text:
\begin{document}
\title{Hyperbolicity of T(6) Cyclically Presented Groups}
\author{Ihechukwu Chinyere and Gerald Williams\thanks{This work was supported by the Leverhulme Trust Research Project Grant RPG-2017-334. Part of the work was carried out during a research visit funded by Leverhulme Trust Emeritus Fellowship EM-2018-023$\backslash$9.}}

\maketitle

\begin{abstract}
We consider groups defined by cyclic presentations where the defining word has length three and the cyclic presentation satisfies the T(6) small cancellation condition. We classify when these groups are hyperbolic. When combined with known results, this completely classifies the hyperbolic T(6) cyclically presented groups.
\end{abstract}

\noindent \textbf{Keywords:} cyclically presented group, hyperbolic group, small cancellation theory.\newline
\noindent \textbf{MSC:} 20F05, 20F06, 20F67.

\section{Introduction}\label{sec:intro}

Groups defined by presentations that satisfy the $C(p)-T(q)$ (non-metric) small cancellation conditions where $1/p+1/q<1/2$ are hyperbolic (\cite[Corollary~3.3]{GerstenShortI}). Therefore the cases $(p,q)=(3,6),(4,4),(6,3)$ present boundary cases and here both hyperbolic and non-hyperbolic groups can arise. For these cases,
in~\cite[Corollary, page~1860]{IvanovSchupp} the $C(p)-T(q)$ presentations that define hyperbolic groups are characterised as those for which there is no minimal flat over the presentation. In this article we consider groups defined by a class of presentations that admit a certain cyclic symmetry and satisfy $C(3)-T(6)$. We classify when the corresponding groups are hyperbolic in terms of the defining parameters of the presentations.

The \em cyclically presented group \em $G_n(w)$ is the group defined by the \em cyclic presentation \em
\[ P_n(w) = \pres{x_0,\ldots, x_{n-1}}{w,\theta(w),\ldots, \theta^{n-1}(w)}\]

where $w(x_0,\ldots ,x_{n-1})$ is a cyclically reduced word in the free group $F_n$ (of length $l(w)$) with generators $x_0,\ldots ,x_{n-1}$ and $\theta : F_n\rightarrow F_n$ is the \em shift automorphism \em of $F_n$ given by $\theta(x_i)=x_{i+1}$ for each $0\leq i<n$ (subscripts mod~$n$, $n\geq 2$).

If a presentation satisfies T(6) then, as observed by Pride (\cite[Section~5]{PrideStarComplexes},\cite[Lemma~3.1]{GerstenShortI}), every piece has length~1 and so if it satisfies T(6) then it satisfies C($l(w)$)-T(6). Thus if $l(w)>3$ then the presentation $P_n(w)$ satisfies C(4)-T(6) and hence $G_n(w)$ is hyperbolic by~\cite[Corollary~4.1]{GerstenShortI} and therefore (with a few exceptions when $n=2$) it is non-elementary hyperbolic by~\cite{Collins73}. If the length $l(w)=1$ then $G_n(w)$ is trivial, and if $l(w)=2$ then $G_n(w)$ is the free product of copies of $\Z$ or $\Z_2$.
Therefore we must consider the case $l(w)=3$ (in which case the $C(3)-T(6)$ condition coincides with the $T(6)$ condition). If $w$ is a positive (or negative) word, then we may assume $w=x_0x_kx_{l}$, and if $w$ is non-positive (and non-negative) then we may assume $w=x_0x_mx_{k}^{-1}$. Our main results consider these cases.

The groups $G_n(x_0x_mx_{k}^{-1})$ are known as the \em groups of Fibonacci-type \em and were introduced independently in~\cite{JohnsonMawdesley} and~\cite{CHR}, for algebraic and topological reasons, respectively. This family of groups contains the \em Fibonacci groups \em $F(2,n)=G_n(x_0x_1x_2^{-1})$ of~\cite{Conway65}, the \em Sieradski groups \em $S(2,n)=G_n(x_0x_2x_1^{-1})$ of~\cite{Sieradski}, and the \em Gilbert-Howie groups \em $H(n,m)=G_n(x_0x_mx_1^{-1})$ of~\cite{GilbertHowie}. They have been subsequently studied in~\cite{BardakovVesnin},\cite{W-CHR},\cite{COS},\cite{HowieWilliams},\cite{HowieWilliamsMFD} -- see \cite{Wsurvey} for a survey. In particular, the T(6) and T(7) presentations $P_n(x_0x_mx_{k}^{-1})$ were classified in~\cite[Theorem~10]{HowieWilliams} (see Corollary~\ref{cor:T6T7nonpositive}, below) and \cite[Theorem~11]{HowieWilliams} records that in the T(7) case the group $G_n(x_0x_mx_{k}^{-1})$ is non-elementary hyperbolic. The groups $G_n(x_0x_kx_{l})$ were introduced in~\cite{CRS}, and studied further in~\cite{EdjvetWilliams},\cite{BogleyWilliamsCoherence},\cite{MohamedWilliams}. The T(6) presentations $P_n(x_0x_kx_l)$ were classified in~\cite[Lemma~5.1]{EdjvetWilliams} (see Lemma~\ref{lem:positiveT6}, below). Moreover, Theorem~3.7 of~\cite{MohamedWilliams} shows that for all but finitely many $n$ the T(6) groups $G_n(x_0x_kx_l)$ are hyperbolic.

Our main results are as follows:

\begin{maintheorem}\label{thm:positive}
Let $n\geq 2$, $0\leq k,l<n$, $(n,k,l)=1$ and suppose that the cyclic presentation $P_n(x_0x_kx_{l})$ is T(6) and let $G=G_n(x_0x_kx_l)$.
If $n=7$ or $8$ or

\begin{itemize}
  \item[(a)] $n=21$ and ($l\equiv 5k$ or $k\equiv 5l$~mod~$n$); or
  \item[(b)] $n=24$ and ($l\equiv 5k$ or $k\equiv -4l$ or $l\equiv -4k$~mod~$n$);
\end{itemize}
then $G$  is not hyperbolic; otherwise $G$ is non-elementary hyperbolic.
\end{maintheorem}

\begin{maintheorem}\label{thm:nonpositive}
Let $n\geq 2$, $0\leq m,k<n$, $(n,m,k)=1$, $m\neq k$, $k\neq 0$, and suppose that the cyclic presentation $P_n(x_0x_mx_k^{-1})$ is T(6) and let $G=G_n(x_0x_mx_k^{-1})$. If $n=8$ or ($n\geq 12$ is even and $2(2k-m)\equiv 0$~mod~$n$) then $G$ is not hyperbolic; otherwise $G$ is non-elementary hyperbolic.
\end{maintheorem}

The coprimality hypotheses of Theorems~\ref{thm:positive},\ref{thm:nonpositive} are imposed to avoid the group decomposing as a free product in a canonical way. Specifically, if $d=(n,k,l)>1$ then $G_n(x_0x_kx_{l})$ decomposes as a free product of $d$ copies of $G_{n/d}(x_0x_{k/d}x_{l/d})$ (\cite[Lemma~2.4]{CRS}) and if $d=(n,m,k)>1$ then $G_n(x_0x_mx_k^{-1})$ decomposes as the free product of $d$ copies of $G_{n/d}(x_0x_{m/d}x_{k/d}^{-1})$ (\cite[Lemma~1.2]{BardakovVesnin}). Since a free product $H*K$ is hyperbolic if and only if $H$ and $K$ are hyperbolic -- see, for example, \cite[Theorem~H]{BaumslagGerstenShapiroShort} -- there is no loss in generality in assuming that such free product decompositions do not arise. The conditions $m\neq k, k\neq 0$ are imposed in Theorem~\ref{thm:nonpositive} to ensure that the relators are cyclically reduced and as otherwise the group is trivial.

A consequence of Theorem~\ref{thm:positive} and the results of~\cite{EdjvetWilliams} (see also~\cite[Example~10, Propositions~7.7,7.8]{MohamedWilliams}) is that the hyperbolicity status of cyclically presented groups $G$ of the form $G_n(x_0x_kx_l)$ is known, except when $G$ is isomorphic to $G_n(x_0x_1x_{n/2-1})$ for even $n\geq 10$, $n\neq 12,18$.

We prove Theorem~\ref{thm:positive} in Section~\ref{sec:positivecase} and Theorem~\ref{thm:nonpositive} in Section~\ref{sec:nonpositivecase}.

