I have a smooth, compact complex surface $X$, and I need an explicit formula for the Euler characteristic
$$\chi(X, \, S^n \Omega^1\_X),$$
where $S^n$ denotes the symmetric product, in terms of $c\_1(X), c\_2(X)$.


I know how to do the computation, by using the splitting principle in order to calculate the Chern classes $c\_i(S^n \Omega^1\_X)$ and then the Hirzebruch-Riemann-Roch formula, and I did it for small values of $n$. However, extracting a formula for general $n$ seems quite tedious.



> 
> **Question.** Is there a reference where I can find the value of $\chi(X, \, S^n \Omega^1\_X)$? Related to this: is there a reference
> for the Chern classes $c\_1(S^n \Omega^1\_X)$, $c\_2(S^n \Omega^1\_X)$?
> 
> 
> 


Notice that I need an *exact* formula, not an asymptotic one (which I already know).



Exact formula for $\chi(X, \, S^n \Omega^1_X)$
As you say, formulae for $c\_1(\Omega\_X^1)$ and $c\_2(\Omega\_X^1)$ can be obtained from the splitting principle. The following is a more general version of the calculation in [this answer](https://mathoverflow.net/a/351571/21564).



> 
> **Lemma:** Let $V \to X$ be a rank two complex vector bundle. Then $c\_1(S^nV) = \frac{1}{2}n(n+1)c\_1(V)$ and $c\_2(S^nV) = \frac{1}{24}(n-1)n(n+1)(3n+2)c\_1(V)^2 + \frac{1}{6}n(n+1)(n+2)c\_2(V)$.
> 
> 
> 


**Proof:** By the splitting principle, there is a map $p : Y \to X$ such that $p^\*$ is injective on integral cohomology and $p^\*V \cong L\_1\oplus L\_2$, so $p^\*(S^nV) \cong S^n(p^\*V) \cong S^n(L\_1\oplus L\_2)$. In general, we have $S^n(E\_1\oplus E\_2) \cong \bigoplus\_{i+j=n} S^i(E\_1)\otimes S^j(E\_2)$, but for a line bundle $L$ we have $S^k(L) = L^k$ (every tensor on a one-dimensional vector space is symmetric). Therefore


$$S^n(L\_1\oplus L\_2) \cong \bigoplus\_{i+j=n} S^i(L\_1)\otimes S^j(L\_2) \cong \bigoplus\_{i+j=n}L\_1^i\otimes L\_2^j \cong \bigoplus\_{i=0}^nL\_1^i\otimes L\_2^{n-i},$$


so we have


\begin{align\*} c\_1(S^n(L\_1\oplus L\_2)) & = c\_1\left(\bigoplus\_{i=0}^nL\_1^i\otimes L\_2^{n-i}\right)\\
&= \sum\_{i=0}^nc\_1(L\_1^i\otimes L\_2^{n-i})\\
&= \sum\_{i=0}^nic\_1(L\_1) + (n-i)c\_1(L\_2)\\
&= \left(\sum\_{i=0}^ni\right)c\_1(L\_1) + \left(\sum\_{i=0}^nn-i\right)c\_1(L\_2)\\
&= \tfrac{1}{2}n(n+1)(c\_1(L\_1) + c\_1(L\_2))\\
&= \tfrac{1}{2}n(n+1)c\_1(L\_1\oplus L\_2).\end{align\*}


That is, $p^\*(c\_1(S^nV)) = p^\*\left(\frac{1}{2}n(n+1)c\_1(V)\right)$ so the claim for $c\_1$ follows by the injectivity of $p^\*$.


For $c\_2$ we have


\begin{align\*} 
&\, c\_2(S^n(L\_1\oplus L\_2))\\ 
&=\, c\_2\left(\bigoplus\_{i=0}^nL\_1^i\otimes L\_2^{n-i}\right)\\
&=\, \sum\_{0 \leq i < j \leq n}c\_1(L\_1^i\otimes L\_2^{n-i})c\_1(L\_1^j\otimes L\_2^{n-j})\\
&=\, \sum\_{0 \leq i < j \leq n}(ic\_1(L\_1) + (n-i)c\_1(L\_2))(jc\_1(L\_1) + (n-j)c\_1(L\_2))\\
&=\, \sum\_{0 \leq i < j \leq n}ijc\_1(L\_1)^2 + [i(n - j) + (n-i)j]c\_1(L\_1)c\_1(L\_2) + (n-i)(n-j)c\_1(L\_2)^2.\end{align\*}


