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Timeline for Hodge numbers of symmetric squares

Current License: CC BY-SA 3.0

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Oct 31, 2013 at 9:22 comment added pokerhodge No, I wasn't asuming $X$ smooth! Thanks to abx and Dan
Oct 31, 2013 at 9:18 comment added Dan Petersen 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 $H^\bullet(X)$.
Oct 31, 2013 at 9:08 history answered abx CC BY-SA 3.0