F4 Closed Branch Vertex Supercharge Index

The domain is dense since $\mathrm{Dom}(D)$ is dense. Suppose $(x_n,e_n)\to (x,e)$ in $H_{+}\oplus E$ and $L(x_n,e_n)\to (y,z)$ in $H_{-}\oplus V$. Boundedness of $S$ gives $D{x}_n={\mathrm{pr}}_{H_{-}}L(x_n,e_n)-S{e}_n\to y-Se$. Closedness of $D$ then gives $x\in \mathrm{Dom}(D)$ and $Dx=y-Se$. Thus $x_n\to x$ in the graph norm. Boundedness of $T$ and $M$ gives $T{x}_n+M{e}_n\to Tx+Me=z$. So $L(x,e)=(y,z)$ and $L$ is closed. On the graph Hilbert space $X\oplus E$, $L$ is bounded. Its graph norm is equivalent to the product graph norm: boundedness gives $\mathrm{\parallel }(x,e){\mathrm{\parallel }}_L\le C_0(\mathrm{\parallel }x{\mathrm{\parallel }}_D^2+\mathrm{\parallel }e{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$ for some $C_0$, while $\mathrm{\parallel }e\mathrm{\parallel }\le \mathrm{\parallel }(x,e){\mathrm{\parallel }}_L$ and $\mathrm{\parallel }Dx\mathrm{\parallel }\le \mathrm{\parallel }{\mathrm{pr}}_{H_{-}}L(x,e)\mathrm{\parallel }+\mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }e\mathrm{\parallel }\le (1+\mathrm{\parallel }S\mathrm{\parallel })\mathrm{\parallel }(x,e){\mathrm{\parallel }}_L$ give the reverse bound. Thus the block argument below computes Fredholmness and index on the graph-norm domain in the Statement.

In $X$, let $K=\ker D$, $R=\mathrm{ran}D$, and $X_1=K^{\perp }$. By the assumed Fredholmness of $D$ and [F5], $K$ is finite and $R$ is closed of finite codimension. By [F3], $X=K\oplus X_1$ and $H_{-}=R\oplus C_0$, where $C_0=R^{\perp }$ is finite. The map $D{\mathrm{\mid }}_{X_1}:X_1\to R$ is a bounded bijection: its kernel is zero, and if $Dx=y\in R$, decomposing $x=k+x_1$ gives $D{x}_1=y$. Both spaces are Banach, so [F2] gives a bounded inverse. Let $P_R$ be the orthogonal projection and $W=\mathrm{span}(P_{R}S(E))$. A basis of the finite dimensional $E$ gives a finite spanning list for $W$; deleting dependent vectors leaves a finite ordered basis (the empty basis if $W=0$). Thus [F6] makes $W$ closed. By [F3], $R=W\oplus R_1$, with $R_1=R\cap W^{\perp }$. Applying the bounded isomorphism $D{\mathrm{\mid }}_{X_1}$ gives $X_1=D^{-1}W\oplus D^{-1}{R}_1$; the first summand is finite-dimensional and both are closed because the isomorphism and its inverse are bounded.

Decompose the source as $X_f\oplus D^{-1}{R}_1$, where $X_f=K\oplus D^{-1}W\oplus E$, and the target as $Y_f\oplus R_1$, where $Y_f=W\oplus C_0\oplus V$. In these coordinates $L=\left(\begin{array}{cc}A& B\\ 0& J\end{array}\right)$, where $J=D{\mathrm{\mid }}_{D^{-1}{R}_1}$ is a bounded isomorphism. The lower-left block vanishes because $D(K)=0$, $D(D^{-1}W)=W$, and $P_{R}S(E)\subset W$. The upper-right block is the bounded map $T{\mathrm{\mid }}_{D^{-1}{R}_1}$. The bounded invertible target change $(y_f,y_1)\mapsto (y_f-B{J}^{-1}{y}_1,y_1)$ conjugates this operator to $A\oplus J$. The finite-dimensional map $A:X_f\to Y_f$ has closed range and index $\dim X_f-\dim Y_f$ by rank-nullity, while $J$ has index zero. Therefore $L$ is Fredholm, and $\mathrm{ind}L=(\dim K-\dim C_0)+\dim E-\dim V=\mathrm{ind}D+\dim E-\dim V$; the first equality uses $H_{-}\mathrm{/}R\cong C_0$.

