Let $X$ be a Seifert fiber space, that is, a 3-manifold which is a circle bundle over a 2-orbifold. Suppose all generic fiber of $X$ is homotopically trivial, can we prove that the universal cover of $X$ is homeomorphic to $S^3$?
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$\begingroup$ I think that lens spaces have Seifert fibrations with this property. They fiber over a football orbifold with two cone points of order $n$, where $n$ is the order of the fundamental group. The generic fiber covers the exceptional fibers $n:1$, and hence lifts to the universal cover $S^3$, so is contractible. $\endgroup$– Ian AgolCommented Jan 27, 2021 at 3:19
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$\begingroup$ @IanAgol Sorry, I forget to mention the universal cover of $X$ in the question. $\endgroup$– TotoroCommented Jan 27, 2021 at 3:28
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2$\begingroup$ Doesn't this follow from your previous question. $\endgroup$– Steve DCommented Jan 27, 2021 at 6:07
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$\begingroup$ @SteveD: Of course, it does... $\endgroup$– Moishe KohanCommented Jan 27, 2021 at 18:14
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1 Answer
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Yes, this follows from classification of Seifert fibered spaces. In fact, you can change the hypothesis to "the fiber is torsion in $\pi_1(X)$" and still get the result. $\newcommand{\RR}{\mathbb{R}}$
Suppose that $X$ is reducible. Thus $X$ has $S^2 \times \RR$ geometry and the fiber is not torsion.
Suppose that $X$ has boundary. The boundary components are tori. By "one-half lives, one-half dies" there is a unique slope $\gamma$ that dies in homology. If the fiber is torsion, a power dies in homology, so in fact $\gamma$ is the fiber slope. The disk theorem tells us that $\gamma$ bounds a disk. Irreducibility tells us that $X$ is a solid torus. We appeal to the classification of fiberings of the solid torus and find that the fiber is not torsion, a contradiction.
Suppose that $X$ is algebraically toroidal. If the immersed torus is a union of fibers then the fibers are not torsion. If the torus is transverse to the fibers then it is embedded. In this case $X$ has $\mathbb{E}^3$ geometry, and the fibers are not torsion.
So, $X$ is Seifert fibered, closed, irreducible, and (algebraically) atoroidal. Thus it has $S^3$ geometry.
