Consider a Seifert fiber space. Is it always possible to find a finite cover that is a circle bundle and the preimage of any fiber is a finite union of circles?
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2 Answers
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If the Seifert fiber space is compact, then this is true, as long as the base orbifold is "good", which means that it has a finite-sheeted manifold cover, which is a compact surface. This induces a cover of the Seifert fiber space which is a circle bundle over the surface. If the base orbifold is bad, then no such covering will exist. This can happen for a Seifert fibering of $S^3$ over a football orbifold with distinct orders of torsion points, or over a teardrop orbifold. If the Seifert fiber space is non-compact, then there may be infinitely many exceptional fibers, and the base orbifold might have torsion of arbitrarily large order, so there is no hope of finding a finite-index cover which is a circle bundle.
See the draft of Thurston's book for more information on orbifolds and Seifert fibered spaces. Exercise 5.7.10 is on the Seifert fibering of $S^3$ over bad orbifolds.
answered Aug 21, 2010 at 19:54
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$\begingroup$ I was thinking about compact case. Thanks! And thanks for the book, it's kinda hard to read online though... $\endgroup$ Aug 21, 2010 at 22:30
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I believe the answer is yes (although I haven't actually checked). Here's the idea: Given a Seifert-fibred space you can think of it as being fibred over a $2$-orbifold. You can de-singularise that $2$-orbifold by taking the appropriate branched cover. Pulling back the Seifert-fibering gives you a genuine $S^1$-bundle. This skirts the issue of whether or not you can de-singularize the $2$-orbifold by an appropriate cover but I believe it's not hard to show such "bad" 2-orbifolds never occur as the base space to a Seifert-fibred $3$-manifold. Right, they're classified here: http://en.wikipedia.org/wiki/Orbifold and you can compare that to the base orbifolds of Seifert-fibred 3-manifolds to answer your question.
answered Aug 21, 2010 at 19:05
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$\begingroup$ I suppose what one needs to do is check what the bad orbifold types are -- if they're only things like the the non-Hopf fibration fiberings of $S^3$, just replace those Seifert fiberings with the Hopf fibering, and you'd be done. $\endgroup$ Aug 21, 2010 at 22:54
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$\begingroup$ Ah, I think my first response (the one that I erased) was the one you were looking for. It's as Ian says, but here is perhaps a slightly more concrete version -- take a lens space like $L_{p,q}$ and its fibering always lifts to a singular fibering of of all of its covering spaces. If you allow yourself the freedom of changing the seifert fibering after you take a covering space, this isn't a problem. $\endgroup$ Aug 21, 2010 at 23:13
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$\begingroup$ How do you know that it's always possible to substitute the fibering over a bad orbifold with a "good" fibering? $\endgroup$ Aug 23, 2010 at 0:59
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$\begingroup$ The caveat is you can only do it after a finite cover of the total space, so it's answering a slightly different question than you asked (the source of my confusion). But the reason is pretty simple. Compare the "bad" orbifolds here: en.wikipedia.org/wiki/Orbifold to the classification of the fiberings of Seifert-fibred manifolds here: math.cornell.edu/~hatcher/3M/3Mdownloads.html and you see the only problems are fiberings of lens spaces, $S^3$, $S^1\times S^2$ all of which have covers where you can replace the Seifert-fibrings with a bundle. $\endgroup$ Aug 27, 2010 at 22:33