j+1 are odd, gives Δ(q)≡∑_j≥0q^1+4j(j+1) =∑_j≥0q^(2j+1)^2pmod2. Distinct nonnegative j give distinct exponents. Therefore τ(n) is odd Longleftrightarrow n is an odd square. In particular, τ(p) is even at every odd prime. This elementary global property will later make the primitive odd-block multiplier odd. It is useful arithmetic, but is consiste…
Exact mathematical passage
\Delta(q)\equiv\sum_{j\ge0}q^{1+4j(j+1)} =\sum_{j\ge0}q^{(2j+1)^2}\pmod2.
q^{10}Use in source · 1 passage
Search aliases: q^10
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
q^{11}Use in source · 1 passage
Search aliases: q^11
Use in source · chapters19_21.tex, line 224
The full size bound survives. At primes other than q, the Hecke roots have absolute value p^11/2 by Deligne. Formula [source reference] therefore gives |τ(p^r)|≤(r+1)p^11r/2. At the modified prime, F_q,2m=(-q^11)^m, F_q,2m+1=0. These obey the same local bound at every exponent, including zero. Multiplying the local estimates gives |f(n)|≤∏_p^r_pVert n(r_p+1)p^11r_p/2 =d(n)n^11/2. A zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified paramet…
Exact mathematical passage
F_{q,2m}=(-q^{11})^m,\qquad F_{q,2m+1}=0.
q^{11m-1}Use in source · 1 passage
Search aliases: q^11m-1
Use in source · chapters19_21.tex, line 244
All primitive squares still survive. Define the arithmetic transform R_f(n)=∑_d^2mid nμ(d)chifive(d)d^10f(n/d^2). It is multiplicative. Its prime blocks are R_f(p)=a_p, and for r≥2, R_f(p^r)=F_p,r-chifive(p)p^10F_p,r-2. For p≠ q, these are the actual R blocks and their even terms are nonzero. At q, direct substitution and chifive(q)=1 give R_f(q^2m)=(-1)^mq^11m-1(q+1)≠0, m≥1. Thus every positive-exponent factor of R_f(s^2) is nonzero, and R_f(1)=1, so the transform is nonzero at every square. Me…
Exact mathematical passage
R_f(q^{2m})=(-1)^mq^{11m-1}(q+1)\ne0,\qquad m\ge1.
\Q^{18}Use in source · 1 passage
Search aliases: Q^18 Q
Use in source · appendix_workbook.tex, line 342
The coarse coefficient vector a∈Q^18 makes x=s_f+M_fa, Ax=(0,0,0,0,0,-52843168)^t. The certificate then chooses a coefficient vector d∈ker M_c. Its final fine vector y=x+M_fd is zero except at index 6, where it equals -52843168.
Exact mathematical passage
a\in\Q^{18}
q^2Use in source · 3 passages
Search aliases: q^2
Use in source · appendix_workbook.tex, line 29
· =-2024+576-24=-1472. The last line comes respectively from the q^3 term in the first factor, the product of its q term with the second factor's q^2 term, and the third factor's q^3 term. Thus it computes τ(4), not τ(3).
Exact mathematical passage
q^2
q^{2m}Use in source · 1 passage
Search aliases: q^2m
Use in source · chapters19_21.tex, line 244
All primitive squares still survive. Define the arithmetic transform R_f(n)=∑_d^2mid nμ(d)chifive(d)d^10f(n/d^2). It is multiplicative. Its prime blocks are R_f(p)=a_p, and for r≥2, R_f(p^r)=F_p,r-chifive(p)p^10F_p,r-2. For p≠ q, these are the actual R blocks and their even terms are nonzero. At q, direct substitution and chifive(q)=1 give R_f(q^2m)=(-1)^mq^11m-1(q+1)≠0, m≥1. Thus every positive-exponent factor of R_f(s^2) is nonzero, and R_f(1)=1, so the transform is nonzero at every square. Me…
Exact mathematical passage
R_f(q^{2m})=(-1)^mq^{11m-1}(q+1)\ne0,\qquad m\ge1.
q^3Use in source · 3 passages
Search aliases: q^3
Use in source · appendix_workbook.tex, line 29
· =-2024+576-24=-1472. The last line comes respectively from the q^3 term in the first factor, the product of its q term with the second factor's q^2 term, and the third factor's q^3 term. Thus it computes τ(4), not τ(3).
Exact mathematical passage
q^3
q^4Use in source · 3 passages
Search aliases: q^4
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
q^5Use in source · 1 passage
Search aliases: q^5
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
q^6Use in source · 1 passage
Search aliases: q^6
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
q^7Use in source · 1 passage
Search aliases: q^7
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
q^8Use in source · 1 passage
Search aliases: q^8
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
q^{8m}Use in source · 1 passage
Search aliases: q^8m
Use in source · chapters01_03.tex, line 106
Parity from the actual product. Modulo 2, the Frobenius identity gives ∏_m≥1(1-q^m)^24 ≡∏_m≥1(1-q^8m)^3. Using [source reference], whose coefficients 2j+1 are odd, gives Δ(q)≡∑_j≥0q^1+4j(j+1) =∑_j≥0q^(2j+1)^2pmod2. Distinct nonnegative j give distinct exponents. Therefore τ(n) is odd Longleftrightarrow n is an odd square. In particular, τ(p) is even at every odd prime. This elementary global property will later make the primitive odd-block multiplier odd. It is useful arithmetic, but is consiste…
Exact mathematical passage
\prod_{m\ge1}(1-q^m)^{24} \equiv\prod_{m\ge1}(1-q^{8m})^3.
q^9Use in source · 1 passage
Search aliases: q^9
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
\Q^hUse in source · 1 passage
Search aliases: Q^h Q
Use in source · chapters13_14.tex, line 117
Mass vectors and aggregation Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
a\in\Q^h
q^jUse in source · 1 passage
Search aliases: q^j
Use in source · appendix_workbook.tex, line 32
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
q\frac{F'(q)}{F(q)}=-24\sum_{j\ge1}\sigma_1(j)q^j
q^{j(j+1)/2}Use in source · 1 passage
Search aliases: q^j(j+1)/2
Use in source · chapters01_03.tex, line 80
Jacobi's triangular series. Jacobi's identity is ∏_m≥1(1-q^m)^3 =∑_j≥0(-1)^j(2j+1)q^j(j+1)/2. Raising this series to the eighth power and multiplying by q gives Δ. The sparse triangular exponents make it useful for coefficient calculations. The source check in [source reference] also uses this identity to establish the parity criterion discussed below. It is an exact formal series identity, not a numerical approximation.
Exact mathematical passage
\label{eq:jacobi} \prod_{m\ge1}(1-q^m)^3 =\sum_{j\ge0}(-1)^j(2j+1)q^{j(j+1)/2}.
q^kUse in source · 3 passages
Search aliases: q^k
Use in source · appendix_workbook.tex, line 12
Write F(q)=fracΔ(q)q=∏_k≥1(1-q^k)^24 =∑_m≥0f_mq^m. Then f_0=1 and τ(n)=f_n-1. This shift is essential. A program that stores τ(0)=0 as a sentinel must not confuse that entry with the constant coefficient f_0.
Exact mathematical passage
F(q)=\frac{\Delta(q)}q=\prod_{k\ge1}(1-q^k)^{24} =\sum_{m\ge0}f_mq^m.
q^{kj}Use in source · 1 passage
Search aliases: q^kj
Use in source · appendix_workbook.tex, line 19
For direct product expansion, terms of degree at most m depend only on factors with k≤ m. The exact binomial expansion (1-q^k)^24=∑_j=0^24(-1)^jbinom24jq^kj therefore supplies a finite algorithm for each desired coefficient. Three short calculations are f_1 · =-24,
Exact mathematical passage
(1-q^k)^{24}=\sum_{j=0}^{24}(-1)^j\binom{24}{j}q^{kj}
q^{\lfloor k/12\rfloor}Use in source · 1 passage
Search aliases: q^lfloor k/12rfloor lfloor rfloor
Use in source · chapters08_10.tex, line 112
Why a finite initial segment suffices. The level-one valence formula implies that a nonzero holomorphic modular form of weight k has order at the cusp at most k/12 [source reference]. A cusp form whose coefficients of q^1,…,q^lfloor k/12rfloor all vanish has order at least lfloor k/12rfloor+1>k/12. It must be zero.
Exact mathematical passage
q^1,\ldots,q^{\lfloor k/12\rfloor}
q^mUse in source · 8 passages
Search aliases: q^m
Use in source · appendices.tex, line 7
The coefficient and its local parameters. The normalized form is Δ(q)=q∏_m≥1(1-q^m)^24 =∑_n≥1τ(n)q^n, τ(1)=1. All coefficient indices in this book are positive integers. A stored value at index zero is a programming sentinel unless a displayed formula explicitly introduces the coefficients of Δ/q. For the latter coefficients, f_m=τ(m+1) and f_0=1.
Exact mathematical passage
\Delta(q)=q\prod_{m\ge1}(1-q^m)^{24} =\sum_{n\ge1}\tau(n)q^n,\qquad \tau(1)=1.
\Q^NUse in source · 1 passage
Search aliases: Q^N Q
Use in source · chapters17_18.tex, line 125
A shell-null interpolant can fail on the first shell. Let m_i be the nonzero mass of the ith orbit. Interpolation makes the map Q↦(m_iQ(v_i)) onto Q^N. Thus any zero-total transfer on this one shell can be produced by a homogeneous polynomial.
Exact mathematical passage
\Q^N
q^NUse in source · 1 passage
Search aliases: q^N
Use in source · chapters01_03.tex, line 3
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag
Exact mathematical passage
q^N
q^rUse in source · 3 passages
Search aliases: q^r
Use in source · chapters01_03.tex, line 59
Exact coefficient engines. The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
(1-q^r)^{24}
q^{rk}Use in source · 1 passage
Search aliases: q^rk
Use in source · chapters01_03.tex, line 60
oefficient engines The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
-24\frac{rq^r}{1-q^r}=-24\sum_{k\ge1}rq^{rk}.
\Q^{s_k}Use in source · 1 passage
Search aliases: Q^s_k Q
Use in source · chapters13_14.tex, line 126
Put H_k=ker(1^T:Q^s_k→Q), T_k=im M_k, E_k=H_k/T_k. The quotient E_k is the vector space of total-preserving changes that are unavailable from the chosen family. It is not a count of failed sign certificates. A sign certificate may exist even when E_k is nonzero.
Exact mathematical passage
H_k=\ker(\mathbf1^T:\Q^{s_k}\to\Q),\qquad T_k=\im M_k,\qquad E_k=H_k/T_k.
q^{T_j}Use in source · 1 passage
Search aliases: q^T_j
Use in source · appendix_workbook.tex, line 48
Jacobi gives a third engine. Put T_j=j(j+1)/2 and a_j=(-1)^j(2j+1). The series A(q)=∏_k≥1(1-q^k)^3=1+∑_j≥1a_jq^T_j satisfies F=A^8. The identity AqF'=8FqA' yields m f_m=∑_substackj≥1
Exact mathematical passage
A(q)=\prod_{k\ge1}(1-q^k)^3=1+\sum_{j\ge1}a_jq^{T_j}
q^{|\lambda|}Use in source · 1 passage
Search aliases: q^|λ| lambda
Use in source · chapters06_07.tex, line 6
The Nekrasov--Okounkov identity in Han's formulation is ∏_m≥1(1-q^m)^z-1 =∑_λ q^|λ|∏_h∈ H(λ)(1-frac zh^2). At z=25, a partition having a hook of length 5 contributes zero. The surviving terms are the 5-cores. Multiplication by q shows that the coefficient τ(n) is the sum of the hook products over 5-cores of size n-1.
Exact mathematical passage
\label{eq:hookidentity} \prod_{m\ge1}(1-q^m)^{z-1} =\sum_\lambda q^{|\lambda|}\prod_{h\in H(\lambda)}\left(1-\frac z{h^2}\right).
Q_1Use in source · 1 passage
Search aliases: Q_1
Use in source · chapters13_14.tex, line 112
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
Q_1,\ldots,Q_h
Q_{10}Use in source · 5 passages
Search aliases: Q_10
Use in source · chapters08_10.tex, line 129
· +69600s_3s_5+45000s_4^2, and Q_10= · 2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2notag · +409600s_2s_3s_5+15000s_2s_4^2notag · -30000s_3^2s_4-264000s_5^2. Both DQ_8 and DQ_10 are harmonic. The first shell has the vector (0,1,2,-2,-1), with s_2=10, s_3=0, s_4=34, s_5=0. Both polynomials vanish there. Their theta weights are 20 and 22. The cusp spaces are respectively spanned by Δ E_4^2 and Δ E_4E_6, each with nonzero first coefficient. The zero first coefficient therefore kills the whole theta form…
Exact mathematical passage
\label{eq:q10} Q_{10}={}&2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2\notag\\ &+409600s_2s_3s_5+15000s_2s_4^2\notag\\ &-30000s_3^2s_4-264000s_5^2.
Q_8Use in source · 7 passages
Search aliases: Q_8
Use in source · chapters08_10.tex, line 124
The first explicit corrections. The construction gives Q_8= · 6273s_2^4-33750s_2^2s_4-56800s_2s_3^2notag · +69600s_3s_5+45000s_4^2, and Q_10= · 2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2notag
Exact mathematical passage
\label{eq:q8} Q_8={}&6273s_2^4-33750s_2^2s_4-56800s_2s_3^2\notag\\ &+69600s_3s_5+45000s_4^2,
Q_dUse in source · 6 passages
Search aliases: Q_d
Use in source · chapters17_18.tex, line 10
Let Q be any finite symmetric polynomial over C, restricted to ∑_i v_i=0. Suppose ∑_v∈VnW(v)Q(v)=0 for every positive integer n. Write Q=∑_dQ_d in homogeneous degrees. Then every Q_d is separately universally null. Moreover, the weighted Q-sum is zero in each exact-content sector of every shell.
Exact mathematical passage
Q=\sum_dQ_d
Q_hUse in source · 1 passage
Search aliases: Q_h
Use in source · chapters13_14.tex, line 112
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
Q_1,\ldots,Q_h
q_iUse in source · 3 passages
Search aliases: q_i
Use in source · chapters08_10.tex, line 245
Completeness retains the total. Suppose a shell has nonzero orbit masses w_1,…,w_N and total T. Even allow every rational correction vector q with ∑_iw_iq_i=0. Then ∃ q: w_i(1+q_i) have one weak sign and are not all zero Longleftrightarrow T≠0. Necessity follows from conservation. For sufficiency, if T≠0, take q_i=fracTNw_i-1. Every corrected mass is T/N and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of T≠0 without independent information …
Exact mathematical passage
\sum_iw_iq_i=0
Q_jUse in source · 7 passages
Search aliases: Q_j
Use in source · chapters08_10.tex, line 69
nThe correction that does not change any coefficient A universally null polynomial is a symmetric polynomial Q with ∑_v∈VnW(v)Q(v)=0 for every n≥1. This is much stronger than being null on one selected shell. If a finite family satisfies [source reference], any rational linear combination of it does too. Thus F(v)=1+∑_j a_jQ_j(v) has the exact property ∑_v∈VnW(v)F(v)=τ(n) at every index, after the coefficients a_j have been fixed. The choice of coefficients may depend on a target shell. The null…
Exact mathematical passage
F(v)=1+\sum_j a_jQ_j(v)
\Q_pUse in source · 8 passages
Search aliases: Q_p Q
Use in source · chapters19_21.tex, line 98
For every prime p, there is a weakly admissible two-dimensional filtered φ-module over Q_p, with monodromy N=0, filtration jumps 0,11 and Frobenius characteristic polynomial X^2+p^11.
Exact mathematical passage
\Q_p
Q_{p,j}Use in source · 2 passages
Search aliases: Q_p,j
Use in source · chapter22.tex, line 82
A sufficient sign input. A second precise route would construct permitted corrections with an independently proved sign. At each remaining prime, choose finitely many rational universal null polynomials Q_p,j and rational coefficients a_p,j, and put F_p(v)=1+∑_j a_p,jQ_p,j(v). The choice may depend on p. Universal nullness must hold after the choice, so its weighted sum is still the actual τ(p).
Exact mathematical passage
Q_{p,j}
\Q_p^\timesUse in source · 1 passage
Search aliases: Q_p^× Q times
Use in source · chapters19_21.tex, line 140
There is no proper nonzero φ-stable Q_p-subspace. In dimension two such a subspace would be a line, and φ on it would have an eigenvalue λ∈Q_p satisfying λ^2=-p^11. This would require 2v_p(λ)=11, impossible for an integer-valued valuation on Q_p^×. Thus the proper-subobject inequalities are vacuous. The zero object imposes equality 0=0, and the whole object satisfies the total equality. The filtered module is weakly admissible.
Exact mathematical passage
\Q_p^\times
\Q_\ellUse in source · 1 passage
Search aliases: Q_ℓ Q ell
Use in source · chapters19_21.tex, line 22
Integral representations and complex conjugation. For each prime ℓ, the representation attached to Δ has the arithmetic-Frobenius normalization ρ_ℓ:G_Q⟶ GL_2(Q_ℓ), is continuous and unramified outside ℓ, and satisfies det(X-ρ_ℓ(Frob_p))=X^2-τ(p)X+p^11, p≠ℓ. Its determinant is the eleventh power of the ℓ-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the prima…
Exact mathematical passage
\rho_\ell:G_{\Q}\longrightarrow GL_2(\Q_\ell),
\RUse in source · 3 passages
Search aliases: R
Use in source · appendix_operator.tex, line 3
The universal null construction depends on an exact differential operator. This appendix derives that operator from the Euclidean metric, rather than treating its power-sum formulas as a black box. It also explains the Newton reductions used by the implementation and gives two low-degree harmonic checks. All polynomial identities are on the hyperplane H={v∈R^5:∑_i v_i=0}, with its induced Euclidean metric. The modularity and valence contracts remain the ones stated in Chapters~[source reference]…
Exact mathematical passage
H=\{v\in\R^5:\sum_i v_i=0\}
\R^5Use in source · 3 passages
Search aliases: R^5 R
Use in source · appendix_operator.tex, line 3
The universal null construction depends on an exact differential operator. This appendix derives that operator from the Euclidean metric, rather than treating its power-sum formulas as a black box. It also explains the Newton reductions used by the implementation and gives two low-degree harmonic checks. All polynomial identities are on the hyperplane H={v∈R^5:∑_i v_i=0}, with its induced Euclidean metric. The modularity and valence contracts remain the ones stated in Chapters~[source reference]…
Exact mathematical passage
H=\{v\in\R^5:\sum_i v_i=0\}
R_fUse in source · 9 passages
Search aliases: R_f
Use in source · chapters19_21.tex, line 182
B. item f(n)≡τ(n)pmod10M for every positive integer n. item Every weight-12 prime-power recurrence holds: f(p^r)=f(p)f(p^r-1)-p^11f(p^r-2), r≥2. item |f(n)|≤ d(n)n^11/2, where d(n) counts the positive divisors of n. item f(n) is odd exactly when n is an odd square, and f(n)≡ nσ(n)pmod5. item The arithmetic primitive transform R_f is nonzero at every square. item Nevertheless f(q)=R_f(q)=0.
Exact mathematical passage
R_f
r_kUse in source · 4 passages
Search aliases: r_k
Use in source · chapters13_14.tex, line 140
The exact refinement sequence. There is a short exact sequence 0⟶ fracker AM_k+1(ker M_k) ⟶ E_k+1⟶ E_k⟶0. Consequently, if r_k=rank M_k, dim E_k+1-dim E_k =(s_k+1-s_k)-(r_k+1-r_k). The aggregation map sends H_k+1 onto H_k. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.
Exact mathematical passage
r_k=\rank M_k
r_{k+1}Use in source · 2 passages
Search aliases: r_k+1
Use in source · chapters13_14.tex, line 141
The exact refinement sequence. There is a short exact sequence 0⟶ fracker AM_k+1(ker M_k) ⟶ E_k+1⟶ E_k⟶0. Consequently, if r_k=rank M_k, dim E_k+1-dim E_k =(s_k+1-s_k)-(r_k+1-r_k). The aggregation map sends H_k+1 onto H_k. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.
Exact mathematical passage
\label{eq:defectgrowth} \dim E_{k+1}-\dim E_k =(s_{k+1}-s_k)-(r_{k+1}-r_k).
r_pUse in source · 4 passages
Search aliases: r_p
Use in source · chapters01_03.tex, line 203
· and τ(p)=0. In particular, Lehmer's conjecture is equivalent to nonvanishing at every prime. Factor n=∏_p p^r_p. Multiplicativity expresses τ(n) as a finite product of local blocks. An integral product is zero exactly when one factor is zero. Apply [source reference] to each factor.
