EDIT: Let a real Lie group $G$ act on a smooth manifold $M$ with finitely many orbits such that each orbit is locally closed ($M$, but not $G$, may be assumed to be compact in my case). Let $\mathcal{O}$ be an orbit of dimension $k$.

Is it true that $\overline{\mathcal{O}}\backslash \mathcal{O}$ is a union of orbits of dimension strictly less that $k$?

(Notice that if $\mathcal{O}$ is not an orbit, but just a locally closed submanifold, then the dimension of $\overline{\mathcal{O}}\backslash \mathcal{O}$ can be arbitrary: take the graph of the function $sin(1/x)$ as a simplest example.)

If the answer to the above question is negative in general, one may assume that $G$, $M$ and the action are real analytic. (In this case I do not know if the orbits are subanalytic subsets.)

A reference would be very helpful.

ADDED: A reference to a precise statement even in the real algebraic case would be helpful too.

UPDATE: It follows from the example by Yves Cornulier below, the answer is negative for $C^0$-action and probably for $C^\infty$-action. The question is still open for real analytic and real algebraic actions.

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    $\begingroup$ Your question is not clear. Are you assuming that there are only finitely many orbits and that each is locally closed or that only finitely many of the orbits are locally closed. If the latter, then the answer is 'No, it is not true, even in the analytic case.' For an example, let $G =\mathbb{R}\subset\mathrm{SO}(4)$ be a connected, noncompact, $1$-parameter subgroup and let $M=S^3\subset\mathbb{R}^4$. With the obvious action of $G$ on $M$, there are only two locally closed orbits in this case, but the closure of the generic orbit is a $2$-torus. $\endgroup$ May 27, 2014 at 15:52
  • $\begingroup$ I meant the former case of finitely many orbits, each locally closed. Thanks. Corrected. $\endgroup$
    – asv
    May 27, 2014 at 16:28

4 Answers 4


Here's an example with a topological action of $\mathbf{R}^2$ on the 2-torus with 3 orbits; I expect it can be made $C^\infty$ but definitely cannot be made real analytic.

If $X$ is a topological space and $f$ a self-homeomorphism, recall that its mapping cylinder $\mathrm{Cyl}(X,f)$ is $X\times [0,1]$ modulo identification $(x,1)\equiv (f(x),0)$. It admits a continuous action of $\mathbf{R}$ given by $t\cdot (x,u)=(f^{-\lfloor t+u\rfloor}(x),u+t-\lfloor u+t\rfloor)$. If $X$ is a smooth manifold and $f^{\pm 1}$ is smooth as well, then so is the mapping cylinder and the action. (Here smooth can mean $C^k$, $C^\infty$, $C^\omega$).

Now consider a continuous action of a topological group $H$ on $X$ commuting with $h$, in the sense that $h\cdot(f(x))=f(h\cdot x)$ for all $x\in X$ and $h\in H$. Then the action of $H$ extends to an action on $\mathrm{Cyl}(X,f)$, simply given by $h\cdot (x,u)=(h(x),u)$, and this action commutes with the flow, so that we get a continuous action of $G=H\times\mathbf{R}$ on $\mathrm{Cyl}(X,f)$. Clearly if the action of $H$ has finitely many orbits, then so does the action of $G$, and if the action of $H$ has locally closed orbits, then so does the action of $G$. Also if the action of both $f$ and $H$ is smooth in any sense then so is the action of $G$.

Now we pick $X$ to be the circle, identified to the projectivized line $\mathbf{R}\cup\{\infty\}$ by sterographic projection, $f(x)=x+1$, $f(\infty)=\infty$. Then we pick $H$ to be the reals, acting in such a way that its orbits are the integers (singletons) and the intervals between integers. [This action can obviously made analytic outside $\{\infty\}$ but behaves badly close to infty]. Anyway the action of $G=\mathbf{R}^2$ on the 2-torus $\mathrm{Cyl}(X,f)$ has 3 orbits: $C=\{\infty\}\times [0,1]$, the orbit $C'$ of $(0,0)$, which is orbit of $(0,0)$ by the flow of the mapping cylinder and is fixed by $H$; it's an immersed line, accumulating on the orbit $C$, and the dense open orbit $C''$, on which $G$ acts freely. Then $C'$ is 1-dimensional and so is the circle $\bar{C'}-C'=C$.

So far this is only a topological action of $\mathbf{R}^2$ on the 2-torus. To get a better smoothness, we need the action of $H=\mathbf{R}$ to be made smooth. To make it real analytic is hopeless because $\infty$ is an accumulation point of isolated fixed points. On the other hand, to make it $C^\infty$ does not sound hard.

