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\begin{document}

\noindent
{\large \bf Additional notes \#8.}

\bigskip

\noindent
{\bf Galois Connections.}

\bigskip

\noindent
{\bf Definition.} Let $S$ and $T$ be classes of
objects, and let $R\subseteq
S\times T$ be a binary relation from $S$ to $T$.
The {\it Galois connection determined by $R$} is
the pair of mappings (both denoted $\perp$) defined by
\[\perp : {\vr P}(S)\rightarrow {\vr P}(T) :
U\mapsto U^{\perp} = \{ t\in T\;|\;\forall u\in U ((u,t)\in R)\}\]
and 
\[\perp : {\vr P}(T)\rightarrow {\vr P}(S) :
V\mapsto V^{\perp} = \{ s\in S\;|\;\forall v\in V ((s,v)\in R)\}.\]

\bigskip

\noindent
{\bf Theorem.} Assume that $U\subseteq V\subseteq S$
and $X\subseteq Y\subseteq T$.

(i) $\perp$ reverses inclusions: $U^{\perp}\supseteq V^{\perp}$
and $X^{\perp}\supseteq Y^{\perp}$.

(ii) $\perp\perp$ is increasing: $U\subseteq U^{\perp\perp}$
and $X\subseteq X^{\perp\perp}$.

(iii) $\perp\perp\perp = \perp$: $U^{\perp} = U^{\perp\perp\perp}$
and $X^{\perp}= X^{\perp\perp\perp}$. 

(iv) The operations
\[\cl : {\vr P}(S)\rightarrow {\vr P}(S) : U\mapsto
U^{\perp\perp}\]
and
\[\cl : {\vr P}(T)\rightarrow {\vr P}(T) : U\mapsto
U^{\perp\perp}\]
are closure operators.

(v) A set is closed if and only if it is in the image of $\perp$.

(vi) If ${\vr L}_S$ is the lattice of closed subsets of
$S$ and ${\vr L}_T$ is the lattice of closed subsets
if $T$, then $\perp: {\vr L}_S\rightarrow {\vr L}_T$
is an order-reversing bijection.

\medskip

{\it Proof.} (i) and (ii) are easy. We show how to derive 
(iii)--(vi) from (i) and (ii).

For (iii), apply $\perp$ to the inclusion in (ii)
and use part(i) to get $U^{\perp}\supseteq U^{\perp\perp\perp}$.
But by part (ii) we have $U^{\perp}\subseteq (U^\perp)^{\perp\perp}$.
Hence $U^{\perp} = U^{\perp\perp\perp}$.

For (iv), we have that $\perp\perp$ is extensive from (ii).
To prove that 
 $\perp\perp$ is isotone we use (i) twice:
\[U\subseteq V \Longrightarrow 
U^{\perp}\supseteq V^{\perp}
\Longrightarrow 
U^{\perp\perp}\subseteq V^{\perp\perp}.\]
For idempotence we $\perp$ the equation from (iii) to get
\[ U^{\perp\perp} = U^{\perp\perp\perp\perp} = 
(U^{\perp\perp})^{\perp\perp}.\]

For (v), note that any closed set is in the image of
$\perp$, since $U = U^{\perp\perp}$ implies that $U$
is the result of applying $\perp$ to $U^{\perp}$.
Conversely, if $U = W^{\perp}$, then 
\[U^{\perp\perp} = W^{\perp\perp\perp} = W^{\perp} = U,\]
so $U$ is closed.

(vi) follows from (v), (iii), and (i). $\Box$

\bigskip\bigskip\bigskip\bigskip


\noindent
{\bf Examples.}

(1) Let $\vr S$ be the class of all algebras defined
with operations $\cdot, {}^{-1}, 1$. Let $\vr T$
be the collection of all equations involving only
these operation symbols. Let $R$ denote the relation
of satisfaction. (This means that $(\m a,\varepsilon)$
is in $R$ if and only if $\m a\models \varepsilon$,
which is a way of writing that the algebra $\m a$ 
satisfies the equation $\varepsilon$.)

In this example, $\{x\cdot(y\cdot z) = (x\cdot y)\cdot z,
x\cdot 1 = x, x\cdot x^{-1} = 1\}^{\perp}$ is
the class of all groups. More generally, 
if $\Sigma\subseteq {\vr T}$, then $\Sigma^{\perp}$
is the variety of all algebras satisfying 
the equations in $\Sigma$. 

If ${\vr K}\subseteq {\vr S}$, then ${\vr K}^{\perp}$
is the set  of all equations that hold in all members
of ${\vr K}$. This set of equations is called
{\it the equational theory of $\vr K$}.

The lattice of closed subclasses of $\vr S$ is the
lattice of all varieties of algebras defined with
operations $\cdot, {}^{-1}, 1$. The lattice of
closed subsets of $\vr T$ is the lattice of 
equational theories in this language. These
lattices are dual to each other.

\medskip

(2) Let $S$ be a set and let $G$ be a group of permutations
of $S$. The relation $R = \{(s,g)\;|\;g(s)=s\}$
determines a Galois connection between $S$ and $G$.

If $s\in S$, then $\{s\}^{\perp}$ is the stabilizer subgroup
of $s$. If $g\in G$, then $\{g\}^{\perp}$ is the 
set of fixed points of $g$.

\bigskip

\noindent
{\bf The Galois Connection of Galois Theory}

\medskip

Let ${\BF}<{\BE}$ be a finite extension. 
Let $G = {\rm Gal}({\BE}/{\BF})$ be the group
of all ${\BF}$-linear automorphisms of ${\BE}$.
Let $R\subseteq {\BE}\times G$ be the relation
$R=\{(e,g)\;|\;g(e)=e\}$. This relation 
determines a Galois connection between 
$\BE$ and $G$.

\bigskip

\noindent
{\bf Exercises.} 

(1) Show that a field automorphism
$\sigma: {\BE}\rightarrow {\BE}$ is ${\BF}$-linear
(i.e., satisfies $\sigma(f\cdot e) = f\cdot \sigma(e)$
for $f\in {\BF}$ and $e\in {\BE}$) if and only if
$\sigma(f) = f$ for all $f\in {\BF}$.

\smallskip

(2) Show that any closed subset of $\BE$ is 
a subfield of $\BE$ containing $\BF$.

\smallskip

(3) Show that any closed subset of $G$ is 
a subgroup.

\bigskip

\noindent
{\bf Fundamental Theorem of Galois Theory.}
If ${\BF}<{\BE}$ is a finite, normal, separable extension,
then every subgroup of $G = {\rm Gal}({\BE}/{\BF})$ 
is closed, and every intermediate subfield
${\BF}<{\BK}<{\BE}$ is closed.
\end{document}

