Let $G$ be a reductive group acting on an affine variety $X$. For simplicity, one may assume $G=SL_n$ or $G=U_n$ and assume the field is $\mathbb C$. Given this one can show $\mathbb C[X]^G$ is finitely generated algebra.


(1) Each of the orbit closures $\overline{Gx}$ is a finite union of $G$-orbits.

(2) Each of the orbit closures $\overline{Gx}$ contains a unique closed orbit which has minimal dimension among these $G$-orbits.

I guess the two questions might be related to a result saying that $$ \pi(x_1)=\pi(x_2) \ \ \ \ \Longleftrightarrow \ \ \ \ \overline{G x_1} \cap \overline{G x_2} \neq \varnothing $$ where $\pi: X\to X //G\equiv Spec \mathbb C[X]^G$ is the GIT quotient.

EDIT: For question (2), suppose $Gx$ is not closed. If $y\in \overline{Gx}$ but $y\not \in Gx$, then what relation can we expect between $\overline{Gx}$ and $\overline{Gy}$? For example, do we have $\dim\overline{Gy} < \dim\overline{Gx}$, or $\dim G_y \ge 1$?

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    $\begingroup$ Question 1 has a negative answer. The counterexample that I love the best is when $G$ equals $\textbf{GL}_5$ and $X$ is the affine cone over a (projective) parameter space of rational normal curves of degree $4$ in $\mathbb{P}^4$, e.g., the Hilbert scheme, the Chow variety, or the space of Kontsevich stable maps of genus $0$. There is one dense orbit parameterizing the rational normal curves. However, these can specialize to a union of 4 concurrent lines. The usual cross-ratio for 4 points on $\mathbb{P}^1$ give continuous moduli of orbits in the orbit closure. $\endgroup$ Jun 22, 2017 at 2:28
  • $\begingroup$ Thank you. But, can we add some further conditions to make it correct? $\endgroup$
    – Hang
    Jun 22, 2017 at 2:36
  • $\begingroup$ You can add stability conditions "to make it correct". The GIT quotient only behaves well on semistable points. In the opposite case of nullcone (for a linear action), very complicated behavior is possible. This is illustrated in the standard counterexample to Q1 presented in my answer, where the nullcone is the whole space $X$. $\endgroup$ Jun 22, 2017 at 4:55

2 Answers 2


The answer to (2) is affirmative: an affine $G$-variety with a dense open orbit (like $\overline{Gx}$) contains a unique closed orbit. The reason for that is that any two closed are orbits are separated by a $G$-invariant. The closed orbit is clearly the unique orbit of minimal dimension since it is contained in the closure of every other orbit.

Question (1) is more delicate. As Victor Protsak's examples show there are, in general, infinitely many orbits in an affine orbit closure. There are some positive results, though:

  1. V.L. Popov has shown (Quasihomogeneous affine algebraic varieties of the group SL(2). Izv. Akad. Nauk SSSR Ser. Mat 37 (1973), 792–832) that $\overline{Gx}$ has finitely many orbits when $G=SL(2)$.

There are also criteria in terms of the isotropy subgroup $H=G_x$:

  1. $\overline{Gx}$ is a finite union of $G$-orbits if $G/H$ is spherical. In that case any embedding (affine or not) of $G/H$ has finitely many orbits. That is even equivalent to $G/H$ being spherical.

  2. $\overline{Gx}$ is a finite union of $G$-orbits if $H$ is a reductive subgroup which is of finite index in its normalizer (e.g. $H=T$, the maximal torus). In that case $G/H=\overline{G/H}$ is already closed in $X$ (by Luna: Adherence d'orbites).

  3. The paper Arzhantsev, I. V.; Timashev, D. A. Affine embeddings with a finite number of orbits. Transform. Groups 6 (2001), no. 2, 101–110 contains more results in this direction. In particular, the authors show that examples 2 and 3 exhaust (almost) all instances when $H$ is reductive.

  • $\begingroup$ Thank you for your answer. For question (2), suppose $Gx$ is not closed. If $Gy$ is the unique closed orbit of minimal dimension, then what relation can we expect between ${Gx}$ and ${Gy}$? For example, do we have $\dim\overline{Gy} < \dim\overline{Gx}$, or $\dim G_y \ge 1$? $\endgroup$
    – Hang
    Jun 27, 2017 at 14:57
  • $\begingroup$ Both. Since $Gy$ is in the closure of $Gx$, one has $\dim Gy<\dim Gx$. In particular, $\dim G_y>0$. Luna's slice theorem tells you even more: one can choose $y$ in such a way that $G_y$ is reductive and contains $G_x$. $\endgroup$ Jun 27, 2017 at 18:46

There is no reason to expect positive answer to Q1, even for a linear action with a dense orbit.

Let $X$ be the space of $n\times m$ matrices over a field $K$ with $G={\rm GL}_n(K)$ acting by left multiplication. By basic linear algebra $A$ and $B$ are in the same orbit iff $\ker A=\ker B$. Now suppose that $n\geq m\geq 2$. Then the matrices with zero kernel (i.e., of maximal rank, $m$) form a single dense orbit in $X$. However, the orbits of $G$ on $X$ are parametrized by pairs $(r, L)$, where $r\leq m$ and $L$ is an $r$-dimensional subspace of $K^m$. In particular, the orbit space is infinite.

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    $\begingroup$ The correct condition is $n\ge m\ge2$. $\endgroup$ Jun 22, 2017 at 11:06
  • $\begingroup$ Another way to see it: the $GL(n)$-orbits in the space of $n\times m$-matrices are classified by matrices in reduced row echelon form. Clearly, there are infinitely many if $n\ge1$ and $m\ge2$. $\endgroup$ Jun 22, 2017 at 11:41
  • $\begingroup$ You are right. I realized it right away, but was a bit tired to update the post. $\endgroup$ Jun 22, 2017 at 15:07

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