  • $\begingroup$ Great! This seems to work. Thanks a lot. I will not mark it yet as a final answer since I still hope to get a positive answer in real analytic or real algebraic case. $\endgroup$
    – asv
    May 28, 2014 at 6:24

Let a reductive group $G$ act on an affine variety $X$ (assume the base field $k$ is algebraically closed and of characteristic zero). Since the algebra of invariants $k[X]^G$ is finitely generated and does not contain nilpotents, the affine variety $X/\!/G:=Spec\, k[X]^G$ is defined. The morphism $\pi:X\to X/\!/G$ induced by the embedding $k[X]^G\to k[X]$ is constant on the orbits of $G$. Some additional, easy to check properties: $\pi$ is surjective, sends invariant closed sets in $X$ to closed sets in $X/\!/G$, and any fiber of $\pi$ is a union of orbits and contains a unique (Zariski-) closed orbit which is of minimum dimension among orbits in the fiber (this is rather elementary, you do not need Luna's slice etale for that). Note that $\pi^{-1}(\pi(x))=\{y\in X| x\in cl(G\cdot y)\}$ if $G\cdot x$ is a closed orbit. Any orbit $\mathcal O$ is open in its closure, so $cl(\mathcal O)\setminus\mathcal O$ is a union of orbis of strictly lower dimension, see J. Humphreys book on Linear Algebraic Groups, Proposition in section 8.3.

In the projective case, results are different, as Allen pointed out in the example.

The results for non-algebraically closed fields, e.g., real algebraic actions, are slightly more complicated. I refer you to the excellent paper (I do not have time now, I will elaborate on that later if necessary)

MR1285780 (95f:14090) Reviewed Bremigan, Ralph J.(1-BLS) Quotients for algebraic group actions over non-algebraically closed fields. J. Reine Angew. Math. 453 (1994), 21–47. 14L30 (14D25) PDF Clipboard Journal Article Make Link

Let $k$ be a field of characteristic zero, not necessarily algebraically closed. The author studies the geometric invariant theory (GIT) quotients over a such a field. Let G be a reductive algebraic group acting on an affine variety V, both defined over the closure of k. The standard GIT quotient $V/\!/G$ exists. Let $(V_k,G_k)$ be the $k$-points of $(V,G)$. The author discusses the space $X=\pi(V_k)$ as a quotient object for the action of $G_k$ on $V_k$, where $\pi:V\to V//G$ is the GIT quotient map. If $k$ is local, there is a second possibility: the space $V_k/\!/G_k$ of closed $G_k$-orbits in $V_k$ is Hausdorff and supports a quotient map $p:V_k\to V_k/\!/G_k$. The main results are in $\S5$.


Let $M$ be the vector space $V$ of quadratic forms on ${\Bbb R}^n$ (or its projectivization), and $G=GL(n)$. The open orbits correspond to non-degenerate quadratic forms, and there are precisely $n+1$ open orbits. This is an example of real algebraic action with $n+1$ open orbits. There are finitely many orbits, because each quadratic form over reals is equivalent to diagonal one with eigenvalues 0 and $\pm 1$.

When $n=2$, and $M={\Bbb P}V$, each orbit is locally closed, because the set of degenerate quadratic forms is a disconnected union of two closed orbits.

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    $\begingroup$ I don't think it answers the question, because the closure of the orbit $O$ of $(p,q,r)$ ($p$ times 1, $q$ times $-1$, $r$ times 0) consists of the union of the orbits $(p',q',r')$ with $p'\le p$ and $q'\le q$, thus $\bar{O}-O$ is the union of the orbits $(p',q',r')$ with $p'\le p$, $q'\le q$, $p'+q'<p+q$. Write $d=p+q$. While $O$ has dimension $f(d)=n^2-d(n-d)-(d-n)^2-d(d-1)/2$, $\bar{O}-O$ has dimension $f(d-1)$ , and $f(d)-f(d-1)=n-d+1$ for all $1\le d\le n$. Thus $\bar{O}-O$ has dimension smaller than the dimension of $O$. $\endgroup$
    – YCor
    May 28, 2014 at 16:07
  • $\begingroup$ Agree with Yves Cornulier. $\endgroup$
    – asv
    May 29, 2014 at 11:25
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    $\begingroup$ I apologize for misreading the question $\endgroup$ May 29, 2014 at 15:19

For reductive groups and smooth actions the answer seems to be positive, if I remember correctly; this follows from Luna's slice theorem:

  • D. Luna, Slices ́etales, Sur les groupes alg ́ebriques, Soc. Math. France, Paris, 1973, pp. 81– 105. Bull. Soc. Math. France, Paris, M ́emoire 33.

There seems to be even the statement: Each orbit contains exactly one closed orbit in its closure.
Added later: At least for a representation. See Claudio Gorodski's more detailed answer.

By the way, an orbit a smooth Lie group action on a smooth manifold is always an initial manifold, see here, or also theorem 6.4 in this book.

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    $\begingroup$ I don't understand the "exactly one closed orbit in its closure" statement. Consider $\mathbb C^\times$ acting on $\mathbb{CP}^1$. $\endgroup$ May 28, 2014 at 20:47

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