Exact mathematical passage
n=\prod_p p^{r_p}
R_QUse in source · 6 passages
Search aliases: R_Q
Use in source · chapters13_14.tex, line 58
Homogeneous universal nulls preserve each content. Let Q be a symmetric homogeneous polynomial of degree d that is universally null. Then its weighted sum is zero separately in every exact-content sector on every shell. Let R_Q(n)=∑_substackv∈Vn g(v)=1W(v)Q(v). Symmetry removes the residue permutation from Q, and homogeneity gives Q(v)=g^dQ(u). The scaling bijection yields ∑_v∈VnW(v)Q(v) =∑_g^2mid nchifive(g)g^10+dR_Q(n/g^2). The left side is zero. At n=1, only g=1 contributes, so R_Q(1)=0. Indu…
Exact mathematical passage
R_Q(n)=\sum_{\substack{v\in\Vn\\g(v)=1}}W(v)Q(v).
\rankUse in source · 3 passages
Search aliases: rank
Use in source · appendix_workbook.tex, line 335
The rational row-reduction results are rank M_c=5, rank M_f=15, rank B=10. An independent check obtains the same lower bounds over both prime fields F_65537 and F_1000003. Every denominator used in a reduction is invertible there. A nonzero minor in either field proves a nonzero rational minor after clearing denominators.
Exact mathematical passage
\rank M_c=5,\qquad\rank M_f=15,\qquad\rank B=10.
\rceilUse in source · 1 passage
Search aliases: rceil
Use in source · chapters11_12.tex, line 20
The sharp threshold. Let k≥1, M=2^k, and set N_k=lceilfrac100M^2+10M+235rceil =frac20M^2+2M+γ_k7, where γ_k= 7 · k≡1pmod3,
Exact mathematical passage
\label{eq:nk} N_k=\left\lceil\frac{100M^2+10M+2}{35}\right\rceil =\frac{20M^2+2M+\gamma_k}{7},
SUse in source · 32 passages
Search aliases: S
Use in source · appendices.tex, line 116
S15: differential equations. Fonseca records the normalized Ramanujan differential relations used in the polynomial computation [source reference]. The project-specific S_4 cancellation is verified algebraically in the main text and by an exact symbolic remainder. Modularity and the level-one dimension input for the cusp identity are separate contracts [source reference]. A finite output check is not substituted for the universal algebraic identity.
Exact mathematical passage
S_4
S^2Use in source · 3 passages
Search aliases: S^2
Use in source · chapters06_07.tex, line 74
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
(x-y)^2=2T-S^2.
s^2Use in source · 7 passages
Search aliases: s^2
Use in source · chapter22.tex, line 33
R(s^2)≠0 for all s · Every even primitive local block survives, including a hypothetical zero prime parameter.
Exact mathematical passage
R(s^2)\ne0
S^3Use in source · 1 passage
Search aliases: S^3
Use in source · chapters08_10.tex, line 49
, since the squared residues sum to 30. It is even, since a_i^2≡ a_ipmod2 and ∑_i a_i=0. The quadratic exponent ∑_i a_i^2/10 is integral. The Weil operator T therefore acts as 1 on the sign line. The Weil action commutes with S_5, so this line is invariant. The rank is even, and the representation descends to SL_2(Z). On the line the relation (ST)^3=S^2 reduces to S^3=S^2, giving S=1.
Exact mathematical passage
S^3=S^2
s_0Use in source · 10 passages
Search aliases: s_0
Use in source · appendix_operator.tex, line 5
Write s_m=∑_i=0^4v_i^m, with s_0=5 and s_1=0, and let D=∏_i<j(v_i-v_j). We distinguish the ordinary Laplacian Δ_H from the conjugated operator L defined by Δ_H(DQ)=D LQ. The factor D changes radial coefficients substantially. Confusing the two operators would invalidate the harmonicity test.
Exact mathematical passage
s_0=5
s_0^2Use in source · 1 passage
Search aliases: s_0^2
Use in source · appendix_operator.tex, line 101
The first values are Ls_2=48, Ls_3=0, Ls_4=frac1885s_2, Ls_5=46s_3. For example, at m=2 the first term is 2s_0^2=50, while the subtraction is 2s_0/5=2. The additional 40 compared with Δ_Hs_2=8 comes from differentiating the degree-ten factor D.
Exact mathematical passage
2s_0^2=50
s_1Use in source · 8 passages
Search aliases: s_1
Use in source · appendix_operator.tex, line 5
Write s_m=∑_i=0^4v_i^m, with s_0=5 and s_1=0, and let D=∏_i<j(v_i-v_j). We distinguish the ordinary Laplacian Δ_H from the conjugated operator L defined by Δ_H(DQ)=D LQ. The factor D changes radial coefficients substantially. Confusing the two operators would invalidate the harmonicity test.
Exact mathematical passage
s_1=0
s_{10}Use in source · 2 passages
Search aliases: s_10
Use in source · chapters04_05.tex, line 114
The resulting first moments are s_2(p) · =p^2-1, s_4(p) · =2p^3-3p-1, s_6(p) · =5p^4-9p^2-5p-1, s_8(p) · =14p^5-28p^3-20p^2-7p-1, s_10(p) · =42p^6-90p^4-75p^3-35p^2-9p-1-τ(p). The integer 42 is the fifth Catalan number. It records the contribution from s_0(p)=p after inversion. The lower-degree terms come from the constant -1 in the trace formula. The coefficient of τ(p) is -1 because the last diagonal entry is 1.
Exact mathematical passage
s_2(p)&=p^2-1,\\ s_4(p)&=2p^3-3p-1,\\ s_6(p)&=5p^4-9p^2-5p-1,\\ s_8(p)&=14p^5-28p^3-20p^2-7p-1,\\ s_{10}(p)&=42p^6-90p^4-75p^3-35p^2-9p-1-\tau(p). \label{eq:s10}
S_{12+d}Use in source · 1 passage
Search aliases: S_12+d
Use in source · chapters08_10.tex, line 55
tionThe modularity contract and the cusp condition The authenticated harmonic theta theorem states that a homogeneous harmonic polynomial of degree ν on a rank-r positive definite even lattice gives a vector-valued modular form of weight ν+r/2 for the Weil representation [source reference]. Here r=4 and ν=10+d. The project-specific sign-line argument above therefore gives Θ_Q∈ S_12+d(SL_2(Z)) for the even-degree harmonic construction. The zero vector contributes no constant term. The full modula…
Exact mathematical passage
\label{eq:thetalevelone} \Theta_Q\in S_{12+d}(SL_2(\Z))
s_2^2Use in source · 10 passages
Search aliases: s_2^2
Use in source · appendix_operator.tex, line 121
Useful instances are L(s_2^2) · =104s_2, L(s_2^3) · =168s_2^2, L(s_2s_4) · =64s_4+frac1885s_2^2, L(s_3^2) · =18s_4-frac185s_2^2. For the first identity, 2s_2 Ls_2+2Γ_22=96s_2+8s_2. For the mixed product, s_4 Ls_2+s_2 Ls_4+2Γ_24 gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.
Exact mathematical passage
\mathcal L(s_2^2)&=104s_2,\label{appop:lradialtwo}\\ \mathcal L(s_2^3)&=168s_2^2,\label{appop:lradialthree}\\ \mathcal L(s_2s_4)&=64s_4+\frac{188}{5}s_2^2,\label{appop:ls2s4}\\ \mathcal L(s_3^2)&=18s_4-\frac{18}{5}s_2^2.\label{appop:ls3square}
s_2^3Use in source · 6 passages
Search aliases: s_2^3
Use in source · appendix_operator.tex, line 121
Useful instances are L(s_2^2) · =104s_2, L(s_2^3) · =168s_2^2, L(s_2s_4) · =64s_4+frac1885s_2^2, L(s_3^2) · =18s_4-frac185s_2^2. For the first identity, 2s_2 Ls_2+2Γ_22=96s_2+8s_2. For the mixed product, s_4 Ls_2+s_2 Ls_4+2Γ_24 gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.
Exact mathematical passage
\mathcal L(s_2^2)&=104s_2,\label{appop:lradialtwo}\\ \mathcal L(s_2^3)&=168s_2^2,\label{appop:lradialthree}\\ \mathcal L(s_2s_4)&=64s_4+\frac{188}{5}s_2^2,\label{appop:ls2s4}\\ \mathcal L(s_3^2)&=18s_4-\frac{18}{5}s_2^2.\label{appop:ls3square}
s_2^4Use in source · 2 passages
Search aliases: s_2^4
Use in source · appendix_operator.tex, line 155
s_7 · =frac710s_2s_5+frac712s_3s_4 -frac724s_2^2s_3. The same rule gives s_8=frac14s_4^2+frac14s_2^2s_4 -frac116s_2^4+frac815s_3s_5. Each term has the required invariant degree. This supplies a useful local check before substitution into a larger operator expression.
Exact mathematical passage
\label{appop:s8} s_8=\frac14s_4^2+\frac14s_2^2s_4 -\frac1{16}s_2^4+\frac8{15}s_3s_5.
s_2^5Use in source · 1 passage
Search aliases: s_2^5
Use in source · chapters08_10.tex, line 129
· +69600s_3s_5+45000s_4^2, and Q_10= · 2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2notag · +409600s_2s_3s_5+15000s_2s_4^2notag · -30000s_3^2s_4-264000s_5^2. Both DQ_8 and DQ_10 are harmonic. The first shell has the vector (0,1,2,-2,-1), with s_2=10, s_3=0, s_4=34, s_5=0. Both polynomials vanish there. Their theta weights are 20 and 22. The cusp spaces are respectively spanned by Δ E_4^2 and Δ E_4E_6, each with nonzero first coefficient. The zero first coefficient therefore kills the whole theta form…
Exact mathematical passage
\label{eq:q10} Q_{10}={}&2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2\notag\\ &+409600s_2s_3s_5+15000s_2s_4^2\notag\\ &-30000s_3^2s_4-264000s_5^2.
s_2^aUse in source · 1 passage
Search aliases: s_2^a
Use in source · chapters08_10.tex, line 109
At each even invariant degree d, one lists monomials s_2^as_3^bs_4^cs_5^e with 2a+3b+4c+5e=d. Writing an unknown rational linear combination, the equation LQ=0 is a finite linear system. The initial theta coefficients supply further rational linear equations. No floating-point approximation is needed to define this kernel.
Exact mathematical passage
s_2^as_3^bs_4^cs_5^e
s_2^{j-1}Use in source · 2 passages
Search aliases: s_2^j-1
Use in source · appendix_operator.tex, line 165
Radial factors and harmonic projection. Let h_ℓ be homogeneous harmonic of degree ℓ on the rank-four space H. Euler's identity is v·∇_Hh_ℓ=ℓ h_ℓ. The product rule for the radial factor s_2^j gives Δ_H(s_2^jh_ℓ) =4j(ℓ+j+1)s_2^j-1h_ℓ, j≥1. At j=0 the left side is zero. At j=1,ℓ=0 the coefficient is 8, agreeing with Δ_Hs_2. At j=1,ℓ=10 it is 48, agreeing with Δ_H(s_2D)=48D.
Exact mathematical passage
\label{appop:radial} \Delta_H(s_2^jh_\ell) =4j(\ell+j+1)s_2^{j-1}h_\ell,\qquad j\ge1.
s_2^kUse in source · 1 passage
Search aliases: s_2^k
Use in source · chapters17_18.tex, line 142
More generally, the top harmonic projection of s_2^kD at degree 10+2k is zero for every k>0. Radial information can disappear. Replacing an interpolant by its top harmonic piece may therefore destroy exactly the values it was designed to have.
Exact mathematical passage
s_2^kD
s_3Use in source · 20 passages
Search aliases: s_3
Use in source · appendix_operator.tex, line 95
The first values are Ls_2=48, Ls_3=0, Ls_4=frac1885s_2, Ls_5=46s_3. For example, at m=2 the first term is 2s_0^2=50, while the subtraction is 2s_0/5=2. The additional 40 compared with Δ_Hs_2=8 comes from differentiating the degree-ten factor D.
Exact mathematical passage
\label{appop:firstoperator} \mathcal Ls_2=48,\qquad \mathcal Ls_3=0,\qquad \mathcal Ls_4=\frac{188}{5}s_2,\qquad \mathcal Ls_5=46s_3.
s_3^2Use in source · 8 passages
Search aliases: s_3^2
Use in source · appendix_operator.tex, line 121
Useful instances are L(s_2^2) · =104s_2, L(s_2^3) · =168s_2^2, L(s_2s_4) · =64s_4+frac1885s_2^2, L(s_3^2) · =18s_4-frac185s_2^2. For the first identity, 2s_2 Ls_2+2Γ_22=96s_2+8s_2. For the mixed product, s_4 Ls_2+s_2 Ls_4+2Γ_24 gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.
Exact mathematical passage
\mathcal L(s_2^2)&=104s_2,\label{appop:lradialtwo}\\ \mathcal L(s_2^3)&=168s_2^2,\label{appop:lradialthree}\\ \mathcal L(s_2s_4)&=64s_4+\frac{188}{5}s_2^2,\label{appop:ls2s4}\\ \mathcal L(s_3^2)&=18s_4-\frac{18}{5}s_2^2.\label{appop:ls3square}
s_3^bUse in source · 1 passage
Search aliases: s_3^b
Use in source · chapters08_10.tex, line 109
At each even invariant degree d, one lists monomials s_2^as_3^bs_4^cs_5^e with 2a+3b+4c+5e=d. Writing an unknown rational linear combination, the equation LQ=0 is a finite linear system. The initial theta coefficients supply further rational linear equations. No floating-point approximation is needed to define this kernel.
Exact mathematical passage
s_2^as_3^bs_4^cs_5^e
S_4Use in source · 14 passages
Search aliases: S_4
Use in source · appendices.tex, line 116
S15: differential equations. Fonseca records the normalized Ramanujan differential relations used in the polynomial computation [source reference]. The project-specific S_4 cancellation is verified algebraically in the main text and by an exact symbolic remainder. Modularity and the level-one dimension input for the cusp identity are separate contracts [source reference]. A finite output check is not substituted for the universal algebraic identity.
Exact mathematical passage
S_4
s_4Use in source · 27 passages
Search aliases: s_4
Use in source · appendix_operator.tex, line 95
The first values are Ls_2=48, Ls_3=0, Ls_4=frac1885s_2, Ls_5=46s_3. For example, at m=2 the first term is 2s_0^2=50, while the subtraction is 2s_0/5=2. The additional 40 compared with Δ_Hs_2=8 comes from differentiating the degree-ten factor D.
Exact mathematical passage
\label{appop:firstoperator} \mathcal Ls_2=48,\qquad \mathcal Ls_3=0,\qquad \mathcal Ls_4=\frac{188}{5}s_2,\qquad \mathcal Ls_5=46s_3.
s_4^2Use in source · 3 passages
Search aliases: s_4^2
Use in source · appendix_operator.tex, line 155
s_7 · =frac710s_2s_5+frac712s_3s_4 -frac724s_2^2s_3. The same rule gives s_8=frac14s_4^2+frac14s_2^2s_4 -frac116s_2^4+frac815s_3s_5. Each term has the required invariant degree. This supplies a useful local check before substitution into a larger operator expression.
Exact mathematical passage
\label{appop:s8} s_8=\frac14s_4^2+\frac14s_2^2s_4 -\frac1{16}s_2^4+\frac8{15}s_3s_5.
s_4^cUse in source · 1 passage
Search aliases: s_4^c
Use in source · chapters08_10.tex, line 109
At each even invariant degree d, one lists monomials s_2^as_3^bs_4^cs_5^e with 2a+3b+4c+5e=d. Writing an unknown rational linear combination, the equation LQ=0 is a finite linear system. The initial theta coefficients supply further rational linear equations. No floating-point approximation is needed to define this kernel.
Exact mathematical passage
s_2^as_3^bs_4^cs_5^e
S_5Use in source · 2 passages
Search aliases: S_5
Use in source · chapters08_10.tex, line 47
If all five residues are distinct, they are 0,1,2,3,4 in some order. Their sum is zero modulo 5, so ∑_i a_i=-t forces t=0. These regular cosets form one free S_5-orbit of size 120. The sign representation occurs exactly once in the corresponding permutation module. It does not occur in an irregular orbit, since an odd stabilizer kills the alternating part. Thus the alternating theta vector lies on a one-dimensional sign line.
Exact mathematical passage
S_5
s_5Use in source · 16 passages
Search aliases: s_5
Use in source · appendix_operator.tex, line 95
The first values are Ls_2=48, Ls_3=0, Ls_4=frac1885s_2, Ls_5=46s_3. For example, at m=2 the first term is 2s_0^2=50, while the subtraction is 2s_0/5=2. The additional 40 compared with Δ_Hs_2=8 comes from differentiating the degree-ten factor D.
Exact mathematical passage
\label{appop:firstoperator} \mathcal Ls_2=48,\qquad \mathcal Ls_3=0,\qquad \mathcal Ls_4=\frac{188}{5}s_2,\qquad \mathcal Ls_5=46s_3.
s_5^2Use in source · 3 passages
Search aliases: s_5^2
Use in source · chapters08_10.tex, line 129
· +69600s_3s_5+45000s_4^2, and Q_10= · 2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2notag · +409600s_2s_3s_5+15000s_2s_4^2notag · -30000s_3^2s_4-264000s_5^2. Both DQ_8 and DQ_10 are harmonic. The first shell has the vector (0,1,2,-2,-1), with s_2=10, s_3=0, s_4=34, s_5=0. Both polynomials vanish there. Their theta weights are 20 and 22. The cusp spaces are respectively spanned by Δ E_4^2 and Δ E_4E_6, each with nonzero first coefficient. The zero first coefficient therefore kills the whole theta form…
Exact mathematical passage
\label{eq:q10} Q_{10}={}&2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2\notag\\ &+409600s_2s_3s_5+15000s_2s_4^2\notag\\ &-30000s_3^2s_4-264000s_5^2.
s_5^eUse in source · 1 passage
Search aliases: s_5^e
Use in source · chapters08_10.tex, line 109
At each even invariant degree d, one lists monomials s_2^as_3^bs_4^cs_5^e with 2a+3b+4c+5e=d. Writing an unknown rational linear combination, the equation LQ=0 is a finite linear system. The initial theta coefficients supply further rational linear equations. No floating-point approximation is needed to define this kernel.
Exact mathematical passage
s_2^as_3^bs_4^cs_5^e
s_6Use in source · 3 passages
Search aliases: s_6
Use in source · appendix_operator.tex, line 129
The metric formula may produce s_6,s_7,s_8. They are not new independent variables. The invariant ring on H is generated by s_2,s_3,s_4,s_5, and Newton's identities reduce higher indices exactly.
Exact mathematical passage
s_6,s_7,s_8
s_7Use in source · 2 passages
Search aliases: s_7
Use in source · appendix_operator.tex, line 129
The metric formula may produce s_6,s_7,s_8. They are not new independent variables. The invariant ring on H is generated by s_2,s_3,s_4,s_5, and Newton's identities reduce higher indices exactly.
Exact mathematical passage
s_6,s_7,s_8
s_8Use in source · 3 passages
Search aliases: s_8
Use in source · appendix_operator.tex, line 129
The metric formula may produce s_6,s_7,s_8. They are not new independent variables. The invariant ring on H is generated by s_2,s_3,s_4,s_5, and Newton's identities reduce higher indices exactly.
Exact mathematical passage
s_6,s_7,s_8
s_aUse in source · 4 passages
Search aliases: s_a
Use in source · appendix_operator.tex, line 80
ed indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. There are m-1 values of a. We obtain 2∇_Hlog D·∇_Hs_m =m∑_a=0^m-2s_as_m-2-a -m(m-1)s_m-2. Adding [source reference] proves Ls_m =m∑_a=0^m-2s_as_m-2-a -fracm…
Exact mathematical passage
s_as_{m-2-a}-s_{m-2}
s_cUse in source · 2 passages
Search aliases: s_c
Use in source · appendix_workbook.tex, line 317
For instance, row 0 aggregates fine indices 0,8,10,14: 2354352-11844448-31156048-7455448=-48101592. The six coarse masses are therefore s_c=(-48101592,-7162848,18972408, -12874848,-15684592,12008304)^t. Their sum, and the sum of the fine masses, is -52843168. These original masses differ in order from the lexicographic orbit bars because the certificate orders the residue-sector keys.
Exact mathematical passage
s_c=(-48101592,-7162848,18972408, -12874848,-15684592,12008304)^t.