\section{The positive case}\label{sec:positivecase}

Let $P=\pres{\mathcal{X}}{\mathcal{R}}$ be a group presentation such that each relator $r\in \mathcal{R}$ is a cyclically reduced word in the generators. Let $\tilde{\mathcal{R}}$ denote the symmetrized closure of $\mathcal{R}$; that is, the set of all cyclic permutations of elements in $\mathcal{R}\cup \mathcal{R}^{-1}$. The \em star graph \em of $P$ is the undirected graph with vertex set $\mathcal{X} \cup \mathcal{X}^{-1}$, and with an edge joining vertices $x,y$ for each word $xy^{-1}u$ in $\tilde{\mathcal{R}}$. These words occur in pairs: $xy^{-1}u\in \tilde{\mathcal{R}}$ implies that $yx^{-1}u^{-1}\in\tilde{\mathcal{R}}$. Such pairs are called \em inverse pairs \em and the two corresponding edges are identified in $\Gamma$~\cite[page~61]{LyndonSchupp}. Thus if $\Gamma$ is the star graph of the cyclic presentation $P_n(x_0x_kx_l)$ then $\Gamma$ has vertices $x_i,x_i^{-1}$ and edges $x_i-x_{i+k}^{-1}$, $x_i-x_{i+l-k}^{-1}$, $x_i-x_{i-l}^{-1}$ ($0\leq i<n$), which we will refer to as edges of type $X,Y,Z$, respectively.

By~\cite{HillPrideVella} a presentation in which each relator has length at least 3 satisfies T($q$) if and only if its star graph has no cycle of length less than~$q$. As we are interested in presentations that satisfy T(6), in Section~\ref{sec:positiveshortcycles} we analyze cycles of length at most 6 in the star graph $\Gamma$ of $P_n(x_0x_kx_l)$. In particular, we note that $\Gamma$ always contains a cycle of length at most 6. We show that if two additional cycle types of length 6 arise then only a few small values of $n$ are possible and $G_n(x_0x_kx_l)$ is isomorphic to one of only a few groups, one of which turns out to be hyperbolic. In Section~\ref{sec:nonhyperbolicgroupspositive} we prove that the remainder are not hyperbolic. In Section~\ref{sec:curvaturepositive} we consider the case when at most one further cycle type of length 6 occurs and perform a detailed analysis of van Kampen diagrams (see~\cite[Chapter~5]{LyndonSchupp}) over the defining presentation to prove that  $G_n(x_0x_kx_l)$ has a linear isoperimetric function, and hence is hyperbolic. We then combine these results to prove Theorem~\ref{thm:positive} in Section~\ref{sec:proofofpositive}.

\begin{table}
\begin{center}
\begin{tabular}{|cr|lll|}\hline
& $j$ & \textbf{0} & \textbf{1} & \textbf{2}\\\hline

$(B.j)$ & congruence &$2k-l\equiv0$ & $2l-k\equiv0$ & $k+l\equiv0$\\
& cycle type& $XY$ & $YZ$ &$ZX$\\\hline

$(C.j)$ & congruence & $l\equiv \pm \frac{n}{3}$ & $k\equiv \pm \frac{n}{3}$ & $k-l\equiv \pm \frac{n}{3}$\\
& cycle type& $XZYZ$  & $YXZX$ & $ZYXY$\\\hline

$(D.j)$ & congruence & $2k-l\equiv \frac{n}{2}$ & $2l-k\equiv \frac{n}{2}$ & $k+l\equiv \frac{n}{2}$\\
& cycle type& $(XY)^2$ or & $(YZ)^2$ or&$(ZX)^2$ or\\
&                 & $XYZYXZ$ & $YZXZYX$ & $ZXYXZY$\\\hline

$(E.j)$ & congruence & $2k-l\equiv \pm \frac{n}{3}$ & $2l-k\equiv \pm \frac{n}{3}$ & $k+l\equiv \pm \frac{n}{3}$\\
& cycle type & $(XY)^3$ & $(YZ)^3$ & $(ZX)^3$\\\hline

$(F1.j)$ & congruence & $5k-l\equiv0$ & $5l-4k\equiv0$ & $k+4l\equiv0$\\
 & cycle type& $(XY)^2XZ$ & $(YZ)^2YX$ & $(ZX)^2ZY$\\\hline

$(F2.j)$ & congruence & $4l-5k\equiv0$ & $4k+l\equiv0$ & $5l-k\equiv0$\\
 & cycle type& $(YX)^2YZ$ & $(XZ)^2XY$ & $(ZY)^2ZX$\\\hline

\end{tabular}
\end{center}
\caption{Congruences (mod~$n$) corresponding to short cycles in the star graph of $P_n(x_0x_kx_l)$.\label{tab:BCDEFG}}
\end{table}

\subsection{Analysis of short cycles in the star graph of $P_n(x_0x_kx_l)$}\label{sec:positiveshortcycles}

The following classification of the T(6) cyclic presentations $P_n(x_0x_kx_l)$ in terms of three types of congruences (B),(C),(D) was obtained in~\cite{EdjvetWilliams}. As indicated in Table~\ref{tab:BCDEFG}, the (B) conditions correspond to cycles (of length 2) of the form $XY$,$YZ$,$ZX$; the (C) conditions correspond to cycles (of length 4) of the form $XZYZ$, $YXZX$, $ZYXY$; and the (D) conditions correspond to cycles (of length 4) of the form $(XY)^2$, $(YZ)^2$, $(ZX)^2$, as well as to cycles (of length 6) of the form $XYZYXZ$, $YZXZYX$, $ZXYXZY$. Replacing parameter $k$ by $l-k$ and $l$ by $-k$ corresponds to replacing edge type $X$ by $Y$, $Y$ by $Z$, and $Z$ by $X$ and to replacing a condition $(*.j)$ of Table~\ref{tab:BCDEFG} by $(*.j+1)$ (mod~$3$), and replacing the group $G_n(x_0x_kx_l)$ by the isomorphic copy $G_n(x_0x_{l-k}x_{-k})$. (To see that $G_n(x_0x_kx_l)\cong G_n(x_0x_{l-k}x_{-k})$ set $j=i+k$ in the relators $x_ix_{i+k}x_{i+l}$ of $P_n(x_0x_kx_l)$ and then cyclically permute to get the relators $x_jx_{j+l-k}x_{j-k}$ of $P_n(x_0x_{l-k}x_{-k})$.) Replacing parameter $k$ by $l-k$ corresponds to interchanging the roles of edge types $X$ and $Y$ and so interchanging the roles of conditions $(F1.j)$ and $(F2.j)$, and replacing the group $G_n(x_0x_kx_l)$ by the isomorphic copy $G_n(x_0x_{l-k}x_l)$. (To see that $G_n(x_0x_kx_l)\cong G_n(x_0x_{l-k}x_l)$ replace the generators $x_i$ by $x_i^{-1}$, negate the subscripts, and set $j=-i-l$ in the relators $x_ix_{i+k}x_{i+l}$ and then invert to get the relators $x_jx_{j+l-k}x_{j+l}$ of $P_n(x_0x_{l-k}x_l)$.)

\begin{lemma}[{\cite[Lemma~5.1]{EdjvetWilliams}}]\label{lem:positiveT6}
Let $n\geq 2$ and suppose that $(n,k,l)=1$, $0\leq k,l<n$. Then $P_n(x_0x_kx_l)$ satisfies T(6) if and only if none of the congruences $(B.j)$, $(C.j)$, $(D.j)$ ($0\leq j\leq 2$)  of Table~\ref{tab:BCDEFG} hold.
\end{lemma}

\begin{observation}[{See~\cite[Theorem~3.4]{MohamedWilliams}}]\label{observation:XYZXYZcycle}
Suppose that $(n,k,l)=1$, $0\leq k,l<n$ and that none of the congruences $(B.j)$,$(C.j)$,$(D.j)$ ($0\leq j\leq 2$) of Table~\ref{tab:BCDEFG} hold. Then for each $0\leq i<n$ the sequence of vertices and edges $x_i-x_{i+k}^{-1}-x_{i+2k-l}-x_{i+2k-2l}^{-1}-x_{i+k-2l}-x_{i-l}^{-1}-x_i$ forms a cycle of length~6 in the star graph $\Gamma$ of the form $(XYZ)^2$.
\end{observation}

We now consider how cycles of length $6$ can arise in $\Gamma$. It follows, using~\cite[Lemma~2.1]{EdjvetWilliams}, that when $n=7$ or $8$, if the cyclic presentation satisfies $T(6)$ then $G_n(x_0x_kx_l)\cong G_n(x_0x_1x_5)$.

\begin{lemma}\label{lem:6cycle}
Let $n\geq 2$ and suppose that $(n,k,l)=1$, $0\leq k,l<n$ and that none of the congruences $(B.j)$,$(C.j)$,$(D.j)$ ($0\leq j\leq 2$) of Table~\ref{tab:BCDEFG} hold.
Then the star graph $\Gamma$ contains a cycle of length 6 of cycle type other than $(XYZ)^2$ if and only if at least one of the congruences $(E.j)$,$(F1.j)$,$(F2.j)$ ($0\leq j\leq 2$) of Table~\ref{tab:BCDEFG} holds, in which case the corresponding entry of the table is a label of the cycle.
\end{lemma}

\begin{proof}
Let $C$ be a cycle of length 6 in $\Gamma$. Then there are no subpaths of $C$ of the form $XX$, $YY$ or $ZZ$. If $C$ involves each of the edge types $X,Y,Z$ twice then $C$ is a cycle of the form $(XYZ)^2$, $XYZYXZ$, $YZXZYX$, or $ZXYXZY$. But these last three cycles only occur if the congruence $(D.j)$ holds, contrary to hypothesis.

