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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 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.

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 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.

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.

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 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.

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 $$d_\chi(S)=\frac1{\chi(e)}\sum_{g\in H}\prod_{i=1}^ns_{ig(i)}.$$ 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.

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.