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](https://mathoverflow.net/questions/105697). 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 only $11$ distinct functions if $n=3$, among $13$ pairs. 



> 
> 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.



A maximal element, where Schur gives a minimal element
This question is better known as the [**permanental dominance conjecture**](http://math.ecnu.edu.cn/~zhan/papers/ZhanICCM.pdf) 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](http://users.monash.edu.au/~iwanless/papers/permsurveyLAA.pdf) by Cheon and Wanless.




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**EDIT** (added 12/10/2015): Incidentally, the closely related [***Soules's conjecture***](http://www.sciencedirect.com/science/article/pii/0024379594901171) whose proof would yield the above permanental dominance (and which states that the largest eigenvalues of the Schur-product matrix of a given Hermitian semidefinite matrix $A$ equals the permanent of $A$), has been very recently shown to be false: [check out this explicit counterexample!](http://www.researchgate.net/profile/Valery_Shchesnovich/publication/280495158_The_permanent_on_top_conjecture_is_false/links/55b6c96108aed621de043bdd.pdf?inViewer=true&pdfJsDownload=true&disableCoverPage=true&origin=publication_detail). (If the first link does not work, try [this link on LAA](http://www.sciencedirect.com/science/article/pii/S0024379515006631))