If $C$ does not involve an edge of type $X$ (resp.\,$Y$, resp.\,$Z$) then $C$ is a cycle of the form $(YZ)^3$ (resp.\,$(XZ)^3$, resp.\,$(XY)^3$), which correspond to the conditions $(E.j)$.

If $C$ involves exactly one edge of type $X$ (resp.\,$Y$, resp.\,$Z$) then $C$ is a cycle of the form $(YZ)^2YX$ or $(ZY)^2ZX$ (resp.\,$(ZX)^2ZY$ or $(XZ)^2XY$, resp.\,$(XY)^2XZ$ or $(YX)^2YZ$), which correspond to the conditions $(F1.j)$ or $(F2.j)$.

Conversely, if any of the congruences $(E.j),(F1.j), (F2.j)$ hold then the corresponding entry of Table~\ref{tab:BCDEFG} is the label of a cycle of length~6 in $\Gamma$.
\end{proof}

\begin{lemma}\label{lem:combinationsofEFG}
Let $n\geq 2$ and suppose that $(n,k,l)=1$, $0\leq k,l<n$ and that none of the congruences $(B.j)$,$(C.j)$,$(D.j)$ ($0\leq j\leq 2$) hold. If more than one of the congruences $(E.j)$,$(F1.j)$, $(F2.j)$ ($0\leq j\leq 2$) hold then one of the following holds:

\begin{itemize}
\item[(a)] $n=7$ and ($l\equiv 5k$ or $k\equiv 5l$~mod~$n$);
\item[(b)] $n=8$ and ($l\equiv 5k$ or $k\equiv 5l$~mod~$n$);
\item[(c)] $n=21$ and ($l\equiv 5k$ or $k\equiv 5l$~mod~$n$);
\item[(d)] $n=24$ and ($l\equiv 5k$ or $k\equiv -4l$ or $l\equiv -4k$~mod~$n$);
\item[(e)] $n=27$ and ($l\equiv 5k$ or $k\equiv 5l$ or $4k\equiv 5l$ or $4l\equiv 5k$ or $k\equiv -4l$ or $l\equiv -4k$~mod~$n$).
\end{itemize}
In each case $G\cong G_n(x_0x_1x_5)$.
\end{lemma}

\begin{proof}
(Throughout this proof, the $j$ value in a condition $(*.j)$ is to be taken mod~$3$.)

If $(E.j)$ and $(F1.j)$ hold then $(B.j+1)$ holds, a contradiction.

If $(E.j)$ and $(F1.j+1)$ hold then $(D.j+2)$ holds.

If $(E.j)$ and $(F2.-j)$ hold then $(B.j+2)$ holds.

If $(E.j)$ and $(F2.1-j)$ hold then $(D.j+1)$ holds.

If $(F1.j)$ and $(F2.-j)$ hold then $(C.j)$ holds.

If $(F1.j)$ and $(F2.1-j)$ hold then $(B.j+1)$ holds.

Suppose now that any two of the $(E.j)$ conditions hold; then all three of them hold. Since $(B.0)$ does not hold condition $(E.0)$ implies $2k-l\equiv \pm n/3$~mod~$n$ and since $(B.2)$ does not hold condition $(E.2)$ implies $k+l\equiv \pm n/3$~mod~$n$. Thus $2k-l\equiv \epsilon (k+l)$ where $\epsilon =\pm 1$. If $\epsilon =+1$ then $(B.1)$ holds, a contradiction; and if $\epsilon =-1$ then $(C.1)$ holds, a contradiction.

Suppose that two of the $(F1.j)$ conditions hold. Then all of them hold so, in particular, $l\equiv 5k$~mod~$n$. Summing the congruences $(F1.0)$ and $(F1.1)$ gives that $k\equiv -4l$~mod~$n$ and so (by $(F1.0)$) $21l\equiv 0$~mod~$n$. Moreover $1=(n,k,l)=(n,-4l,l)=(n,l)$ so $n|21$. If $n=3$ then $(F1.0)$ implies that $(B.0)$ holds, so $n=7$ or $21$. An analogous argument shows that if two of the $(F2.j)$ conditions hold then $k\equiv 5l$ and $n=7$ or $21$, thus giving cases (a),(c) of the statement.

Suppose $(F1.j)$ and $(F2.2-j)$ hold. We claim that $n=8$ or $24$; it then follows from one of the congruences that $l\equiv 5k$ or $k\equiv 5l$~mod~$n$ (by multiplying by 5, if necessary), giving cases (b),(d). We prove this in the case $(F1.0)$ and $(F2.2)$, the other cases being similar. The congruence $(F1.0)$ implies $l\equiv 5k$~mod~$n$, so substituting into $(F2.2)$ gives $24k\equiv 0$~mod~$n$, but $1=(n,k,l)=(n,k)$ so $n|24$. If $n\leq 6$ then some condition $(B.j),(C.j),(D.j)$ holds, and if $n=12$ then $(D.2)$ holds, a contradiction; thus $n=8$ or $24$, as claimed.

Suppose that either ($(E.j)$ and $(F1.j+2)$) or ($(E.j)$ and $(F2.2-j)$) hold. We claim that $n=27$. We prove this in the case where $(E.0)$ and $(F1.2)$ hold, the other cases being proved analogously. The congruence $(F1.2)$ implies $k\equiv -4l$~mod~$n$ so $(E.0)$ implies $27l\equiv 0$~mod~$n$, but $1=(n,k,l)=(n,l)$ so $n|27$. If $n=3$ or $9$ then $(B.0)$ holds, and hence $n=27$, as claimed.

The final assertion that $G_n(x_0x_kx_l)\cong G_n(x_0x_1x_5)$ in each case follows from~\cite[Lemma~2.1]{EdjvetWilliams}.
\end{proof}

We now deal with the group arising in case~(e) of Lemma~\ref{lem:combinationsofEFG}.

\begin{example}[The group $G_{27}(x_0x_1x_5)$]\label{ex:n=27}
\em Using KBMAG~\cite{KBMAG} it is straightforward to show that the group $G_{27}(x_0x_1x_5)$ is hyperbolic, and since it contains a non-abelian free subgroup (by~\cite[Corollary~5.2]{EdjvetWilliams}), it is non-elementary hyperbolic.\em
\end{example}

\subsection{Non-hyperbolic groups $G_n(x_0x_kx_l)$}\label{sec:nonhyperbolicgroupspositive}

In this section we consider the groups arising in cases (a)-(d) of Lemma~\ref{lem:combinationsofEFG}. First we recall that the group $G_7(x_0x_1x_5)$ is not hyperbolic; see~\cite[Example~3.8]{MohamedWilliams} for a discussion.

\begin{lemma}[{\cite{EdjvetVdovina},\cite{Cartwrightetal}}]\label{lem:n=7nonhyp}
The group $G_7(x_0x_1x_5)$ is not hyperbolic.
\end{lemma}

We now show that the group $G_8(x_0x_1x_5)$ is not hyperbolic. We do this by an application of the Flat Plane Theorem~\cite{Bridson} (an alternative approach would be to use~\cite[Corollary, page 1860]{IvanovSchupp}).

\begin{lemma}\label{lem:n=8nonhyp}
The group $G_8(x_0x_1x_5)$ is not hyperbolic.
\end{lemma}

\begin{proof}
Since the presentation $P_8(x_0x_1x_5)$ satisfies C(3)-T(6) and each relator has length~3, each face in the geometric realisation $\tilde{C}$ of the Cayley complex of $P$ (obtained by assigning length~1 to each edge) is an equilateral triangle, and so $\tilde{C}$ satisfies the CAT(0) inequality. Consider the geometric realisation $\Delta_0$ of the reduced van Kampen diagram given in Figure~\ref{fig:positiveflat} and for each $0\leq i<n$ let $\Delta_i$ be obtained from $\Delta_0$ by applying the shift $\theta^i$ to each edge. Then placing $\Delta_0,\Delta_2,\Delta_4,\Delta_6$ one above the other gives the geometric realisation $\Delta$ of a reduced van Kampen diagram. Copies of $\Delta$ tile the Euclidean plane without cancellation of faces. Thus there is an isometric embedding of the Euclidean plane in $\tilde{C}$, and so the result follows from the Corollary to Theorem~A in~\cite{Bridson}.
\end{proof}

\begin{figure}
\begin{center}

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\end{center}
  \caption{A van Kampen diagram over the presentation $P_8(x_0x_{1}x_5)$ with boundary label $(x_2x_0)(x_3x_5x_7x_1)(x_2x_0)^{-1}(x_1x_3x_5x_7)^{-1}$.\label{fig:positiveflat}}
\end{figure}

