Skip to content

Commit

Permalink
starting working 5.3 over
Browse files Browse the repository at this point in the history
  • Loading branch information
UlrikBuchholtz committed Jan 9, 2025
1 parent 536efb2 commit d837aad
Showing 1 changed file with 19 additions and 11 deletions.
30 changes: 19 additions & 11 deletions actions.tex
Original file line number Diff line number Diff line change
Expand Up @@ -652,18 +652,26 @@ \section{Subgroups}
\subsection{Subgroups through $G$-sets}

The idea is that a $G$-set $X$ picks out those symmetries in $G$
that keep the point of $X(\sh_G)$ in place. For this to work well
that keep a chosen point of $X(\sh_G)$ in place. For this to work well
we need to point $X(\sh_G)$ and $X$ must be transitive.
%so that the set of symmetries that are picked out is closed under composition and reverse.

\begin{definition}\label{def:set-of-subgroups}
For any group $G$, define the type of \emph{subgroups of $G$} as
\index{type!of subgroups of a group}
% \glossary(SubG){$\protect{\Sub_G}$}{type of subgroups of $G$} ERROR???
$$\typesubgroup_G\defequi\sum_{X:\BG\to\Set}{\,}X(\sh_G)
\times\istrans(X).$$
The \emph{underlying group} of the subgroup $(X,x,!) : \Sub_G$ is
$$\mkgroup \bigl(\sum_{z:BG}X(z),(\sh_G,x)\bigr).\qedhere$$
For any group $G$, define the type of \emph{subgroups of $G$} as%
\index{type!of subgroups of a group}%
\glossary(SubG){$\protect\Sub(G)$}{type of subgroups of $G$}
\[
\Sub(G)\defequi\sum_{X:\BG\to\Set}{\,}X(\sh_G)
\times\istrans(X).
\]
The \emph{underlying group} of the subgroup $(X,x) : \Sub(G)$ is\footnote{%
To lighten the notation, we leave out the proof that $X$ is transitive.
(Otherwise, we would write $(X,x,!):\Sub(G)$.)
In~\cref{rem:notationsubgroup} below we'll set out further notational conveniences
regarding subgroups.}
\[
\mkgroup \biggl(\sum_{z:BG}X(z),(\sh_G,x)\biggr).\qedhere
\]
\end{definition}

\begin{xca}\label{xca:group-Xx!}
Expand All @@ -687,18 +695,18 @@ \subsection{Subgroups through $G$-sets}
$R(\Sloop) \defis \etop\zs$, see \cref{def:RtoS1}.
Again we point by $0: R(\base)$ and transitivity of $R$ is obvious.
The only symmetry that keeps $0$ in place is $\refl{\base}$,
since $R(\Sloop)= \zs$ if and only if $k=0$.
since $R(\Sloop^k)(0) = \zs^k(0) = k = 0$ if and only if $k=0$.
Again, no surprise in view of the results in \cref{sec:symcirc}
identifying $R$ as the universal \covering over $\Sc$.

The following result is analogous to the fact that $\Sub_T$ is
The following result is analogous to the fact that $\Sub(T)$ is
a set for any type $T$, see \cref{xca:subtypes-set}. It captures
that the essence of picking out symmetries (or picking out elements
of a type), is a predicate, like $R_m(p)(0)=0$ above.

\begin{lemma}
\label{lem:SubGisset}%
The type $\typesubgroup_G$ is a set, for any group $G$.
The type $\Sub(G)$ is a set, for any group $G$.
\end{lemma}
\begin{proof}
Let $G$ be a group, and let $X$ and $X'$ be transitive $G$-sets with
Expand Down

0 comments on commit d837aad

Please sign in to comment.