Example 95.17.1. Let $S$ be a scheme. Let $G$ be a commutative group. Let $X \to S$ be a scheme over $S$. Let $a : G \times X \to X$ be an action of $G$ on $X$. For $g \in G$ we denote $g : X \to X$ the corresponding automorphism. In this case the inertia stack of $[X/G]$ (see Remark 95.15.5) is given by

where, given an element $g$ of $G$, the symbol $X^ g$ denotes the scheme $X^ g = \{ x \in X \mid g(x) = x\} $. In a formula $X^ g$ is really the fibre product

Indeed, for any $S$-scheme $T$, a $T$-point on the inertia stack of $[X/G]$ consists of a triple $(P/T, \phi , \alpha )$ consisting of an fppf $G$-torsor $P\to T$ together with a $G$-equivariant morphism $\phi : P \to X$, together with an automorphism $\alpha $ of $P\to T$ over $T$ such that $\phi \circ \alpha = \phi $. Since $G$ is a sheaf of *commutative* groups, $\alpha $ is, locally in the fppf topology over $T$, given by multiplication by some element $g$ of $G$. The condition that $\phi \circ \alpha = \phi $ means that $\phi $ factors through the inclusion of $X^ g$ in $X$, i.e., $\phi $ is obtained by composing that inclusion with a morphism $P \to X^\gamma $. The above discussion allows us to define a morphism of fibred categories $I_{[X/G]} \to \coprod _{g\in G} [X^ g/G]$ given on $T$-points by the discussion above. We omit showing that this is an equivalence.

## Comments (2)

Comment #3560 by Neeraj Deshmukh on

Comment #3684 by Johan on