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.


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Basic facts


The polarization identity


Any symmetric bilinear form ()()(-) \cdot (-) defines a quadratic form () 2(-)^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 22 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 RR be a commutative ring. Let VV and WW be RR-modules, and let m:V×VWm\colon V \times V \to W be a bilinear map; that is, we have an RR-module homomorphism VVWV \otimes V \to W. Let Q:VWQ\colon V \to W be the quadratic map given by Q(x)=m(x,x)Q(x) = m(x,x); this is not an RR-module homomorphism.

Then we have:

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

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

  • 2x 2+2y 2=(x+y) 2+(xy) 22 x^2 + 2 y^2 = (x + y)^2 + (x - y)^2,
  • 2xy+2yx=(x+y) 2(xy) 22 x y + 2 y x = (x + y)^2 - (x - y)^2.

The polarization identity also has these alternative forms:

  • xy+yx=(x+y) 2x 2y 2x y + y x = (x + y)^2 - x^2 - y^2,
  • xy+yx=x 2+y 2(xy) 2x 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 22 is cancellable in WW; they can also both be proved directly without any such assumption.

The most general form of such expressions is

kxy+kyx= ic i(a ix+b iy) 2 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)(a,b,c) from RR such that

ia i 2c i= ib i 2c i=0, \sum_i a_i ^ 2 c_i = \sum_i b_i ^ 2 c_i = 0 ,


k= ia ib ic i. k = \sum_i a_i b_i c_i .

In the three specific versions of the polarization identity listed above, the values of aa, bb, and cc are all always 00, 11, or 1-1, 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 RR). (If we want to allow RR 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 mm is symmetric?, so that xy=yxx y = y x. And suppose that multiplication by 22 is (not merely cancellable but also) invertible in WW. (If 1/2R1/2 \in R, then that is more than sufficient.) Then the polarization identities read:

  • xy=14(x+y) 214(xy) 2x y = \frac{1}{4} (x + y)^2 - \frac{1}{4} (x - y)^2,
  • xy=12(x+y) 212x 212y 2x y = \frac{1}{2} (x + y)^2 - \frac{1}{2} x^2 - \frac{1}{2} y^2,
  • xy=12x 2+12y 212(xy) 2x y = \frac{1}{2} x^2 + \frac{1}{2} y^2 - \frac{1}{2} (x - y)^2.

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

xy= ic i(a ix+b iy) 2 i2a ib ic i 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 2ca^2 c and b 2cb^2 c are zero (as above) and the sum in the denominator is invertible. (In that case, one could also normalize the ccs so that the denominator becomes 11.) Notice that 22 must be invertible to have any hope of recovering mm or its symmetrization; but we could recover xy+yxx y + y x without that.

We can also go the other direction: given a quadratic map QQ, if 22 is invertible, then any polarization identity defines a symmetric bilinear map mm; these all agree iff QQ obeys the parallelogram law, and then QQ may be recovered from this mm once more.

If RR is a **-ring, then mm could be sesquilinear instead of bilinear. Then QQ would satisfy Q(tx)=t *tQ(x)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(tx)=t 2Q(x)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 mm is symmetric (or conjugate-symmetric or anything else), we can recover mm from QQ if RR has (in addition to 1/21/2) an imaginary unit: an element ii such that i+i *=0i + i^* = 0 and i *i=1i^* i = 1. Specifically,

m(x,y)=14Q(x+y)14Q(xy)14iQ(x+iy)+14iQ(xiy) 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 mm is conjugate-linear in the first variable and linear in the second. (Swap the signs on the two terms involving ii if mm is linear in the first variable and conjugate-linear in the second.) The more general version of this is

m(x,y)= ic iQ(a ix+b iy) ia i *b ic i m(x,y) = \frac{\sum_i c_i Q(a_i x + b_i y)} {\sum_i a_i^* b_i c_i}


ia i *a ic i= ib i *b ic i= ia ib i *c i=0 \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 RR is the trivial ring.)


This is best known in the case of bilinear and quadratic forms, where WW is the ground ring RR. Here, mm is an inner product, making VV into an inner product space, and QQ is (the square of) the norm, making VV into a normed space. Instead of abbreviating m(x,y)m(x,y) as xyx y, we might abbreviate it as x,y\langle x, y \rangle (or any other common notation for an inner product) and Q(x)Q(x) as x 2\|x\|^2. (In the general algebraic context that we are assuming here, QQ may not have a square root, much less a canonical ‘principal’ square root to give meaning to x\|x\| alone. Even assuming that RR is the field \mathbb{R} of real numbers, that Q(x)0Q(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 WW is VV. Actually, there is no need for mm 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 22).

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

Higher order

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

Q(tv)=t pQ(v). Q(t v) = t^p Q(v) .

We wish to turn QQ into a symmetric multilinear map mm of rank pp (so m:Sym pVWm\colon \Sym_p V \to W is linear) as follows:

  • If p=0p = 0, m()=Q(0)m() = Q(0);
  • If p=1p = 1, m(v 1)=Q(v 1)m(v_1) = Q(v_1);
  • If p=2p = 2, m(v 1,v 2)=Q(v 1+v 2)Q(v 1)Q(v 2)m(v_1, v_2) = Q(v_1 + v_2) - Q(v_1) - Q(v_2);
  • If p=3p = 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)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=4p = 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)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 ±Q(0)\pm Q(0), but Q(0)=0Q(0) = 0 for any homogeneous map QQ of positive degree (since Q(0)=Q(0v)=0 pQ(v)=0Q(v)=0Q(0) = Q(0 v) = 0^p Q(v) = 0 Q(v) = 0 when p>0p \gt 0), so it is convenient to leave off this term except for p=0p = 0. The p=0p = 0 case is instead Q(0)=Q(0v)=0 pQ(v)=1Q(v)=Q(v)Q(0) = Q(0 v) = 0^p Q(v) = 1 Q(v) = Q(v), so a homogeneous map of degree 00 is simply a constant map, and we are defining mm to be this constant.)

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

To get the same mm as in earlier sections of this text, the expressions here must be divided by the factorial p!p!. Of course, this only works if p!p! is invertible in RR; even if RR is a field, we need its characteristic to be 00 (or at least greater than pp if we are only interested in certain values of pp). The expressions above work regardless of this to define what a homogeneous polynomial is. However, if you wish to recover QQ from mm, then you do need to divide by the factorial; otherwise, the only rule that holds in general is that

m(v,,v)=p!Q(v). 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.