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Feb 27, 2016 at 23:47 history bounty ended CommunityBot
Feb 23, 2016 at 18:11 comment added HYL A nonzero rational section need not exist. Take for example a finite group $G$ acting trivially on $X$ but non-trivially on its trivial bundle. In this case, any rational section of the quotient bundle over $[X/G]$ is $0$ since it is identified with a $G$-invariant rational section of $\mathcal{O}_X$.
Feb 22, 2016 at 17:58 comment added O-Ren Ishii I agree that the pullback of a rational section is then well-defined. However, the existence of a single nonzero rational section is not clear due to the required compatibility with $p$.
Feb 20, 2016 at 20:42 history edited HYL CC BY-SA 3.0
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Feb 20, 2016 at 20:28 history edited HYL CC BY-SA 3.0
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Feb 20, 2016 at 19:26 comment added HYL You're right, what we can only define if we take all possible $\mathscr{X} \to S$ into consideration is the class of $\mathscr{L}$ inside $\mathrm{Pic}(\mathscr{M})_\mathbf{Q}$, but this will be enough for us to define the degree of $\mathscr{L}$. To define what rational sections are, as you said we need to restrict to those $\mathscr{X} \to S$ such that are the induced map $S \to \mathcal{M}$ is étale. In this case, the morphism $p : S' \to S$ in every base change diagram above is étale as well, so the pullback of a rational section on $S$ to $S'$ is well-defined.
Feb 20, 2016 at 18:19 comment added O-Ren Ishii Thank you for your answer. Why does a compatible system of $s_f$ exist? The way you phrase it, it seems that every $s_f$ needs to be everywhere defined (take $S'$ to be a variable point of $S$). To avoid this problem you may restrict to $S$ that are etale over $\mathscr{M}$, but then I still do not see how to get the $s_f$ (this is related to the pushforward of $\mathscr{L}$ to $M$ being generically a line bundle).
Feb 20, 2016 at 14:29 history edited HYL CC BY-SA 3.0
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Feb 20, 2016 at 14:24 history answered HYL CC BY-SA 3.0