Polarization identity

If the vector space is over the real numbers then the polarization identities are:[4]

$${\displaystyle {\begin{alignedat}{4}\langle x,y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)\\[3pt]&={\frac {1}{2}}\left(\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}\right)\\[3pt]&={\frac {1}{2}}\left(\|x\|^{2}+\|y\|^{2}-\|x-y\|^{2}\right).\\[3pt]\end{alignedat}}}$$

These various forms are all equivalent by the parallelogram law:[proof 1]

$${\displaystyle 2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}.}$$

This further implies that ${\displaystyle L^{p}}$ class is not a Hilbert space whenever ${\displaystyle p\neq 2}$, as the parallelogram law is not satisfied. For the sake of counterexample, consider ${\displaystyle x=1_{A}}$ and ${\displaystyle y=1_{B}}$ for any two disjoint subsets ${\displaystyle A,B}$ of general domain ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ and compute the measure of both sets under parallelogram law.

For vector spaces over the complex numbers, the above formulas are not quite correct because they do not describe the imaginary part of the (complex) inner product. However, an analogous expression does ensure that both real and imaginary parts are retained. The complex part of the inner product depends on whether it is antilinear in the first or the second argument. The notation ${\displaystyle \langle x|y\rangle ,}$ which is commonly used in physics will be assumed to be antilinear in the first argument while ${\displaystyle \langle x,\,y\rangle ,}$ which is commonly used in mathematics, will be assumed to be antilinear its the second argument. They are related by the formula:

$${\displaystyle \langle x,\,y\rangle =\langle y\,|\,x\rangle \quad {\text{ for all }}x,y\in H.}$$

The real part of any inner product (no matter which argument is antilinear and no matter if it is real or complex) is a symmetric bilinear map that for any ${\displaystyle x,y\in H}$ is always equal to:[4][proof 1]

$${\displaystyle {\begin{alignedat}{4}R(x,y):&=\operatorname {Re} \langle x\mid y\rangle =\operatorname {Re} \langle x,y\rangle \\&={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)\\&={\frac {1}{2}}\left(\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}\right)\\[3pt]&={\frac {1}{2}}\left(\|x\|^{2}+\|y\|^{2}-\|x-y\|^{2}\right).\\[3pt]\end{alignedat}}}$$

It is always a symmetric map, meaning that[proof 1]

$${\displaystyle R(x,y)=R(y,x)\quad {\text{ for all }}x,y\in H,}$$

and it also satisfies:[proof 1]

$${\displaystyle R(y,ix)=-R(x,iy)\quad {\text{ for all }}x,y\in H.}$$

Thus ${\displaystyle R(ix,y)=-R(x,iy),}$ which in plain English says that to move a factor of ${\displaystyle i}$ to the other argument, introduce a negative sign.

Proof of properties of ${\displaystyle R}$

Let $${\displaystyle R(x,y):={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right).}$$ Then ${\displaystyle 2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}}$ implies $${\displaystyle R(x,y)={\frac {1}{4}}\left(\left(2\|x\|^{2}+2\|y\|^{2}-\|x-y\|^{2}\right)-\|x-y\|^{2}\right)={\frac {1}{2}}\left(\|x\|^{2}+\|y\|^{2}-\|x-y\|^{2}\right)}$$ and $${\displaystyle R(x,y)={\frac {1}{4}}\left(\|x+y\|^{2}-\left(2\|x\|^{2}+2\|y\|^{2}-\|x+y\|^{2}\right)\right)={\frac {1}{2}}\left(\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}\right).}$$ Moreover, $${\displaystyle 4R(x,y)=\|x+y\|^{2}-\|x-y\|^{2}=\|y+x\|^{2}-\|y-x\|^{2}=4R(y,x),}$$ which proves that ${\displaystyle R(x,y)=R(y,x).}$ From ${\displaystyle 1=i(-i)}$ it follows that ${\displaystyle y-ix=i(-iy-x)=-i(x+iy)}$ and ${\displaystyle y+ix=i(-iy+x)=i(x-iy)}$ so that $${\displaystyle -4R(y,ix)=\|y-ix\|^{2}-\|y+ix\|^{2}=\|(-i)(x+iy)\|^{2}-\|i(x-iy)\|^{2}=\|x+iy\|^{2}-\|x-iy\|^{2}=4R(x,iy),}$$ which proves that ${\displaystyle R(y,ix)=-R(x,iy).}$ ${\displaystyle \blacksquare }$

Unlike its real part, the imaginary part of a complex inner product depends on which argument is antilinear.

Antilinear in first argument

The polarization identities for the inner product ${\displaystyle \langle x\,|\,y\rangle ,}$ which is antilinear in the first argument, are

${\displaystyle {\begin{alignedat}{4}\langle x\,|\,y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}-i\|x+iy\|^{2}+i\|x-iy\|^{2}\right)\\&=R(x,y)-iR(x,iy)\\&=R(x,y)+iR(ix,y)\\\end{alignedat}}}$

where ${\displaystyle x,y\in H.}$ The second to last equality is similar to the formula expressing a linear functional ${\displaystyle \varphi }$ in terms of its real part: ${\displaystyle \varphi (y)=\operatorname {Re} \varphi (y)-i(\operatorname {Re} \varphi )(iy).}$

Antilinear in second argument

The polarization identities for the inner product ${\displaystyle \langle x,\ y\rangle ,}$ which is antilinear in the second argument, follows from that of ${\displaystyle \langle x\,|\,y\rangle }$ by the relationship: ${\displaystyle \langle x,\ y\rangle$ :=\langle y\,|\,x\rangle ={\overline {\langle x\,|\,y\rangle }}\quad {\text{ for all }}x,y\in H.} So for any ${\displaystyle x,y\in H,}$[4]

${\displaystyle {\begin{alignedat}{4}\langle x,\,y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}+i\|x+iy\|^{2}-i\|x-iy\|^{2}\right)\\&=R(x,y)+iR(x,iy)\\&=R(x,y)-iR(ix,y).\\\end{alignedat}}}$

This expression can be phrased symmetrically as:[5]

$${\displaystyle \langle x,y\rangle ={\frac {1}{4}}\sum _{k=0}^{3}i^{k}\left\|x+i^{k}y\right\|^{2}.}$$

Summary of both cases

Thus if ${\displaystyle R(x,y)+iI(x,y)}$ denotes the real and imaginary parts of some inner product's value at the point ${\displaystyle (x,y)\in H\times H}$ of its domain, then its imaginary part will be:

$${\displaystyle I(x,y)~=~{\begin{cases}~R({\color {red}i}x,y)&\qquad {\text{ if antilinear in the }}{\color {red}1}{\text{st argument}}\\~R(x,{\color {blue}i}y)&\qquad {\text{ if antilinear in the }}{\color {blue}2}{\text{nd argument}}\\\end{cases}}}$$

where the scalar ${\displaystyle i}$ is always located in the same argument that the inner product is antilinear in.

Using ${\displaystyle R(ix,y)=-R(x,iy),}$ the above formula for the imaginary part becomes:

$${\displaystyle I(x,y)~=~{\begin{cases}-R(x,{\color {black}i}y)&\qquad {\text{ if antilinear in the }}{\color {black}1}{\text{st argument}}\\-R({\color {black}i}x,y)&\qquad {\text{ if antilinear in the }}{\color {black}2}{\text{nd argument}}\\\end{cases}}}$$

If ${\displaystyle A:H\to Z}$ is a linear isometry between two Hilbert spaces (so ${\displaystyle \|Ah\|=\|h\|}$ for all ${\displaystyle h\in H}$) then

$${\displaystyle \langle Ah,Ak\rangle _{Z}=\langle h,k\rangle _{H}\quad {\text{ for all }}h,k\in H;}$$

that is, linear isometries preserve inner products.

