Timeline for If Derivative at each point $x \in \Bbb R^n$ is an orthogonal matrix then $f(x) = Ox +b $ [duplicate]
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 13, 2018 at 1:45 | history | closed |
Igor Rivin abx Loïc Teyssier Chris Godsil Francois Ziegler |
Duplicate of functions with orthogonal Jacobian | |
| Aug 13, 2018 at 1:13 | comment | added | Marco | This only shows the map is distance non-increasing. Since the volumes are preserved, would this be sufficient to finish the problem? | |
| Aug 12, 2018 at 23:06 | comment | added | Pietro Majer | @Marco: don't get the conclusion. Every line segment is mapped to a curve of equal length, but why a line segment? The endpoints are not fixed. | |
| Aug 12, 2018 at 22:56 | answer | added | Igor Rivin | timeline score: 3 | |
| Aug 12, 2018 at 21:34 | comment | added | Marco | If $\gamma(t)$ is a curve in $\mathbb{R}^n$, then the image of $\gamma(t)$ under $f$ has the same length as $\gamma(t)$. This shows every line segment is mapped to a line segment of equal length. | |
| Aug 12, 2018 at 20:31 | review | Suggested edits | |||
| Aug 12, 2018 at 23:16 | |||||
| Aug 12, 2018 at 19:21 | comment | added | Christian Remling | While the question clearly is a duplicate of the linked one, I would still like to see a direct elementary answer if there is one (Alexandre's refers to theory, and David's makes the extra assumption that $f\in C^2$). It could of course be posted as an answer to the earlier question. | |
| Aug 12, 2018 at 16:45 | review | Close votes | |||
| Aug 13, 2018 at 1:50 | |||||
| Aug 12, 2018 at 15:21 | history | asked | user119602 | CC BY-SA 4.0 |