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Sep 29, 2014 at 14:50 vote accept CommunityBot
Sep 28, 2014 at 18:19 comment added abx On the hyperelliptic curve $X$, there is no $f$-invariant 1-form (such a form would descend to $\mathbb{P}^1$), hence $f^*\alpha =-\alpha $ for all $\alpha \in H^0(X,\Omega ^1_X)$. For an analogous reason, $g^*\omega =-\omega $ on $S$. Thus $pr_1^*\alpha $ and $pr_2^*\omega $ are anti-invariant under $f\times g$, hence their product is invariant.
Sep 26, 2014 at 21:24 comment added user58018 Thanks a lot for the answer. I'm just missing one thing. Why is the 3-form $pr_1^*\alpha\wedge pr_2^{*}\omega$ on $X\times S$ invariant under $f\times g$ ?
Sep 23, 2014 at 4:29 history answered abx CC BY-SA 3.0