Timeline for Does this condition imply a polynomial is a product of linear factors
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2018 at 2:25 | vote | accept | user44191 | ||
| Apr 24, 2018 at 2:24 | answer | added | user44191 | timeline score: 1 | |
| Nov 21, 2014 at 8:12 | history | edited | user44191 | CC BY-SA 3.0 |
Added a group cohomology version
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| Nov 14, 2014 at 19:31 | comment | added | user44191 | I've manged to come up with a sketch of a proof for this; the proof goes through the fact that for any irreducible $a_i(\lambda)$ that divides $b_\mu(\lambda)$, there is some $\alpha_i$ such that for any $\gamma$ with $\langle \alpha_i, \gamma\rangle = 0$, $a_i(\lambda + \gamma) = a_i(\lambda)$, which shows that $a_i$ is divisible by $\alpha_i + k_i$ for some $k_i$. | |
| Nov 14, 2014 at 17:12 | comment | added | user44191 | Per: the dimension formula appears in some of the research that led to this question, but isn't the sole basis for it. | |
| Nov 14, 2014 at 17:09 | comment | added | user44191 | Pietro: any polynomial in one variable will split; for yours, $b_1(\mu) = (\lambda + 1)^2 + 1 = (\lambda + 1 + i)(\lambda + 1 - i)$. | |
| Nov 14, 2014 at 17:06 | comment | added | user44191 | Alex: I don't just have one polynomial, I have many with a certain compatibility condition; I think that that compatibility condition implies factoring. | |
| Nov 14, 2014 at 12:49 | comment | added | Per Alexandersson | Is this related to the Weyl dimension formula? That is, in the end of properties on en.wikipedia.org/wiki/Schur_polynomial#Properties ? This formula is essentially a lattice point counting number. | |
| Nov 14, 2014 at 11:10 | comment | added | Pietro Majer | Also, what if $H(\lambda):=\prod_{j=1}^\lambda (j^2+1)$, for $\lambda\in\mathbb{Z}_+$ ? | |
| S Nov 14, 2014 at 10:48 | history | suggested | gaoxinge | CC BY-SA 3.0 |
body change
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| Nov 14, 2014 at 10:23 | review | Suggested edits | |||
| S Nov 14, 2014 at 10:48 | |||||
| Nov 14, 2014 at 9:35 | comment | added | Alex Degtyarev | I don't quite get it. You have a multivariate (if $\operatorname{rank}\Lambda>1$) polynomial; why should it split into linear factors? | |
| Nov 14, 2014 at 8:57 | history | asked | user44191 | CC BY-SA 3.0 |