2007
DOI: 10.1016/j.jalgebra.2006.07.033
|View full text |Cite
|
Sign up to set email alerts
|

Positivity results for Stanley's character polynomials

Abstract: In Stanley [R.P. Stanley, Irreducible symmetric group characters of rectangular shape, Sém. Lothar. Combin. 50 (B50d) (2003) 11 pp.] the author introduces expressions for the normalized characters of the symmetric group and states some positivity conjectures for these expressions. Here, we give an affirmative partial answer to Stanley's positivity conjectures about the expressions using results on Kerov polynomials. In particular, we use new positivity results in Goulden and Rattan [I.P. Goulden, A. Rattan, An… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1

Citation Types

0
10
0

Year Published

2008
2008
2010
2010

Publication Types

Select...
4

Relationship

1
3

Authors

Journals

citations
Cited by 4 publications
(10 citation statements)
references
References 11 publications
0
10
0
Order By: Relevance
“…Thus, I p;q (x) − 1 and −G p;−q (−x) both satisfy (5). Combining this with (9) and Proposition 2.1 gives the result.…”
mentioning
confidence: 56%
See 2 more Smart Citations
“…Thus, I p;q (x) − 1 and −G p;−q (−x) both satisfy (5). Combining this with (9) and Proposition 2.1 gives the result.…”
mentioning
confidence: 56%
“…This conjecture has only been proved in the case m = 1 for general µ (see [8,Theorem 1.1]); in particular, the conjecture is not known to be true for m > 1 even when µ has one part; that is, the conjecture is unknown even for (−1) k F k (p; −q). Some partial results showing positivity of the coefficients of (−1) k F k (p; −q) were given in Rattan [5], but otherwise little is known about these polynomials. Recently, Stanley [9] has a conjectured combinatorial interpretation for F µ (p; q), which we now explain.…”
Section: Introductionmentioning
confidence: 99%
See 1 more Smart Citation
“…As suggested by Rattan in [6] where he gives a new easier proof of the case m = 1, we will use the following theorem:…”
Section: Link With Charactersmentioning
confidence: 99%
“…In this situation, we prove that (−1) k z µ θ λ µ,1 n−k (α) is a polynomial in (p, −q, β), with nonnegative rational coefficients. The proof is much more cumbersome and lenghty than in the case α = 1, studied in [15,13].…”
Section: Introductionmentioning
confidence: 99%