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 $S^1$.

## Best Answer

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.

Applying all of this to your specific situation will answer all your questions:

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.