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(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.

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  • $\begingroup$ There must be a typo in the formula for $P_{i,k}$: the variable $y_{kj}$ should be $x_{kj}$. $\endgroup$ – Friedrich Knop Nov 20 '17 at 17:57
  • $\begingroup$ I will check that :) $\endgroup$ – Hector Blandin Nov 20 '17 at 18:04
  • $\begingroup$ This seems rather similar to your question on math.SE: Polarization Operators identity and $GL_{\ell}(\mathbb{R})$. $\endgroup$ – Martin Sleziak Nov 20 '17 at 18:18
  • $\begingroup$ @FriedrichKnop: Thank you for the answer, this give me a good idea how to continue :) $\endgroup$ – Hector Blandin Nov 20 '17 at 18:19
  • $\begingroup$ @MartinSleziak: Can we put the same question in both sites ? or probably I did something I cannot ? $\endgroup$ – Hector Blandin Nov 20 '17 at 18:20
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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.

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