# Closed reductive sub-orbits

Let $G$ be a reductive affine algebraic $\mathbb{C}$-group (not necessarily connected). Suppose $X$ is an irreducible affine algebraic set over $\mathbb{C}$ where $G$ acts rationally. Suppose that $H$ is a reductive subgroup of $G$ (again not necessarily connected). Let $x\in X$.

If the orbit $G\cdot x$ is closed in $X$ (in the ball topology), is the sub-orbit $H\cdot x$ also closed?

NOTE: Originally, I left off the assumption that $H$ is a reductive subgroup, and had not emphasized that I was allowing the adjective "reductive" to include disconnected groups (connected component of identity has trivial unipotent radical). I have editted the problem to reflect my original intentions.

• I just added a tag. The question itself could use more precision: I'm assuming the group is connected, while the subgroup is any closed (and maybe also connected?) subgroup, relative to the Zariski topology. Jul 6 '12 at 15:48
• First, thanks for the interest Jim. I am not assuming $G$ is connected. I am assuming that $G$ is an algebraic $\mathbb{C}$-group that is the Zariski closure of a maximal compact subgroup (same for $H$ but as a subgroup). Jul 6 '12 at 19:27
• @Sean "... same for $H$ but as a subgroup ..." What precisely do you mean by this. For instance, in the corrected example I give, $H$ is not a reductive subgroup of $G$. Jul 6 '12 at 20:10
• I mean that $H$ is the Zariski closure of a compact subgroup of K. Sorry, it appears I left that off originally. I will edit if possible. Jul 6 '12 at 20:26
• Sorry, $K$ is a maximal compact subgroup of $G$. Jul 6 '12 at 20:26

$$\def\smat#1{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}$$

The answer to your question is "no". Let $$G=GL(2,\mathbb C) \times GL(2,\mathbb C)$$ act on $$X=GL(2, \mathbb C)$$ by $$(A,B)\cdot C= ACB^{-1}.$$ The orbit of the matrix $$x=\smat{ 1 & 1 \\ 0 & 1 \\\ }$$ is $$X$$. (Hence it is closed.) Let $$H$$ be the subgroup of matrices of the form $$\smat{ a & 0 \\ 0 & a^{-1}\\\ } \times \smat{ a & 0 \\\ 0 & a^{-1}\\\ } \subset G$$. Then $$Hx= \smat{ 1 & b \\\ 0 & 1\\\ }$$, for $$b\ne 0$$. Hence $$Hx$$ is not closed.

No, $H\cdot x$ need not be closed. Consider the standard action of $G=\mathbf{PGL}_2$ on the projective line $X=\mathbb{P}^1$. Consider the point $x=[1,1]$. Then $G\cdot x$ is all of $X$. Yet for a maximal torus $H$, the subgroup of diagonal matrices, the orbit $H\cdot x$ is $\mathbb{P}^1 \setminus \{ [1,0],[0,1] \}$, which is open yet not closed.

EDIT: The OP asked for $X$ to be affine, whereas the example above is projective.

FIX: Here is a fix (modeled on the previous example). Let $G$ be $\mathbf{PGL}_3$, the group of automorphisms of $\mathbb{P}^2$. Let $X$ be the parameter space for smooth conics in $\mathbb{P}^2$; this is an open affine inside the projective space $\mathbb{P}^5$ of all plane conics. Let $x$ parameterize a smooth conic $C$; then $G\cdot x$ equals all of $X$. Let $L$ be a tangent line to $C$, tangent at a point $p$. Let $q$ be a point of $L$ distinct from $p$. Let $H$ be the subgroup of $\mathbf{PGL}_3$ parameterizing automorphism that both fix $q$ and map $L$ to itself. Then $H\cdot x$ is the set of smooth plane conics which are tangent to $L$ at some point different from $q$. This is not closed in $X$; the closure is the set of all smooth plane conics which are tangent to $L$ at some point, including those tangent at $q$.

• I guess similar reasoning applies whether the Zariski or complex topology is used here. From the algebraic viewpoint, a simple algebraic group of dimension 3 is acting on its flag variety (a homogeneous space) while a standard maximal torus has only two fixed points (corresponding to the Weyl group). Jul 6 '12 at 15:46
• Technically this counter-example doesn't fit the question as the OP asks for $X$ to be affine. I'd suggest taking the group of affine transformations $G=\mathbb G_a^n \rtimes GL_n$, letting $X=\mathbb A^n$ be affine $n$-space and taking $H=GL_n$. Then the orbit of any point $0\neq x \in X$ under $H$ is $X-\{ 0 \}$ (i. e. not closed), but $G\cdot x = X$, because $G$ is transitive. Jul 6 '12 at 16:27
• Okay, on second thought, the $G$ in my previous comment is not reductive. So I guess that isn't a counter-example either. Jul 6 '12 at 16:37
• @Florian -- Thank you for pointing out the mistake. I gave another example where $X$ is affine. Jul 6 '12 at 20:13

Adam has given the simplest type of concrete example of what can go wrong, But since the formulation of the question has caused some confusion, it may be worth commenting at length on the broader algebraic group background involved. Here the given group $G$ can be any connected reductive algebraic group over an algebraically closed field $K$ (not just in characteristic 0), while $H$ is a (closed but not necessarily connected) reductive subgroup. The only assumption is that $G$ acts (as an algebraic group) on some affine variety $X$ with a closed orbit $G \cdot x$. The question is whether the orbit $H \cdot x$ must also be closed. [Some other language involving maximal compact subgroups in the question is irrelevant.] The answer is no, though the approaches of Adam and Jason both look somewhat ad hoc.

In Adam's example, $G$ is taken to be the product of two copies of the general linear group $X=\mathrm{GL}(2,K)$. Since $X$ is a homogeneous space for its natural left and right multiplication actions (taking inverses on the right so that multiplication becomes a left action), the combined left action of $G$ on $X$ makes $X$ into a homogeneous space. It is the orbit of any invertible matrix $x$, for instance $I$ whose isotropy group is the diagonally embedded copy of $X$ in $G$ acting by conjugation on itself (other isotropy groups being conjugate to this one). In particular, $X$ is just the affine homogeneous space of $G$ modulo that reductive subgroup. This construction is quite general.

In the example, the point $x \in X$ is chosen to be the standard unipotent Jordan block (a regular unipotent element of the group $X$), while $H$ is taken to be the standard maximal torus in $\mathrm{SL}(2,K)$ inside the diagonally embedded copy of the group $X$ in $G$. Now $H$ is reductive but has a mixture of orbits on the variety $X$, a closed orbit being that of $I$. But the orbit of $x$ fails to be closed (in the Zariski topology, or the analytic topology when that's appropriate). However, this orbit is a perfectly good affine variety, being in fact the quotient of $H$ by a subgroup of order 2.

In general, closed orbits always exist for such algebraic group actions (e.g., orbits of smallest dimension). But for a semisimple group, an orbit under the conjugation action is closed precisely when it is the orbit of a semisimple element. Working with unipotent elements is a natural way to get non-closed orbits. For example, if we chose $H$ instead to be the full diagonally embedded copy of $X$ in $G$, the orbit of $x$ would be its non-closed conjugacy class in $X$ while the isotropy group (= centralizer) would in fact be non-reductive. For more background on homogeneous spaces, see the old papers:

R.W. Richardson, Affine coset spaces of reductive algebraic groups, Bull. London Math. Soc. 9 (1977), 38-41

E. Cline, B. Parshall, and L. Scott, Induced modules and affine quotients, Math. Ann. 230 (1977), 1-14