Atiyah–Singer index formula for a closed even-dimensional geometric Clifford module

If $m=0$, compactness makes the discrete manifold $M$ finite. The $H^1$ and $L^2$ spaces are the same finite direct sums of fibers, and $D_V^{+}=0:{⨁}_x{V}_x^{+}\to {⨁}_x{V}_x^{-}$ because $T^{\ast }M=0$. Its kernel and cokernel are those two finite direct sums, so its index is ${\sum }_x(\dim V_x^{+}-\dim V_x^{-})$. The reference grading in the Statement gives ${\mathrm{str}}_{W_x}I={\epsilon }_x(\dim V_x^{+}-\dim V_x^{-})$, while ${\widehat{A}}_0=1$ and $F^{V\mathrm{/}S}=0$. The signed integration rule [F6] therefore gives ${\int }_M[\widehat{A}\mathrm{ch}(V\mathrm{/}S){]}_0={\sum }_x{\epsilon }_x^2(\dim V_x^{+}-\dim V_x^{-})=\mathrm{ind}(D_V^{+})$. This includes negative points, mixtures of signs, arbitrary fiber gradings, rank-zero fibers and the empty manifold.

Suppose $m\ge 1$. By [F1], $D_V^{+}$ is a Fredholm operator on the stated exact domain. For each $t>0$, [F2] makes $e^{-t{D}_V^2}$ trace class and gives $\mathrm{ind}(D_V^{+})=\mathrm{Tr}({\mathrm{\Gamma }}_V{e}^{-t{D}_V^2})={\int }_M\mathrm{str}{K}_t(x,x)d{\mathrm{vol}}_g(x)$, where ${\mathrm{\Gamma }}_V$ is $+1$ on $V^{+}$ and $-1$ on $V^{-}$.

In each normal-coordinate neighborhood, [F3] replaces $K_t$ by its finite heat parametrix with a remainder $O(t^N)$ in the diagonal $C^0$ norm for any prescribed $N$, after increasing the truncation order. Since $M$ is compact, the volume is finite, so the integral of that remainder tends to zero. By [F4], $\mathrm{str}{u}_j(x,x)=0$ for $j<m$. In $(4\pi t{)}^{-m}{\sum }_j{t}^j{u}_j(x,x)$ each smooth coefficient with $j>m$ is multiplied by $t^{j-m}$ and tends uniformly to zero. Within $u_m$, only Clifford degree $2m$ contributes to supertrace. Thus the limit of the diagonal supertrace integral as $t\downarrow 0$ is the integral of its single degree-$2m$ oscillator coefficient.

By [F5], that coefficient is the form $[\widehat{A}(TM,{\mathrm{\nabla }}^{LC})\mathrm{ch}(V\mathrm{/}S){]}_{2m}$. Taking the limit in the $t$-independent identity of step 1.2 proves the displayed index formula. The result is computed over the complex Hilbert spaces $L^2(V^{\pm })$, so no conclusion about real KO-index groups follows. The empty manifold has zero domain and codomain and both sides are zero; the zero-dimensional case is computed in step 1.1, and no boundary condition is needed because $M$ has no boundary.