Timeline for Characterization of extrinsic distance prevserving embedding (see the definition given!) from low dimensional Euclidean spaces to high dimensions
Current License: CC BY-SA 4.0
15 events
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| Apr 7, 2020 at 7:27 | comment | added | Learning math | @DeaneYang Thank you - this is a useful trick indeed! | |
| Apr 5, 2020 at 11:31 | answer | added | Gael Meigniez | timeline score: 2 | |
| Mar 30, 2020 at 22:58 | comment | added | Deane Yang | The trick is to observe that you can recover the coefficients of a vector with respect to an orthonormal basis by using the inner product. That plus the fact the inner product is preserved under $f$ implies the coefficients are preserve by $f$. | |
| Mar 30, 2020 at 22:24 | comment | added | Learning math | @DeaneYang Thank you again! The first argument is a direct consequence of polarization identity, the second is clear from the drefintion of orthonormal basis (ONB) itself. I'm currently stuck at the third: showing that a map that sends ONB to ONB has to be linear. Hmm, may be I'll show that it's derivative is constant? (I'm thinking!) The rest is clear about the translating it in order to fix the origin. | |
| Mar 30, 2020 at 22:18 | comment | added | Deane Yang | By the way, no assumption on $n$ and $m$ is needed. That $m \le n$ is a consequence of the proof. | |
| Mar 30, 2020 at 21:51 | comment | added | Deane Yang | Oops. Sorry about that. So here's how the argument for your question goes. First show that, given inner product spaces $V$ and $W$ if a map $f: V \rightarrow W$ satisfies $|f(v_1)-f(v_2)| = |v_1-v_2|$, for all $v_1, v_2 \in V$, then $f(v_1)\cdot f(v_2) = v_1\cdot v_2$, for all $v_1, v_2 \in V$. This implies that $f$ maps an orthonormal basis to an orthonormal basis. You can prove that $f$ is linear and therefore a rotation. Finally, given an extrinsic distance preserving map, compose it with a translation so the origin is fixed, This map satisfies the assumptions of $f$ above. | |
| Mar 30, 2020 at 19:47 | history | edited | Learning math | CC BY-SA 4.0 |
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| Mar 30, 2020 at 19:40 | history | edited | Learning math | CC BY-SA 4.0 |
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| Mar 30, 2020 at 19:38 | comment | added | Learning math | @DeaneYang Sorry but that's not the kind of isometry i'm looking for, as specified in the question. What you described is an isometry, if we consider the intrinsic distance on the image, not the extrinsic Euclidean distance. For example, if $m=1, n=2,$ then $d(\phi(t_1), \phi(t_2))= |t_1 - t_2|$ if $|\phi'|=1$, fir sure; but the intrisic distance $d(\phi(t_1), \phi(t_2)) \ne ||\phi(t_1) - \phi(t_2)||_{\mathbb{R}^2}$. | |
| Mar 30, 2020 at 19:28 | comment | added | Deane Yang | This is incorrect, even for $m=1$ and $n-3$, as well as $m=2$ and $n-3$. In the first case, any arclength parameterization $c: \mathbb{R}\rightarrow\mathbb{R}^2$ is isometric. In the second case, any map of the form $(x,y) \mapsto (a(x),y,b(x))$, where $x\mapsto c(x)=(a(x),b(x))$ is an arclength parameterization of a curve in $\mathbb{R}^2$, is an isometric embedding of $\mathbb{R}^2$ in $\mathbb{R}^3$. | |
| Mar 30, 2020 at 19:23 | history | edited | Learning math | CC BY-SA 4.0 |
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| Mar 30, 2020 at 19:18 | history | edited | Learning math | CC BY-SA 4.0 |
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| Mar 30, 2020 at 19:03 | history | edited | Learning math | CC BY-SA 4.0 |
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| Mar 30, 2020 at 18:57 | history | edited | Learning math | CC BY-SA 4.0 |
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| Mar 30, 2020 at 18:52 | history | asked | Learning math | CC BY-SA 4.0 |