Top Chern Class = Euler Characteristic

Question

Let $X$ be a (quasi-)projective, nonsingular, complex variety. Denote by $\mathcal{T}_X$ its tangent sheaf and by $X^{\mathrm{an}}$ its analytification. I am looking for a proof for the equality

        $\displaystyle \int_X c_n(\mathcal{T}_X) = \chi(X^{\mathrm{an}})$,

i.e. the degree of the top chern class is equal to the topological Euler characteristic of $X$. There's Example 3.2.13 in Fulton's book on intersection theory which briefly mentions this, but it does not give a reference. Can someone help me out with one? Thanks in advance.

Answer 1

As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says $$ \chi(V)=\int_{X}{\rm Td}({\rm T}X){\rm ch}(V) $$ where $$\chi(V):=\sum_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes $$ {\rm ch}(\sum_{r}(-1)^r\Omega_X^r){\rm Td}(\Omega^\vee_X)=c^{\rm top}(\Omega^\vee_X) $$ (called the Borel-Serre identity). Here $\Omega_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee_X={\rm T}X$. Plugging the element $\sum_{r}(-1)^r\Omega_{X}^r$ into the HRR theorem, one gets $$ \sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int_{X}c^{\rm top}(TX) $$ and by the Hodge decomposition theorem $$ \sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C})) $$ where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group. The quantity $\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want. The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.

Answer 2

I'll convert my comments to an answer.

Characteristic classes were originally defined as obstruction classes, going back to the work of Whitney and Stiefel. The top Chern class of a complex bundle is the obstruction to an everywhere non-zero vector field, and frequently called the Euler class.

The reason for the name "Euler class" comes from the case where you're looking at the tangent bundle. When you pair this class against the fundamental class of the manifold, via Poincare duality you can interpret this as the signed intersection number of the 0-section of the tangent bundle with (a transverse perturbation of) itself. Since this number does not depend on the perturbation, there are various ways to compute it and check it's the classical Euler characteristic of the base manifold. I like to use a triangulation, and adapting a vector field to it so that the zeros of the vector field correspond to the (barycentres) of the cells of the triangulation.

The core geometric argument here is the Poincare-Hopf index theorem.

Perhaps there are more adapted ways to get the result in the context you like (I don't know), but I think this is the historical route to the result.