S_kUse in source · 1 passage
Search aliases: S_k
Use in source · chapters04_05.tex, line 101
The Eichler--Selberg trace identities give, for r≥1, ∑_j=0^r(-1)^r-jbinomr+jr-jp^r-js_2j(p) =-1-Tr T_2r+2(p), together with s_0(p)=p. Here T_k(p) is the Hecke operator at p on S_k(SL_2(Z)). It is distinct from the scalar recurrence notation T_r=τ(p^r) at a fixed prime. The trace normalization and Catalan inversion are checked against [source reference], Sections 4--6. The cusp spaces in weights 4,6,8,10 are zero. The weight-12 cusp space is spanned by Δ, so its Hecke trace is τ(p).
Exact mathematical passage
S_k(SL_2(\Z))
s_kUse in source · 6 passages
Search aliases: s_k
Use in source · appendix_operator.tex, line 145
Every coordinate satisfies its characteristic polynomial. Multiplying that equality by v_i^k-5 and summing, for k≥6, gives s_k=-e_2s_k-2+e_3s_k-3-e_4s_k-4+e_5s_k-5. This recurrence uses only lower indices. Substitution at k=6,7 yields s_6 · =frac34s_2s_4+frac13s_3^2-frac18s_2^3,
Exact mathematical passage
\label{appop:newtonrec} s_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.
s_{k+1}Use in source · 2 passages
Search aliases: s_k+1
Use in source · chapters13_14.tex, line 141
The exact refinement sequence. There is a short exact sequence 0⟶ fracker AM_k+1(ker M_k) ⟶ E_k+1⟶ E_k⟶0. Consequently, if r_k=rank M_k, dim E_k+1-dim E_k =(s_k+1-s_k)-(r_k+1-r_k). The aggregation map sends H_k+1 onto H_k. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.
Exact mathematical passage
\label{eq:defectgrowth} \dim E_{k+1}-\dim E_k =(s_{k+1}-s_k)-(r_{k+1}-r_k).
S_{k,r}Use in source · 1 passage
Search aliases: S_k,r
Use in source · chapters13_14.tex, line 113
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
t_{k,r}=\sum_{v\in S_{k,r}}W(v),\qquad (M_k)_{rj}=\sum_{v\in S_{k,r}}W(v)Q_j(v).
s_{k-2}Use in source · 1 passage
Search aliases: s_k-2
Use in source · appendix_operator.tex, line 145
Every coordinate satisfies its characteristic polynomial. Multiplying that equality by v_i^k-5 and summing, for k≥6, gives s_k=-e_2s_k-2+e_3s_k-3-e_4s_k-4+e_5s_k-5. This recurrence uses only lower indices. Substitution at k=6,7 yields s_6 · =frac34s_2s_4+frac13s_3^2-frac18s_2^3,
Exact mathematical passage
\label{appop:newtonrec} s_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.
s_{k-3}Use in source · 1 passage
Search aliases: s_k-3
Use in source · appendix_operator.tex, line 145
Every coordinate satisfies its characteristic polynomial. Multiplying that equality by v_i^k-5 and summing, for k≥6, gives s_k=-e_2s_k-2+e_3s_k-3-e_4s_k-4+e_5s_k-5. This recurrence uses only lower indices. Substitution at k=6,7 yields s_6 · =frac34s_2s_4+frac13s_3^2-frac18s_2^3,
Exact mathematical passage
\label{appop:newtonrec} s_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.
s_{k-4}Use in source · 1 passage
Search aliases: s_k-4
Use in source · appendix_operator.tex, line 145
Every coordinate satisfies its characteristic polynomial. Multiplying that equality by v_i^k-5 and summing, for k≥6, gives s_k=-e_2s_k-2+e_3s_k-3-e_4s_k-4+e_5s_k-5. This recurrence uses only lower indices. Substitution at k=6,7 yields s_6 · =frac34s_2s_4+frac13s_3^2-frac18s_2^3,
Exact mathematical passage
\label{appop:newtonrec} s_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.
s_{k-5}Use in source · 1 passage
Search aliases: s_k-5
Use in source · appendix_operator.tex, line 145
Every coordinate satisfies its characteristic polynomial. Multiplying that equality by v_i^k-5 and summing, for k≥6, gives s_k=-e_2s_k-2+e_3s_k-3-e_4s_k-4+e_5s_k-5. This recurrence uses only lower indices. Substitution at k=6,7 yields s_6 · =frac34s_2s_4+frac13s_3^2-frac18s_2^3,
Exact mathematical passage
\label{appop:newtonrec} s_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.
s_mUse in source · 11 passages
Search aliases: s_m
Use in source · appendix_operator.tex, line 5
Write s_m=∑_i=0^4v_i^m, with s_0=5 and s_1=0, and let D=∏_i<j(v_i-v_j). We distinguish the ordinary Laplacian Δ_H from the conjugated operator L defined by Δ_H(DQ)=D LQ. The factor D changes radial coefficients substantially. Confusing the two operators would invalidate the harmonicity test.
Exact mathematical passage
s_m=\sum_{i=0}^4v_i^m
s_{m+n-2}Use in source · 2 passages
Search aliases: s_m+n-2
Use in source · appendix_operator.tex, line 105
The chain-rule metric and polynomial assembly. The same projected gradients give Γ_mn:=∇_Hs_m·∇_Hs_n =mn(s_m+n-2-frac15s_m-1s_n-1). Expanding the product in [source reference] gives the first power sum once and the coordinate-mean product with coefficient -1/5. This identity includes m,n≥1, with the evident zero tangential gradient for s_1.
Exact mathematical passage
\label{appop:chainmetric} \Gamma_{mn}:=\nabla_Hs_m\cdot\nabla_Hs_n =mn\left(s_{m+n-2}-\frac15s_{m-1}s_{n-1}\right).
s_{m-1}Use in source · 3 passages
Search aliases: s_m-1
Use in source · appendix_operator.tex, line 64
Deriving the power-sum operator. For m≥2, differentiation and projection give (∇_Hs_m)_i · =m(v_i^m-1-frac15s_m-1), Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two co…
Exact mathematical passage
(\nabla_Hs_m)_i &=m\left(v_i^{m-1}-\frac15s_{m-1}\right),\label{appop:gradsm}\\ \Delta_Hs_m &=\frac45m(m-1)s_{m-2}.\label{appop:lapsm}
s_{m-2}Use in source · 5 passages
Search aliases: s_m-2
Use in source · appendix_operator.tex, line 64
Deriving the power-sum operator. For m≥2, differentiation and projection give (∇_Hs_m)_i · =m(v_i^m-1-frac15s_m-1), Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two co…
Exact mathematical passage
(\nabla_Hs_m)_i &=m\left(v_i^{m-1}-\frac15s_{m-1}\right),\label{appop:gradsm}\\ \Delta_Hs_m &=\frac45m(m-1)s_{m-2}.\label{appop:lapsm}
s_{m-2-a}Use in source · 4 passages
Search aliases: s_m-2-a
Use in source · appendix_operator.tex, line 80
ed indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. There are m-1 values of a. We obtain 2∇_Hlog D·∇_Hs_m =m∑_a=0^m-2s_as_m-2-a -m(m-1)s_m-2. Adding [source reference] proves Ls_m =m∑_a=0^m-2s_as_m-2-a -fracm…
Exact mathematical passage
s_as_{m-2-a}-s_{m-2}
s_nUse in source · 3 passages
Search aliases: s_n
Use in source · appendix_operator.tex, line 105
The chain-rule metric and polynomial assembly. The same projected gradients give Γ_mn:=∇_Hs_m·∇_Hs_n =mn(s_m+n-2-frac15s_m-1s_n-1). Expanding the product in [source reference] gives the first power sum once and the coordinate-mean product with coefficient -1/5. This identity includes m,n≥1, with the evident zero tangential gradient for s_1.
Exact mathematical passage
\label{appop:chainmetric} \Gamma_{mn}:=\nabla_Hs_m\cdot\nabla_Hs_n =mn\left(s_{m+n-2}-\frac15s_{m-1}s_{n-1}\right).
s_{n-1}Use in source · 2 passages
Search aliases: s_n-1
Use in source · appendix_operator.tex, line 105
The chain-rule metric and polynomial assembly. The same projected gradients give Γ_mn:=∇_Hs_m·∇_Hs_n =mn(s_m+n-2-frac15s_m-1s_n-1). Expanding the product in [source reference] gives the first power sum once and the coordinate-mean product with coefficient -1/5. This identity includes m,n≥1, with the evident zero tangential gradient for s_1.
Exact mathematical passage
\label{appop:chainmetric} \Gamma_{mn}:=\nabla_Hs_m\cdot\nabla_Hs_n =mn\left(s_{m+n-2}-\frac15s_{m-1}s_{n-1}\right).
S_rUse in source · 3 passages
Search aliases: S_r
Use in source · chapters11_12.tex, line 116
Why strict localization and automatic valuation fail. Conserved sectors can preserve opposite signs. Partition a shell into sectors S_r, with masses t_r=∑_v∈ S_rW(v). Suppose every permitted correction is null inside every sector: ∑_v∈ S_rW(v)Q_j(v)=0. Then F=1+∑ a_jQ_j leaves each corrected sector mass equal to t_r. If two sector masses have opposite nonzero signs, all corrected packets cannot have one weak sign. Summing within those two sectors proves the obstruction immediately.
Exact mathematical passage
S_r
\sgnUse in source · 3 passages
Search aliases: sgn
Use in source · chapters01_03.tex, line 121
figplaceholdercoefficients512The exact coefficients through 512 on a signed logarithmic scale. The plotted ordinate is sgn(τ(n))log_10(1+|τ(n)|). Sign changes and large values are visible, while every plotted coefficient is nonzero. This finite plot does not establish the conjecture.
Exact mathematical passage
\sgn(\tau(n))\log_{10}(1+|\tau(n)|)
\mathrm{st}Use in source · 1 passage
Search aliases: st mathrm
Use in source · chapters19_21.tex, line 145
The realization theorem and conventions. Colmez--Fontaine's theorem makes a weakly admissible filtered module admissible [source reference]. In the present setting K=K_0=Q_p and the base Frobenius is the identity. The theorem supplies a two-dimensional semistable Q_p-representation realizing the filtered module. Since N=0, the comparison is crystalline, using B_cris=B_st^N=0.
Exact mathematical passage
B_{\mathrm{cris}}=B_{\mathrm{st}}^{N=0}
T^2Use in source · 1 passage
Search aliases: T^2
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
T^3Use in source · 1 passage
Search aliases: T^3
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
t^3Use in source · 1 passage
Search aliases: t^3
Use in source · chapters17_18.tex, line 102
An explicit homogeneous interpolation basis. Let v_1,…,v_N represent the distinct J-orbits. For an integer t≥0, define the degree-10 symmetric polynomial E_t=s_4s_2^3+t s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2. For two distinct orbits, E_t(v_i)-E_t(v_j) is a nonzero polynomial in t of degree at most 3, by Proposition~[source reference]. It has at most three roots. Among all pairs there are at most 3N(N-1)/2 forbidden values. Hence some integer 0≤ t≤frac3N(N-1)2 separates all e_i=E_t(v_i).
Exact mathematical passage
\label{eq:et} E_t=s_4s_2^3+t\,s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2.
T^5Use in source · 1 passage
Search aliases: T^5
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
T_{-1}Use in source · 1 passage
Search aliases: T_-1
Use in source · chapters15_16.tex, line 17
∑_r≥0R(p^r)X^r =frac1-cX^21-aX+bX^2. It follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular, R(1)=1, R(p)=a, R(p^r)=T_r-cT_r-2 (r≥2). The subtraction begins at r=2. No undefined T_-1 or T_-2 is needed.
Exact mathematical passage
T_{-1}
T_{-2}Use in source · 1 passage
Search aliases: T_-2
Use in source · chapters15_16.tex, line 17
local ∑_r≥0R(p^r)X^r =frac1-cX^21-aX+bX^2. It follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular, R(1)=1, R(p)=a, R(p^r)=T_r-cT_r-2 (r≥2). The subtraction begins at r=2. No undefined T_-1 or T_-2 is needed.
Exact mathematical passage
T_{-2}
T_0Use in source · 3 passages
Search aliases: T_0
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
T_0=1
T_1Use in source · 4 passages
Search aliases: T_1
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
T_1=a
T_2Use in source · 2 passages
Search aliases: T_2
Use in source · appendix_workbook.tex, line 57
T_j≤ m a_j(9T_j-m)f_m-T_j. At m=3, only T_1=1, T_2=3 contribute. Its numerator is (-3)(9-3)(252)+5(27-3)(1)=-4536+120=-4416. Division by 3 again gives -1472. The direct product, divisor recurrence and triangular recurrence reach the same integer through different algebraic operations.
Exact mathematical passage
T_2=3
T_{2m}Use in source · 3 passages
Search aliases: T_2m
Use in source · chapters01_03.tex, line 187
uadτ(3)=252. Their quotients a^2/p^11 are not integers. Thus no prime power of 2 or 3 vanishes. We obtain the exact local criterion: τ(p^r)=0 Longleftrightarrow τ(p)=0 and r is odd. The forward implication is what we just proved. For the reverse implication and the even case, set a=0 in the recurrence: T_2m=(-p^11)^m, T_2m+1=0.
Exact mathematical passage
T_{2m}=(-p^{11})^m,\qquad T_{2m+1}=0.
T_{2m-2}Use in source · 3 passages
Search aliases: T_2m-2
Use in source · chapters15_16.tex, line 63
First suppose a≠0, and write t=v_p(a). In T_2m, the term indexed by j has valuation at least 2mt+j(11-2t). If t≤4, the j=0 term a^2m has strictly least valuation 2mt. Likewise T_2m-2 has valuation (2m-2)t. The correction p^10T_2m-2 has valuation 2mt+(10-2t)>2mt. Thus v_p(R(p^2m))=2mt, t≤4.
Exact mathematical passage
T_{2m-2}
T_{2r+2}Use in source · 1 passage
Search aliases: T_2r+2
Use in source · chapters04_05.tex, line 97
The Eichler--Selberg trace identities give, for r≥1, ∑_j=0^r(-1)^r-jbinomr+jr-jp^r-js_2j(p) =-1-Tr T_2r+2(p), together with s_0(p)=p. Here T_k(p) is the Hecke operator at p on S_k(SL_2(Z)). It is distinct from the scalar recurrence notation T_r=τ(p^r) at a fixed prime. The trace normalization and Catalan inversion are checked against [source reference], Sections 4--6. The cusp spaces in weights 4,6,8,10 are zero. The weight-12 cusp space is spanned by Δ, so its Hecke trace is τ(p).
Exact mathematical passage
\label{eq:traceidentity} \sum_{j=0}^r(-1)^{r-j}\binom{r+j}{r-j}p^{r-j}s_{2j}(p) =-1-\Tr T_{2r+2}(p),
T_3Use in source · 1 passage
Search aliases: T_3
Use in source · chapters01_03.tex, line 144
The first local blocks are T_2=a^2-b, T_3=a^3-2ab, T_4=a^4-3a^2b+b^2. Even if a≠0, it is not formally obvious from these expressions that none vanishes. The reduction to primes needs a short arithmetic argument.
Exact mathematical passage
T_2=a^2-b,\qquad T_3=a^3-2ab,\qquad T_4=a^4-3a^2b+b^2.
T_4Use in source · 1 passage
Search aliases: T_4
Use in source · chapters01_03.tex, line 144
The first local blocks are T_2=a^2-b, T_3=a^3-2ab, T_4=a^4-3a^2b+b^2. Even if a≠0, it is not formally obvious from these expressions that none vanishes. The reduction to primes needs a short arithmetic argument.
Exact mathematical passage
T_2=a^2-b,\qquad T_3=a^3-2ab,\qquad T_4=a^4-3a^2b+b^2.
T_gUse in source · 5 passages
Search aliases: T_g
Use in source · chapters17_18.tex, line 147
The strongest conceivable content fullness. Let the orbits be grouped by exact content and let T_g be the mass of each group. Suppose, as an unproved hypothesis, that all-degree universal null transfers span every vector with zero sum in each content class. Then a strictly positive corrected mass on every orbit exists exactly when every nonempty T_g is positive. Necessity is conservation. For sufficiency, distribute T_g equally among its orbits, then use fullness to realize the difference from t…
Exact mathematical passage
T_g
t_HUse in source · 4 passages
Search aliases: t_H
Use in source · chapters19_21.tex, line 127
Checking weak admissibility. For the chosen cohomological convention, define t_H(D)=∑_i idimgr^iD, t_N(D)=v_p(detφ). Weak admissibility requires t_H(D)=t_N(D) and t_H(D')≤ t_N(D') for every φ-stable subobject D', with the induced filtration. The total equality is t_H(D)=11=t_N(D).
Exact mathematical passage
t_H(D)=\sum_i i\dim\operatorname{gr}^iD,\qquad t_N(D)=v_p(\det\varphi).
T_jUse in source · 3 passages
Search aliases: T_j
Use in source · appendix_workbook.tex, line 47
Jacobi gives a third engine. Put T_j=j(j+1)/2 and a_j=(-1)^j(2j+1). The series A(q)=∏_k≥1(1-q^k)^3=1+∑_j≥1a_jq^T_j satisfies F=A^8. The identity AqF'=8FqA' yields m f_m=∑_substackj≥1
Exact mathematical passage
T_j=j(j+1)/2
t_kUse in source · 3 passages
Search aliases: t_k
Use in source · chapters13_14.tex, line 117
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
t_k
T_{k+1}Use in source · 3 passages
Search aliases: T_k+1
Use in source · chapters13_14.tex, line 149
Equation [source reference] gives A(T_k+1)=T_k, so aggregation induces a surjective map between the quotients. Its kernel is the quotient of ker A by T_k+1∩ker A. A transfer M_k+1a lies in ker A exactly when AM_k+1a=M_ka=0. Thus T_k+1∩ker A=M_k+1(ker M_k), which proves [source reference]. Finally dim E_k=s_k-1-r_k, giving [source reference].
Exact mathematical passage
A(T_{k+1})=T_k
t_{k+1}Use in source · 1 passage
Search aliases: t_k+1
Use in source · chapters13_14.tex, line 120
Every depth-k+1 sector belongs to exactly one parent at depth k. Let A be the corresponding 0--1 aggregation matrix. Summing child contributions gives the exact identities t_k=At_k+1, M_k=AM_k+1. Every occupied parent has a child. Each correction column has global total zero.
Exact mathematical passage
\label{eq:aggregate} t_k=At_{k+1},\qquad M_k=AM_{k+1}.
t_{k,r}Use in source · 1 passage
Search aliases: t_k,r
Use in source · chapters13_14.tex, line 113
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
t_{k,r}=\sum_{v\in S_{k,r}}W(v),\qquad (M_k)_{rj}=\sum_{v\in S_{k,r}}W(v)Q_j(v).
t_NUse in source · 4 passages
Search aliases: t_N
Use in source · chapters19_21.tex, line 127
Checking weak admissibility. For the chosen cohomological convention, define t_H(D)=∑_i idimgr^iD, t_N(D)=v_p(detφ). Weak admissibility requires t_H(D)=t_N(D) and t_H(D')≤ t_N(D') for every φ-stable subobject D', with the induced filtration. The total equality is t_H(D)=11=t_N(D).
Exact mathematical passage
t_H(D)=\sum_i i\dim\operatorname{gr}^iD,\qquad t_N(D)=v_p(\det\varphi).
t_rUse in source · 2 passages
Search aliases: t_r
Use in source · chapters11_12.tex, line 116
Why strict localization and automatic valuation fail. Conserved sectors can preserve opposite signs. Partition a shell into sectors S_r, with masses t_r=∑_v∈ S_rW(v). Suppose every permitted correction is null inside every sector: ∑_v∈ S_rW(v)Q_j(v)=0. Then F=1+∑ a_jQ_j leaves each corrected sector mass equal to t_r. If two sector masses have opposite nonzero signs, all corrected packets cannot have one weak sign. Summing within those two sectors proves the obstruction immediately.
Exact mathematical passage
t_r=\sum_{v\in S_r}W(v)
T_{r-1}Use in source · 1 passage
Search aliases: T_r-1
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
T_r=aT_{r-1}-bT_{r-2}
T_{r-2}Use in source · 3 passages
Search aliases: T_r-2
Use in source · appendices.tex, line 13
At a prime p, the default local parameters are a=τ(p) and b=p^11. The scalar recurrence has T_0=1, T_1=a and T_r=aT_r-1-bT_r-2 for r≥2. Thus T_r=τ(p^r). A construction that permits an arbitrary integer a states that change explicitly. Such a recurrence need not be the local sequence of Δ.
Exact mathematical passage
T_r=aT_{r-1}-bT_{r-2}
\TrUse in source · 3 passages
Search aliases: Tr
Use in source · chapters04_05.tex, line 97
The Eichler--Selberg trace identities give, for r≥1, ∑_j=0^r(-1)^r-jbinomr+jr-jp^r-js_2j(p) =-1-Tr T_2r+2(p), together with s_0(p)=p. Here T_k(p) is the Hecke operator at p on S_k(SL_2(Z)). It is distinct from the scalar recurrence notation T_r=τ(p^r) at a fixed prime. The trace normalization and Catalan inversion are checked against [source reference], Sections 4--6. The cusp spaces in weights 4,6,8,10 are zero. The weight-12 cusp space is spanned by Δ, so its Hecke trace is τ(p).
