5
$\begingroup$

Let $G$ be a compact Lie group acting effectively on a compact, Hausdorff topological space $X$. I am looking for results of the type

If $X$ is a ... and the action is ... then $\dim(X/G)\leq \dim(X)-\dim(G)$.

Here $\dim$ denotes the covering dimension.

For example, $X$ is a smooth manifold and $G$ acts smoothly and almost-freely (all isotropy groups finite).

Can anyone point me to a more general theorem of this type?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Something to do with positive-dimensional stabilisers, I'm sure would help here. $\endgroup$
    – David Roberts ♦
    Commented Feb 2, 2011 at 7:22

3 Answers 3

Reset to default
6
$\begingroup$

I'm loathe to answer my own question, but...

Theorem IV.3.8 of Bredon's book Introduction to compact transformation groups (which sadly is on the page google books decided not to include!) seems satisfying.

Let $G$ be a compact Lie group acting locally smoothly on the manifold $M$, such that $M/G$ is connected. If $P$ is a principal orbit (an orbit of maximum dimension) then

$$\dim(M/G)=\dim(M)-\dim(P).$$

Improve this answer
$\endgroup$
4
$\begingroup$

The following statement is not quite the inequality that you want, but it is for more general spaces than orbifolds. (Note that so-called "good" orbifolds are the quotients of locally linear actions on manifolds with finite stabilizers.)

COROLLARY 1.7.32 of R Palais' 1960 book, Classification of $G$-spaces, p41:

Let $G$ be a compact Lie group. Let X be a separable metric $G$-space. Then $\dim(X/G) = \sup_{H\leqslant G} \dim(X_H/W_G H) - \dim(G/H)$. In particular, $\dim(X/G) \leqslant \dim(X)$.

I adapted the statement slightly to be less terse. Here, $W_G H := N_G H/H$ is the Weyl group of $H$ in $G$, $G_x := \{g \in G ~|~ gx=x\}$ is the isotropy group of $x$, and $X_H := \{x \in X ~|~ G_x = H \}$ is the $H$-stratum.

Improve this answer
$\endgroup$
2
$\begingroup$

There is a finer approach to the dimension of the quotient when the action of $G$ on $M$ doesn't define a fibration, but has singular orbits. You'll find it here. Of course for the principal orbit you have your formula, but on singular orbits it gives something else. I didn't give a general formula (I should do it however) but I treated the examples $\Delta_n = {\bf R}^n/{\rm SO}(n)$. Topologically, for any $n$ these quotients are homeomorphic to the half line $[0,\infty[$, but not diffeologically, and if we have well (according to your formula) ${\rm dim}_x(\Delta_n) = 1$ for $x \neq 0$, we have however ${\rm dim}_0(\Delta_n) = n$; which proves by the way, that for 2 different quotients, if they are homeomorphic, they are not diffeomorphic. I'm not sure it will help you very much, because it seems that you are interested only in the principal orbits, but your question was too tempting to not underline this construction.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Pattrick, thanks for your answer. Diffeological spaces are new to me! Are they related to the notion of stratifold (as described in Matthias Kreck's book on Geometric Algebraic Topology)? Actually, I am really only interested in the topological (covering) dimension of the quotient, but its always good to learn about new notions of dimension. (Also, its not that I'm only interested in principal orbits, just that I can't think of examples with $\dim(P)\neq\dim(G)$!?) $\endgroup$
    – Mark Grant
    Commented Feb 3, 2011 at 6:51
  • $\begingroup$ Or rather I can't think of effective actions. $\endgroup$
    – Mark Grant
    Commented Feb 3, 2011 at 6:52
  • $\begingroup$ Hi Mark, diffeology is more general than stratifolds. Because also, it is not aimed to answer the same questions. A diffeological space $X$ is defined by claiming what parametrizations (maps defined on open subset of numerical vector spaces) in $X$ are "smooth". These smooth parametrizations (called plots) must satisfy 3 axioms (covering, smooth compatibility, locality). For $\dim(P)\neq\dim(G)$ what about the example os $\Delta_n$, you want something more singular? $\endgroup$
    – Patrick I-Z
    Commented Feb 3, 2011 at 8:01
  • $\begingroup$ Ah yes of course, in your example the principal orbits are spheres, which checks out with Bredon's theorem :) I have been too busy thinking about circle actions I forgot about the classical symmetry groups. Thanks. $\endgroup$
    – Mark Grant
    Commented Feb 3, 2011 at 8:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .