# linearization

Remarks.

1. 1.
2. 2.

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.

3. 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

 $\sum_{\sigma\in S_{n}}x_{\sigma(1)}\cdots x_{\sigma(n)}=\sum_{\sigma\in S_{n}}% \prod_{i=1}^{n}x_{\sigma(i)},$

where $S_{n}$ is the symmetric group  on $\{1,\ldots,n\}$. If in addition all the indeterminates commute with each other and $n!\neq 0$ in the ground ring, then the linearization becomes

 $n!x_{1}\cdots x_{n}=\prod_{i=1}^{n}ix_{i}.$
• $f(x,y)=x^{3}y^{2}+xyxyx$. 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 $(ax_{1}+bx_{2}+cx_{3},y)$, we get

 $g(x_{1},x_{2},x_{3},y):=\sum x_{i}x_{j}x_{k}y^{2}+x_{i}yx_{j}yx_{k},$

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

 $h(x_{1},x_{2},x_{3},y_{1},y_{2}):=\sum x_{i}x_{j}x_{k}(y_{1}y_{2}+y_{2}y_{1})+% (x_{i}y_{1}x_{j}y_{2}x_{k}+x_{i}y_{2}x_{j}y_{1}x_{k}).$
Title linearization Linearization 2013-03-22 14:53:52 2013-03-22 14:53:52 CWoo (3771) CWoo (3771) 5 CWoo (3771) Definition msc 15A63 msc 15A69 msc 16R99 msc 17A99 polarization linearized