Example 94.17.1. 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 94.15.5) is given by

$I_{[X/G]} = \coprod \nolimits _{g\in G} [X^ g/G],$

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

$X^ g = X \times _{(1, 1), X \times _ S X, (g, 1)} X.$

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.

Comment #4554 by Leo on

Typo on 9th line : "together with a $G$-equivariant isomorphism $\phi$" should just be a morphism, not an isomorphism.

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