quadratic map

Given a commutative ring $K$ and two $K$ -modules $M$ and $N$ then a map $q:M\\rightarrow N$ 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 singlevariable in a formula with $x+y$ and comparing the result with the formula over $x$ and $y$ separately.) Continuingwithout $2$ -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 symmetricbilinear 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 ispossible for $b(T,T)=0$ yet $q(T)\eq 0$ . For instance, we can have $q(x)\eq 0$ yet $b(x,x)=q(2x)-q(x)-q(x)=0$ . This is summed up in the followingdefinition:

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