Exact mathematical passage
\label{eq:traceidentity} \sum_{j=0}^r(-1)^{r-j}\binom{r+j}{r-j}p^{r-j}s_{2j}(p) =-1-\Tr T_{2r+2}(p),
\operatorname{tr}Use in source · 1 passage
Search aliases: tr operatorname
Use in source · appendix_operator.tex, line 37
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\Delta_Hs_2=\operatorname{tr}(G^{-1}2G)=8.
uUse in source · 21 passages
Search aliases: u
Use in source · chapters11_12.tex, line 87
[3pt] 2, k≥2 · (-25M-5,10M+2,-15M+4,15M-4,15M+3) · (10M-5,10M+2,20M+4,-20M-4,-20M+3) [3pt] 0, k≥3 · (-20M-15,10M-3,10M+4,20M+1,-20M+13) · (15M-15,-25M-3,10M+4,-15M+1,15M+13) The appropriate powers of 2 make each entry divisible by 7. After division, each vector has the standard residues, sum zero and norm squared 10N_k. The norm identities before division are ∑_i(7v_i)^2=∑_i(7w_i)^2 =1400M^2+140M+70γ_k=490N_k. In the first two rows, (w-v)/(5M)=(1,0,1,-1,-1). In the third it is (1,-1,0,-1,1). Thu…
Exact mathematical passage
3pt] 2, $k\ge2$&$(-25M-5,10M+2,-15M+4,15M-4,15M+3)$&$(10M-5,10M+2,20M+4,-20M-4,-20M+3)$\\[3pt] 0, $k\ge3$&$(-20M-15,10M-3,10M+4,20M+1,-20M+13)$&$(15M-15,-25M-3,10M+4,-15M+1,15M+13)$\\ \bottomrule \end{tabular}\end{center} The appropriate powers of $2$ make each entry divisible by $7$. After division, each vector has the standard residues, sum zero and norm squared $10N_k$. The norm identities before division are \[ \sum_i(7v_i)^2=\sum_i(7w_i)^2 =1400M^2+140M+70\gamma_k=490N_k.
U^2Use in source · 5 passages
Search aliases: U^2
Use in source · appendix_workbook.tex, line 145
G_499(1) · =-250854265203840. Reducing the normalized coefficients gives G_499(U) · ≡2+3U^2pmod5, G_499(U) · ≡2+8U^2+15U^3+10U^4pmod25. The first polynomial has the simple roots 1 and 4 modulo 5. At U=1, the exact sum is 10 modulo 25 and its derivative is 1 modulo 5. Consequently G_499(1+5h)≡10+5hpmod25. The five possibilities are 10,15,20,0,5 as h runs from 0 to 4. The root in the branch U≡1pmod5 therefore lifts to U=16pmod25.
Exact mathematical passage
G_{499}(U)&\equiv2+3U^2\pmod5,\\ G_{499}(U)&\equiv2+8U^2+15U^3+10U^4\pmod{25}.
u^2Use in source · 3 passages
Search aliases: u^2
Use in source · chapters15_16.tex, line 97
a=p^5u with pnmid u, and set H_r(u)=p^-5rT_r. Equation [source reference] shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading term comes from u^2m. The constant term comes from H_2m(0)=(-p)^m and H_2m-2(0)=(-p)^m-1.
Exact mathematical passage
x=u^2
u^{2m}Use in source · 1 passage
Search aliases: u^2m
Use in source · chapters15_16.tex, line 102
H_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading term comes from u^2m. The constant term comes from H_2m(0)=(-p)^m and H_2m-2(0)=(-p)^m-1.
Exact mathematical passage
u^{2m}
U^3Use in source · 4 passages
Search aliases: U^3
Use in source · appendix_workbook.tex, line 145
G_499(1) · =-250854265203840. Reducing the normalized coefficients gives G_499(U) · ≡2+3U^2pmod5, G_499(U) · ≡2+8U^2+15U^3+10U^4pmod25. The first polynomial has the simple roots 1 and 4 modulo 5. At U=1, the exact sum is 10 modulo 25 and its derivative is 1 modulo 5. Consequently G_499(1+5h)≡10+5hpmod25. The five possibilities are 10,15,20,0,5 as h runs from 0 to 4. The root in the branch U≡1pmod5 therefore lifts to U=16pmod25.
Exact mathematical passage
G_{499}(U)&\equiv2+3U^2\pmod5,\\ G_{499}(U)&\equiv2+8U^2+15U^3+10U^4\pmod{25}.
U^4Use in source · 1 passage
Search aliases: U^4
Use in source · appendix_workbook.tex, line 145
G_499(1) · =-250854265203840. Reducing the normalized coefficients gives G_499(U) · ≡2+3U^2pmod5, G_499(U) · ≡2+8U^2+15U^3+10U^4pmod25. The first polynomial has the simple roots 1 and 4 modulo 5. At U=1, the exact sum is 10 modulo 25 and its derivative is 1 modulo 5. Consequently G_499(1+5h)≡10+5hpmod25. The five possibilities are 10,15,20,0,5 as h runs from 0 to 4. The root in the branch U≡1pmod5 therefore lifts to U=16pmod25.
Exact mathematical passage
G_{499}(U)&\equiv2+3U^2\pmod5,\\ G_{499}(U)&\equiv2+8U^2+15U^3+10U^4\pmod{25}.
U^5Use in source · 1 passage
Search aliases: U^5
Use in source · appendix_workbook.tex, line 226
G_16249(1) · =61339951998700724370240. The first equals 288τ(16249), and the second equals the first divided by 625. Modulo 5, the normalized polynomial is 3+U^2+U^3=(U-1)^2(U-2). Modulo 25, the table gives G_16249(U)≡13+15U+11U^2+11U^3+15U^5. Thus G_16249(1)≡15pmod25 while G'_16249(1)≡0pmod5. For every integer or 5-adic integer h, polynomial Taylor expansion gives G_16249(1+5h)≡ G_16249(1)+5hG'_16249(1) ≡15pmod25. Terms of order at least two have a factor 25. This is a uniform obstruction throu…
Exact mathematical passage
G_{16249}(U)\equiv13+15U+11U^2+11U^3+15U^5.
U^jUse in source · 1 passage
Search aliases: U^j
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
u_1Use in source · 1 passage
Search aliases: u_1
Use in source · chapters11_12.tex, line 176
Take the prime indices (u_1,p_1)=(15,245759), (u_2,p_2)=(47,770047). The exact values are τ(245759) · =77900595298037436891536424960,
Exact mathematical passage
(u_1,p_1)=(15,245759),\qquad(u_2,p_2)=(47,770047).
u_2Use in source · 1 passage
Search aliases: u_2
Use in source · chapters11_12.tex, line 176
Take the prime indices (u_1,p_1)=(15,245759), (u_2,p_2)=(47,770047). The exact values are τ(245759) · =77900595298037436891536424960,
Exact mathematical passage
(u_1,p_1)=(15,245759),\qquad(u_2,p_2)=(47,770047).
u_jUse in source · 2 passages
Search aliases: u_j
Use in source · chapters11_12.tex, line 144
The all-even class scales by 2. The bijection u_j=v_2jbmod5/2 has a coordinate permutation of sign -1, so A(q)=-2^10Δ(q^4). Using the Hecke recurrence at 2 gives B(q) · =-24Δ(q^2)-2^10Δ(q^4),
Exact mathematical passage
u_j=v_{2j\bmod5}/2
\mathcal VUse in source · 8 passages
Search aliases: V mathcal
Use in source · chapter22.tex, line 19
On each such prime shell, all packets have content one. The residual is therefore the full signed sum τ(p)=∑_v∈ V_pW(v). The summands are nonzero integers, and for p>20 they are units at the index prime. Neither property forbids their sum from being zero. The unresolved statement concerns the exact balance of their signed masses.
Exact mathematical passage
\label{eq:primepacketobligation} \tau(p)=\sum_{v\in\mathcal V_p}W(v).
v_{(-i\bmod5)}Use in source · 1 passage
Search aliases: v_(-ibmod5) bmod
Use in source · chapters08_10.tex, line 4
Symmetry and the harmonic theta mechanism. An involution that preserves the weight. The coordinate labels are residue labels, so negation alone does not preserve the standard shell. The correct involution is J(v)_i=-v_(-ibmod5). It preserves the residues, sum and norm. Its coordinate permutation is (1 4)(2 3), which is even. Negation contributes (-1)^10=1 to the degree-10 Vandermonde. Hence W(Jv)=W(v). Write s_m(v)=∑_i v_i^m. Then s_m(Jv)=(-1)^ms_m(v). In particular, every odd homogeneous symmet…
Exact mathematical passage
\label{eq:j} J(v)_i=-v_{(-i\bmod5)}.
v_{-i\bmod5}Use in source · 2 passages
Search aliases: v_-ibmod5 bmod
Use in source · appendices.tex, line 42
The corresponding five-core partition has size n-1. In particular, the coefficient at n=1 corresponds to the empty partition. The involution is J(v)_i=-v_-ibmod5. A J-orbit may contain one or two packets, and its mass is the sum of their weights.
Exact mathematical passage
J(v)_i=-v_{-i\bmod5}
v_0Use in source · 2 passages
Search aliases: v_0
Use in source · chapters06_07.tex, line 68
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
v_0,v_1,v_2
v_0^2Use in source · 1 passage
Search aliases: v_0^2
Use in source · chapters06_07.tex, line 69
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
x+y=S=-v_0-v_1-v_2,\qquad x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.
v_1Use in source · 3 passages
Search aliases: v_1
Use in source · chapters06_07.tex, line 68
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
v_0,v_1,v_2
\mathcal V_{13}Use in source · 1 passage
Search aliases: V_13 mathcal
Use in source · chapters11_12.tex, line 79
Witnesses attaining the bound. At k=1, take v=(-5,-4,2,-2,9), w=(5,6,-8,-2,-1). They belong to V_13, have opposite weights and differ coordinatewise by multiples of 10. Thus they lie in one depth-one sector.
Exact mathematical passage
\mathcal V_{13}
v_1^2Use in source · 1 passage
Search aliases: v_1^2
Use in source · chapters06_07.tex, line 69
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
x+y=S=-v_0-v_1-v_2,\qquad x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.
\mathcal V_2Use in source · 1 passage
Search aliases: V_2 mathcal
Use in source · chapters06_07.tex, line 81
figplaceholdershell_coordinatesThe shell V_2 is a concrete intersection of a zero-sum hyperplane, an integral residue condition and a sphere. Coordinates are labeled by their residues modulo 5. A shell point is a packet, with its separate nonzero integral Vandermonde weight.
Exact mathematical passage
\mathcal V_2
v_2^2Use in source · 1 passage
Search aliases: v_2^2
Use in source · chapters06_07.tex, line 69
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
x+y=S=-v_0-v_1-v_2,\qquad x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.
v_{2j\bmod5}Use in source · 1 passage
Search aliases: v_2jbmod5 bmod
Use in source · chapters11_12.tex, line 144
The all-even class scales by 2. The bijection u_j=v_2jbmod5/2 has a coordinate permutation of sign -1, so A(q)=-2^10Δ(q^4). Using the Hecke recurrence at 2 gives B(q) · =-24Δ(q^2)-2^10Δ(q^4),
Exact mathematical passage
u_j=v_{2j\bmod5}/2
v_3Use in source · 1 passage
Search aliases: v_3
Use in source · chapters06_07.tex, line 68
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
x=v_3
v_4Use in source · 2 passages
Search aliases: v_4
Use in source · appendix_operator.tex, line 221
The retained programs organize these same operations using sparse rational polynomials. A direct four-coordinate implementation can instead substitute v_4=-∑_a=0^3x_a into DQ and apply [source reference]. Agreement between these organizations checks the symbolic construction through different algebraic representations. The universal justification remains the metric calculation, the exact polynomial identities and the stated harmonic theta and valence contracts. None of those statements alone exc…
Exact mathematical passage
v_4=-\sum_{a=0}^3x_a
v_5Use in source · 18 passages
Search aliases: v_5
Use in source · appendix_workbook.tex, line 94
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
v_5(0)=+\infty
v_{gj\bmod5}Use in source · 1 passage
Search aliases: v_gjbmod5 bmod
Use in source · chapters13_14.tex, line 6
The content is compatible with the shell in a precise way. If g=g(v), define u_j=fracv_gjbmod5g, j=0,…,4. The coordinate permutation restores the standard residues after division. The vector u has sum zero and gcd 1. Its norm squared is 10n/g^2. For every integral standard-residue vector of sum zero, the norm is divisible by 5, because its squared residues sum to 30, and by 2, because squares and coordinates agree modulo 2. Thus 10n/g^2 is 10 times an integer, and g^2mid n. The map is a bijectio…
Exact mathematical passage
\label{eq:contentscale} u_j=\frac{v_{gj\bmod5}}g,\qquad j=0,\ldots,4.
v_i^2Use in source · 4 passages
Search aliases: v_i^2
Use in source · appendices.tex, line 30
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
\Vn=\left\{v\in\Z^5: v_i\equiv i\pmod5,\quad \sum_i v_i=0,\quad \sum_i v_i^2=10n\right\}.
v_i^aUse in source · 1 passage
Search aliases: v_i^a
Use in source · appendix_operator.tex, line 76
Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. …
Exact mathematical passage
\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j} =\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.
v_i^{k-5}Use in source · 1 passage
Search aliases: v_i^k-5
Use in source · appendix_operator.tex, line 144
Every coordinate satisfies its characteristic polynomial. Multiplying that equality by v_i^k-5 and summing, for k≥6, gives s_k=-e_2s_k-2+e_3s_k-3-e_4s_k-4+e_5s_k-5. This recurrence uses only lower indices. Substitution at k=6,7 yields s_6 · =frac34s_2s_4+frac13s_3^2-frac18s_2^3,
Exact mathematical passage
v_i^{k-5}
v_i^mUse in source · 2 passages
Search aliases: v_i^m
Use in source · appendix_operator.tex, line 5
Write s_m=∑_i=0^4v_i^m, with s_0=5 and s_1=0, and let D=∏_i<j(v_i-v_j). We distinguish the ordinary Laplacian Δ_H from the conjugated operator L defined by Δ_H(DQ)=D LQ. The factor D changes radial coefficients substantially. Confusing the two operators would invalidate the harmonicity test.
Exact mathematical passage
s_m=\sum_{i=0}^4v_i^m
v_i^{m-1}Use in source · 4 passages
Search aliases: v_i^m-1
Use in source · appendix_operator.tex, line 64
Deriving the power-sum operator. For m≥2, differentiation and projection give (∇_Hs_m)_i · =m(v_i^m-1-frac15s_m-1), Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two co…
Exact mathematical passage
(\nabla_Hs_m)_i &=m\left(v_i^{m-1}-\frac15s_{m-1}\right),\label{appop:gradsm}\\ \Delta_Hs_m &=\frac45m(m-1)s_{m-2}.\label{appop:lapsm}
v_jUse in source · 18 passages
Search aliases: v_j
Use in source · appendices.tex, line 37
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
D(v)=\prod_{0\le i<j\le4}(v_i-v_j),\qquad W(v)=D(v)/288.
v_j^2Use in source · 1 passage
Search aliases: v_j^2
Use in source · chapters06_07.tex, line 85
The index-prime unit obstruction. If p>20 is prime and v∈ V_p, then for each pair 0<(v_i-v_j)^2≤2(v_i^2+v_j^2)≤20p<p^2. Thus p divides none of the differences and does not divide 288. We obtain For every prime p>20 and every v∈ V_p, one has v_p(W(v))=0. All packets are units at the index prime. There is no increasing termwise p-adic depth automatically built into this representation. At p=2411, for example, all 2412 packet weights are 2411-adic units while their sum has valuation 1. The divisibi…
Exact mathematical passage
0<(v_i-v_j)^2\le2(v_i^2+v_j^2)\le20p<p^2.
v_j^{m-1}Use in source · 3 passages
Search aliases: v_j^m-1
Use in source · appendix_operator.tex, line 71
Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. …
Exact mathematical passage
2\nabla_H\log D\cdot\nabla_Hs_m =2m\sum_{i<j}\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.
v_j^{m-2-a}Use in source · 1 passage
Search aliases: v_j^m-2-a
Use in source · appendix_operator.tex, line 76
Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. …
Exact mathematical passage
\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j} =\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.
v_NUse in source · 1 passage
Search aliases: v_N
Use in source · chapters17_18.tex, line 101
An explicit homogeneous interpolation basis. Let v_1,…,v_N represent the distinct J-orbits. For an integer t≥0, define the degree-10 symmetric polynomial E_t=s_4s_2^3+t s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2. For two distinct orbits, E_t(v_i)-E_t(v_j) is a nonzero polynomial in t of degree at most 3, by Proposition~[source reference]. It has at most three roots. Among all pairs there are at most 3N(N-1)/2 forbidden values. Hence some integer 0≤ t≤frac3N(N-1)2 separates all e_i=E_t(v_i).
Exact mathematical passage
v_1,\ldots,v_N
\mathcal V_pUse in source · 6 passages
Search aliases: V_p mathcal
Use in source · chapter22.tex, line 19
On each such prime shell, all packets have content one. The residual is therefore the full signed sum τ(p)=∑_v∈ V_pW(v). The summands are nonzero integers, and for p>20 they are units at the index prime. Neither property forbids their sum from being zero. The unresolved statement concerns the exact balance of their signed masses.
Exact mathematical passage
\label{eq:primepacketobligation} \tau(p)=\sum_{v\in\mathcal V_p}W(v).
v_pUse in source · 11 passages
Search aliases: v_p
Use in source · chapters01_03.tex, line 39
For a nonzero actual coefficient, the same comparison gives v_p(τ(p))≤5. A hypothetical assertion v_p(τ(p))≤3 would be stronger than needed, and hence sufficient, but requires its own proof. One must not obtain such a valuation bound merely by rewriting the coefficient as another exact sum.
Exact mathematical passage
v_p(\tau(p))\le5
v_\ellUse in source · 10 passages
Search aliases: v_ℓ ell
Use in source · appendices.tex, line 17
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d,
Exact mathematical passage
v_\ell(x)
\VertUse in source · 3 passages
Search aliases: Vert
Use in source · chapters19_21.tex, line 45
The actual joint image. Factor M=∏_ℓ^eVert Mℓ^e and form the single joint finite map Ψ:G_Q⟶ (∏_ℓ^eVert MGL_2(Z/ℓ^eZ)) ×(Z/5MZ)^×. Its components are the integral representation reductions and the cyclotomic character modulo 5M. Let G be its actual image. The kernel defines a finite Galois extension of Q. The image need not be the whole displayed product. Relations between different representations and determinants are allowed.
Exact mathematical passage
M=\prod_{\ell^e\Vert M}\ell^e
WUse in source · 39 passages
Search aliases: W
Use in source · appendices.tex, line 37
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
D(v)=\prod_{0\le i<j\le4}(v_i-v_j),\qquad W(v)=D(v)/288.
wUse in source · 26 passages
Search aliases: w
Use in source · chapters08_10.tex, line 245
Completeness retains the total. Suppose a shell has nonzero orbit masses w_1,…,w_N and total T. Even allow every rational correction vector q with ∑_iw_iq_i=0. Then ∃ q: w_i(1+q_i) have one weak sign and are not all zero Longleftrightarrow T≠0. Necessity follows from conservation. For sufficiency, if T≠0, take q_i=fracTNw_i-1. Every corrected mass is T/N and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of T≠0 without independent information …
Exact mathematical passage
w_1,\ldots,w_N
w_1Use in source · 1 passage
Search aliases: w_1
Use in source · chapters08_10.tex, line 245
Completeness retains the total. Suppose a shell has nonzero orbit masses w_1,…,w_N and total T. Even allow every rational correction vector q with ∑_iw_iq_i=0. Then ∃ q: w_i(1+q_i) have one weak sign and are not all zero Longleftrightarrow T≠0. Necessity follows from conservation. For sufficiency, if T≠0, take q_i=fracTNw_i-1. Every corrected mass is T/N and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of T≠0 without independent information …
Exact mathematical passage
w_1,\ldots,w_N
w_iUse in source · 6 passages
Search aliases: w_i
Use in source · chapters08_10.tex, line 245
Completeness retains the total. Suppose a shell has nonzero orbit masses w_1,…,w_N and total T. Even allow every rational correction vector q with ∑_iw_iq_i=0. Then ∃ q: w_i(1+q_i) have one weak sign and are not all zero Longleftrightarrow T≠0. Necessity follows from conservation. For sufficiency, if T≠0, take q_i=fracTNw_i-1. Every corrected mass is T/N and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of T≠0 without independent information …
Exact mathematical passage
\sum_iw_iq_i=0
w_jUse in source · 1 passage
Search aliases: w_j
Use in source · chapters11_12.tex, line 52
If d ^2≥6, [source reference] gives n≥frac154M^2>frac100M^2+10M+235, M≥2. The remaining case is d ^2=4. Then d has two coordinates 1, two -1 and one 0. Of the ten coordinate pairs, two have equal d entries and eight have unequal entries. For a pair with δ=d_i-d_j, (v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2δ^2. The two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of D(v) and D(w). N…
Exact mathematical passage
(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\delta^2.
w_NUse in source · 1 passage
Search aliases: w_N
Use in source · chapters08_10.tex, line 245
Completeness retains the total. Suppose a shell has nonzero orbit masses w_1,…,w_N and total T. Even allow every rational correction vector q with ∑_iw_iq_i=0. Then ∃ q: w_i(1+q_i) have one weak sign and are not all zero Longleftrightarrow T≠0. Necessity follows from conservation. For sufficiency, if T≠0, take q_i=fracTNw_i-1. Every corrected mass is T/N and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of T≠0 without independent information …
Exact mathematical passage
w_1,\ldots,w_N
XUse in source · 12 passages
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Use in source · appendices.tex, line 65
Local and global conventions. For ℓ≠ p, the arithmetic-Frobenius convention in the integral ℓ-adic representation gives the characteristic polynomial X^2-τ(p)X+p^11. The cohomological conventions in the filtered-module discussion are stated there separately. An abstract trace-zero filtered module and a representation realizing it are local objects. A comparison identifying its normalized Frobenius trace with the actual coefficient τ(p) is additional mathematical data.
