quadratic map

Given a commutative ring $K$ and two $K$-modules $M$ and $N$ then a map $q:M\to N$ is called quadratic if

1. $q(\alpha x)={\alpha }^{2}q(x)$ for all $x\in M$ and $\alpha \in K$.

2. $b(x,y):=q(x+y)-q(x)-q(y)$, for $x,y\in M$, is a bilinear map.

The only difference between quadratic maps and quadratic forms is the insistence on the codomain $N$ instead of a $K$. So in this way every quadratic form is a special case of a quadratic map. Most of the properties for quadratic forms apply to quadratic maps as well. For instance, if $K$ has no 2-torsion ($2x=0$ implies $x=0$) then

$$2c(x,y)=q(x+y)-q(x)-q(y).$$

defines a symmetric $K$-bilinear map $c:M\times M\to N$ with $c(x,x)=q(x)$. In particular if $1/2\in K$ then $c(x,y)=\frac{1}{2}b(x,y)$. This definition is one instance of a polarization (i.e.: substituting a single variable in a formula with $x+y$ and comparing the result with the formula over $x$ and $y$ separately.) Continuing without $2$-torsion, if $b$ is a symmetric $K$-bilinear map (perhaps not a form) then defining $q_b(x)=b(x,x)$ determines a quadratic map since

$$q_b(\alpha x)=b(\alpha x,\alpha x)={\alpha }^{2}b(x,x)={\alpha }^{2}q(x)$$

and

$$q_b(x+y)-q_b(x)-q_b(y)=b(x+y,x+y)-b(x,x)-b(y,y)=b(x,y)+b(y,x)=2b(x,y).$$

Have have no $2$-torsion we can recover $b$ form $q_b$. So in odd and 0 characteristic rings we find symmetric bilinear maps and quadratic maps are in 1-1 correspondence.

An alternative understanding of $b$ is to treat this as the obstruction to $q$ being an additive homomorphism. Thus a submodule $T$ of $M$ for which $b(T,T)=0$ is a submodule of $M$ on which ${q|}_T$ is an additive homomorphism. Of course because of the first condition, $q$ is semi-linear on $T$ only when $\alpha \mapsto {\alpha }^2$ is an automorphism of $K$, in particular, if $K$ has characteristic 2. When the characteristic of $K$ is odd or 0 then $q(T)=0$ if and only if $b(T,T)=0$ simply because $q(x)=b(x,x)$ (or up to a $1/2$ multiple depending on conventions). However, in characteristic 2 it is possible for $b(T,T)=0$ yet $q(T)\ne 0$. For instance, we can have $q(x)\ne 0$ yet $b(x,x)=q(2x)-q(x)-q(x)=0$. This is summed up in the following definition:

A subspace $T$ of $M$ is called totally singular if $q(T)=0$ and totally isotropic if $b(T,T)=0$. In odd or 0 characteristic, totally singular subspaces are precisely totally isotropic subspaces.

Title	quadratic map
Canonical name	QuadraticMap
Date of creation	2013-03-22 16:27:55
Last modified on	2013-03-22 16:27:55
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	9
Author	Algeboy (12884)
Entry type	Derivation
Classification	msc 11E08
Classification	msc 11E04
Classification	msc 15A63
Related topic	QuadraticJordanAlgebra
Related topic	IsotropicQuadraticSpace
Defines	quadratic map
Defines	totally singular
Defines	totally isotropic
Defines	polarization formula
Defines	polarization identity