# On the Euler Class of a manifold

Let $M$ be an orientable, compact $n$ dimensional differentiable manifold and $e \in H^n(M, \mathbb{Z})$ the Euler class of the tangent bundle of $M$, defined via the Thom Isomorphism. Also, let $[M] \in H_n(M, \mathbb{Z})$ be the fundamental class of $M$. How does one prove that $e([M]) = \chi(M)$, where $\chi(M) = \sum_{i=0}^n (-1)^i \text{dim } H^i(M, \mathbb{Z})$ is the Euler characteristic of $M$?

• Here's a solution mathoverflow.net/questions/696/… at least if you know about (geometric) cup product and Poincaré duality Apr 1, 2016 at 8:04

The pair $(M \times M, M \times M \setminus \operatorname{diag}(M))$ has a fundamental class $u'$, where $\operatorname{diag}(M) = \{ (x,y) \in M \times M \mid x = y \}$. You can then restrict this class through $$H^n(M \times M, M \times M \setminus \operatorname{diag}(M)) \to H^n(M \times M)$$ to get a class $u''$ called the diagonal class. Let $\Delta : M \to M \times M$ eb given by $x \mapsto (x,x)$. You then get the relation $\Delta^*(u'') = e(M)$ using the Thom isomorphism definition of $e(M)$ (there are a few facts hidden there).
Now let $\{b_i\}$ be a graded basis for $H^*(M)$, and let $\{b_i'\}$ be a Poincaré dual basis, i.e. $$\langle b_i \smile b_j', [M] \rangle = \delta_{ij}.$$ Then using the relation $u'' / [M] = 1 \in H^0(M)$ (this is the slant product), you get that: $$u'' = \sum_i (-1)^{\deg b_i} b_i \times b_i'.$$ Evaluating this on $[M]$, you get $\langle e(M), [M] \rangle = \langle \Delta^*(u''), [M] \rangle = \sum_i (-1)^{\deg b_i}$, and a simple degree counting argument shows that this last sum is equal to $\chi(M) = \sum_n (-1)^n \dim H^n(M)$.