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Cyclic category

We know, that $\B\bZ$ is homotopy equivalent to $S^1$. Consider a question: what is the simplicial set $C_{\bullet}$ whose geometric realization is circle with the cell structure consisting of one 0-cell and one 1-cell (not up to homotopy)?

The 0-cell $*\in C_0$ generate only one element, still denoted by $*$ in each $C_n$. Suppose we add additional element $\tau$ to $C_1$. Then we get
\begin{align*}
C_0 &= \{*\} \\
C_1 &= \{*, \tau\}\\
C_1 &= \{*, s_0\tau, s_1\t...
...= \{*, \hdots, s_{n-1}\hdots \Hat{s_i}, \hdots s_0\tau,
\hdots\}\\
\end{align*}
The faces are obvious to find. In particular $d_0(\tau)=*=d_1(\tau)$. Then $\vert C_{\bullet}\vert$ is a circle with its simplest cell structure. We can identify

\begin{displaymath}
C_n = \{*, \hdots, s_{n-1}\hdots \Hat{s_i}, \hdots s_0\tau, \hdots\}
\end{displaymath}

with the cyclic group $\bZ/ (n+1)\bZ =: C_n$ by sending $*$ to 0, and $s_{n-1}\hdots \Hat{s_i}, \hdots s_0\tau$ to $i+1$. Denote the generator of $C_n$ by $t_n$.

There exists a cyclic category $\Dl C$ whose objects are finite ordered sets $[n]=\{0, \hdots, n\}$, and morphism $\Mor([n],
[m])$ are generated by $\dl_i$, $\sg_j$ as in simplicial category, and additional morphisms $\tau_n\:[n]\to [n]$ for all $n\geq 0$ satisfying the relations
\begin{align*}
\tau_n^{n+1} &= \id_{[n]} \\
\tau_n\dl_i &= \dl_{i-1}\tau_{n-1},...
...au_{n+1}, \quad 1\leq j \leq n \\
\tau_n\sg_0 &= \sg_n\tau_{n+1}^2
\end{align*}
If in this presentation we omit the relation $\tau_n^{n+1}=\id_{[n]}$, then we get a different category, denoted $\Dl \bZ$.
\begin{defn}
A cyclic set is a functor $\Dl C^{op} \to \Sets$.
\end{defn}

\begin{prop}
$C_{\bullet}$\ is a cyclic set.
\end{prop}

\begin{prop}
\begin{displaymath}
\Aut_{\Dl}([n])=\{1\}
\end{displaymath}\begin{displaymath}
\Aut_{\Dl C}([n])=C_n
\end{displaymath}\end{prop}
Every morphism of $\Dl C$ can be written uniquely as $\phi\circ
g$, where $\phi\in\Mor_{\Dl}([n], [m])$, $g\in C_n=\Mor_{\Dl C}([n],
[n])$. As sets

\begin{displaymath}
\Hom_{\Dl C}([n], [m])\isom \Hom_{\Dl}([n], [m])\times C_n
\end{displaymath}

The compositon of two morphisms $(g\circ\phi)$ and $(h\circ \psi)$ is in $\Dl C$, so there exist $\phi^*(h)\in C_n$ and $h_*(\phi)\in \Mor_{\Dl}([n], [m])$ such that the following diagram commutes.

\begin{displaymath}
\xymatrix{ [n] \ar[r]^{g} & [n] \ar[r]^{\phi}
\ar[rd]^{\phi^...
...^h & [m] \ar[r]^{\psi} & [r] \\ &&
[n] \ar[ru]_{h_*(\phi)} &&}
\end{displaymath}

Analogously, suppose we have two subgroups $A,B<G$ such that every element of $G$ can be written uniquely as $g=ab$, $a\in A$, $b\in B$. In this situation

\begin{displaymath}
gg'=aba'b'=a\underbrace{b^*(a')}_{\in A} \underbrace{a_*'(b)}_{\in
B}b'
\end{displaymath}

The relations satisfied by $\phi^*$ and $h_*$ are exactly the same as the relations satisfied by $b^*\: A\to A$ and $a_*\: B\to B$.


\begin{remark}
% latex2html id marker 340There is a way of constructing a cate...
...in{displaymath}
\Dl\times \bZ \to \Dl\bZ \to \Dl C
\end{displaymath}\end{remark}
We can ask what kind of structure on the geometric realization of the underlying simplicial set $X_{\bullet}$, that is $\vert X_{\bullet}\vert$, does the cyclic structure give? The answer is a structure of $S^1$-space. An open question is can we discretize analogously $S^3=\SU(2)$?


next up previous contents
Next: Noncommutative sets Up: Cyclic category Previous: Fibrations   Contents
Pawel Witkowski 2006-11-07