Question

Let $X$ be a smooth Calabi-Yau 3-fold with Kahler form $w$,

It is true that $\int c_2(TX) \wedge w \geq 0$ (for any Kahler form $w$ on $X$).

Proof via algebraic geometry is rather difficult. Some where It was saying that for such $X$

$$\int c_2(TX) \wedge w = \int \left\| R \right\|^2 dvol $$

where $R$ is the curvature tensor; so it is non negative and zero only if $X$ is abelian variety .

Can any body write down in local coordinates or ...(or refer to somewhere) why the identity above is true.

Answer 1

The calculation in local coordinates is not too hard and it works in any dimension $n$, namely if $c_1(M)=0$ then $\int_M c_2(M)\wedge\omega^{n-2}\geq 0$ for any Kahler metric $\omega$. Of course you are free to assume that the Kahler metric $\omega$ is Ricci-flat (by Yau's theorem), and in this case the integral is equal to the $L^2$ norm of the Riemann curvature tensor up to a factor.

To see this, let $\Omega_i^j=\frac{\sqrt{-1}}{2\pi} R^j_{i k\overline{\ell}}dz^k\wedge d\overline{z}^\ell$ denote the curvature form, then by Chern-Weil theory the form $Tr(\Omega\wedge\Omega)=\sum_{k,i} \Omega_i^k\wedge\Omega_k^i =\frac{(\sqrt{-1})^2}{4\pi^2}R^k_{ip\overline{q}} R^i_{kr\overline{s}}dz^p\wedge d\overline{z}^q\wedge dz^r\wedge d\overline{z}^s$ represents $c_1^2(M)-2c_2(M)=-2c_2(M)$. You can assume that the metric $\omega$ is the identity at one point so $\omega=\sqrt{-1}\sum_i dz^i\wedge d\overline{z}^i$, and then at that point you compute $$n(n-1)Tr(\Omega\wedge\Omega)\wedge\omega^{n-2}=\sum_{p\neq r}(R^k_{ip\overline{p}} R^i_{kr\overline{r}}-R^k_{ip\overline{r}} R^i_{kr\overline{p}})\omega^n$$ $$=\sum_{p,r}(R^k_{ip\overline{p}} R^i_{kr\overline{r}}-R^k_{ip\overline{r}} R^i_{kr\overline{p}})\omega^n$$ $$=(|\textrm{Ric}|^2-|\textrm{Rm}|^2)\omega^n=- |\textrm{Rm}|^2\omega^n.$$ Integrating this you get what you want, with equality if and only if $\omega$ is flat and so $M$ is finitely covered by a complex torus.

A similar calculation works for any compact Kahler-Einstein manifold, and it can be used to prove the Miyaoka-Yau inequality for manifolds with ample canonical bundle. A reference for a general statement (I think this is not the earliest paper where this result appears) is:

Chen, Bang-yen; Ogiue, Koichi, Some characterizations of complex space forms in terms of Chern classes, Quart. J. Math. Oxford Ser. (2) 26 (1975), no. 104, 459–464.

Answer 2

The reason is that, for a Calabi-Yau $3$-fold, the curvature tensor $R$ is an irreducible representation of $SU(3)$. Since the form representing the Chern class $c_2(TX)$, when computed with respect to the Calabi-Yau metric, is clearly quadratic in the curvature tensor and since $c_2(TX)\wedge\omega$ is a $(3,3)$-form that is a quadratic form in the curvature times the metric volume form, the irreducibility of the curvature implies that there is a universal constant $\lambda$ such that $c_2(TX)\wedge\omega = \lambda\ \|R\|^2\ dvol$ for all Calabi-Yau $3$-folds. Checking on a single example (say, a quintic hypersurface in $\mathbb{CP}^4$) shows that $\lambda >0$.

The irreducibility comes from the fact (due, I believe, to Bochner) that, for a general Kähler metric on an $n$-fold, the curvature tensor breaks up into $3$ irreducible parts under $U(n)$: the scalar curvature, the traceless Ricci curvature, and the Bochner curvature. For a Calabi-Yau metric, the first two vanish. Thus, the above formula actually generalizes to $c_2(TX)\wedge\omega^{n-2} = \lambda_n\ \|R\|^2\ dvol$ for all Calabi-Yau $n$-folds, where $\lambda_n>0$ for all $n\ge 2$.