Equivariant Sheaf: Explanation on Stalks

Question

I have a question about the explanation of the data defining a so called equivariant sheaf $F$ on a scheme X from wiki: https://en.wikipedia.org/wiki/Equivariant_sheaf. Let denote by $\sigma: G \times_S X \to X$ an action of a group scheme $G$ on $X$ . Then a $O_X$-module $F$ is called equivariant if there exist in isomorphism $\phi: \sigma^* F \simeq p_2^*F$ of $\mathcal{O}_{G \times_S X}$-modules and additionally the "cocycle" condition $p_{23}^* \phi \circ (1_G \times \sigma)^* \phi = (m \times 1_X)^* \phi$ is satisfied where $p_{23}, 1_G \times \sigma, m \times 1_X$ a maps between $G \times G \times X$ and $G \times X$.

FOLLOWING EXPLANATION I DON'T UNDERSTAND: Then there is said that the cocycle condition tells that on level of stalks the isomorphism $F_{gh \cdot x} \simeq F_x$ is the same as $F_{g \cdot h \cdot x} \simeq F_{h \cdot x} \simeq F_x$; namely it reflects the associativity. (*)

What I don't understand is why the induced isomorphism $(m \times 1_X)^* \phi$ provides a map $F_{gh \cdot x} \to F_x$ on the level of stalks?

Namely I don't see why $F_{gh \cdot x} $ and $ F_x$ are correct domain and codomain of this map $(m \times 1_X)^* \phi$.

Indeed, the induced isomorphism $(m \times 1_X)^* \phi$ is a sheaf iso $$(m \times 1_X)^*\sigma^* F= (\sigma \circ (m \times 1_X))^*F \to (m \times 1_X)^*p_2^*F=(p_2 \circ (m \times 1_X))^*F.$$

Fix a point $(g,h,x) \in G \times G \times X$.

Then $\sigma \circ (m \times 1_X)(g,h,x)= gh \cdot x$ and $p_2 \circ (m \times 1_X)(g,h,x)= x$

It's not clear to me why the stalks at $(g,h,x) $ of the domain and codomain are given by $$((\sigma \circ (m \times 1_X))^*F)_{(g,h,x)}= F_{gh \cdot x}$$ and $$((p_2 \circ (m \times 1_X))^*F)_{(g,h,x)}= F_x$$ as stated in (*)?

I learned that generally for a morphism $f:X \to Y$ and a sheaf $F$ on $Y$ we have following formula for the stalk in $z \in X$ of the pullback sheaf:

$$(f^*F)_z= O_{X,z} \otimes_{O_{Y,f(z)}} F_{f(z)}.$$

Now we apply this formula it to our situation, therefore $f=\sigma \circ (m \times 1_X)$ and $z=(g,h,z)$ then we obtain

$$((\sigma \circ (m \times 1_X))^*F)_{(g,h,x)}= O_{G \times G \times X,(g,h,x)} \otimes_{O_{X,gh \cdot x}} F_{gh \cdot x}$$

But (*) says that it should equal $F_{gh \cdot x}$. Why? Or where is the error in my reasonings? What happens with the left "factor"?

Answer 1

I am just posting my comment as an answer. For a scheme $S$, one definition of a group $S$-scheme is a datum of $S$-schemes, $$(\pi:G\to S, m:G\times_S G\to G, i:G\to G, e:S\to G),$$ of an $S$-scheme $G$ and $S$-morphisms $m$, $i$, and $e$ such that for every $S$-scheme $T$, the induced datum of sets, $$(G(T),m(T):G(T) \times G(T) \to G(T), i(T):G(T) \to G(T), e(T): \{\text{Id}_T\} \to G(T)),$$ is a group with its group operation, with its group inverse, and with its specified group identity element. In particular, setting $T$ equal to $G\times_S G\times_S G$ with its three projections to $G$, associativity of the group operation implies equality of the compositions $$G\times_S G\times_S G \xrightarrow{\text{Id}_G \times m} G\times_S G \xrightarrow{m} G, \ \ \ \ G\times_S G\times_S G \xrightarrow{m\times_S \text{Id}_G} G\times_S G \xrightarrow{m} G.$$ Similarly, the following compositions are equal, $$G\xrightarrow{\pi} S \xrightarrow{e} G, \ \ G \xrightarrow{\Gamma_i} G\times_S G \xrightarrow{m} G,$$ and the following compositions are equal, $$G\xrightarrow{\text{Id}_G} G, \ \ G \xrightarrow{\Gamma_{\pi\circ e}} G\times_S G \xrightarrow{m} G.$$ Conversely, if each of these pairs of compositions are equal, then each datum of sets above is a group.

