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.
No comments:
Post a Comment