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