(**Also in Mathematics stack Exchange:** <https://math.stackexchange.com/questions/2528216/polarization-operators-identity-and-gl-ell-mathbbr>) 


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.


Let $Q$ be the following diagonal matrix:
\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 undertand 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.
> 
> 
> 


***An example is the following :***


Let $f({\bf y}\_1,{\bf y}\_2)=y\_{\_{11}}y\_{\_{21}}+y\_{\_{12}}y\_{\_{22}}$. This polynomial $f({\bf y}\_1,{\bf y}\_2)\in\mathcal{R}\_{2}^{(2)}(Y)$ is homogeneous of multidegree $(1,1)$ in the matrix variable $Y$ bellow:


\begin{equation\*}
Y:=\left(\begin{array}{cc}
y\_{\_{11}}&y\_{\_{12}}\\
y\_{\_{21}}&y\_{\_{22}}
\end{array}\right)
\end{equation\*}


We write the rows of $X$ as ${\bf x}\_1=(x\_{\_{11}},x\_{\_{12}})$ et ${\bf x}\_2=(x\_{\_{21}},x\_{\_{22}})$ and the same for $Y$.


Let $M$ be the following matrix:
\begin{equation\*}
M:=\left(\begin{array}{cc}
m\_{\_{11}}&m\_{\_{12}}\\
m\_{\_{21}}&m\_{\_{22}}
\end{array}\right)
\end{equation\*}
Notice that $f(Y)=f({\bf y}\_1,{\bf y}\_2)=f(y\_{\_{11}},y\_{\_{12}};y\_{\_{21}},y\_{\_{22}})=y\_{\_{11}}y\_{\_{21}}+y\_{\_{12}}y\_{\_{22}}$, then
\begin{align\*}
&f\left(MX\right)=f\left(\left(\begin{array}{cc}
m\_{\_{11}}&m\_{\_{12}}\\
m\_{\_{21}}&m\_{\_{22}}
\end{array}\right)\left(\begin{array}{c} {\bf x}\_1\\{\bf x}\_2 \end{array}\right)\right)
=f\big(m\_{\_{11}}{\bf x}\_1+m\_{\_{12}}{\bf x}\_2\,;\,m\_{\_{21}}{\bf x}\_1+m\_{\_{22}}{\bf x}\_2\big)\\
&=f\big(m\_{\_{11}}x\_{\_{11}}+m\_{\_{12}}x\_{\_{21}},m\_{\_{11}}x\_{\_{12}}+m\_{\_{12}}x\_{\_{22}}\,;\,m\_{\_{21}}x\_{\_{11}}+m\_{\_{22}}x\_{\_{21}},m\_{\_{21}}x\_{\_{12}}+m\_{\_{22}}x\_{\_{22}}\big)\\
&=\big(m\_{\_{11}}x\_{\_{11}}+m\_{\_{12}}x\_{\_{21}}\big)\big(m\_{\_{21}}x\_{\_{11}}+m\_{\_{22}}x\_{\_{21}}\big)+\big(m\_{\_{11}}x\_{\_{12}}+m\_{\_{12}}x\_{\_{22}}\big)\big(m\_{\_{21}}x\_{\_{12}}+m\_{\_{22}}x\_{\_{22}}\big)\\
&=m\_{\_{11}}m\_{\_{21}}x\_{\_{11}}^{2}+m\_{\_{11}}m\_{\_{21}}x\_{\_{12}}^{2}
+m\_{\_{11}}m\_{\_{22}}x\_{\_{11}}x\_{\_{21}}+m\_{\_{11}}m\_{\_{22}}x\_{\_{12}}x\_{\_{22}}
+m\_{\_{12}}m\_{\_{21}}x\_{\_{11}}x\_{\_{21}}\\
&+m\_{\_{12}}m\_{\_{21}}x\_{\_{12}}x\_{\_{22}}+m\_{\_{12}}m\_{\_{22}}x\_{\_{21}}^{2}+m\_{\_{12}}m\_{\_{22}}x\_{\_{22}}^{2}.\\
&=m\_{\_{21}}m\_{\_{12}} \left( x\_{\_{11}}x\_{\_{21}}+x\_{\_{12}}x\_{\_{22}}
 \right) +m\_{\_{11}}m\_{\_{22}} \left( x\_{\_{11}}x\_{\_{21}}+x\_{\_{12}}x\_{\_{22}} \right)+m\_{\_{11}}m\_{\_{21}} \left( x\_{\_{11}}^{2}+x\_{\_{12}}^{2} \right) +m\_{\_{12}}m\_{\_{22}} \left( x\_{\_{21}}^{2}+x\_{\_{22}}^{2}
 \right)\\
&=m\_{\_{21}}m\_{\_{12}}P\_{1,2}P\_{2,1}(f(Y))+m\_{\_{11}}m\_{\_{22}}P\_{11}P\_{22}(f(Y))
+m\_{\_{11}}m\_{\_{21}}P\_{12}P\_{11}(f(Y))+m\_{\_{12}}m\_{\_{22}}P\_{21}P\_{22}(f(Y)).
\end{align\*}


Thank for any hint on this.



Polarization operators and the action of $GL_{\ell}(\mathbb{R})$ on $\mathcal{R}_{n}^{(\ell)}$
I don't have the time to work out precisely how to derive Procesi's claim from the formula for $f(MX)$ but probably it is a cute exercise along the following lines: let $E\_{k,i}$ be the elementary matrix whose $(k,i)$-entry equals $1$ and all other entries are $0$. Now calculate $f(MX)$ for $M={\bf1}\_\ell+tE\_{k,i}$ and take the derivative with respect to $t$ at $t=0$. This should be $P\_{i,k}f(X)$ from which the claim follows.


But much more directly, Procesi's claim follows simply from the fact the $P\_{i,k}$ generate the action of the Lie algebra ${\rm Lie}\, GL\_\ell(\mathbb R)$. Since $GL\_\ell(\mathbb R)$ is connected (as an algebraic group!) and the ground field $\mathbb R$ is of characteristic zero, a subspace is $GL\_\ell(\mathbb R)$-closed iff it is closed for the action of its Lie algebra.


