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\newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{group T-complex} \hypertarget{contents}{}\section*{{Contents}}\label{contents} \noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak \noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak \noindent\hyperlink{results}{Results}\dotfill \pageref*{results} \linebreak \noindent\hyperlink{the_groupoid_tcomplex_associated_to_a_simplicial_groupoid}{The group(oid) T-complex associated to a simplicial group(oid)}\dotfill \pageref*{the_groupoid_tcomplex_associated_to_a_simplicial_groupoid} \linebreak \noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak \hypertarget{idea}{}\subsection*{{Idea}}\label{idea} A \textbf{group $T$-complex} is `'a [[simplicial T-complex]] internal to the category of groups''. They were first studied in Nick Ashley's thesis (see references below for the published version) More precisely: \hypertarget{definition}{}\subsection*{{Definition}}\label{definition} A \textbf{group $T$-complex} is a pair, $(G,T)$, in which $G$ is a [[simplicial group]] and $T$ is a graded subgroup of $G$ consisting of \emph{thin} elements, and which satisfies the conditions: \begin{itemize}% \item Every degenerate element is thin. \item Every [[horn]] in $G$ has a unique thin filler. \item A thin filler of a thin box also has its last face thin. \end{itemize} \hypertarget{results}{}\subsection*{{Results}}\label{results} \begin{ulemma} Let $D = (D_n)_{n\geq 1}$ be the graded subgroup of $G$ generated by the images of the degeneracy maps, $s_i :G_n \rightarrow G_{n+1}$, for all $i$ and $n$, then any box in $G$ has a filler in $D$. \end{ulemma} \begin{proof} The algorithmic formulae used when proving that any simplicial group is a Kan complex (cf., entry on [[simplicial group]]) give a filler defined as a product of degenerate copies of the faces of the given box, so is in $D_n$. \end{proof} \begin{uproposition} If $(G,T)$ is a group $T$-complex then $T = D$. \end{uproposition} \begin{proof} To see this, we note that as every degenerate element is this, $D \subseteq T$. Conversely if $t \in T_n$, then it fills the box made up of $( - , d_1t, \ldots, d_n t )$. This, in turn, has a filler, $d$, in $D$, but, as this filler is also thin, it must be that $t = d$, since thin fillers are uniquely determined. \end{proof} This is neat. It says there is basically only one possible group $T$-complex structure on a given simplicial group. The next result (again by Ashley) shows that not all simplicial groups carry such a structure. \begin{uproposition} If $G$ is a simplicial group, then $(G,D)$ is a group $T$-complex if and only if $NG\cap D$ is the trivial graded subgroup. \end{uproposition} \begin{proof} One way around, this is nearly trivial. If $(G,D)$ is a group $T$-complex and $x\in NG_n$, then $x$ fills a box $(-, 1, \ldots, 1)$, so if $x\in NG_n\cap D_n$, $x$ must itself be the thin filler, however 1 is also a thin filler for this box, so $x = 1$ as required. Conversely if $NG\cap D = \{1\}$, then we must check the other two axioms as the first is trivial. As any box has a standard filler in $D$, we only have to check uniqueness, but if $x$ and $y$ are in $D_n$, and both fill the same box (with the $k^{th}$ face missing) then $z = xy^{-1}$ fills a box with 1s on all faces (and the $k^{th}$ face missing). If $k = 0$, then as $z \in NG_n\cap D_n$, we have $z = 1$ and $x$ and $y$ are equal. If $k \gt 0$, assume that if $\ell \lt k$ and $z \in D_n \cap \bigcap_{i\neq \ell} Ker\, d_i$ then $z = 1$, (i.e, that we have uniqueness up to at least the $(k-1)^{st}$ case). Consider $w = zs_{k-1}d_k z^{-1}$. This is still in $D_n$ and $d_i w = 1$ unless $i = k-1$, hence by assumption $w = 1$. Of course, this implies that $z = s_{k-1}d_kz$, but then $d_{k-1} z = d_k z$. We know that $d_{k-1} z = 1$, so $d_k z = 1$ and $z = 1$, i.e., $x = y$ and we have uniqueness at the next stage. To verify the third axiom, assume that $x \in D_{n + 1}$ and each $d_i x \in D_n$ for $i \neq k$, then we can assume that $k = 0$, since otherwise we can skew the situation around as before to get that to be true, verify it in that case and `skew' it back again later. Suppose therefore that $d_i x \in D_n$ for all $0 \lt i \lt n$. As $x$ must be the degenerate filler given by the standard method, we can calculate $x$ as follows: let $w_n = s_{n-1}d_n x$, $w_i = w_{i+1}(s_{i-1}d_i w_{i+1})^{-1}s_{i-1}y_i$ for $i = 1$, then $x = w_1$. We can therefore check that $d_0x \in D_n$ as required. \end{proof} \hypertarget{the_groupoid_tcomplex_associated_to_a_simplicial_groupoid}{}\subsection*{{The group(oid) T-complex associated to a simplicial group(oid)}}\label{the_groupoid_tcomplex_associated_to_a_simplicial_groupoid} This suggests that, given an arbitrary simplicial group(oid), $G$, we could form a quotient which would be a group(oid) $T$-complex, simply by dividing out by the subgroupoids, $NG_n\cap D_n$. We would need check that the face and degeneracy maps worked correctly, that the result did not somehow generate some new `thin' elements, etc. In fact the idea does not work because of a much more elementary problem, which is already apparent, even in the case of simplicial groups, \begin{itemize}% \item $NG_n\cap D_n$ need not be a normal subgroup of $G_n$! \end{itemize} A variant of the idea does however work. We concentrate on the case of simplicial groups. The extension to simplicial groupoids is then straightforward. We need the Conduch\'e{} non-Abelian version of the [[decomposition theorem for simplicial groups]] that makes up an important part of the Dold-Kan correspondence. With that we note the fairly obvious point that when we divide out by a graded subgroup in a simplicial group, then it has effects in all dimensions due to the face and degeneracy maps, so if we kill elements in $NG_n\cap D_n$ in all dimensions, we must also kill $d_0(NG_{n+1}\cap D_{n+1})$ and all the $s_k(NG_{n-1}\cap D_{n-1})$. The end result of this is a group(oid) $T$-complex whose Moore complex has a [[crossed complex]] structure in a natural way. It is easier to give the corresponding formula for the equivalent crossed complex to this group $T$-complex. Explicitly we can form \begin{displaymath} {C}(G)_{n+1} = \frac{\mathcal{N}G_n}{(\mathcal{N}G_n\cap D_n)d_0(\mathcal{N}G_{n+1}\cap D_{n+1})}, \end{displaymath} in higher dimensions with at its `bottom end', the crossed module, \begin{displaymath} \frac{\mathcal{N}G_1}{d_0(\mathcal{N}G_2\cap D_2)} \to \mathcal{N}G_0 \end{displaymath} with $\partial$ induced from the boundary in the [[Moore complex]]. This gives a crossed complex. \hypertarget{references}{}\subsection*{{References}}\label{references} \begin{itemize}% \item N. Ashley, \emph{Simplicial T-Complexes: a non abelian version of a theorem of Dold-Kan}, Dissertations Math., 165, (1989), 11 -- 58. [[!redirects group T - complex]] \end{itemize} \end{document}