MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

A fiber bundle over a circle $M^{3} \longrightarrow S^{1}$ with fiber a surface $F_{g}$ is characterized via a homeomorphism $\varphi \colon F_{g} \to F_{g}$. It can be one of the following: periodic, reducible or psevdo-anosov. In which cases $M^{3}$ is a graph-manifold?

share|cite|improve this question
up vote 6 down vote accepted

$M$ is a graph manifold if and only if $\phi$ is not pseudo-Anosov and, in the reducible case, no $\phi$-orbit of components of the complete reduction of $\phi$ is pseudo-Anosov.

To prove this cut $M$ along the torus system $T$ obtained by suspending the canonical reducing system for $\phi$. If no component of the complete reduction of $\phi$ is pseudo-Anosov then each component of $M-T$ is Seifert fibered. If some $\phi$-orbit of components of the complete reduction is pseudo-Anosov then the corresponding component of $M-T$ is hyperbolic.

share|cite|improve this answer
Thanks a lot!!! – Andrew Jun 29 '12 at 15:34

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.