Exact mathematical passage
X^2-\tau(p)X+p^{11}.
xUse in source · 42 passages
Search aliases: x
Use in source · appendices.tex, line 17
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d,
Exact mathematical passage
v_\ell(x)
X^2Use in source · 9 passages
Search aliases: X^2
Use in source · appendices.tex, line 65
Local and global conventions. For ℓ≠ p, the arithmetic-Frobenius convention in the integral ℓ-adic representation gives the characteristic polynomial X^2-τ(p)X+p^11. The cohomological conventions in the filtered-module discussion are stated there separately. An abstract trace-zero filtered module and a representation realizing it are local objects. A comparison identifying its normalized Frobenius trace with the actual coefficient τ(p) is additional mathematical data.
Exact mathematical passage
X^2-\tau(p)X+p^{11}.
x^2Use in source · 1 passage
Search aliases: x^2
Use in source · chapters06_07.tex, line 69
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
x+y=S=-v_0-v_1-v_2,\qquad x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.
x^{m-1}Use in source · 1 passage
Search aliases: x^m-1
Use in source · chapters15_16.tex, line 98
refeq:trexpansion shows that H_r is an integral polynomial. Its recurrence is H_0=1, H_1=u, H_r=uH_r-1-pH_r-2. The primitive block becomes R(p^2m)=p^10m(H_2m(u)-H_2m-2(u)). The expression in parentheses is a monic polynomial F_m(x) in x=u^2. Its constant term and reduction modulo p are F_m(0)=(-1)^mp^m-1(p+1), F_m(x)≡ x^m-1(x-1)pmod p. The leading term comes from u^2m. The constant term comes from H_2m(0)=(-p)^m and H_2m-2(0)=(-p)^m-1.
Exact mathematical passage
\label{eq:fm} F_m(0)=(-1)^mp^{m-1}(p+1),\qquad F_m(x)\equiv x^{m-1}(x-1)\pmod p.
X^rUse in source · 3 passages
Search aliases: X^r
Use in source · chapters01_03.tex, line 138
For a fixed prime put T_r=τ(p^r) and b=p^11. The generating series is ∑_r≥0T_rX^r=frac11-aX+bX^2. Multiplication by the denominator gives T_0=1, T_1=a and the recurrence. Thus this is a formal identity over Z[[X]]. No analytic continuation in the prime index is involved.
Exact mathematical passage
\label{eq:heckegen} \sum_{r\ge0}T_rX^r=\frac1{1-aX+bX^2}.
x^TUse in source · 1 passage
Search aliases: x^T
Use in source · appendix_operator.tex, line 36
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
s_2=x^TGx
x_0Use in source · 2 passages
Search aliases: x_0
Use in source · appendix_operator.tex, line 20
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
x_0,\ldots,x_3
x_1Use in source · 1 passage
Search aliases: x_1
Use in source · appendix_operator.tex, line 21
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3).
x_2Use in source · 1 passage
Search aliases: x_2
Use in source · appendix_operator.tex, line 21
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3).
x_3Use in source · 2 passages
Search aliases: x_3
Use in source · appendix_operator.tex, line 20
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
x_0,\ldots,x_3
x_5Use in source · 3 passages
Search aliases: x_5
Use in source · appendix_workbook.tex, line 349
One parent gives a concrete view of this second stage. Parent 5 has fine children 5,6. Just after the coarse stage their values are x_5 · =-frac161598180629860124657534386744 23300823161076013875625, x_6 · =-frac1069691132209170737342448593256 23300823161076013875625. Their exact sum is -52843168. The detail stage adds -x_5 to child 5 and x_5 to child 6, giving (0,-52843168) in that parent. All other parents retain their zero totals. This is a rational conservation calculation, not a numerical n…
Exact mathematical passage
x_5&=-\frac{161598180629860124657534386744} {23300823161076013875625},\\ x_6&=-\frac{1069691132209170737342448593256} {23300823161076013875625}.
x_6Use in source · 1 passage
Search aliases: x_6
Use in source · appendix_workbook.tex, line 349
One parent gives a concrete view of this second stage. Parent 5 has fine children 5,6. Just after the coarse stage their values are x_5 · =-frac161598180629860124657534386744 23300823161076013875625, x_6 · =-frac1069691132209170737342448593256 23300823161076013875625. Their exact sum is -52843168. The detail stage adds -x_5 to child 5 and x_5 to child 6, giving (0,-52843168) in that parent. All other parents retain their zero totals. This is a rational conservation calculation, not a numerical n…
Exact mathematical passage
x_5&=-\frac{161598180629860124657534386744} {23300823161076013875625},\\ x_6&=-\frac{1069691132209170737342448593256} {23300823161076013875625}.
x_aUse in source · 3 passages
Search aliases: x_a
Use in source · appendix_operator.tex, line 32
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\label{appop:chartlap} \Delta_H=\sum_{a,b=0}^3 \left(\delta_{ab}-\frac15\right)\frac{\partial^2}{\partial x_a\partial x_b}.
x_bUse in source · 1 passage
Search aliases: x_b
Use in source · appendix_operator.tex, line 32
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\label{appop:chartlap} \Delta_H=\sum_{a,b=0}^3 \left(\delta_{ab}-\frac15\right)\frac{\partial^2}{\partial x_a\partial x_b}.
X_HUse in source · 1 passage
Search aliases: X_H
Use in source · bibliography.tex, line 25
bibitemDHZ Maarten Derickx, Mark van Hoeij, and Jinxiang Zeng, Computing Galois representations and equations for modular curves X_H(ℓ), https://arxiv.org/pdf/1312.6819v2arXiv:1312.6819v2, 18 March 2014. Corollary~1.2, p.~2; its computational proof in S4, p.~11. The cited nonvanishing range has the strict endpoint n<816212624008487344127999. This is a historical source result, not an edition replay or a verified latest record.
Exact mathematical passage
X_H(\ell)
yUse in source · 16 passages
Search aliases: y
Use in source · appendix_workbook.tex, line 346
The coarse coefficient vector a∈Q^18 makes x=s_f+M_fa, Ax=(0,0,0,0,0,-52843168)^t. The certificate then chooses a coefficient vector d∈ker M_c. Its final fine vector y=x+M_fd is zero except at index 6, where it equals -52843168.
Exact mathematical passage
y=x+M_fd
y^2Use in source · 2 passages
Search aliases: y^2
Use in source · chapters06_07.tex, line 69
A complete shell enumerator. The shell is finite because every coordinate has absolute value at most sqrt10n. Enumeration can reduce its dimension without discarding any solution. Choose v_0,v_1,v_2 in their residue classes within the bound. The remaining coordinates x=v_3, y=v_4 must satisfy x+y=S=-v_0-v_1-v_2, x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2. Then (x-y)^2=2T-S^2. The right side must be a nonnegative square. For each permitted sign of its square root, the formulas x=(S+(x-y))/2 and y=(S-(x-y))/…
Exact mathematical passage
x+y=S=-v_0-v_1-v_2,\qquad x^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.
y_iUse in source · 3 passages
Search aliases: y_i
Use in source · chapters08_10.tex, line 206
An exact obstruction in a specified smaller space. At n=31, the seven retained corrections through degree 16 do not admit a nontrivial common-sign certificate. The shell has sixteen J-orbits. Let ϵ_i=sgn W(v_i) for their representatives. The retained exact dual has strictly positive rational entries y_i satisfying ∑_i y_i=1, ∑_i y_iϵ_i=0, ∑_i y_iϵ_iQ_j(v_i)=0 for all seven correction columns. Consequently, for every coefficient vector, ∑_i y_iϵ_i(1+∑_j a_jQ_j(v_i))=0. If every corrected packet h…
Exact mathematical passage
y_i
zUse in source · 11 passages
Search aliases: z
Use in source · appendices.tex, line 120
Cores, harmonic theta series and valence. S02: Han's packet identity. Han's Definition~1.1 and Theorems~1.1--1.2 in the checked first arXiv version provide the five-core specialization and Vandermonde product [source reference]. The specialization t=5, z=25 gives the shell norm 10n after the coefficient shift. With the orientation used here, the normalization is D/288. The identity permits cancellation and does not imply that its terms have one sign.
Exact mathematical passage
z=25
\Z^5Use in source · 3 passages
Search aliases: Z^5 Z
Use in source · appendices.tex, line 30
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
\Vn=\left\{v\in\Z^5: v_i\equiv i\pmod5,\quad \sum_i v_i=0,\quad \sum_i v_i^2=10n\right\}.
\Z_5Use in source · 4 passages
Search aliases: Z_5 Z
Use in source · chapters06_07.tex, line 145
=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)≠0. Thus v_5(G_499(U))=v_5(U-ρ) on that residue class. At U=1 the difference from ρ has valuation…
Exact mathematical passage
\Z_5
\Z_\ellUse in source · 3 passages
Search aliases: Z_ℓ Z ell
Use in source · chapters19_21.tex, line 32
A continuous representation of the compact group G_Q has a stable Z_ℓ lattice. To see this directly, take an initial lattice L_0. Compactness bounds all its translates inside ℓ^-aL_0 for some a. The module L=∑_g∈ G_Qρ_ℓ(g)L_0 is contained in ℓ^-aL_0 and contains L_0. It is finitely generated, since a submodule of a finite free Z_ℓ module is finitely generated, and it is invariant by construction. Reductions modulo ℓ^e are therefore defined. No image-surjectivity theorem is required.
Exact mathematical passage
\Z_\ell
|_HUse in source · 1 passage
Search aliases: |_H
Use in source · appendix_operator.tex, line 13
projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:ambientlap} \Delta_H(F|_H) =\left(\sum_{i=0}^4\partial_i^2F -\frac15\left(\sum_{i=0}^4\partial_i\right)^2F\right)\Big|_H.
\pmUse in source · 3 passages
Search aliases: ± pm
Use in source · chapters11_12.tex, line 45
in one sector. Replace w by Jw if needed so that w-v=5Md, d∈ A_4∖{0}. Set a=5M/2 and m=(v+w)/2. Since M is even, both are integral. Equal norms give m· d=0,qquad10n= m ^2+a^2 d ^2. The squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be 2 here. Such a vector is ±(e_i-e_j), and m· d=0 would force v_i-v_j=±5M. This is divisible by 5, contrary to the distinct coordinate residues.
Exact mathematical passage
\pm(e_i-e_j)
\alphaUse in source · 4 passages
Search aliases: α alpha
Use in source · chapters01_03.tex, line 151
The two roots and a root of unity. Let α,β be the complex roots of X^2-aX+p^11=0. They are nonzero because their product is p^11. They are distinct: equality would require a^2=4p^11, but the p-adic valuation of the right side is odd for odd p, and is 13 for p=2.
Exact mathematical passage
\alpha,\beta
\alpha^{r+1}Use in source · 1 passage
Search aliases: α^r+1 alpha
Use in source · chapters01_03.tex, line 158
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\label{eq:rootrec} T_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta} =\sum_{j=0}^r\alpha^{r-j}\beta^j.
\alpha^{r-j}Use in source · 1 passage
Search aliases: α^r-j alpha
Use in source · chapters01_03.tex, line 158
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\label{eq:rootrec} T_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta} =\sum_{j=0}^r\alpha^{r-j}\beta^j.
\betaUse in source · 4 passages
Search aliases: β beta
Use in source · chapters01_03.tex, line 151
The two roots and a root of unity. Let α,β be the complex roots of X^2-aX+p^11=0. They are nonzero because their product is p^11. They are distinct: equality would require a^2=4p^11, but the p-adic valuation of the right side is odd for odd p, and is 13 for p=2.
Exact mathematical passage
\alpha,\beta
\beta^jUse in source · 1 passage
Search aliases: β^j beta
Use in source · chapters01_03.tex, line 158
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\label{eq:rootrec} T_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta} =\sum_{j=0}^r\alpha^{r-j}\beta^j.
\beta^{r+1}Use in source · 1 passage
Search aliases: β^r+1 beta
Use in source · chapters01_03.tex, line 158
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\label{eq:rootrec} T_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta} =\sum_{j=0}^r\alpha^{r-j}\beta^j.
\GammaUse in source · 4 passages
Search aliases: Γ Gamma
Use in source · appendix_operator.tex, line 105
The chain-rule metric and polynomial assembly. The same projected gradients give Γ_mn:=∇_Hs_m·∇_Hs_n =mn(s_m+n-2-frac15s_m-1s_n-1). Expanding the product in [source reference] gives the first power sum once and the coordinate-mean product with coefficient -1/5. This identity includes m,n≥1, with the evident zero tangential gradient for s_1.
Exact mathematical passage
\label{appop:chainmetric} \Gamma_{mn}:=\nabla_Hs_m\cdot\nabla_Hs_n =mn\left(s_{m+n-2}-\frac15s_{m-1}s_{n-1}\right).
\gammaUse in source · 6 passages
Search aliases: γ gamma
Use in source · chapters11_12.tex, line 20
The sharp threshold. Let k≥1, M=2^k, and set N_k=lceilfrac100M^2+10M+235rceil =frac20M^2+2M+γ_k7, where γ_k= 7 · k≡1pmod3,
Exact mathematical passage
\label{eq:nk} N_k=\left\lceil\frac{100M^2+10M+2}{35}\right\rceil =\frac{20M^2+2M+\gamma_k}{7},
\Gamma_{22}Use in source · 1 passage
Search aliases: Γ_22 Gamma
Use in source · appendix_operator.tex, line 127
L(s_3^2) · =18s_4-frac185s_2^2. For the first identity, 2s_2 Ls_2+2Γ_22=96s_2+8s_2. For the mixed product, s_4 Ls_2+s_2 Ls_4+2Γ_24 gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.
Exact mathematical passage
2s_2\mathcal Ls_2+2\Gamma_{22}=96s_2+8s_2
\Gamma_{24}Use in source · 1 passage
Search aliases: Γ_24 Gamma
Use in source · appendix_operator.tex, line 127
L(s_3^2) · =18s_4-frac185s_2^2. For the first identity, 2s_2 Ls_2+2Γ_22=96s_2+8s_2. For the mixed product, s_4 Ls_2+s_2 Ls_4+2Γ_24 gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.
Exact mathematical passage
s_4\mathcal Ls_2+s_2\mathcal Ls_4+2\Gamma_{24}
\gamma_kUse in source · 4 passages
Search aliases: γ_k gamma
Use in source · chapters11_12.tex, line 20
The sharp threshold. Let k≥1, M=2^k, and set N_k=lceilfrac100M^2+10M+235rceil =frac20M^2+2M+γ_k7, where γ_k= 7 · k≡1pmod3,
Exact mathematical passage
\label{eq:nk} N_k=\left\lceil\frac{100M^2+10M+2}{35}\right\rceil =\frac{20M^2+2M+\gamma_k}{7},
\Gamma_{mn}Use in source · 2 passages
Search aliases: Γ_mn Gamma
Use in source · appendix_operator.tex, line 105
The chain-rule metric and polynomial assembly. The same projected gradients give Γ_mn:=∇_Hs_m·∇_Hs_n =mn(s_m+n-2-frac15s_m-1s_n-1). Expanding the product in [source reference] gives the first power sum once and the coordinate-mean product with coefficient -1/5. This identity includes m,n≥1, with the evident zero tangential gradient for s_1.
Exact mathematical passage
\label{appop:chainmetric} \Gamma_{mn}:=\nabla_Hs_m\cdot\nabla_Hs_n =mn\left(s_{m+n-2}-\frac15s_{m-1}s_{n-1}\right).
\deltaUse in source · 9 passages
Search aliases: δ delta
Use in source · appendix_operator.tex, line 32
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\label{appop:chartlap} \Delta_H=\sum_{a,b=0}^3 \left(\delta_{ab}-\frac15\right)\frac{\partial^2}{\partial x_a\partial x_b}.
\delta^2Use in source · 3 passages
Search aliases: δ^2 delta
Use in source · chapters11_12.tex, line 52
If d ^2≥6, [source reference] gives n≥frac154M^2>frac100M^2+10M+235, M≥2. The remaining case is d ^2=4. Then d has two coordinates 1, two -1 and one 0. Of the ten coordinate pairs, two have equal d entries and eight have unequal entries. For a pair with δ=d_i-d_j, (v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2δ^2. The two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of D(v) and D(w). N…
Exact mathematical passage
(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\delta^2.
\delta_{ab}Use in source · 1 passage
Search aliases: δ_ab delta
Use in source · appendix_operator.tex, line 32
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\label{appop:chartlap} \Delta_H=\sum_{a,b=0}^3 \left(\delta_{ab}-\frac15\right)\frac{\partial^2}{\partial x_a\partial x_b}.
\delta_{ij}Use in source · 1 passage
Search aliases: δ_ij delta
Use in source · chapters17_18.tex, line 117
For that t, set L_i(v)=∏_j≠ i fracE_t(v)-e_j(s_2(v)/C)^5e_i-e_j. Each factor is homogeneous of degree 10, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree 10(N-1). On the shell, s_2/C=1, so L_i(v_j)=δ_ij. Consequently, any prescribed rational orbit values a_1,…,a_N are interpolated by Q=∑_ia_iL_i. If N=1, an ordinary constant suffices.
Exact mathematical passage
L_i(v_j)=\delta_{ij}.
\epsilonUse in source · 4 passages
Search aliases: ϵ epsilon
Use in source · chapters08_10.tex, line 206
An exact obstruction in a specified smaller space. At n=31, the seven retained corrections through degree 16 do not admit a nontrivial common-sign certificate. The shell has sixteen J-orbits. Let ϵ_i=sgn W(v_i) for their representatives. The retained exact dual has strictly positive rational entries y_i satisfying ∑_i y_i=1, ∑_i y_iϵ_i=0, ∑_i y_iϵ_iQ_j(v_i)=0 for all seven correction columns. Consequently, for every coefficient vector, ∑_i y_iϵ_i(1+∑_j a_jQ_j(v_i))=0. If every corrected packet h…
Exact mathematical passage
\epsilon_i=\sgn W(v_i)
\varepsilonUse in source · 3 passages
Search aliases: ε varepsilon
Use in source · chapter22.tex, line 88
Suppose the construction proves a sign ε_p∈{-1,1} for which ε_p W(v)F_p(v)≥0 for every v∈ V_p, and proves that at least one of these finitely many quantities is strictly positive. Then summing [source reference] gives ε_pτ(p)>0. This would be a valid all-prime certificate theorem.
Exact mathematical passage
\varepsilon_p\in\{-1,1\}
\epsilon_iUse in source · 4 passages
Search aliases: ϵ_i epsilon
Use in source · chapters08_10.tex, line 206
An exact obstruction in a specified smaller space. At n=31, the seven retained corrections through degree 16 do not admit a nontrivial common-sign certificate. The shell has sixteen J-orbits. Let ϵ_i=sgn W(v_i) for their representatives. The retained exact dual has strictly positive rational entries y_i satisfying ∑_i y_i=1, ∑_i y_iϵ_i=0, ∑_i y_iϵ_iQ_j(v_i)=0 for all seven correction columns. Consequently, for every coefficient vector, ∑_i y_iϵ_i(1+∑_j a_jQ_j(v_i))=0. If every corrected packet h…
Exact mathematical passage
\epsilon_i=\sgn W(v_i)
\varepsilon_pUse in source · 3 passages
Search aliases: ε_p varepsilon
Use in source · chapter22.tex, line 88
Suppose the construction proves a sign ε_p∈{-1,1} for which ε_p W(v)F_p(v)≥0 for every v∈ V_p, and proves that at least one of these finitely many quantities is strictly positive. Then summing [source reference] gives ε_pτ(p)>0. This would be a valid all-prime certificate theorem.
