polarization identity

This entry is about the notion in linear algebra relating bilinear and quadratic forms. For the notion in symplectic geometry see at

polarization. For polarization of light, seewave polarization.

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Any symmetric bilinear form $(-) \cdot (-)$ defines a quadratic form $(-)^2$. The *polarization identity* reconstructs the bilinear form from the quadratic form. More generally, starting from any bilinear form, the polarization identity reconstructs its symmetrization. A slight variation applies this also to sesquilinear forms, which don't require any kind of symmetry to work. The whole business actually applies to bilinear maps, not just forms (that is, taking arbitrary values, not just values in the base field or some other line). Some kind of linearity is crucial: polarization doesn't work without addition (and subtraction); we must also be able to divide by $2$ for the best results. It is also possible to generalize from quadratic forms to higher-order homogenous form?s if we can divide by larger integers.

Let $R$ be a commutative ring. Let $V$ and $W$ be $R$-modules, and let $m\colon V \times V \to W$ be a bilinear map; that is, we have an $R$-module homomorphism $V \otimes V \to W$. Let $Q\colon V \to W$ be the quadratic map given by $Q(x) = m(x,x)$; this is *not* an $R$-module homomorphism.

Then we have:

- the
*parallelogram law*: $2 Q(x) + 2 Q(y) = Q(x + y) + Q(x - y)$, and - the
**polarization identity**: $2 m(x,y) + 2 m(y,x) = Q(x + y) - Q(x - y)$.

Writing $Q(x)$ as $x^2$ and $m(x,y)$ as $x y$, these read:

- $2 x^2 + 2 y^2 = (x + y)^2 + (x - y)^2$,
- $2 x y + 2 y x = (x + y)^2 - (x - y)^2$.

The polarization identity also has these alternative forms:

- $x y + y x = (x + y)^2 - x^2 - y^2$,
- $x y + y x = x^2 + y^2 - (x - y)^2$;

these may derived by adding or subtracting the parallelogram law and the first polarization identity assuming that multiplication by $2$ is cancellable in $W$; they can also both be proved directly without any such assumption.

The most general form of such expressions is

$k x y + k y x = \sum_i c_i (a_i x + b_i y)^2$

for any finite sequence of triples $(a,b,c)$ from $R$ such that

$\sum_i a_i ^ 2 c_i = \sum_i b_i ^ 2 c_i = 0 ,$

if

$k = \sum_i a_i b_i c_i .$

In the three specific versions of the polarization identity listed above, the values of $a$, $b$, and $c$ are all always $0$, $1$, or $-1$, but there is no reason to restrict ourselves to this or even assume that they are integers (or even rational numbers or more generally members of the prime field of $R$). (If we want to allow $R$ to be a rig, then we can use a version of this with sums on both sides of the equation, but that seems like centipede mathematics since I don't know any application of that case.)

Now suppose that $m$ is symmetric?, so that $x y = y x$. And suppose that multiplication by $2$ is (not merely cancellable but also) invertible in $W$. (If $1/2 \in R$, then that is more than sufficient.) Then the polarization identities read:

- $x y = \frac{1}{4} (x + y)^2 - \frac{1}{4} (x - y)^2$,
- $x y = \frac{1}{2} (x + y)^2 - \frac{1}{2} x^2 - \frac{1}{2} y^2$,
- $x y = \frac{1}{2} x^2 + \frac{1}{2} y^2 - \frac{1}{2} (x - y)^2$.

That is, we may recover $m$ from $Q$ (in any of these ways). The second one is probably the most widely used. Regardless of the original symmetry of $m$, we may recover its symmetrization $x \circ y = (x y + y x)/2$ (the Jordan product?) in any these ways. The most general formula for $x \circ y$ is

$x \circ y
= \frac{\sum_i c_i (a_i x + b_i y)^2}
{\sum_i 2 a_i b_i c_i}$

when the sums of $a^2 c$ and $b^2 c$ are zero (as above) and the sum in the denominator is invertible. (In that case, one could also normalize the $c$s so that the denominator becomes $1$.) Notice that $2$ must be invertible to have any hope of recovering $m$ or its symmetrization; but we could recover $x y + y x$ without that.

We can also go the other direction: given a quadratic map $Q$, if $2$ is invertible, then any polarization identity defines a symmetric bilinear map $m$; these all agree iff $Q$ obeys the parallelogram law, and then $Q$ may be recovered from this $m$ once more.

If $R$ is a $*$-ring, then $m$ could be sesquilinear instead of bilinear. Then $Q$ would satisfy $Q(t x) = t^* t Q(x)$ (which I don't know a term for, maybe ‘sesqui-quadratic’? or using whatever's Latin for 3/4?) instead of $Q(t x) = t^2 Q(x)$ as for a quadratic map. Since everything is still bilinear or quadratic over the integers, the parallelogram identity still follows, as do the polarization identities in which only integers appear and before assuming symmetry. Then *regardless* of whether $m$ is symmetric (or conjugate-symmetric or anything else), we can recover $m$ from $Q$ if $R$ has (in addition to $1/2$) an imaginary unit: an element $i$ such that $i + i^* = 0$ and $i^* i = 1$. Specifically,

$m(x,y) = \frac{1}{4} Q(x + y) - \frac{1}{4} Q(x - y) - \frac{1}{4} i Q(x + i y) + \frac{1}{4} i Q(x - i y)$

when $m$ is conjugate-linear in the first variable and linear in the second. (Swap the signs on the two terms involving $i$ if $m$ is linear in the first variable and conjugate-linear in the second.) The more general version of this is

$m(x,y)
= \frac{\sum_i c_i Q(a_i x + b_i y)}
{\sum_i a_i^* b_i c_i}$

when

$\sum_i a_i^* a_i c_i = \sum_i b_i^* b_i c^i = \sum_i a_i b_i^* c_i = 0$

and the denominator is invertible. (Notice that the last two of these conditions are contradictory if the involution is trivial; or rather, they imply that $R$ is the trivial ring.)

