A chiral geometric Dirac operator is Fredholm

Since $D$ is odd and selfadjoint, its blocks satisfy $(D^{+}{)}^{\ast }=D^{-}$ on the $H^1$ domain. If $D^{+}u=0$, then $P_{+}u=D^{-}{D}^{+}u=0$ distributionally, so [F2] makes $u$ a smooth element of the finite-dimensional $\ker P_{+}$. Conversely $P_{+}u=0$ implies $\mathrm{\parallel }{D}^{+}u{\mathrm{\parallel }}_2^2=⟨P_{+}u,u⟩=0$, hence $D^{+}u=0$. The same argument applies to $D^{-}$.

Define $Q=D^{-}{G}_{-}:L^2(V^{-})\to H^1(V^{+})$. The Green bound gives $G_{-}:L^2\to H^2$ and the first-order local formula gives $D^{-}:H^2\to H^1$ bounded. Therefore $D^{+}Q=P_{-}{G}_{-}=I-{\mathrm{\Pi }}_{-}$. The range of $D^{+}$ contains $\ker P_{-}^{\perp }$ and is orthogonal to $\ker P_{-}$ because $⟨D^{+}u,h_{-}⟩=⟨u,D^{-}{h}_{-}⟩=0$. Hence its range is exactly $\ker P_{-}^{\perp }$, which is closed, and its cokernel is $\ker P_{-}$.

For $u\in H^1(V^{+})$, approximate by smooth sections. For smooth $u$, $D^{-}{G}_{-}{D}^{+}u-u$ lies in $\ker D^{+}$ because applying $D^{+}$ gives $D^{+}u-{\mathrm{\Pi }}_{-}{D}^{+}u-D^{+}u=0$; here ${\mathrm{\Pi }}_{-}{D}^{+}u=0$ by the orthogonality in step 1.2. The term $D^{-}{G}_{-}{D}^{+}u$ is orthogonal to $\ker D^{+}$, because $D^{-}$ is the adjoint of $D^{+}$ and $D^{+}{h}_{+}=0$. The difference lies in that kernel and its projection is $-{\mathrm{\Pi }}_{+}u$, so it equals $-{\mathrm{\Pi }}_{+}u$. Therefore $D^{-}{G}_{-}{D}^{+}u=u-{\mathrm{\Pi }}_{+}u$. Boundedness and density extend the identity to $H^1$. The two-sided inverses modulo finite-rank projections and the finite kernel/cokernel prove Fredholmness and the index formula. Full AC is inherited through [A1] and the Green-operator supplier.