3
$\begingroup$

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, $u(y)=[u,(\sum_k y_kx_k)^d]$.

Improve this question
$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – T. Le
    Commented Jul 5, 2017 at 15:53

2 Answers 2

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ does the domain of integration matter here? $\endgroup$
    – Dima Pasechnik
    Commented Jul 5, 2017 at 18:16
  • $\begingroup$ 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. $\endgroup$
    – T. Le
    Commented Jul 5, 2017 at 18:47
  • $\begingroup$ 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... $\endgroup$
    – Dima Pasechnik
    Commented Jul 6, 2017 at 22:35
  • $\begingroup$ 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 $e^{-|z|^2}dV(z)$. $\endgroup$
    – T. Le
    Commented Jul 7, 2017 at 0:36
  • $\begingroup$ Thanks, I was too careless reading your comments. Now I see that it is crucial to integrate over $\mathbb{C}^n$ rather than $\mathbb{R}^n$... $\endgroup$
    – Dima Pasechnik
    Commented Jul 7, 2017 at 9:48
1
$\begingroup$

Yes, this seems like a right track. In fact this kind inner product appeared many times in representation theory. See e.g. chapter two of Differential operators and highest weight representations by Davidson, Enright and Stanke. This inner product is related (in the hermitian symmetric case) to the Shapovalov form.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .