21
$\begingroup$

Given a fiber bundle $S\hookrightarrow M \rightarrow S^1$ with $M$ (suppose compact closed connected and oriented) 3-manifold and $S$ a compact connected surface, it follows form the exact homotopy sequence that $\pi_1(S)\hookrightarrow \pi_1(M)$. Does this imply that the "fiber" of a 3-manifold which fibers over $S^1$ is well defined?

The answer should be NO, so I am asking: are there simple examples of 3-manifolds which are the total space of two fiber bundles over $S^1$ with fibers two non homeomorphic surfaces?

EDIT: the answer is NO (see Autumn Kent's answer). I'm just seeking for a "practical" example to visualize how this phoenomenon can happen.

Improve this question
$\endgroup$
1
  • $\begingroup$ I don't know an answer to this, but why do you think that there could only be one such $S$? $\endgroup$
    – Olivier Bégassat
    Commented Apr 7, 2011 at 19:49

2 Answers 2

Reset to default
30
$\begingroup$

There are simple examples with $M = F \times S^1$ for $F$ a closed surface of genus $2$ or more. Choose a nonseparating simple closed curve $C$ in $F$, then take $n$ fibers $F_1,\cdots,F_n$ of $F\times S^1$, cut these fibers along the torus $T=C\times S^1$, and reglue the resulting cut surfaces so that $F_i$ connects to $F_{i+1}$ when it crosses $T$, with subscripts taken mod $n$. The resulting connected surface is an $n$-sheeted cover of $F$ and is a fiber of a new fibering of $M$ over $S^1$. The monodromy of this fibering is a periodic homeomorphism of the new fiber, of period $n$.

Improve this answer
$\endgroup$
19
$\begingroup$

As you suspect, the answer is no. There are $3$-manifolds that fiber over the circle in infinitely many ways, with fibers of unbounded genera.

Thurston constructed a seminorm on $H_2(M,\partial M; \mathbb{R})$, now called the Thurston norm. The unit ball is a polyhedron, and Thurston showed that fibers of fibrations of $M$ over the circle are those integral classes which lie in the open cone on a collection of certain top dimensional faces (the ``fibered faces") of this ball. The norm is defined by extending the absolute value of the Euler characteristic of integral classes, and so any fibered manifold whose second homology has rank at least two gives you an example.

EDIT:

For a discussion of some explicit examples, see:

Hilden, Lozano, Montesinos-Amilibia, On hyperbolic 3-manifolds with an infinite number of fibrations over $S^1$. Math. Proc. Cambridge Philos. Soc. 140 (2006), no. 1, 79–93.

Improve this answer
$\endgroup$
4
  • $\begingroup$ You're right, I've read Thurston's original paper for a seminar but I focused on the foliations part. So I reformulate my question: instead of this argument, is there any simple "practical" example to visualize this phenomenon? $\endgroup$
    – Francesco Lin
    Commented Apr 7, 2011 at 20:34
  • 1
    $\begingroup$ I added a reference for some nice examples. $\endgroup$
    – Autumn Kent
    Commented Apr 7, 2011 at 20:59
  • 2
    $\begingroup$ As a follow up, Geometrization implies that any example for the question you asked must be hyperbolic. (Euclidean manifolds like $T^3$ often fiber in infinitely many ways, but the fibers will always be tori.) The examples in the cited paper are indeed pretty nice; they come from taking a branched covering of $T^3$ so that many of the different fibrations of $T^3$ lift to the cover, with different fibers. $\endgroup$
    – Dylan Thurston
    Commented Apr 7, 2011 at 23:46
  • 4
    $\begingroup$ What I wrote above about Geometrization is wrong, see Allen Hatcher's answer. $\endgroup$
    – Dylan Thurston
    Commented Apr 8, 2011 at 3:26

You must log in to answer this question.

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