Timeline for Step by step construction of 3-manifolds in $R^4$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2018 at 21:15 | comment | added | user21230 | We can conjecture following. For $M$ embedded in $R^4$ there is different embedding such that $N$ retracts to 2-complex ($\delta N=M$). We can call it "unknotting the tunnels". | |
| Jun 9, 2018 at 21:12 | comment | added | Bruno Martelli | Yes, it's enough. But in case you need a genus-two surface in S^3 that does not bound a handlebody in neither side... | |
| Jun 9, 2018 at 10:47 | comment | added | user21230 | That's amazing ! Thank you ! Isn't enough to have just knotted tunnel ? | |
| Jun 9, 2018 at 7:02 | comment | added | Bruno Martelli | Take a ball, add a knotted 1-handle, and dig a knotted tunnel. | |
| Jun 8, 2018 at 19:15 | comment | added | user21230 | Regarding 4-dim, I am not sure, just testing what would be the shape of 3- dim submanifold M of 4-dim space. Assuming that I am observer, I don't distinguish whether there is 4-dim interior or not. That's why I start thinking of the retract. The shape of the retract is the same as shape of M. So I thought of classifying the retracts. Next thing I thought is relation between fundamental group of retract K and fundamental group of surrounding manifold M. Finite 2-complex K can hav any finitely generated group as fundamental group. This is not the case for 3-manifold. This is maybe too much here | |
| Jun 8, 2018 at 18:39 | comment | added | user21230 | Really ?! Can you please give example of compact 3-dimensional smooth submanifold of $R^3$ with connected boundary which is not handlebody ? It can be knotted, but it doesn't matter - it is still handlebody. Regardin | |
| Jun 8, 2018 at 13:27 | comment | added | Bruno Martelli | 3-dimensional compact submanifolds of R^3 do not necessarily retract onto a bouquet of circles (ie are not necessarily handlebodies). | |
| Jun 7, 2018 at 18:35 | comment | added | user21230 | Ok, but in my case the boundary is connected. For 3-dimensional compact submanifold of $R^3$ with boundary in one piece the retract is bouquet of circles. Therefore one would expect 2-dimensional retract in case of 4-dim submanilold of $R^4$ with one component boundary. | |
| Jun 7, 2018 at 17:43 | comment | added | Ryan Budney | Retracts of compact 4-dimensional submanifolds of $\mathbb R^4$ are generally 3-complexes. | |
| Jun 7, 2018 at 12:31 | comment | added | user21230 | One argument which come to my mind is consider the Morse function $f$ from your comment. Cut in function regular point $f^{-1}(p)$ of $N$ is 3-manifold with boundary in $R^3$ which can be retracted to 1-complex. | |
| Jun 7, 2018 at 9:59 | comment | added | user21230 | @RyanBudney If we consider 4-dimensional manifold $N$ with boundary $M$ in $R^4$. Do we know what are possible retracts of $N$ ? Is it always 2-complex ? I saw this mathoverflow.net/questions/32787/… question which you asked 8 years ago. | |
| Jun 7, 2018 at 7:05 | comment | added | user21230 | Thank you for this response. I will take a look into "11 tetrahedron census" again. I should probably install SnapPea and Regina software. On wikipedia I read that 3-manifolds can be classified algorithmically (Matveev 2003) ! | |
| Jun 6, 2018 at 16:04 | comment | added | Ryan Budney | You can construct all 3-manifolds in $\mathbb R^4$ via such surgery constructions. Roughly the proof is to put a standard linear height function on $\mathbb R^4$ which is a Morse function on the 3-manifold (any 3-manifold in $\mathbb R^4$). This gives you the corresponding surgery description of the 3-manifold. This is theoretically nice but if you want to reverse the process it is difficult, i.e. given a 3-manifold determine if it has a surgery description corresponding to an embedding. There are a few simple cases of this construction that appear in my preprint, ref in the prev. thread. | |
| Jun 6, 2018 at 10:11 | history | edited | user21230 |
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| Jun 6, 2018 at 9:05 | history | asked | user21230 | CC BY-SA 4.0 |