MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here reducible means that the mapping class for the fiber is a reducible auto-homeomorph in the sense of Nielsen-Thruston. So,

could anyone give me a hint to classify them?

In contrast, do you agree that -in the sense of connected sum- all these bundles are irreducible?

share|cite|improve this question
Are you asking for a classification of reducible elements in the mapping class group of the two-torus? – Sam Nead Dec 8 '09 at 20:00
well, modulo fiber isomorphisms of bundles, YES. do I remember well that an isotopy class give the same bundle? – janmarqz Dec 8 '09 at 21:28
up vote 4 down vote accepted

Yes - surface bundles over the circle are irreducible (*) as long as the fiber is not a two-sphere. This follows from the fact that the universal cover of such a surface bundle is homeomorphic to $\mathbb{R}^3$ and Proposition 1.6 of Hatcher's three-manifold notes.

(*) in the sense of connect sum.

EDIT: To answer the question posed by Juan in the comment. A orientation preserving homeomorphism $h$ of $T^2$ is reducible (that is, preserves the isotopy class of some essential multicurve) if and only if $h$ is a power of a Dehn twist or is the power of a Dehn twist followed by the hyperelliptic element. Here is a "cut-and-paste" proof: if $h$ preserves a multicurve then it preserves a curve, say $\alpha$. Now, $h$ either preserves or reverses the orientation of $\alpha$. If the latter case replace $h$ by $h$ followed by the hyperelliptic, to reduce to the former case. Isotope $h$ so that $\alpha$ is fixed pointwise. Note that, as $h$ preserves orientation of $T^2$, the sides of $\alpha$ are preserved as well. Thus $h$ restricts to a homeomorphism of the annulus, fixing the boundary pointwise. By the classification of mapping classes in the annulus, $h$ is a Dehn twist.

As a bit of an advertisement: Farb and Margalit have written a primer on the maping class group. You can find a discussion of the mapping class groups of the disk, annulus, and torus in Section 2.4, on the "Alexander Method". (In particular they give the usual proof that the group of orientation oreserving classes on $T^2$ is $SL(2,Z)$. They also give as an exercise the characterization of Dehn twists.)

I'll end by pointing out that there is not a contradiction between my answer and Hatcher's. If the monodromy is reducible then the resulting torus bundle is Seifert fibered and in fact has Nil geometry.

share|cite|improve this answer
non trivial F-surface bundles M have $\pi_1(M)=\pi_(F)*_{Z}Z$ which isn't a free product so it can't be reducible (in the sense of connected sum) and were the amalgamation is the group generated by the isotopy class of the auto-homeomorph... – janmarqz Dec 8 '09 at 21:37
@Juan - I would guess that there is also a proof along these lines. I would also guess that there are some details to fill-in here. For example, there certainly are HNN extensions that decompose as free products... – Sam Nead Dec 8 '09 at 21:50
for a torus bundle $T\subset E\to S^1$ we have that $\pi_1(T)$ is a normal subgroup of $\pi_1(E)=\pi_1(T)*_{Z}Z$ and if $\pi_1(E)$ could split freely then $\pi_1(T)$ would split freely too, which is ridiculous. So $\pi_1(E)$ can't freely splitted and then $E$ can't be a connected sum... – janmarqz Dec 11 '09 at 1:23
@Sam: 10-4, gonna check, super-thanks – janmarqz Dec 14 '09 at 0:58

OK, time to give some references for this classical material. Orientable 3-manifolds that are torus bundles are classified up to diffeomorphism by the conjugacy class of their monodromy map in SL(2,Z). [More precisely: two monodromy matrices A and B in SL(2,Z) determine diffeomorphic torus bundles if and only if A is conjugate in GL(2,Z) to either B or the inverse of B.] If the monodromy is elliptic (finite order) or parabolic (having an eigenvalue 1 or -1) then the torus bundle is Seifert fibered, so the classification in these cases is covered by the general classification of Seifert manifolds. If the monodromy is hyperbolic (two distinct real eigenvalues) then the torus bundle has a solvgeometry structure. One possible reference for the more topological parts of this classification is my notes on 3-manifolds, available on my webpage, where Section 2.2 gives the topological classification of orientable torus bundles in some detail, including the conjugacy classification of elements of SL(2,Z). For the "eight geometries" aspects of the classification there is the well-known expository article of Peter Scott, "The Geometries of 3-Manifolds", in the 1983 Bull.L.M.S. (a scanned pdf version is available on his webpage). See pp. 470-472 in particular.

The case of nonorientable torus bundles should be similar. Scott actually treats the nonorientable case, but I was lazy and didn't go that far. There are probably other references too that I'm forgetting at the moment. Can anyone else remember?

In particular, to answer the original question, torus bundles are irreducible both in the Thurston geometries sense and in the embedded spheres sense.

share|cite|improve this answer

I think the hint you need is that the (orientation-preserving) mapping class group of the torus is SL_2(Z). So you need to figure out which elements are "reducible", and what the corresponding bundles look like.

share|cite|improve this answer
Prof Wilton: do you know that there are only seven periodic elements? – janmarqz Dec 8 '09 at 21:38
I also believe that this will work. Another much more machine dependent proof would go through the Nielsen-Thurston classification of mapping classes and our previous discussion (on this website!) of the classification of periodic mappings of the two-torus. – Sam Nead Dec 8 '09 at 21:40
Maybe if we could see an algebraic condition for pseudo Anosov bundles... – janmarqz Dec 8 '09 at 21:41
They are called Anosov maps when the surface in question is the two-torus. They are exactly the elements of SL(2, Z) having an eigenvalue greater than one. Perhaps you would be interested in the relevant Wikipedia page: – Sam Nead Dec 8 '09 at 21:46
No. There's also reducible elements, things like $\begin{pmatrix} 1 & 0 \\ n & 1\end{pmatrix}$ for $n \neq 0$. – Ryan Budney Dec 14 '09 at 2:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.