Timeline for Coefficients of the monomials appearing in a Schubert polynomial
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 23, 2018 at 13:10 | comment | added | Oliver | Darij's suspicion is correct: The leading coefficient is $x^{L(w)}$. This is transparent to see from the Kohnert formula for Schubert polynomials (e.g. arxiv.org/abs/1703.00088). (Some people don't believe some proofs of this formula, but I think everyone believes the formula!) | |
| Jun 23, 2018 at 7:44 | answer | added | Per Alexandersson | timeline score: -1 | |
| S Jun 23, 2018 at 2:58 | history | suggested | J.J. Green | CC BY-SA 4.0 |
tidier link, minor typography fixes
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| Jun 22, 2018 at 23:55 | review | Suggested edits | |||
| S Jun 23, 2018 at 2:58 | |||||
| Jun 22, 2018 at 17:14 | comment | added | Sam Hopkins | See also Lemma 4.11 here: ams.org/journals/jams/1995-08-02/S0894-0347-1995-1290232-1/… | |
| Jun 22, 2018 at 17:11 | comment | added | Sam Hopkins | Look at Corollary 3.9 of "RC-Graphs and Schubert Polynomials" by Bergeron and Billey: projecteuclid.org/download/pdf_1/euclid.em/1048516036. It describes the (unique) leading term of a Schubert polynomial. | |
| Jun 22, 2018 at 17:10 | comment | added | darij grinberg | @misao: I said that the coefficient of $\mathbf{x}^{L\left(w\right)}$ in $\mathfrak{S}_w$ is $1$ (in particular, this monomial must thus occur in $\mathfrak{S}_w$), and that all other monomials are smaller in lex order. But I also said that this is all speculation :) Maybe Sam's reference helps (I regret I don't speak its language). | |
| Jun 22, 2018 at 16:59 | comment | added | Sam Hopkins | I think this is claimed at the beginning of the proof of Proposition 1.4 of the Billey-Jockush-Stanley paper (bottom of pg. 357): link.springer.com/content/pdf/10.1023/A:1022419800503.pdf | |
| Jun 22, 2018 at 16:58 | comment | added | user111492 | @darij grinberg : umm ... I thought you said the coefficient of $x^{L(w)}$ is $1$ ... does $x^{L(w)}$ always occur in the Schubert polynomial of $w$ ? | |
| Jun 22, 2018 at 16:55 | comment | added | darij grinberg | As I said, I suspect that the leading coefficient is $1$. | |
| Jun 22, 2018 at 16:55 | comment | added | user111492 | @Sam Hopkins: I thought so initially too ... but I don't quite see any pattern ... please do feel free to give your inputs | |
| Jun 22, 2018 at 16:54 | comment | added | user111492 | @darij grinberg : thanks for clarifying your lexicographic order ... but I don't quite see how the coefficient $1$ conjecture would follow from your conjecture ... could you elaborate here please ? | |
| Jun 22, 2018 at 15:53 | comment | added | Sam Hopkins | Probably it should not be too hard to see this from the "pipe dreams" definition of Schubert polynomials. | |
| Jun 22, 2018 at 15:44 | comment | added | darij grinberg | The lexicographic order in which $x_1^{a_1} x_2^{a_2} x_3^{a_3} \cdots > x_1^{b_1} x_2^{b_2} x_3^{b_3} \cdots$ if and only if the smallest $i$ satisfying $a_i \neq b_i$ satisfies $a_i > b_i$. Sorry for not being precise before; there are indeed several lexicographic orders. | |
| Jun 22, 2018 at 15:42 | comment | added | user111492 | @darij grinberg : can you please explicitly state what lexicographic order you are talking about here ? | |
| Jun 22, 2018 at 15:40 | comment | added | darij grinberg | I suspect that for each permutation $w \in S_n$, each monomial appearing in the Schubert polynomial $\mathfrak{S}_w$ is lexicographically smaller-or-equal to the monomial $\mathbf{x}^{L\left(w\right)} := \prod\limits_{i=1}^n x_i^{\ell_i\left(w\right)}$, where $\ell_i\left(w\right)$ denotes the number of all $j > i$ satisfying $w\left(i\right) > w\left(j\right)$. Moreover, I suspect that the coefficient of this monomial $\mathbf{x}^{L\left(w\right)}$ is $1$. | |
| Jun 22, 2018 at 15:31 | history | edited | darij grinberg | CC BY-SA 4.0 |
deleted 138 characters in body
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| Jun 22, 2018 at 15:07 | history | asked | user111492 | CC BY-SA 4.0 |