Let $H$ denote an $n$ by $n$ hermitian positive semidefinite matrix. Let $G$ and $K$ be two subgroups of the symmetric group $\Sigma_n$. Define $$ f_{G, K}(H) = \sum_{(\sigma, \tau) \in G \times K} \prod_{i = 1}^n h_{\sigma(i), \tau(i)}, $$ where $H = (h_{ij})$. Does there exist a positive constant $c(n, G, K)$ depending on $n$, $G$ and $K$ such that the following holds? $$ \Re(f_{G, K}(H)) \geq c(n, G, K) \prod_{i=1}^n h_{ii}$$ In the previous formula, $\Re(.)$ denotes the real part.

I am interested in a more special case of the above, where $G$ and $K$ are each products of symmetric groups on disjoint subsets of $\{1, \ldots, n\}$ (like for example the symmetric group on $\{1, 3 \}$ times the symmetric group on $\{2, 4\}$). Moreover, in the special case I am interested in, $H$ has rank $2$. That being said, the question does make sense in general. I did not yet run numerical tests. I wonder if anyone has seen such inequalities, whether proved or conjectured. If $G$ is trivial, then these have definitely been studied. They include the known Marcus's inequality for the permanent if $K$ is the full symmetric group $\Sigma_n$. I will have to dig in the literature to see if they have been proved in general or not if $K$ is an arbitrary subgroup of $\Sigma_n$. But what about these two-sided cousins? Have they been studied too, please?

Motivation: they do occur when studying some special functions on the configuration space of $n$ distinct points that are associated to graphs, in the special case where the graph is a tree, though the groups $G$ and $K$ are each products of symmetric groups (on disjoint sets of indices between $1$ and $n$, inclusively).

Additional questions:

  1. Are the inequalities true, known to be true/false or conjectured to be true if one of the two groups is trivial?
  2. Is it true that the real part of $f_{G, K}$ is positive on the cone of all positive semidefinite matrices with positive entries on their diagonal?

Note that on the previous cone, the ratio of the real part of $f_{G, K}$ divided by the product of the diagonal elements of $H$ is bounded, since it descends to a continuous (actually smooth) function on the product of $n$ copies of $\mathbb{C}P^{n-1}$. Indeed, if we write $H = V^*V$, for some complex $n$ by $n$ matrix $V$, then both $f_{G, K}$ and the product of the diagonal elements of $H$ have the same homogeneity as functions of the columns of $V$. So by a continuity/compactness argument, we get that the ratio is bounded (so the real part of the ratio is also bounded).

So the main inequality would follow for a given $(G, K)$ provided the real part of $f_{G, K}$ does not vanish anywhere on the cone of $n$ by $n$ hermitian positive semidefinite matrices with positive diagonal elements.

Edit: the problem can be reformulated using the same ideas that show that the permanent of an hermitian positive semidefinite matrix is the norm squared of a certain tensor. I hope that this remark will excuse me from writing up all the details. If someone wants me to write more, please let me know.

Here is the equivalent problem. Let $v_i \in \mathbb{C}^n$ be nonzero vectors, for $i = 1, \ldots, n$ and let $G$ and $K$ be two subgroups of the symmetric group $\Sigma_n$. Define $$\psi_G = \sum_{\sigma \in G} \bigotimes_{i=1}^n v_{\sigma(i)}$$ and similarly $$\psi_K = \sum_{\sigma \in K} \bigotimes_{i=1}^n v_{\sigma(i)}.$$ If $\langle ., . \rangle$ denotes the standard hermitian inner product on $\mathbb{C}^n$, my question is whether or not $$ \Re \langle \psi_G, \psi_K \rangle > 0, $$ no matter what non-zero vectors $(v_i)$ in $\mathbb{C}^n$ we start with.

This is a more coordinate-free and geometric (sort of) formulation of my question.

Note that the case where, say, one of $G$ and $K$ is trivial and the other is a subgroup, say $H$, of $\Sigma_n$ are equivalent, and are also equivalent to the case of $G$ and $K$ both equal to $H$. This can be easily proved by "transferring" one index to the other side and using that the left translate of $H$ using an element of $H$ is nothing but a permutation of $H$.

As a remark, the questions asked in this post led to a kind of interesting other post (in my humble opinion): A question regarding symmetrizing the tensor product of vectors in two different ways.

  • $\begingroup$ Why did these go from cousins of immanants to cousins of permanents? $\endgroup$
    – LSpice
    Oct 13, 2023 at 23:38
  • 1
    $\begingroup$ Hi @LSpice. It is just that immanents also include a character of the group inside the summation symbol. The character in my case is trivial. So they are say, closer cousins to permanents and more distant cousins to immanants, or something like that :). $\endgroup$
    – Malkoun
    Oct 13, 2023 at 23:40


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