This question is certainly not research level and in fact quite elementary which is why I asked it on math.stackexchange before: math.stackexchange. However it doesn't seem to get much attention there and I thought I would try it here. My questions are regarding actions of a group scheme $G$ on a scheme $X$. I'm fine with assuming $G$ affine. $\newcommand{\IG}{\mathbb{G}} \newcommand{\pmo}{{\pm 1}} \newcommand{\IZ}{\mathbb{Z}} \newcommand{\Spec}{\mathrm{Spec}} \newcommand{\tensor}{\otimes} \newcommand{\into}{\hookrightarrow} \newcommand{\iso}{\cong} \newcommand{\onto}{\twoheadrightarrow} \newcommand{\sheaf}{\mathcal} \newcommand{\inv}{{-1}}$

Originally, I was thinking about actions of the multiplicative group $\IG_m = \Spec(\IZ[T^\pmo])$ on affine schemes $X = \Spec(R)$. The category of affine schemes with a $\IG_m$-action is equivalent to the category of $\IZ$-graded rings as follows:

To a $\IZ$-graded ring $R$ associate the action $\IG_m \times \Spec(R) \to \Spec(R)$ which is given by the map of rings $R \to \IZ[T^\pmo] \tensor R$, $f = \sum f_d \mapsto \sum f_d \tensor T^d$, where $f_d$ are the homogenous components of $f$. Conversely, if the action $\IG_m \times \Spec(R) \to \Spec(R)$ is given, call $f_d \in R$ homogenous of degree $d$ if it is sent to a homogenous element of rank $d$ by the morphism $R \to \IZ[T^\pmo] \tensor R \cong R[T^\pmo]$, where we regard the latter ring as graded by the powers of $T$.

If $Y \subseteq X$ is a closed subscheme, I call $Y$ invariant under the action if the morphism $\IG_m \times Y \into \IG_m \times X \to X$ factors over $Y \into X$.

Proposition: The correspondence $$\{\text{Closed subschemes of }X\} \iso \{\text{Ideals of }R\}$$ restricts to a bijective correspondence $$\{\IG_m\text{-invariant closed subschemes of }X\} \iso \{\text{Homogeneous ideals of }R\}.$$

For example if $Y$ is $\IG_m$-invariant, then by definition the given action restricts to an action of $\IG_m$ on $Y$ and the inclusion $Y \into X$ becomes a morphism of affine schemes with a $\IG_m$-action. Hence the surjective map $R=\sheaf{O}(X) \onto \sheaf{O}(Y)$ is a morphism of graded rings. But its kernel is $I$ which is, hence, homogeneous.

I was trying to prove the corresponding result for open subschemes:

Conjecture 1: The correspondence $$\{\text{Open subschemes of }X\} \iso \{\text{Radical ideals of }R\}$$ restricts to a bijective correspondence $$\{\IG_m\text{-invariant open subschemes of }X\} \iso \{\text{Homogeneous radical ideals of }R\}.$$

It is easy to see that a homogeneous radical ideal defines an invariant open subscheme. First note that the union of invariant open subschemes is again invariant. If the open subschemes $U_j$ are defined by $I_j$, then $\bigcup U_j$ is defined by $\sqrt{\sum I_j}$. Hence it suffices to consider $U = D(f)$ for $f$ a homogeneous element. But then it is easy to see that $U$ is invariant. In fact, the restricted action $\IG_m \times U \to U$ makes $U = \Spec(R[f^\inv])$ into an affine schemes with a $\IG_m$-action. The associated grading on $R[f^\inv]$ is the natural grading that you would expect on a localisation at a homogeneous element.

However I cannot manage to prove the converse, i.e. if $I$ is a radical ideal of $R$ such that the open subscheme $X_I \into X$ defined by $I$ is $\IG_m$-invariant then $I$ is homogeneous. By the proposition it would suffice to prove

Conjecture 2: If $G$ is an (affine) group scheme acting on a scheme $X$ then the bijection $$\{\text{Open subschemes of }X\} \iso \{\text{Reduced closed subschemes of }X\}$$ given by “reduced closed complement” and “open complement” restricts to a bijection $$\{G\text{-invariant open subschemes of }X\} \iso \{G\text{-invariant reduced closed subschemes of }X\}.$$

This certainly sounds reasonable if one thinks for example of an action of a topological group on a topological space, where the complement of an invariant subset is clearly invariant again. However I cannot prove it in the context of schemes and I'm not quite sure that it is correct. If $X = \Spec(R)$ is reduced and of finite type over an algebraically closed field and $G = \IG_m$ (so that $\IG_m \times X$ is again reduced and of finite type), then it suffices to consider $k$-valued points, so that the reduced complement of an invariant open subscheme is indeed invariant again.

If $G = \IG_m$ and $X$ is affine then this conjecture is equivalent to the first one.

My conjecture is also equivalent to the claim that if $U \subseteq X$ is an open $\IG_m$-invariant subscheme then $U$ is a union of subschemes of the form $D(f) \subseteq X$ where $f$ in $R$ is a homogeneous element. In particular, this would imply (a special case of) the following conjecture:

Conjecture 3: If $G$ is an (affine) group scheme acting on a scheme $X$ and if $U \subseteq X$ is an open $G$-invariant subscheme then $U$ is a union of $G$-invariant affine open subschemes.

Unfortunately I really don't have a good intuition for actions of groups schemes and I would be glad about some clarification.

  • $\begingroup$ In Conj. 3 you mean that $U$ is such a union? $\endgroup$ Jan 26, 2020 at 8:20
  • $\begingroup$ @MartinBrandenburg Yes, sure. $\endgroup$ Jan 26, 2020 at 10:35

2 Answers 2


Your definition

$\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X \to X$ factors over $Y \hookrightarrow X$

can be simplified a bit. A better one for our purposes is

The two subschemes of $\mathbb G_m \times X$ defined as the inverse image of $Y$ under the right projection $\mathbb G_m \times X \to X$ and the action map $\mathbb G_m \times X \to X$ are equal.

To see they are equivalent, note that the universal property of a fiber product implies your factorization is equivalent to the claim that the immersion $\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X $ factors through the map from $\left(\mathbb G_m \times X \right) \times_X Y $ to $\mathbb G_m \times X $. That fiber product is the inverse image of $Y$ under the action map, and $\mathbb G_m \times Y$ is the inverse image of $Y$ under the right projection. So your factorization is equivalent to $p^* Y \subseteq a^* Y$, where $p$ is projection and $a$ is action. But $p^* Y \subseteq a^* Y$ is equivalent to $a^* Y \subseteq p^* Y$, and therefore equivalent to $p^* Y =a^* Y$, because we can swap the projection and action maps using the automorphism of $\mathbb G_m \times X$ that sends $(g,x)$ to $(g^{-1}, gx)$.

Now this one is equivalent for open subschemes and their closed complements because the operation of taking closed complement is compatible with pullback under smooth morphisms.

  • $\begingroup$ Where does this use the special case $\mathbb{G}_m$? Angelo wrote that it is false for general group schemes. $\endgroup$ Jan 28, 2020 at 9:49
  • $\begingroup$ @MartinBrandenburg Then the maps $\mathbb G_m \times X\to X$ are not smooth, and in particular the inverse image of a reduced closed subscheme can clearly be non-reduced. I think this approach does work for general smooth group schemes. $\endgroup$
    – Will Sawin
    Jan 28, 2020 at 13:24

Your conjecture 2 is false if you don't assume that $G$ is reduced (in positive characteristic there are affine group schemes that are not reduced).

As to conjecture 3, it is hopelessly wrong (think of the action of $\mathrm{GL}_n$ on $\mathbb P^{n-1}$). The only non-trivial case I know is Sumihiro's theorem: a normal algebraic variety with an action of a torus can be covered by invariant affine open subsets. Being normal is essential: consider the standard action of $\mathbb{G}_{\mathrm m}$ on $\mathbb P^1$, and glue together the origin and the point at infinity.


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