Exact formula for $\chi(X, \, S^n \Omega^1_X)$

Question

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).

Answer 1

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.

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 $\int_Xc_2(X) = \chi(X)$:

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

Answer 2

Just for reference: using the Schubert2 package from Macaulay2 this can be quite effortlessly done.

i1 : needsPackage "Schubert2";

i2 : X = abstractVariety(2, QQ[n,c1,c2,Degrees=>{0,1,2}]);

i3 : T = abstractSheaf(X, Rank=>2, ChernClass=>1+c1+c2);

i4 : X.TangentBundle = T;

i5 : chi symmetricPower(n, cotangentBundle X)

              1 3  2   1 3      1    2   1 2      1  2   1        1
o5 = integral(-n c1  - -n c2 - --n*c1  - -n c2 + --c1  - -n*c2 + --c2)
              6        6       12        2       12      4       12

If one looks at the implementation, the $n$-th symmetric power of a bundle $\mathcal F$ is computed using the pushforward of the relative $\mathcal{O}(n)$ on the projectivization $\mathbf{P}(\mathcal F^\vee)$ (as already mentioned in the other answer).

Answer 3

Following the method you indicate in your question, if we let $x,y$ be the chern roots then determining $c_2(Sym^{n} \Omega)$ corresponds to expanding $$\sum_{0 \leq i\neq i' \leq n} (ix + (n-i)y)(i'x + (n-i')y)$$ in terms of the symmetric functions $xy$ (corresponding to $c_2(X)$) and $(x+y)^2$ (corresponding to $c_1(X)^2$). It seems easier to use the function $x^2 + y^2$ instead of $(x+y)^2$ corresponding to $c_1(X)^2 - 2c_2(X)$.

Then we have to determine the two sums $$\sum_{0\leq i\neq i' \leq n} ii'$$ and $$\sum_{0 \leq i \neq i' \leq n} i(n-i')+ i'(n-i).$$

There must be more elegant ways of doing this, but the brute force one is to argue that these are polynomials in $n$ of degree $4$ and compute enough values to determine them.

Answer 4

Concerning the first part of the question, my initial recollection was that Miyaoka's famous paper On the Chern numbers of surfaces of general type has the formulas you want. In fact, he doesn't quite have them, nevertheless his basic approach should give you the formula. Set $Y= \mathbb{P}(\Omega_X^1)$, with $L= \mathcal{O}_{Y}(1)$ the relative tautological bundle. Then the Leray spectral sequence, together with Hirzebruch-Riemann-Roch and lemma 5 of Miyaoka yields $$\chi(X,S^n\Omega_X^1)= \chi(Y, L^n)= \int ch(L^n)\cdot todd(Y)$$ $$=\frac{n^3}{6}(c_1(X)^2-c_2(X)) + \ldots$$ Sorry, I'm too lazy to compute the rest, but hopefully that should get you started.