# What is the dimension of $M/G$ if it is a manifold?

Let $G$ be a Lie group acting smoothly and freely on a smooth manifold $M$. If the action is proper, then we know that $M/G$ is a smooth manifold of dimension $$\dim (M/G)=\dim M-\dim G.$$ (See e.g. Lee, Introduction to Smooth Manifolds, Theorem 21.10.)

Now suppose that $G$ does not necessarily act properly on $M$ (but still smoothly and freely), and we know that $M/G$ is a topological manifold. Is the dimension necessarily also equal to $\dim M-\dim G$?

• Posted on mathoverflow.
– SHP
Commented Aug 31, 2016 at 12:49

Yes. This is easily seen as follows: let $m\in M$, then the maps $$*\stackrel{e}{\longrightarrow}G\xrightarrow{g\mapsto g\cdot m}M\xrightarrow{\text{quotient}}M/G\longrightarrow*$$ induce a short exact sequence of vector spaces for the tangents: $$0\longrightarrow\mathfrak{g}\longrightarrow T_mM\longrightarrow T_{[m]}(M/G)\longrightarrow0,$$ where $\mathfrak{g} = T_eG$ is the Lie algebra of $G$. As the dimension of a manifold is the same as the dimension of its tangent space, we get what you want. Notice that this is independent from the properness of the action (as long the quotient is a manifold).
• Thanks for your answer. You are implicitly saying that we can always find a smooth structure on $M/G$ such that $M\to M/G$ is a smooth submersion. Do you have any hint for that? Commented Aug 22, 2016 at 17:20
• @Peter Ah, you are only asking for $M/G$ to be a topological manifold... I have to think about it. I'll leave my answer here for the moment. Commented Aug 23, 2016 at 1:35