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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 $S^1$?

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The "conjecture" is certainly true: you always get a particular Seifert manifold whose universal covering is known to be homeomorphic to three-space since the early work of Seifert - 80 years ago. – Bruno Martelli May 31 '11 at 7:35
The story is completely worked out in the Orlik reference in the Wikipedia page Bruno links to. You might also like to read the Peter Scott paper linked to on that page, to see the description these manifolds have from geometrization. – Ryan Budney May 31 '11 at 8:45
Thank Bruno & Ryan for telling me these useful references´╝ü – Yuchen Liu Aug 17 '11 at 3:57

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