Linearization 2004-12-14 20:30:10 2004-12-14 22:57:54 Definition linearized % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \emph{Linearization} is the process of reducing a homogeneous polynomial into a multilinear map over a commutative ring. There are in general two ways of doing this: \begin{itemize} \item \textbf{Method 1.} Given any homogeneous polynomial $f$ of degree $n$ in $m$ indeterminates over a commutative scalar ring $R$ (scalar simply means that the elements of $R$ commute with the indeterminates). \begin{enumerate} \item[Step 1] If all indeterminates are linear in $f$, then we are done. \item[Step 2] Otherwise, pick an indeterminate $x$ such that $x$ is not linear in $f$. Without loss of generality, write $f=f(x,X)$, where $X$ is the set of indeterminates in $f$ excluding $x$. Define $g(x_1,x_2,X):=f(x_1+x_2,X)-f(x_1,X)-f(x_2,X)$. Then $g$ is a homogeneous polynomial of degree $n$ in $m+1$ indeterminates. However, the highest degree of $x_1,x_2$ is $n-1$, one less that of $x$. \item[Step 3] Repeat the process, starting with Step 1, for the homogeneous polynomial $g$. Continue until the set $X$ of indeterminates is enlarged to one $X^{'}$ such that each $x\in X^{'}$ is linear. \end{enumerate} \item \textbf{Method 2.} This method applies only to homogeneous polynomials that are also homogeneous in each indeterminate, when the other indeterminates are held constant, i.e., $f(tx,X)=t^nf(x,X)$ for some $n\in\mathbb{N}$ and any $t\in R$. Note that if all of the indeterminates in $f$ commute with each other, then $f$ is essentially a monomial. So this method is particularly useful when indeterminates are non-commuting. If this is the case, then we use the following algorithm: \begin{enumerate} \item[Step 1] If $x$ is not linear in $f$ and that $f(tx,X)=t^nf(x,X)$, replace $x$ with a formal linear combination of $n$ indeterminates over $R$: $$r_1x_1+\cdots+r_nx_n\mbox{, where }r_i\in R.$$ \item[Step 2] Define a polynomial $g\in R\langle x_1,\ldots,x_n \rangle$, the non-commuting free algebra over $R$ (generated by the non-commuting indeterminates $x_i$) by: $$g(x_1,\ldots,x_n):=f(r_1x_1+\cdots+r_nx_n).$$ \item[Step 3] Expand $g$ and take the sum of the monomials in $g$ whose coefficent is $r_1\cdots r_n$. The result is a linearization of $f$ for the indeterminate $x$. \item[Step 4] Take the next non-linear indeterminate and start over (with Step 1). Repeat the process until $f$ is completely linearized. \end{enumerate} \end{itemize} \textbf{Remarks.} \begin{enumerate} \item The method of linearization is used often in the studies of Lie algebras, Jordan algebras, PI-algebras and quadratic forms. \item If the characteristic of scalar ring $R$ is 0 and $f$ is a monomial in one indeterminate, we can recover $f$ back from its linearization by setting all of its indeterminates to a single indeterminate $x$ and dividing the resulting polynomial by $n!$: $$f(x)=\frac{1}{n!}\operatorname{linearization}(f)(x,\ldots,x).$$ Please see the first example below. \item If $f$ is a homogeneous polynomial of degree $n$, then the linearized $f$ is a multilinear map in $n$ indeterminates. \end{enumerate} \textbf{Examples}. \begin{itemize} \item $f(x)=x^2$. Then $f(x_1+x_2)-f(x_1)-f(x_2)=x_1x_2+x_2x_1$ is a linearization of $x^2$. In general, if $f(x)=x^n$, then the linearization of $f$ is $$\sum_{\sigma\in S_n}x_{\sigma(1)}\cdots x_{\sigma(n)}= \sum_{\sigma\in S_n}\prod_{i=1}^{n}x_{\sigma(i)},$$ where $S_n$ is the symmetric group on $\lbrace1,\ldots,n\rbrace$. If in addition all the indeterminates commute with each other and $n!\neq0$ in the ground ring, then the linearization becomes $$n!x_1\cdots x_n=\prod_{i=1}^{n}ix_i.$$ \item $f(x,y)=x^3y^2+xyxyx$. Since $f(tx,y)=t^3f(x,y)$ and $f(x,ty)=t^2f(x,y)$, $f$ is homogeneous over $x$ and $y$ separately, and thus we can linearize $f$. First, collect all the monomials having coefficient $abc$ in $(ax_1+bx_2+cx_3,y)$, we get $$g(x_1,x_2,x_3,y):=\sum x_ix_jx_ky^2+x_iyx_jyx_k,$$ where $i,j,k\in {1,2,3}$ and $(i-j)(j-k)(k-i)\neq0$. Repeat this for $y$ and we have $$h(x_1,x_2,x_3,y_1,y_2):=\sum x_ix_jx_k(y_1y_2+y_2y_1)+(x_iy_1x_jy_2x_k+x_iy_2x_jy_1x_k).$$ \end{itemize}