# Hadamard-like inequalites for positive definite symmetric matrices

Let $S$ be any positive semi-definite symmetric matrix (Hermitian psd matrices work as well). The Hadamard inequality is that $$\det S\le\prod_{i=1}^n s_{ii}.$$ My question is whether there are some other upper bounds of $\det S$ in terms of a partial sum $$\sum_{\sigma\in F}\epsilon(\sigma)\prod_{i=1}^n s_{i\sigma(i)},$$ where $F$ is some subset of ${\frak S}_n$. The Hadamard inequality corresponds to the case $F=({\rm id})$.

More precisely, I am interested in the validity of $$\det S\le\sum_{\epsilon(\sigma)=1}\prod_{i=1}^n s_{i\sigma(i)}.\qquad\qquad(1)$$

This latter inequality is true at least when $n\le4$, essentially because the difference $$\sum_{\epsilon(\sigma)=-1}\prod_{i=1}^n s_{i\sigma(i)}$$ contains enough many non-negative terms (the half if $n=4$, all of them if $n=2$ or $3$). This reason (which can be combined to the Cauchy-Schwarz inequality) fails if $n\ge5$, and therefore I suspect that the inequality (1) could be false if $n\ge5$.

Edit. Felix Goldberg's answer tells us that inequality (1) is true. More generally, Schur proved that if $G$ is a subgroup of ${\frak S}$, and $\chi$ is a character of $G$, then $$\chi(e)\det S\le\sum_{\sigma\in G}\chi(\sigma)\prod_{i=1}^n s_{i\sigma(i)}=:d_\chi(S).$$ This gives (1) when $G={\frak A}_n$ and $\chi={\bf 1}$.

This leads me raising a second question:

If $G$ is a subgroup of ${\frak S}_n$, let $K_G$ be the set of class functions $f$ over $G$ with the property that, for every positive definite symmetric matrix $S$, one has $$f(e)\det S\le\sum_{\sigma\in G}\chi(\sigma)\prod_{i=1}^n s_{i\sigma(i)}=:d_\chi(S).$$ Clearly, $K_G$ is a convex cone, which contains all the characters. Is it equal to the convex cone spanned by the ireeducible characters of $G$ ?

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It may be trivial, but neverhteless: there exist many different inequalities of this kind (for the sets $F$ different from the one you need). Actually, if you split $[n]=\sqcup_i N_i$ and denote by $S_i$ the principal minor on the set of indices $N_i$, then $\det S\leq\prod_i \det S_i$. This is true because of the interpretation of $S$ as the Gram matrix of some vectors, and of $\det S_i$ as the squares of the respective volumes. –  Ilya Bogdanov Aug 28 '12 at 11:13
@Ilya. Shame on me, I should have think to it. This is block-wise Hadamard inequality, and its proof is essentially the same as that for the entrywise inequality. –  Denis Serre Aug 28 '12 at 11:34