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Let $C$ be a smooth projective curve. Let $G$ be a finite group which acts on $C$. Let $C'=C/G$ the quotient of the action, which is a smooth curve. Then $f:C\rightarrow C'$ is a finite, possibly ramified morphism.

1) Let $L$ be a line bundle on $C$ which admits a $G$-action. What is $(p_*L)^{G}$? Suppose $L=p^*L'$ for some line bundle $L'$, then $(p_*L)^{G}=L'$.

2) Suppose $L$ does not come from below, is $(p_*L)^G$=0?

Do we have a characterization as to when $(p_*L)^G=0$?

I would like to understand what happens both when $f$ is unramified and $f$ is ramified.

Edit: It looks like $p^*(p_*L)^G\subset L$. Also that $(p_*L)$ and $(p_*L)^G$ are locally free. By the above inclusion, $(p_*L)^G$ can be at most of rank one.

3) Consider the quotient $p_*L/(p_*L)^G$, is that locally free as well?

It would be great if someone can direct me to a reference where such things are explained.

user52991
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  • First you can find an open over $(U_i){i\in I}$ of your curve downstairs such that $p^{-1}(U_i)=\coprod{g} gV_i$ with $V_i\to U_i$ being a homeomorphism (recall that the algebraic quotient commutes with the topological one).

    To define $(p_*L)^G$ it suffices to define its restriction to each $U_i$ and gives a glueing condition.

    For any $W\subset U_i$ open, you have a natural action of $G$ on the sections of $L$ on $p^{-1}(W)$ given by $(g.s)(w)=g.s(g^{-1}.W)$. Then $(p_*L)^G(W)$ is defined to be the subspace of invariant sections by this action.

     – Ahr May 18 '17 at 11:01
  • You can check that this defines a sheaf $(p_*L)^G_{|U_i}$ on $U_i$.

    You have an obvious isomorphism of sheaves $(p_L)^G_{|U_i\cap U_j}\to (p_L)^G_{|U_j\cap U_i}$ that glue together, and you thus get a global sheaf $(p_*L)^G$.

    Now, in the case where $G$ acts freely the projection will be étale and $L$ will always come from below as $p^(p_L)^G\to L$ is an isomorphism.

     – Ahr May 18 '17 at 11:01
  • No, I don't think so. Unramified points will be mapped to unramified point by the group action. So you can find an invariant open affine subset that is not ramified if you denote $u:U\to C$ the open immersion we obvisouly have $p'(u^L)^G=u'^*(pL^G)$ with obvious notations. Thus $p_L^G$ has no reason to be zero. – Ahr May 21 '17 at 10:20
  • @A.Rod, we always have the inclusion, $p^(p_L)^G\hookrightarrow L$ right? This will tell us $(p_*L)^G$ has rank at most one. Isn't that right? – user52991 May 21 '17 at 18:39
  • In fact I realized that as $L$ is locally free then $L$ is always isomorphic to $p^(p_L)^G$ whether the projection is étale or not. – Ahr May 22 '17 at 08:55
  • @A.Rod, your comment means that every vector bundle descends. I don't think that's true. – user52991 May 22 '17 at 12:19
  • You may asume that $C=\text{Spec }A$ is affine and then $C'=\text{Spec }A^G$ and $L$ is a free module over $A$ generated by $s$. As the action on $L$ and $C$ are compatible then $L$ is actually isomorphic to $A$ as a $G$-module and then $\pi_(L)^G$ is $L^G\simeq A^G$ as an $A^G$-module and $\pi^\pi_*L^G$ is simply $L^G\otimes_{A^G} A$ isomorphic to $L$. This may fail for a general equivariant sheaf. – Ahr May 22 '17 at 13:04

1 Answers1

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I assume you suppose the action of $G$ to be faithful. If the action is free, a structure of $G$-equivariant line bundle on $L$ is the same as a descent data for $$C \to C':= C/G : L \text{ is the pullback of } L':=(p_*L)^G.$$ I would need to think more when $C \to C'$ is not tame. Assuming tame, I like to think of $L$ as a line bundle on $C'$, viewed as an orbifold. One will have $L'$ is a line bundle on $C'$ and $L$ is $(p^*L')(D)$, where $D$ is on the ramification locus and: for each $x'$ in $C'$ above which $C$ ramifies with ramification $d(x')$ there is $n(x')$,$$0 \le n(x') < d(x')$$such that at $x$ above $x'$, $D$ is $n(x')x$. Note that as such a divisor $D$ is stable by $G$, a $L$ so constructed is clearly $G$-equivariant. Given $L'$, all the $L$ obtained from different $D$ as above have the same $(p_*)^G$. In the orbifold language, as a line bundle on the orbifold $C'$, $L$ is$$L'\left(\sum n(x')/d(x').x'\right).$$

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