Timeline for Projecting the unit cube onto subspaces
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 7, 2011 at 8:14 | vote | accept | Aaron Meyerowitz | ||
| Apr 7, 2011 at 8:10 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
added 3759 characters in body; edited title; added 1 characters in body
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| Apr 6, 2011 at 17:51 | comment | added | Seva | @Aaron: there is a typo in the title of the question. @Everybody else: if you like the present question, you may wish to check the question it originated from (mathoverflow.net/questions/60604) and consider voting to re-open it. | |
| Apr 6, 2011 at 13:38 | history | edited | Gerry Myerson |
added arXiv tag
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| Apr 6, 2011 at 7:38 | answer | added | Denis Serre | timeline score: 3 | |
| Apr 6, 2011 at 7:32 | answer | added | Seva | timeline score: 4 | |
| Apr 6, 2011 at 7:08 | comment | added | Gerhard Paseman | If n=2, I imagine that the subspace (t, -t) gives the minimum value. For n > 2, and d=1, the subspace (t, -t, 0, ...,0) sets the limbo bar pretty low. For arbitrary d, looking at n = d+1 might help with determining the minimum, and will be something like the subspace orthogonal to (1,1,...,1) (d+1) ones, which you can generalize. This is not a proof so much as a goal. Gerhard "How Low Can You Go" Paseman, 2011.04.06 | |
| Apr 6, 2011 at 7:01 | comment | added | Aaron Meyerowitz | Yes, that was my intention. | |
| Apr 6, 2011 at 6:58 | comment | added | Denis Serre | I presume that the norm is the standard Euclidian and the projection is orthogonal. | |
| Apr 6, 2011 at 6:51 | history | asked | Aaron Meyerowitz | CC BY-SA 2.5 |