Exact mathematical passage
\varepsilon_p\in\{-1,1\}
\zetaUse in source · 3 passages
Search aliases: ζ zeta
Use in source · chapters01_03.tex, line 162
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\zeta=\alpha/\beta
\zeta^{-1}Use in source · 1 passage
Search aliases: ζ^-1 zeta
Use in source · chapters01_03.tex, line 163
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\label{eq:rootunity} \frac{a^2}{p^{11}} =\frac{(\alpha+\beta)^2}{\alpha\beta} =2+\zeta+\zeta^{-1}.
\zeta^{r+1}Use in source · 1 passage
Search aliases: ζ^r+1 zeta
Use in source · chapters01_03.tex, line 162
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\zeta^{r+1}=1
\ThetaUse in source · 3 passages
Search aliases: Θ Theta
Use in source · chapters08_10.tex, line 29
More generally, if Q is symmetric and DQ is harmonic homogeneous of degree 10+d, its weighted shell theta series is Θ_Q(q)=∑_n≥1(∑_v∈VnW(v)Q(v))q^n. The harmonic theta theorem initially supplies a vector-valued modular form. Obtaining a scalar full-level cusp form requires the following lattice specialization, rather than an unexplained assertion that every shell theta series has level one.
Exact mathematical passage
\label{eq:thetaq} \Theta_Q(q)=\sum_{n\ge1}\left(\sum_{v\in\Vn}W(v)Q(v)\right)q^n.
\thetaUse in source · 2 passages
Search aliases: θ theta
Use in source · chapters17_18.tex, line 30
Radial factors become derivatives. For an alternating harmonic h_ℓ, write θ_h(τ)=∑_substackv_i≡ i (5) ∑ v_i=0 h(v)exp(2π iτfrac v ^210). Chapter~[source reference] establishes that it is a scalar cusp form of weight ℓ+2. The normalization includes only a fixed scalar if one divides by 288.
Exact mathematical passage
\theta_h(\tau)=\sum_{\substack{v_i\equiv i\ (5)\\\sum v_i=0}} h(v)\exp\left(2\pi i\tau\frac{\|v\|^2}{10}\right).
\theta_hUse in source · 2 passages
Search aliases: θ_h theta
Use in source · chapters17_18.tex, line 30
Radial factors become derivatives. For an alternating harmonic h_ℓ, write θ_h(τ)=∑_substackv_i≡ i (5) ∑ v_i=0 h(v)exp(2π iτfrac v ^210). Chapter~[source reference] establishes that it is a scalar cusp form of weight ℓ+2. The normalization includes only a fixed scalar if one divides by 288.
Exact mathematical passage
\theta_h(\tau)=\sum_{\substack{v_i\equiv i\ (5)\\\sum v_i=0}} h(v)\exp\left(2\pi i\tau\frac{\|v\|^2}{10}\right).
\Theta_QUse in source · 3 passages
Search aliases: Θ_Q Theta
Use in source · chapters08_10.tex, line 29
More generally, if Q is symmetric and DQ is harmonic homogeneous of degree 10+d, its weighted shell theta series is Θ_Q(q)=∑_n≥1(∑_v∈VnW(v)Q(v))q^n. The harmonic theta theorem initially supplies a vector-valued modular form. Obtaining a scalar full-level cusp form requires the following lattice specialization, rather than an unexplained assertion that every shell theta series has level one.
Exact mathematical passage
\label{eq:thetaq} \Theta_Q(q)=\sum_{n\ge1}\left(\sum_{v\in\Vn}W(v)Q(v)\right)q^n.
\theta_{s_2^jh}Use in source · 1 passage
Search aliases: θ_s_2^jh theta
Use in source · chapters17_18.tex, line 37
Each application of DQ=q d/dq multiplies the shell coefficient by v ^2/10. Therefore θ_s_2^jh=10^jDQ^jθ_h. Differentiating a modular transformation law shows that DQ^jf, for modular weight w, is quasimodular of weight w+2j and depth at most j. Combining [source reference] and [source reference], the homogeneous shell series F_d(τ)=∑_n≥1(∑_v∈VnW(v)Q_d(v))q^n is quasimodular of weight (M-2j+2)+2j=12+d, independent of the radial index j. Its depth is at most d/2. Polynomial growth and Gaussian deca…
Exact mathematical passage
\label{eq:radialderivative} \theta_{s_2^jh}=10^j\DQ^j\theta_h.
\lambdaUse in source · 6 passages
Search aliases: λ lambda
Use in source · chapters06_07.tex, line 6
The Nekrasov--Okounkov identity in Han's formulation is ∏_m≥1(1-q^m)^z-1 =∑_λ q^|λ|∏_h∈ H(λ)(1-frac zh^2). At z=25, a partition having a hook of length 5 contributes zero. The surviving terms are the 5-cores. Multiplication by q shows that the coefficient τ(n) is the sum of the hook products over 5-cores of size n-1.
Exact mathematical passage
\label{eq:hookidentity} \prod_{m\ge1}(1-q^m)^{z-1} =\sum_\lambda q^{|\lambda|}\prod_{h\in H(\lambda)}\left(1-\frac z{h^2}\right).
\lambda^2Use in source · 1 passage
Search aliases: λ^2 lambda
Use in source · chapters19_21.tex, line 137
There is no proper nonzero φ-stable Q_p-subspace. In dimension two such a subspace would be a line, and φ on it would have an eigenvalue λ∈Q_p satisfying λ^2=-p^11. This would require 2v_p(λ)=11, impossible for an integer-valued valuation on Q_p^×. Thus the proper-subobject inequalities are vacuous. The zero object imposes equality 0=0, and the whole object satisfies the total equality. The filtered module is weakly admissible.
Exact mathematical passage
\lambda^2=-p^{11}.
\muUse in source · 16 passages
Search aliases: μ mu
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
\mu_{16249}Use in source · 1 passage
Search aliases: μ_16249 mu
Use in source · appendix_workbook.tex, line 190
At 16249 the shell contains 16250 packets, with 8284 positive and 7966 negative weights. The exact coefficient is τ(16249)=133116215274958169206250. Here μ_16249=4. The full coefficient table is as follows.
Exact mathematical passage
\mu_{16249}=4
\mu_{499}Use in source · 2 passages
Search aliases: μ_499 mu
Use in source · appendix_workbook.tex, line 113
At n=499 the shell has 500 packets. Of their weights, 262 are positive and 238 negative. Their sum is τ(499)=-108877719272500. The minimum layer valuation is μ_499=3.
Exact mathematical passage
\mu_{499}=3
\mu_nUse in source · 5 passages
Search aliases: μ_n mu
Use in source · appendix_workbook.tex, line 95
For a nonzero P_n, use the convention v_5(0)=+∞ and define μ_n=min_j(j+v_5(B_j)), G_n(U)=5^-μ_nP_n(5U)=∑_j=0^10C_jU^j. Every C_j=B_j5^j-μ_n is an integer. The power of 5 should be removed only after verifying divisibility for each coefficient. Since 288 is a 5-adic unit, v_5(τ(n))=μ_n+v_5(G_n(1)). All coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a la…
Exact mathematical passage
\mu_n=\min_j\bigl(j+v_5(B_j)\bigr),\qquad G_n(U)=5^{-\mu_n}P_n(5U)=\sum_{j=0}^{10}C_jU^j.
\mu_pUse in source · 2 passages
Search aliases: μ_p mu
Use in source · chapters06_07.tex, line 160
A double residual root with no lift. At p=16249, the shell has 16250 packets. The weighted coefficient valuations of P_p(5U) are (4,5,4,4,6,5,7,8,9,9,11), so μ_p=4. The reduction is overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)pmod5. The residual root at 1 is double. Exact coefficient reduction gives G_p(1)≡15pmod25, G'_p(1)≡0pmod5. For every y∈Z_5, polynomial Taylor expansion has an integral remainder: G_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y), H∈Z_5[y]. Consequently G_p(1+5y)≡15pmod25 for every y∈Z_5. No …
Exact mathematical passage
\mu_p=4
\nuUse in source · 5 passages
Search aliases: ν nu
Use in source · appendices.tex, line 130
S03: harmonic theta modularity. Ibukiyama and Williams, Section~3.4 of the checked version, give the vector-valued harmonic theta theorem with weight ν+r/2 [source reference]. Here the rank is four and the harmonic polynomial is alternating of degree 10+d. The scalar full-level transformation follows from the separate sign-line descent in Chapter~8. It is not silently imported as part of the general theorem.
Exact mathematical passage
\nu+r/2
\piUse in source · 1 passage
Search aliases: π pi
Use in source · chapters17_18.tex, line 30
Radial factors become derivatives. For an alternating harmonic h_ℓ, write θ_h(τ)=∑_substackv_i≡ i (5) ∑ v_i=0 h(v)exp(2π iτfrac v ^210). Chapter~[source reference] establishes that it is a scalar cusp form of weight ℓ+2. The normalization includes only a fixed scalar if one divides by 288.
Exact mathematical passage
\theta_h(\tau)=\sum_{\substack{v_i\equiv i\ (5)\\\sum v_i=0}} h(v)\exp\left(2\pi i\tau\frac{\|v\|^2}{10}\right).
\rhoUse in source · 9 passages
Search aliases: ρ rho
Use in source · chapters06_07.tex, line 145
luations are (3,5,3,4,4,5,7,11,8,9,12). Hence μ_499=3 and reduction gives overline G_499(U)=2+3U^2=3(U-1)(U+1) in F_5[U]. At the relevant value U=1, G_499(1)≡10pmod25, G'_499(1)≡1pmod5. Substitute U=1+5d. The integral Taylor identity gives G_499(1+5d)≡10+5dpmod25. The root in this residue class is therefore ρ≡16pmod25. Hensel's lemma lifts it uniquely to Z_5. The divided difference (G(U)-G(ρ))/(U-ρ) is a unit on U≡1pmod5, since its reduction is G'(1)≠0. Thus v_5(G_499(U))=v_5(U-ρ) on that residu…
Exact mathematical passage
\rho\equiv16\pmod{25}
\rho_\ellUse in source · 5 passages
Search aliases: ρ_ℓ rho ell
Use in source · chapters19_21.tex, line 22
Integral representations and complex conjugation. For each prime ℓ, the representation attached to Δ has the arithmetic-Frobenius normalization ρ_ℓ:G_Q⟶ GL_2(Q_ℓ), is continuous and unramified outside ℓ, and satisfies det(X-ρ_ℓ(Frob_p))=X^2-τ(p)X+p^11, p≠ℓ. Its determinant is the eleventh power of the ℓ-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the prima…
Exact mathematical passage
\rho_\ell:G_{\Q}\longrightarrow GL_2(\Q_\ell),
\sigma_1Use in source · 4 passages
Search aliases: σ_1 sigma
Use in source · appendix_workbook.tex, line 31
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
\sigma_1(j)=\sum_{d\mid j}d
\sigma_7Use in source · 1 passage
Search aliases: σ_7 sigma
Use in source · appendix_workbook.tex, line 275
These four coefficients were checked independently by the Eisenstein convolution. If σ_r(n)=∑_dmid nd^r, the coefficient of (E_4E_8-E_6^2)/1728 is obtained from 1728τ(n)= · 240σ_3(n)+480σ_7(n)+1008σ_5(n) · +115200∑_k=1^n-1σ_3(k)σ_7(n-k) · -254016∑_k=1^n-1σ_5(k)σ_5(n-k). This computes the selected integers without the triangular recurrence. The numerator is divided exactly by 1728. The replay checks this independent formula for 1≤ n≤128 and at the four displayed primes, giving 132 checks.
Exact mathematical passage
1728\tau(n)={}&240\sigma_3(n)+480\sigma_7(n)+1008\sigma_5(n)\\ &+115200\sum_{k=1}^{n-1}\sigma_3(k)\sigma_7(n-k)\\ &-254016\sum_{k=1}^{n-1}\sigma_5(k)\sigma_5(n-k).
\sigma_jUse in source · 1 passage
Search aliases: σ_j sigma
Use in source · chapters01_03.tex, line 54
Exact coefficient engines. The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
\sigma_j(n)=\sum_{d\mid n}d^j
\sigma_rUse in source · 1 passage
Search aliases: σ_r sigma
Use in source · appendix_workbook.tex, line 274
These four coefficients were checked independently by the Eisenstein convolution. If σ_r(n)=∑_dmid nd^r, the coefficient of (E_4E_8-E_6^2)/1728 is obtained from 1728τ(n)= · 240σ_3(n)+480σ_7(n)+1008σ_5(n)
Exact mathematical passage
\sigma_r(n)=\sum_{d\mid n}d^r
\varphiUse in source · 8 passages
Search aliases: φ varphi
Use in source · chapters19_21.tex, line 98
For every prime p, there is a weakly admissible two-dimensional filtered φ-module over Q_p, with monodromy N=0, filtration jumps 0,11 and Frobenius characteristic polynomial X^2+p^11.
Exact mathematical passage
\varphi
\varphi^2Use in source · 2 passages
Search aliases: φ^2 varphi
Use in source · chapters19_21.tex, line 109
The matrix and filtration. Take D=Q_pe_1+Q_pe_2 and define φ(e_1)=e_2, φ(e_2)=-p^11e_1, N=0. In the displayed basis, φ=0 · -p^11 1 · 0, Trφ=0, detφ=p^11, φ^2=-p^11I. Choose any Q_p-line L⊂ D and put Fil^iD= D · i≤0,
Exact mathematical passage
\varphi=\begin{pmatrix}0&-p^{11}\\1&0\end{pmatrix},\qquad \Tr\varphi=0,\qquad\det\varphi=p^{11},\qquad \varphi^2=-p^{11}I.
\chiUse in source · 10 passages
Search aliases: χ chi
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
\chi^2Use in source · 1 passage
Search aliases: χ^2 chi
Use in source · chapters15_16.tex, line 44
A residue eliminates three prime classes. Suppose p≠5 and put χ=chifive(p). Modulo 5, χ≡ p^2, so the two recurrence roots may be taken as p and χ: p+χ≡ a, pχ≡ p^11pmod5. The complete homogeneous recurrence coefficient is T_r≡ C_r=∑_i=0^r p^r-iχ^ipmod5. Also χ p^10≡1pmod5. Since χ^2=1, for r≥2 the sum telescopes: R(p^r)≡ C_r-C_r-2 =p^r-1(p+χ)pmod5. The same expression holds at r=1. It is nonzero for p≡1,2,3pmod5. It proves nonvanishing for all positive exponents in those prime classes. Only p=5 a…
Exact mathematical passage
\chi^2=1
\chi^iUse in source · 1 passage
Search aliases: χ^i chi
Use in source · chapters15_16.tex, line 41
A residue eliminates three prime classes. Suppose p≠5 and put χ=chifive(p). Modulo 5, χ≡ p^2, so the two recurrence roots may be taken as p and χ: p+χ≡ a, pχ≡ p^11pmod5. The complete homogeneous recurrence coefficient is T_r≡ C_r=∑_i=0^r p^r-iχ^ipmod5. Also χ p^10≡1pmod5. Since χ^2=1, for r≥2 the sum telescopes: R(p^r)≡ C_r-C_r-2 =p^r-1(p+χ)pmod5. The same expression holds at r=1. It is nonzero for p≡1,2,3pmod5. It proves nonvanishing for all positive exponents in those prime classes. Only p=5 a…
Exact mathematical passage
T_r\equiv C_r=\sum_{i=0}^r p^{r-i}\chi^i\pmod5.
\chi_5Use in source · 1 passage
Search aliases: χ_5 chi
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
\PsiUse in source · 2 passages
Search aliases: Ψ Psi
Use in source · chapters19_21.tex, line 46
The actual joint image. Factor M=∏_ℓ^eVert Mℓ^e and form the single joint finite map Ψ:G_Q⟶ (∏_ℓ^eVert MGL_2(Z/ℓ^eZ)) ×(Z/5MZ)^×. Its components are the integral representation reductions and the cyclotomic character modulo 5M. Let G be its actual image. The kernel defines a finite Galois extension of Q. The image need not be the whole displayed product. Relations between different representations and determinants are allowed.
Exact mathematical passage
\label{eq:jointimage} \Psi:G_{\Q}\longrightarrow \left(\prod_{\ell^e\Vert M}GL_2(\Z/\ell^e\Z)\right) \times(\Z/5M\Z)^\times.
\ldotsUse in source · 9 passages
Search aliases: … ldots
Use in source · appendix_operator.tex, line 20
For direct symbolic computation, use four coordinates x_0,…,x_3 and substitute v=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3). The coordinate tangent vectors are b_a=e_a-e_4. Their Gram matrix and its inverse are G=I_4+11^T, G^-1=I_4-frac1511^T, now with a four-entry vector 1. Matrix multiplication verifies the inverse. The factor 1/5 remains because the missing fifth coordinate is determined by the four chosen coordinates.
Exact mathematical passage
x_0,\ldots,x_3
\ellUse in source · 59 passages
Search aliases: ℓ ell
Use in source · appendices.tex, line 17
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d,
Exact mathematical passage
v_\ell(x)
\ell^{-a}Use in source · 2 passages
Search aliases: ℓ^-a ell
Use in source · chapters19_21.tex, line 32
A continuous representation of the compact group G_Q has a stable Z_ℓ lattice. To see this directly, take an initial lattice L_0. Compactness bounds all its translates inside ℓ^-aL_0 for some a. The module L=∑_g∈ G_Qρ_ℓ(g)L_0 is contained in ℓ^-aL_0 and contains L_0. It is finitely generated, since a submodule of a finite free Z_ℓ module is finitely generated, and it is invariant by construction. Reductions modulo ℓ^e are therefore defined. No image-surjectivity theorem is required.
Exact mathematical passage
\ell^{-a}L_0
\ell^eUse in source · 4 passages
Search aliases: ℓ^e ell
Use in source · chapters19_21.tex, line 36
l lattice. To see this directly, take an initial lattice L_0. Compactness bounds all its translates inside ℓ^-aL_0 for some a. The module L=∑_g∈ G_Qρ_ℓ(g)L_0 is contained in ℓ^-aL_0 and contains L_0. It is finitely generated, since a submodule of a finite free Z_ℓ module is finitely generated, and it is invariant by construction. Reductions modulo ℓ^e are therefore defined. No image-surjectivity theorem is required.
Exact mathematical passage
\ell^e
\toUse in source · 1 passage
Search aliases: → to
Use in source · chapters13_14.tex, line 126
Put H_k=ker(1^T:Q^s_k→Q), T_k=im M_k, E_k=H_k/T_k. The quotient E_k is the vector space of total-preserving changes that are unavailable from the chosen family. It is not a count of failed sign certificates. A sign certificate may exist even when E_k is nonzero.
Exact mathematical passage
H_k=\ker(\mathbf1^T:\Q^{s_k}\to\Q),\qquad T_k=\im M_k,\qquad E_k=H_k/T_k.
\mapstoUse in source · 2 passages
Search aliases: ↦ mapsto
Use in source · chapters13_14.tex, line 15
The permutation j↦ gjbmod5 has sign chifive(g). For the four units modulo 5, multiplication by 1 is the identity, multiplication by 2 or 3 is a four-cycle on the nonzero residues, and multiplication by 4 is two transpositions. Scaling the degree-10 Vandermonde therefore gives W(v)=chifive(g)g^10W(u). The character is not an arbitrary normalization. It records the sign of the residue-restoring permutation.
Exact mathematical passage
j\mapsto gj\bmod5
\forallUse in source · 2 passages
Search aliases: ∀ forall
Use in source · chapters19_21.tex, line 78
The assertion is ∀ M ∃ a positive-density set of suitable p. It is not ∃ p ∀ M: Mmidτ(p). The surviving primes may change with M. No passage to an exact integer zero is available by reversing the quantifiers.
Exact mathematical passage
\forall M\ \exists\text{ a positive-density set of suitable }p.
\partialUse in source · 4 passages
Search aliases: ∂ partial
Use in source · appendix_operator.tex, line 13
projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:ambientlap} \Delta_H(F|_H) =\left(\sum_{i=0}^4\partial_i^2F -\frac15\left(\sum_{i=0}^4\partial_i\right)^2F\right)\Big|_H.
