As Mumford pointed out in his paper 'Topology of Normal Singularities and a Criterion for Simplicity'(1961), every point $p$ on a normal complex surface $V$ has an associated 3-manifold $M$ which is obtained by taking a small sphere $S_\varepsilon$ around $p$ and define $M=S_\varepsilon \cap V$, and $p$ is a non-singular point if and only if $\pi_1(M)$ is trivial.

I want to get some feeling about this theorem, and I've found some useful examples in Milnor's book 'Singular Points of Complex Surfaces'. Denote $V(m,n,k)$ to be the zero set of $x^m+y^ n+z^k$, where $x,y,z$ are all complex numbers, then some surprisingly beautiful results appear:

$\bullet V(2,2,k)\cap S_\varepsilon$ is diffeomorphic to the lens space $S^3/G$, where $G$ is cyclic of order $k$;

$\bullet V(2,3,3)\cap S_\varepsilon$ is diffeomorphic to the space $S^3/G$, where $G$ is the quaternion group;

$\bullet V(2,3,4)\cap S_\varepsilon$ is diffeomorphic to the space $S^3/G$, where $G$ is the binary tetrahedral group;

$\bullet V(2,3,5)\cap S_\varepsilon$ is diffeomorphic to the space $S^3/G$, where $G$ is the binary icosahedral group, hence is Poincare sphere.

All the examples above have some common properties: $1/m+1/n+1/k>1$, and $M$ has universal covering space $S^3$.

When $1/m+1/n+1/k=1$, Brieskorn pointed out that $M$ must have infinite fundamental group, and has an open 3-cell as universal covering space. When $1/m+1/n+1/k<1$, it is conjectured that the same results will happen.

I've been trying to calculate some examples, and I noticed that the action of $S^1$ on $M$: $(x,y,z)\mapsto(e^{i\theta/m}x, e^{i\theta/n}y, e^{i\theta/k}z)$ is a foliation for all $(m,n,k)$, and it's actually a fibration when $m=n$ and $k$ is a multiple of $m$. For those cases, I can use Riemann-Hurwitz formula to calculate the genus of the base under the fibration, and show that the base is a closed oriented surface with $g\geqslant 1$, and hence prove the conjecture for those particular cases. I want to know, if this foliation isn't a fibration (it happens in almost every case), what can we say about the topology of $M$ by checking the action of $S^1$?