# How do the invariants of a group scheme action compare to the invariants of the group action by global sections

If $$G$$ is a group scheme over $$S$$ acting on an $$S$$-scheme $$X$$, I'd like to understand the algebra of invariants $$(\mathcal{O}_X)^G$$. Specifically, I'd like to understand its relation to invariants $$(\mathcal{O}_X)^{G(S)}$$.

To simplify notation, say everything is affine: $$G = \operatorname{Spec}R$$, $$X = \operatorname{Spec}A$$, and $$S = \operatorname{Spec}k$$, where $$k$$ is an arbitrary ring (not necessarily a field). If it helps we can assume $$G$$ is smooth. We work in the category of $$k$$-schemes.

The action is given by a map $$\sigma : G\times X\rightarrow X$$. Let $$p : G\times X\rightarrow X$$ be the projection map. Then there is a natural bijection $$A = \operatorname{Hom}(X,\mathbb{A}^1)$$, and by definition the subalgebra of invariants $$A^G$$ is the set of $$f\in A$$ whose corresponding map $$F : X\rightarrow\mathbb{A}^1$$ satisfies $$F\circ\sigma = F\circ p$$ Via $$\sigma$$, the group $$G(k)$$ acts on $$X(k)$$, and for any $$k$$-scheme $$T$$, $$G(k)$$ maps to $$G(T)$$ and hence also acts on $$X(T)$$, so $$G(k)$$ acts on $$X$$. Thus, we may also consider the ring of invariants $$A^{G(k)}$$. Certainly we have $$A^G\subset A^{G(k)}$$ My main question is: What is the clearest way to express this relationship? I'm looking for a statement of the form $$f\in A$$ is $$G$$-invariant if and only if it is fixed by $$G(k)$$ and some other conditions.

I think one can say that $$A^G = \{f\in A| f\otimes_k 1\in A\otimes_k B \text{ is fixed by G(B) for every k-algebra B}\}$$ Is this correct? Is it possible to further restrict the class of $$B$$'s that you have to consider? Are there other ways of thinking about this?

• A simple and unsatisfactory one: if $k$ is not an algebraic extension of a finite field, and $G$ is smooth, then $G(k)$ is Zariski dense, so $A^{G(k)} = A^G$. – LSpice Oct 27 '20 at 12:51
• I agree that the key thing to say is that, if $G(k)$ is Zariski dense in $G$, then $A^{G(k)} = A^G$. But I don't think your finite field criterion is the right one. For example, let $G = \mu_3$, the group of $3$-rd roots of unity, and let $k = \mathbb{R}$. @LSpice – David E Speyer Oct 27 '20 at 15:20
• @DavidESpeyer is right; I definitely meant smooth connected, and there is a slight possibility I meant reductive. – LSpice Oct 27 '20 at 15:52
• Ah, found it. If (1) $k$ is infinite, (2) $G$ is connected and either (3a) $G$ is reductive or (3b) $k$ is perfect, then $G(k)$ is Zariski dense in $G$. mathoverflow.net/q/56192/297 – David E Speyer Oct 27 '20 at 16:36
• While I'm here, I just wanted to say that this is not at all a stupid question, and neither are any of your others! – David E Speyer Jan 14 at 15:48

Combining LSpice's (1 2) and my (1 2) comments into an answer: If $$G(k)$$ is Zariski dense in $$G$$, then $$A^{G(k)} = A^G$$. It is very common that $$G(k)$$ is Zariski dense in $$G$$: This happens whenever (1) $$k$$ is infinite and (2) $$G$$ is connected and either (3a) $$G$$ is reductive or else (3b) $$k$$ is perfect. See Density question in algebraic group.