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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 of degree in indeterminates over a commutative scalar ring (scalar simply means that the elements of commute with the indeterminates).

Step 1

If all indeterminates are linear in , then we are done.

Step 2

Otherwise, pick an indeterminate such that is not linear in . Without loss of generality, write , where is the set of indeterminates in excluding . Define . Then is a homogeneous polynomial of degree in indeterminates. However, the highest degree of is , one less that of .

Step 3

Repeat the process, starting with Step 1, for the homogeneous polynomial . Continue until the set of indeterminates is enlarged to one $X^{'}$ such that each $x\in X^{'}$ is linear.
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., for some $n\in\mathbb{N}$ and any $t\in R$ . Note that if all of the indeterminates in commute with each other, then 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:

Step 1

If is not linear in and that , replace with a formal linear combination of indeterminates over :

$\displaystyle r_1x_1+\cdots+r_nx_n$ , where $\displaystyle r_i\in R.$

Step 2

Define a polynomial $g\in R\langle x_1,\ldots,x_n \rangle$ , the non-commuting free algebra over (generated by the non-commuting indeterminates ) by:

$\displaystyle g(x_1,\ldots,x_n):=f(r_1x_1+\cdots+r_nx_n).$

Step 3

Expand and take the sum of the monomials in whose coefficent is $r_1\cdots r_n$ . The result is a linearization of for the indeterminate .

Step 4

Take the next non-linear indeterminate and start over (with Step 1). Repeat the process until is completely linearized.

Remarks.

The method of linearization is used often in the studies of Lie algebras, Jordan algebras, PI-algebras and quadratic forms.
If the characteristic of scalar ring is 0 and is a monomial in one indeterminate, we can recover back from its linearization by setting all of its indeterminates to a single indeterminate and dividing the resulting polynomial by :

$\displaystyle f(x)=\frac{1}{n!}\operatorname{linearization}(f)(x,\ldots,x).$
Please see the first example below.
If is a homogeneous polynomial of degree , then the linearized is a multilinear map in indeterminates.

Examples.

. Then is a linearization of . In general, if , then the linearization of is

$\displaystyle \sum_{\sigma\in S_n}x_{\sigma(1)}\cdots x_{\sigma(n)}= \sum_{\sigma\in S_n}\prod_{i=1}^{n}x_{\sigma(i)},$
where 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

$\displaystyle n!x_1\cdots x_n=\prod_{i=1}^{n}ix_i.$
. Since and , is homogeneous over and separately, and thus we can linearize . First, collect all the monomials having coefficient in , we get

$\displaystyle g(x_1,x_2,x_3,y):=\sum x_ix_jx_ky^2+x_iyx_jyx_k,$
where $i,j,k\in {1,2,3}$ and $(i-j)(j-k)(k-i)\neq0$ . Repeat this for and we have

$\displaystyle h(x_1,x_2,x_3,y_1,y_2):=\sum x_ix_jx_k(y_1y_2+y_2y_1)+(x_iy_1x_jy_2x_k+x_iy_2x_jy_1x_k).$

1	Scalar
1	Basic Polynomial
1	Summation
1	Function
1	Commutative Ring
1	Ring
1	Addition
1	Generating Set Of A Group
1	Commutative
2	Algorithm
2	Indeterminate
2	Lie Algebra
3	Linear Combination
3	Brouwer Degree
3	Characteristic
3	Polynomials In Algebraic Systems
3	W L O G
3	Symmetric Group
5	Quadratic Form
6	Multilinear
6	Ground Fields And Rings
8	Homogeneous Linear Problem
8	Homogeneous Polynomial
9	Jordan Algebra
12	Free Algebra
14	Polynomial Identity Algebra