5
$\begingroup$

The following is motivated by a (now-deleted) MSE-question by @aglearner.

Suppose that $X\subset {\mathbb C}^n$ is an affine subvariety, equipped with the classical (Euclidean) topology. Consider the group $G= {\mathbb C}^\times$, and suppose that $G\times X\to X$ is an algebraic action. Assume, in addition, that each $G$-orbit in $X$ is closed.

Question 1. Is the quotient $X/G$ (in the sense of general topology) Hausdorff?

(Note that I am not taking the GIT quotient here.)

Hausdorffness of the quotient fails if I relax the assumption on $X$ to that of a quasi-affine subvariety in the following standard example: Take $X={\mathbb C}^2\setminus \{(0,0)\}$ and consider the action given by $$ (t, (x,y)) \mapsto (tx, t^{-1}y), \quad t\in G, (x,y)\in X. $$

The next question is a complex-analytic version of Question 1:

Question 2. Suppose that $X\subset {\mathbb C}^n$ is a Stein submanifold, $G$ is as above and $G\times X\to X$ is a holomorphic action with closed orbits. Is $X/G$ Hausdorff?

I expect answers to both questions to be negative (and counter-examples given by a smooth affine subvariety $X$) but cannot think of any examples.

$\endgroup$

2 Answers 2

3
$\begingroup$

I think the answer to Question 1 is yes, since under your hypothesis the set-theoretical quotient is essentially the GIT quotient. This follows from the fact that $G$-invariant regular functions on $X$ separate orbits (when these are closed).

There is an analytic proof of this fact. Let $V$ and $W$ be two disjoint closed orbits. Then 1 can be written as the sum of two regular functions $f_V$ and $f_W$ vanishing respectively on $V$ and $W$ (because $I(V) + I(W) = \mathbb C[X]$). In particular, $f_V = 0$ on $V$ and $f_V = 1$ on $W$. After averaging $f_V$ under the action of the unit circle $U_1\subset \mathbb C^\times$, one can assume that $f_V$ is $U_1$-invariant. Since $f_V$ is holomorphic, the action of $\mathbb C^\times$ is holomorphic and $U_1$ is a real form of $\mathbb C^\times$, this implies that $f_V$ is in fact $\mathbb C^\times$-invariant.

My impression is that one could do the same with holomorphic functions on Stein manifolds. My only concern is about the existence of a holomorphic function which is $0$ on $V$ and $1$ on $W$. (This condition ensures that, in the averaging process, $f_V$ remains non-zero on $W$.)

$\endgroup$
3
  • $\begingroup$ If the argument in the Stein manifold case only relies on the existence of a function which is 0 on W and 1 on V, then this is indeed possible, cf., for example mathoverflow.net/q/383638 (which is stated for X=C^n, but the same argument applies for any Stein manifold X) $\endgroup$ Nov 9, 2021 at 14:11
  • $\begingroup$ Great! Then I think we're good. After averaging, you can make this function $\mathbb U_1$-invariant, hence $\mathbb C^\times$-invariant. So invariant holomorphic functions separate points, implying that the largest Hausdorff quotient is the set theoretical quotient. $\endgroup$
    – Nicolast
    Nov 9, 2021 at 14:57
  • $\begingroup$ Thank you Nicolas and Richard! $\endgroup$ Nov 16, 2021 at 22:18
2
$\begingroup$

Let $X$ be a complex affine algebraic set with the analytic topology and let $G$ be a complex affine algebraic group acting rationally on $X$.

$\newcommand\sslash{/\hspace{-0.2ex}/}$Let $X\sslash G=\operatorname{Spec}_\text{max}\mathbb{C}[X]^G$ be the affine GIT quotient of $X$ by $G$. Give $X\sslash G$ the analytic topology. So $X\sslash G$ is necessarily Hausdorff.

Let $X^*\subset X$ be the subspace of points with closed $G$-orbits (so $X^*=X$ if all orbits are closed). Give $X^*/G$ the quotient topology, as with $X/G$.

Then $X\sslash G$ is homeomorphic to $X^*/G$ (this follows from work of Luna), and homotopic to $X/G$ (Proposition 3.4 of Florentino, Lawton, and Ramras - Homotopy Groups of Free Group Character Varieties).

So, in particular, the answer to your first question is yes. I am not sure about your second question.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you, Sean! I wish I could accept both proofs, but I will accept Nicolas' answer since, as explained by Richard, it can be adopted to the general Stein setting. $\endgroup$ Nov 16, 2021 at 22:17
  • $\begingroup$ I completely understand :) If you need to cite something though in the affine case, I have an elementary proof here (Theorem 2.1): arxiv.org/pdf/1301.7616.pdf. $\endgroup$ Nov 16, 2021 at 23:53

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.