# Quotient by a non-free action of a Lie group and manifolds with corners

The quotient manifold theorem says that

If $G$ is a Lie group acting freely and properly on a smooth manifold $M$ then $M/G$ has a (unique) smooth structure such that the projection $\pi:M\to M/G$ is a submersion.

I was wondering what happens when the action is not free. My intuition suggests that we get corners, I have in mind this example: The action of $\frac{\mathbb{Z}}{2\mathbb{Z}}$ over $\mathbb{S}^2$ induced by the reflection wrt the $zy$-plane. The quotient manifold obtained is $\mathbb{D}^2$ and the boundary $\partial\mathbb{D}^2$ can be identified with the fixed points of the action i.e. $\mathbb{S}^2\cap zy \text{-plane}$.

Does anyone know a theorem that covers the non-free case? Where can I read about it?

• There is no uniform treatment of the non-free case; the quotient is not in any natural way a smooth manifold (when G is finite, though, it is an orbifold). When the orbit space is 1- or 2-dimensional it is topologically a manifold with corners; there is a proof in Bredon's book on transformation groups. – Mike Miller Jul 27 '18 at 10:09
• I should say manifold with boundary in the above because topologically there is no difference. – Mike Miller Jul 27 '18 at 10:16
• If the action is proper, then the quotient is a Whitney stratified space. Each stratum is a manifold, and restricted to these, the quotient is a submersion. Also these glue together well, to give what is arguably the stratified version of a submersion. – David Roberts Jul 27 '18 at 10:54
• @DavidRoberts interesting, any reference? – Warlock of Firetop Mountain Jul 27 '18 at 11:10
• @WarlockofFiretopMountain, see theorem 15 in faculty.math.illinois.edu/~ruiloja/Math519/michiels.pdf. It refers to theorem 2.7.4 on page 113 of J.J. Duistermaat and J.A.C. Kolk. Lie Groups. Universitext. Springer, 2000. – Igor Belegradek Jul 27 '18 at 17:13