Are the minors of this Hadamard product Schur positive?

Question

Let $h_i (x)$ denote the complete symmetric function of degree $i$ in some set of variables $x = (x_1 , x_2 , \dots)$. Then the minors of the Toeplitz matrix $T (x) = \left(h_{i-j} (x) \right)_{i,j}$ are all Schur positive, in the sense that every minor can be expanded as a nonnegative linear combination of Schur polynomials $s_\lambda (x)$ (this is a direct consequence of the Jacobi--Trudi identity).

In a similar vein, one can consider the Hadamard product of the above matrix in two different sets of variables: $$T(x) * T(y) = \left( h_{i-j} (x) \cdot h_{i-j} (y) \right)_{i,j}.$$ Then my question is the following: are the minors of this matrix also Schur positive, in the sense that every minor is a nonnegative linear combination of products of Schur polynomials $s_\lambda (x) \cdot s_\mu (y)$ for some partitions $\lambda, \mu$? By some older results of Wagner, I know that if I plug in $1$ for all of the variables, then the resulting product has nonnegative minors, but I don't think this is enough to deduce Schur positivity.

For certain classes of minors I know the answer to be true, since I can identify these as being the characters of some special classes of representations associated to Segre products of the polynomial ring. That being said, the Schur expansion is very complicated in these cases so I suspect that a closed form Schur expansion would likely be pretty difficult to write down. For my purposes a closed form isn't really necessary, but could be nice.

Answer 1

The answer to this question is no, with a quick counterexample being given by the determinant of the matrix $$\begin{pmatrix} h_{2} (x) h_{2} (y) & h_{3} (x) h_{3} (y) & h_{4} (x) h_{4} (y)\\\\ h_{1} (x) h_{1} (y) & h_{2} (x) h_{2} (y) & h_{3}(x) h_{3} (y)\\\\ h_{0} (x) h_{0} (y) & h_{1} (x) h_{1} (y) & h_{2} (x)h_{2} (y) \end{pmatrix}$$ One can verify using a CAS (or directly by hand, if one is so inclined ..) that the Schur expansion of this determinant is: $$s_{2,2,2}(x)s_{6} (y)+\left(-s_{3,3}(x)+2\,s_{2,2,2} (x)\right)s_{5,1} (y)+\left(2\,s_{4,2 } (x)+2\,s_{3,2,1} (x)+3\,s_{2,2,2} (x) \right)s_{4,2}(y)+\left(-s_{3,3} (x)+s_{2,2,2} (x)\right) s_{4,1,1} (y)+\left(-s_{5,1} (x)-s_{4,1,1} (x)+s_{2,2,2} (x)\right)s_{3,3} (y)+\left(2\,s_{4, 2} (x)+2\,s_{3,2,1}(x)+2\,s_{2,2,2} (x)\right)s_{3,2,1} (y)+\left(s_{6} (x)+2\,s_{5,1} (x)+3\,s _{4,2} (x)+s_{4,1,1} (x)+s_{3,3} (x)+2\,s_{3,2,1} (x)+s_{2,2,2} (x)\right)s_{2,2,2} (y)$$

I also wanted to give a conceptual reason (which I will credit to Steven Sam) for why we didn't need to do any computations in the first place to deduce that this statement is false. Note first that there are well-known counterexamples to the statement that the Hadamard product of totally positive sequences is also totally positive.

Now, if the Hadamard product of the Toeplitz matrix stated in the question above were totally positive, then this would actually imply the clearly false conclusion that the Hadamard product of arbitrary totally positive Toeplitz matrices would still be totally positive. This is because we could consider the specialization map $\Lambda \otimes \Lambda \to \mathbb{Z}$, where $\Lambda$ denotes the ring of symmetric functions (which is just a polynomial ring on the functions $h_i (x)$), that sends $h_i (x) \mapsto a_i$ and $h_i (y) \mapsto b_i$ for all $i$. This is an algebra map, and so if all minors of the Toeplitz matrix were Schur positive, then the image of these minors would be a nonnegative linear combination of nonnegative integers, hence itself nonnegative (here we are using the fact that the Jacobi-Trudi identity for Schur polynomials is a minor of a Toeplitz matrix).