Graph views, holonomy, and reconstruction

Proof

Write \(U_0=I\) and \(U_k=\tau_{i_0i_1}\cdots\tau_{i_{k-1}i_k}\) for \(1\le k\le m\). The transport convention makes every \(U_k\) a map from the fiber at \(i_k\) to that at \(i_0\). Since \(U_{k+1}=U_k\tau_{i_ki_{k+1}}\), the \(k\)-th transported defect is

Summing over \(k=0,\ldots,m-1\) cancels each interior value once with each sign. The remaining terms are \(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\), proving the identity with its actual ordered products. For an empty walk both sides are zero.

The triangle inequality and the operator norm bound now give

The bracket is \(\gamma(q)\), with the empty product equal to one. Repeated vertices or edges do not affect the cancellation. Reordering the edge maps would, in general, invalidate it.

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Proof

Every \(T_i\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \((i,j)\), apply \(T_i\) to its defect:

Because \(T_i\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \(z\). Thus \(dR_pz=0\) holds exactly for \(z\in K\). The root route is empty, so \((R_pz)_o=z\).

Conversely, let \(x\in\ker d\). The path telescope along \(p_i\) has all zero summands, hence \(T_ix_i=x_o\) for every vertex. Applying \(T_i^{-1}\) gives \(x_i=T_i^{-1}x_o\), so \(x=R_p(x_o)\). The first part then places \(x_o\) in \(K\). We have proved both inverse equations and uniqueness of an exact extension of a root value.

The maps are linear. On the finite vertex product,

These bounds also hold on the subspaces \(K\) and \(\ker d\), proving continuity of both directions. No completeness is needed for these finite bounds.

For route independence, the proved equivalence identifies \(K\) with the image of root evaluation from the intrinsically defined space \(\ker d\). That image does not refer to any chosen routes, so every route system yields the same \(K\). For \(z\in K\), reconstruction using either system is an exact family with root \(z\), and uniqueness makes the families equal. For \(z\notin K\), this uniqueness argument does not assert equality of the two generally inconsistent reconstructed families.

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Proof

Boundedness of \(D\) makes \(K=\ker D\) closed. Lemma E.1 constructs \(q=P_K\) and gives the orthogonal decomposition \(x=qx+(x-qx)\). The first term belongs to \(K\), so \(Dqx=0\). Projection fixes every point of \(K\), giving \(q(qx)=qx\).

Pythagoras applied to that decomposition gives \({\left\lVert x\right\rVert}^2={\left\lVert qx\right\rVert}^2+{\left\lVert x-qx\right\rVert}^2\), hence \({\left\lVert qx\right\rVert}\le{\left\lVert x\right\rVert}\). For any \(y\in K\), the vector \(qx-y\) lies in \(K\), whereas \(x-qx\) lies in its orthogonal complement. Consequently

Equality holds only when \({\left\lVert qx-y\right\rVert}=0\), or \(y=qx\). This proves nearest-point uniqueness without any closedness assumption on \({\operatorname{ran}} D\). Closedness of the kernel and of the range serve different purposes.

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Proof

Let \(K=\ker D\) and \(K^\perp\) carry the induced Hilbert norm. Define \(D_\perp u=Du\), but regard its target as \({\operatorname{ran}} D\) with the norm inherited from \(R\). It is injective: \(Du=0\) puts \(u\) in both \(K\) and \(K^\perp\), so \(u=0\). It is surjective onto that target: if \(r=Dx\), then \(u=x-qx\in K^\perp\) and \(Du=Dx=r\). Thus it is a bounded linear bijection before any continuity of its inverse is claimed.

Suppose first that the range is closed. Because \(R\) is Banach, \({\operatorname{ran}} D\) is then Banach. The domain \(K^\perp\) is also complete. The bounded inverse theorem, proved in Appendix E, gives a bounded inverse \(A=D_\perp^{-1}\). Applying it to \(Dx=D(x-qx)\) yields

Hence condition 2 holds. If any \(C\ge0\) satisfies condition 2, then for every \(r\in{\operatorname{ran}} D\) apply it to \(u=Ar\in K^\perp\). Since \(qu=0\), it gives \({\left\lVert Ar\right\rVert}\le C{\left\lVert r\right\rVert}\). Taking the operator norm proves \({\left\lVert A\right\rVert}\le C\). This establishes the least constant, not just a sufficient one. If \(D=0\), both \(K^\perp\) and the range are zero spaces and \({\left\lVert A\right\rVert}=0\).

Now assume condition 2. To prove closedness, take an arbitrary sequence \(r_n\in{\operatorname{ran}} D\) with \(r_n\to r\) in \(R\). Choose \(x_n\) with \(Dx_n=r_n\) and put \(u_n=x_n-qx_n\). Then \(u_n\in K^\perp\) and \(Du_n=r_n\). Applying the estimate to \(u_n-u_m\) gives

Thus \((u_n)\) is Cauchy in the complete space \(K^\perp\). Let its limit be \(u\). Continuity gives \(Du=\lim_n r_n=r\), so \(r\in{\operatorname{ran}} D\). In a normed space, a set containing the limits of all its convergent sequences is closed. This proves condition 1 without presupposing that the initially unbounded inverse is continuous.

Condition 2 implies condition 3 by choosing \(y=qx\). Conversely, if condition 3 supplies an exact \(y\) for a given \(x\), the nearest-point property of \(qx\) gives

The same \(C\) works for all \(x\), proving condition 2. This last implication does not require the selected answers \(y\) to depend linearly or continuously on \(x\).

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Proof

Feasibility gives some \(h_0=x-y_0\in M\) with \(Bh_0=Dx\). Conversely, every \(h\in M\) satisfying \(Bh=Dx\) produces an exact protected answer \(y=x-h\). Thus the answer problem and the correction problem are equivalent, with correction set

This is a nonempty closed affine subset of the Hilbert space \(M\). Closedness of \(\ker B\) follows from continuity of \(B\), independently of whether its image is closed.

Let \(k_0=P^M_{\ker B}h_0\) and \(h_*=h_0-k_0\). Then \(Bh_*=Bh_0=Dx\) and \(h_*\perp\ker B\) in \(M\). If a different initial correction is chosen, it differs from \(h_0\) by an element of \(\ker B\). Its orthogonal-complement component is therefore again \(h_*\). This proves that the construction is independent of the feasibility witness.

Every other correction can be written uniquely as \(h_*+k\) with \(k\in\ker B\). Orthogonality gives

Writing \(y_*=x-h_*\) and \(y=x-h_*-k\), we have \(x-y=h_*+k\) and \(y_*-y=k\). Substitution proves the stated Pythagorean identity. Its last term vanishes only when \(y=y_*\), establishing the unique closest answer.

If \({\operatorname{ran}} B\) is closed, Theorem 18.2 applied on \(M\) gives a bounded inverse of \(B\) on \((\ker B)^{\perp_M}\). The vector \(h_*\) is exactly that inverse applied to \(Dx\), so its norm obeys the required bound.

For the converse, suppose the bound holds for every feasible raw candidate. Every \(x\in M\) is feasible, because \(y=0\) has \(Dy=0\) and \(py=px=0\). Its closest protected exact answer is \(P^M_{\ker B}x\). Thus the assumed estimate reduces on \(M\) to

Theorem 18.2 forces \({\operatorname{ran}} B\) to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.

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Proof

First quantify the residual seen at the root. The identity \(\mathcal H=N_pdR_p\), applied to \(x_o\), gives

Closed range of \(\mathcal H\) lets Theorem 18.2 control the anchor correction by \({\left\lVert x_o-qx_o\right\rVert}\le\kappa{\left\lVert\mathcal H(x_o)\right\rVert}_\infty\). Insert the uncorrected reconstruction between \(x\) and \(Px=R_pqx_o\):

Substituting the two preceding bounds yields (19.1). The finite maximum-norm residual product is a Banach space because its fiber is Hilbert, so the target completeness needed in Theorem 18.2 is available.

