Stallings' fibration theorem states that if we have a compact irreducible $3$-manifold $M^3,$ with $G\rightarrow \pi_1(M^3) \rightarrow \mathbb{Z},$ and $G$ is finitely generated and is not of order $2,$ then $M^3$ fibers over a circle. The question is whether the last condition (that $G\neq \mathbb{Z}/2\mathbb{Z}$) is actually necessary (the answer is probably in Stallings' original paper, but I can't find it online).


This should follow from geometrization. The fundamental group of the manifold is $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}$, and geometrization should tell you that the manifold is then $\mathrm{RP}^2 \times S^1$.

I looked in Stallings's paper, and he says that it is a hard open problem, so this might be the only way.

| cite | improve this answer | |
  • $\begingroup$ That's what I was afraid of :( $\endgroup$ – Igor Rivin Nov 15 '11 at 20:04
  • $\begingroup$ Without geometrization, you can show it is homotopy equivalent to $\mathrm{RP}^2 \times S^1$, but I agree that you probably need the full force of geometrization to get diffeomorphism. $\endgroup$ – Steve D Nov 16 '11 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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