The Heat Kernel — The Hirzebruch Signature Theorem

4 The Hirzebruch Signature Theorem

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

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

is an elliptic operator whose index is exactly the signature of MM.

In Section 1, we have established that

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

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

tretDDk=0tkd/2M1(4π)d/2tr(ukΩ+(x,x))  dx. \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

indexDk=0tkd/2Mak(x)  dx. \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

indexD=Mad/2(x)  dx. \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 ad/2a_{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 truk\operatorname{tr}u_ k is a rational function of the components of the metric and its derivatives. The same is not true for truiΩ±\operatorname{tr}u_ i|_{\Omega _{\pm }}. However, the change of basis matrix to an eigenbasis of * can only be a function of gg, as it is performed fiberwise. So we can write

truiΩ±=bi(g)mi, \operatorname{tr}u_ i|_{\Omega _{\pm }} = \sum b_ i(g) m_ i,

where bib_ i are arbitrary smooth functions of the component of the metric, and mim_ i are monomials in the derivatives of gijg_{ij}.

Theorem 4.1 (Gilkey)

Suppose ω\omega associates to each Riemannian manifold a differential form that, in local coordinates, can be expressed in the form ai(g)mi\sum a_ i(g) m_ i as above.

Moreover, suppose ω\omega has weight w0w \geq 0. That is, if we replace the metric gg by λg\lambda g, then ωλwω\omega \mapsto \lambda ^ w \omega . Then ω\omega is a polynomial function of the Pontryagin forms, and in fact has weight 00.

We need to calculate the weight of ak(x)  dxa_ k(x) \; \mathrm{d}x. Remembering that dx\mathrm{d}x also depends on the metric, careful bookkeeping reveals that aka_ k is of weight d2kd - 2k. So we deduce that indexD\operatorname{index}D is the integral over MM of a polynomial function of the Pontryagin forms!

Once we know that ak(x)  dxa_ 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 LL-genus:

Theorem 4.2 (Hirzebruch signature formula)

If MM is a closed Riemannian manifold of dimension dd (with 4d4 \mid d), then

signM=ML(p1,,pd/4). \operatorname{sign}M = \int _ M L(p_1, \ldots , p_{d/4}).\fakeqed