For necessity, suppose a constant \(C\) satisfies \({\operatorname{dist}}(w,\ker d)\le C{\left\lVert dw\right\rVert}_\infty\) for every vertex family \(w\). Fix \(z\in K^\perp\) and take \(w=R_pz\). Every exact family \(e\) has \(e_o\in K\), and root evaluation has norm at most one. Therefore

Taking the infimum over \(e\in\ker d\) is legitimate without assuming that this global nearest point is attained. It gives

Here \(N_p\) is a continuous linear equivalence on the finite edge product because every component \(T_i\) is one. For a general root vector, remove its projection onto \(K\); this leaves the same \(\mathcal H\)-image and puts it in \(K^\perp\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem 18.2, which forces \({\operatorname{ran}}\mathcal H\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.

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Proof

The subspace \(L=\ker D\) is closed, hence Hilbert with the inherited inner product. The restriction \(B=C|_L\) is continuous, so its kernel is a closed subspace of \(L\). Under the inclusion into \(E\), this kernel is exactly \(M=L\cap\ker C\). Thus both projections used to define \(P\) exist.

For an input \(x\), write \(\ell=P_Lx\) and let \(Px\) be the projection of \(\ell\) onto \(\ker B\) inside \(L\). Then \(Px\in M\), \(u=x-\ell\in L^\perp\), and \(v=\ell-Px\in L\) is orthogonal to \(M\). Since \(M\subseteq L\), the vector \(u\) is also orthogonal to \(M\). Their sum \(x-Px=u+v\) is therefore orthogonal to \(M\). A vector in \(M\) with that error is the unique orthogonal projection of \(x\), so \(P=P_M\).

The two errors themselves are orthogonal because \(u\in L^\perp\) and \(v\in L\). Expanding the squared norm of their sum gives \({\left\lVert x-Px\right\rVert}^2={\left\lVert u\right\rVert}^2+{\left\lVert v\right\rVert}^2\). Finally \(\ker(D,C)=\ker D\cap\ker C=M\), so this is also the projection onto the kernel of the joint defect. The proof uses a second projection inside \(L\). Replacing it by the ambient projection onto \(\ker C\) would invalidate the membership step and need not preserve local exactness.

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Proof

Suppose the range of the joint map is closed. To test closedness of \({\operatorname{ran}} B\), take a sequence \(Bk_n\to y\) with \(k_n\in\ker D\). The joint images are

Closedness supplies \(x\in E\) with \((Dx,Cx)=(0,y)\). The first coordinate gives \(x\in\ker D\), and the second gives \(Bx=y\). Hence the matching range is closed. This direction did not require closedness of \({\operatorname{ran}} D\).

For the reverse direction, assume both \({\operatorname{ran}} D\) and \({\operatorname{ran}} B\) are closed, and take joint images converging to \((r,y)\). Thus \(Dx_n\to r\) and \(Cx_n\to y\). Decompose each input orthogonally as

Because \(r\in{\operatorname{ran}} D\) and the transverse inverse \(A\) of \(D\) is bounded by Theorem 18.2, the first components determine

Continuity of \(C\) now gives

Closedness of \({\operatorname{ran}} B\) supplies \(k\in\ker D\) with \(Bk=y-Cu\). There is no need for the sequence \(k_n\) itself to converge. The single vector \(u+k\) satisfies

Thus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem 18.2 applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.

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Proof

For \(z\in\ker D'\), the old residual bound gives

Interchanging the roles gives

We need a dual form of (*). For \(v\in(\ker D')^\perp\) and any \(y\in E\), self-adjointness of \(q\) and \(v=(I-q')v\) give

The Hilbert norm equals the supremum of these pairings on the unit ball, so \({\left\lVert qv\right\rVert}\le C'\varepsilon{\left\lVert v\right\rVert}\). This proves the precise restriction estimate needed on \((I-q')x\), not just a coarser bound in terms of \({\left\lVert x\right\rVert}\).