If ${\displaystyle A:H\to Z}$ is instead an antilinear isometry then

$${\displaystyle \langle Ah,Ak\rangle _{Z}={\overline {\langle h,k\rangle _{H}}}=\langle k,h\rangle _{H}\quad {\text{ for all }}h,k\in H.}$$

The second form of the polarization identity can be written as

$${\displaystyle \|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).}$$

This is essentially a vector form of the law of cosines for the triangle formed by the vectors ${\displaystyle {\textbf {u}},{\textbf {v}},}$ and ${\displaystyle {\textbf {u}}-{\textbf {v}}.}$ In particular,

$${\displaystyle {\textbf {u}}\cdot {\textbf {v}}=\|{\textbf {u}}\|\,\|{\textbf {v}}\|\cos \theta ,}$$

where ${\displaystyle \theta }$ is the angle between the vectors ${\displaystyle {\textbf {u}}}$ and ${\displaystyle {\textbf {v}}.}$

The equation is numerically unstable if u and v are similar because of catastrophic cancellation and should be avoided for numeric computation.

The basic relation between the norm and the dot product is given by the equation

$${\displaystyle \|{\textbf {v}}\|^{2}={\textbf {v}}\cdot {\textbf {v}}.}$$

Then

$${\displaystyle {\begin{aligned}\|{\textbf {u}}+{\textbf {v}}\|^{2}&=({\textbf {u}}+{\textbf {v}})\cdot ({\textbf {u}}+{\textbf {v}})\\[3pt]&=({\textbf {u}}\cdot {\textbf {u}})+({\textbf {u}}\cdot {\textbf {v}})+({\textbf {v}}\cdot {\textbf {u}})+({\textbf {v}}\cdot {\textbf {v}})\\[3pt]&=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}+2({\textbf {u}}\cdot {\textbf {v}}),\end{aligned}}}$$

and similarly

$${\displaystyle \|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).}$$

Forms (1) and (2) of the polarization identity now follow by solving these equations for ${\displaystyle {\textbf {u}}\cdot {\textbf {v}},}$ while form (3) follows from subtracting these two equations. (Adding these two equations together gives the parallelogram law.)

The polarization identities are not restricted to inner products. If ${\displaystyle B}$ is any symmetric bilinear form on a vector space, and ${\displaystyle Q}$ is the quadratic form defined by

$${\displaystyle Q(v)=B(v,v),}$$

then

$${\displaystyle {\begin{aligned}2B(u,v)&=Q(u+v)-Q(u)-Q(v),\\2B(u,v)&=Q(u)+Q(v)-Q(u-v),\\4B(u,v)&=Q(u+v)-Q(u-v).\end{aligned}}}$$

The so-called symmetrization map generalizes the latter formula, replacing ${\displaystyle Q}$ by a homogeneous polynomial of degree ${\displaystyle k}$ defined by ${\displaystyle Q(v)=B(v,\ldots ,v),}$ where ${\displaystyle B}$ is a symmetric ${\displaystyle k}$-linear map.[7]

The formulas above even apply in the case where the field of scalars has characteristic two, though the left-hand sides are all zero in this case. Consequently, in characteristic two there is no formula for a symmetric bilinear form in terms of a quadratic form, and they are in fact distinct notions, a fact which has important consequences in L-theory; for brevity, in this context "symmetric bilinear forms" are often referred to as "symmetric forms".

These formulas also apply to bilinear forms on modules over a commutative ring, though again one can only solve for ${\displaystyle B(u,v)}$ if 2 is invertible in the ring, and otherwise these are distinct notions. For example, over the integers, one distinguishes integral quadratic forms from integral symmetric forms, which are a narrower notion.

More generally, in the presence of a ring involution or where 2 is not invertible, one distinguishes ${\displaystyle \varepsilon }$-quadratic forms and ${\displaystyle \varepsilon }$-symmetric forms; a symmetric form defines a quadratic form, and the polarization identity (without a factor of 2) from a quadratic form to a symmetric form is called the "symmetrization map", and is not in general an isomorphism. This has historically been a subtle distinction: over the integers it was not until the 1950s that relation between "twos out" (integral quadratic form) and "twos in" (integral symmetric form) was understood – see discussion at integral quadratic form; and in the algebraization of surgery theory, Mishchenko originally used symmetric L-groups, rather than the correct quadratic L-groups (as in Wall and Ranicki) – see discussion at L-theory.

Finally, in any of these contexts these identities may be extended to homogeneous polynomials (that is, algebraic forms) of arbitrary degree, where it is known as the polarization formula, and is reviewed in greater detail in the article on the polarization of an algebraic form.