0
$\begingroup$

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"?

$\endgroup$
  • 5
    $\begingroup$ The wikipedia article does not claim that the stalk of a sheaf on $G\times G\times X$ equals the stalk of a sheaf on $X$. The wikipedia article claims that the cocycle condition, formulated in terms of sheaves on $G\times G\times X$, implies a stalk condition for sheaves on $X$. $\endgroup$ – Jason Starr Mar 27 '19 at 12:55
  • $\begingroup$ @JasonStarr: how does the argument that this stalk condition (*) for sheaves on $X$ comes from the cocycle condition work? I don't think that here one can just "neglect the left factor" to obtain this. Is this done by a "composition" argument to come back to $X$ or is it a bit deeper application of descent? $\endgroup$ – KarlPeter Mar 28 '19 at 0:37
  • $\begingroup$ The morphism $\sigma$ induces local homomorphisms $\sigma_{g,x}^*:\mathcal{O}_{X_R,g\cdot x} \to \mathcal{O}_{X_R,x}$ for every local $\mathcal{O}_S$-algebra $(R,\mathfrak{m},k)$ and $g,x\in G(k)$. Associativity of $\sigma$ implies $\sigma^*_{h,x}\circ \sigma^*_{g,h\cdot x}$ equals $\sigma^*_{gh,x}$ for every $g,h\in G(k)$, $x\in X(k)$. You can reduce (*) to this associativity. Use the cocycle condition to extend the $G$-group action from $X$ to the scheme $X_F$ whose underlying topological space equals $X$ and with structure sheaf $\mathcal{O}_X\oplus F\cdot \epsilon$, $\epsilon^2=0$. $\endgroup$ – Jason Starr Mar 28 '19 at 9:26
3
$\begingroup$

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 (*).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you a lot for your detailed answer. Two points are still unclear: The first one is how does the compatibility with closed immersions contribute to the conclusion that the morphism $(G\times_S X)_{\text{pr}_2^*\mathcal{F}} \to (G\times_S X)_{\sigma^*\mathcal{F}}$ arises from a morphism of quasi-coherent sheaves $\phi:\sigma^*\mathcal{F} \to \text{pr}_2^*\mathcal{F}$? And secondsly concerning the "epsilon"-argument providing (*): Why in the induced iso $\mathcal{F}_{g\cdot x}\cdot \epsilon \xrightarrow{\cong} \mathcal{F}_x \cdot \epsilon$ we can "forget" the epsilon? $\endgroup$ – KarlPeter Mar 28 '19 at 23:53
  • $\begingroup$ The factor $\epsilon$ is a placeholder that reminds us of the algebra structure. The stalk at $x$ of the sheaf of $\mathcal{O}_X$-algebras equals $\mathcal{O}_{X,x}\oplus \mathcal{F}_x\cdot \epsilon$. As a module over the stalk $\mathcal{O}_{X,x}$, this is just the direct sum of modules: $\epsilon$ plays no role. Regarding the closed immersions, please write this out for yourself. $\endgroup$ – Jason Starr Mar 29 '19 at 7:01
  • $\begingroup$ Could you give a reference where the universal property/ the natural equivalence of $X_{\mathcal{F}}$ you pointed out is worked out detailed $\endgroup$ – KarlPeter Sep 19 '19 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.