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[parent] quadratic map (Derivation)

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

$\displaystyle 2c(x,y)=q(x+y)-q(x)-q(y).$
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
$\displaystyle q_b(\alpha x)=b(\alpha x,\alpha x)=\alpha^2 b(x,x)=\alpha^2 q(x)$
and
$\displaystyle q_b(x+y)-q_b(x)-q_b(y) =b(x+y,x+y)-b(x,x)-b(y,y)=b(x,y)+b(y,x)=2 b(x,y).$
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\vert _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.



"quadratic map" is owned by Algeboy.
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See Also: quadratic Jordan algebra, isotropic quadratic space

Also defines:  quadratic map, totally singular, totally isotropic, polarization formula, polarization identity

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Cross-references: subspace, multiple, automorphism, submodule, homomorphism, additive, 1-1 correspondence, rings, characteristic, odd, variable, polarization, symmetric, implies, properties, codomain, quadratic forms, difference, bilinear map, map, commutative ring
There are 5 references to this entry.

This is version 6 of quadratic map, born on 2006-12-14, modified 2007-07-30.
Object id is 8625, canonical name is QuadraticMap2.
Accessed 2827 times total.

Classification:
AMS MSC: 15A63 (Linear and multilinear algebra; matrix theory :: Quadratic and bilinear forms, inner products)
 11E04 (Number theory :: Forms and linear algebraic groups :: Quadratic forms over general fields)
 11E08 (Number theory :: Forms and linear algebraic groups :: Quadratic forms over local rings and fields)
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