Let $X$ be an algebraic variety (i.e. an integral separated scheme of finite type over an algebraically closed field $k$) and let $G$ be a finite group of automorphisms of $X$. Suppose (as we may in the case of quasi-projective varieties) that for any $x$ in $X$, the orbit $G_x$ of $x$ is contained in an affine open subset of $X$. A classical result states that there exists a (unique) $k$-variety $Y$ together with a finite, surjective and separable morphism $\pi \colon X\to Y$ such that:

1) As topological space $(Y,\pi)$ is the quotient of $X$ for the action of $G$.

2)There is a natural isomorphism $\mathcal{O}_Y \to \pi_{*}(\mathcal{O}_X)^{G}$.

In this setting, let $\mathcal{F}$ be a coherent sheaf on $Y$. Since for any $g$ in $G$ we have a commutative diagram:

\begin{array}{ccc} X & \to^{g} & X \\ \downarrow^{\pi} & & \downarrow^{\pi}\\ Y & \to^{id} & Y \end{array}

there should be a map between $\pi^{*}\mathcal{F}$ and $\pi^{*}\mathcal{F}$ induced by $g$, that is a natural automorphism. However, I can't really understand what the map is supposed to do... The conclusion is that $G$ acts on the pullback of the sheaf $\mathcal{F}$ in a compatible manner (with respect to the action on $X$), but I can hardly imagine what is really happening here, beyond the formal arguments. Can anybody explain this in a more concrete way?

P.S. Sorry for the bad $\TeX$ typesetting: I suppose that the above diagram should be understood as a commutative triangle.


1 Answer 1


The slogan here is that questions about the (non-equivariant) geometry of $Y$ have answers coming from the $G$-equivariant geometry of $X$.

It's probably helpful to consider first the case of a topological space $X$ with a free action of a finite group $G$, and a vector bundle $E$ on $Y=X/G$. In this case, the pullback is by definition the vector bundle of pairs $(x,e)$ where $x \in X$ and $e$ is in the fiber of $e$ over the orbit of $x$. The group $G$ evidently acts on $X \times E$ via its action on $X$, and this set of pairs is stable.

Now, to motivate the definition of a $G$-equivariant coherent sheaf, let's begin with a fiber bundle $E$ on a topological space $X$ with a continuous action of a topogical group $G$. There are two maps $\pi,a:G \times X \rightarrow X$, the projection $\pi$ onto the second factor and the action map $a$. We can pull $E$ back by these to obtain two bundles $\pi^*E$ and $a^*E$ on $G\times X$. Concretely, we have $$\pi^* E=\{(g,x,e) \ | \ e \in \text{fiber through} \ x \} \quad \text{and} \quad a^* E=\{(g,x,e) \ | \ e \in \text{fiber through} \ gx \},$$ and the map $\alpha:(g,x,e) \mapsto (g,x,ge)$ defines an isomorphism from $\pi^* E$ onto $a^*E$. The associativity for the action of $G$ on $E$ and the fact that $1 \in G$ acts by the identity force this map $\alpha$ to satisfy some additional requirements, which I will be happy to tell you about if this is what you want to know.

This is the official definition of $G$-equivariant sheaf on $X$, which works in any geometric category (e.g., schemes, or algebraic spaces): it is a sheaf on $X$ together with an isomorphism $\alpha:\pi^* E \rightarrow a^* E$ satisfying additional conditions corresponding to $g(h(x))=(gh)(x)$ and $1 x=x$.

At the level of sections, given a $G$-equivariant sheaf $E$ and a section $f$ of $E$ over an open subset $U$ of $X$, we obtain a section $gf$ of $E$ over $gU$. But of course we want this to vary continuously in $g$, $f$, and $U$, and the definition we have indicated above packages all this conveniently.

  • $\begingroup$ In the case of a topological space without extra structure (that is the case of a scheme) this is probably a good example. My problems arise when I want to understand the relations between the two structure sheaves. In particular, what do we mean when we say that the action of $G$ on the pullback of $\mathcal{F}$ is compatible with the action of $G$ on $X$? What happens at the level of sections? $\endgroup$
    – FedeB
    Commented Mar 4, 2012 at 21:40
  • $\begingroup$ Dear @FedeB, Well, I have given more information, but it's still not clear to me from the text of your question what you really want to know. Please let me know if I can be of further assistance. $\endgroup$
    – Stephen
    Commented Mar 4, 2012 at 23:21
  • $\begingroup$ Thank you, this is much better! Would you mind writing down the additional requirements you mentioned in the answer? $\endgroup$
    – FedeB
    Commented Mar 5, 2012 at 8:58
  • $\begingroup$ Dear @Steve, do you have any readable reference for the subject? I also have a small related question: when you write $G\times X\to X$ I suppose that you mean the product as topological spaces (with $G$ endowed with a discrete topology. In the scheme setting (say, $X$ a $k$-scheme), do you put a kind of standard $k$-scheme structure on $G$ in order to consider the fiber product $G \times_k X$ as schemes over $k$? $\endgroup$
    – FedeB
    Commented Mar 5, 2012 at 11:03

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