"The index is independent of the Dirac operator"

Question

Fix a Clifford module bundle $E$ on a compact Riemannian manifold $M$ and let $D_0$ and $D_1$ be two Dirac operators (compatible with the Clifford action). The proof of the Atiyah-Singer index theorem in Heat Kernels and Dirac operators makes use of the fact that in this situation $$\mathrm{ind}(D_0)=\mathrm{ind}(D_1).$$ The proof of the equality is discussed on page 125. However I feel like there is gap in the proof:

We seem so to assume that we can find a family $(D_t)_{t\in [0,1]}$ of Dirac operators such that the function $$[0,1]\ni t\mapsto\mathrm{ind}(D_t)\in\mathbb{R}$$ is continuous. Since the image is contained in $\mathbb{Z}$, it would be constant. In fact there is a very obvious candidate: As explained on page 117 we have that $A=D_1-D_0$ is function-linear, i.e. $A\in \Gamma(M,\mathrm{End}(E))$ and $$D_t=D_0+tA$$ is a Dirac operator for all $t\in[0,1]$. But how do we know that the above-mentioned function is continuous?

The authors suggest to use the McKean-Singer formula, but it was only proven for a Dirac operator which happens to be symmetric w.r.t. some metric. I assume that this is not the case for all Dirac operators (?), so can we avoid using McKean Singer? Or does the latter actually hold for all Dirac operators?

Answer 1

First, for a Dirac operator $D$, it extends to a Fredholm operator \begin{equation} D^+\colon H^1(M,E^+)\to L^2(M,E^-), \end{equation} see Lawson-Michelsohn "Spin Geometry", Page 193, Theorem 5.2.

Also, a classical result in functional analysis says the index of the Fredholm operator is locally constant, see for instance Lemma 16.18 of https://ocw.mit.edu/courses/18-965-geometry-of-manifolds-fall-2004/8a7e4dd837d1bdd6988e0330babb8c5e_lecture16_17.pdf

Therefore, using the above lemma for the family $t\in [0,1]\mapsto tD_0^++(1-t)D_1^+$, the index is constant as required.

To see that the Fredholm index \begin{equation} Ind(D^+)=\dim \mathrm{kernel}(D^+)-\mathrm{codim\ range}(D^+) \end{equation} coincides with the one you want, we only need to prove that \begin{equation} \mathrm{codim\ range}(D^+)=\dim \ker(D^-). \end{equation} Now we consider the dual $D^- $ of $D^+$ \begin{equation} D^-\colon L^2(M,E^-)\to H^{-1}(M,E^+), \end{equation} Using Rudin Functional Analysis Theorem 4.12, we get \begin{equation} L^2(M,E^-)=\ker D^-\oplus \overline{\mathrm{range}(D^+)}, \end{equation} and since $D^+$ is Fredholm, its image is closed, $\overline{\mathrm{range}(D^+)} =\mathrm{range}(D^+)$, so \begin{equation} L^2(M,E^-)=\ker D^-\oplus\mathrm{range}(D^+), \end{equation} from which we clearly get $\mathrm{codim\ range}(D^+)=\dim \ker(D^-)$ as required.

A subtle point is to distinguish different adjoints, first, \begin{equation} (D^+)^*\colon L^2(M,E^-)\to H^{1}(M,E^+), \end{equation} is defined by \begin{equation} ((D^+)^*s,t)_{H^{1}(M,E^+)}=(s,D^+t)_{L^2(M,E^-)} \end{equation} for $s\in L^2(M,E^-),t\in H^{1}(M,E^+)$. But remember that this adjoint is NOT the way we define $D^-$, because $D^-$ is defined via the $L^{2}(M,E^+)$ product, not the $H^{1}(M,E^+)$ product: \begin{equation} ((D^+)^*s,t)_{L^{2}(M,E^+)}=(s,D^+t)_{L^2(M,E^-)}, \end{equation} that is why $(D^+)^*s$ is in $H^{-1}(M,E^+)$, not $H^{1}(M,E^+)$.