This is best known in the case of bilinear and quadratic *forms*, where $W$ is the ground ring $R$. Here, $m$ is an inner product, making $V$ into an inner product space, and $Q$ is (the square of) the norm, making $V$ into a normed space. Instead of abbreviating $m(x,y)$ as $x y$, we might abbreviate it as $\langle x, y \rangle$ (or any other common notation for an inner product) and $Q(x)$ as $\|x\|^2$. (In the general algebraic context that we are assuming here, $Q$ may not have a square root, much less a canonical ‘principal’ square root to give meaning to $\|x\|$ alone. Even assuming that $R$ is the field $\mathbb{R}$ of real numbers, that $Q(x) \geq 0$ is an additional assumption, the positive (semi)-definiteness of the inner product.)

Besides forms, the general theory also applies to commutative algebras, where $W$ is $V$. Actually, there is no need for $m$ to be associative; although one rarely studies commutative but non-associative algebras, we have an exception with Jordan algebras. Although the Jordan identity is simpler to express in terms of the multiplication operation (as usual), the application to quantum mechanics may be more easily motivated through the squaring operation (since the square of an observable has a more obvious meaning than the Jordan product of two observables), and the polarization identities allow us to recover multiplication from squaring (and subtraction and division by $2$).

When $V = W$ is a $*$-algebra over $R$, then although the usual multiplication is bilinear, we can also define a (nonassociative) sesquilinear operator $m$ by explicitly taking the adjoint of one argument: $m(x,y) \coloneqq x^* y$. When $R$ is a complete normed field and $V = W$ is a $C^*$-algebra, the resulting 3/4-quadratic operator $Q$ must have a principal square root, but now we write $Q(x) \coloneqq x^* x$ as $|x|^2$ rather than $\|x\|^2$, which is taken. (However, $|{-}|$ is not an absolute value, since it's not multiplicative, only submultiplicative.)

Let $R, V, W$ be as before. Let $p$ be a natural number, and let $Q\colon V \to W$ be homogeneous of degree $p$; that is,

$Q(t v) = t^p Q(v) .$

We wish to turn $Q$ into a symmetric multilinear map $m$ of rank $p$ (so $m\colon \Sym_p V \to W$ is linear) as follows:

- If $p = 0$, $m() = Q(0)$;
- If $p = 1$, $m(v_1) = Q(v_1)$;
- If $p = 2$, $m(v_1, v_2) = Q(v_1 + v_2) - Q(v_1) - Q(v_2)$;
- If $p = 3$, $m(v_1, v_2, v_3) = Q(v_1 + v_2 + v_3) - Q(v_1 + v_2) - Q(v_1 + v_3) - Q(v_2 + v_3) + Q(v_1) + Q(v_2) + Q(v_3)$;
- If $p = 4$, $m(v_1, v_2, v_3, v_4) = Q(v_1 + v_2 + v_3 + v_4) - Q(v_1 + v_2 + v_3) - Q(v_1 + v_2 + v_4) - Q(v_1 + v_3 + v_4) - Q(v_2 + v_3 + v_4) + Q(v_1 + v_2) + Q(v_1 + v_3) + Q(v_1 + v_4) + Q(v_2 + v_3) + Q(v_2 + v_4) + Q(v_3 + v_4) - Q(v_1) - Q(v_2) - Q(v_3) - Q(v_4)$;
- etc.

(Morally, the expressions on the right-hand side of each item above should end with $\pm Q(0)$, but $Q(0) = 0$ for any homogeneous map $Q$ of positive degree (since $Q(0) = Q(0 v) = 0^p Q(v) = 0 Q(v) = 0$ when $p \gt 0$), so it is convenient to leave off this term except for $p = 0$. The $p = 0$ case is instead $Q(0) = Q(0 v) = 0^p Q(v) = 1 Q(v) = Q(v)$, so a homogeneous map of degree $0$ is simply a constant map, and we are defining $m$ to be this constant.)

So defined, $m$ is manifestly symmetric, but it might *not* be multilinear just because $Q$ is homogeneous (for $p \gt 0$); instead, we *define* a **homogeneous polynomial** of degree $p$ from $V$ to $W$ be such a homogeneous map $Q$ such that $m$ *is* multilinear. Then a **polynomial** from $V$ to $W$ is a sum of homogeneous polynomials of various degrees. Note that if $V$ is the free module $R^n$ and $W$ is $R$, then this is the usual notion of a polynomial function on $n$ variables over $R$. (But different polynomials can give rise to the same polynomial function; see there for examples.)

To get the same $m$ as in earlier sections of this text, the expressions here must be divided by the factorial $p!$. Of course, this only works if $p!$ is invertible in $R$; even if $R$ is a field, we need its characteristic to be $0$ (or at least greater than $p$ if we are only interested in certain values of $p$). The expressions above work regardless of this to define what a homogeneous polynomial is. However, if you wish to recover $Q$ from $m$, then you do need to divide by the factorial; otherwise, the only rule that holds in general is that

$m(v,\ldots,v) = p!\, Q(v) .$

Last revised on March 21, 2021 at 17:23:01. See the history of this page for a list of all contributions to it.