# Polarization operators and the action of $GL_{\ell}(\mathbb{R})$ on $\mathcal{R}_{n}^{(\ell)}$

(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 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.

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.

• There must be a typo in the formula for $P_{i,k}$: the variable $y_{kj}$ should be $x_{kj}$. Nov 20, 2017 at 17:57
• This seems rather similar to your question on math.SE: Polarization Operators identity and $GL_{\ell}(\mathbb{R})$. Nov 20, 2017 at 18:18
• @FriedrichKnop: Thank you for the answer, this give me a good idea how to continue :) Nov 20, 2017 at 18:19

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.