4 The Hirzebruch Signature Theorem

We now apply our theory to the case of the Hirzebruch signature theorem. Fix a Riemannian manifold $M$. Recall that the Hodge star operator acts on $\bigoplus _p \Omega ^p$ with $\pm 1$ eigenspaces $\Omega _{\pm }$. The operator

$$D = \mathrm{d}+ \mathrm{d}^*\colon \Omega _+ \to \Omega _-$$

is an elliptic operator whose index is exactly the signature of $M$.

In Section 1, we have established that

$$\operatorname{index}D = \operatorname{tr}e^{-t D^*D} - \operatorname{tr}e^{-t DD^*}$$

for all $t > 0$. In Section 2, we constructed a heat kernel for $\mathrm{d}+ \mathrm{d}^*\colon \Omega ^* \to \Omega ^*$, and we observe that this commutes with the Hodge star. So we get an asymptotic series

$$\operatorname{tr}e^{-t D^*D} \sim \sum _{k = 0}^\infty t^{k - d/2}\int _M \frac{1}{(4\pi )^{d/2}} \operatorname{tr}(u_k|_{\Omega _+}(x, x))\; \mathrm{d}x.$$

Putting everything together, we can write

$$\operatorname{index}D \sim \sum _{k = 0}^\infty t^{k - d/2} \int _M a_k(x) \; \mathrm{d}x.$$

However, we also know the left-hand side is a constant. So the non-constant terms must in fact cancel out. So we know that

$$\operatorname{index}D = \int _M a_{d/2}(x) \; \mathrm{d}x.$$

At this point, you might think it is rather hopeless to trace through the construction to obtain an explicit identification of $a_{d/2}$, and you would be right. We will cheat as Hirzebruch did. However, it turns out for purely formal reasons, there aren't too many possibilities for it.

Recall that $\operatorname{tr}u_k$ is a rational function of the components of the metric and its derivatives. The same is not true for $\operatorname{tr}u_i|_{\Omega _{\pm }}$. However, the change of basis matrix to an eigenbasis of $*$ can only be a function of $g$, as it is performed fiberwise. So we can write

$$\operatorname{tr}u_i|_{\Omega _{\pm }} = \sum b_i(g) m_i,$$

where $b_i$ are arbitrary smooth functions of the component of the metric, and $m_i$ are monomials in the derivatives of $g_{ij}$.

We need to calculate the weight of $a_k(x) \; \mathrm{d}x$. Remembering that $\mathrm{d}x$ also depends on the metric, careful bookkeeping reveals that $a_k$ is of weight $d - 2k$. So we deduce that $\operatorname{index}D$ is the integral over $M$ of a polynomial function of the Pontryagin forms!

Once we know that $a_k (x) \; \mathrm{d}x$ is a polynomial function of the Pontryagin forms, we just do as Hirzebruch did and evaluate both sides on enough spaces to show that it must be the $L$-genus: