Timeline for Can we perturb a map $\mathbb{R}^n \to \mathbb{R}^n$ to have high rank?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 11, 2020 at 5:10 | vote | accept | Asaf Shachar | ||
| Dec 9, 2019 at 17:26 | comment | added | Asaf Shachar | Thank you. I hope to have some time soon to read thoroughly your answer. | |
| Dec 1, 2019 at 19:32 | comment | added | Dap | @AsafShachar: I finally got round to ironing out most of the details, which I hope answer your previous questions. I have decided to stay anonymous, unfortunately, to help me stay focused on other commitments. Thanks for the interesting questions. | |
| Dec 1, 2019 at 19:28 | history | edited | Dap | CC BY-SA 4.0 |
minor fixes
|
| Nov 30, 2019 at 16:40 | history | edited | Dap | CC BY-SA 4.0 |
added 3209 characters in body
|
| Nov 5, 2019 at 6:32 | comment | added | Asaf Shachar | If I will ever write a paper which uses most of them (which I am not really sure at the moment...) I will invite you to be a coauthor, if you would be interested, or give you credit in any other way you may want (that is, only if you would want any. I see you are anonymous here...) | |
| Nov 5, 2019 at 6:24 | comment | added | Asaf Shachar | cont': (3) I don't understand the last sentence "And the Lipschitz constant of $\phi$ will be $O(1/\epsilon)$ because this $g$ has Lipschitz constant at most $1/\epsilon.$"The definition of $g$ does not involve $\epsilon$ at all. Should this be related to the relation between $\psi$ and $\epsilon$ in the definition of $\psi$?BTW, I just wanted to say that the idea you suggested is very nice, and that you have answered many many question of mine lately. | |
| Nov 5, 2019 at 6:22 | comment | added | Asaf Shachar | Thanks. I have some more questions on the second part: (1) Why $g(v)=\infty$ except on a set of dimension at most $n-4,$ implies that $h(v)=2$ except on a set of $(n-1)$-dimensional measure at most $O(\epsilon^3)$? In particular, do you claim that $g(v)=\infty \Rightarrow h(v)=2$? I don't see why... (2) I don't understand how $\phi(rv)$ is defined when $g(v)=\infty$. (It should be "$(\psi(-\infty)+\infty)v$". Should I interpret this as $v$?... | |
| Nov 4, 2019 at 11:07 | comment | added | Dap | @AsafShachar: that's right - in other words in Sard's theorem take "critical point" to mean "derivative is not surjective" to get a stronger notion of "regular value". (I just learnt from Wikipedia that this is not the standard meaning of "critical point".) | |
| Nov 2, 2019 at 6:58 | comment | added | Asaf Shachar | Thank you. Just to be sure I understand: For small ranks $r$, $\phi_r^{-1}(N)$ will be empty for generic $N $, right? We have $\dim(\phi_r^{-1}(N))=n-(n-r)^2 \ge 0 \iff n \ge (n-r)^2 \iff r \ge n-\sqrt{n}$. (So, $r < n-\sqrt{n}$ corresponds to the case of Sard's theorem where the dimension of the source is smaller than the dimension of the target ). | |
| Oct 31, 2019 at 12:26 | history | answered | Dap | CC BY-SA 4.0 |