Timeline for Isometric embedding as a graph
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2013 at 15:16 | comment | added | Willie Wong | (3) The third question is tied to the first where $N,f$ can both be freely specified. As you showed in the answer if $k$ is constant then indeed the local problem just need dimension $k$. But is it true that for $k$ non-constant the local problem can be solved in dimension $\max k$? If not, is there an explicit counterexample so I can see how it fails? I know that this is asking a lot of questions and I really appreciate your spending time thinking about this. | |
| Aug 21, 2013 at 15:09 | comment | added | Willie Wong | (2) This question is in fact inspired by Nash's embedding. In particular, consider the case where $(M,g)$ is a standard Euclidean space $\mathbb{R}^m$, and $(N,h)$ is the standard Euclidean space $\mathbb{R}^n$ (with $n$ allowed to be chosen) I was wondering if there is an "easy" case of the embedding theorem for some metrics $\bar{g}$ so that (a) the construction is simpler and (b) the dimension bound is lower. | |
| Aug 21, 2013 at 15:03 | comment | added | Willie Wong | I had three main questions, and your previous comment partially solved the first. Let me list the questions. (1) Is the construction always possible if we allow sufficiently high dimensions? In the case $k = \dim M$ your previous comment indicates that it is indeed the case, and that is very helpful indeed! I had that part in the back of my mind already but I forgot to mention it in the question statement. I was wondering particular in the case whether non-constant $k$ will introduce additional obstructions. | |
| Aug 21, 2013 at 14:56 | comment | added | Anton Petrunin | In the case $k=\dim M$, anyway, you have the standard embedding problem for $(M,\bar g -g)$ --- if you need $C^\infty$-smooth solution use Nash's theorem, it will only make the dimension of $N$ higher. | |
| Aug 21, 2013 at 14:50 | comment | added | Anton Petrunin | @WillieWong Yes, I consider this case only. I may try to think about general case, but tell me little more; say do you want to construct such $N$ and $f$, or $N$ is given and you want to check if such $f$ exists, or do you want to show that there is no $f$ and $N$... | |
| Aug 21, 2013 at 7:08 | comment | added | Willie Wong | A couple questions: (1) $f^{-1}(q)$ is only a submanifold if $q$ is a regular point, no? Or are you only answering the question when $\bar{g} - g$ has constant rank? Do you have any insights when the rank is not constant? (2) I am sorry that I didn't make it clear, I intend for smooth $C^\infty$ case, and not the $C^1$ case that Nash-Kuiper applies. | |
| Aug 21, 2013 at 2:02 | comment | added | Anton Petrunin | P.S. If $k=\dim M$ then by Nash--Kuiper theorem you can do it for any $N$ if $\dim N>k$ and if $\dim N=k$ you have to take $N=(M,\bar g-g)$. | |
| Aug 21, 2013 at 1:25 | history | answered | Anton Petrunin | CC BY-SA 3.0 |