For later reference (in Section~\ref{sec:nonpositivecase}) we note that the relabelling of generators $y_0=x_0, y_1=x_7^{-1}, y_2=x_2, y_3=x_1^{-1}, y_4=x_4, y_5=x_3^{-1}, y_6=x_6, y_7=x_5^{-1}$ shows that $G_8(x_0x_1x_5)\cong G_8(y_0y_4y_1^{-1})$, and so we have:

\begin{corollary}\label{cor:G8nonposnothyp}
The group $H(8,4)=G_8(x_0x_4x_1^{-1})$ is not hyperbolic.
\end{corollary}

\begin{remark}\label{rk:Z+Zin(n,n/2+2)}
\em The van Kampen diagram arising in the proof of Lemma~\ref{lem:n=8nonhyp}, and later the one in the proof of Lemma~\ref{lem:H(n,n/2+2)nothyp}, provides a pair of commuting elements whose axes in the geometric realisation $\Delta$ meet at an angle $2\pi/3$. It follows that the groups considered in these results contain a free abelian subgroup of rank~2 (see, for example, \cite[page~446]{WiseApproximating}).\em
\end{remark}

In Corollaries~\ref{cor:n=21nonhyp},\ref{cor:n=24nonhyp}  we use Lemmas~\ref{lem:n=7nonhyp},\ref{lem:n=8nonhyp} respectively to prove that the groups in cases~(c),(d) of Lemma~\ref{lem:combinationsofEFG} are not hyperbolic. To do this we first recall the shift extension of a cyclically presented group. The shift automorphism $\theta$ of a cyclically presented group $G_n(w)$ has exponent $n$ and the resulting $\Z_n$-action on $G_n(w)$ determines the \em shift extension \em $E_n(w) = G_n(w) \rtimes_\theta \Z_n$, which admits a two-generator two-relator presentation of the form

\[E_n(W)=\pres{x,t}{t^n,W(x,t)} \]

where $W=W(x, t)$ is obtained by rewriting $w$ in terms of the substitutions $x_i=t^ixt^{-i}$ (see, for example, \cite[Theorem~4]{JWW}). Thus there is a retraction $\nu^0: E_n(W) \rightarrow \Z_n$ given by $\nu^0(t)=t$, $\nu^0(x)=t^0=1$ with kernel $G_n(w)$. Moreover, as shown in~\cite[Section~2]{BogleyShift}, for certain values of $f$ ($0\leq f<n$) there may be further retractions $\nu^f$. Specifically, by~\cite[Theorem~2.3]{BogleyShift} the kernel of a retraction $\nu^f:E_n(W) \rightarrow \Z_n$ given by $\nu^f(t)=t$, $\nu^f(x)=t^f$ is cyclically presented, generated by the elements $y_i=t^ixt^{-(i+f)}$ ($0\leq i<n$). Since (non-elementary) hyperbolicity is preserved under taking finite index subgroups and finite extensions, the group $E_n(W)$ is (non-elementary) hyperbolic if and only if the kernel of any, and hence all, of its retractions $\nu^f$ is (non-elementary) hyperbolic.

In the case $w=x_0x_kx_{l}$ we have

\begin{alignat*}{1}
E_n(W)=G_n(w) \rtimes_\theta \pres{t}{t^n}=\pres{x,t}{t^n, xt^{k}xt^{l-k}xt^{-l}}\label{eq:positiveextension}
\end{alignat*}

which admits a retraction $\nu^f:E_n\rightarrow \pres{t}{t^n}$ given by $\nu^f(t)=t, \nu^f(x)=t^f$ if and only if $3f\equiv 0$~mod~$n$; the kernel of such a retraction is the cyclically presented group $G_n(x_0x_{f+k}x_{2f+l})$ (see~\cite[page~158]{BogleyShift}).

\begin{corollary}\label{cor:n=21nonhyp}
The group $G_{21}(x_0x_1x_5)$ is not hyperbolic.
\end{corollary}

\begin{proof}
The free product of three copies of $G_7(x_0x_1x_5)$ is the cyclically presented group $G_{21}(x_0x_3x_{15})$  with shift extension $E=\pres{x,t}{t^{21}, xt^3xt^{12}xt^{-15}}$. The kernel of the retraction $\nu^7:E\rightarrow \Z_n=\pres{t}{t^{21}}$ given by $\nu^7(t)=t$, $\nu^7(x)=t^7$ is the group $G_{21}(x_0x_{10}x_8)$ which, by~\cite[Lemma~2.1(iv),(v)]{EdjvetWilliams}, is isomorphic to $G_{21}(x_0x_1x_5)$. Since $G_7(x_0x_1x_5)$ is not hyperbolic, neither is $G_{21}(x_0x_3x_{15})$, nor $E$, and hence, nor is $G_{21}(x_0x_1x_5)$.
\end{proof}

\begin{corollary}\label{cor:n=24nonhyp}
The group $G_{24}(x_0x_1x_5)$ is not hyperbolic.
\end{corollary}

\begin{proof}
The free product of three copies of $G_8(x_0x_1x_5)$ is the cyclically presented group $G_{24}(x_0x_3x_{15})$ with shift extension $E=\pres{x,t}{t^{24}, xt^3xt^{12}xt^{-15}}$. The kernel of the retraction $\nu^8:E\rightarrow \Z_{24}=\pres{t}{t^{24}}$ given by $\nu^8(t)=t$, $\nu^8(x)=t^8$ is the group $G_{24}(x_0x_{11}x_7)$ which, by~\cite[Lemma~2.1(v),(ii)]{EdjvetWilliams}, is isomorphic to $G_{24}(x_0x_1x_5)$. Since $G_8(x_0x_1x_3)$ is not hyperbolic, neither is $G_{24}(x_0x_3x_{15})$, nor $E$, and hence, nor is $G_{24}(x_0x_1x_5)$.
\end{proof}

\subsection{Analysis of van Kampen diagrams over $P_n(x_0x_kx_l)$}\label{sec:curvaturepositive}

In this section we show that if the cyclic presentation $P=P_n(x_0x_kx_l)$ is $T(6)$ and at most one of the congruences $(E.j),(F1.j)$, $(F2.j)$ hold then $G=G_n(x_0x_kx_l)$ is hyperbolic. Following the method of proof of~\cite[Theorem~13]{HowieWilliams} we show that $G$ has a linear isoperimetric function~\cite[Theorem~3.1]{GerstenIsoperimetric}. That is, we show that there is a linear function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for all $N\in \mathbb{N}$ and all freely reduced words $W\in F_n$ with length at most $N$ that represent the identity of $G$ we have $\mathrm{Area}(W)\leq f(N)$, where $\mathrm{Area} (W)$ denotes the minimum number of faces in a reduced van Kampen diagram over $P$ with boundary label $W$. Without loss of generality we may assume that the boundary of such a van Kampen diagram $D$ is a simple closed curve. Note that each face in $D$ is a triangle, as shown in Figure~\ref{fig:positiveface}, where the corner labels $X,Y,Z$ correspond to the edge types of the star graph of $P$. We say that a vertex of $D$ is a \em boundary vertex \em if it lies on $\partial D$, and is an \em interior vertex \em otherwise. In order to obtain a linear isoperimetric function (in Lemma~\ref{lem:curvaturepositivecase}) we first carefully analyse degrees of vertices within $D$.

\begin{figure}
\begin{center}

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\rput[tl](11.8,15.7){$X$}
\rput[tl](10.6,14.4){$Y$}
\rput[tl](13.2,14.4){$Z$}
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  \caption{A typical face in a van Kampen diagram over the presentation $P_n(x_0x_kx_l)$.\label{fig:positiveface}}
\end{figure}

\begin{figure}

\begin{center}

\begin{tabular}{c@{}c}
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&

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\end{center}
  \caption{Possible configurations when two vertices have label $(XYZ)^2$.\label{fig:ICFig2}}
\end{figure}

\begin{lemma}\label{lem:twolabelled(XYZ)^2}
Let $\Delta$ be an interior face of $D$ in which two of the vertices have label $(XYZ)^2$. Then the label of the third vertex contains a subword of the form $aba$, where $b$ is the label of the corner of $\Delta$ at this vertex, and $a,b \in \{X,Y,Z\}$, $a\neq b$.
\end{lemma}

\begin{proof}
Without loss of generality we may assume that the edges of $\Delta$ are oriented in an anticlockwise manner. We name its vertices $v_1,v_2,v_3$, read in an anticlockwise manner and suppose $v_1,v_2$ are labelled $(XYZ)^2$. If the corner of $\Delta$ at $v_1$ has label $X$ (resp. $Y$, resp. $Z$) then the corner of $\Delta$ at $v_2$ has label $Y$ (resp. $Z$, resp. $X$), in which case the label of $v_3$ has a subword $YZY$ or $XZX$ (resp. $ZXZ$ or $YXY$, resp. $XYX$ or $ZYZ$), as shown in Figure~\ref{fig:ICFig2}.
\end{proof}

\begin{figure}
\begin{center}

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  \caption{Neighbourhood of an interior vertex labelled $(XYZ)^2$.\label{fig:ICFig3}}
\end{figure}

\begin{figure}[ht]
\begin{center}

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\end{scriptsize}
\end{pspicture*}

