Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
    
Can you just say what is $R_{ik\bar{l}}^{j}$ in your notation? because in mine that is trivial (in my notattion $R(\frac{\partial}{\partial{z}_i},\frac{\partial}{\partial{\bar{z}}_j}) \frac{\partial}{\partial{z}_k}= \sum R_{i\bar{j}k}^{l} \frac{\partial}{\partial{z}_l}$ ) –  Mohammad F. Tehrani Jul 8 '11 at 17:01
    
I am using the notation from Kobayashi's book "Differential geometry of complex vector bundles", for example (page 12), which is $R(\partial_k, \overline{\partial}_\ell)\partial_i=R^j_{ik\overline{\ell}}\partial_j.$ –  YangMills Jul 10 '11 at 9:21
add comment

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 $n\ge 2$.

share|improve this answer
    
Is not it possible to write down every thing explicitly in a local coordinate and prove it this way. When I write it down I see a huge summation from which I have no idea how to get that identity. I don't want to use any of those non trivial facts you mentioned. Can you say why the irreducibility of the curvature implies that there is a universal constant $\lambda$ such that $c_2(TX)∧ω=\lambda \left\|R\right\|^2$ dvol$ for all Calabi-Yau 3-folds. –  Mohammad F. Tehrani Jul 7 '11 at 17:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.