Timeline for what else is in $\prod_{j=1}^n(1+q^j)$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2017 at 13:05 | vote | accept | T. Amdeberhan | ||
| Dec 18, 2016 at 3:33 | comment | added | T. Amdeberhan | @TimothyChow: Now that you mentioned $M(n)$, let me point out that I learned this from Stanley: the poset $M(n)$, in addition to its usual definition, is also the "greedy (weak) Bruhat order" on $\mathfrak{S}_n$. This is Exercise 187(b) in Chapter 3 of Enumerative Combinatorics, Volume 1, second edition. | |
| Dec 17, 2016 at 17:47 | comment | added | Timothy Chow | Proctor (Amer. Math. Monthly 89 (1982), 721-734) did give a more elementary proof of unimodality but his argument was "secretly" motivated by the Lie-algebraic proof. He phrased his result in terms of the rank-unimodality of a certain poset $M(n)$ whose rank generating function is $\prod_{j=1}^n (1+q^j)$. | |
| Dec 17, 2016 at 17:13 | comment | added | Timothy Chow | In Richard's expository paper on log-concave and unimodal sequences (Ann. New York Acad. Sci., 1989), he mentions that by considering the principal $\mathfrak{sl}(2)$ inside $\mathfrak{so}(2n+1)$ we can prove the unimodality of the coefficients of $\prod_{j=1}^n (1+q^j)$. See Equation (23). It seems to be very difficult to prove this unimodality result otherwise. | |
| Dec 17, 2016 at 7:05 | comment | added | T. Amdeberhan | It is the orthogonal Lie algebra in $2n+1$ dimensions. | |
| Dec 17, 2016 at 6:11 | comment | added | Włodzimierz Holsztyński | What is this parasite on the left of the formula $\mathfrak{so}(2n+1)$ ? | |
| Dec 17, 2016 at 2:59 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 73 characters in body
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| Dec 17, 2016 at 2:08 | comment | added | T. Amdeberhan | Cool, up-voted. | |
| Dec 17, 2016 at 1:05 | history | answered | Richard Stanley | CC BY-SA 3.0 |