Let $M$ be a 3-manifold with boundary. If $M$ has an orientable finite cover that is a Seifert fiber space, then is $M$ also a Seifert fiber space?
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4$\begingroup$ I think you must have scrambled the question. Since every manifold has an orientable double-cover, it appears that right now you are asking if every 3-manifold is seifert-fibered (clearly false). $\endgroup$– Andy PutmanCommented Oct 5, 2013 at 5:28
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$\begingroup$ Sorry, M have finite covering orientable that is seifert then M is 3-manifold seifert $\endgroup$– jhoelCommented Oct 5, 2013 at 12:58
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$\begingroup$ A 3-manifold is a Seifert fibration if and only if the center of its fundamental group contains an infinite cyclic subgroup (Casson-Jungreis, Gabai). So you would have to check that this cyclic group is still central in the group extension which is the fundamental group of your original manifold, but this seems not obvious. $\endgroup$– ThiKuCommented Oct 8, 2013 at 5:24
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2 Answers
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This is certainly true, though one needs to be sufficiently careful about one's definition of Seifert fibred---it's important to allow fibres with a neighbourhood that looks like a fibred solid Klein bottle. See p. 429 of P. Scott, 'The geometries of 3-manifolds', Bull. LMS 15(5), 1983, pp. 401--487 .
I don't think I know a reference in the literature, but one can prove it using the following result (see Theorem 3.9 of Scott's paper).
Theorem: Let $M$ be a compact Seifert fibre space and let $f: M \to N$ be a homeomorphism. Then $f$ is homotopic to a fibre-preserving homeomorphism (and hence an isomorphism of Seifert bundles) unless one of the following occurs.
- $M$ is covered by $S^3$ or $S^2\times\mathbb{R}$,
- $M$ is covered by $S^1\times S^1\times S^1$,
- $M$ is $S^1 \times D^2$ or an I-bundle over the torus or Klein bottle.
It follows that, except in the above three cases, the Seifert structure on the orientable double cover is invariant under the covering transformation, and hence descends.
All closed 3-manifolds with finite fundamental group are orientable (by work of Epstein), so there are only finitely many manifolds left to check. I'll leave them as an exercise (which I confess I've never done myself).
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$\begingroup$ Thanks Henry, I'm studying his work Residually Free 3-Manifolds, I have some questions :). would you be kind enough to answer $\endgroup$– jhoelCommented Oct 5, 2013 at 19:34
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$\begingroup$ @jhoel - sure. Please feel free to contact me by e-mail. $\endgroup$– HJRWCommented Oct 5, 2013 at 20:51
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2$\begingroup$ One should add here that one needs the Spherical Space Form part of Perelman's geometrization theorem in order to make the conclusion in the spherical case. (The exercise left to the reader.) $\endgroup$– MishaCommented Oct 6, 2013 at 5:52
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$\begingroup$ @Misha - I don't think this is necessary. Epstein proved that every compact, non-orientable 3-manifold with finite fundamental group is homotopy equivalent to $\mathbb{P}^2\times I$. (The Poincar\'e Conjecture implies that the homotopy equivalence can be upgraded to homeomorphism, but we don't need that here.) In particular, it follows that there are no closed, non-orientable 3-manifolds with finite fundamental group, independent of Geometrization. $\endgroup$– HJRWCommented Oct 6, 2013 at 8:15
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$\begingroup$ Henry: Maybe I misunderstood the question, but I do not think the manifold was assumed to be nonorientable. $\endgroup$– MishaCommented Oct 6, 2013 at 12:08
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You need geometrisation to prove this fact. See Corollary 12.9.5 here for a reference.
You can't prove this without Perelman, at least with our present knowledge. For instance, if the orientable cover is $S^3$, then you must ensure that $M$ be elliptic, and that's precisely the space form conjecture, which is "one third" of geometrisation. But even when the finite cover is some other Seifert space, I don't see an easy argument to conclude without using geometrisation.
Edit. I overlooked the "with boundary" hypothesis. In that case Thurston's proof of geometrisation suffices.
answered Jan 20, 2017 at 12:54
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$\begingroup$ I think that, with a bit of care, in the non-spherical case one can get by with just using Nielsen Realization (and similar results for the Euclidean case). But I'm not sure this has been carefully written down anywhere. $\endgroup$– HJRWCommented Jan 20, 2017 at 13:11
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$\begingroup$ I am perhaps missing something here, but the question is asking about a 3-manifold with boundary. Doesn't that allow us to use Thurston's geometrization of cusped 3-manifolds? I don't see how a manifold with boundary can be covered by $S^3$. $\endgroup$ Commented Jan 20, 2017 at 16:28