Geometrization for 3-manifolds that contain two-sided projective planes - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:39:58Z http://mathoverflow.net/feeds/question/38098 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38098/geometrization-for-3-manifolds-that-contain-two-sided-projective-planes Geometrization for 3-manifolds that contain two-sided projective planes Igor Belegradek 2010-09-08T20:03:30Z 2012-04-05T03:38:22Z <p>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 <a href="http://front.math.ucdavis.edu/0809.4040" rel="nofollow">paper</a> 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</p> <p><b> Question 1.</b> Is the above a correct interpretation of what is proved in Morgan-Tian's paper?</p> <p>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).</p> <p><b>Question 2.</b> What is the status of the geometrization conjecture for manifolds that contain two-sided projective planes?</p> <p>Note: on the last two pages of Scott's wonderful survey <a href="http://www.math.lsa.umich.edu/~pscott/8geoms.pdf" rel="nofollow">"The Geometries of 3-manifolds"</a> he describes a version of the geometrization conjecture that makes sense in the presence of two-sided projective planes. </p> <p>UPDATE: </p> <ol> <li><p>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. </p></li> <li><p>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. </p></li> </ol> http://mathoverflow.net/questions/38098/geometrization-for-3-manifolds-that-contain-two-sided-projective-planes/38146#38146 Answer by Ian Agol for Geometrization for 3-manifolds that contain two-sided projective planes Ian Agol 2010-09-09T06:02:53Z 2010-09-10T15:09:49Z <p>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 <a href="http://www.ams.org/mathscinet-getitem?mr=595203" rel="nofollow">Meeks-Yau</a>), 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 <a href="http://front.math.ucdavis.edu/0801.0803%20" rel="nofollow">Dinkelbach and Leeb</a> 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.</p> <p>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 <a href="http://www.ams.org/mathscinet-getitem?mr=420620" rel="nofollow">theorem of Hatcher</a> 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). </p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/38098/geometrization-for-3-manifolds-that-contain-two-sided-projective-planes/93185#93185 Answer by jonathan for Geometrization for 3-manifolds that contain two-sided projective planes jonathan 2012-04-05T03:38:22Z 2012-04-05T03:38:22Z <p>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.</p>