Timeline for What's the Lipschitz constant of the exponential map for $\mathrm{SO}(n,R)$?
Current License: CC BY-SA 4.0
13 events
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| Mar 21, 2022 at 0:35 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
fixed arxiv front-end link, gave title and author
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| May 12, 2020 at 13:31 | comment | added | ABIM | Do you happen to have a reference to this fact? | |
| Oct 8, 2011 at 20:57 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Oct 8, 2011 at 17:19 | comment | added | Suvrit | My approach was to use the Daleckii-Krein theorem to write the Frechet derivative $De_X(Y)$ as a Hadamard product of two matrices, and then bound every unitarily invariant norm appropriately. But you already updated your answer using the nice formula above, saving me the trouble of typing an anwer :-) | |
| Oct 8, 2011 at 17:05 | comment | added | Deane Yang | This is a really nice answer. I'm very happy that you've joined MO. | |
| Oct 8, 2011 at 16:54 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Oct 8, 2011 at 15:36 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Oct 8, 2011 at 1:02 | vote | accept | John Jiang | ||
| Oct 8, 2011 at 1:02 | comment | added | John Jiang | Hi Prof. Kapovitch, Thank you for your detailed solution. I tried to prove the map is 1-Lipschitz using matrix exponentiation but I get terms of the form $X^n Y + X^{n-1}Y X + \ldots + Y X^n$ for $d \exp_X (Y)$, when $X$ and $Y$ do not commute, which I cannot bound easily. I am not sure why the local 1-Lipschitz argument is needed. But it must be since otherwise the same argument using Jacobi field could be pushed to noncompact Lie groups. | |
| Oct 8, 2011 at 0:40 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Oct 8, 2011 at 0:23 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Oct 7, 2011 at 14:55 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Oct 7, 2011 at 14:47 | history | answered | Vitali Kapovitch | CC BY-SA 3.0 |