# nLab polarization identity

The 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, see wave polarization (if we ever write it).

# The polarization identity

## Idea

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. 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). The linearity is crucial: polarization doesn't work without addition (and subtraction); we must also be able to divide by $2$. It is also possible to generalize from quadratic forms to higher-order homogenous form?s.

## Statement

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 $Q(v) = m(v,v)$; 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; although the derivation requires cancelling $2$, the alternative polarization identities remain valid regardless of whether $2$ is cancellable in $W$.

Now suppose that $m$ is symmetric?, so that $x y = y x$. And suppose that $2 \coloneqq 1 + 1$ is (not merely cancellable but also) invertible in $W$. 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). Regardless of the original symmetry of $m$, we may recover its symmetrization $x \circ y = (x y + y x)/2$.

We can 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 if $Q$ obeys the parallelogram law, and then $Q$ may be recovered from this $m$ once more.

If $R$ is a $*$-ring, then $m$ could be conjugate-symmetric (Hermitian?). Then $Q$ would satisfy $Q(t x) = t^* t Q(x)$ 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 does the polarization identity in its general forms (before assuming that $m$ is symmetric). We can still recover $m$ from $Q$ if $R$ has an imaginary unit: an element $i$ such that $i + i^* = 0$ and $i i^* = 1$; we do this as follows:

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

## Examples

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.

This 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).

## Higher order

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.

So defined, $m$ is manifestly symmetric, but it might not be multilinear just because $Q$ is homogeneous (except for $p = 1$); instead, we define a homogeneous polynomial of degree $p$ from $V$ to $W$ be such a homogeneous $Q$ 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 $R^n$, then this is the usual notion of a polynomial with $n$ variables.

Morally, the expressions on the right-hand side of each item above should end with $\pm Q(0)$, but this ends up not mattering for the definition of a homogeneous polynomial (except when $p = 0$), in which case $Q(0) = 0$ (including when $p = 0$). If one were ever to consider the polarization of a non-polynomial homogeneous map, however, then presumably this term would need to be restored.

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 greater than $p$ (or $0$). 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 polynomial; otherwise, the only rule that holds in general is that

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

Last revised on August 7, 2019 at 19:23:30. See the history of this page for a list of all contributions to it.