Question

This is probably quite easy, but how do you show that the Euler characteristic of a manifold M (defined for example as the alternating sum of the dimensions of integral cohomology groups) is equal to the self intersection of M in the diagonal (of M × M)?

The few cases which are easy to visualise (ℝ in the plane, S¹ in the torus) do not seem to help much.

The Wikipedia article about the Euler class mentions very briefly something about the self-intersection and that does seem relevant, but there are too few details.

Answer 1

The normal bundle to M in MxM is isomorphic to the tangent bundle of M, so a tubular neighborhood N of M in MxM is isomorphic to the tangent bundle of M. A section s of the tangent bundle with isolated zeros thus gives a submanifold M' of N \subset MxM with the following properties:

1) M' is isotopic to M.

2) The intersections of M' with M are in bijection with the zeros of s (and their signs are given by the indices of the zeros).

The desired result then follows from the Hopf index formula.

Answer 2

Here's an attempt at a sheaf theoretic argument that I always thought would work, but never actually tried:

The intersection # of two transversal submanifolds A and B of complimentary dimension inside a 3rd manifold C can be computed as chi(A (x) B), where I'm using A and B to denote the structure sheaves of the corresponding manifolds, and the tensor product is taking place in C-mod. In the case that the intersection is not transversal, this presumably still works provided you take a derived tensor product (take a flat family moving one of the intersectands to a general position, and use invariance of chi under flat deformation for a perfect complex representing the other intersectand, perhaps?)).

Assuming the above, the self-intersection M.M of the diagonal M in M x M is chi(M (x)^L M). As M is smooth, Tor^i(M,M) = Omega^i. By the additivity of chi, you get:

M.M = \sum_i chi(Omega^i) (-1)^i

On the other hand, de Rham's theorem (or the Poincare lemma?) identifies the right hand side with chi(M,constant sheaf) = chi(M), so we are done.