Write $\mathrm{\Gamma }(L)=\{(x,Lx):x\in \mathrm{Dom}(L)\}\subset H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}\oplus H_{\mathrm{o}\mathrm{d}\mathrm{d}}$ and define the unitary $U:H_{\mathrm{o}\mathrm{d}\mathrm{d}}\oplus H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}\to H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}\oplus H_{\mathrm{o}\mathrm{d}\mathrm{d}}$ by $U(y,z)=(-z,y)$. By the cross-space adjoint definition in the Statement, $U\mathrm{\Gamma }(L^{\ast })=\mathrm{\Gamma }(L{)}^{\perp }$: indeed, $(y,L^{\ast }y)$ is sent to $(-L^{\ast }y,y)$, which is perpendicular to every $(x,Lx)$ by the adjoint identity; conversely, if $(-z,y)\in \mathrm{\Gamma }(L{)}^{\perp }$, that orthogonality is exactly $⟨Lx,y⟩=⟨x,z⟩$ for every $x\in \mathrm{Dom}(L)$, so $y\in \mathrm{Dom}(L^{\ast })$ and $z=L^{\ast }y$. If $w\in H_{\mathrm{o}\mathrm{d}\mathrm{d}}$ is perpendicular to $\mathrm{Dom}(L^{\ast })$, then $⟨(0,w),(-L^{\ast }y,y)⟩=0$ for every $y\in \mathrm{Dom}(L^{\ast })$; the identity just proved gives $(0,w)\in \mathrm{\Gamma }(L{)}^{\perp \perp }=\mathrm{\Gamma }(L)$, since $L$ is closed and [F3] applies. A graph contains only one output over each input, so $(0,w)=(0,L0)=(0,0)$ and $w=0$. Thus $\mathrm{Dom}(L^{\ast })$ is dense in $H_{\mathrm{o}\mathrm{d}\mathrm{d}}$. Apply the same graph identity to $L^{\ast }$, using $U^{\mathrm{\prime }}:H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}\oplus H_{\mathrm{o}\mathrm{d}\mathrm{d}}\to H_{\mathrm{o}\mathrm{d}\mathrm{d}}\oplus H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$, $U^{\mathrm{\prime }}(x,z)=(-z,x)$. Since $\mathrm{Dom}(L^{\ast })$ is dense, $L^{\ast \ast }$ is defined, and $\mathrm{\Gamma }(L^{\ast \ast })=U^{\mathrm{\prime }-1}\mathrm{\Gamma }(L^{\ast }{)}^{\perp }=U^{\mathrm{\prime }-1}{U}^{-1}\mathrm{\Gamma }(L)=-\mathrm{\Gamma }(L)=\mathrm{\Gamma }(L)$. Here $\mathrm{\Gamma }(L)$ is closed, and the middle equality uses $U^{\mathrm{\prime }-1}{U}^{-1}(x,y)=(-x,-y)$; a linear subspace equals its negative. Hence $L^{\ast \ast }=L$ as operators with equal domains. The domain of $Q$ is dense because $\mathrm{Dom}(L)$ is dense by step 1.1 and $\mathrm{Dom}(L^{\ast })$ was just shown dense, so $Q^{\ast }$ is defined. For $Q^{\ast }(a,b)=(c,d)$, the adjoint identity is $⟨L^{\ast }y,a{⟩}_{H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}}+⟨Lx,b{⟩}_{H_{\mathrm{o}\mathrm{d}\mathrm{d}}}=⟨x,c{⟩}_{H_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}}+⟨y,d{⟩}_{H_{\mathrm{o}\mathrm{d}\mathrm{d}}}$ for every $x\in \mathrm{Dom}(L)$ and $y\in \mathrm{Dom}(L^{\ast })$. Taking $x=0$ and then $y=0$, the cross-space adjoint definition and $L^{\ast \ast }=L$ give respectively $a\in \mathrm{Dom}(L)$, $d=La$, and $b\in \mathrm{Dom}(L^{\ast })$, $c=L^{\ast }b$. Conversely these conditions make the identity hold, so $\mathrm{Dom}(Q^{\ast })=\mathrm{Dom}(L)\oplus \mathrm{Dom}(L^{\ast })$ and $Q^{\ast }=Q$. The grading is odd by the block formula. The even and odd zero spaces are $\ker L$ and $\ker L^{\ast }$; the latter is $(\mathrm{ran}L{)}^{\perp }$, which has dimension $\dim (H_{\mathrm{o}\mathrm{d}\mathrm{d}}\mathrm{/}\mathrm{ran}L)$ because $\mathrm{ran}L$ is closed. Their signed dimension is $\mathrm{ind}L$.

For finitely many supplied local sectors set $E={⨁}_p{E}_p$ and $V={⨁}_p{V}_p$. Finite-dimensionality gives $\dim E-\dim V={\sum }_p(\dim E_p-\dim V_p)$, so the index corrections add. This bookkeeping does not assert an actual singular transfer is finite dimensional or that a physical Fredholm regulator exists.