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This question arose from my last question, which I considered answered - from the comments, however, it is obvious that the answer is only complete in characteristic zero, and I am trying to understand why.

Let $A$ be a $\Bbbk$-algebra1, where $\Bbbk$ is some algebraically closed field. Let $G$ be a reductive algebraic group. If $G$ acts algebraically on $X=\newcommand{\Spec}{\mathrm{Spec}}\Spec(A)$, then it induces an action of $G$ on $A$.

On the other hand, assume that $G$ acts on $A$ by $\Bbbk$-algebra automorphisms. For any maximal ideal $\mathfrak m\subset A$ and any $g\in G$, the image $g.\mathfrak m$ is again a maximal ideal. This defines an action of $G$ on the closed points of $X$ and for any $f\in A$, note that the image of $f$ in $A/g.\mathfrak m=\Bbbk$ is the same as the image of $g^{-1}.f$ in $A/\mathfrak m=\Bbbk$, $$\begin{matrix} g^{-1}.f & \in & A & \xrightarrow{\quad g\quad} & A & \ni & f \\ && \downarrow && \downarrow && \\ g^{-1}.f+\mathfrak m & \in & A/\mathfrak m & \xrightarrow{\quad \sim\quad} & A/g.\mathfrak m & \ni & f+g.\mathfrak m \end{matrix} $$

so if $\mathfrak m$ is viewed as a closed point of $X$, we have the familiar formula $(g^{-1}.f)(\mathfrak m) = f(g.\mathfrak m)$. This is good, but I forgot to ask myself (and now I am asking you):

When is this action algebraic?

By this, I mean that there is a morphism $G\times X\to X$ of $\Bbbk$-schemes (or varieties) which gives the above action on closed points.

The first assumption should probably be that each $f\in A$ is contained in a finite-dimensional $G$-module. because this property holds when the action on $A$ comes from an algebraic action on $X$. On the other hand, I suspect that one will need at least characteristic zero (or more generally, some separability condition). However, I don't know exactly how to put this together.

1 You may assume $A$ finitely generated over $\Bbbk$ and reduced, or even a domain, but I have a feeling that it won't matter much whether we deal with varieties or $\Bbbk$-schemes.

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  • $\begingroup$ What is your definition of "algebraic" action? $\endgroup$
    – S. Carnahan
    Oct 16, 2013 at 14:58
  • $\begingroup$ @S.Carnahan: I edited an explanation into the post. $\endgroup$ Oct 16, 2013 at 15:44
  • $\begingroup$ The action is nothing more than an algebra homomorphism ${\cal O}(X)\to{\cal O}(G)\otimes{\cal O}(X)$ subject to certain conditions (that involve also the homomorphism ${\cal O}(G)\to{\cal O}(G)\otimes{\cal O}(G)$ induced by the multiplication in $G$), if I undersood correctly the question. $\endgroup$ Oct 16, 2013 at 16:30

1 Answer 1

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If I understand the question correctly, you have a map of schemes $G \times X \to X$, and the corresponding map of $k$-points is a group action (meaning that the obvious two maps $G(k) \times G(k) \times X(k) \to X(k)$ coincide), but you are not sure that it is a group action in the category of schemes. In other words, you fear that you may have two maps $G \times G \times X \to X$ which coincide on $k$ points but not as maps of schemes.

This certainly can't happen if $G$ and $X$ are reduced. So, if you are talking about varieties, there is no issue. It's not obvious to me what happens when $G$ is reduced (which is automatic in characteristic zero) but $X$ isn't.


Based on comments below, and on reading the motivating question, I didn't understand right. The question is, given an action $G(k) \times X(k) \to X(k)$, so that $g \times X(k) \to X(k)$ is algebraic for every $k \in G(k)$, can we conclude that it comes from an algebraic map $G \times X \to X$. But I don't think there is any good way to force this. For example, suppose that $k$ has a nontrivial automorphism $\sigma$ and $G$ is defined over the fixed field of $\sigma$. (Think of complex conjugation.) Then $\sigma$ induces an automorphism of $G(k)$ as an abstract group. Take any algebraic action $G \times X \to X$ and compose with the automorphism of $k$ to get a very nonalgebraic action of $G(k)$ on $X(k)$.

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  • $\begingroup$ Oh, hm, actually I think that I have a map $G(k)\times X(k)\to X(k)$ of sets, and I am wondering if it comes from a morphism $G\times X\to X$ of schemes. $\endgroup$ Oct 16, 2013 at 22:01
  • $\begingroup$ Looking back at your motivating question, I agree. The problem there is getting a map of schemes in the first place. $\endgroup$ Oct 16, 2013 at 22:33
  • $\begingroup$ But then, the anwswer to my previous question would not work at all, because simply talking about actions on the coordinate rings will not give you an algebraic action on the varieties. At least in that very special case where $A$ is integral over some ring $R$ and the restricted action is algebraic, will it work? $\endgroup$ Oct 17, 2013 at 5:17
  • $\begingroup$ I agree, the currently selected answer doesn't work. Did you look at the second half of the answer I left there? $\endgroup$ Oct 17, 2013 at 13:36
  • $\begingroup$ That looks great! Thanks a lot. I'd be really thankful if you could also see to my comment over there. $\endgroup$ Oct 18, 2013 at 10:11

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