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.

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    $\begingroup$ 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

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 $$\int_\Delta\eta = \int_M \iota^*\eta = \sum_{i,j} (-1)^i \sum_{j=1}^{\dim H^i(M)}\alpha_{i,j}\wedge\beta_{n-i,j} = \sum_i (-1)^i\dim H^i(M).$$

  • $\begingroup$ 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

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