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.
Best Answer
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.