F4 Finite Vertex Coupling And Net Index

When $E=V=\mathbb{C}$ and $S=T=0$, the operator is the direct sum $D\oplus m$. If $m\mathrm{\ne }0$, scalar multiplication by $m$ is bijective, so the vertex kernel and cokernel both vanish. If $m=0$, its kernel and cokernel are each $\mathbb{C}$. The two zero spaces of the odd operator therefore contain respectively one even and one odd vertex state in the zero case, and none in the nonzero case. Formula [F1] gives $\mathrm{ind}L=\mathrm{ind}D+1-1=\mathrm{ind}D$ in both cases.

If $E=0$, $V=\mathbb{C}$, and $T=0$, then $Lx=(Dx,0)$. Its kernel is $\ker D$, its range is $\mathrm{ran}D\oplus \{0\}$, and its cokernel is $(H_{-}\mathrm{/}\mathrm{ran}D)\oplus \mathbb{C}$. The range is closed because $\mathrm{ran}D$ is closed. Thus its index is $\mathrm{ind}D-1$, also as [F1] gives. The additional odd cokernel copy of $\mathbb{C}$ is present regardless of any physical interpretation.