\end{center}
  \caption{Neighbourhood of an interior vertex labelled $(E.0)$ and no boundary neighbours.\label{fig:ICFig4}}
\end{figure}

\begin{lemma}\label{lem:INT(XYZ)^2twoneighbours}
If an interior vertex $v$ of $D$ of degree 6 has label $(XYZ)^2$ then two adjacent neighbours of $v$ have $XY$ as a cyclic subword of their labels, two adjacent neighbours have $XZ$ as a cyclic subword of their labels, and two adjacent neighbours have $YZ$ as a cyclic subword of their labels.
\end{lemma}

\begin{proof}
If the label of $v$ is $(XYZ)^2$ oriented clockwise, then the neighbourhood of $v$ is as given in Figure~\ref{fig:ICFig3}, from which the conclusion can be observed. A similar figure deals with the case when the label of $v$ is $(XYZ)^2$ oriented anticlockwise.
\end{proof}

\begin{lemma}\label{lem:label(E.j)}
Suppose that all interior vertices of $D$ have degree at least 6 and all labels of interior vertices of degree 6 are either $(XYZ)^2$ or label $(E.j)$ for precisely one $j\in \{0,1,2\}$.
If $v$ is an interior vertex of degree 6 in $D$ with label $(E.j)$ and where all the neighbours of $v$ are interior vertices of degree 6 then every neighbour of $v$ has two neighbours which are either boundary vertices or have degree at least~8.
\end{lemma}

\begin{proof}
Consider first the case $(E.0)$, that is, a vertex label $(XY)^3$. As shown in Figure~\ref{fig:ICFig4} all the neighbours of $v$ must have label $(XYZ)^2$. Then each of the vertices $u_1,\ldots ,u_6$ has a corner labelled $Z$. If a vertex $u_i$ ($1\leq i\leq 6$) is interior then if it is of degree 6, its label is not $(XYZ)^2$, by Lemma~\ref{lem:twolabelled(XYZ)^2}, and so it must be $(XY)^3$, a contradiction. Therefore $u_i$ is either interior of degree at least 8, or a boundary vertex, as required. The cases $(E.1),(E.2)$ are dealt with by replacing $X$ by $Y$, $Y$ by $Z$, and $Z$ by $X$, as explained in Section~\ref{sec:positiveshortcycles}.
\end{proof}

\begin{figure}[ht]
\begin{center}

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\rput[tl](7.4,10.6){$Y$}
\rput[tl](7.9,9.8){$Z$}
\rput[tl](7.3,9.7){$X$}
\rput[tl](6.8,9.8){$Y$}
\rput[tl](7.8,10.4){$X$}

\rput[tl](8.3,8.4){$Y$}

\rput[tl](5.9,8.6){$X$}
\rput[tl](6.4,8.4){$Z$}
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\rput[tl](5.9,11.6){$Z$}
\rput[tl](5.3,9.9){$Z$}
\rput[tl](8.8,11.6){$Y$}
\rput[tl](8.7,8.7){$X$}
\rput[tl](8.3,11.8){$Z$}

\rput[tl](10.2,10.1){$v_1$}
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\rput[tl](5.3,10.4){$Y$}

\rput[tl](8.8,12.5){$v_2$}

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\end{center}
  \caption{Neighbourhood of an interior vertex labelled $(F1.0)$.\label{fig:ICFig5}}
\end{figure}

\begin{lemma}\label{lem:label(F1F2.j)}
Suppose that all interior vertices of $D$ have degree at least 6 and all labels of interior vertices of degree 6 are either
$(XYZ)^2$ or $(F1.j)$ (resp. $(F2.j)$) for precisely one $j\in \{0,1,2\}$.
If $v$ is an interior vertex of degree 6 in $D$ with label $(F1.j)$ (resp. $(F2.j)$) and where all the neighbours of $v$ are interior vertices then $v$ has a neighbour of degree at least 8.
\end{lemma}

\begin{proof}
Consider the case $(F1.0)$, that is, $v$ has label $(XY)^2XZ$ and suppose that all neighbours of $v$ have degree~6. Then Figure~\ref{fig:ICFig5} shows one of the two possible labellings of neighbours that can occur. Since two adjacent neighbours have $YZ$ as a cyclic subword of their label, these must each be labelled $(XYZ)^2$, but this is impossible by Lemma~\ref{lem:twolabelled(XYZ)^2}; therefore $v$ has a neighbour of degree at least~8. The same conclusion can be obtained if the second possible labelling of neighbours occurs. The cases $(F1.1),(F1.2)$ are dealt with by replacing $X$ by $Y$, $Y$ by $Z$, and $Z$ by $X$. The cases $(F2.j)$ are obtained from the cases $(F1.j)$ by interchanging the roles of $X$ and $Y$, as described in Section~\ref{sec:positiveshortcycles}.
\end{proof}

We are now in a position to be able to establish the existence of a suitable isoperimetric function.

\begin{lemma}\label{lem:curvaturepositivecase}
Let $n\geq 2$ and suppose that none of the congruences $(B.j),(C.j),(D.j)$ hold ($0\leq j\leq 2)$ and that at most one congruence $(E.j)$, $(F1.j)$, $(F2.j)$ holds ($j\in \{0,1,2\}$). Then $G_n(x_0x_kx_{l})$ has a linear isoperimetric function.
\end{lemma}

\begin{proof}
As at the beginning of this section, let $N\in \mathbb{N}$ and let $W$ be a freely reduced word in the free group $F_n$ (with generators $x_0,\ldots ,x_{n-1})$ of length at most $N$ that represents the identity of $G$ and let $D$ be a reduced van Kampen diagram whose boundary $\partial D$ is a simple closed curve with label $W$. We let $I$ denote the set of interior vertices of $D$, $B$ denote the set of boundary vertices of $D$, and $F$ denote the set of faces of $D$. Then $\mathrm{Area}(W)\leq |F|$. Writing $\pi$ to denote $180$, we define the curvature of a face $f$ by $\kappa(f)=-\pi + (\mathrm{sum~of~angles~in}~f)$, the curvature of an interior vertex $v$ by $\kappa (v)=2\pi - (\mathrm{sum~of~angles~at}~v)$, and the curvature of a boundary vertex $\hat{v}$ by $\kappa(\hat{v})=\pi - (\mathrm{sum~of~angles~at}~\hat{v})$. It follows from the Gauss-Bonnet Theorem that

\begin{equation}
  \sum_{v\in I} \kappa(v) + \sum_{\hat{v}\in B} \kappa (\hat{v}) +\sum_{f\in F} \kappa(f)=2\pi. \label{eq:GaussBonnet}
\end{equation}

(see \cite[Section~4]{MccammondWise} and the references therein).

Since none of the congruences $(B.j),(C.j),(D.j)$ hold ($0\leq j\leq 2)$ every interior vertex of $D$ is of degree at least 6, and since at most one congruence $(E.j)$, $(F1.j)$, $(F2.j)$ holds, the label of an interior vertex of degree~6 is either $(XYZ)^2$ or it is the label corresponding to that congruence, given in Table~\ref{tab:BCDEFG}.

We assign angles to the corners of the faces in $D$ as follows. If $v$ is a boundary vertex or an interior vertex of degree at least 8 then assign angle 47 to every corner at $v$. Assume now that $v$ is an interior vertex of degree~6 and consider a face $f$ with vertices $v$ and $u,w$: if $u,w$ are interior of degree 6 then assign angle 59 to the corner of $f$ at $v$; otherwise assign 66 to the corner of $f$ at $v$.

Then, if a face $f$ contains a boundary vertex then $\kappa(f)\leq -\pi +(47+66+66)=-1$; if a face $f$ has all its vertices interior and one vertex of degree at least $8$ then $\kappa(f)\leq -\pi +(47+66+66)=-1$; if all the vertices of a face $f$ are interior of degree 6 then $\kappa(f)\leq -\pi+(59+59+59)=-3$. Therefore $\kappa(f)\leq -1$ for all $f\in F$.

We now consider curvature of the vertices. If $v$ is an interior vertex of degree at least 8 then $\kappa(v)\leq 2\pi -8(47) =-16$. If $v$ is an interior vertex of degree 6 with a neighbour that is either on the boundary $\partial D$ or has degree at least 8, then $\kappa(v)\leq 2\pi -2(66) -4(59)=-8$.

