Timeline for A question about homogenous polynomials of degree $\frac{n(n-1)}{2}$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2016 at 15:02 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
added 25 characters in body
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| Mar 26, 2016 at 8:07 | comment | added | Fedor Petrov | Sometimes it does work. I do not understand which specific polynomial do you consider for large $n$. Antisymmetrization is your $\alpha (f)$. | |
| Mar 26, 2016 at 7:52 | comment | added | user173856 | Fedor Petrov: Thanks for your answer! But I think your method is not suitable for big $n$. Do you still have any method to deal with similar problems when $n$ is big? By the way, is the conception "antisymmetrization" for a polynomial you mentioned above a defined mathematical concept? | |
| Mar 26, 2016 at 7:25 | vote | accept | user173856 | ||
| Mar 22, 2016 at 17:54 | comment | added | Fedor Petrov | @Denis In other words, you mention cyclic invariance, but really there is dihedral $D_5$-invariance, and up to this invariance there is unique non-zero summand. | |
| Mar 22, 2016 at 16:04 | comment | added | Fedor Petrov | of course, these are the same cycle 10234 in two orders, totally 10 summands | |
| Mar 22, 2016 at 15:56 | comment | added | Denis Serre | Nice idea. I wish however bring a correction. After the remark that we may always restrict to permutations fixing $0$, because of the cyclic invariance of $f$, there remains two permutations $\pi$ for with $g^\pi$ does not vanish at the point $(0,1,2,3,4)$, namely the transposition $(24)$ and the $4$-cycle $(1,2,3,4)$. Fortunately, they give the same contribution, with the same signature. | |
| Mar 22, 2016 at 14:58 | history | answered | Fedor Petrov | CC BY-SA 3.0 |