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In dimension 2, there are two remarkable non-orientable closed manifolds, the projective plane (from synthetic geometry; has the fixed point property; algebraic compactification of the plane etc) and the Klein bottle (nowhere vanishing vector field; with immersions sold in your nearest nonorientable store). There is also a classification of all closed non-orientable surfaces, as connected sums of projective planes.

I am looking for examples of non-orientable 3 dimensional closed (compact, boundaryless) manifolds. Any with some special properties or arising from interesting geometrical problems? Is there a simple classification for them?

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    $\begingroup$ Is only assuming 'compact' important for you? If not, I think it would be good to make it 'non-orientable 3-dimensional closed manifold'. Then, I think, people knowing much about Thurston's geometrization conjecture could, hopefully, relate your question to the current state of knowledge about geometrization. (For compact yet non-closed manifolds there is, I think, 'less' uniqueness in the geometric model structures, though much is understood for compact non-closed manifolds as well.) I am just suggesting you place your question 'squarely' into the best-understood context of all. $\endgroup$ – Peter Heinig Sep 20 '17 at 13:39
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    $\begingroup$ Yes indeed, I am interested in boundaryless manifolds - edited. $\endgroup$ – coudy Sep 20 '17 at 13:53
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    $\begingroup$ There is a theorem of Postnikov that if $V$ is a finite-dimensional $\mathbb F_2$-vector space, $\theta$ is a trilinear form on $V$, and $w\in V$ is such that $\theta(w, w, v) = \theta(w, v, v)$ for all $v$, then there's a closed 3-manifold $M$ and an isomorphism $H^1(M;\mathbb F_2)\cong V$ carrying $w_1(M)\mapsto w$ and the cup product $(a, b,c)\mapsto a\smile b\smile c$ to $\theta$. The upshot is that as soon as you've satisfied the Wu formula and Poincaré duality, you have complete freedom in choosing the mod 2 cohomology ring of a closed 3-manifold, and if $w_1\ne 0$, it's unorientable. $\endgroup$ – Arun Debray Sep 20 '17 at 14:15
  • $\begingroup$ Thus even a classification of homotopy classes of closed, unoriented 3-manifolds is likely to be complicated. $\endgroup$ – Arun Debray Sep 20 '17 at 14:16
  • $\begingroup$ One contrast with the orientable case is that non-orientable closed 3-manifolds always have infinite $\pi_{1}$. See here math.stackexchange.com/questions/421303/…. $\endgroup$ – Nick L Sep 20 '17 at 15:19
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One gets non-orientable closed 3-manifolds by taking a non-orientable surface, and crossing with $S^1$, such as $P^2\times S^1$.

In fact, the geometrization theorem hasn't been proven completely for non-orientable 3-manifolds. Of course, a 2-fold cover has a connect sum and geometric decomposition. But the problem is that no one has shown that this decomposition can be made equivariant with respect to the covering translation. One expects that a careful analysis of the proof using Ricci flow could be made equivariant (at least Ricci flow preserves symmetries). One can't do the initial connect sum decomposition, because 1-sided projective planes must be cut along and coned off, resulting in an orbifold with isolated cone points.

In any case, I think that the most interesting non-orientable 3-manifold is the smallest known volume closed manifold, which has volume $2.0298...$, the same as the figure 8 complement, and fibers over the circle. This was discovered by Weeks in his census. I think it is known to be arithmetic. See Table 2 of a paper by Hodgson and Weeks.

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  • $\begingroup$ For the last paragraph: you are probably referring to the Gieseking manifold, whose volume is half of that of the figure eight knot complement and which in fact is 2-fold covered by the figure eight knot complement. $\endgroup$ – ThiKu Sep 20 '17 at 14:50
  • $\begingroup$ en.wikipedia.org/wiki/Gieseking_manifold $\endgroup$ – ThiKu Sep 20 '17 at 14:50
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    $\begingroup$ About Ricci Flow: it seems to me that the work of Dinkelbach-Leeb on equivariant Ricci flow doesn't make orientability assumptions: arxiv.org/pdf/0801.0803.pdf So it should prove geometrization in the nonorientable case or am I missing something? $\endgroup$ – ThiKu Sep 20 '17 at 14:55
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    $\begingroup$ @ThiKu: It's possible that one can deduce this from their paper, but for some reason they don't formulate a theorem of this sort. I suppose one would just need an equivariant connect-sum decomposition of the 2-fold orientable cover, which ought to follow from Meeks-Yau. One also needs to know that involutions of Haken manifolds preserve the geometric decomposition, which should be okay. Then deduce that it is an isometry on each geometric piece from their theorems, and glue up along the JSJ and connect sum decompositions. $\endgroup$ – Ian Agol Sep 20 '17 at 15:22
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    $\begingroup$ @ThiKu: the Gieseking manifold isn't closed, and its volume is half the figure-8 exterior. The figure-8 exterior is the double-cover of the Gieseking.. . although, the Gieseking is also quite an interesting 3-manifold. $\endgroup$ – Ryan Budney Sep 21 '17 at 4:48
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A curiosity is obtained as follows: take a solid cube $[-1,1]^3$ and identify one pair of two opposite faces by a symmetry with respect to a coordinate axis, while identifying the other two pairs of opposite faces by translations. The resulting manifold contains an embedded Klein bottle which has two sides (in the mathematical sense that its normal bundle is trivial, i.e. removing its zero section disconnects it), i.e. you can paint one side of the Klein bottle blue and the other side red without the two color meeting.

This shows that it is not an intrinsic property of non-orientable surface to "have only one side", it is a property of some of their codimension $1$ embeddings (including all embeddings in orientable $3$-manifold). In fact, having only one side means not being co-orientable rather than being not orientable.

The above example also has an embedded $2$-torus which is not coorientable (and thus has only one face!), of course.

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    $\begingroup$ There's some discussion of this phenomenon in this old question mathoverflow.net/questions/130660 $\endgroup$ – j.c. Sep 20 '17 at 15:59
  • $\begingroup$ Very interesting example! Thank you for pointing this out. $\endgroup$ – Gro-Tsen Sep 20 '17 at 18:01

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