Timeline for A particular specialization of symmetric polynomials: is it bijective?
Current License: CC BY-SA 3.0
12 events
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| Oct 3, 2014 at 18:36 | answer | added | Mark Wildon | timeline score: 1 | |
| Sep 29, 2014 at 21:39 | answer | added | David Hill | timeline score: 1 | |
| Sep 29, 2014 at 15:13 | comment | added | David Hill | Conjugating $\mathcal{C}$ by the automorphism that interchanges the elementary and homogeneous symmetric functions may be helpful. Doing this in your example yields a triangular matrix (after deleting the first column). | |
| Sep 29, 2014 at 13:33 | history | edited | eddy ardonne | CC BY-SA 3.0 |
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| Sep 29, 2014 at 13:20 | comment | added | Mark Wildon | My comment above has a serious error: as the example in the question shows, the $p_{r_1}(y)\ldots p_{r_k}(y)$ are not necessarily linearly independent. I've deleted an incorrect answer that built on this comment. | |
| Sep 29, 2014 at 12:34 | history | edited | eddy ardonne | CC BY-SA 3.0 |
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| Sep 23, 2014 at 13:47 | comment | added | eddy ardonne | @MarkWildon This method certainly works in the case of an infinite number of variables, but seems more complicated in the case of a finite number of variables. In that case, we need to take the 'partial degree' restriction into account. The power sums $p_\lambda$ with $\lambda$ a partition of at most $n$ parts, each part $\leq d$ do not form a basis of $\Lambda^d_n$, which complicates matters. Do you mean to say that we do not have to worry about this? If so, could you explain why this isn't a problem? | |
| Sep 23, 2014 at 13:41 | comment | added | eddy ardonne | @per we tried this with several types of polynomials in several combinations, but we didn't find a combination that gives a triangular structure. | |
| Sep 22, 2014 at 12:51 | comment | added | Mark Wildon | Let $p_r$ be the power sum symmetric function of degree $r$. Since $\mathcal{C}(p_r(x_1,...,x_{mN})) = mp_r(y_1,..y_N)$ we have $\mathcal{C}(p_r(x)) = mp_r(y)$. Since plethystic substitution is a ring homomorphism, $\mathcal{C}(p_{r_1}(x)\ldots p_{r_k}(x)) = m^k p_{r_1}(y) \ldots p_{r_k}(y)$. So $\mathcal{C}$ is diagonal with non-zero eigenvalues in the power-sum basis of $\Lambda^d_n$. | |
| Sep 22, 2014 at 12:40 | comment | added | Per Alexandersson | Have you managed to compute this map for say, powersum, elementary, or monomial basis? If you do this, express the result in the same basis (or different), and look at the transition matrix. With some luck, and some smart ordering of the elements, you might hope for some triangular system. | |
| Sep 22, 2014 at 12:11 | review | First posts | |||
| Sep 22, 2014 at 12:12 | |||||
| Sep 22, 2014 at 12:10 | history | asked | eddy ardonne | CC BY-SA 3.0 |