Timeline for I was wondering if the set of singular loops is a (somewhere) submanifold of loop space?
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| Sep 19, 2011 at 4:38 | comment | added | Nikita Kalinin | in any event Ryan have answered to my question. Thank you very much, too. Yes, I have asked about exactly one self-intersection, see "So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)?" Maybe, ununcomprehension was caused by my English and by numerous revisions, I'm sorry. | |
| Sep 18, 2011 at 18:53 | comment | added | Andrew Stacey | First comment: That's Ryan's point. The space of loops with a single self-intersection and linearly independent tangent vectors is open in the space of all loops with at least one self-intersection. But the space of loops with a single self-intersection and no other conditions is not. As $S^1$ is compact, I don't think that the topology matters. I'll have to think about the second comment. You want to restrict to loops with exactly one self-intersection? | |
| Sep 17, 2011 at 23:10 | comment | added | Nikita Kalinin | and what I mean by "stratification" is a segmentation of loop space by complexity of singularity. First stratum is a set of loops with one simple self-intersection and without any additional singularities Second stratum is a set of loops with $f'(x)=0$ only. And so on. There are some loops which is not contained in any startum. But we can throw qway them from space of loops. | |
| Sep 17, 2011 at 23:03 | comment | added | Nikita Kalinin | It seems strange for me: "by taking a sequence of loops with a double coincidence (one of which is labelled) which converges to a loop with a single coincidence". In which topology do you consider this limit? In standard Whitney topology the limit would be a loop with two self-intersection, or loop with triple intersection, or loop with selfintersection and $f'(x)=0$ in some $x$ or loop with $f'(x)=f''(x)=0$ for some $x$. You can't obtain a loop with simple single self-intersection. Marking can't change type of convergency, I think... | |
| Sep 16, 2011 at 13:00 | history | answered | Andrew Stacey | CC BY-SA 3.0 |