Now suppose that $v$ is an interior vertex of degree 6 with all its neighbours interior of degree~6. Then $\kappa(v)=2\pi -6(59)=6$ and by Lemma~\ref{lem:label(F1F2.j)} the label of $v$ is either $(XYZ)^2$ or $(E.j)$ for some $j\in \{0,1,2\}$. If the label of $v$ is $(XYZ)^2$ then, since precisely one other label of degree 6 vertices is possible, Lemma~\ref{lem:INT(XYZ)^2twoneighbours} implies that $v$ must have two adjacent neighbours, each labelled $(XYZ)^2$, but this is impossible by Lemma~\ref{lem:twolabelled(XYZ)^2}.
If the label of $v$ is $(E.j)$ (for some $j\in \{0,1,2\}$) then Lemma~\ref{lem:label(E.j)} implies that every neighbour $v_i$ ($1\leq i \leq 6$) of $v$ has two neighbours which are either boundary vertices or have degree at least 8. Therefore, for each $i\in\{1,\dots,6\}$ the curvature $\kappa (v_i)\leq 2\pi - 4(59)-2(66)=-8$. In this situation, transfer curvature of $-1$ from each vertex $v_i$ to vertex $v$; the resulting curvatures are $\kappa(v_i)\leq -8+1=-7$ ($1\leq i\leq 6$) and $\kappa(v)=6-6(1)=0$. Since each vertex $v_i$ has degree 6, the maximum number of times curvature can be transferred away from $v_i$ is 6, so its curvature cannot exceed $\kappa(v)=-8+6(1)=-2$. Therefore for each interior vertex $v$ we have $\kappa(v)\leq 0$.

Now~(\ref{eq:GaussBonnet}) implies

\begin{alignat*}{1}
  2\pi &= \sum_{v\in I} \kappa(v) + \sum_{\hat{v}\in B} \kappa (\hat{v}) +\sum_{f\in F} \kappa(f)\\
  &\leq \sum_{v\in I} 0 + \sum_{\hat{v}\in B} (\pi - \mathrm{sum~of~angles~at}~\hat{v}) +\sum_{f\in F} (-1)\\
  &= |B|\pi - \sum_{\hat{v}\in B} (\mathrm{sum~of~angles~at}~\hat{v}) - |F|
\end{alignat*}
 so
\[  \sum_{\hat{v}\in B} (\mathrm{sum~of~angles~at}~\hat{v}) \leq (|B|-2)\pi - |F|.\]

On the other hand, the corner angle at any boundary vertex is 47, so the sum of angles over the boundary vertices is bounded below by $47|B|$. Therefore $47|B|\leq (|B|-2)\pi - |F|$, so $|F| \leq 133|B| -360$.

But $\mathrm{Area}(W)\leq |F|$ and $|B|\leq N$ so $\mathrm{Area}(W)\leq 133N -360$, so $f(N)=133N-360$ is a linear isoperimetric function.
\end{proof}

We now have all the ingredients to prove Theorem~\ref{thm:positive}.

\subsection{Proof of Theorem~\ref{thm:positive}}\label{sec:proofofpositive}

Suppose that $n\geq 2$, $0\leq k,l<n$, $(n,k,l)=1$ and that the cyclic presentation $P_n(x_0x_kx_l)$ satisfies T(6). Then Lemma~\ref{lem:positiveT6} implies that none of the congruences $(B.j),(C.j),(D.j)$ ($0\leq j\leq 2)$ (of Table~\ref{tab:BCDEFG}) hold and so $n\geq 7$. If $n=7$ or $8$ then $G\cong G_n(x_0x_1x_5)$ (see, for example, \cite[Table~2]{MohamedWilliams}) so is not hyperbolic by Lemmas~\ref{lem:n=7nonhyp},\ref{lem:n=8nonhyp}. Assume then that $n>8$. If more than one of the congruences $(E.j),(F1.j)$, $(F2.j)$ ($0\leq j\leq 2)$ hold then one of the cases (c),(d),(e)  of Lemma~\ref{lem:combinationsofEFG} hold. In cases (c),(d) $G$ is not hyperbolic by Corollaries~\ref{cor:n=21nonhyp},\ref{cor:n=24nonhyp} and in case (e) $G$ is non-elementary hyperbolic, by Example~\ref{ex:n=27}. Thus we may assume that at most one of $(E.j),(F1.j)$, $(F2.j)$ ($0\leq j\leq 2$) hold, in which case Lemma~\ref{lem:curvaturepositivecase} implies that $G_n(x_0x_kx_l)$ has a linear isoperimetric function, and hence is hyperbolic.  By~\cite[Corollary~5.2]{EdjvetWilliams} $G$ contains a non-abelian free subgroup so it is non-elementary hyperbolic.

\section{The non-positive case}\label{sec:nonpositivecase}

As in~\cite{HowieWilliams}, we express our arguments in terms of parameters $A=k, B=k-m$.

Let $\Gamma$ be the star graph of the cyclic presentation $P_n(x_0x_mx_k^{-1})$. Then $\Gamma$ has vertices $x_i,x_i^{-1}$ and edges $x_i-x_{i+m}^{-1}$, $x_i-x_{i+B}$, $x_i^{-1}-x_{i+A}^{-1}$ ($0\leq i<n$), which we will refer to as edges of type $X,Y,Z$, respectively. Replacing parameter $k$ by $m-k$ corresponds to interchanging the roles of edges of types $Y$ and $Z$, and so will correspond to interchanging the roles of conditions $(*.0)$ and $(*.1)$ in Table~\ref{tab:nonposshortcycles}, and replacing the group $G_n(x_0x_mx_k^{-1})$ by the isomorphic copy $G_n(x_0x_mx_{m-k}^{-1})$. (To see that  $G_n(x_0x_mx_{k}^{-1})\cong G_n(x_0x_mx_{m-k}^{-1})$ replace generators $x_i$ by $x_i^{-1}$, invert the relators, negate the subscripts and set $j=-i-m$ to get the relators $x_jx_{j+m}x_{j+m-k}^{-1}$ of $G_n(x_0x_mx_{m-k}^{-1})$.)

As in the positive case, we are interested in cycles of length at most $6$ in $\Gamma$, so we analyze these in Section~\ref{sec:nonpositiveshortcycles}. We observe that if a particular cycle type of length~6 (which we refer to as $(\gamma+)$) occurs then $G=G_n(x_0x_mx_k^{-1})$ is isomorphic to $G_n(x_0x_{n/2+2}x_1^{-1})=H(n,n/2+2)$ which (in Section~\ref{sec:nonhyperbolicgroupsnonpositive}) we show is non-hyperbolic whenever its presentation satisfies $T(6)$. We then show that if two of the remaining cycle types of length~6 occur, then $G_n(x_0x_mx_k^{-1})$ is isomorphic to one of a few groups with low values of $n$, all but one of which turn out to be hyperbolic (the other, $G_8(x_0x_4x_1^{-1})=H(8,4)$, being non-hyperbolic). In Section~\ref{sec:curvaturenonpositive} we consider the case when exactly one cycle type of length~6 occurs and perform a detailed analysis of van Kampen diagrams over the defining presentation to prove that $G_n(x_0x_mx_k^{-1})$ has a linear isoperimetric function, and hence is hyperbolic. We then combine these results to prove Theorem~\ref{thm:nonpositive} in Section~\ref{sec:proofofnonpositive}.

\subsection{Analysis of short cycles in the star graph of $P_n(x_0x_mx_k^{-1})$}\label{sec:nonpositiveshortcycles}

Short cycles  in $\Gamma$ were analyzed in~\cite{HowieWilliams}:

\begin{lemma}[{\cite[Theorem~10]{HowieWilliams}}]\label{lem:nonposshortcycles}
Let $n\geq 2$, $0\leq m,k<n$, $m\neq k$, $k\neq 0$ and set $A=k,B=k-m$, and let $\Gamma$ be the star graph of $P_n(x_0x_mx_k^{-1})$.
\begin{itemize}
  \item[(a)] $\Gamma$ has a cycle of length less than $6$ if and only if at least one of the congruences $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$ of Table~\ref{tab:nonposshortcycles} holds, in which case a label of the cycle is the corresponding entry of the table.

