linearization

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:

• Method 1. Given any homogeneous polynomial $f$ of degree $n$ in $m$ indeterminates over a commutativescalar ring $R$ (scalar simply means that the elements of $R$ commute with the indeterminates).

  1. If all indeterminates are linear in $f$, then we are done.

  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 $x$.

  3. Repeat the process, starting with Step 1, for the homogeneous polynomial $g$. Continue until the set $X$of indeterminates is enlarged to one $X^{{}^{\prime }}$ such that each $x\in X^{{}^{\prime }}$ is linear.

• Method 2. This method applies only to homogeneous polynomials that are also homogeneous in eachindeterminate, when the other indeterminates are held constant, i.e., $f(tx,X)=t^{n}f(x,X)$ for some $n\in \mathbb{N}$ andany $t\in R$. Note that if all of the indeterminates in $f$ commute with each other, then $f$ is essentially amonomial. So this method is particularly useful when indeterminates are non-commuting. If this is the case, then weuse the following algorithm:

  1. If $x$ is not linear in $f$ and that $f(tx,X)=t^{n}f(x,X)$, replace $x$ with a formal linear combination of$n$ indeterminates over $R$:

     $$r_1{x}_1+\mathrm{\cdots }+r_n{x}_n\text{, where }{r}_i\in R.$$

  2. Define a polynomial $g\in R⟨x_1,\mathrm{\dots },x_n⟩$, the non-commuting free algebra over $R$(generated by the non-commuting indeterminates $x_i$) by:

     $$g(x_1,\mathrm{\dots },x_n):=f(r_1{x}_1+\mathrm{\cdots }+r_n{x}_n).$$

  3. Expand $g$ and take the sum of the monomials in $g$ whose coefficent is $r_1\mathrm{\cdots }{r}_n$. The result is alinearization of $f$ for the indeterminate $x$.

  4. Take the next non-linear indeterminate and start over (with Step 1). Repeat the process until $f$ iscompletely linearized.

Remarks.

1. The method of linearization is used often in the studies of Lie algebras, Jordan algebras, PI-algebras andquadratic forms.

2. If the characteristic of scalar ring $R$ is 0 and $f$ is a monomial in one indeterminate, we can recover $f$ backfrom its linearization by setting all of its indeterminates to a single indeterminate $x$ and dividing the resultingpolynomial by $n!$:

   $$f(x)=\frac{1}{n!}\mathrm{linearization}(f)(x,\mathrm{\dots },x).$$

   Please see the first example below.

3. If $f$ is a homogeneous polynomial of degree $n$, then the linearized $f$ is a multilinear map in $n$indeterminates.

Examples.

• $f(x)=x^2$. Then $f(x_1+x_2)-f(x_1)-f(x_2)=x_1{x}_2+x_2{x}_1$ is a linearization of $x^2$. In general, if $f(x)=x^n$,then the linearization of $f$ is

  $$\sum _{\sigma \in S_n}{x}_{\sigma (1)}\mathrm{\cdots }{x}_{\sigma (n)}=\sum _{\sigma \in S_n}\prod _{i=1}^n{x}_{\sigma (i)},$$

  where $S_n$ is the symmetric group on $\{1,\mathrm{\dots },n\}$.If in addition all the indeterminates commute with each other and $n!\ne 0$ in the ground ring, then the linearizationbecomes

  $$n!x_1\mathrm{\cdots }{x}_n=\prod _{i=1}^{n}i{x}_i.$$

• $f(x,y)=x^3{y}^2+xyxyx$. Since $f(tx,y)=t^{3}f(x,y)$ and $f(x,ty)=t^{2}f(x,y)$, $f$ is homogeneous over $x$ and $y$separately, and thus we can linearize $f$. First, collect all the monomials having coefficient $abc$ in$(a{x}_1+b{x}_2+c{x}_3,y)$, we get

  $$g(x_1,x_2,x_3,y):=\sum x_i{x}_j{x}_k{y}^2+x_{i}y{x}_{j}y{x}_k,$$

  where $i,j,k\in 1,2,3$ and $(i-j)(j-k)(k-i)\ne 0$. Repeat this for $y$ and we have

  $$h(x_1,x_2,x_3,y_1,y_2):=\sum x_i{x}_j{x}_k(y_1{y}_2+y_2{y}_1)+(x_i{y}_1{x}_j{y}_2{x}_k+x_i{y}_2{x}_j{y}_1{x}_k).$$