In the same way, for an $S$-scheme $$\rho:X\to S,$$ for an $S$-morphism, $$\sigma:G\times_S X \to X,$$ for every $S$-scheme $T$, the induced datum of sets, $$\sigma(T):G(T)\times X(T) \to X(T),$$ satisfies the axioms for the action of the group $G(T)$ on the set $X(T)$ if and only if the following compositions are equal, $$G\times_S G \times_S X \xrightarrow{m\times\text{Id}_X} G\times_S X \xrightarrow{\sigma} X, \ \ G\times_S G \times_S X \xrightarrow{\text{Id}_G \times \sigma} G\times_S X \xrightarrow{\sigma} X,$$ $$X\xrightarrow{\text{Id}_X} X, \ \ X\xrightarrow{\Gamma_{e\circ \rho}}G\times_S X \xrightarrow{\sigma} X.$$

Now consider the special case that $S$ equals $\text{Spec}\ R$ for a local ring $(R,\mathfrak{m},k)$, e.g., $R$ might equal the residue field $k$ with maximal ideal $\mathfrak{m}=\{0\}$. For every $g\in G(S)$, there is an induced isomorphism of $S$-schemes, $$\sigma_g:X \xrightarrow{\Gamma_{g\circ \rho}}G\times_S X \xrightarrow{\sigma} X.$$ For every point $x$ of $X$, denote the image point $\sigma_g(x)$ by $g\cdot x$. Then this isomorphism of schemes induces an isomorphism of stalks, $$\sigma_{g,x}^*:\mathcal{O}_{X,g\cdot x} \to \mathcal{O}_{X,x}.$$ For every pair $(g,h)\in G(S)\times G(S)$, the first equality of compositions in the previous paragraph implies that $\sigma_{gh}$ equals $\sigma_g\circ \sigma_h$. Thus, for every point $x$ of $X$, also $\sigma_{gh,x}^*$ equals $\sigma_{h,x}^*\circ \sigma_{g,h\cdot x}^*$.

Finally, for a fixed group $S$-scheme $G$, the $S$-schemes together with an $S$-action by $G$ form a category (with a forgetful functor to the category of $S$-schemes). The $G$-equivariant morphisms are defined in the usual way. For every $S$-scheme $X$, for every quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$, there is an associated $S$-scheme $X_\mathcal{F}$ with a closed immersion and a retraction, both of which are universal homeomorphisms, $$j_{\mathcal{F}}:X\hookrightarrow X_{\mathcal{F}}, \ \ r_{\mathcal{F}}:X_\mathcal{F} \to X,$$ with $j^{-1}\mathcal{O}_{X_\mathcal{F}}$ such that the structure sheaf of $X_{\mathcal{F}}$ equals the commutative, unital $\mathcal{O}_X$-algebra $$\mathcal{O}_X \xrightarrow{\text{Id}\oplus 0} \mathcal{O}_X\oplus \mathcal{F}\cdot \epsilon \xrightarrow{(\text{Id},0)} \mathcal{O}_X.$$ For every $X$-scheme $$t:T\to X,$$ the lifts of $t$ to an $X$-morphism $$\widetilde{t}:T\to X_{\mathcal{F}}$$ are naturally equivalent to the $\mathcal{O}_T$-module homomorphisms $$\theta:t^*\mathcal{F}\to \mathcal{O}_T,$$ such that $\text{Image}(\theta)\cdot \text{Image}(\theta)$ is the zero ideal sheaf in $\mathcal{O}_T$.

The $G$-linearizations of $\mathcal{F}$ are equivalent to the lifts of the $G$-action on $X$ to a $G$-action on $X_{\mathcal{F}}$ such that both the closed immersion and the retraction are $G$-equivariant. Indeed, chasing universal properties, the fiber product $G\times_S X_{\mathcal{F}}$ as a scheme with a morphism to $G\times_S X$ is equivalent to $(G\times_S X)_{\text{pr}_2^*\mathcal{F}}$. On the other hand, the pullback of $X_{\mathcal{F}}$ by $\sigma:G\times_S X \to X$ as a scheme with a projections to $G\times_S X$ is equivalent to $(G\times_S X)_{\sigma^*\mathcal{F}}$. Thus, a pair of morphisms from $G\times_S X_{\mathcal{F}}$ to $G\times_S X$ and to $X_{\mathcal{F}}$ whose compsitions with $\sigma$ and with $r_{\mathcal{F}}$ commute is equivalent to a morphism from $(G\times_S X)_{\text{pr}_2^*\mathcal{F}}$ to $(G\times_S X)_{\sigma^*\mathcal{F}}$. Compatibility with closed immersions forces this morphism to arise from a morphism of quasi-coherent sheaves $$\phi:\sigma^*\mathcal{F} \to \text{pr}_2^*\mathcal{F}.$$ The axioms from a group action hold if and only if $\phi$ satisfies the usual axioms for a $G$-linearization.

Finally, for the lifted $G$-action $\sigma_\phi$ on $X_{\mathcal{F}}$ associated to a $G$-linearization $\phi$, the maps of stalks $(\sigma_\phi)_{g,x}$ are local homomorphisms $$\mathcal{O}_{X,g\cdot x} \oplus \mathcal{F}_{g\cdot x}\cdot \epsilon \xrightarrow{\cong} \mathcal{O}_{X,x} \oplus \mathcal{F}_x \cdot \epsilon.$$ For these local homomorphisms, the associativity identity $$(\sigma_{\phi})_{gh,x}^* = (\sigma_{\phi})_{h,x}^*\circ (\sigma_{\phi})_{g,h\cdot x}^*,$$ gives the associativity in (*).