## 77.10 Equivariant quasi-coherent sheaves

Please compare with Groupoids, Section 39.12.

Definition 77.10.1. Let $B \to S$ as in Section 77.3. Let $(G, m)$ be a group algebraic space over $B$, and let $a : G \times _ B X \to X$ be an action of $G$ on the algebraic space $X$ over $B$. An $G$-equivariant quasi-coherent $\mathcal{O}_ X$-module, or simply a equivariant quasi-coherent $\mathcal{O}_ X$-module, is a pair $(\mathcal{F}, \alpha )$, where $\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ X$-module, and $\alpha$ is a $\mathcal{O}_{G \times _ B X}$-module map

$\alpha : a^*\mathcal{F} \longrightarrow \text{pr}_1^*\mathcal{F}$

where $\text{pr}_1 : G \times _ B X \to X$ is the projection such that

1. the diagram

$\xymatrix{ (1_ G \times a)^*\text{pr}_2^*\mathcal{F} \ar[r]_-{\text{pr}_{12}^*\alpha } & \text{pr}_2^*\mathcal{F} \\ (1_ G \times a)^*a^*\mathcal{F} \ar[u]^{(1_ G \times a)^*\alpha } \ar@{=}[r] & (m \times 1_ X)^*a^*\mathcal{F} \ar[u]_{(m \times 1_ X)^*\alpha } }$

is a commutative in the category of $\mathcal{O}_{G \times _ B G \times _ B X}$-modules, and

2. the pullback

$(e \times 1_ X)^*\alpha : \mathcal{F} \longrightarrow \mathcal{F}$

is the identity map.

For explanation compare with the relevant diagrams of Equation (77.8.1.1).

Note that the commutativity of the first diagram guarantees that $(e \times 1_ X)^*\alpha$ is an idempotent operator on $\mathcal{F}$, and hence condition (2) is just the condition that it is an isomorphism.

Lemma 77.10.2. Let $B \to S$ as in Section 77.3. Let $G$ be a group algebraic space over $B$. Let $f : X \to Y$ be a $G$-equivariant morphism between algebraic spaces over $B$ endowed with $G$-actions. Then pullback $f^*$ given by $(\mathcal{F}, \alpha ) \mapsto (f^*\mathcal{F}, (1_ G \times f)^*\alpha )$ defines a functor from the category of quasi-coherent $G$-equivariant sheaves on $Y$ to the category of quasi-coherent $G$-equivariant sheaves on $X$.

Proof. Omitted. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).