semilinear transformationSemilinearTransformation2006-04-15 12:20:332006-06-19 12:06:43Definitionsemilinear transformGamma Lfield automorphismlinearGamma L\usepackage{latexsym}
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\providecommand{\bigintersect}{\bigcap}Let $K$ be a field and $k$ its prime subfield. For example, if $K$ is $\mathbb{C}$ then $k$ is $\mathbb{Q}$, and if $K$ is the finite field of order $q=p^i$, then $k$ is $\mathbb{Z}_p$.
\begin{defn}
Given a field automorphism $\theta$ of $K$, a function
$f:V\rightarrow W$ between two $K$ vector spaces $V$ and $W$ is
$\theta$-semilinear, or simply semilinear, if for all $x,y\in V$ and $l\in K$
it follows: (shown here first in left hand notation and then in the preferred right hand notation.)
\begin{enumerate}
\item $f(x+y)=f(x)+f(y)$, (in right hand notation: $(x+y)f=xf+yf$.)
\item $f(lx)=l^\theta f(x)$, (in right hand notation: $(lx)f=l^\theta xf$.)
\end{enumerate}
where $l^\theta$ denotes the image of $l$ under $\theta$.
\end{defn}
\begin{remark}
$\theta$ must be a field automorphism for $f$ to remain additive, for example,
$\theta$ must fix the prime subfield as
\[n^\theta xf=(nx)f=(x+\cdots +x)f=n(xf).\]
Also
\[(l_1+l_2)^\theta xf=((l_1+l_2)x)f=(l_1 x)f+(l_2 x)f=(l_1^\theta + l_2^\theta)xf\]
so $(l_1+l_2)^\theta=l_1^\theta+l_2^\theta$. Finally,
\[(l_1 l_2)^\theta xf=((l_1 l_2 x)f=l_1^\theta (l_2 x)f=l_1^\theta l_2^\theta xf.\]
\end{remark}
Every linear transformation is semilinear, but the converse is generally not true. If we treat $V$ and $W$ as vector spaces over $k$, (by considering $K$ as vector space over $k$ first) then every $\theta$-semilinear map is a $k$-linear map, where $k$ is the prime subfield of $K$.
\textbf{Example}
\begin{itemize}
\item
Let $K=\mathbb{C}$, $V=\mathbb{C}^n$ with standard basis $e_1,\dots,e_n$. Define the map $f:V\rightarrow V$ by
\[f\left(\sum_{i=1}z_i e_i\right)=\sum_{i=1}^n \bar{z}_i e_i.\]
$f$ is semilinear (with respect to the complex conjugation field automorphism)
but not linear.
\item
Let $K=GF(q)$ -- the Galois field of order $q=p^i$, $p$ the characteristic.
Let $l^\theta=l^p$, for $l\in K$. By the Freshman's dream it is known that this is a field automorphism. To every linear map $f:V\rightarrow W$ between
vector spaces $V$ and $W$ over $K$ we can establish a $\theta$-semilinear map
\[\left(\sum_{i=1}l_i e_i\right)\tilde{f}=\sum_{i=1}^n l_i^\theta e_i f.\]
\end{itemize}
$\Box$
Indeed every linear map can be converted into a semilinear map in such a way.
This is part of a general observation collected into the following result.
\begin{defn}
Given a vector space $V$, the set of all invertible semilinear maps (over all field automorphisms) is the group $\Gamma L(V)$.
\end{defn}
\begin{prop}
Given a vector space $V$ over $K$, and $k$ the prime subfield of $K$, then
$\Gamma L(V)$ decomposes as the semidirect product
\[\Gamma L(V)=GL(V)\rtimes Gal(K/k)\]
where $Gal(K/k)$ is the Galois group of $K/k$.
\end{prop}
\begin{remark}
We identify $Gal(K/k)$ with a subgroup of $\Gamma L(V)$ by fixing a basis
$B$ for $V$ and defining the semilinear maps:
\[\sum_{b\in B} l_b b\mapsto \sum_{b\in B} l_b^\sigma b\]
for any $\sigma\in Gal(K/k)$. We shall denoted this subgroup by $Gal(K/k)_B$. We also see these complements to $GL(V)$ in $\Gamma L(V)$ are acted on regularly by $GL(V)$ as they correspond to a change of basis.
\end{remark}
\begin{proof}
Every linear map is semilinear thus $GL(V)\leq \Gamma L(V)$. Fix a basis
$B$ of $V$. Now given any semilinear map $f$ with respect to a field automorphism $\sigma\in Gal(K/k)$, then define $g:V\rightarrow V$ by \[\left(\sum_{b\in B} l_b b\right)g=\sum_{b\in B} (l_b^{\sigma^{-1}} b)f
=\sum_{b\in B} l_b (b)f.\]
As $(B)f$ is also a basis of $V$, it follows $g$ is simply a basis exchange
of $V$ and so linear and invertible: $g\in GL(V)$.
Set $h:=g^{-1}f$. For every $v=\sum_{b\in B}l_b\neq 0$ in $V$,
\[vh=vg^{-1}f=\sum_{b\in B} l_b^{\sigma} b\]
thus $h$ is in the $Gal(K/k)$ subgroup relative to the fixed basis $B$.
This factorization is unique to the fixed basis $B$. Furthermore, $GL(V)$ is normalized by the action of $Gal(K/k)_B$, so $\Gamma L(V)=GL(V)\rtimes Gal(K/k)$.
\end{proof}
The $\Gamma L(V)$ groups extend the typical classical groups in $GL(V)$. The importance in considering such maps follows from the consideration of projective geometry.
The projective geometry of a vector space $V$, denoted $PG(V)$, is the lattice of all subspaces of $V$. Although the typical semilinear map is not a linear map, it does follow that every semilinear map $f:V\rightarrow W$ induces an order-preserving map $f:PG(V)\rightarrow PG(W)$. That is, every semilinear map induces a projectivity. The converse of this observation is the Fundamental Theorem of Projective Geometry. Thus semilinear maps are useful because they define the automorphism group of the projective geometry of a vector space.
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\begin{thebibliography}{10}
\bibitem{HP}
Gruenberg, K. W. and Weir, A.J.
\emph{Linear Geometry 2nd Ed.} (English)
[B] Graduate Texts in Mathematics. 49. New York - Heidelberg - Berlin: Springer-Verlag. X, 198 p. DM 29.10; \$ 12.80 (1977).
\end{thebibliography}