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1 \hypertarget{v11-coupled_structure_full-constant-linear-coordinates-and-coupled-dilation}{%
2 \chapter[Constant linear coordinates]{Constant linear coordinates and coupled
3 dilation}\label{app:coupled-structure}\label{v11-coupled_structure_full-constant-linear-coordinates-and-coupled-dilation}}
4 
5 \hypertarget{v11-coupled_structure_full-exact-meaning-of-decoupling}{%
6 \section{Exact meaning of
7 decoupling}\label{v11-coupled_structure_full-exact-meaning-of-decoupling}}
8 
9 Let \(q,r,a,b>0\), with contraction \(<1\) imposed when discussing the
10 normalized entire solution. For an arbitrary analytic vector input
11 \(X=(u,v)\) with \(u(0)=v(0)=0\), define the functional vector field
12 
13 \[
14 \mathcal F[X](z)=
15 \begin{pmatrix}1+u(qz)v(rz)\\1+u(az)v(bz)\end{pmatrix}.
16 \tag{S1}
17 \]
18 
19 A nonzero constant row \(c=(c_1,c_2)\in\mathbb C^2\) is an
20 \textbf{autonomous scalar projection} if some scalar functional
21 \(\mathcal H\) satisfies
22 
23 \[
24 c\mathcal F[X]=\mathcal H[cX]
25 \quad\text{for every such analytic input }X.
26 \tag{S2}
27 \]
28 
29 The functional \(\mathcal H\) may use the whole scalar input. Thus this
30 definition allows any scalar dilation equation, not only a prescribed
31 quadratic ansatz. Equality (S2) means that two vector inputs with the
32 same scalar projection have the same projected vector field. It is an
33 identity of functional differential operators. It does not assert an
34 identity only along the distinguished normalized solution.
35 
36 A constant invertible change \(W=PX\) completely decouples the operator
37 if both rows of \(P\) are autonomous scalar projections. No assumption
38 about the form of their scalar functionals is needed.
39 
40 \hypertarget{v11-coupled_structure_full-classification-of-autonomous-linear-projections}{%
41 \section{Classification of autonomous linear
42 projections}\label{v11-coupled_structure_full-classification-of-autonomous-linear-projections}}
43 
44 \textbf{Theorem H.1.} For positive nonzero scales, a nonzero row \(c\)
45 satisfies (S2) if and only if
46 
47 \[
48 q=a,\qquad r=b,\qquad c=\gamma(1,-1),\qquad\gamma\ne0.
49 \tag{S3}
50 \]
51 
52 In that exceptional case the projected vector field is identically zero.
53 Consequently, no invertible constant linear change completely decouples
54 (S1), including when some or all scales coincide. If \((q,r)\ne(a,b)\),
55 there is not even one nonzero autonomous scalar projection. Four
56 pairwise distinct contraction scales are a special case of this stronger
57 result.
58 
59 \textbf{Proof.} Subtract the constant forcing. Define
60 
61 \[
62 Q_c[u,v](z)=c_1u(qz)v(rz)+c_2u(az)v(bz).
63 \]
64 
65 If (S2) holds, \(Q_c\) is constant on every fiber of
66 \((u,v)\mapsto c_1u+c_2v\). Fix \(z_0\ne0\). Polynomial interpolation
67 gives a scalar polynomial \(f\) with \(f(0)=0\) and \(f(sz_0)=1\) at
68 every distinct scale \(s\in\{q,r,a,b\}\). For inputs \(u=xf,v=yf\),
69 where \(x,y\in\mathbb C\),
70 
71 \[
72 Q_c[xf,yf](z_0)=(c_1+c_2)xy.
73 \tag{S4}
74 \]
75 
76 It must depend only on \(c_1x+c_2y\). If \(c_1+c_2\ne0\),
77 differentiating the polynomial \(xy\) along the fiber direction
78 \((c_2,-c_1)\) would give
79 
80 \[
81 c_2y-c_1x=0\quad\text{for every }x,y,
82 \]
83 
84 which forces \(c=0\), a contradiction. Therefore \(c_2=-c_1\), and
85 scaling the row reduces the question to \(c=(1,-1)\).
86 
87 For this row, put \(u=w+g,v=g\). This input and \((w,0)\) have the same
88 projection \(w\). The second input has \(Q_c[w,0]=0\), so autonomy
89 forces
90 
91 \[
92 w(qz)g(rz)-w(az)g(bz)
93 +g(qz)g(rz)-g(az)g(bz)=0
94 \tag{S5}
95 \]
96 
97 for all zero-constant analytic \(w,g\). Setting \(w=0\) and subtracting
98 that instance of (S5) leaves
99 
100 \[
101 w(qz)g(rz)=w(az)g(bz).
102 \tag{S6}
103 \]
104 
105 Taking successively \((w,g)=(z,z),(z^2,z),(z,z^2)\) yields \(qr=ab\),
106 \(q^2r=a^2b\), and \(qr^2=ab^2\). Nonzero scales then give \(q=a\) and
107 \(r=b\). Conversely these equalities make the two components of (S1)
108 identical on every input, so \((1,-1)\mathcal F[X]=0\). The only
109 possible autonomous rows therefore span a one-dimensional space, which
110 cannot be the two rows of an invertible matrix. \(\square\)
111 
112 The same proof works on odd analytic inputs for positive scales: choose
113 \(f(z)=z p(z^2)\) by interpolation at the distinct positive scale
114 squares, and use \((z,z),(z^3,z),(z,z^3)\) in (S6). Positivity then
115 resolves the equalities of squares. The theorem does not depend on
116 allowing even inputs.
