Polarization of an algebraic form

The fundamental ideas are as follows. Let ${\displaystyle f(\mathbf {u} )}$ be a polynomial in ${\displaystyle n}$ variables ${\displaystyle \mathbf {u} =\left(u_{1},u_{2},\ldots ,u_{n}\right).}$ Suppose that ${\displaystyle f}$ is homogeneous of degree ${\displaystyle d,}$ which means that

$${\displaystyle f(t\mathbf {u} )=t^{d}f(\mathbf {u} )\quad {\text{ for all }}t.}$$

Let ${\displaystyle \mathbf {u} ^{(1)},\mathbf {u} ^{(2)},\ldots ,\mathbf {u} ^{(d)}}$ be a collection of indeterminates with ${\displaystyle \mathbf {u} ^{(i)}=\left(u_{1}^{(i)},u_{2}^{(i)},\ldots ,u_{n}^{(i)}\right),}$ so that there are ${\displaystyle dn}$ variables altogether. The polar form of ${\displaystyle f}$ is a polynomial

$${\displaystyle F\left(\mathbf {u} ^{(1)},\mathbf {u} ^{(2)},\ldots ,\mathbf {u} ^{(d)}\right)}$$

which is linear separately in each ${\displaystyle \mathbf {u} ^{(i)}}$ (that is, ${\displaystyle F}$ is multilinear), symmetric in the ${\displaystyle \mathbf {u} ^{(i)},}$ and such that

$${\displaystyle F\left(\mathbf {u} ,\mathbf {u} ,\ldots ,\mathbf {u} \right)=f(\mathbf {u} ).}$$

The polar form of ${\displaystyle f}$ is given by the following construction

$${\displaystyle F\left({\mathbf {u} }^{(1)},\dots ,{\mathbf {u} }^{(d)}\right)={\frac {1}{d!}}{\frac {\partial }{\partial \lambda _{1}}}\dots {\frac {\partial }{\partial \lambda _{d}}}f(\lambda _{1}{\mathbf {u} }^{(1)}+\dots +\lambda _{d}{\mathbf {u} }^{(d)})|_{\lambda =0}.}$$

In other words, ${\displaystyle F}$ is a constant multiple of the coefficient of ${\displaystyle \lambda _{1}\lambda _{2}\ldots \lambda _{d}}$ in the expansion of ${\displaystyle f\left(\lambda _{1}\mathbf {u} ^{(1)}+\cdots +\lambda _{d}\mathbf {u} ^{(d)}\right).}$

A quadratic example. Suppose that ${\displaystyle \mathbf {x} =(x,y)}$ and ${\displaystyle f(\mathbf {x} )}$ is the quadratic form

$${\displaystyle f(\mathbf {x} )=x^{2}+3xy+2y^{2}.}$$

Then the polarization of ${\displaystyle f}$ is a function in ${\displaystyle \mathbf {x} ^{(1)}=\left(x^{(1)},y^{(1)}\right)}$ and ${\displaystyle \mathbf {x} ^{(2)}=\left(x^{(2)},y^{(2)}\right)}$ given by

$${\displaystyle F\left(\mathbf {x} ^{(1)},\mathbf {x} ^{(2)}\right)=x^{(1)}x^{(2)}+{\frac {3}{2}}x^{(2)}y^{(1)}+{\frac {3}{2}}x^{(1)}y^{(2)}+2y^{(1)}y^{(2)}.}$$

More generally, if ${\displaystyle f}$ is any quadratic form then the polarization of ${\displaystyle f}$ agrees with the conclusion of the polarization identity.

A cubic example. Let ${\displaystyle f(x,y)=x^{3}+2xy^{2}.}$ Then the polarization of ${\displaystyle f}$ is given by

$${\displaystyle F\left(x^{(1)},y^{(1)},x^{(2)},y^{(2)},x^{(3)},y^{(3)}\right)=x^{(1)}x^{(2)}x^{(3)}+{\frac {2}{3}}x^{(1)}y^{(2)}y^{(3)}+{\frac {2}{3}}x^{(3)}y^{(1)}y^{(2)}+{\frac {2}{3}}x^{(2)}y^{(3)}y^{(1)}.}$$

The polarization of a homogeneous polynomial of degree ${\displaystyle d}$ is valid over any commutative ring in which ${\displaystyle d!}$ is a unit. In particular, it holds over any field of characteristic zero or whose characteristic is strictly greater than ${\displaystyle d.}$

The polarization isomorphism (by degree)

For simplicity, let ${\displaystyle k}$ be a field of characteristic zero and let ${\displaystyle A=k[\mathbf {x} ]}$ be the polynomial ring in ${\displaystyle n}$ variables over ${\displaystyle k.}$ Then ${\displaystyle A}$ is graded by degree, so that

$${\displaystyle A=\bigoplus _{d}A_{d}.}$$

The polarization of algebraic forms then induces an isomorphism of vector spaces in each degree

$${\displaystyle A_{d}\cong \operatorname {Sym} ^{d}k^{n}}$$

where ${\displaystyle \operatorname {Sym} ^{d}}$ is the ${\displaystyle d}$-th symmetric power of the ${\displaystyle n}$-dimensional space ${\displaystyle k^{n}.}$

These isomorphisms can be expressed independently of a basis as follows. If ${\displaystyle V}$ is a finite-dimensional vector space and ${\displaystyle A}$ is the ring of ${\displaystyle k}$-valued polynomial functions on ${\displaystyle V}$ graded by homogeneous degree, then polarization yields an isomorphism

$${\displaystyle A_{d}\cong \operatorname {Sym} ^{d}V^{*}.}$$

The algebraic isomorphism

Furthermore, the polarization is compatible with the algebraic structure on ${\displaystyle A,}$so that

$${\displaystyle A\cong \operatorname {Sym} ^{\cdot }V^{*}}$$

where ${\displaystyle \operatorname {Sym} ^{\cdot }V^{*}}$ is the full symmetric algebra over ${\displaystyle V^{*}.}$

• Homogeneous function – Function with a multiplicative scaling behaviour

• Claudio Procesi (2007) Lie Groups: an approach through invariants and representations, Springer, ISBN 9780387260402 .