Euler-Poincare characteristic of $L \otimes Sym^k(\Omega^1_S)$

Question

Let $S$ be a smooth complex rational surface, and $L$ be an ample line bundle on $S$.

I want to compute the Euler-Poincare characteristic of $L \otimes Sym^k(\Omega^1_S)$, i.e., $$\chi(S, L \otimes Sym^k(\Omega^1_S)), \ \ k > 0.$$

How can I calculate this value?

Thanks in advance.

Answer 1

From your comment, you need to know how to calculate Chern classes of symmetric powers. The surefire way to do this (albeit, sometimes tedious) is to use splitting principle and Whitney's theorem Let me illustrate.

If $E$ is a vector bundle, say of rank $r$, splitting principle says, you can pretend it is a direct sum of line bundles, $L_i$ and thus the Chern polynomial is, $c_t(E)=\prod c_t(L_i)$ and $c_t(L_i)=1+c_1(L_i)t$. (Be careful, that $L_i$'s are virtual).

So, if you want to calculate, say the Chern polynomial of $S^2E$ of a rank 2 vector bundle on a surface (as an illustration), by our pretense, we assume $E$ is a direct sum of two line bundles, $L,M$ with $c_1(L)=a, c_1(M)=b$. Then $c_1(E)=a+b, c_2(E)=ab$.

$S^2E$ is then a direct sum of $L^2, L\otimes M, M^2$, whose first Chern classes are $2a, a+b, 2b$.Thus $c_t(S^2E)=(1+2at)(1+(a+b)t)(1+2bt)$. You can expand this and write it in terms of $c_i(E)$ (no $a,b$). I hope the method is clear.