Timeline for Nash isometric embedding theorem with keeping the symplectic structures of our ambient spaces
Current License: CC BY-SA 4.0
19 events
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| S Oct 4, 2018 at 21:01 | history | bounty ended | CommunityBot | ||
| S Oct 4, 2018 at 21:01 | history | notice removed | CommunityBot | ||
| Sep 26, 2018 at 20:34 | history | edited | Ali Taghavi |
edited tags
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| S Sep 26, 2018 at 19:46 | history | bounty started | Ali Taghavi | ||
| S Sep 26, 2018 at 19:46 | history | notice added | Ali Taghavi | Draw attention | |
| Sep 26, 2018 at 5:16 | comment | added | Willie Wong | I hope I am not mistaken, but I think this then follows from the fact that embedded Riemannian submanifolds have normal tubular neighborhoods. This I think implies you have local coordinates $(x^1, \ldots, x^n)$ of $N$, inducing coordinates $(x^1, \ldots, x^n, y^1, \ldots, y^n)$ of $T^*N$, such that the embedding $\sigma$ sends $T^*M$ (locally) to the set $\{x^{m+1}= \cdots = x^{n} = y^{m+1} = \cdots =y^{n} = 0\}$. | |
| Sep 26, 2018 at 4:46 | comment | added | Willie Wong | The isometric embedding and the introduction of $TM$ are probably a bit of red herring here. I think what you are asking is this: Given a Riemannian manifold $(N,h)$ and a Riemannian submanifold $(M,g)$, the musical isomorphisms allows you to define a a smooth map $\sigma: T^*M \to T^*N$. Your question is whether that the pull-back by $\sigma$ of the canonical symplectic structure on $T^*N$ is equal to the canonical symplectic structure on $T^*M$. Or, in other words, you are asking whether the embedding $\sigma$ makes $T^*M$ a symplectic submanifold of $T^*N$. | |
| Sep 25, 2018 at 21:08 | answer | added | Tobias Diez | timeline score: 8 | |
| Sep 25, 2018 at 7:50 | comment | added | Ali Taghavi | @PaulBryan Ok, Have a nice tripe. The naturality property you mentioned is amazing(and somehow a chalenging question, at least for me). Any way I thank you for your attention to my question and I look forward to hear from you about this subject. | |
| Sep 24, 2018 at 7:53 | comment | added | Paul Bryan | I see. Still seems plausible to me, perhaps naively. I'm in transit so I don't have time to check it properly right now sorry. | |
| Sep 24, 2018 at 7:46 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 213 characters in body; edited title
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| Sep 24, 2018 at 7:36 | comment | added | Ali Taghavi | For an isometric embedding $j$ I do not see why $j^*(\omega)=\omega_{j^*\delta}$?May be the question is obvious? | |
| Sep 24, 2018 at 7:35 | comment | added | Ali Taghavi | @PaulBryan $\omega_g$ is constructed as follows: the metric gives a dieffeomorphism between $TM$ and $T^* M$, on the other hand $T^* M$ has a canonical symplectic structure (independent of any Riemannian metric). We pull back this canonical symplectic structure via the above mentioned diffeomorphism. The resulting structure is denoted by $\omega_g$. With such construction, we have the naturality property you mentioned when we have DIFFEOMORPHISM isometry but I am not sure it works for isometric embedding(not necessarilly surjective). | |
| Sep 24, 2018 at 6:19 | comment | added | Paul Bryan | I don't know if it works. It was jus a first guess. To be sure, can't you just check directly from the definitions? I suppose the answer depends on exactly what "natural" means here. Often in this sort of context where some tensor/form in DG is constructed from another, natural means "in some canonical way" but this may not be necessarily be the case here. I guess you construct $\omega_g$ by some lift of $g$ to $TM$ by splitting $TTM$ into $VM$ and $TM$ via the Levi Civita connection and the use a metric contraction of the canonical sympletic form on $T^{\ast}M$. Chase it through now and see. | |
| Sep 24, 2018 at 5:55 | comment | added | Ali Taghavi | @PaulBryan may be I am missing some thing but the problem is that the symplectic structure originally comes from cotangent bundle but the "cotangent bundle" is a contravariant (and not covariant) functor. Does realy your argument work? | |
| Sep 24, 2018 at 3:03 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
I fixed some typos.
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| Sep 24, 2018 at 1:15 | comment | added | Paul Bryan | For the question as stated, I would have thought that the "naturality" implies $j^{\ast} \omega = \omega_{j^{\ast} \delta}$. Here $\delta$ is the Euclidean metric and $\omega = \omega_{\delta}$. Then the solution is simply that there exists an isometric embedding into some $\mathbb{R}^n$ since then $g = j^{\ast} \delta$. | |
| Sep 24, 2018 at 0:28 | answer | added | L.F. Cavenaghi | timeline score: 2 | |
| Sep 23, 2018 at 23:22 | history | asked | Ali Taghavi | CC BY-SA 4.0 |