78.15 Groupoids and group spaces
Please compare with Groupoids, Section 39.16.
Lemma 78.15.1. Let $B \to S$ as in Section 78.3. Let $(G, m)$ be a group algebraic space over $B$ with identity $e_ G$ and inverse $i_ G$. Let $X$ be an algebraic space over $B$ and let $a : G \times _ B X \to X$ be an action of $G$ on $X$ over $B$. Then we get a groupoid in algebraic spaces $(U, R, s, t, c, e, i)$ over $B$ in the following manner:
We set $U = X$, and $R = G \times _ B X$.
We set $s : R \to U$ equal to $(g, x) \mapsto x$.
We set $t : R \to U$ equal to $(g, x) \mapsto a(g, x)$.
We set $c : R \times _{s, U, t} R \to R$ equal to $((g, x), (g', x')) \mapsto (m(g, g'), x')$.
We set $e : U \to R$ equal to $x \mapsto (e_ G(x), x)$.
We set $i : R \to R$ equal to $(g, x) \mapsto (i_ G(g), a(g, x))$.
Proof.
Omitted. Hint: It is enough to show that this works on the set level. For this use the description above the lemma describing $g$ as an arrow from $v$ to $a(g, v)$.
$\square$
Lemma 78.15.2. Let $B \to S$ as in Section 78.3. Let $(G, m)$ be a group algebraic space over $B$. Let $X$ be an algebraic space over $B$ and let $a : G \times _ B X \to X$ be an action of $G$ on $X$ over $B$. Let $(U, R, s, t, c)$ be the groupoid in algebraic spaces constructed in Lemma 78.15.1. The rule $(\mathcal{F}, \alpha ) \mapsto (\mathcal{F}, \alpha )$ defines an equivalence of categories between $G$-equivariant $\mathcal{O}_ X$-modules and the category of quasi-coherent modules on $(U, R, s, t, c)$.
Proof.
The assertion makes sense because $t = a$ and $s = \text{pr}_1$ as morphisms $R = G \times _ B X \to X$, see Definitions 78.10.1 and 78.12.1. Using the translation in Lemma 78.15.1 the commutativity requirements of the two definitions match up exactly.
$\square$
Comments (0)