Now write

The two displayed terms lie in \(\ker D\) and \((\ker D)^\perp\), respectively, so they are orthogonal. The input components \((I-q')x\) and \(q'x\) are also orthogonal. Applying the dual estimate to the first and the original cross estimate to the second gives

The bound \({\left\lVert D-D'\right\rVert}\le\varepsilon\) implies \(\varepsilon\ge0\), so taking square roots yields the operator estimate. If the common gain is \(M\), substitute \(C=C'=M\) and \(\varepsilon={\left\lVert D-D'\right\rVert}\). This is exactly the Lipschitz inequality on the class carrying that gain. The use of two orthogonal decompositions explains the maximum coefficient instead of a sum.

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Proof

The map \(T=(D-D')A\) acts on the Banach target \(R\) and satisfies \({\left\lVert T\right\rVert}\le a\varepsilon<1\). The operator series

converges in operator norm and obeys \({\left\lVert Z\right\rVert}\le(1-a\varepsilon)^{-1}\). Multiplying its finite partial sums on either side by \(I-T\) gives \(I-T^{N+1}\). Passing to the operator-norm limit proves \((I-T)Z=Z(I-T)=I\). Appendix E supplies the full series argument, including completeness of the operator space.

Since \(DA=I\), one has \(D'A=DA-(D-D')A=I-T\). Therefore \(A'=AZ\) satisfies

The first equation explicitly supplies a preimage of every target vector, proving surjectivity.

For any \(x\), put \(y=x-A'D'x\). Then \(D'y=D'x-D'A'D'x=0\), so \(y\) is a valid exact competitor for orthogonal projection. Its error gives

The same argument using \(A\) gives \({\left\lVert x-qx\right\rVert}\le a{\left\lVert Dx\right\rVert}\). Both coefficients are nonnegative and \(a/(1-a\varepsilon)\ge a\). Substitution into Theorem 21.1 yields

No assertion that \(I-A'D'\) is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.

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Proof

Let \(R_0=I-AD\) and \(R_1=I-A'D'\). From \(DA=D'A'=I\) we obtain \(DR_0=0\) and \(D'R_1=0\). Each \(R_i\) fixes its corresponding kernel. Thus \(R_1\) restricts to a map from \(\ker D\) to \(\ker D'\), while \(R_0\) restricts in the other direction.

The inverse identities require two additional cancellations. First,

Second, using \(D'A=I-T\) and \(Z(I-T)=I\),

For \(x\in\ker D\), these identities give

For \(y\in\ker D'\), they give

Both restricted maps are bounded and linear as restrictions of bounded linear operators. They are therefore inverse continuous linear maps, establishing the asserted equivalence.

Finally, if \(Dx=0\), then \(D'x=(D'-D)x\). Hence

This derivation uses the selected right inverse \(A\) throughout. It proves a controlled transport of exact vectors, not a claim that the selected map is the nearest-point projection onto the new kernel.

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Proof

The vector \(v_t=(-t,1)\) spans \(\ker D_t\) and has squared norm \(1+t^2>0\). Projection onto this line is

which gives the matrix in (21.2). In particular it is the orthogonal projector, not merely a map whose image is contained in the kernel.

Subtract \(q_0=\operatorname{diag}(0,1)\), obtaining the symmetric matrix

The two off-diagonal entries of its square are zero, since \(-t^3+t^3=0\). Both diagonal entries are \((t^4+t^2)/(1+t^2)^2=t^2/(1+t^2)\). Thus

Pairing this identity with an arbitrary vector proves the all-vector norm formula. It bounds the operator norm above by \(|t|/\sqrt{1+t^2}\), and evaluating any unit vector gives equality. The argument includes \(t=0\), where both sides vanish.

For the common section \(A(r)=(r,0)\), the perturbation inverse leaves \(A\) unchanged because \(D_tA=I\). The constructed transport is therefore \(Jx=x-AD_tx\) on the old kernel. At \(x=(0,b)\), it gives \((-tb,b)\). Substitution into the projection formula gives instead \((-tb/(1+t^2),b/(1+t^2))\). The strict-radius theorem applies when \(|t|<1\), but these explicit line and projection calculations are valid for every real \(t\).

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