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 an $n$-dimensional compact Kahler manifold with negative first Chern class. Are its Chern numbers $\prod_{i=1}^{n-1} c_i^{k_i}$, over $k _i \geq 0$ with $\sum ik_i = n$, bounded in terms of $(-1)^n e(X) = (-1)^nc_n = c_n(\Omega_X^1)$?

Are there, moreover, only finitely many deformation types for $|e(X)| = (-1)^{n}e(X)$ bounded?

For $n = 2$, we have of course the famous Bogomolov-Miyaoka-Yau bound $c_1^2 \leq 3c_2$.

Remark. One may also include the convention $c_0 := 1$, for which the answer is positive since $(-1)^nc_n > 0$, in order to cover the prototypical observation that the Euler number of a hyperbolic surface is negative. In view of this example, I would also like to extend the original question to the realm of log manifolds and hyperbolicity.

share|improve this question
    
Should your product be from $i=1$ until $i=n$, and your sum read $\sum_i k_i =n$? –  Ari Feb 9 '13 at 17:39
    
I ask about bounding the degree-$n$ combinations of the $c_i$ in terms of $c_n$. Now, $c_i$ has degree $i$. Since the only degree-$n$ combination involving $c_n$ is $c_n$ itself, I removed $i=n$ from the product. –  Vesselin Dimitrov Feb 9 '13 at 22:40
    
Ow I misread your question. Thanks for the clarification. –  Ari Feb 10 '13 at 10:37
1  
In this related question mathoverflow.net/questions/26586/… Dmitri points out that the answer is definitely NO if you drop the assumption of negative first Chern class. –  YangMills Feb 11 '13 at 20:35
2  
Proposition 3.13 in this paper of Catanese-Schneider dx.doi.org/10.1007/BF01444736 gives universal bounds for the Chern numbers (assuming $K_X$ ample like you want) in terms of $(−1)^n c^n_1=K^n_X$. Does this help? When $n=3$, you also have the Yau inequality which bounds −$c^3_1\leq(8/3)(−c_1)c_2$, but then? –  YangMills Feb 12 '13 at 1:46
show 1 more comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.