117 
118 \hypertarget{v11-coupled_structure_full-tensor-explanation-for-four-distinct-scales}{%
119 \section{Tensor explanation for four distinct
120 scales}\label{v11-coupled_structure_full-tensor-explanation-for-four-distinct-scales}}
121 
122 The preceding proof also makes a coordinate invariant obstruction
123 explicit. Under an invertible state change, write the original
124 coordinates as \(u=\ell_1(W),v=\ell_2(W)\), where \(\ell_1,\ell_2\) are
125 independent linear forms. At the pair of distinct nodes \((qz,rz)\), the
126 first quadratic term has bilinear coefficient
127 
128 \[
129 B=\ell_1\otimes\ell_2.
130 \]
131 
132 It has rank one, but is not symmetric:
133 \(B-B^{\mathsf T}=\ell_1\wedge\ell_2\ne0\). At four distinct nonzero
134 nodes, polynomial interpolation lets their vector values vary
135 independently, so terms supported at \((az,bz)\) cannot cancel this
136 coefficient. A scalar autonomous channel at the same node pair would
137 require \(B=\kappa\ell\otimes\ell\), a symmetric rank-one tensor, after
138 projecting the output onto that channel. For any output row with a
139 nonzero coefficient on this term, the alternating part remains nonzero.
140 The other node pair gives the same obstruction if the first output
141 coefficient vanishes.
142 
143 This is not a rank-only argument: the tensor \(\ell_1\otimes\ell_2\)
144 already has rank one. Its nonzero alternating part is the decisive
145 invariant at distinct nodes. When the nodes of a product coincide, the
146 symmetric quadratic form \(uv\) has matrix
147 
148 \[
149 \begin{pmatrix}0&1/2\\1/2&0\end{pmatrix},
150 \]
151 
152 of rank two, whereas a square of one linear coordinate has rank one.
153 Rank is then the appropriate obstruction. The fiber proof in Theorem H.1
154 handles node coincidences and cancellations without mixing these two
155 cases.
156 
157 \hypertarget{v11-coupled_structure_full-particular-solutions-and-degenerate-parameters}{%
158 \section{Particular solutions and degenerate
159 parameters}\label{v11-coupled_structure_full-particular-solutions-and-degenerate-parameters}}
160 
161 Theorem H.1 is an operator statement. It does not exclude coincident
162 components for particular initial data, algebraic relations along a
163 solution, elimination into a scalar equation of higher order, nonlinear
164 coordinate changes, or identities with other named special functions.
165 
166 For contraction scales, the normalized analytic germ is unique. If
167 \((a,b)=(q,r)\), its components coincide because their derivatives are
168 identical and their initial values agree. The common component is the
169 scalar family \(T'=1+T(qz)T(rz),T(0)=0\). The same coincidence holds if
170 \((a,b)=(r,q)\): \((T,T)\) solves both equations because multiplication
171 is commutative, and uniqueness identifies the normalized solution. For
172 \(q\ne r\), this reversed-pair case has no autonomous scalar operator
173 projection by Theorem H.1, despite having \(U=V\) along its normalized
174 solution. This is a concrete reason the two notions must remain
175 distinct.
176 
177 In fact, equality of normalized components occurs exactly when the
178 unordered scale pairs agree:
179 
180 \[
181 U=V\quad\Longleftrightarrow\quad\{q,r\}=\{a,b\}.
182 \tag{S7}
183 \]
184 
185 To prove necessity, write \(U=V=T=z+c_1z^3+\cdots\). Then
186 \(3c_1=qr=ab>0\). Comparing the \(z^4\) coefficients of
187 \(T(qz)T(rz)=T(az)T(bz)\) gives \(c_1qr(q^2+r^2)=c_1ab(a^2+b^2)\), so
188 both products and sums of squares agree. Positive scales then have equal
189 sums and are the roots of the same quadratic. Sufficiency is the
190 uniqueness argument above. For example, \((q,r,a,b)=(1/2,1/3,1/3,1/2)\)
191 has coincident normalized components and no nonzero autonomous operator
192 projection.
193 
194 At \(q=r=a=b=s>0\), the variables \(w=u+v,d=u-v\) give
195 
196 \[
197 w'=2+\tfrac12\bigl(w(sz)^2-d(sz)^2\bigr),\qquad d'=0.
198 \]
199 
200 This triangular system is not a completely decoupled operator. The
201 normalized solution lies in the invariant subspace \(d=0\).
202 
203 Zero scales lie outside Theorem H.1. On zero-constant inputs, a product
204 vanishes identically if either of its scales is zero. If both products
205 vanish, both equations are already \(u'=v'=1\). If only the first
206 product vanishes, the projection \(u\) is autonomous, regardless of
207 whether the second pair agrees with the first. Thus extending Theorem H.1
208 to zero scales would be false. Scale one and positive scales greater
209 than one do not affect the algebraic operator theorem, although the
210 contraction-domain entire growth theorems do not automatically extend to
211 them. Negative or complex scales and odd-input coincidences caused by
212 signs are not classified here.
213 
214 \hypertarget{v11-coupled_structure_full-claim-status}{%
215 \section{Claim status}\label{v11-coupled_structure_full-claim-status}}
216 
217 Theorem H.1 and (S7) are proved above. The proof is elementary and
218 independent of the coupled growth theorem. Historical priority for these
219 precise operator statements is not asserted. The theorem provides a
220 specified nondecoupling result, not a universal assertion that the
221 normalized functions cannot reduce to any previously studied function
222 class.

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