# "The index is independent of the Dirac operator"

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?

• You are right, from the proof of Theorem 3.51, we can see that the author implicitly assume that each $D^z$ is self-adjoint with respect to certain metric data (depend on $z$). There is a very general theorem in functional analysis: the index of Fredholm operator is locally constant. I think the authors did not use this result for the sake of self-containment. Commented Feb 23 at 8:12
• @Local Thank you for the comment, but I don't quite see how this resolves the issue. Could you please elaborate a bit? Commented Feb 23 at 10:52

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

A subtle point is to distinguish different adjoints, first, $$$$(D^+)^*\colon L^2(M,E^-)\to H^{1}(M,E^+),$$$$ is defined by $$$$((D^+)^*s,t)_{H^{1}(M,E^+)}=(s,D^+t)_{L^2(M,E^-)}$$$$ for $$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^+)$$ product: $$$$((D^+)^*s,t)_{L^{2}(M,E^+)}=(s,D^+t)_{L^2(M,E^-)},$$$$ that is why $$(D^+)^*s$$ is in $$H^{-1}(M,E^+)$$, not $$H^{1}(M,E^+)$$.

• Thank you for elaborating. But I am still not sure that this resolves the issue: The index of$$D=\begin{pmatrix}0&D^1\\D^0&0\end{pmatrix}$$is defined by$$\operatorname{ind}(D)=\dim\ker{D^0}-\dim\ker{D^1}$$ and does not equal the Fredholm index of $D$. I know that proposition 3.48 says that the index of $D$ equals the Fredholm index of $D^0$, but in that proposition it is assumed that $D$ is symmetric. Commented Feb 23 at 13:52
• @Filippo I modified the notation to solve the confusion of $0$ and $1$ being used to represent odd and even parts or the value of $t$. Commented Feb 23 at 19:30
• Hmm given two Fredholm operators $A_0$ and $A_1$ I can of course not just use the lemma for the family $t\mapsto A_t=A_0+t(A_1-A_0)$. Are you using that in our case 1) $A_t$ is Fredholm for all $t$ and 2) the function $t\mapsto A_t$ is a continuous function to the topological space of Fredholm operators? Commented Feb 23 at 19:45
• @Filippo Yes, indeed the set of Fredholm operators is open in the space of bounded linear operators (with respect to the operator norm), and we easily check that $t\mapsto A_t$ is continuous in the topological space of bounded linear operators. Commented Feb 23 at 20:41
• Yes the Fredholm index of $D^+$ probably is more intrinsic. The authors of this book are world-class experts on index theory...I would hardly say they are wrong, I think they just work under the implicit assumptions of symmetric case, so their definition is consistent with the Fredholm index. Commented Feb 26 at 14:00