Seifert Fibered Spaces with Boundary are $\mathbb{P}^2$-irreducible

Question

I'm reading Peter Scott's The Geometry of 3-Manifolds and am trying to understand the argument behind this statement, which arises in the proof of Corollary 3.3:

If $M$ is a Seifert fibered 3-manifold with boundary, then $M$ is $\mathbb{P}^2$-irreducible.

I understand why $M$ is irreducible: the interior of the universal cover is $\mathbb{R}^3$ (as opposed to $S^3$ or $S^2 \times \mathbb{R}$), so any sphere in $M$ lifts to $\mathbb{R}^3$ wherein it bounds a ball. But why are 2-sided $\mathbb{P}^2$'s forbidden?

And if there are better hypotheses for the statement I've presented, please do share.

Answer 1

Here's an argument that doesn't involve covering spaces that illustrates some important facts about Seifert fibered spaces. I'm basically following Hatcher's "Notes on basic 3-manifold topology", Propositions 1.11 and 1.12, but I've reconsolidated my understanding. (See the edit history for one that follows Hatcher closer.)

Let $M$ be a connected compact irreducible Seifert fibered space, and let $S$ be a closed embedded incompressible surface. It turns out $S$ is isotopic to a surface which is either vertical (a union of regular fibers) or horizontal (transverse to all fibers), which we will see. The first thing we will do is describe a vertical 2D subcomplex $A\subset M$ such that the complement of a tubular neighborhood of $A$ is a disjoint union of vertical solid tori. Collapsing each fiber of $M$ yields a 2D orbifold $B$ called the orbit space, which is topologically a compact connected 2-manifold, and the exceptional fibers correspond to branch points of the orbifold. Choose an embedded 1-complex $\Gamma\subset B$ such that (1) every branch point is a vertex, (2) there is at least one vertex and at least one edge, and (3) the complement of a regular neighborhood of $\Gamma$ in $B$ is a disjoint union of disks. Now, let $A$ be the preimage of $\Gamma$, let $V$ be the preimage of the vertices of $\Gamma$, and let $A'=A-V$, which is a disjoint union of open annuli.

Perturb $S$ so that it is in general position with respect to $A$. If $S$ intersects $A'$ in a loop that bounds a disk, then take the innermost such and use the incompressibility of $S$ and the irreducibility of $M$ to construct an isotopy that removes this loop of intersection. After finitely many such moves and another isotopy, $S$ meets $A'$ only in arcs and vertical loops. If any of these arcs bounds a lune (this happens when both endpoints of the arc meet the same component of $V$), then take the innermost such. Since $S$ does not meet $\partial M$, we can push $S$ along this lune through $V$, reducing the number of points of intersection between $S$ and $V$ by two. Hence, after finitely many such moves, $S\cap A'$ consists of vertical loops and horizontal arcs.

Suppose there is a compressing disk $D\subset M-A$ for $S-A$ that does not bound a disk in $S-A$. Let $D'\subset S$ be a disk with $\partial D'=\partial D$. Since $M$ is irreducible, $D\cup D'$ bounds a ball so there is an isotopy carrying $D'$ to $D$. Because $D'$ intersected $V$, the new surface has fewer intersections with $V$, so after finitely many such moves we may assume $S-A$ is incompressible in $M-A$. Each component of $M-A$ is an open solid torus, and incompressible surfaces in solid tori are isotopic to disjoint unions of vertical annuli and horizontal disks, hence after an isotopy the surface is a collection of vertical annuli or horizontal disks within each component of $M-A$. Since the base surface $B$ is connected, having vertical annuli and having horizontal disks are mutually exclusive. In the first case $S$ is a vertical surface, and in the second $S$ is horizontal. This completes the argument

That was assuming $M$ was irreducible. Suppose $M$ is a connected compact Seifert fibered space with nonempty boundary, and suppose for sake of contradiction $S\subset M$ is an embedded sphere that does not bound a ball. Follow the steps of the preceding argument, but with three changes. First, when we eliminate loops of $S\cap A'$ that bound disks, instead use the disk to compress $S$, yielding two spheres; replace $S$ by whichever one does not bound a ball. Second, we assume $S-A$ is not sphere, since if it were it would bound a ball in the solid torus it meets of $M-A$ and hence in $M$. Third, if $S-A$ is not incompressible in $M-A$, then compressing $S$ with the compressing disk yields two spheres, at least one of which does not bound a ball in $M$; replace $S$ with this sphere. Therefore, we end up with a sphere $S$ that does not bound a ball that is isotopic to a vertical or horizontal surface. It can't be vertical since $2$-spheres are not $S^1$ bundles, so it's horizontal, but $M$ has nonempty boundary and $S$ does not.

Putting this all together: if $M$ is a connected compact Seifert fibered space with nonempty boundary, then it is irreducible. Hence, if $M$ had a two-sided $\mathbb{P}^2$, which is incompressible since it is $H_1$-injective, it is isotopic to a vertical or horizontal surface. It cannot be vertical since $\mathbb{P}^2$ is not an $S^1$ bundle, but it cannot be horizontal since it has no boundary but $M$ does. Therefore $M$ is $\mathbb{P}^2$-irreducible.

Answer 2

I know this is an old question, but just wanted to note that Kyle Miller's answer misses a subtlety -- the correct statement is that an orientable closed embedded incompressible surface in an orientable irreducible Seifert fibered space is vertical or horizontal. It isn't true in the one-sided case.

You need to assume that the surface $S$ is orientable as this allows you to assume that its intersection with the solid tori is too. The only incompressible orientable surfaces in the solid torus are indeed the meridian disc and boundary-parallel annuli; however there are lots of incompressible one-sided surfaces (all of which are non-orientable). See, for example, the results in 2.3-2.5 in Przytycki's nice paper for a summary of the classification; as he notes, it's implicitly proved in Rubinstein's paper in section 2. Rubinstein explicitly constructs a large family of incompressible surfaces in lens spaces that aren't horizontal or vertical. Lens spaces don't contain any horizontal surfaces, and don't contain any incompressible vertical ones. However $L(2n, 1)$ contains an incompressible connect sum of $n$ projective planes, constructed in this manner. Rannard gives the best possible correct classification of the nonorientable case in the main theorem of his paper.

For the $\mathbb{R}P^2$-irreducible question, it's enough to note that we generally require Seifert fibered spaces to be orientable by definition. For orientable 3-manifolds, irreducible and $\mathbb{R}P^2$-irreducible are equivalent (except for $RP^3$). (If we don't require orientability, it's not true -- take $\mathbb{R}P^2\times S^1$, which has a two-sided $\mathbb{R}P^2$ sitting in it as a section.)

Przytycki, Józef H., Nonorientable, incompressible surfaces in punctured-torus bundles over (S^1), Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat., RACSAM 113, No. 3, 1975-2000 (2019). ZBL1476.57022.

Rannard, Richard, Incompressible surfaces in Seifert fibered spaces, Topology Appl. 72, No. 1, 19-30 (1996). ZBL0859.57019.

Rubinstein, J. H., One-sided Heegaard splittings of 3-manifolds, Pac. J. Math. 76, 185-200 (1978). ZBL0394.57013.