# The orbit space of a principal $S^1$ group action

geometric-topologylie-groupslow-dimensional-topologysmooth-manifolds

I was reading the classification of the circle actions on closed $$2$$-surfaces (which is connected, compact with no boundaries), but the author left some key details to the reader, here is a problem I met (where we denoted by $$M$$ the surface):

The orbit space $$M/S^1$$ of a principal $$S^1$$ action is a circle, and the equivariant tubular neighborhoods have the form $$S^1\times\mathbb{R}$$

I have no idea how to prove this claim, I know that since the action is principal, each orbit is a $$S^1$$, and I guess the $$M$$ could only be a torus $$\mathbb{T}^2$$ in this case.

Here is the explanation of the "principal action":

In the case of compact Lie group action on a connected compact manifold $$G\times M\longrightarrow M$$, each orbit $$G.x$$ will correspond to a conguation class of subgroups, called the orbit type, namely $$(G_x)=\{gG_xg^{-1}\}$$, which means all the isotropy Groups of that orbit are contained in that class, where $$G_x$$ is the isotropy group of point $$x$$.

A well-known theorem states that, the union of all orbits which have same orbit type $$(H)$$ is a submanifold, namely $$M_{(H)}$$, if it is dense in $$M$$, then we call an orbit of this type a principal orbit, another theorem states that any action of a compact Lie group acts on a connected compact manifold has a principal orbit, if all the orbits in the action are principal, we call it a principal action.

In our case, $$S^1$$ is an Abelian compact Lie group, and a principal action implies all the orbits are homeomorphic to $$S^1$$.

We begin with the slice theorem (which you seem to already be aware of). It asserts that if $$G$$ is a compact Lie group acting on a smooth manifold $$M$$ and $$p\in M$$, then there exists a tubular neighborhood of the orbit $$G\cdot p$$ which is $$G$$-equivariantly diffeomorphic to $$G\times_{G_p} S$$ where $$G_p$$ is the isotropy group at $$p$$ and where $$S$$ is a so called slice at $$p$$. Namely, $$S\subseteq (T_p G\cdot p)^\bot \subseteq (T_p M)$$ is diffeomorphic to a ball in the normal space of the orbit. Moreover, the action of $$G_p$$ on $$S$$ is linear, and $$G$$ acts on $$G\times_{G_p} S$$ by left multiplication on the first factor.

From my my answer here, an orbit is principal if and only if the $$G_p$$ action on $$S$$ is trivial. This has an important corollary: If $$M^0$$ denote the subset of $$M$$ given by unioning all the principal orbits, then the quotient space $$M^0/G$$ has the structure of a manifold for which the projection $$M^0\rightarrow M^0/G$$ is a submersion. The point is that the quotient of $$G\times_{G_p} S$$ under the $$G$$-action is naturally identified with $$S/G_p = S$$, which gives the Euclidean neighborhoods in the quotient.

First, in your particular situation, because all orbits are principal, neighborhoods of orbits have the form $$S^1\times_{S^1_p} S\cong (S^1/S^1_p)\times S\cong S^1\times \mathbb{R}$$. Second, since, $$M^0 = M$$ by assumption, the quotient $$M/S^1$$ inherits the structure of a manifold in such a way that the natural projection $$M\rightarrow M/S^1$$ is a subermsion.
Note that $$M/S^1$$ is $$1$$-dimensional. $$1$$-dimensional closed manifolds have been completely classified: the only option is that $$M/S^1\cong S^1$$. (I know Milnor's book "Topology from the differentiable viewpoint" has a proof of this classification result.)
By the way, you're right that $$M$$ has to be diffeomorphic to $$T^2$$ in this case. Here's a sketch. Since the action is principal, all isotropy groups are conjugate. But $$S^1$$ abelian, so conjugate subgroups are identical. That means, we can consider the action of $$S^1/S^1_p$$ and that this will have trivial isotropy group. In other words, the action will be free. This in turn tells you that the projection $$M\rightarrow S^1$$ is a principal $$S^1$$-bundle. Principal $$S^1$$ bundles over anything are classified: they are in bijective correspondence with elements of the second cohomology group. This group vanishes for $$S^1$$, so there is a unique principal bundle - the trivial one.