Euler characteristic of closed orientable surface of dimension $4n + 2$.

Question

I'm considering the Euler characteristic of closed orientable surface of dimension $4n + 2$. The goal is to show that it is even.

By Poincare Duality, and applying the Universal Coefficient Theorem, I was able to show that $H_i(M, \mathbb{Z}) \cong H^{4n + 2- i}(M, \mathbb{Z})$. So $\chi(M) = (-1)^{2k + 1}H_{2k + 1}(M, \mathbb{Z})$. However, I have no information about homology group on this level. It being $2k + 1$ also makes me think that I should be using some odd-ness of the homology group here.

I searched online and saw that there are some cup product argument but there are some gaps that I couldn't follow.

Any help will be appreciated!

Answer 1

It's not true that $\chi(M)=(-1)^{2k+1} \dim H_{2k+1}(M,\Bbb Z)$, rather $\chi(M)\equiv\dim H_{2k+1}(M,\Bbb Z)\pmod2$, but that's all you need.

It's easier to work with real coefficients. Poincare duality means that the cup product $H^{2k+1}(M,\Bbb R) \times H^{2k+1}(M,\Bbb R)\to H^{4k+2}(M,\Bbb R)\cong\Bbb R$ is a perfect pairing. But since $2k+1$ is odd, $\alpha\smile\alpha=0$ for all $\alpha\in H^{2k+1}(M,\Bbb R)$. This cup product is therefore symplectic, and one cannot have a perfect symplectic pairing on an odd-dimensional space. Therefore $\dim H^{2k+1}(M,\Bbb R)$ is even, and so is $\chi(M)$.

Answer 2

It's even easier if you use field coefficients. But you're exactly on the right track -- the Euler characteristic is exactly $\pm$ the dimension of the middle homology vector space.

(What you've written doesn't quite make sense: $\chi(M)$ is a number, but the right hand side is a group. Perhaps you meant the $2i+1$ Betti number.)

The only question, then, is "What ranks are possible?"

The answer for $k = 0$ is well known -- that's the classification theorem for surfaces. But that makes finding examples in higher dimensions easy: in dimension $6$, for instance, you can take $S^4 \times F$, where $F$ is a surface of dimension $2$ to get lots of examples. And you can take examples of $F_1 \times F_2 \times F_3$ as well (using the Kunneth formula to help compute homology groups).

At any rate, once you've done the first step-- the cancellation of low- and high-index homology groups' Betti numbers, to see that the only thing remaining is the middle Betti number, you've pretty much finished up the problem.

Answer 3

Hint: Poincare duality and universal coefficients imply that the parity of the Euler characteristic is determined by the rank of the middle cohomology group. The graded commutative property of the cup product implies that the intersection form $H_{2n-1} \times H_{2n-1} \to \Bbb R$ is a symplectic (non-degenerate an skew-symmetric) pairing. In which dimension do such pairings exist?