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Often when people write about the geometrization conjecture they assume (for simplicity) that the manifold is orientable. I never seriously thought of non-orientable 3-manifolds, but while reading Morgan-Tian's paper Completion of the Proof of the Geometrization Conjecture I realized that they prove the geometrization for every compact 3-manifold $M$ with no embedded two-sided projective planes, which was news to me. There is a lot of notation in their paper so I ask

Question 1. Is the above a correct interpretation of what is proved in Morgan-Tian's paper?

Note: one way to rule out two-sided projective planes is to assume that $\pi_1(M)$ has no 2-torsion (because then a two-sided projective plane would lift to the orientation cover where it cannot exist). In particular, this gives the geometrization conjecture when $M$ is aspherical (in which case $\pi_1(M)$ is torsion free).

Question 2. What is the status of the geometrization conjecture for manifolds that contain two-sided projective planes?

Note: on the last two pages of Scott's wonderful survey "The Geometries of 3-manifolds" he describes a version of the geometrization conjecture that makes sense in the presence of two-sided projective planes.

UPDATE:

  1. Looks like my reading of Morgan-Tian was hasty, and I no longer think they prove the geometrization for non-orientable manifolds that contain no two-sided projective planes. They only prove it for manifolds that become extinct in finite time under Ricci flow.

  2. As we discussed with Ryan in comments the geometrization for manifolds that contain two-sided projective planes reduces to geometrization of certain non-orientable orbifolds with isolated singular points. However, in contrast with what Ryan says, I was unable to find the geometrization for such orbifolds in the literature. Again, lots of particular cases are known, but I could not find it claimed (let alone proved) in full generality.

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  • $\begingroup$ The geometrization conjecture applies to all compact 3-manifolds. My understanding is that it tends to be presented in simplified forms for various reasons. Perhaps the biggest one is a precise statement in full generality takes a lot of space to write up! Francis Bonahon's paper "Geometric Structures on 3-Manifolds" is a great source: www-bcf.usc.edu/~fbonahon/Research/Preprints/Preprints.html $\endgroup$
    – Ryan Budney
    Commented Sep 8, 2010 at 20:13
  • $\begingroup$ In particular once you get to 3-manifolds with boundary, geometric structures tend to not be unique, in general. For example, the trefoil complement can be considered to have both $H^2 \times \mathbb R$ geometry and $PSL_2 \mathbb R$ geometry. I think it's for reasons like this that simplified versions tend to be stated most often. $\endgroup$
    – Ryan Budney
    Commented Sep 8, 2010 at 20:18
  • $\begingroup$ Ryan, doesn't the statement of the geometrization conjecture (conjecture 4.1) in Bonahon's survey exclude manifolds with projective plane on the boundary? Is there a version of the conjecture for such manifolds? $\endgroup$
    – Igor Belegradek
    Commented Sep 8, 2010 at 20:44
  • $\begingroup$ It's okay if they're on the boundary. In particular this is exactly the kind of manifold you get after doing the $\mathbb RP^2$-decomposition -- see Theorem 3.2 (page 34). $\endgroup$
    – Ryan Budney
    Commented Sep 8, 2010 at 20:57
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    $\begingroup$ Oh, that's interesting. The Boileau, Maillot and Porti paper in the beginning (Theorem 3.27) announce orbifold geometrization but Theorem 9.1 is restricted to orientable orbifolds, and none of the other references they mention seem to cover the non-orientable case -- if they do it's not clear from the statements of the theorems, anyhow. I've never had a direct interest in the non-orientable case so I've never chased down this story. I see what your concern is, now. $\endgroup$
    – Ryan Budney
    Commented Sep 9, 2010 at 6:43

2 Answers 2

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Most 3-manifold topologists tend to hypothesize away 2-sided projective planes. If a 3-manifold contains a 2-sided projective plane, then it must be non-orientable, and the preimage of the projective plane in the orientable double cover must be essential. Cutting along a maximal collection of disjoint essential embedded 2-spheres in the orientable cover (which may be made equivariant by Meeks-Yau), and capping off with spheres, one obtains an irreducible 3-manifold with an orientation reversing involution. Corresponding to projective planes will be spheres coned off which are acted on by the involution as the suspension on the antipodal map. If the components of this 3-manifold are atoroidal, then this is covered by a result of Dinkelbach and Leeb for spherical, hyperbolic, and $S^2\times R$ metrics. The case of the other homogeneous metrics was taken care of before by Meeks and Scott.

As far as I can tell, the toroidal case may still be open in general. One may argue that the involution preserves the JSJ tori, so it preserves the decomposition into geometric pieces. But I'm not sure it is now completely proven that the action on each geometric piece is standard, so that the quotient is geometric on each JSJ piece, because the results quoted above only cover involutions on closed geometric manifolds. A theorem of Hatcher implies that it is isotopic to such an involution, but I'm not sure his theorem implies that the isotopy is achieved by a path of involutions (with fixed points).

Kleiner and Lott are working on a proof of the orbifold theorem using Ricci flow, and I think their argument ought to give a unified argument for all of these things.

For aspherical non-orientable 3-manifolds, I think the proof via Ricci flow still works, even though Perelman didn't formulate it this way. Or Thurston's original proof should work, since these are Haken manifolds.

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    $\begingroup$ I corrected a typo in the above answer. Thanks for thinking about this! I do not think orientability assumption is natural and it would be nice if eventually the geometrization is stated (and proven) without it. Hopefully, Kleiner-Lott will do it this way. $\endgroup$
    – Igor Belegradek
    Commented Sep 10, 2010 at 15:14
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    $\begingroup$ So that I do not forget, here is the reference for the assertion that non-orientable closed aspherical manifolds are Haken: lemma 6.7 in Hempel's "3-manifolds" where it is actually proved that a compact non-orientable 3-manifold is sufficiently large if its boundary does not contain projective planes. $\endgroup$
    – Igor Belegradek
    Commented Sep 13, 2010 at 22:31
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    $\begingroup$ Unfortunately, the paper by Kleiner and Lott deals only with the orientable case. There is a paper by Bruno Zimmermann B. Zimmermann, Finite group actions on Haken $3$-manifolds. Quart. J. Math. Oxford Ser. (2) 37 (1986), no. 148, 499–511, which claims that the finite group actions on geometric pieces (with toral boundary) are all standard. However, pdf file of this paper linked by the journal is unreadable. $\endgroup$
    – Misha
    Commented Apr 5, 2012 at 16:28
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If $N$ is a non-orientable closed 3-manifold then $H_3(N;Q)=0$. Since $\chi(N)=0$ the first rational Betti number must be strictly positive. Therefore there is an essential map $f:N\to{S^1}$, and so $N$ contains an essential 2-sided surface. Haken-ness then follows easily.

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    $\begingroup$ @jonathan: Haken manifolds have to be sufficiently large, but also irreducible and $RP^2$-irreducible, i.e., contain no essential spheres and no projective planes (see e.g. Hempel's book "3-manifolds"). For instance, $RP^2\times S^1$ is not Haken. $\endgroup$
    – Misha
    Commented Apr 5, 2012 at 14:32

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