\partial^2Use in source · 2 passages
Search aliases: ∂^2 partial
Use in source · appendix_operator.tex, line 32
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\label{appop:chartlap} \Delta_H=\sum_{a,b=0}^3 \left(\delta_{ab}-\frac15\right)\frac{\partial^2}{\partial x_a\partial x_b}.
\partial_iUse in source · 2 passages
Search aliases: ∂_i partial
Use in source · appendix_operator.tex, line 13
projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:ambientlap} \Delta_H(F|_H) =\left(\sum_{i=0}^4\partial_i^2F -\frac15\left(\sum_{i=0}^4\partial_i\right)^2F\right)\Big|_H.
\partial_i^2Use in source · 1 passage
Search aliases: ∂_i^2 partial
Use in source · appendix_operator.tex, line 13
projection from R^5 to H is P=I_5-frac1511^T, where 1=(1,1,1,1,1)^T. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies P^2=P, and has image H. For an ambient polynomial F, its gradient along H is P∇ F. Its Laplacian along H is Δ_H(F|_H) =(∑_i=0^4∂_i^2F -frac15(∑_i=0^4∂_i)^2F)|_H. Only tangential derivatives enter. Thus this expression is independent of the polynomial extension away from H.
Exact mathematical passage
\label{appop:ambientlap} \Delta_H(F|_H) =\left(\sum_{i=0}^4\partial_i^2F -\frac15\left(\sum_{i=0}^4\partial_i\right)^2F\right)\Big|_H.
\existsUse in source · 3 passages
Search aliases: ∃ exists
Use in source · chapters08_10.tex, line 246
Completeness retains the total. Suppose a shell has nonzero orbit masses w_1,…,w_N and total T. Even allow every rational correction vector q with ∑_iw_iq_i=0. Then ∃ q: w_i(1+q_i) have one weak sign and are not all zero Longleftrightarrow T≠0. Necessity follows from conservation. For sufficiency, if T≠0, take q_i=fracTNw_i-1. Every corrected mass is T/N and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of T≠0 without independent information …
Exact mathematical passage
\label{eq:fullsignequiv} \exists q:\ w_i(1+q_i)\text{ have one weak sign and are not all zero} \quad\Longleftrightarrow\quad T\ne0.
\prod_{0\le i<j\le4}Use in source · 3 passages
Search aliases: ∏_0≤ i<j≤4 prod le le
Use in source · appendices.tex, line 37
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
D(v)=\prod_{0\le i<j\le4}(v_i-v_j),\qquad W(v)=D(v)/288.
\prod_{h\in H(\lambda)}Use in source · 1 passage
Search aliases: ∏_h∈ H(λ) prod in lambda
Use in source · chapters06_07.tex, line 6
The Nekrasov--Okounkov identity in Han's formulation is ∏_m≥1(1-q^m)^z-1 =∑_λ q^|λ|∏_h∈ H(λ)(1-frac zh^2). At z=25, a partition having a hook of length 5 contributes zero. The surviving terms are the 5-cores. Multiplication by q shows that the coefficient τ(n) is the sum of the hook products over 5-cores of size n-1.
Exact mathematical passage
\label{eq:hookidentity} \prod_{m\ge1}(1-q^m)^{z-1} =\sum_\lambda q^{|\lambda|}\prod_{h\in H(\lambda)}\left(1-\frac z{h^2}\right).
\prod_iUse in source · 1 passage
Search aliases: ∏_i prod
Use in source · appendix_operator.tex, line 133
Newton reductions with zero first power sum. Let e_j be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is ∏_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5, because e_1=s_1=0. The first Newton identities give e_2 · =-frac12s_2, · e_3 · =frac13s_3,
Exact mathematical passage
\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,
\prod_{j=0}^4Use in source · 1 passage
Search aliases: ∏_j=0^4 prod
Use in source · chapters06_07.tex, line 34
Integrality follows from an integer-valued basis. For every integer x, including negative integers, the generalized binomial coefficient binomxj is an integer. Its leading coefficient as a polynomial in x is 1/j!. Thus det[binomv_ij]_0≤ i,j≤4 =frac∏_i<j(v_j-v_i)∏_j=0^4j!. There are ten pairs, so reversing every difference changes the sign by (-1)^10=1. The determinant is W(v) and is integral. This proof is independent of the fact that the total in [source reference] is an integer.
Exact mathematical passage
\det\left[\binom{v_i}{j}\right]_{0\le i,j\le4} =\frac{\prod_{i<j}(v_j-v_i)}{\prod_{j=0}^4j!}.
\prod_{j\ne i}Use in source · 1 passage
Search aliases: ∏_j≠ i prod ne
Use in source · chapters17_18.tex, line 112
For that t, set L_i(v)=∏_j≠ i fracE_t(v)-e_j(s_2(v)/C)^5e_i-e_j. Each factor is homogeneous of degree 10, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree 10(N-1). On the shell, s_2/C=1, so L_i(v_j)=δ_ij. Consequently, any prescribed rational orbit values a_1,…,a_N are interpolated by Q=∑_ia_iL_i. If N=1, an ordinary constant suffices.
Exact mathematical passage
\label{eq:li} L_i(v)=\prod_{j\ne i} \frac{E_t(v)-e_j(s_2(v)/C)^5}{e_i-e_j}.
\prod_{k\ge1}Use in source · 2 passages
Search aliases: ∏_k≥1 prod ge
Use in source · appendix_workbook.tex, line 12
Write F(q)=fracΔ(q)q=∏_k≥1(1-q^k)^24 =∑_m≥0f_mq^m. Then f_0=1 and τ(n)=f_n-1. This shift is essential. A program that stores τ(0)=0 as a sentinel must not confuse that entry with the constant coefficient f_0.
Exact mathematical passage
F(q)=\frac{\Delta(q)}q=\prod_{k\ge1}(1-q^k)^{24} =\sum_{m\ge0}f_mq^m.
\prod_{m\ge1}Use in source · 5 passages
Search aliases: ∏_m≥1 prod ge
Use in source · appendices.tex, line 7
The coefficient and its local parameters. The normalized form is Δ(q)=q∏_m≥1(1-q^m)^24 =∑_n≥1τ(n)q^n, τ(1)=1. All coefficient indices in this book are positive integers. A stored value at index zero is a programming sentinel unless a displayed formula explicitly introduces the coefficients of Δ/q. For the latter coefficients, f_m=τ(m+1) and f_0=1.
Exact mathematical passage
\Delta(q)=q\prod_{m\ge1}(1-q^m)^{24} =\sum_{n\ge1}\tau(n)q^n,\qquad \tau(1)=1.
\prod_pUse in source · 3 passages
Search aliases: ∏_p prod
Use in source · chapters01_03.tex, line 203
· and τ(p)=0. In particular, Lehmer's conjecture is equivalent to nonvanishing at every prime. Factor n=∏_p p^r_p. Multiplicativity expresses τ(n) as a finite product of local blocks. An integral product is zero exactly when one factor is zero. Apply [source reference] to each factor.
Exact mathematical passage
n=\prod_p p^{r_p}
\prod_{p^{r_p}\Vert n}Use in source · 1 passage
Search aliases: ∏_p^r_pVert n prod Vert
Use in source · chapters19_21.tex, line 228
The full size bound survives. At primes other than q, the Hecke roots have absolute value p^11/2 by Deligne. Formula [source reference] therefore gives |τ(p^r)|≤(r+1)p^11r/2. At the modified prime, F_q,2m=(-q^11)^m, F_q,2m+1=0. These obey the same local bound at every exponent, including zero. Multiplying the local estimates gives |f(n)|≤∏_p^r_pVert n(r_p+1)p^11r_p/2 =d(n)n^11/2. A zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified paramet…
Exact mathematical passage
|f(n)|\le\prod_{p^{r_p}\Vert n}(r_p+1)p^{11r_p/2} =d(n)n^{11/2}.
\prod_{\ell^e\Vert M}Use in source · 2 passages
Search aliases: ∏_ℓ^eVert M prod ell Vert
Use in source · chapters19_21.tex, line 45
The actual joint image. Factor M=∏_ℓ^eVert Mℓ^e and form the single joint finite map Ψ:G_Q⟶ (∏_ℓ^eVert MGL_2(Z/ℓ^eZ)) ×(Z/5MZ)^×. Its components are the integral representation reductions and the cyclotomic character modulo 5M. Let G be its actual image. The kernel defines a finite Galois extension of Q. The image need not be the whole displayed product. Relations between different representations and determinants are allowed.
Exact mathematical passage
M=\prod_{\ell^e\Vert M}\ell^e
\sum_{a,b=0}^3Use in source · 1 passage
Search aliases: ∑_a,b=0^3 sum
Use in source · appendix_operator.tex, line 32
The chart Laplacian is consequently Δ_H=∑_a,b=0^3 (δ_ab-frac15)frac∂^2∂ x_a∂ x_b. The coordinates x_a are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, s_2=x^TGx, so its Hessian is 2G and Δ_Hs_2=tr(G^-12G)=8. The rank of the Euclidean space is four, giving twice that rank. Later we obtain Ls_2=48, an intentionally different number.
Exact mathematical passage
\label{appop:chartlap} \Delta_H=\sum_{a,b=0}^3 \left(\delta_{ab}-\frac15\right)\frac{\partial^2}{\partial x_a\partial x_b}.
\sum_{a=0}^3Use in source · 1 passage
Search aliases: ∑_a=0^3 sum
Use in source · appendix_operator.tex, line 221
The retained programs organize these same operations using sparse rational polynomials. A direct four-coordinate implementation can instead substitute v_4=-∑_a=0^3x_a into DQ and apply [source reference]. Agreement between these organizations checks the symbolic construction through different algebraic representations. The universal justification remains the metric calculation, the exact polynomial identities and the stated harmonic theta and valence contracts. None of those statements alone exc…
Exact mathematical passage
v_4=-\sum_{a=0}^3x_a
\sum_{a=0}^{m-2}Use in source · 4 passages
Search aliases: ∑_a=0^m-2 sum
Use in source · appendix_operator.tex, line 76
Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. …
Exact mathematical passage
\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j} =\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.
\sum_dUse in source · 1 passage
Search aliases: ∑_d sum
Use in source · chapters17_18.tex, line 10
Let Q be any finite symmetric polynomial over C, restricted to ∑_i v_i=0. Suppose ∑_v∈VnW(v)Q(v)=0 for every positive integer n. Write Q=∑_dQ_d in homogeneous degrees. Then every Q_d is separately universally null. Moreover, the weighted Q-sum is zero in each exact-content sector of every shell.
Exact mathematical passage
Q=\sum_dQ_d
\sum_{d^2\mid n}Use in source · 4 passages
Search aliases: ∑_d^2mid n sum mid
Use in source · CURRENT_STATUS.md, line 11
R(n)=∑_d^2mid nμ(d)χ_5(d)d^10τ(n/d^2).
Exact mathematical passage
R(n)=\sum_{d^2\mid n}\mu(d)\chi_5(d)d^{10}\tau(n/d^2).
\sum_{d\mid j}Use in source · 1 passage
Search aliases: ∑_dmid j sum mid
Use in source · appendix_workbook.tex, line 31
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
\sigma_1(j)=\sum_{d\mid j}d
\sum_{d\mid m}Use in source · 1 passage
Search aliases: ∑_dmid m sum mid
Use in source · chapters13_14.tex, line 37
The square-supported arithmetic kernel in [source reference] has inverse d^2⟼μ(d)chifive(d)d^10. Indeed, at a square index m^2, the convolution of the two kernels is chifive(m)m^10∑_dmid mμ(d), which is 1 at m=1 and 0 otherwise. At nonsquare indices both kernels have no support. Consequently R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). Both formulas are exact for every positive integer. They are not fitted from the first few shells.
Exact mathematical passage
\chifive(m)m^{10}\sum_{d\mid m}\mu(d),
\sum_{d\mid n}Use in source · 4 passages
Search aliases: ∑_dmid n sum mid
Use in source · appendices.tex, line 124
S13: packet cardinality. Garvan, Kim and Stanton's Theorem~4 and equation~(5.3) give the five-core count [source reference]. In the present conventions this reads a_5(n-1)=∑_dmid nchifive(d)frac nd. For a prime different from 5, the shell cardinality is p+chifive(p). This is the number of packets, not their weighted sum.
Exact mathematical passage
a_5(n-1)=\sum_{d\mid n}\chifive(d)\frac nd.
\sum_gUse in source · 1 passage
Search aliases: ∑_g sum
Use in source · chapters17_18.tex, line 149
The negative version is identical. For a weakly nonnegative certificate with at least one strict mass, the criterion is T_g≥0 for every class and ∑_gT_g>0.
Exact mathematical passage
\sum_gT_g>0
\sum_{g^2\mid n}Use in source · 3 passages
Search aliases: ∑_g^2mid n sum mid
Use in source · appendices.tex, line 45
The content of a packet is the positive gcd of its five coordinates. Content-one packets are primitive. Multiplication followed by the prescribed residue reindexing gives τ(n)=∑_g^2mid nchifive(g)g^10R(n/g^2), R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). The boundary values are R(1)=1 and R(p)=τ(p). At a prime shell the only possible content is one, so the extraction has not created a new prime coefficient.
Exact mathematical passage
\tau(n)=\sum_{g^2\mid n}\chifive(g)g^{10}R(n/g^2), \qquad R(n)=\sum_{d^2\mid n}\mu(d)\chifive(d)d^{10}\tau(n/d^2).
\sum_{g\in G_{\Q}}Use in source · 1 passage
Search aliases: ∑_g∈ G_Q sum in Q
Use in source · chapters19_21.tex, line 33
A continuous representation of the compact group G_Q has a stable Z_ℓ lattice. To see this directly, take an initial lattice L_0. Compactness bounds all its translates inside ℓ^-aL_0 for some a. The module L=∑_g∈ G_Qρ_ℓ(g)L_0 is contained in ℓ^-aL_0 and contains L_0. It is finitely generated, since a submodule of a finite free Z_ℓ module is finitely generated, and it is invariant by construction. Reductions modulo ℓ^e are therefore defined. No image-surjectivity theorem is required.
Exact mathematical passage
L=\sum_{g\in G_{\Q}}\rho_\ell(g)L_0
\sum_iUse in source · 17 passages
Search aliases: ∑_i sum
Use in source · appendices.tex, line 30
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
\Vn=\left\{v\in\Z^5: v_i\equiv i\pmod5,\quad \sum_i v_i=0,\quad \sum_i v_i^2=10n\right\}.
\sum_{i<j}Use in source · 1 passage
Search aliases: ∑_i<j sum
Use in source · appendix_operator.tex, line 71
Δ_Hs_m · =frac45m(m-1)s_m-2. In the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices i,j and j,i to obtain 2∇_Hlog D·∇_Hs_m =2m∑_i<jfracv_i^m-1-v_j^m-1v_i-v_j. The divided difference is the finite polynomial fracv_i^m-1-v_j^m-1v_i-v_j =∑_a=0^m-2v_i^av_j^m-2-a. The sum over a is symmetric in the two coordinates. Hence twice the sum over i<j equals the sum over i≠ j. For each a, that ordered sum is s_as_m-2-a-s_m-2. …
Exact mathematical passage
2\nabla_H\log D\cdot\nabla_Hs_m =2m\sum_{i<j}\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.
\sum_{i=0}^4Use in source · 3 passages
Search aliases: ∑_i=0^4 sum
Use in source · appendix_operator.tex, line 5
Write s_m=∑_i=0^4v_i^m, with s_0=5 and s_1=0, and let D=∏_i<j(v_i-v_j). We distinguish the ordinary Laplacian Δ_H from the conjugated operator L defined by Δ_H(DQ)=D LQ. The factor D changes radial coefficients substantially. Confusing the two operators would invalidate the harmonicity test.
Exact mathematical passage
s_m=\sum_{i=0}^4v_i^m
\sum_{i=0}^rUse in source · 1 passage
Search aliases: ∑_i=0^r sum
Use in source · chapters15_16.tex, line 41
A residue eliminates three prime classes. Suppose p≠5 and put χ=chifive(p). Modulo 5, χ≡ p^2, so the two recurrence roots may be taken as p and χ: p+χ≡ a, pχ≡ p^11pmod5. The complete homogeneous recurrence coefficient is T_r≡ C_r=∑_i=0^r p^r-iχ^ipmod5. Also χ p^10≡1pmod5. Since χ^2=1, for r≥2 the sum telescopes: R(p^r)≡ C_r-C_r-2 =p^r-1(p+χ)pmod5. The same expression holds at r=1. It is nonzero for p≡1,2,3pmod5. It proves nonvanishing for all positive exponents in those prime classes. Only p=5 a…
Exact mathematical passage
T_r\equiv C_r=\sum_{i=0}^r p^{r-i}\chi^i\pmod5.
\sum_jUse in source · 3 passages
Search aliases: ∑_j sum
Use in source · chapter22.tex, line 83
A sufficient sign input. A second precise route would construct permitted corrections with an independently proved sign. At each remaining prime, choose finitely many rational universal null polynomials Q_p,j and rational coefficients a_p,j, and put F_p(v)=1+∑_j a_p,jQ_p,j(v). The choice may depend on p. Universal nullness must hold after the choice, so its weighted sum is still the actual τ(p).
Exact mathematical passage
F_p(v)=1+\sum_j a_{p,j}Q_{p,j}(v).
\sum_{j=0}^{24}Use in source · 1 passage
Search aliases: ∑_j=0^24 sum
Use in source · appendix_workbook.tex, line 19
For direct product expansion, terms of degree at most m depend only on factors with k≤ m. The exact binomial expansion (1-q^k)^24=∑_j=0^24(-1)^jbinom24jq^kj therefore supplies a finite algorithm for each desired coefficient. Three short calculations are f_1 · =-24,
Exact mathematical passage
(1-q^k)^{24}=\sum_{j=0}^{24}(-1)^j\binom{24}{j}q^{kj}
\sum_{j=0}^{\lfloor r/2\rfloor}Use in source · 1 passage
Search aliases: ∑_j=0^lfloor r/2rfloor sum lfloor rfloor
Use in source · chapters15_16.tex, line 53
The recurrence polynomial and untied valuations. For p≡4pmod5, χ=1 and a≡0pmod5. The exact recurrence expansion is T_r=∑_j=0^lfloor r/2rfloor (-1)^jbinomr-jjb^ja^r-2j. It follows by induction: multiplying the previous row by a and subtracting b times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are 1, so a strictly least valuation at either end cannot be raised by a coefficient divisible by p.
Exact mathematical passage
\label{eq:trexpansion} T_r=\sum_{j=0}^{\lfloor r/2\rfloor} (-1)^j\binom{r-j}{j}b^ja^{r-2j}.
\sum_{j=0}^rUse in source · 2 passages
Search aliases: ∑_j=0^r sum
Use in source · chapters01_03.tex, line 158
Solving the recurrence gives T_r=fracα^r+1-β^r+1α-β =∑_j=0^rα^r-jβ^j. If T_r=0 for some r≥1, then ζ=α/β is a root of unity with ζ^r+1=1. Consequently fraca^2p^11 =frac(α+β)^2αβ =2+ζ+ζ^-1. The left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between 0 and 4. Thus fraca^2p^11∈{0,1,2,3,4}. This argument uses the root-of-unity consequence of a zero, rather than assuming that a…
Exact mathematical passage
\label{eq:rootrec} T_r=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta} =\sum_{j=0}^r\alpha^{r-j}\beta^j.
\sum_{j=1}^{m}Use in source · 1 passage
Search aliases: ∑_j=1^m sum
Use in source · appendix_workbook.tex, line 36
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
m f_m=-24\sum_{j=1}^{m}\sigma_1(j)f_{m-j} \qquad(m\ge1). \label{eq:workbook-divisor-recurrence}
\sum_{j\ne i}Use in source · 1 passage
Search aliases: ∑_j≠ i sum ne
Use in source · appendix_operator.tex, line 55
Translation invariance of D gives ∑_ifrac∂_iDD=0, frac∂_iDD=∑_j≠ ifrac1v_i-v_j. Its ambient logarithmic gradient is already tangential. This removes a potential extra projection term in the derivation below.
Exact mathematical passage
\sum_i\frac{\partial_iD}{D}=0, \qquad \frac{\partial_iD}{D}=\sum_{j\ne i}\frac1{v_i-v_j}.
\sum_{j\ge0}Use in source · 3 passages
Search aliases: ∑_j≥0 sum ge
Use in source · chapters01_03.tex, line 80
Jacobi's triangular series. Jacobi's identity is ∏_m≥1(1-q^m)^3 =∑_j≥0(-1)^j(2j+1)q^j(j+1)/2. Raising this series to the eighth power and multiplying by q gives Δ. The sparse triangular exponents make it useful for coefficient calculations. The source check in [source reference] also uses this identity to establish the parity criterion discussed below. It is an exact formal series identity, not a numerical approximation.
