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 2torsion ($2x=0$ 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 11 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 semilinear 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  20130322 16:27:55 
Last modified on  20130322 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 