Seifert fiberings of zero euler number which are semi-bundles

Question

Let M be a closed oriented manifold which has the structure of a "semi-bundle" (See Section 1.2. of Hatcher's notes on three-manifolds) over an interval I. Assume that M is Seifert fibered over a base B and that the Euler number of the Seifert fibering is zero. Suppose finally that M has a unique Seifert fibering, so B is uniquely determined. Is it always the case that B is non-orientable?

Answer 1

Let $\Sigma$ be the orientable generic fiber of the semi-bundle structure on $M$. I only consider the case $\chi(\Sigma) < 0$ for simplicity: the cases $\chi(\Sigma) \geq 0$ should be worked out by hand; in many cases the fibration is not unique there.

In that case the answer is yes, because of a stronger fact:

If an orientable Seifert manifold $M \to B$ contains an orientable separating incompressible surface $\Sigma$ with $\chi(\Sigma)<0$, then the base orbifold $B$ is non-orientable.

The proof goes as follows. Since $\Sigma$ is incompressibile, it is isotopic to a horizontal or vertical surface. Since $\chi(\Sigma)<0$, it cannot be vertical, so it is horizontal. The fibration now induces an orbifold covering map $\pi\colon\Sigma \to B$. Since $\Sigma$ is separating in $M$, one sees easily that some deck transformation of $\pi$ must be orientation-reversing.

The fact that I stated is actually equivalent your assertion, since every horizontal surface is a fiber in a (semi-)bundle structure for $M$.

Answer 2

I think that's right. Given your hypotheses, such a manifold should be constructed by taking a finite dihedral subgroup of the homeomorphism group of a surface $\Sigma$ with two (fixed-point free) involution generators $\alpha, \beta$, and creating a manifold by taking $M\cong \Sigma\times[0,1]/\{ (x,0)\sim (\alpha(x), 0), (x,1)\sim (\beta(x),1)\}$. They need to be orientation-reversing so that $M$ is orientable, and by construction this fibers over a mirrored interval by projecting to the second coordinate (so is a semi-bundle) and is Seifert-fibered by projecting to the (quotient of) the first coordinate. Then the base of the Seifert-fibration will be homeomorphic to $\Sigma / \langle \alpha, \beta \rangle$, a non-orientable 2-orbifold.

As an example, if $\Sigma \cong S^2$, then one should have the group $\mathbb{Z}/2\mathbb{Z}$ realized by $\alpha=\beta$ the antipodal involution of $S^2$. Then the manifold will be homeomorphic to $\mathbb{RP}^2\#\mathbb{RP}^2$ fibering over $\mathbb{RP}^2$. Notice that the fibration will be non-orientable to match the non-orientability of the base.

For $\Sigma \cong T^2$, I think that there are 3 possible conjugacy classes of finite dihedral subgroups of the mapping class group generated by non-orientable involutions. However, I think only two of these are realized by a manifold, and without a unique Seifert-fibering. One is the orientable circle bundle over the Klein bottle. The other has base a projective plane with two cone points of order 2 ($22\times$ in orbifold notation). One may determine this by the classification of wallpaper groups. The desired manifold is the unit tangent bundle to these orbifolds, with Seifert-fibering given by the unit tangent circles. But one may also "pull back" fiberings from the base orbifold to get Seifert fiberings of the manifold. However the base orbifold should be the same for each of the three fibrations.

For $\Sigma$ of higher genus, the model geometry will be $\mathbb{H}^2 \times \mathbb{R}$, and the Seifert fibering will be unique.