Timeline for Bivariate polynomials with special properties
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2011 at 3:17 | comment | added | ARupinski | What is the sequence? If it is related to these polynomials, there is a chance some of the recurrences related to SU(3) might help in developing a recurrence (or other properties) for your other sequence. | |
| Mar 25, 2011 at 9:49 | comment | added | Per Alexandersson | Thanks for the answer! There is a polynomial in $v$, such that all roots of $v$ admits a solution to the bivariate system (one may find these by using groebner bases). The sequence one gets is a sequence of polynomials of degree n(n+1)/2. A recurrence for this sequence would be very nice, however, I doubt there is a reasonable nice one... | |
| Mar 25, 2011 at 9:34 | vote | accept | Per Alexandersson | ||
| Mar 23, 2011 at 21:44 | history | edited | ARupinski | CC BY-SA 2.5 |
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| Mar 23, 2011 at 4:56 | history | edited | ARupinski | CC BY-SA 2.5 |
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| Mar 22, 2011 at 18:14 | history | edited | ARupinski | CC BY-SA 2.5 |
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| Mar 21, 2011 at 18:51 | comment | added | ARupinski | On the other hand, the characters of SU(3) carry a natural $\mathbb{Z}/3\mathbb{Z}$-grading: give the variable $u$ a grade of 1, $v$ a grade of 2, and constants a grade of 0. Each of the polynomials listed is homogenous with respect to this grading (i.e. all terms have the same grade mod 3). Perhaps this grading might be what is needed to make a polynomial "remarkable" in the sense of the question. | |
| Mar 21, 2011 at 18:28 | comment | added | ARupinski | Offhand, without knowing the algorithm or motivation behind the polynomials, its hard to say whether one can easily see SU(3). As for whether every polynomial satisfying the properties is a character of SU(3), a partial answer is that the irreducible characters of SU(3) form a $\mathbb{Z}$-basis of $\mathbb{Z}[u,v]$, so on the one hand every 2-variable polynomial does correspond to the character of some SU(3) representation. | |
| Mar 21, 2011 at 18:12 | comment | added | Daniel Litt | This is awesome! Can one see SU(3) in the original problem itself somehow? Also, is every polynomial satisfying the required properties a character of SU(3)? | |
| Mar 21, 2011 at 17:20 | history | edited | ARupinski | CC BY-SA 2.5 |
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| Mar 21, 2011 at 17:15 | comment | added | ARupinski | Incidentally, once you know the first few of the $P_{a,b}$ you can use Klimyk's formula to recursively derive many more, for example one has: $P_{a+1,b} = P_{1,0}\cdot P_{a,b} - P_{a-1,b+1} - P_{a,b-1}$ | |
| Mar 21, 2011 at 17:08 | history | answered | ARupinski | CC BY-SA 2.5 |