By writing $\displaystyle\sum\_{0 \leq i < j \leq n}$ as $\displaystyle\sum\_{j=1}^n\sum\_{i=0}^{j-1}$ and using formulae for the sum of the first $k$ integers, squares, and cubes, one can show that


$$\sum\_{0 \leq i < j \leq n}ij = \sum\_{0\leq i < j \leq n}(n-i)(n-j) = \tfrac{1}{24}(n-1)n(n+1)(3n+2)$$


and


$$\sum\_{0 \leq i < j \leq n}i(n-j)+(n-i)j = \tfrac{1}{12}n(n+1)(3n^2+n+2),$$


so


\begin{align\*}
&\, c\_2(S^n(L\_1\oplus L\_2))\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)(c\_1(L\_1)^2 + c\_1(L\_2)^2)\\ 
&\, \ \quad + \tfrac{1}{12}n(n+1)(3n^2+n+2)c\_1(L\_1)c\_1(L\_2)\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)(c\_1(L\_1)^2 + 2c\_1(L\_1)c\_1(L\_2) + c\_1(L\_2)^2)\\ 
&\, \ \quad + \left[\tfrac{1}{12}n(n+1)(3n^2+n+2) - 2\tfrac{1}{24}(n-1)n(n+1)(3n+2)\right]c\_1(L\_1)c\_1(L\_2)\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)(c\_1(L\_1) + c\_1(L\_2))^2\\
&\, \ \quad + \tfrac{1}{12}n(n+1)[3n^2+n+2-(n-1)(3n+2)]c\_1(L\_1)c\_1(L\_2)\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)c\_1(L\_1\oplus L\_2)^2 + \tfrac{1}{6}n(n+1)(n+2)c\_2(L\_1\oplus L\_2).\\
\end{align\*}
As before, the claim for $c\_2$ follows from the injectivity of $p^\*$.$\quad\square$


Let $x\_n$, $y\_n$, and $z\_n$ denote the expressions of $n$ in the formulae above so that $c\_1(S^nV) = x\_nc\_1(V)$ and $c\_2(S^nV) = y\_nc\_1(V)^2 + z\_nc\_2(V)$. Then we have


\begin{align\*}
\operatorname{ch}(S^n\Omega\_X^1) &= \operatorname{rank}(S^n\Omega\_X^1) + c\_1(S^n\Omega\_X^1) + \tfrac{1}{2}(c\_1(S^n\Omega\_X^1)^2 - 2c\_2(S^n\Omega\_X^1))\\
&= \tbinom{n+2-1}{n} + x\_nc\_1(\Omega\_X^1) + \tfrac{1}{2}(x\_n^2c\_1(\Omega\_X^1)^2 - 2y\_nc\_1(\Omega\_X^1)^2 -2z\_nc\_2(\Omega\_X^1))\\
&= \tbinom{n+1}{n} - x\_nc\_1(X) + \tfrac{1}{2}(x\_n^2c\_1(X)^2 - 2y\_nc\_1(X)^2 -2z\_nc\_2(X))\\
&= n + 1 - x\_nc\_1(X) + \tfrac{1}{2}((x\_n^2 - 2y\_n)c\_1(X)^2 - 2z\_nc\_2(X))
\end{align\*}


so


\begin{align\*}
&\, \chi(X, S^n\Omega\_X^1)\\ 
&=\, \int\_X\operatorname{ch}(S^n\Omega\_X^1)\operatorname{Td}(X)\\
&=\, \int\_X\left(n + 1 - x\_nc\_1(X) + \tfrac{1}{2}((x\_n^2-2y\_n)c\_1(X)^2 - 2z\_nc\_2(X))\right)\cdot\\
&\qquad\qquad\qquad\left(1 + \tfrac{1}{2}c\_1(X) + \tfrac{1}{12}(c\_1(X)^2+c\_2(X))\right)\\
&=\, \int\_X\tfrac{1}{12}(n+1)(c\_1(X)^2 + c\_2(X)) -\tfrac{1}{2}x\_nc\_1(X)^2 + \tfrac{1}{2}((x\_n^2-2y\_n)c\_1(X)^2 - 2z\_nc\_2(X))\\
&=\, \left(\tfrac{1}{12}(n+1) - \tfrac{1}{2}x\_n + \tfrac{1}{2}(x\_n^2-2y\_n)\right)\int\_Xc\_1(X)^2 + \left(\tfrac{1}{12}(n+1) - z\_n\right)\int\_Xc\_2(X)\\
&=\, \tfrac{1}{12}(n+1)(2n^2-2n+1)\int\_Xc\_1(X)^2 - \tfrac{1}{12}(n+1)(2n^2+4n-1)\int\_Xc\_2(X).\end{align\*}


This can further be expressed in terms of the Euler characteristic $\chi(X)$ and signature $\sigma(X)$ using the fact that $\int\_Xc\_1(X)^2 = 2\chi(X) + 3\sigma(X)$ and $\int\_Xc\_2(X) = \chi(X)$:


$$\chi(X, S^n\Omega\_X^1) = \tfrac{1}{12}(n+1)(2n^2-8n+3)\chi(X) + \tfrac{1}{4}(n+1)(2n^2-2n+1)\sigma(X).$$


