Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let me recall a result due to I. Schur, which I learnt from F. Goldberg's answer to my MO question Hadamard-like inequalites for positive definite symmetric matrices. If $H$ is a subgroup of $\frak S_n$ and $\chi$ is an irreducible complex character over $H$, define $$d_\chi(S)=\frac1{\chi(e)}\sum_{g\in H}\chi(g)\prod_{i=1}^ns_{ig(i)}.$$ Then for every $S\in SPD_n$, we have $$\det(S)\le d_\chi(S).$$ Notice that if $H=\frak S_n$ and $\chi$ is the signature, then $d_\chi$ is the determinant. Thus $\det$ is the smallest element among the $d_\chi$'s. If instead $\chi={\bf1}$, then $d_\chi$ is the permanent. If $H=(e)$, Schur's inequality is just the Hadamard inequality $$\det S\le\prod_is_{ii}.$$

Given $n$, there are many distinct $d_\chi$'s, even though several choices of the pair $(H,\chi)$ yield the same function. For instance, there are $11$ distinct functions if $n=3$.

My question is whether the permanent is the largest element among the $d_\chi$'s. In other words, is it true that for every $S\in SPD_n$, we have $$d_\chi(S)\le{\rm per}(S)\quad ?$$

I checked the truth of this assertion if $n=2$, $n=3$, and also in quite a complicated case of $n=4$, where $H={\frak A}_4$ and $\chi\ne{\bf1}$ is a linear character.

share|cite|improve this question
Dear Denis, I fixed the representation theory tag and added an open problem tag. I hope that is acceptable to you. Regards, –  Suvrit Oct 1 '12 at 13:31

1 Answer 1

up vote 8 down vote accepted

This question is better known as the permanental dominance conjecture and is still an open problem.

According to Zhan's survey, it has been confirmed for every irreducible character of $S_n$ for $n \le 13$. Another reference cited for this conjecture is this survey on open problems about permanents by Cheon and Wanless.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.