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'quadratic map'
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| Title of object: |
quadratic map |
| Canonical Name: |
QuadraticMap2 |
| Type: |
Derivation |
| Created on: |
2006-12-14 11:50:42 |
| Modified on: |
2006-12-14 11:52:13 |
| Classification: |
msc:15A63, msc:11E04, msc:11E08 |
| Defines: |
quadratic map, totally singular, totally isotropic |
Revision comment (for changes between this and next version):
| Changes for correction #11071 ('polarization identity'). |
Preamble:
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Content:
Given a commutative ring $K$ and two $K$-modules $M$ and $N$ then a map
$q:M\rightarrow N$ is called \emph{quadratic} if
\begin{enumerate}
\item $q(\alpha x)=\alpha^2 q(x)$ for all $x\in M$ and $\alpha\in K$.
\item $b(x,y):=q(x+y)-q(x)-q(y)$, for $x,y\in M$, is a bilinear map.
\end{enumerate}
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, using
\[b(x,y):=\frac{1}{2}(q(x+y)-q(x)-q(y)).\]
makes $b(x,x)=q(x)$, but requires odd or 0 characteristic on $K$. Also when
$b$ is a symmetric 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
\[\frac{1}{2}(q_b(x+y)-q_b(x)-q_b(y))
=\frac{1}{2}(b(x+y,x+y)-b(x,x)-b(y,y))=\frac{1}{2}(b(x,y)+b(y,x))
=b(x,y).\]
So in odd and 0 characteristic 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 \emph{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.
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