4
$\begingroup$

What is a quotient of an affine scheme that is not a universal quotient? Let's recall some terminology.

Suppose that $k$ is an algebraically closed field and $G$ is a reductive group acting on an affine scheme $X$. Theorem 1.1 of Geometric Invariant Theory states that the uniform categorical quotient $X//G$ of $X$ exists.

In other words, $X \to X//G$ is universal with respect to $G$-invariant morphisms out of $X$ and this property persists under base change by a flat morphism $T \to X//G$.

When $\text{char}(k)=0$, the theorem states that $X \to X//G$ is a universal categorical quotient, so that the universal property persists under base change by an arbitrary morphism $T \to X//G$.

What is an example where $X \to X//G$ is not a universal quotient?

I'd be particularly interested in the case where the stabilizers of the action on $X$ are all linearly reductive.

$\endgroup$

1 Answer 1

3
$\begingroup$

Here is an example, which is in some sense the simplest one. Suppose that $k$ has characteristic $p > 0$; set $X := \mathop{\rm Spec} k[x,y]$. Let $G$ be a cyclic group of order $p$ acting via $(x,y) \mapsto (x, x+y)$. The ring of invariants is $k[u,v] := k[x, y^p - x^{p-1}y]$. Consider the point $\mathop{\rm Spec} k = \mathop{\rm Spec} k[u,v]/(u,v)$ of $X/G = \mathop{\rm Spec} k[u,v]$; the inverse image $Y$ in $X$ is $\mathop{\rm Spec} k[x,y]/(x, y^p)$; it is immediate to check that the action of $G$ on $Y$ is trivial, so $Y/G = Y \neq \mathop{\rm Spec} k$.

If you want an example with a connected group, embed $G$ into $\mathrm{GL}_n$ and take the induced action.

I don't know any example with linearly reductive stabilizers, and I suspect that they don't exist.

$\endgroup$
8
  • 2
    $\begingroup$ Angelo's suspicion is right. By thm of Nagata (Ch.IV, Thm. 3.6 in Demazure-Gabriel), in char. $p > 0$ a smooth affine $k$-gp is lin. red. iff its comp. group has order not divisible by $p$ and id. component is torus. Hence, enough to treat tori and finite gps of order prime to $p$. The latter is universal for affine $X$ in char. $p$ via averaging. For tori $S$, can restrict to noetherian base change and then by noetherian induction $S$-invariants in coordinate rings upstairs and downstairs are $S[n]$-invariants for sufficiently divisible $n$ coprime to $p$. So reduced back to the first case. $\endgroup$
    – BCnrd
    Sep 13, 2010 at 19:05
  • 1
    $\begingroup$ Brian, this is for the case of actions of linearly reductive stabilizers, but jlk was asking for the case when the stabilizers are linearly reductive, and I think this is much harder $\endgroup$
    – Angelo
    Sep 13, 2010 at 20:28
  • $\begingroup$ @Angelo: Thanks. Your interpretation is correct: I would particularly like a proof/example in the case where the stabilizers are all linearly reductive, but the group $G$ is not. $\endgroup$
    – jlk
    Sep 13, 2010 at 21:16
  • 1
    $\begingroup$ Dear Torsten, of course the approach would be the one you mention. There is a formal slice theorem for group actions with linearly reductive stabilizers, but it is very hard to get consequences from it. This is discussed in a paper of Jarod Alper, On the local quotient structure of Artin stacks <arxiv.org/abs/0904.2050>. $\endgroup$
    – Angelo
    Sep 14, 2010 at 6:13
  • 2
    $\begingroup$ Consider the action of $\mathrm{PGL}_2$ on the space of unordered 4-tuples of points of $\mathbb P^1$ (a.k.a. $\mathbb P^4$). The generic stabilizer is isomorphic to the product of two cyclic groups of order 2 (the Klein group); but the stabilizer of a point corresponding to a double points and two distinct point of $\mathbb P^1$ is cyclic of order 2. Clearly, this means that you can't have a slice around this point, since the generic stabilizer is not conjugate to a subgroup of the special stabilizer. $\endgroup$
    – Angelo
    Sep 15, 2010 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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