  \item[(b)] $\Gamma$ has a cycle  of length 6 if and only if at least one of the congruences ($\alpha.j$),($\beta+.j$),($\beta-.j$), ($\gamma+$),($\gamma-$) of Table~\ref{tab:nonposshortcycles} holds, in which case a label of the cycle  is the corresponding entry of the table.
\end{itemize}
\end{lemma}

\begin{corollary}\label{cor:T6T7nonpositive}
Let $n\geq 2$ and suppose $(n,m,k)=1$, $0\leq m,k<n$, $m\neq k$, $k\neq 0$. Then
\begin{itemize}
  \item[(a)] $P_n(x_0x_mx_k^{-1})$ satisfies $T(6)$ if and only if none of the congruences $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$ of Table~\ref{tab:nonposshortcycles} holds;
  \item[(b)] $P_n(x_0x_mx_k^{-1})$ satisfies $T(7)$ if and only if none of the congruences $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$, ($\alpha.j$),($\beta+.j$),($\beta-.j$), ($\gamma+$),($\gamma-$) of Table~\ref{tab:nonposshortcycles} holds.
\end{itemize}
\end{corollary}

\begin{table}
\begin{center}
\begin{tabular}{|cr|ll|}\hline
& $j$ & \textbf{0} & \textbf{1} \\\hline

($\rho$.$j$) & $m,k$ congruence & {$k-m\equiv \frac{n}{2}; \pm \frac{n}{3}; \pm \frac{n}{4}; \pm \frac{n}{5}, \pm \frac{2n}{5}$} & { $k\equiv \frac{n}{2}; \pm \frac{n}{3}; \pm \frac{n}{4}; \pm \frac{n}{5}, \pm \frac{2n}{5}$ }\\
& $A,B$ congruence&{\color{black} $B\equiv \frac{n}{2}; \pm \frac{n}{3}; \pm \frac{n}{4}; \pm \frac{n}{5}, \pm \frac{2n}{5}$} & {$A\equiv \frac{n}{2}; \pm \frac{n}{3}; \pm \frac{n}{4}; \pm \frac{n}{5}, \pm \frac{2n}{5}$}\\
& cycle type& {$Y^2$; $Y^3$; $Y^4$; $Y^5$} & {$Z^2$; $Z^3$; $Z^4$; $Z^5$}\\\hline

($\sigma+$) & $m,k$ congruence &$2k-m\equiv0$ & $2k-m\equiv 0$\\
& $A,B$ congruence&$A+B\equiv 0$ & $B+A\equiv 0$\\
& cycle type& $XYXZ$ & $XZXY$\\\hline

($\sigma-$) & $m,k$ congruence &$m\equiv0$ & $m\equiv 0$\\
& $A,B$ congruence&$A-B\equiv 0$ & $B-A\equiv 0$\\
& cycle type& $XYXZ$ & $XZXY$\\\hline

($\tau+.j$) & $m,k$ congruence &$3k-2m\equiv0$ & $3k-m\equiv 0$\\
& $A,B$ congruence&$A+2B\equiv 0$ & $B+2A\equiv 0$\\
& cycle type& $XZXYY$ & $XYXZZ$\\\hline

($\tau-.j$) & $m,k$ congruence &$2m-k\equiv0$ & $m+k\equiv 0$\\
& $A,B$ congruence&$A-2B\equiv 0$ & $B-2A\equiv 0$\\
& cycle type& $XZXYY$ & $XYXZZ$ \\\hline

($\alpha.j$) & $m,k$ congruence &{$k-m\equiv \pm \frac{n}{6}$} & $k\equiv \pm \frac{n}{6}$\\
& $A,B$ congruence&{\color{black} $B\equiv \pm \frac{n}{6}$} & $A\equiv \pm \frac{n}{6}$\\
& cycle type& $Y^6$ & $Z^6$\\\hline

($\beta+.j$) & $m,k$ congruence &$4k-3m\equiv0$ & $4k-m\equiv 0$\\
& $A,B$ congruence&$A+3B\equiv 0$ & $B+3A\equiv 0$\\
& cycle type& $XZXYYY$ & $XYXZZZ$\\\hline

($\beta-.j$) & $m,k$ congruence &$3m-2k\equiv0$ & $2k+m\equiv 0$\\
& $A,B$ congruence&$A-3B\equiv 0$ & $B-3A\equiv 0$\\
& cycle type& $XZXYYY$ & $XYXZZZ$\\\hline

($\gamma+$) & $m,k$ congruence & $2k-m\equiv n/2$ & $2k-m\equiv n/2$\\
& $A,B$ congruence&$A+B\equiv n/2$ & $B+A\equiv n/2$\\
& cycle type& $XZZXYY$ & $XYYXZZ$\\\hline

($\gamma-$) & $m,k$ congruence &$m\equiv n/2$ & $m\equiv n/2$\\
& $A,B$ congruence&$A-B\equiv n/2$ & $B-A\equiv n/2$\\
& cycle type& $XZZXYY$ & $XYYXZZ$\\\hline

\end{tabular}
\end{center}
\caption{Congruences (mod~$n$) corresponding to short cycles in the star graph of $P_n(x_0x_mx_k^{-1})$. Here $A=k, B=k-m$. \label{tab:nonposshortcycles}}
\end{table}

Note that the two ($\gamma+$) conditions in Table~\ref{tab:nonposshortcycles} are identical conditions and the two ($\gamma-$) conditions are identical; for this reason we do not add the `$.j$' to these conditions. We first identify the groups $G_n(x_0x_mx_k^{-1})$ in the presence of a cycle of type $(\gamma+)$ of Table~\ref{tab:nonposshortcycles}.

\begin{lemma}\label{lem:2(A-B)=0groups}
Let $n\geq 2$, $0\leq m,k<n$ and let $A=k,B=k-m$. Suppose $A+B\equiv n/2$~mod~$n$ and that $(n,m,k)=1$ and let $G=G_n(x_0x_mx_k^{-1})$. Then $G\cong G_n(x_0x_{n/2+2}x_1^{-1})=H(n,n/2+2)$.
\end{lemma}

\begin{proof}
The hypotheses imply $m\equiv 2k+n/2$~mod~$n$ so $1=(n,m,k)=(n,n/2+2k,k)$ which implies $(n/2,k)=1$ so either $(n,k)=1$ or ($n/2$ is odd and $(n,k)=2$). In the former case $G\cong G_n(x_0x_{n/2+2k}x_k^{-1})\cong G_n(x_0x_{n/2+2}x_1^{-1})$ (by~\cite[Lemma~1.3]{BardakovVesnin}); in the latter case $G\cong G_n(x_0x_{n/2+2k}x_k^{-1})\cong G_n(x_0x_{n/2+4}x_2^{-1})$ which is isomorphic to $G_n(x_0x_{n/2+2}x_1^{-1})$ by~\cite[Lemma~1.3]{BardakovVesnin}, \cite[Lemma~7]{W-CHR}.
\end{proof}

In Lemma~\ref{lem:H(n,n/2+2)nothyp} we will show that the groups $H(n,n/2+2)$ are not hyperbolic for any even $n\geq 8$, $n \neq 10$. We now consider the groups that arise when more than one of the remaining length~6 cycle cases hold.

\begin{lemma}\label{lem:nonposmorethanonelength6NewVersion}
Let $n\geq 2$, $0\leq m,k<n$, $m\neq k$, $k\neq 0$ where $(n,m,k)=1$, and set $A=k,B=k-m$, and let $\Gamma$ be the star graph of $P_n(x_0x_mx_k^{-1})$. Suppose that none of the congruences $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$, hold for $j\in \{0,1\}$ and that $(\gamma+)$ does not hold. If more than one of the congruences $(\alpha.j)$, $(\beta+.j)$, $(\beta-.j)$, $(\gamma-)$ holds, then $G_n(x_0x_mx_k^{-1})$ is isomorphic to one of $H(8,4)$, $H(8,6)$, $H(10,4)$, $H(18,4)$, $H(18,16)$.
\end{lemma}

\begin{proof}
If $(\alpha.0)$ and $(\alpha.1)$ hold then $A\equiv \pm n/6$ and $B\equiv \pm n/6$~mod~$n$ so either $(\sigma+)$ or $(\sigma-)$ holds, a contradiction.

If $(\alpha.0)$ and $(\gamma-)$ hold then $(\alpha.1)$ holds, a contradiction.
If $(\alpha.1)$ and $(\gamma-)$ hold then $(\alpha.0)$ holds.

If $(\alpha.0)$ and ($(\beta+.0)$ or $(\beta-.0)$) hold then $2A\equiv 0$~mod~$n$, so $(\rho.1)$ holds, a contradiction; if $(\alpha.1)$ and ($(\beta+.1)$ or $(\beta-.1)$) hold then $2B\equiv 0$~mod~$n$, so $(\rho.0)$ holds.

If $(\beta+.0)$ and $(\beta-.0)$ hold then $2A\equiv 0$~mod~$n$ so $(\rho.1)$ holds, a contradiction; if $(\beta+.1)$ and $(\beta-.1)$ hold then $2B\equiv 0$~mod~$n$ so $(\rho.0)$ holds.

If $(\beta-.0)$ and $(\gamma-)$ hold then $A\equiv 3B$~mod~$n$ and $4B\equiv 0$~mod~$n$ so $(\rho.0)$ holds, a contradiction. Similarly, if $(\beta-.0)$ and $(\gamma-)$ hold then $(\rho.1)$ holds.

If $(\beta+.0)$ and $(\beta+.1)$ hold then $B\equiv -3A$~mod~$n$ and $8A\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,A,-3A)=(n,A)$ so $n|8$, and if $n<8$ then $(\rho.0)$ holds, a contradiction, so $n=8$ and $G_n(x_0x_mx_k^{-1})$ is isomorphic to either $H(8,4)$ or $H(8,6)$. Similarly, if $(\beta+.1)$ and $(\beta-.1)$ hold then $G_n(x_0x_mx_k^{-1})\cong H(8,4)$ or $H(8,6)$.

