The Stacks project

59.12 The example of G-sets

Let $G$ be a group and define a site $\mathcal{T}_ G$ as follows: the underlying category is the category of $G$-sets, i.e., its objects are sets endowed with a left $G$-action and the morphisms are equivariant maps; and the coverings of $\mathcal{T}_ G$ are the families $\{ \varphi _ i : U_ i \to U\} _{i\in I}$ satisfying $U = \bigcup _{i\in I} \varphi _ i(U_ i)$.

There is a special object in the site $\mathcal{T}_ G$, namely the $G$-set $G$ endowed with its natural action by left translations. We denote it ${}_ G G$. Observe that there is a natural group isomorphism

\[ \begin{matrix} \rho : & G^{opp} & \longrightarrow & \text{Aut}_{G\textit{-Sets}}({}_ G G) \\ & g & \longmapsto & (h \mapsto hg). \end{matrix} \]

In particular, for any presheaf $\mathcal{F}$, the set $\mathcal{F}({}_ G G)$ inherits a $G$-action via $\rho $. (Note that by contravariance of $\mathcal{F}$, the set $\mathcal{F}({}_ G G)$ is again a left $G$-set.) In fact, the functor

\[ \begin{matrix} \mathop{\mathit{Sh}}\nolimits (\mathcal{T}_ G) & \longrightarrow & G\textit{-Sets} \\ \mathcal{F} & \longmapsto & \mathcal{F}({}_ G G) \end{matrix} \]

is an equivalence of categories. Its quasi-inverse is the functor $X \mapsto h_ X$. Without giving the complete proof (which can be found in Sites, Section 7.9) let us try to explain why this is true.

  1. If $S$ is a $G$-set, we can decompose it into orbits $S = \coprod _{i\in I} O_ i$. The sheaf axiom for the covering $\{ O_ i \to S\} _{i\in I}$ says that

    \[ \xymatrix{ \mathcal{F}(S) \ar[r] & \prod _{i\in I} \mathcal{F}(O_ i) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod _{i, j \in I} \mathcal{F}(O_ i \times _ S O_ j) } \]

    is an equalizer. Observing that fibered products in $G\textit{-Sets}$ are induced from fibered products in $\textit{Sets}$, and using the fact that $\mathcal{F}(\emptyset )$ is a $G$-singleton, we get that

    \[ \prod _{i, j \in I} \mathcal{F}(O_ i \times _ S O_ j) = \prod _{i \in I} \mathcal{F}(O_ i) \]

    and the two maps above are in fact the same. Therefore the sheaf axiom merely says that $\mathcal{F}(S) = \prod _{i\in I} \mathcal{F}(O_ i)$.

  2. If $S$ is the $G$-set $S= G/H$ and $\mathcal{F}$ is a sheaf on $\mathcal{T}_ G$, then we claim that

    \[ \mathcal{F}(G/H) = \mathcal{F}({}_ G G)^ H \]

    and in particular $\mathcal{F}(\{ *\} ) = \mathcal{F}({}_ G G)^ G$. To see this, let's use the sheaf axiom for the covering $\{ {}_ G G \to G/H \} $ of $S$. We have

    \begin{eqnarray*} {}_ G G \times _{G/H} {}_ G G & \cong & G \times H \\ (g_1, g_2) & \longmapsto & (g_1, g_1^{-1} g_2) \end{eqnarray*}

    is a disjoint union of copies of ${}_ G G$ (as a $G$-set). Hence the sheaf axiom reads

    \[ \xymatrix{ \mathcal{F} (G/H) \ar[r] & \mathcal{F}({}_ G G) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod _{h\in H} \mathcal{F}({}_ G G) } \]

    where the two maps on the right are $s \mapsto (s)_{h \in H}$ and $s \mapsto (hs)_{h \in H}$. Therefore $\mathcal{F}(G/H) = \mathcal{F}({}_ G G)^ H$ as claimed.

This doesn't quite prove the claimed equivalence of categories, but it shows at least that a sheaf $\mathcal{F}$ is entirely determined by its sections over ${}_ G G$. Details (and set theoretical remarks) can be found in Sites, Section 7.9.

Comments (2)

Comment #7779 by Reimundo Heluani on

Hi, I think the map can be given by or

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