Atiyah Singer index theorem and Hodge de Rham operator

Question

When I read about Atiyah Singer index theorem I met the following example: let $M$ is (orientable closed smooth) Riemannian manifold and consider Hodge-de Rham Dirac operator defined by $d+d^*$ (adjoint takes into account the metric). This operator is self adjoint and interchanges odd and even degree forms: therefore is of antidiagonal form. The index of $(d+d^*)^+$ is equal to the Euler characteristic of $M$: $$ind(d+d^*)^+=\chi(M)$$

1. Can we derive this formula from the right hand side of the Atiyah Singer index formula (involving Todd genus and Chern character)?

Moreover there is so called generalized Gauss Bonnet theorem which relates Euler characteristic to the geometric quantity $\int_{M}Pf(\Omega)$ (Pfaffian of the curvature form): $$\chi(M)=(2\pi)^{-n}\int_MPf(\Omega).$$ So we have at all three quantities: the index, Euler characteristic and the integral of Pfaffian.

2. Is it possible to obtain the equality $$ind(d+d^*)^+=(2\pi)^{-n}\int_MPf(\Omega)$$ without invoking Gauss-Bonnet (which as far as I know requires some work).

The first question is something like "first step" in order to get into index theory and to get some better understanding of the background of Atiyah Singer index theorem. Second question is inspired by the following: $\chi(M)$ is a topological quantity: from the other hand $Pf(\Omega)$ is of geometric flavour. Gauss-Bonnet formula relates these two quantities: but as $d^*$ requires choice of the metric, $ind(d+d^*)^+$ should also be viewed as geometric (or maybe better, analytical quantity) so maybe it would be easier to relate this two geometric quantities and then refer to the Euler characteristic.

Answer 1

The Euler operator $d+d^*$ of a spin manifold is the spin Dirac operator twisted by the $\mathbb Z/2$-graded spinor bundle $W=S^+\ominus S^-$, as explained in Heat Kernels and Dirac Operators by Berline, Getzler, Vergne [BGV, chapter 4.1]. The Chern character $\mathrm{ch}(W)$ equals $e(TM)\hat A(TM)^{-1}$, see [BGV] or Nicolaescu's notes. The cohomological index theorem therefore gives $$\mathrm{ind}(d+d^*)^+=(\hat A(TM)\wedge\mathrm{ch}(W))[M]=e(TM)[M]$$

The de Rham representative of $e(TM)$ is given by the Pfaffian, and you get your second formula [BGV, Theorem 4.6].

The formulas above are pairings of "characteristic cohomology classes" with "fundamental homology classes". If you regard the Dirac operator as a fundamental $K$-homology class for the $K$-orientation of the tangent bundle given by the chosen spin structure, then the left hand side also constitutes a pairing of the $K$-theory class of the spinor bundle with the corresponding fundamental class.

To avoid the spin condition, you define the ``twist Chern character'' $\mathrm{tr}(\mathrm{exp}(-F^{W/S}))$ as explained in [BGV, chapter 4.1]. The formulas above still hold.

You can even get rid of orientability by regarding $e(TM)\in H^{\dim M}(M,o(TM))$ as a cohomology class twisted by the orientation bundle. Similarly, the Pfaffian is in fact a differential form twisted by the orientation bundle. Then all formulas above make sense without an orientation of $TM$.

On the other hand, the Euler number is the "trace" of id$_M$ in a sense explained nicely here (Def 2.2) - no matter if you count cells or dimensions of (co-) homology groups. To relate it to the pairings above, you may use either Hodge theory, or the Gauss-Bonnet-Chern theorem, or deduce $e(TM)[M]=\chi(M)$ from the Poincar'e-Hopf theorem, e.g. using Morse theory. It seems that because $\chi(M)$ and $e(TM)[M]$ are different kinds of objects, some work is needed for this step.

Answer 2

The first equality,

$$ {\rm ind}\; (d+d^*)^+=\chi(M), $$

follows easily from Hodge theory

$$\dim \ker (d+d^*)^+=\sum_{k\geq 0}b_{2k}(M), $$

$$\dim {\rm coker}\; (d+d^*)^+=\sum_{k\geq 0}b_{2k+1}(M). $$

The 2nd equality follow from the index theorem. For details, see Section 3.2 of these notes.