Exact mathematical passage
\label{eq:jacobi} \prod_{m\ge1}(1-q^m)^3 =\sum_{j\ge0}(-1)^j(2j+1)q^{j(j+1)/2}.
\sum_{j\ge1}Use in source · 2 passages
Search aliases: ∑_j≥1 sum ge
Use in source · appendix_workbook.tex, line 32
The logarithmic derivative provides a different exact engine. With σ_1(j)=∑_dmid jd, comparison of coefficients in qfracF'(q)F(q)=-24∑_j≥1σ_1(j)q^j gives m f_m=-24∑_j=1^mσ_1(j)f_m-j (m≥1). For m=3, the divisor sums are 1,3,4. Hence f_3=-8(252+3(-24)+4)=-8·184=-1472. An implementation should check that the numerator in [source reference] is divisible by m, rather than rounding the quotient.
Exact mathematical passage
q\frac{F'(q)}{F(q)}=-24\sum_{j\ge1}\sigma_1(j)q^j
\sum_kUse in source · 2 passages
Search aliases: ∑_k sum
Use in source · chapters17_18.tex, line 58
c · d]∈ SL_2(Z). Constants from the usual quasimodular convention can be absorbed into the coefficient functions. Assume ∑_kF_k=0 identically.
Exact mathematical passage
\sum_kF_k=0
\sum_{k=1}^{n-1}Use in source · 2 passages
Search aliases: ∑_k=1^n-1 sum
Use in source · appendix_workbook.tex, line 275
These four coefficients were checked independently by the Eisenstein convolution. If σ_r(n)=∑_dmid nd^r, the coefficient of (E_4E_8-E_6^2)/1728 is obtained from 1728τ(n)= · 240σ_3(n)+480σ_7(n)+1008σ_5(n) · +115200∑_k=1^n-1σ_3(k)σ_7(n-k) · -254016∑_k=1^n-1σ_5(k)σ_5(n-k). This computes the selected integers without the triangular recurrence. The numerator is divided exactly by 1728. The replay checks this independent formula for 1≤ n≤128 and at the four displayed primes, giving 132 checks.
Exact mathematical passage
1728\tau(n)={}&240\sigma_3(n)+480\sigma_7(n)+1008\sigma_5(n)\\ &+115200\sum_{k=1}^{n-1}\sigma_3(k)\sigma_7(n-k)\\ &-254016\sum_{k=1}^{n-1}\sigma_5(k)\sigma_5(n-k).
\sum_{k\ge1}Use in source · 1 passage
Search aliases: ∑_k≥1 sum ge
Use in source · chapters01_03.tex, line 60
oefficient engines The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
-24\frac{rq^r}{1-q^r}=-24\sum_{k\ge1}rq^{rk}.
\sum_{m,n=2}^5Use in source · 1 passage
Search aliases: ∑_m,n=2^5 sum
Use in source · appendix_operator.tex, line 112
If Q=F(s_2,s_3,s_4,s_5), the ordinary second-order chain rule and [source reference] yield LQ =∑_m=2^5frac∂ F∂ s_m Ls_m +∑_m,n=2^5 frac∂^2F∂ s_m∂ s_nΓ_mn. The second sum runs over ordered pairs. Thus a mixed derivative contributes twice when m≠ n. If one instead sums over m≤ n, the off-diagonal terms require a factor two. These are two equivalent conventions, but their factors must not be mixed.
Exact mathematical passage
\label{appop:chainrule} \mathcal LQ =\sum_{m=2}^5\frac{\partial F}{\partial s_m}\mathcal Ls_m +\sum_{m,n=2}^5 \frac{\partial^2F}{\partial s_m\partial s_n}\Gamma_{mn}.
\sum_{m=1}^{n-1}Use in source · 1 passage
Search aliases: ∑_m=1^n-1 sum
Use in source · chapters01_03.tex, line 66
Comparing coefficients of q^n in [source reference], for n>1, yields (n-1)τ(n)=-24∑_m=1^n-1σ(m)τ(n-m), τ(1)=1. This recurrence uses only smaller indices. Division by n-1 is exact because the recurrence came from the integral product, although an implementation should still check divisibility rather than silently round a quotient. It is a convenient independent coefficient engine at a moderate range.
Exact mathematical passage
\label{eq:logrec} (n-1)\tau(n)=-24\sum_{m=1}^{n-1}\sigma(m)\tau(n-m), \qquad \tau(1)=1.
\sum_{m=2}^5Use in source · 1 passage
Search aliases: ∑_m=2^5 sum
Use in source · appendix_operator.tex, line 112
If Q=F(s_2,s_3,s_4,s_5), the ordinary second-order chain rule and [source reference] yield LQ =∑_m=2^5frac∂ F∂ s_m Ls_m +∑_m,n=2^5 frac∂^2F∂ s_m∂ s_nΓ_mn. The second sum runs over ordered pairs. Thus a mixed derivative contributes twice when m≠ n. If one instead sums over m≤ n, the off-diagonal terms require a factor two. These are two equivalent conventions, but their factors must not be mixed.
Exact mathematical passage
\label{appop:chainrule} \mathcal LQ =\sum_{m=2}^5\frac{\partial F}{\partial s_m}\mathcal Ls_m +\sum_{m,n=2}^5 \frac{\partial^2F}{\partial s_m\partial s_n}\Gamma_{mn}.
\sum_{m\ge0}Use in source · 1 passage
Search aliases: ∑_m≥0 sum ge
Use in source · appendix_workbook.tex, line 12
Write F(q)=fracΔ(q)q=∏_k≥1(1-q^k)^24 =∑_m≥0f_mq^m. Then f_0=1 and τ(n)=f_n-1. This shift is essential. A program that stores τ(0)=0 as a sentinel must not confuse that entry with the constant coefficient f_0.
Exact mathematical passage
F(q)=\frac{\Delta(q)}q=\prod_{k\ge1}(1-q^k)^{24} =\sum_{m\ge0}f_mq^m.
\sum_{m\ge1}Use in source · 1 passage
Search aliases: ∑_m≥1 sum ge
Use in source · chapters01_03.tex, line 55
Exact coefficient engines. The product and a logarithmic recurrence. Let σ_j(n)=∑_dmid nd^j and write σ=σ_1. Logarithmic differentiation of the product gives DQΔ=(1-24∑_m≥1σ(m)q^m)Δ, DQ=qfracddq. Indeed, the contribution from the factor (1-q^r)^24 is -24fracrq^r1-q^r=-24∑_k≥1rq^rk. Collecting the terms with rk=m gives -24σ(m). The leading factor q contributes 1.
Exact mathematical passage
\label{eq:logdelta} \DQ\Delta=\left(1-24\sum_{m\ge1}\sigma(m)q^m\right)\Delta, \qquad \DQ=q\frac{d}{dq}.
\sum_rUse in source · 1 passage
Search aliases: ∑_r sum
Use in source · chapters17_18.tex, line 61
Fix z in the upper half-plane. The polynomial P_z(c,d)=∑_k∑_r f_k,r(z)c^r(cz+d)^k-r vanishes at every relatively prime integer pair (c,d), because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime c. Infinitely many integers d are coprime to it, so the polynomial in d vanishes identically. Each of its coefficient polynomials in c then vanishes at infinitely many primes, so is identically zero. Hen…
Exact mathematical passage
\label{eq:weightpolynomial} P_z(c,d)=\sum_k\sum_r f_{k,r}(z)c^r(cz+d)^{k-r}
\sum_{r=0}^{p_k}Use in source · 1 passage
Search aliases: ∑_r=0^p_k sum
Use in source · chapters17_18.tex, line 54
Separating the quasimodular weights. We need only an elementary weight-separation statement. Suppose finitely many quasimodular functions F_k of distinct nonnegative integer weights satisfy F_k(γ z)=∑_r=0^p_kf_k,r(z)c^r(cz+d)^k-r, f_k,0=F_k, p_k≤ k, for γ=[a · b
Exact mathematical passage
F_k(\gamma z)=\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r}, \qquad f_{k,0}=F_k,\qquad p_k\le k,
\sum_{r\ge0}Use in source · 3 passages
Search aliases: ∑_r≥0 sum ge
Use in source · chapters01_03.tex, line 138
For a fixed prime put T_r=τ(p^r) and b=p^11. The generating series is ∑_r≥0T_rX^r=frac11-aX+bX^2. Multiplication by the denominator gives T_0=1, T_1=a and the recurrence. Thus this is a formal identity over Z[[X]]. No analytic continuation in the prime index is involved.
Exact mathematical passage
\label{eq:heckegen} \sum_{r\ge0}T_rX^r=\frac1{1-aX+bX^2}.
\sum_{\substack{j\ge1\\T_j\le m}}Use in source · 1 passage
Search aliases: ∑_substackj≥1 T_j≤ m sum substack ge T le
Use in source · appendix_workbook.tex, line 52
Jacobi gives a third engine. Put T_j=j(j+1)/2 and a_j=(-1)^j(2j+1). The series A(q)=∏_k≥1(1-q^k)^3=1+∑_j≥1a_jq^T_j satisfies F=A^8. The identity AqF'=8FqA' yields m f_m=∑_substackj≥1 T_j≤ m a_j(9T_j-m)f_m-T_j. At m=3, only T_1=1, T_2=3 contribute. Its numerator is (-3)(9-3)(252)+5(27-3)(1)=-4536+120=-4416. Division by 3 again gives -1472. The direct product, divisor recurrence and triangular recurrence reach the same integer through different algebraic operations.
Exact mathematical passage
m f_m=\sum_{\substack{j\ge1\\T_j\le m}} a_j(9T_j-m)f_{m-T_j}. \label{eq:workbook-triangular-recurrence}
\sum_{\substack{v_i\equiv i\ (5)\\\sum v_i=0}}Use in source · 1 passage
Search aliases: ∑_substackv_i≡ i (5) ∑ v_i=0 sum substack equiv sum
Use in source · chapters17_18.tex, line 30
Radial factors become derivatives. For an alternating harmonic h_ℓ, write θ_h(τ)=∑_substackv_i≡ i (5) ∑ v_i=0 h(v)exp(2π iτfrac v ^210). Chapter~[source reference] establishes that it is a scalar cusp form of weight ℓ+2. The normalization includes only a fixed scalar if one divides by 288.
Exact mathematical passage
\theta_h(\tau)=\sum_{\substack{v_i\equiv i\ (5)\\\sum v_i=0}} h(v)\exp\left(2\pi i\tau\frac{\|v\|^2}{10}\right).
\sum_{\substack{v\in\mathcal V_p\\g(v)=1}}Use in source · 1 passage
Search aliases: ∑_substackv∈ V_p g(v)=1 sum substack in mathcal g
Use in source · chapters15_16.tex, line 207
At a prime shell there is no imprimitive part, so the universal target is still ∑_substackv∈ V_p g(v)=1W(v) =R(p)=τ(p)≠0. The content geometry has clarified the radial arithmetic. The prime total still needs a global or uniform arithmetic mechanism excluding cancellation to zero.
Exact mathematical passage
\sum_{\substack{v\in\mathcal V_p\\g(v)=1}}W(v) =R(p)=\tau(p)\ne0.
\sum_{\substack{v\in\Vn\\g(v)=1}}Use in source · 2 passages
Search aliases: ∑_substackv∈Vn g(v)=1 sum substack in Vn g
Use in source · chapters13_14.tex, line 23
The primitive residual and its inverse. Define R(n)=∑_substackv∈Vn g(v)=1W(v). Partitioning the packets by content and using [source reference] yields τ(n)=∑_g^2mid nchifive(g)g^10R(n/g^2). Terms with 5mid g have character zero and correspond to no actual content sector. They may be included in the formula without adding contributions.
Exact mathematical passage
\label{eq:residualdef} R(n)=\sum_{\substack{v\in\Vn\\g(v)=1}}W(v).
\sum_{v\in S_{k,r}}Use in source · 1 passage
Search aliases: ∑_v∈ S_k,r sum in
Use in source · chapters13_14.tex, line 113
Refinement and exact transfer spaces. Mass vectors and aggregation. Fix a shell, a finite rational family Q_1,…,Q_h of universally null corrections, and the occupied sectors at depth k. Let their number be s_k. Define t_k,r=∑_v∈ S_k,rW(v), (M_k)_rj=∑_v∈ S_k,rW(v)Q_j(v). The vector t_k contains sector masses. The columns of M_k are the effects of the permitted corrections. Coefficients a∈Q^h change the mass vector to t_k+M_ka.
Exact mathematical passage
t_{k,r}=\sum_{v\in S_{k,r}}W(v),\qquad (M_k)_{rj}=\sum_{v\in S_{k,r}}W(v)Q_j(v).
\sum_{v\in S_r}Use in source · 2 passages
Search aliases: ∑_v∈ S_r sum in
Use in source · chapters11_12.tex, line 116
Why strict localization and automatic valuation fail. Conserved sectors can preserve opposite signs. Partition a shell into sectors S_r, with masses t_r=∑_v∈ S_rW(v). Suppose every permitted correction is null inside every sector: ∑_v∈ S_rW(v)Q_j(v)=0. Then F=1+∑ a_jQ_j leaves each corrected sector mass equal to t_r. If two sector masses have opposite nonzero signs, all corrected packets cannot have one weak sign. Summing within those two sectors proves the obstruction immediately.
Exact mathematical passage
t_r=\sum_{v\in S_r}W(v)
\sum_{v\in\mathcal V_p}Use in source · 1 passage
Search aliases: ∑_v∈ V_p sum in mathcal
Use in source · chapter22.tex, line 19
On each such prime shell, all packets have content one. The residual is therefore the full signed sum τ(p)=∑_v∈ V_pW(v). The summands are nonzero integers, and for p>20 they are units at the index prime. Neither property forbids their sum from being zero. The unresolved statement concerns the exact balance of their signed masses.
Exact mathematical passage
\label{eq:primepacketobligation} \tau(p)=\sum_{v\in\mathcal V_p}W(v).
\sum_\lambdaUse in source · 1 passage
Search aliases: ∑_λ sum lambda
Use in source · chapters06_07.tex, line 6
The Nekrasov--Okounkov identity in Han's formulation is ∏_m≥1(1-q^m)^z-1 =∑_λ q^|λ|∏_h∈ H(λ)(1-frac zh^2). At z=25, a partition having a hook of length 5 contributes zero. The surviving terms are the 5-cores. Multiplication by q shows that the coefficient τ(n) is the sum of the hook products over 5-cores of size n-1.
Exact mathematical passage
\label{eq:hookidentity} \prod_{m\ge1}(1-q^m)^{z-1} =\sum_\lambda q^{|\lambda|}\prod_{h\in H(\lambda)}\left(1-\frac z{h^2}\right).
\setminusUse in source · 1 passage
Search aliases: ∖ setminus
Use in source · chapters11_12.tex, line 38
The geometric lower bound. Take opposite-sign packets v,w in one sector. Replace w by Jw if needed so that w-v=5Md, d∈ A_4∖{0}. Set a=5M/2 and m=(v+w)/2. Since M is even, both are integral. Equal norms give m· d=0,qquad10n= m ^2+a^2 d ^2. The squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be 2 here. Such a vector is ±(e_i-e_j), and m· d=0 would force v_i-v_j=±5M. This is divisible by 5, contrary to the distinct coordinate residues.
Exact mathematical passage
w-v=5Md,\qquad d\in A_4\setminus\{0\}.
\inftyUse in source · 4 passages
Search aliases: ∞ infty
Use in source · appendices.tex, line 17
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d,
Exact mathematical passage
v_\ell(0)=\infty
\capUse in source · 3 passages
Search aliases: ∩ cap
Use in source · chapters08_10.tex, line 35
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
A_4=H\cap\Z^5
\congUse in source · 1 passage
Search aliases: ≅ cong
Use in source · chapters08_10.tex, line 36
The regular discriminant cosets. Let A_4=H∩Z^5 and L=sqrt5 A_4. Its discriminant module is L^*/L≅ A_4^*/5A_4, |L^*/L|=5^5=3125. A representative of L^* can be written x=frac1sqrt5(a_i+t/5)_i=0^4, qquad0≤ t<5, ∑_i a_i=-t. If two a_i have the same residue modulo 5, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.
Exact mathematical passage
L^*/L\cong A_4^*/5A_4, \qquad |L^*/L|=5^5=3125.
\equivUse in source · 66 passages
Search aliases: ≡ equiv
Use in source · appendices.tex, line 18
The valuation v_ℓ(x) is the exponent of ℓ in a nonzero rational number x, with v_ℓ(0)=∞. The quadratic character used in the radial extraction is chifive(d)= 0 · 5mid d, 1 · d≡1,4pmod5, -1 · d≡2,3pmod5. The character value and the tenth power in the scaling formula are both essential.
Exact mathematical passage
\chifive(d)= \begin{cases} 0&5\mid d,\\ 1&d\equiv1,4\pmod5,\\ -1&d\equiv2,3\pmod5. \end{cases}
\leUse in source · 37 passages
Search aliases: ≤ le
Use in source · appendices.tex, line 37
Packets, content and the coefficient shift. Coordinates have indices 0,1,2,3,4. A shell is Vn={v∈Z^5: v_i≡ ipmod5, ∑_i v_i=0, ∑_i v_i^2=10n}. The standard orientation is D(v)=∏_0≤ i<j≤4(v_i-v_j), W(v)=D(v)/288. The identity is τ(n)=∑_v∈VnW(v). No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of W is proved separately from the core identity.
Exact mathematical passage
D(v)=\prod_{0\le i<j\le4}(v_i-v_j),\qquad W(v)=D(v)/288.
\subsetUse in source · 1 passage
Search aliases: ⊂ subset
Use in source · chapters19_21.tex, line 114
1 · 0, Trφ=0, detφ=p^11, φ^2=-p^11I. Choose any Q_p-line L⊂ D and put Fil^iD= D · i≤0,
Exact mathematical passage
L\subset D
\perpUse in source · 2 passages
Search aliases: ⊥ perp
Use in source · chapters11_12.tex, line 60
Project e_i-e_j orthogonally to d^⊥. Its squared norm is 2-fracδ^24. Because m⊥ d, Cauchy's inequality gives m ^2≥frac(a|δ|+1)^22-δ^2/4. For |δ|=1 this is 4(a+1)^2/7. For |δ|=2 it is (2a+1)^2, a larger quantity. Therefore 10n≥4a^2+frac47(a+1)^2, which simplifies to the lower bound inside the ceiling in [source reference]. Taking the ceiling proves that no earlier shell can be mixed. Since powers of 2 modulo 7 cycle as 2,4,1, direct division gives the displayed γ_k values.
Exact mathematical passage
d^\perp
\cdotsUse in source · 1 passage
Search aliases: ⋯ cdots
Use in source · chapters01_03.tex, line 4
A fixed modular form and an exact question. The coefficient that must not disappear. For |q|<1 the product defining Δ converges. As a formal power series it is even simpler: the coefficient of q^N depends only on the first N product factors. Thus each τ(n) is an unambiguously defined integer. The opening terms are Δ(q)= · q-24q^2+252q^3-1472q^4+4830q^5-6048q^6notag · -16744q^7+84480q^8-113643q^9-115920q^10+⋯. The signs already change repeatedly. Nonvanishing cannot be inferred from positivity of…
Exact mathematical passage
\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\notag\\ &-16744q^7+84480q^8-113643q^9-115920q^{10}+\cdots.
\longrightarrowUse in source · 3 passages
Search aliases: ⟶ longrightarrow
Use in source · chapters13_14.tex, line 135
The exact refinement sequence. There is a short exact sequence 0⟶ fracker AM_k+1(ker M_k) ⟶ E_k+1⟶ E_k⟶0. Consequently, if r_k=rank M_k, dim E_k+1-dim E_k =(s_k+1-s_k)-(r_k+1-r_k). The aggregation map sends H_k+1 onto H_k. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.
Exact mathematical passage
\label{eq:refinement} 0\longrightarrow \frac{\ker A}{M_{k+1}(\ker M_k)} \longrightarrow E_{k+1}\longrightarrow E_k\longrightarrow0.
\longmapstoUse in source · 1 passage
Search aliases: ⟼ longmapsto
Use in source · chapters13_14.tex, line 33
The square-supported arithmetic kernel in [source reference] has inverse d^2⟼μ(d)chifive(d)d^10. Indeed, at a square index m^2, the convolution of the two kernels is chifive(m)m^10∑_dmid mμ(d), which is 1 at m=1 and 0 otherwise. At nonsquare indices both kernels have no support. Consequently R(n)=∑_d^2mid nμ(d)chifive(d)d^10τ(n/d^2). Both formulas are exact for every positive integer. They are not fitted from the first few shells.
Exact mathematical passage
d^2\longmapsto\mu(d)\chifive(d)d^{10}.
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