polarization/linearization as in jordan forms

Question

I am new to this branch of math, so bear with me.

This question started when reading Kevin McCrimmon's "A Taste of Jordan Algebras" It talks about polarization and gives a general description.

the general process of linearization (often called polarization, espe- cially in analysis in dealing with quadratic mappings on a complex space). This is an important technique in nonassociative algebras ... Given a homogeneous polynomial p(x) of degree n, the process of linearization is designed to create a symmetric multilinear polynomial p'(x1,... ,xn) in n variables such that the original polynomial arises by specializing all the variables to the same value x : p'(x,... ,x) = p(x). For example, the full linearization of the square x^2= xx is 1/2 (x1x2+x2x1), and the full linearization of the cube x3= xx*x of degree 3 is1/6(x1x2x3+ x1x3x2+ x2x1x3+ x2x3x1+ x3x1x2+ x3x2x1).

I understand the first part

Full linearization is usually achieved one step at a time by a series of partial linearizations, in which a polynomial homogeneous of degree n in a particular variable x is replaced by one of degree n − 1 in x and linear in a new variable y. Intuitively, in an expression with n occurrences of x we simply replace each occurrence of x, one at a time, by a y, add up the results

Ok to here...But then he says

we often have no very explicit expression for p, and must describe linearization in a more intrinsic way. The clearest formulation is to take p(x+ λy) for an indeterminate scalar λ and expand this out as a polynomial in λ: p(x + λy) = p(x) + λp1(x;y) + λ2p2(x;y) + ··· + λnp(y). Here pi(x;y) is homogeneous of degree n − i in x and i in y (intuitively, we obtain it by replacing i of the x’s in p(x) by y’s in all possible ways)

This is where i almost get it, but not quite. where does the y come from here. Is it the same y as above, and so one... and wish I could look at another description of it... How would I symmetrize/linearize/polarize something like 3x^2yz + 2y^2x^2 + z^2x + z^3y

Also later he suggests this method for cases where p is quadratic ( ===q)

we take the value [of q[x]] on the sum x + y of two elements, and then subtract the pure x and y terms to obtain q(x,y) := q(x + y) − q(x) − q(y).

This only works then for squares of one variable then? ( ax^2?). not very useful...

I am looking for some more details about this, but links such as the following are confusing me as much as they are enlightening me because the notation and context are different. Is there a good textbook that explains this clearly?

• http://planetmath.org/polarizationbydifferentialoperators
• https://en.wikipedia.org/wiki/Polarization_of_an_algebraic_form
• polarization formula for homogeneous polynomials

Answer 1

Alternatively, you can polarise right away as follows: if $p(x)$ is homogeneous of degree $n$ (here $x$ may be a variable with values in $\mathbb{R}^k$, e.g. $p(x)=\mathop{\mathrm{tr}}(x^4)$, where $x$ is a matrix), then you can look at $$ p(\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n), $$ where $\lambda_i$ are elements of the ground field (e.g. real numbers), and $x_i$ are variables of the same type as $x$ (e.g. matrices like in my example above). In that expression, extract the coefficient of $\lambda_1\lambda_2\cdots\lambda_n$ there. That coefficient is the result of polarisation (multilinearisation).

Answer 2

The principle of polarizing is given by what you wrote yourself in the third paragraph. If $p$ is a homogeneous polynomial of degree $n$, then:

The clearest formulation is to take $p(x+ \lambda y)$ for an indeterminate scalar $\lambda$ and expand this out as a polynomial in $\lambda$: $$ p(x + \lambda y) = p(x) + \lambda p_1(x;y) + \lambda^2 p_2(x;y) + \dotsm + \lambda^n p(y). $$

Formally speaking, you could say that you extend the ring of scalars from $R$ to $R[\lambda]$ (where $\lambda$ commutes with $R$), so that the expressions $p_i(x;y)$ are indeed uniquely determined by expanding the left hand side $p(x + \lambda y)$.

If you want to obtain a full linearization of $p$, then you have to apply this procedure repeatedly, by polarizing $p_1(x; y)$ in the variable $x$ and continuing the process in the same manner. (Note that $p_1(x; y)$ is homogeneous of degree $n-1$ in $x$.)

For example, if $p(x) = x^3$ over some non-commutative ring $R$, then $$ p(x + \lambda y) = x^3 + \lambda (xxy + xyx + yxx) + \lambda^2 (xyy + yxy + yyx) + \lambda^3 y^3,$$ so $p_1(x; y) = x^2 y + xyx + yx^2$. We then write $$ p_1(x + \lambda y; z) = p_1(x; z) + \lambda p_{11}(x, y, z) + \lambda^2 p_1(y; z),$$ which yields $p_{11}(x, y, z) = xyz + xzy + yxz + yzx + zxy + zyx$. Observe that indeed $p_{11}(x,x,x) = 3! \cdot p(x)$.

(Some authors prefer to assume that $n!$ is invertible in $R$, and instead define the linearization of $p$ as $1/(n!)$ times the linearization defined above.)

Also note that the variable $x$ could be a multi-variable to start with, e.g. if $p(x_1,\dots,x_n)$ is a quadratic form in $n$ variables, then its linearization is obtained by expanding $p(x_1 + \lambda y_1, \dots, x_n + \lambda y_n)$.