Timeline for Symmetric basis of harmonic homogeneous polynomials
Current License: CC BY-SA 4.0
11 events
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| Apr 5 at 6:35 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Nov 15, 2018 at 20:23 | comment | added | Dima Pasechnik | Harmonic $n$-variate polynomials of a given degree are $S_n$-invariant (as a set), and thus one could come up with an $S_n$-invariant basis, but the corresponding action won't be so simple. | |
| Apr 20, 2018 at 7:42 | vote | accept | Pietro Majer | ||
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 4, 2013 at 20:03 | history | edited | Yemon Choi | CC BY-SA 3.0 |
as we are fixing broken rendering on very old posts, I've fixed a typo in the title and cleaned up some English
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| Sep 4, 2013 at 13:59 | history | edited | Pietro Majer | CC BY-SA 3.0 |
deleted 1 characters in body
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| Oct 20, 2011 at 22:42 | answer | added | Qiaochu Yuan | timeline score: 7 | |
| Oct 20, 2011 at 22:25 | comment | added | Qiaochu Yuan | Sorry if I am being silly, but I don't see how the result you cite can be true for $n = 2$. | |
| Oct 20, 2011 at 19:55 | comment | added | Pietro Majer | Thank you... So, the existence of an invariant basis may really depend on the pair $(n,m)$ ? | |
| Oct 20, 2011 at 19:45 | comment | added | Qiaochu Yuan | If the result is false it can be disproved with a finite, albeit tedious, calculation. A finite-dimensional real vector space on which $S_n$ acts admits a permutation-invariant basis if and only if it is a direct sum of permutation representations, and for fixed $n$ this can be checked by comparing the character of the representation against the characters of all permutation representations (this is the tedious part). In particular the number of copies of the trivial representation is the number of orbits of the action of $S_n$, so if there are no such copies then the result is false. | |
| Oct 20, 2011 at 19:12 | history | asked | Pietro Majer | CC BY-SA 3.0 |