Timeline for Is it true $\left\|\log(RS)\right\|≤\left\|\log(R)+\log(S)\right\|$ for all $R,S \in \mathrm{SO}(3)$, where $\|\cdot\|$ is the Frobenius norm?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| May 8, 2023 at 16:46 | comment | added | Spencer Kraisler | @RodrigodeAzevedo I deleted it since this was essentially the same question, but with a good discussion. | |
| May 8, 2023 at 16:46 | comment | added | Spencer Kraisler | @BenMcKay I agree, I marked it. | |
| S May 8, 2023 at 16:43 | vote | accept | Spencer Kraisler | ||
| May 8, 2023 at 14:37 | comment | added | Ben McKay | I prefer the answer that covers all compact Lie groups; maybe you might change your choice of accepted answer. | |
| May 8, 2023 at 12:30 | answer | added | Robert Bryant | timeline score: 7 | |
| May 8, 2023 at 9:25 | comment | added | Rodrigo de Azevedo | Is your question on Math SE still available? | |
| May 7, 2023 at 17:35 | vote | accept | Spencer Kraisler | ||
| S May 8, 2023 at 16:43 | |||||
| May 6, 2023 at 17:46 | history | edited | Michael Hardy | CC BY-SA 4.0 |
In cases like this, \left and \right influence horizontal spacing.
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| May 5, 2023 at 14:24 | answer | added | Robert Bryant | timeline score: 7 | |
| May 4, 2023 at 16:46 | comment | added | Spencer Kraisler | @YCor I said in my post that I'm restricting only to pairs $R,S$ such that the inequality above is well-defined. | |
| May 4, 2023 at 16:44 | comment | added | Spencer Kraisler | @DanielAsimov Are you familiar with Lie theory? I cannot give an entire course in a small comment, but in my particular case the Lie and matrix logarithm coincide. So, it is simply the matrix logarithm, which is well-defined. Wikipedia has a good page on it. | |
| May 4, 2023 at 15:25 | comment | added | YCor | Actually, OP requires rather a condition of "expanding distances". | |
| May 4, 2023 at 8:32 | comment | added | Robert Bryant | If you remove the $\mathbb{RP}^2$ of elements in $\mathrm{SO}(3)$ with trace equal to the mimimum value of $-1$, there is a well-defined smooth logarithm on the open ball that remains. It satisfies $\exp(\log A) = A$. For unit quaternions (or, equivalently, $\mathrm{SU}(2)$), there is a well-defined, smooth logarithm after you remove the single element $-1$ (equivalently, $-I_2$). | |
| May 4, 2023 at 5:07 | comment | added | YCor | If one could assign to every matrix $\begin{pmatrix}\cos t&-\sin t&0\\\sin t&\cos t & 0\\0&0&1\end{pmatrix}$ a Lie logarithm in a continuous way, one could lift continuously the circle to the line, and this is not possible. | |
| May 4, 2023 at 5:05 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| May 4, 2023 at 4:51 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 19 characters in body; edited title
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| May 4, 2023 at 2:47 | comment | added | Daniel Asimov | How do you define "the Lie logarithm"? (Of course, for M in SO(3), there are infinitely many Lie algebra elements v with exp(v) = M.) | |
| May 4, 2023 at 0:47 | history | asked | Spencer Kraisler | CC BY-SA 4.0 |