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

1. 1.

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

2. 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 $q(T)\neq 0$. For instance, we can have $q(x)\neq 0$ 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