# Are Seifert fibered spaces with a horizontal surface exactly the surface bundles over the circle with periodic monodromy?

Are Seifert fibered spaces with a horizontal surface exactly the surface bundles over a circle with periodic monodromy?

I am unsure of my arguments for this: If a SFS has a horizontal surface then splitting along this surface gives a $$(surface \times I)$$ with a periodic monodromy. Conversely given a surface bundle over a circle with periodic monodromy the flow lines of the monodromy give a foliation of the manifold with circles, with the surface fiber as a horizontal surface.

## 1 Answer

This is not quite true, because after splitting along the horizontal surface you may get interval bundles over non-orientable surfaces, and these are not products (I am assuming your 3-manifold is orientable). This holds for instance if the manifold is a circle bundle over a non-orientable surface (with Euler number zero, so that it has indeed a horizontal surface).

If I am not missing something, the following facts should be equivalent on an orientable Seifert fibered space $$M$$ without boundary:

1. The fibration has a horizontal surface
2. The fibration has Euler number zero
3. The manifold $$M$$ either fibers over $$S^1$$ with periodic monodromy, or it fibers over the interval orbifold, in such a way that its natural double cover fibers over $$S^1$$ with periodic monodromy
4. The manifold $$M$$ has some cover (of any finite degree) that fibers over $$S^1$$ with periodic monodromy
5. The manifold $$M$$ is finitely covered by $$S\times S^1$$ for some closed orientable surface $$S$$.
6. The manifold $$M$$ has a geometry of type $$S^2 \times \mathbb R$$, $$\mathbb R^3$$, or $$\mathbb H^2 \times \mathbb R$$

There are few cases where the fibration is not unique, but "having a horizontal surface" or "having Euler number zero" are properties that hold for some fibration if and only if they hold for any fibration.