Recently I've been reading the Jankins/Neumann notes Lectures on Seifert Manifolds (PDF). In the notes it is claimed that Seifert spaces are almost fiber bundles (with $S^1$ fiber), except around their exceptional fibers.

The example given is of taking a a solid cylinder $D^2\times I$, and gluing the ends with a rational $\frac{p}{q}$ turn; then the central fiber $0\times I$ becomes a circle in the resulting space that doesn't have a locally trivial neighborhood. But this space is still a solid torus, so of course it still is a fiber bundle $S^1\rightarrow M\rightarrow D^2$.

I think the point is that it isn't a fiber bundle in the "obvious" way, but I'm now curious about finding a connected Seifert fiber space $M$ that is not a fiber bundle $S^1\rightarrow M\rightarrow F$ for some surface $F$. Does such an $M$ exist? Ideally $M$ would have as many of the following qualifiers as possible: compact; orientable; without boundary.

  • 1
    $\begingroup$ The most basic examples will be lens spaces which, with few exceptions, are not circle bundles. This one can see by looking at their fundamental groups (just make sure the order is $\ge 3$). $\endgroup$ Commented Mar 28, 2023 at 1:59
  • $\begingroup$ @MoisheKohan: thank you for your comment! It does seem that only the $L(p,1)$ lens spaces can fit into the fiber bundle $S^1\rightarrow M\rightarrow S^2$. But I don't understand the fundamental group part: it seems like $L(5,2)$ would be a lens space without being a fiber bundle, but its fundamental group is order $5$, same as the fiber bundle $L(5,1)$. Is there something extra to do here to distinguish the two cases? $\endgroup$ Commented Mar 28, 2023 at 6:25

1 Answer 1


Definitely in many cases there is no actual fiber bundle structure. In fact, in most cases the Seifert fibration on a closed orientable Seifert fibered 3-manifold is unique, so if it isn't already a fiber bundle then it doesn't have an alternate structure as an honest $S^1$-bundle over a surface. For a precise statement of this uniqueness that lists all the exceptions, see Theorem 2.3 of Hatcher's notes on 3-manifolds, which can be found here.

  • $\begingroup$ Sorry, are you saying "definitely" to the question of such an $M$ existing? Is there an easy such example? I think I understand what you're saying, but my go-to examples of Seifert spaces are all listed in the exceptions of theorem 2.3 you referenced. $\endgroup$ Commented Mar 28, 2023 at 1:04
  • 2
    $\begingroup$ @Hempelicious: Sorry, I changed the first sentence to clarify what I mean. It's easy enough to construct Seifert fibrations that aren't on Hatcher's list -- just start with an actual $S^1$-bundle over a surface (you can even use the trivial one!), and then do a series of Dehn surgeries along fibers to introduce enough singular fibers that you aren't on Hatcher's list of exceptions. $\endgroup$ Commented Mar 28, 2023 at 1:06
  • $\begingroup$ Ah, OK that makes a ton of sense, thanks! Can you tell me if my rephrasing captures what you're saying correctly? If $M$ is a Seifert fiber space with exceptional fibers, then if it is a fiber bundle $S^1\rightarrow M\rightarrow F$, that gives us two non-isomorphic fiberings of $M$, so if $M$ is not on the list in Hatcher's notes, then this is a contradiction. $\endgroup$ Commented Mar 28, 2023 at 6:23
  • 1
    $\begingroup$ @Hempelicious: That's right! $\endgroup$ Commented Mar 28, 2023 at 14:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .