Timeline for Non-orientable 3-manifolds
Current License: CC BY-SA 4.0
27 events
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| S Sep 19, 2020 at 3:03 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Corrected false statement
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| Sep 19, 2020 at 0:40 | review | Suggested edits | |||
| S Sep 19, 2020 at 3:03 | |||||
| Sep 3, 2018 at 20:48 | comment | added | user21230 | Still for me it is the question is there any relation between orientable and non-orientable ones. Connected sum with fixed non-orientable gives mapping in one direction. What could be the mapping in opposite direction ? Or what are the left-overs in non-orientable world (not obtained this way) ? | |
| Sep 3, 2018 at 19:06 | comment | added | Paul | @MarekMitros ok, then he is right and this is not true in dimension 3. I understood it in a different way: just if you can “produce” non orientable manifolds by taking connected sums with a non- orientable one. And this is true in every dimension | |
| Sep 3, 2018 at 11:08 | history | edited | user21230 | CC BY-SA 4.0 |
added 1 character in body
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| Sep 3, 2018 at 9:01 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
complexity-6 is an adjective; also it was confusing to have "...6 manifolds are listed. There are 5 of them". Expanded reference and gave arXiv link
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| Sep 3, 2018 at 8:25 | history | edited | user21230 | CC BY-SA 4.0 |
clarification of A
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| Sep 3, 2018 at 8:20 | comment | added | user21230 | @Paul thank you for discussion on my question. What I was hoping for in (A) was: every non orientable closed 3-manifold can be obtained as connected sum of closed orientable 3-manifold and $S^2\widetilde\times S^1$ (or any other basic non-orientable). According to Ryan Budney this is not true. My knowledge in area is too poor to add anything here. | |
| Aug 31, 2018 at 14:00 | comment | added | Paul | @RyanBudney I really don’t understand this discussion. Let the OP clarify what they want. You said (A) is false as is states and I say is (trivially) true. | |
| Aug 31, 2018 at 13:55 | comment | added | Ryan Budney | @Paul, for surfaces that's an if and only if statement. For 3-manifolds it's not an if and only if statement. Taking connect-sum with a non-orientable manifold is non-orientable, but that's a trivial statement. | |
| Aug 31, 2018 at 5:17 | comment | added | Paul | @RyanBudney ok, but if I understood correctly, as stated, the OP is asking for an analogy “I take connected sums with RP2 to generate non orientable surfaces” but in 3 dimensions substituting RP2 with this twisted product. And it is still true that you generate non-orientable 3 manifolds. | |
| Aug 31, 2018 at 5:11 | comment | added | Ryan Budney | @Paul: not every non-orientable 3-manifold has such a summand. That's the point I was making. | |
| Aug 30, 2018 at 22:23 | comment | added | Paul | @RyanBudney or if you are not satisfied I can try to elaborate more on the argument I suggested. But Jason DeVito's argument suffices I think. | |
| Aug 30, 2018 at 22:21 | comment | added | Paul | @RyanBudney I don't understand why your comment invalidates my argument. (A) suggests that by considering any orientable 3-manifold and taking the connected sum with $S^2 \widetilde{\times} S^1$ we get an non-orientable manifold. See for example the second anwser in this MS question that says that if a connected sum is orientable so are both of its summands math.stackexchange.com/questions/50002/… | |
| Aug 30, 2018 at 22:13 | comment | added | Ryan Budney | @Paul: the boundary of a tubular neighbourhood of a non-orientable loop in a surface is a circle. In a 3-manifold that boundary is a Klein bottle. That's ultimately the reason -- there's more diversity among co-dimension one submanifolds in 3-manifolds than in 2-manifolds. | |
| Aug 30, 2018 at 19:30 | comment | added | Paul | @RyanBudney Why the answer to (A) is no? I think that "if the connected sum is orientable" implies that both pieces are orientable. This is easy to prove by taking the "holonomic" definition of orientability. i.e. taking close paths and checking if a framing changes orientation when you run a full loop. | |
| Jul 10, 2018 at 7:42 | comment | added | user21230 | @BrunoMartelli Thank you for clarification. In your work there is written that Stiefel Whitney torus is incompressible. Can we think of following procedure producing non-orientable manifold ? Let $\Sigma$ be incompressible torus in oriented closed 3-manifold $M$ such that boundary of $R(\Sigma)$ is two tori. Remove neighborhood $R(\Sigma)$ and glue it back by changing orientation on one end. Resulting manifold is not orientable, because each path intersecting $\Sigma$ change orientation. (I wonder whether we had essential path intersecting $\Sigma$ it in one point in manifold $M$ ?) | |
| Jul 9, 2018 at 5:47 | history | edited | user21230 | CC BY-SA 4.0 |
deleted 46 characters in body
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| Jul 9, 2018 at 4:43 | comment | added | user21230 | Sorry for my language. I mean $D^1\times D^2$ with top and bottom $D^2$ glued with reflection homeomorphism. I guess this is called "solid Klein bottle". | |
| Jul 9, 2018 at 2:13 | comment | added | janmarqz | by a full Kleinbottle you mean solid Kleinbottle? | |
| Jul 8, 2018 at 12:59 | answer | added | YCor | timeline score: 7 | |
| Jul 8, 2018 at 11:10 | history | edited | user21230 | CC BY-SA 4.0 |
added 526 characters in body
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| Jul 6, 2018 at 16:22 | comment | added | Bruno Martelli | The notion of P2-irreducible contains in particular irreducible. So there are no non-trivial spheres. | |
| Jul 6, 2018 at 6:00 | comment | added | user21230 | Thank you again. At least we can sort irreducible non-orientable 3-manifolds by genus of Stiefel-Whitney surface. | |
| Jul 5, 2018 at 20:33 | history | edited | j.c. | CC BY-SA 4.0 |
add ref link, attempt at fixing grammar
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| Jul 5, 2018 at 14:55 | comment | added | Ryan Budney | (A) No. What you see in dimension 2 is a very special case. (B) No again. There are obstructions to embedding punctured 3-manifolds in $\mathbb R^4$, for example there is one using the torsion linking form. (C) The boundary of a regular neighbourhood, yes. The neighbourhood itself is the associated disc bundle to the Klein bottle. | |
| Jul 5, 2018 at 14:12 | history | asked | user21230 | CC BY-SA 4.0 |