# 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?

## 1 Answer

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 \begin{equation} \sum_{p=0}^d (-1)^p\cdot\operatorname{ch}(\Omega_X^p)\cdot\operatorname{td}(\mathcal T_X) = c_d(\mathcal T_X). \end{equation} 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 \begin{equation} \int_X \operatorname{ch}(\Omega_X^p) \cdot \operatorname{td}(\mathcal T_X) = \chi(X,\Omega_X^p). \end{equation}

Finally, we require the Hodge Decomposition Theorem, which we quote from Corollaries 2.6.21 and 3.2.12 in Huybrecht's book as \begin{equation} H^r(X^{\mathrm{an}},\mathbb C) = \bigoplus_{p+q=r} \mathcal H^q(X,\Omega_X^p). \end{equation} 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
• Dear @user125763: I have proved this once for nonsingular complex projective varieties using a couple of big hammers: Borel-Serre identity, Hirzebruch-Riemann-Roch and Hodge Decomposition. If you want, I can show you how that works. I deal with varieties much more than with manifolds, but in the above case the two notions overlap. Concerning complex conjugation, I unfortunately do not know. This sounds like a basic fact though, if it is true. Have you checked standard literature? – Jesko Hüttenhain Jan 25 '15 at 19:24
• Regarding conjugation: I haven't checked properly, I'll have another look at Huybrechts maybe. Regarding the proof: It would be nice if you sketched it for me (and the rest of the community) but no worries – user125763 Jan 25 '15 at 20:20
• @user125763: Not a big problem, I did this in my diplom thesis and this is (almost) a straight copy and paste from the source. Hope it is of any use to you. – Jesko Hüttenhain Jan 25 '15 at 23:00