Let $X$ be a projective variety.
Consider $Sym^2X$, the quotient of $X \times X$ by the involution $(x, x') \mapsto (x', x)$.
What is the relation between the (mixed) Hodge numbers of $Sym^2 X$ and the ones of $X$? Is there a simple formula?
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1 Answer
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For a $V$-manifold the Hodge structure in each degree is pure (see Peters-Steenbrink, § 2.5 in Mixed Hodge structures, Springer). Thus $H^{p,q}(Sym^2X)$ is just the part of $H^{p,q}(X\times X)$ invariant under the involution. It is easy to compute the dimension using the Künneth formula. For instance, if I didn't make a mistake, you find $h^{2,0}(Sym^2X)= h^{2,0}(X)+\frac{1}{2} h^{1,0}(X)(h^{1,0}(X)-1)$, $h^{1,1}(Sym^2X)= h^{1,1}(X)+ h^{1,0}(X)^2 $.
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3$\begingroup$ I think you're assuming $X$ smooth (but maybe "pokerhodge" is doing that too without saying it). What you're saying is true also in the context of mixed Hodge theory, though. If $X$ is any complex algebraic variety and $G$ is a finite group then the rational MHS on $H^\bullet(X/G)$ coincides with the MHS on the $G$-invariants on $H^\bullet(X)$. $\endgroup$ Commented Oct 31, 2013 at 9:18
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$\begingroup$ No, I wasn't asuming $X$ smooth! Thanks to abx and Dan $\endgroup$ Commented Oct 31, 2013 at 9:22