M2 Protected Block Index Comparison

Write $X=\mathrm{dom}(A)$ with its graph norm and $Y=H_P$. In case (i), the compact boundaryless manifold hypothesis and [F2], under full AC [F1], apply to the explicitly specified $A\in {\mathrm{\Psi }}^1(C;E,E)$ and make $A:H^1(C,E)\to L^2(C,E)$ Fredholm. Its Sobolev norm and graph norm are equivalent: the graph norm is bounded by the Sobolev norm by [F2], and the identity between these two Banach domains has bounded inverse by [F6]. In case (ii), density of $\mathrm{dom}(A)$ and finite dimensionality give $\mathrm{dom}(A)=H_P$; hence $A$ is a map between finite-dimensional spaces and is Fredholm.

In case (i), the finite-chart Sobolev identification and partition [F7] reduce the inclusion $H^1(C,E)\hookrightarrow L^2(C,E)$ to finitely many bounded smooth extension domains. Rellich compactness [F3] on each chart, combined over the finite cover, makes $X\hookrightarrow Y$ compact. In case (ii), this inclusion has finite-dimensional target and is compact. The use of the weaker Countable/Dependent Choice hypotheses in [F3, F7] is supplied by [F8].

Put $K:=B{D}^{-1}{B}^{\ast }$. Since $D=D^{\ast }$ has a bounded everywhere-defined inverse, $(D^{-1}{)}^{\ast }=D^{-1}$. Thus $K^{\ast }=K$. Each of $B,D^{-1},B^{\ast }$ reverses grading, so their product $K$ reverses grading. A bounded self-adjoint perturbation of $A$ is self-adjoint on $\mathrm{dom}(A)$: the adjoint-domain identity $(A-K{)}^{\ast }=A^{\ast }-K^{\ast }$ follows by moving the bounded term across the Hilbert pairing. Therefore $S=A-K$ is odd and self-adjoint on the stated domain.

The operator $K:=B{D}^{-1}{B}^{\ast }:Y\to Y$ is bounded by the hypotheses. Its restriction $X\to Y$ is compact as the composition of the compact inclusion $X\hookrightarrow Y$ from step 1.2 with $K$.

For every $t\in [0,1]$, $A-tK:X\to Y$ is a compact perturbation of the Fredholm operator $A$. Atkinson's theorem [F4] implies that each $A-tK$ is Fredholm. Its odd block decomposition is a direct sum of the two chiral maps, so each chiral map is Fredholm as well.

The index is locally constant along this norm-continuous path. To verify this, fix a Fredholm map $L:X\to Y$, put $N=\ker L$, $R=\mathrm{ran}L$, and choose the orthogonal splittings $X=N\oplus X_1$ and $Y=R\oplus R^{\perp }$ by [F5]. The restriction $L:X_1\to R$ is a bounded bijection; [F6] gives a bounded inverse. For every sufficiently small perturbation $E:X\to Y$, the $X_1\to R$ block $G$ of $L+E$ remains invertible by the Neumann series. In the decompositions above, write $L+E=\left(\begin{array}{cc}{E}_{C,N}& E_{C,X_1}\\ E_{R,N}& G\end{array}\right),C:=R^{\perp }\mathrm{.}$ Bounded invertible row and column operations reduce this block matrix to $\mathrm{diag}(F,G)$, where $F=E_{C,N}-E_{C,X_1}{G}^{-1}{E}_{R,N}:N\to C$. The finite-dimensional map $F$ has index $\dim N-\dim C$, independent of its rank, and $G$ has index zero. Thus $\mathrm{ind}(L+E)=\dim N-\dim C=\mathrm{ind}L$ near $L$. Applying this argument on the Hilbert spaces $X^{+}$ and $Y^{-}$ to $L=(A-tK{)}_{+}$, connectedness of $[0,1]$ gives $\mathrm{ind}{S}_{+}=\mathrm{ind}{A}_{+}$.

On $X_Q:=\mathrm{dom}(A)\oplus \mathrm{dom}(D)$, let $Q_0=A\oplus D$ and $C_B=\left(\begin{array}{cc}0& B\\ B^{\ast }& 0\end{array}\right)$, so $Q=Q_0+C_B$ with $C_B$ bounded on $H_P\oplus H_H$. For every $x\in X_Q$,

Consequently the graph norms, for $x=(x_P,x_H)$, $\mathrm{\parallel }x{\mathrm{\parallel }}_Q=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Qx{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$ and $\mathrm{\parallel }x{\mathrm{\parallel }}_{Q_0}=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }A{x}_P{\mathrm{\parallel }}^2+\mathrm{\parallel }D{x}_H{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$ are equivalent. Thus the identity identifies the graph domain used to define the Fredholm index of $Q$ with the direct-sum graph domain. Define the even triangular map $T=\left(\begin{array}{cc}I& 0\\ D^{-1}{B}^{\ast }& I\end{array}\right)\mathrm{.}$ The map $D^{-1}{B}^{\ast }:H_P\to \mathrm{dom}(D)$ is bounded for the graph norm because $\mathrm{\parallel }{D}^{-1}{B}^{\ast }x\mathrm{\parallel }+\mathrm{\parallel }D(D^{-1}{B}^{\ast }x)\mathrm{\parallel }\le (\mathrm{\parallel }{D}^{-1}\mathrm{\parallel }+1)\mathrm{\parallel }{B}^{\ast }\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Thus $T$ and $T^{-1}=\left(\begin{array}{cc}I& 0\\ -D^{-1}{B}^{\ast }& I\end{array}\right)$ are bounded inverses on $X_Q$ with its direct-sum graph norm, and hence on the equivalent $Q$ graph norm. Direct block multiplication gives $Q=T^{\ast }\left(\begin{array}{cc}S& 0\\ 0& D\end{array}\right)T\mathrm{.}$ Since $S=A-K$ with bounded $K$, the graph norm of $S$ is equivalent to that of $A$. The outer maps are therefore bounded even isomorphisms on the graph domain and Hilbert target. Right composition by $T$ identifies kernels, and left composition by $T^{\ast }$ identifies ranges and cokernels; both preserve parity. The block-diagonal index is the sum of the two block indices. Since $D$ is invertible, its chiral index is zero, and $\mathrm{ind}{Q}_{+}=\mathrm{ind}{S}_{+}$.

If $u\in \ker S$, direct substitution gives $Q(u,-D^{-1}{B}^{\ast }u)=(Su,0)=0$. Conversely, if $Q(u,v)=0$, its second block equation and invertibility of $D$ force $v=-D^{-1}{B}^{\ast }u$, and its first equation then gives $Su=0$. The two assignments are inverse. The product $D^{-1}{B}^{\ast }$ is even, so the isomorphism preserves parity. For an odd self-adjoint Fredholm operator, the even-to-odd index is $\dim \ker Q{\mathrm{\mid }}_{H^{+}}-\dim \ker Q{\mathrm{\mid }}_{H^{-}}$, because the cokernel of its even-to-odd block is the kernel of its odd-to-even adjoint block. The parity-preserving kernel isomorphism therefore gives the claimed kernel comparison as well.