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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Mar 3, 2014 at 12:00 vote accept Heitor
Nov 29, 2013 at 17:48 comment added abx This is true is up to multiplication by 2 : there is a homomorphism $Nm: \mathrm{Pic}(X)\rightarrow \mathrm{Pic}(Y)$ such that $\pi ^*Nm(L)=L\otimes \theta ^*L$. Thus if $L$ is $\theta $-invariant, $L^2$ belongs to $\pi ^*S_Y$.
Nov 29, 2013 at 17:32 comment added Heitor Dear abx, if I understand correctly what you are saying is that in general $S(\theta)=\pi^\ast S_Y$, where $S_Y$ is the Picard group of $Y$ and $S(\theta)$ are the $\theta$-fixed elements of $S_X$. Could you please explain me why is this true?
Nov 29, 2013 at 17:14 history undeleted abx
Nov 29, 2013 at 17:12 history deleted abx via Vote
Nov 29, 2013 at 17:12 history answered abx CC BY-SA 3.0