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).


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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. 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 22. It is also possible to generalize from quadratic forms to higher-order homogenous form?s.


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(v)=m(v,v)Q(v) = m(v,v); 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; although the derivation requires cancelling 22, the alternative polarization identities remain valid regardless of whether 22 is cancellable in WW.

Now suppose that mm is symmetric?, so that xy=yxx y = y x. And suppose that 21+12 \coloneqq 1 + 1 is (not merely cancellable but also) invertible in WW. 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). Regardless of the original symmetry of mm, we may recover its symmetrization xy=(xy+yx)/2x \circ y = (x y + y x)/2.

We can 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 if 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 conjugate-symmetric (Hermitian?). Then QQ would satisfy Q(tx)=t *tQ(x)Q(t x) = t^* t Q(x) 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 does the polarization identity in its general forms (before assuming that mm is symmetric). We can still recover mm from QQ if RR has an imaginary unit: an element ii such that i+i *=0i + i^* = 0 and ii *=1i i^* = 1; we do this as follows:

xy=14Q(x+y)14Q(xy)+14iQ(x+iy)14iQ(xiy). 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) .


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.

This 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).

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.

So defined, mm is manifestly symmetric, but it might not be multilinear just because QQ is homogeneous (except for p=1p = 1); instead, we define a homogeneous polynomial of degree pp from VV to WW be such a homogeneous 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, then this is the usual notion of a polynomial with nn variables.

Morally, the expressions on the right-hand side of each item above should end with ±Q(0)\pm Q(0), but this ends up not mattering for the definition of a homogeneous polynomial (except when p=0p = 0), in which case Q(0)=0Q(0) = 0 (including when p=0p = 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 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 greater than pp (or 00). 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 polynomial; 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 August 7, 2019 at 19:23:30. See the history of this page for a list of all contributions to it.