Timeline for What are the matrices preserving the $\ell^1$-norm?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 12, 2018 at 12:19 | answer | added | Ilya Bogdanov | timeline score: 5 | |
| Dec 10, 2017 at 14:10 | history | edited | YCor | CC BY-SA 3.0 |
improved formatting, changed tags
|
| Dec 9, 2017 at 18:59 | comment | added | YCor | (If $p$ is not an integer, for $k> p$ we have $\partial_i^kf(x)=\lambda x_i^{p-k}$ for $x_i\neq 0$ for some scalar $\lambda\neq 0$, which does not extend by continuity. If $p$ is an odd integer, $\partial_i^pf(x)=\lambda\mathrm{sign}(x_i)$ for $x_i\neq 0$, for some scalar $\lambda\neq 0$, which again does not extend by continuity. So, as soon as $1\le p<\infty$ is not an even integer, the smoothness locus of $f$ is the complement of the union of coordinate hyperplanes.) | |
| Dec 9, 2017 at 18:51 | comment | added | YCor | @Gro-Tsen The approach with smoothness will probably work with reasonable efforts when it can work, namely when $p$ is not an even integer... but cannot work when $p$ is an even integer, since in this case $f:x\mapsto\|x\|_p^p$ is real analytic outside 0. For other values of $p$ the smoothness locus is indeed the complement of the union of coordinate hyperplanes. | |
| Dec 9, 2017 at 18:47 | answer | added | YCor | timeline score: 14 | |
| Dec 9, 2017 at 18:35 | comment | added | YCor | @JochenGlueck you're right. I got confused with the notion of smooth Banach space: that $x\mapsto\|x\|_p^p$ is of class $C^1$ with non-vanishing gradient outside 0 is enough to make $\ell^p$ a smooth Banach space, for $1<p<\infty$. | |
| Dec 9, 2017 at 18:00 | comment | added | Jochen Glueck | @Ycor Why is the mapping $x \mapsto \|x\|_p^p$ smooth on the $p$-unit sphere? For instance, let $p =3/2$ and let the dimension be two. Then the mapping $x \mapsto \|x\|_p^p = |x_1|^p + |x_2|^p$ is not twice differentiable at $x = (0, 1)$. | |
| Dec 9, 2017 at 16:53 | comment | added | YCor | PS my curvature approach, anyway, is a priori not sufficient, since it would only show that among $\ell^2$-isometries, only signed permutations are also $\ell^p$-isometries. The group of $\ell^p$-isometries indeed preserves a Euclidean metric, but possibly not the standard one (yes it does, but only a posteriori). | |
| Dec 9, 2017 at 16:47 | comment | added | YCor | @Gro-Tsen for $1<p<\infty$ the $\ell^p$- unit sphere is smooth (this is obvious, since $x\mapsto \|x\|_p^p$ is smooth and its gradient vanishes only at 0) . | |
| Dec 9, 2017 at 12:24 | comment | added | Gro-Tsen | @YCor I suspect the result for $p\neq 2$ might be provable by considering the points at which the $L^p$-unit sphere is not a $C^\infty$-submanifold, which should be the skeleton of the cross-polytope (for $p<\infty$). Whether this approach is simple enough to be interesting is another question, however (I'm afraid the details might be pretty painful to fill in). | |
| Dec 9, 2017 at 11:51 | vote | accept | D. Rusin | ||
| Dec 9, 2017 at 11:19 | answer | added | Jochen Glueck | timeline score: 26 | |
| S Dec 9, 2017 at 10:04 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits
|
| Dec 9, 2017 at 9:25 | comment | added | YCor | Yes it's true for all $p\neq 2$, although tricks using extremal points work only for $p=1,\infty$. It's classical but I have no optimal reference (projecteuclid.org/download/pdf_1/euclid.bams/1183538497 sounds too general) Maybe one can use extremal properties of some curvature function on the unit sphere (for $1<p<2$ one could expect the scalar curvature to be maximal on vertices of the cross-polytope, and for $p>2$ idem in the dual). | |
| Dec 9, 2017 at 9:23 | review | Suggested edits | |||
| S Dec 9, 2017 at 10:04 | |||||
| Dec 9, 2017 at 0:48 | review | Close votes | |||
| Dec 9, 2017 at 16:14 | |||||
| Dec 9, 2017 at 0:44 | comment | added | D. Rusin | @Gro-Tsen Just one more question: Is it also true that the signed permutations are exactly the matrices preserving the L_p-norm for all p > 2? Because the unit sphere for the p-norm for p > 2 is the surface of something with (more or less) sharp edges as well. | |
| Dec 8, 2017 at 23:10 | comment | added | Gro-Tsen | Yes, a signed permutation is what you said, and yes, there are only finitely many of them. The geometric argument with the cross-polytope is meant to explain why they are the only $L^1$-norm-preserving matrices. | |
| Dec 8, 2017 at 23:03 | comment | added | D. Rusin | I assume, a signed permutation is just a permutation matrix where some ones can be negative? Hmm they definitely preserve this, but are those really all preserving matrices? 'Cause there are only finitely many such matrices for each dimension of the vector space. | |
| Dec 8, 2017 at 22:46 | comment | added | Gro-Tsen | The unit sphere for the $1$ norm is the surface of a cross-polytope. Trying to map it linearly onto itself has to preserve the extremal points (=vertices) of the cross-polytope. So we end up with just a signed permutation. There are details to be filled in, but I think this should more or less work. | |
| Dec 8, 2017 at 22:36 | history | asked | D. Rusin | CC BY-SA 3.0 |