39.10 Actions of group schemes
Let $(G, m)$ be a group and let $V$ be a set. Recall that a (left) action of $G$ on $V$ is given by a map $a : G \times V \to V$ such that
(associativity) $a(m(g, g'), v) = a(g, a(g', v))$ for all $g, g' \in G$ and $v \in V$, and
(identity) $a(e, v) = v$ for all $v \in V$.
We also say that $V$ is a $G$-set (this usually means we drop the $a$ from the notation – which is abuse of notation). A map of $G$-sets $\psi : V \to V'$ is any set map such that $\psi (a(g, v)) = a(g, \psi (v))$ for all $v \in V$.
Definition 39.10.1. Let $S$ be a scheme. Let $(G, m)$ be a group scheme over $S$.
An action of $G$ on the scheme $X/S$ is a morphism $a : G \times _ S X \to X$ over $S$ such that for every $T/S$ the map $a : G(T) \times X(T) \to X(T)$ defines the structure of a $G(T)$-set on $X(T)$.
Suppose that $X$, $Y$ are schemes over $S$ each endowed with an action of $G$. An equivariant or more precisely a $G$-equivariant morphism $\psi : X \to Y$ is a morphism of schemes over $S$ such that for every $T/S$ the map $\psi : X(T) \to Y(T)$ is a morphism of $G(T)$-sets.
In situation (1) this means that the diagrams
39.10.1.1
\begin{equation} \label{groupoids-equation-action} \vcenter { \xymatrix{ G \times _ S G \times _ S X \ar[r]_-{1_ G \times a} \ar[d]_{m \times 1_ X} & G \times _ S X \ar[d]^ a \\ G \times _ S X \ar[r]^ a & X } } \quad \quad \vcenter { \xymatrix{ G \times _ S X \ar[r]_-a & X \\ X\ar[u]^{e \times 1_ X} \ar[ru]_{1_ X} } } \end{equation}
are commutative. In situation (2) this just means that the diagram
\[ \xymatrix{ G \times _ S X \ar[r]_-{\text{id} \times \psi } \ar[d]_ a & G \times _ S Y \ar[d]^ a \\ X \ar[r]^\psi & Y } \]
commutes.
Definition 39.10.2. Let $S$, $G \to S$, and $X \to S$ as in Definition 39.10.1. Let $a : G \times _ S X \to X$ be an action of $G$ on $X/S$. We say the action is free if for every scheme $T$ over $S$ the action $a : G(T) \times X(T) \to X(T)$ is a free action of the group $G(T)$ on the set $X(T)$.
Lemma 39.10.3. Situation as in Definition 39.10.2, The action $a$ is free if and only if
\[ G \times _ S X \to X \times _ S X, \quad (g, x) \mapsto (a(g, x), x) \]
is a monomorphism.
Proof.
Immediate from the definitions.
$\square$
Comments (2)
Comment #5569 by Lucas das Dores on
Comment #5750 by Johan on