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= x*x is 1/2 (x1*x2+x2*x1), and the full linearization of the cube x3= xxx 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/encyclopedia/Polarization.html http://en.wikipedia.org/wiki/Polarization_of_an_algebraic_form polarization formula for homogeneous polynomials

  • $\begingroup$ Tell us something about your background. $\endgroup$ – Will Jagy Jan 8 '12 at 22:43
  • $\begingroup$ you mean mathematically? A bit hard...education so far in math and physics at the undergraduate level, but from there a smattering of knowledge from many areas, e.g. mathematical physics, e.g. hilbert spaces, quantum theory, linear and abstract algebra, real and complex analysis, basic and advanced calculus, topology and topological spaces, differential geometry, a bit of number theory, even a bit of knot theory...but still just piecing it all together...more broad than deep. I can recognize most mathematical concepts, but not always full up on the details. $\endgroup$ – asllearner Jan 9 '12 at 0:18
  • $\begingroup$ I guess the core of my questions is two things: he says > 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). in a case like f(x,y,z) = 3x^2yz + 2y^2x^2 + z^2x + z^3y does he mean define p(x)=f(x|y,z) compute p(x+λx')=f2(x,x',y,z) then define p2(y)=f2() compute p2(y+λy')=f3(x,x',y,y',z) define p2(y)=f3() compute p2(z+λz')=f3(x,x',y,y',z,z') $\endgroup$ – asllearner Jan 9 '12 at 0:33
  • $\begingroup$ 2. I really have had a hard time finding any links/references about "linearization" (too close to linearizing a function... polarizing (too close to polarized light, etc) symmetrizing (to many other meanings) so I am trying to find out what theory he is talking about and where it was developed...is it a branch of abstract algebra, what? I can understand what a quadratic mapping of a complex space is, but don't know where "linearizing" fits in it??? I don'T mind trying to figure things out myself, but I think I need to be pointed in a direction...or at least confirmed in my thinking. $\endgroup$ – asllearner Jan 9 '12 at 0:33
  • $\begingroup$ eventual goal is to figure out relationship if any to De Casteljau's algorithm $\endgroup$ – asllearner Jan 9 '12 at 0:37

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).

  • $\begingroup$ thank you. I get your general point. I have never seen this discussed in any textbook I have read, so I am wondering where it comes from. This is probably a really stupid question...if $x$ is a matrix with entries in $\mathbb{R}$ is it an element of $\mathbb{R}^4$? isn't an element of $\mathbb{R}^4$ a vector (e.g. four-vector with real entries)--yes technically a matrix, but you cant take powers of it...or can you!? $\endgroup$ – asllearner Jan 9 '12 at 16:42
  • $\begingroup$ 1. In various contexts related to identities in the noncommutative/nonassociative settings this is quite standard. Check, e.g. the MO question mathoverflow.net/questions/61884/… and the Wikipedia article en.wikipedia.org/wiki/Homogeneous_function#Polarization - to name just two of zillions. $\endgroup$ – Vladimir Dotsenko Jan 9 '12 at 17:30
  • $\begingroup$ 2. A $4$-dimensional vector space $V$ over real numbers is isomorphic to $\mathbb{R}^4$. If, incidentally, the vector space $V$ is an algebra, you can take the algebra structure with you via an isomorphism of your choice, right? ;-) $\endgroup$ – Vladimir Dotsenko Jan 9 '12 at 17:33
  • $\begingroup$ I actually overlooked that you complained about the available links and asked for textbooks. A reference I like is Procesi, "Lie groups: an approach through invariants and representations", section on Aronhold method. (It is available on Google Books, tinyurl.com/aronhold - check it out.) $\endgroup$ – Vladimir Dotsenko Jan 9 '12 at 17:48
  • $\begingroup$ I see. wow. Thanks. grand hat tip. Sorry to be so peevish....just what I needed. I will check out procesi... $\endgroup$ – asllearner Jan 10 '12 at 0:47

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 $p(x_1 + \lambda y_1, \dots, x_n + \lambda y_n)$.

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    $\begingroup$ OK, I need to work it out in detail for myself (e.g. see if I can get mathematica to do it), but I think I got it. The worked out example makes clear how to use the formula for $p(x+ \gamma y)$. And the reference to ring extensions was helpful...I was misreading where the formula came from... Where would I go to find out more about the theory of "polar forms/polarizing"...What (kind of) textbook would I find it in? $\endgroup$ – asllearner Jan 9 '12 at 16:11

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