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 waves (light), see wave polarization.
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Any symmetric bilinear form defines a quadratic form . 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 for the best results. It is also possible to generalize from quadratic forms to higher-order homogenous forms if we can divide by larger integers.
Let
be a commutative ring;
be -modules;
be a bilinear map,
(an -module homomorphism out of their tensor product);
be the quadratic map given by
(NB: this is not an -module homomorphism).
Then we have:
the parallelogram law:
and
the polarization identity:
Writing as and as , these read:
,
.
The polarization identity also has these alternative forms:
,
;
these may derived by adding or subtracting the parallelogram law and the first polarization identity assuming that multiplication by is cancellable in ; they can also both be proved directly without any such assumption.
The most general form of such expressions is
for any finite sequence of triples from such that
if
In the three specific versions of the polarization identity listed above, the values of , , and are all always , , or , 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 ). (If we want to allow 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 is symmetric?, so that . And suppose that multiplication by is (not merely cancellable but also) invertible in . (If , then that is more than sufficient.) Then the polarization identities read:
,
,
.
That is, we may recover from (in any of these ways). The second one is probably the most widely used. Regardless of the original symmetry of , we may recover its symmetrization (the Jordan product) in any these ways. The most general formula for is
when the sums of and are zero (as above) and the sum in the denominator is invertible. (In that case, one could also normalize the s so that the denominator becomes .) Notice that must be invertible to have any hope of recovering or its symmetrization; but we could recover without that.
We can also go the other direction: given a quadratic map , if is invertible, then any polarization identity defines a symmetric bilinear map ; these all agree iff obeys the parallelogram law, and then may be recovered from this once more.
If is a -ring, then could be sesquilinear instead of bilinear. Then would satisfy (which I don't know a term for, maybe ‘sesqui-quadratic’? or using whatever's Latin for 3/4?) instead of 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 is symmetric (or conjugate-symmetric or anything else), we can recover from if has (in addition to ) an imaginary unit: an element such that and . Specifically,
when is conjugate-linear in the first variable and linear in the second. (Swap the signs on the two terms involving if is linear in the first variable and conjugate-linear in the second.) The more general version of this is
when
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 is the trivial ring.)
This is best known in the case of bilinear and quadratic forms, where is the ground ring . Here, is an inner product, making into an inner product space, and is (the square of) the norm, making into a normed space. Instead of abbreviating as , we might abbreviate it as (or any other common notation for an inner product) and as . (In the general algebraic context that we are assuming here, may not have a square root, much less a canonical ‘principal’ square root to give meaning to alone. Even assuming that is the field of real numbers, that is an additional assumption, the positive (semi)-definiteness of the inner product.)
Besides forms, the general theory also applies to commutative algebras, where is . Actually, there is no need for 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 ).
When is a -algebra over , then although the usual multiplication is bilinear, we can also define a (nonassociative) sesquilinear operator by explicitly taking the adjoint of one argument: . When is a complete normed field and is a -algebra, the resulting 3/4-quadratic operator must have a principal square root, but now we write as rather than , which is taken. (However, is not an absolute value, since it's not multiplicative, only submultiplicative.)
Let be as before. Let be a natural number, and let be homogeneous of degree ; that is,
We wish to turn into a symmetric multilinear map of rank (so is linear) as follows:
(Morally, the expressions on the right-hand side of each item above should end with , but for any homogeneous map of positive degree (since when ), so it is convenient to leave off this term except for . The case is instead , so a homogeneous map of degree is simply a constant map, and we are defining to be this constant.)
So defined, is manifestly symmetric, but it might not be multilinear just because is homogeneous (for ); instead, we define a homogeneous polynomial of degree from to be such a homogeneous map such that is multilinear. Then a polynomial from to is a sum of homogeneous polynomials of various degrees. Note that if is the free module and is , then this is the usual notion of a polynomial function on variables over . (But different polynomials can give rise to the same polynomial function; see there for examples.)
To get the same as in earlier sections of this text, the expressions here must be divided by the factorial . Of course, this only works if is invertible in ; even if is a field, we need its characteristic to be (or at least greater than if we are only interested in certain values of ). The expressions above work regardless of this to define what a homogeneous polynomial is. However, if you wish to recover from , then you do need to divide by the factorial; otherwise, the only rule that holds in general is that
Last revised on May 24, 2024 at 09:14:35. See the history of this page for a list of all contributions to it.