12
$\begingroup$

The Hirzebruch signature theorem tells us that for a smooth compact oriented 4-manifold, the signature $\sigma(M)$ is proportional to the first Pontryagin number of $M$: $$ 3\sigma(M)= p_1(M) = k \int_M \mathrm{tr}(R^2) $$ for some appropriate constant $k$ (for the purposes of this question, take the integral to be the definition of $p_1$, though properly this uses some Chern–Weil theory to prove). The usual proof of this fact goes via a general theorem involving the $L$-genus. The question was recently raised in conversations as to whether this result had a more elementary geometric proof, for instance at the level of Chern's intrinsic proof of the Gauss–Bonnet theorem, with additional specialisations that are specific to dimension 4.

Some simple-minded observations:

  1. We can (apparently) simplify the integral to one of the form $\int_M |W^+|^2 - |W^-|^2$ (up to a proportionality constant), using the decomposition of the curvature tensor $R$ via Hodge theory and the self- and anti-self-dual parts of the Weyl curvature, $W^\pm$.

  2. Likewise, we can write the signature as the difference $b_+ - b_-$ of dimensions of a decomposition of $H^2$.

So one might conceivably be able to prove $3(b_+ - b_-) = \frac{1}{4\pi^2}\int_M |W^+|^2 - |W^-|^2$ more directly using geometric methods, but I am completely unfamiliar with this area and existing research. Clearly one can prove the result using even fancier methods, but this is not what I'm looking for.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ There is a heat-kernel proof based on the same idea as Atiyah-Singer index theorem. The proof is quite elementary, but it is really long. I am not sure if this is what you are looking for. $\endgroup$
    – Bombyx mori
    Commented Aug 30, 2019 at 6:50
  • 1
    $\begingroup$ I was thinking more geometric than analytic. And, if possible, taking maximum advantage of working on a 4-manifold, rather than just proving the general case with the additional assumption that $d=4$. $\endgroup$
    – David Roberts ♦
    Commented Aug 30, 2019 at 7:05
  • 1
    $\begingroup$ @Bombyxmori the equality as stated is not true for arbitrary $4k$-manifolds, because you need more terms involving the other Pontryagin numbers. Also, the question was specifically about $4$-manifolds in the discussion I was having, with someone whose speciality is 4-manifolds and complex surfaces. $\endgroup$
    – David Roberts ♦
    Commented Aug 30, 2019 at 8:22
  • 5
    $\begingroup$ This probably isn't quite what you're looking for, but there's a mostly elementary proof using bordism: both $\sigma(M)$ and $p_1(M)$ are oriented bordism invariants valued in $\mathbb Z$, so it suffices to compute them on a generating set for $\Omega_4^{\mathrm{SO}}\otimes\mathbb Q$. Determining that this is one-dimensional requires some input from algebraic topology but isn't all that bad; then, showing $\mathbb{CP}^2$ is a generator is straightforward, and computing $\sigma(\mathbb{CP}^2)$ and $p_1(\mathbb{CP}^2)$ is also straightforward. $\endgroup$
    – Arun Debray
    Commented Aug 30, 2019 at 13:47
  • 2
    $\begingroup$ @ArunDebray yes, it's not quite what the person I was talking to wanted. Checking two bordism invariants are equal on the generator is far from geometric, and I believe a big part of the desiderata was knowing that these are indeed bordism invariants by purely geometric means. $\endgroup$
    – David Roberts ♦
    Commented Aug 31, 2019 at 13:08

0

Reset to default

You must log in to answer this question.