(This question is also posted in Mathoverflow: https://mathoverflow.net/questions/286539/polarization-operators-and-the-action-of-gl-ell-mathbbr-on-mathcalr)
Let $X$ be a matrix of variables $x_{ij}$ of size $\ell\times n$: \begin{equation*} X=\left(\begin{array}{cccc} x_{11}&x_{12} &\dots & x_{1n}\\ x_{21}&x_{22} &\dots & x_{2n}\\ \vdots& \vdots & \ddots & \vdots \\ x_{\ell 1}&x_{\ell 2} &\dots & x_{\ell n}\\\end{array}\right). \end{equation*} The ring $\mathcal{R}_{n}^{(\ell)}=\mathbb{R}[X]$ of polynomials in $\ell$ sets of $n$ variables is defined as the $\mathbb{R}$-vector space generated by the monomials: $$X^{A}=\prod_{i=1}^{\ell}\prod_{j=1}^{n}x_{ij}^{a_{ij}}$$ The multidegree of a monomial in this ring is given by: \begin{equation*} \deg\left(X^{A}\right):=\left(\sum_{j=1}^{n}a_{1j},\sum_{j=1}^{n}a_{2j},\dots,\sum_{j=1}^{n}a_{\ell j}\right) \end{equation*} So an element $f\in\mathcal{R}_{n}^{(\ell)}$ have the form \begin{equation*} \displaystyle{f(X)=\sum_{A\in\mathbb{N}^{\ell\times n}}f_{A}X^{A}} \end{equation*} The multidegree of $f$ is given by: \begin{equation*} \deg\left(f(X)\right):={\max}_{{\rm grlex}}\left\{\deg\left(X^{A}\right):\ f_{A}\neq 0\right\}. \end{equation*} where the maximum is taken w.r.t. the graded lexicographic order in $\mathbb{N}^{\ell}$. So the ring $\mathcal{R}_{n}^{(\ell)}$ is $\mathbb{N}^{\ell}$ graded.
Soit $Q$ la matrice diagonale suivante: \begin{equation*} Q=\left(\begin{array}{cccc} q_1&0&\dots&0\\ 0&q_2&\dots&0\\ \vdots&\vdots&\dots&\vdots\\ 0&0&\dots&q_{\ell}\\ \end{array}\right) \end{equation*} A polynomial $f(X)\in\mathcal{R}_{n}^{(\ell)}$ is said to be homogeneous of multidegree ${\bf d}$ if the following condition holds: \begin{equation*} f(QX)={\bf q}^{\bf d}f(X). \end{equation*} where ${\bf q}=(q_1,\dots,q_{\ell})$, ${\bf d}=(d_1,\dots,d_{\ell})$ , ${\bf q}^{\bf d}=q_1^{d_1}\dots q_{\ell}^{d_{\ell}}$. It's well known that
For a matrix of variables $Y=(y_{{ij}})$ of size $\ell\times n$ and $f\in\mathcal{R}_{n}^{(\ell)}$ an homogeneous polynomial of multidegree ${\bf d}$, then for every matrix $M$ of size $\ell\times\ell$ we have \begin{equation*} f(MX)=\sum_{\big\{K\in\mathbb{N}^{\ell\times\ell}:\ \big\vert{K}\big\vert={\bf d}\big\}} \frac{M^{K}}{K!}\,\prod_{i=1}^{\ell}\prod_{j=1}^{\ell}P_{j,i}^{k_{ij}} \big(f(Y)\big), \end{equation*} where $$K!:=\displaystyle{\prod_{i=1}^{\ell}\prod_{j=1}^{\ell}k_{ij}!}$$
$$\displaystyle{M^{K}=\prod_{i=1}^{\ell}\prod_{j=1}^{\ell}m_{i,j}^{k_{ij}}}$$
and the polarization operator $P_{ik}$ is given by $$P_{i,k}:=\displaystyle{\sum_{j=1}^{n}x_{ij}\frac{\partial\ }{\partial y_{kj}}}.$$
The notation $\ \big\vert{K}\big\vert={\bf d}$ represents the set of all squares matrices $K$ of order $\ell$ such that $\displaystyle{\sum_{j=1}^{n}k_{ij}=d_{i}}$, for all $i$ such that $1\leq i\leq\ell$.
My question: Reading Claudio Procesi book I saw that the ring $\mathcal{R}_{n}^{(\ell)}$ is closed under polarization operators $P_{ik}$ if and only if is closed under the action of the general linear group $GL_{\ell}(\mathbb{R})$. Using the formula for $f(MX)$ above I can understand why geing closed under the $P_{ik}$ implies that $\mathcal{R}_{n}^{(\ell)}$ is closed under the right side action of $GL_{\ell}(\mathbb{R})$. But, how to find the matrices $M$ to show the reciprocal of this result.
Thank for any hint on this.