3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In Conj. 3 you mean that $U$ is such a union? $\endgroup$ – Martin Brandenburg Jan 26 at 8:20
  • $\begingroup$ @MartinBrandenburg Yes, sure. $\endgroup$ – Jakob Werner Jan 26 at 10:35
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Where does this use the special case $\mathbb{G}_m$? Angelo wrote that it is false for general group schemes. $\endgroup$ – Martin Brandenburg Jan 28 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 at 13:24
2
$\begingroup$

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.

$\endgroup$

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.