3
$\begingroup$

It is well-known that the Euler characteristic of a closed manifold $M^n$, which can be defined as $\chi(M)=\sum_{k=0}^n (-1)^k \operatorname{dim}H^k(M)$, equals the intersection number $I(\Delta,\Delta)$ of the diagonal with itself in $M\times M$.

Is there a way to prove this without constructing triangulations? I am looking for a proof which uses only basic facts such as Poincaré Duality. Also notice that this identity is equivalent to the Poincaré-Hopf theorem.

$\endgroup$
  • 1
    $\begingroup$ you could use the euler class of the normal bundle of $\Delta \cong M$ in $M\times M$ $\endgroup$ – Daniel Valenzuela Nov 23 '14 at 10:24
2
$\begingroup$

Here's what you want to check. I'm thinking of deRham cohomology, but still writing $H^i(M)$ for $H^i(M,\Bbb R)\cong H^i_{\text{deRham}}(M)$; but it all works fine with cup product, too. We assume $M$ compact and oriented.

Let $\alpha_{i,j}$ ($j=1,\dots,\dim H^i(M)$) be a basis for $H^i(M)$, and let $\beta_{n-i,j}$ be the dual basis for $H^{n-i}(M)$, in the sense that $$\int_M \alpha_{i,j}\wedge\beta_{n-i,j} = 1 \text{ for all }i,j.$$ Note that $(-1)^{i(n-i)}\alpha_{i,j}$ is the dual of $\beta_{n-i,j}$.

By definition, the Poincaré dual of $\Delta\subset M\times M$ will be an $n$-form $\eta$ on $M\times M$ with the property that $$\int_\Delta \phi = \int_{M\times M} \phi\wedge\eta \text{ for every closed $n$-form $\eta$ on $M\times M$}.$$ Now you just have to do some careful sign-checking (keeping track of skew-commutativity) to verify that $$\eta = \sum_{i,j} (-1)^i \pi_1^*\alpha_{i,j}\wedge\pi_2^*\beta_{n-i,j},$$ where $\pi_i$ are the obvious projections from $M\times M$ to $M$. Since $\Delta\cdot\Delta = \displaystyle\int_\Delta \eta$, we just have to then check that $$\int_\Delta\eta = \sum_i (-1)^i \dim H^i(M).$$ Letting $\iota\colon M\to M\times M$ be the diagonal inclusion, we have $$\int_\Delta\eta = \int_M \iota^*\eta = \sum_{i,j} (-1)^i \sum_{j=1}^{\dim H^i(M)}\alpha_{i,j}\wedge\beta_{n-i,j} = \sum_i (-1)^i\dim H^i(M).$$

$\endgroup$
  • $\begingroup$ That can be found in Bott and Tu's Differential forms in algebraic topology, p.126 $\endgroup$ – user99914 Nov 23 '14 at 9:52
  • $\begingroup$ Ah, of course. When I thought through all this decades ago, it was before Bott and Tu. And I didn't have it here with me. $\endgroup$ – Ted Shifrin Nov 23 '14 at 12:18
  • $\begingroup$ Thank you, that's exactly what I was looking for! $\endgroup$ – Mizar Nov 23 '14 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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