Examples of Lie group actions on manifolds with singular quotients Let G be a Lie group acting on a manifold M And say that there exists a point $m \in M$ such that the stabilizer of the point is non trivial. I am unable to understand how the quotient M/G fails to be a manifold. More precisely, how is a singularity created at the point $m \in M$ while constructing the quotient. Can anybody help me understand this?
 A: A general nonsense answer to your question would be the orbit decomposition theorem.  Given a compact lie group $G$ acting on a manifold $M$, it describes a stratification of $M$ by orbit types. 
The main theorem is that if $p \in M$, and $G_p$ is the stabilizer of $p$, then there is a $G$-equivariant diffeomorphism between a neighbourhood of $G.p$ (the orbit of $p$) in $M$ and 
$$V \times_{G_p} G$$
where $V$ is a vector space and $G_p$ acts linearly on $V$, i.e. there is a representation $G \to Aut(V)$.  So $V \times_{G_p} G$ denotes the quotient of $V \times G$ by the diagonal action of $G_p$.  $V \times_{G_p} G$ is an equivariant tubular neighbourhood of $G.p$, and the projection map $V \times_{G_p} G \to G.p$ is simply projection onto the 2nd factor $V \times_{G_p} G \to G/G_p$ followed by the identification $G/G_p \equiv G.p$ given by the action.  
So the underlying reason for $M/G$ to have singularities is that $(V\times_{G_p} G)/G \simeq V/G_p$ i.e. the action of $G_p$ on $V$ tends to have a singular quotient.  Igor gave an example of this in his comment. 
A: Complementing Ryan's answer: we can assume that $G$ preserves a Riemannian metric on $M$ (since $G$ is compact, we can take just any metric and average it with respect to $G$). Then $V$ from Ryan's posting is the the orthogonal complement in $T_p M$ of the tangent space of the orbit.
The quotient $M/G$ locally looks like $V/G_p$. This will often be singular, but when $M$ is a complex 1-manifold and $G$ is a finite group that acts by holomorphic transformations, the quotient will be a smooth manifold: any finite subgroup $H\subset GL_1(\mathbb{C})=\mathbb{C}^*$ is generated by some root of 1, and so $\mathbb{C}/H\cong \mathbb{C}$.
