MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 13 down vote accepted

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.

share|cite|improve this answer
Thanks, that's very pretty! – Sam Derbyshire Oct 16 '09 at 2:34

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.