# Majorization and Schur Polynomials

Let me first define the majorization order (or dominance order) on partitions as $\lambda \succeq \mu$ iff $$\sum _{i=1}^{k}\lambda_i \geq \sum_{i=1}^{k}\mu_i$$ for all $1\le k\le l-1$ and $$\lambda_1+\cdots+\lambda_l=\mu_1+\cdots+\mu_l.$$

While playing around with some variation on Muirhead's inequality which says that $$m_{\lambda}(x_1,x_2,\dots,x_n)\geq m_{\mu}(x_1,x_2,\dots,x_n)$$ for all $x_i\geq 0$ when $\lambda \succeq \mu$ (here the $m_{\lambda}$'s are the monomial symmetric polynomials) I ended up investigating if such an inequality holds for Schur polynomials, which are another famous basis of the ring of symmetric functions. I tried a few special cases, and the following seems to hold, but I don't have a proof or a counterexample. So the question is:

Let $x_1,\dots,x_n \geq 0$ and $\lambda\succeq \mu$, is it always true that $$\frac{s _{\lambda}(x _1,x _2,\dots,x _n)}{s _{\lambda}(1,1,\dots,1)}\geq \frac{s _{\mu}(x _1,x _2,\dots,x _n)}{s _{\mu}(1,1,\dots,1)}?$$

If not, does majorization induce some similarly nice order on values of Schur polynomials?

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Perhaps it is possible to introduce a second set of variables, and look at double Schur functions as well... – Per Alexandersson Jul 27 '15 at 0:39

This question is listed as a conjecture (conjecture 7.4 in the section "Open questions") in a recent paper of Cuttler, Greene and Skandera.

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Thank you! I will have to wait till tomorrow to gain access to the paper. About a year ago I had proofs for the corresponding inequalities on elementary, complete or power symmetric functions, and I thought I had proved the inequality in the OP as well, but I had made a few silly mistakes. I guess now I'm less hopeful that it can be fixed (i.e. that it can be approached by elementary methods)... – Gjergji Zaimi May 17 '11 at 9:42
FYI, there are several online versions on author's pages, e.g. haverford.edu/math/cgreene/papers/smean10-15.pdf - in case you don't care if you get the journal version. – Vladimir Dotsenko May 17 '11 at 9:59
Oh no, that's perfect. It looks like an interesting read! – Gjergji Zaimi May 17 '11 at 10:13

This was recently proven by Suvrit Sra, on the arXiv. (Make sure to read version 3.)

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