# the top chern class of the holomorphic tangent bundle is the euler class

Is the following true?

Let X be a complex manifold of complex dimension d and let V denote its holomorphic tangent bundle (ie it's $T^{1,0} \subset T \otimes_R C$, where T is the tangent bundle of X as a smooth manifold). Is $c_d(V) = e(X) := e(T)$?

If so, do you have a reference?

The answer is affirmative if $X$ is compact and compactness is also necessary. A reference for the first statement can be found in the discussion after Corollary 5.1.4, the item (ii) (page 235) in the book Complex Geometry by Huybrechts. This is also known as the Gauss-Bonnet-Formula and this Wikipedia Link contains a counterexample for the noncompact case.
Edit. I give a proof for the Gauss-Bonnet formula. I use some standard terminology: $X$ is now a smooth complex projective variety over $\mathbb C$ and we denote by $\Omega_X$ its sheaf of relative differentials. We denote sheaf cohomology by $\mathcal H^\bullet$ and singular cohomology by $H^\bullet$. Also, $\mathcal T_X$ is the tangent sheaf of $X$, $\operatorname{ch}$ is the exponential Chern character, $\operatorname{td}$ the Todd class and $\chi$ the Euler characteristic. Finally, $X^{\mathrm{an}}$ denotes the analytification of $X$, i.e. the GAGA-associated complex manifold.
Write $\Omega_X^p:=\bigwedge^p\Omega_X$ for the $p$-th exterior power of $\Omega_X$. We will use the Borel-Serre-Identity, given in Fulton's Intersection Theory as Example 3.2.5. It says $$\sum_{p=0}^d (-1)^p\cdot\operatorname{ch}(\Omega_X^p)\cdot\operatorname{td}(\mathcal T_X) = c_d(\mathcal T_X).$$ Note that $d$ is the rank of $\Omega_X$ and $\Omega_X^\vee=\mathcal T_X$ by definition. As a second tool, we require the Hirzebruch-Riemann-Roch Theorem to conclude $$\int_X \operatorname{ch}(\Omega_X^p) \cdot \operatorname{td}(\mathcal T_X) = \chi(X,\Omega_X^p).$$
Finally, we require the Hodge Decomposition Theorem, which we quote from Corollaries 2.6.21 and 3.2.12 in Huybrecht's book as $$H^r(X^{\mathrm{an}},\mathbb C) = \bigoplus_{p+q=r} \mathcal H^q(X,\Omega_X^p).$$ Putting it all together, we obtain \begin{align*} \int_X c_d(\mathcal T_X) &= \sum_{p=0}^d (-1)^p\cdot \int_X \operatorname{ch}(\Omega_X^p)\cdot\operatorname{td}(\mathcal T_X) && \text{by Borel-Serre} \\ &= \sum_{p=0}^d (-1)^p \cdot \chi(X,\Omega_X^p) && \text{by HRR} \\ &= \sum_{p,q} (-1)^{p+q}\cdot\operatorname{rank}(\mathcal H^q(X,\Omega_X^p)) && \\ &= \sum_{r=0}^d (-1)^r \cdot H^r(X^{\mathrm{an}},\mathbb C) && \text{by the Hodge Decomposition} \\ &= \chi(X^{\mathrm{an}}). \end{align*}
• Thanks. The way I was attempting to do it was by using the formula $c_{top}(E \otimes C) = e(E) \cup e(E)$ for real vector bundles and the Hodge decomposition of $T \otimes \C$ but no luck. By the way, if $V$ is a complex vector bundle and $\bar{V}$ is its conjugate, is $c_k(\bar{V}) = (-1)^kc_k(V)$? – user125763 Jan 24 '15 at 17:00