Timeline for Elementary-ish geometric proof of Hirzebruch signature theorem for Riemannian 4-manifolds?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2019 at 19:58 | comment | added | Arun Debray | @DavidRoberts if you're willing to consider spin bordism, there are certainly bordism invariants arising from geometry, namely the index (or mod 2 index) of the Dirac operator. | |
| Aug 31, 2019 at 13:08 | comment | added | David Roberts♦ | @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. | |
| Aug 30, 2019 at 20:17 | comment | added | Bombyx mori | @ArunDebray: Yes this is a sensible approach. I recall this is one of the proofs I learned many years ago. | |
| Aug 30, 2019 at 13:47 | comment | added | Arun Debray | 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. | |
| Aug 30, 2019 at 8:22 | comment | added | David Roberts♦ | @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. | |
| Aug 30, 2019 at 8:11 | comment | added | Bombyx mori | I think you do not really need the 4 manifold assumption here, the theorem holds for any dimension 4k manifolds. But I could be wrong. You may be right that for n=4 the proof simplifies. | |
| Aug 30, 2019 at 7:05 | comment | added | David Roberts♦ | 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 $d=4$. | |
| Aug 30, 2019 at 6:50 | comment | added | Bombyx mori | 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. | |
| Aug 30, 2019 at 6:44 | history | asked | David Roberts♦ | CC BY-SA 4.0 |