# inner product on matrix spaces of multivariate polynomials?

Let $H_{n,d}=\mathbb{R}_d[x_1,..,x_n]$ be the space of $n$-variate homogeneous degree $d$ polynomials, $D=D^\top\in \mathbb{N}^{m\times m}$ a symmetric $m\times m$ matrix. Consider the space $P_D$ of symmetric $m\times m$ matrices with $(i,j)$-entries in $H_{n,D_{ij}}$. Is there a natural (matrix?) inner product structure on $P_D$? In my application I have cone $C$ of positive semidefinite (globally, for any value of $x=(x_1,\dots,x_n)$) matrices in $P_D$, and I'd like to find defining inequalities for it - i.e. the dual cone $C^*$.

One possibility might be to use Fischer-Fock inner product $[,]$ on the entries, so that for $f,g\in P_D$ one has $p=\langle f,g\rangle\in\mathbb{R}^{m\times m}$ with $p_{ij}=[f_{ij},g_{ij}]$, and nonnegativity (resp. positivity) of $p$ understood as $p$ being positive semidefinite (resp. definite).

Is true that $\langle f,f\rangle$ is p.s.d. whenever $f$ is p.s.d.?

Or am I on wrong track?

Note: for $u,v\in H_{n,d}$ the Fischer-Fock product (see below) is defined as $$[u,v]:=\sum_{|\alpha|=d}\binom{|\alpha|}{\alpha} u_\alpha v_\alpha,$$ where the usual multinomial notation $\alpha:=(\alpha_1,\dots,\alpha_n)$, $|\alpha|:=\sum_k\alpha_k$, $\binom{|\alpha|}{\alpha}:=\frac{|\alpha|!}{\prod_k\alpha_k!}$, $x_\alpha:=\prod_k x_k^{\alpha_k}$, $u(x)=\sum_\alpha \binom{|\alpha|}{\alpha}u_\alpha x^\alpha$ is used. In particular, $u(y)=[u,(\sum_k y_kx_k)^d]$.

• If you modify your definition of Fischer-Fock inner product on $H_{n,d}$ by dividing by d! then the inner product can be given by an integral with respect to the Gaussian measure over $\mathbb{C}^n$. I then believe that $\langle f,f\rangle$ is p.s.d whenever $f$ is p.s.d. pointwise as you asked. – T. Le Jul 5 '17 at 15:53

Claim: Suppose $[u,v]=\int u\bar{v}\,d\mu$. Then whenever the matrix $(f_{ij})_{1\leq i,j\leq m}$ is p.s.d. pointwise, the matrix $([f_{ij},f_{ij}])_{1\leq i,j\leq m}$ is p.s.d. as well.
To see this, notice that the matrix $(\bar{f_{ij}})_{1\leq i,j\leq m}$ is also p.s.d. pointwise, which implies $(|f_{ij}|^2)_{1\leq i,j\leq m}$ is p.s.d. pointwise since it is the Hadamard product of two p.s.d. matricies. Taking integral would give the desired conclusion.
• No, the domain of integration does not matter. The claim holds for functions $f_{ij}$ in $L^2(X,\mu)$, where $\mu$ is a positive measure on $X$. But note that the same measure must be used for all entries. – T. Le Jul 5 '17 at 18:47
• Using $\int u\overline{v}$ as the scalar product seems a bit too coarse in the case the difference between the cone and its dual is measure 0. E.g. consider the cone $C=H_{3,(4)}$, i.e. the nonnegative ternary quartics in $H_{3,4}$, and $m=1$. Its dual $C^*$, if I use my product, is spanned by 4th powers of linear forms, a proper subcone (look at possible sets of zeros), whereas your product would give that $C=C^*$. Probably some topology is needed to make this make sense... – Dima Pasechnik Jul 6 '17 at 22:35
• On each $H_{n,d}$, your Fischer-Fock inner product in fact comes from an integral, not over $\mathbb{R}^n$ but over $\mathbb{C}^n$ with respect to a multiple of the Gaussian measure. That is, for any $u,v\in H_{n,d}$, the Fischer-Fock product $$[u,v]=\text{(constant depending only on n and d)}*\int_{\mathbb{C}^n}u(z)\overline{v(z)}e^{-|z|^2}dV(z).$$ As I mentioned in a comment, if you are willing to change your product on each $H_{n,d}$ by a constant depending on $n$ and $d$ then the new product will be given by an integral over $\mathbb{C}^n$ with respect to $e^{-|z|^2}dV(z)$. – T. Le Jul 7 '17 at 0:36
• Thanks, I was too careless reading your comments. Now I see that it is crucial to integrate over $\mathbb{C}^n$ rather than $\mathbb{R}^n$... – Dima Pasechnik Jul 7 '17 at 9:48
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