Timeline for A reference to a characterization of metric spaces admitting an isometric embedding into a Hilbert space
Current License: CC BY-SA 4.0
23 events
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| Apr 22 at 5:22 | history | edited | Taras Banakh | CC BY-SA 4.0 |
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| Jun 2, 2023 at 18:47 | comment | added | ABIM | Aha, I see. Interesting trick; thanks :) | |
| Jun 2, 2023 at 9:47 | comment | added | Taras Banakh | @ABIM The problem of embedding into sheres can be reduced to the problem of embedding into the Hilbert space: given a metric spaces $X$, consider the space $X\cup\{o\}$ with added point $o$ at the constant distances $d(o,x)=1+diam(X)$ from the points $x$ of $X$. | |
| Jun 1, 2023 at 21:57 | comment | added | ABIM | Is there a variant of such a result, characterizing metric spaces which can be isometrically embedded into spheres? | |
| Jan 22, 2023 at 17:32 | vote | accept | Taras Banakh | ||
| Jan 22, 2023 at 12:08 | answer | added | Anton Petrunin | timeline score: 7 | |
| Apr 13, 2018 at 7:42 | answer | added | Uri Bader | timeline score: 4 | |
| Apr 12, 2018 at 13:22 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Apr 12, 2018 at 13:13 | comment | added | Nik Weaver | For what it's worth, this characterization is new to me. | |
| Apr 12, 2018 at 11:23 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Apr 12, 2018 at 10:13 | comment | added | Taras Banakh | @erz For characterization of negative definiteness of $d^2$ I need only the symmetry and real-valuedness of $d^2$. I hope the same argument works for positive real-valued kernels. | |
| Apr 12, 2018 at 10:06 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Apr 12, 2018 at 9:46 | comment | added | erz | Do I understand correctly, that the only properties of $d$ that you use are the symmetry and real-valueness? You don't need neither the triangle inequality, nor non-negativity? Also, the same arguement allows a characterization of positive semi-definite real-valued kernels, right? | |
| Apr 12, 2018 at 8:56 | comment | added | Wlod AA | (You have missed "t" in Joram's name). | |
| Apr 12, 2018 at 8:52 | comment | added | YCor | (Answer to previous comment by Taras) Yes of course. But dealing with all finite $X$ altogether is meaningful for infinite $X$, since each given condition in Schoenberg's theorem deals with finitely many points. | |
| Apr 12, 2018 at 8:52 | comment | added | Wlod AA | I remember that Hilbert had something to do with this topic in a basic way. (Sorry, I . don't have my books from the past). | |
| Apr 12, 2018 at 8:43 | comment | added | Taras Banakh | @YCor For infinite $X$ the cone is infinite-dimensional. So your comment about faces concerns only finite $X$? | |
| Apr 12, 2018 at 8:41 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Apr 12, 2018 at 8:19 | comment | added | YCor | Possibly there have been people looking at a description of the cone of conditionally negative (semi)definite kernels (i.e., the conditions satisfied by $d^2$ in Schoenberg's criterion). What are its facets? your characterization shows that it's a polyhedral cone and that the facets are among the hyperplanes occurring in it. | |
| Apr 12, 2018 at 8:06 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Apr 12, 2018 at 8:05 | comment | added | Taras Banakh | @YCor Thank you for the comment. I have made an edit, including the change of terminology to "bipartite". Writing "geometric" I had in mind that the Schoenberg criterion is "analytic". | |
| Apr 12, 2018 at 7:56 | comment | added | YCor | Your first sentence has a double possible meaning. After checking in the book, "Which can be found" refers to Schoenberg's theorem, not to what you call its geometric version. (Is it more geometric?? for me it rather sounds like a bipartite version.) | |
| Apr 12, 2018 at 7:27 | history | asked | Taras Banakh | CC BY-SA 3.0 |