If $(\beta+.0)$ and $(\gamma-)$ hold then $A\equiv -3B$~mod~$n$ and $8B\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,-3B,B)=(n,B)$ so $n|8$, and if $n<8$ then $(\rho.0)$ holds, a contradiction, so $n=8$. Similarly, if $(\beta+.1)$ and $(\gamma-)$ hold then $n=8$.

If $(\beta+.0)$ and $(\beta-.1)$ hold then $B\equiv 3A$~mod~$n$ and $10A\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,A)$ so $n|10$, and if $n<10$ then $(\rho.0)$ holds, a contradiction, so $n=10$. Then $(k,n)=1$ so by~\cite[Lemma~1.3]{BardakovVesnin} we may assume $k=1$, so $A=1$ and $k-m=B=3$, and hence $m=8$; thus $G_n(x_0x_mx_k^{-1})\cong G_{10}(x_0x_8x_1^{-1})$. Similarly, if $(\beta+.1)$ and $(\beta-.0)$ hold then $n=10$ and $G_n(x_0x_mx_k^{-1})\cong G_{10}(x_0x_4x_1^{-1})$. By~\cite[Theorem~2]{COS} we have $G_{10}(x_0x_8x_1^{-1})\cong G_{10}(x_0x_4x_1^{-1})= H(10,4)$.

If $(\alpha.0)$ and $(\beta+.1)$ hold then $B\equiv -3A$ and $18A\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,A)$ so $n|18$ and if $n\leq 9$ then $(\rho.0)$ holds so $n=18$. Then $(k,n)=1$ so we may assume $k=1$ so $k-m=B=-3$ and hence $m=4$ so $G_n(x_0x_mx_k^{-1})\cong G_{18}(x_0x_4x_1^{-1})=H(18,4)$.

If $(\alpha.1)$ and $(\beta+.0)$ hold then $A\equiv -3B$ and $18B\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,B)$ so $n|18$ and again $n=18$. Then $(k-m,n)=1$ so we may assume that $B=k-m=1$ so $k=A=-3$, and hence $m=-4$, so $G_n(x_0x_mx_k^{-1})=G_{18}(x_0x_{-4}x_{-3}^{-1})\cong G_{18}(x_0x_4x_1^{-1})$ by~\cite[Lemma~1.3]{BardakovVesnin}, \cite[Lemma~7]{Wsurvey}.

If $(\alpha.0)$ and $(\beta-.1)$ hold then $B\equiv 3A$ and $18A\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,A)$ so $n|18$ and if $n\leq 9$ then $(\rho.0)$ holds so $n=18$. Then $(k,n)=1$ so we may assume $k=1$ so $k-m=B=3$ and hence $m=-2$ so $G_n(x_0x_mx_k^{-1})\cong G_{18}(x_0x_{16}x_1^{-1})=H(18,16)$.

If $(\alpha.1)$ and $(\beta-.0)$ hold then $A\equiv 3B$ and $18B\equiv 0$~mod~$n$; moreover $1=(n,A,B)=(n,B)$ so $n|18$ and if $n\leq 9$ then $(\rho.1)$ holds so $n=18$. Then $(k-m,n)=1$ so we may assume $k-m=1$ so $k=A=3$ and hence $k=3, m=2$ so $G_n(x_0x_mx_k^{-1})\cong G_{18}(x_0x_{2}x_3^{-1})\cong H(18,16)$ by~\cite[Lemma~1.3]{BardakovVesnin}, \cite[Lemma~7]{Wsurvey}.
\end{proof}

In Corollary~\ref{cor:G8nonposnothyp} we showed that $H(8,4)$ is not hyperbolic; in Lemma~\ref{lem:H(n,n/2+2)nothyp} we will show that $H(8,6)$ is not hyperbolic. We now show that the remaining groups arising in Lemma~\ref{lem:nonposmorethanonelength6NewVersion} are hyperbolic:

\begin{example}\label{ex:hypgroupswithn=9,10,18}
\em Using KBMAG~\cite{KBMAG} it is straightforward to show that the groups $H(10,4)$, $H(18,4)$, $H(18,16)$ are hyperbolic, and since they contain a non-abelian free subgroup (by~\cite[Corollary~11]{HowieWilliams}), they are non-elementary hyperbolic.\em
\end{example}

\subsection{Non-hyperbolic groups $G_n(x_0x_mx_k^{-1})$}\label{sec:nonhyperbolicgroupsnonpositive}

We now show that the $T(6)$ groups $H(n,n/2+2)$ arising in Lemma~\ref{lem:2(A-B)=0groups} are not hyperbolic. (Note that if $n=2,4,6$ or $10$ then the presentation of $H(n,n/2+2)$ does not satisfy $T(6)$, by Corollary~\ref{cor:T6T7nonpositive}.) As in the proof of Lemma~\ref{lem:n=8nonhyp} we do this by an application of the Flat Plane Theorem.

\begin{figure}[ht]
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  \caption{A van Kampen diagram over the presentation $P_n(x_0x_{n/2+2}x_1^{-1})$ with boundary label $(x_0x_{n/2}) (x_1x_{n/2+5}x_{n/2+7}x_{11})(x_{12}x_{n/2+12})^{-1} (x_1x_{n/2+5}x_{n/2+7}x_{11})^{-1}$.\label{fig:nonpositiveflat}}
\end{figure}

\begin{lemma}\label{lem:H(n,n/2+2)nothyp}
Suppose $n\geq 8$ is even, $n\neq 10$. Then $H(n,n/2+2)=G_n(x_0x_{n/2+2}x_{1}^{-1})$ is not hyperbolic.
\end{lemma}

\begin{proof}
Since the presentation $P_n(x_0x_{n/2+2}x_1^{-1})$ satisfies C(3)-T(6) and each relator has length~3, each face in the geometric realisation $\tilde{C}$ of the Cayley complex of $P$ (obtained by assigning length~1 to each edge) is an equilateral triangle, and so $\tilde{C}$ satisfies the CAT(0) inequality. Consider the geometric realisation $\Delta_0$ of the reduced van Kampen diagram given in Figure~\ref{fig:nonpositiveflat} and for each $0\leq i<n$ let $\Delta_i$ be obtained from $\Delta_0$ by applying the shift $\theta^i$ to each edge. Then placing $\Delta_0,\Delta_{12},\Delta_{24},\ldots ,\Delta_{6n-12}$ side by side gives the geometric realisation $\Delta$ of a reduced van Kampen diagram. Copies of $\Delta$ tile the Euclidean plane without cancellation of faces. Thus there is an isometric embedding of the Euclidean plane in $\tilde{C}$, and so the result follows from the Corollary to Theorem~A in~\cite{Bridson}.
\end{proof}

\subsection{Analysis of van Kampen diagrams over $P_n(x_0x_mx_k^{-1})$}\label{sec:curvaturenonpositive}

In this section we show that if the cyclic presentation $P=P_n(x_0x_mx_k^{-1})$ is $T(6)$ and precisely one of the
congruences ($\alpha.j$),($\beta+.j$),($\beta-.j$),($\gamma-$) hold then $G = G_n(x_0x_mx_k^{-1})$ is hyperbolic. As in Section~\ref{sec:curvaturepositive} we do this by showing that there is a linear function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for all $N\in \mathbb{N}$ and all freely reduced words $W\in F_n$ with length at most $N$ that represent the identity of $G$ we have $\mathrm{Area}(W)\leq f(N)$. Note that each face in $D$ is a triangle, as shown in Figure~\ref{fig:nonpositiveface}, where the corner labels $X,Y,Z$ correspond to the edge types of the star graph of $P$. In order to obtain a linear isoperimetric function (in Lemma~\ref{lem:curvaturenonpositivecase}) we first rule out certain configurations in $D$.

\begin{figure}
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  \caption{A typical face in a van Kampen diagram over the presentation $P_n(x_0x_mx_k^{-1})$.\label{fig:nonpositiveface}}
\end{figure}

\begin{lemma}\label{lem:nonposalldegree6}
Suppose that all interior vertices of $D$ have degree at least 6 and that all interior vertices of degree 6 of $D$ have label precisely one of $(\alpha.j)$,$(\beta+.j)$, $(\beta-.j)$, $(\gamma-)$ for $j\in \{0,1\}$. If $v$ is an interior vertex of degree~6 where all the neighbours of $v$ are interior vertices then $v$ has a neighbour of degree at least 7.
\end{lemma}

\begin{proof}
If $v$ is labelled $Y^6$ (resp. $Z^6$) then clearly none of it neighbours can be labelled $Y^6$ (resp. $Z^6$), so they must each have degree at least 7.
If $v$ is labelled $XZXYYY$ (resp. $XYXZZZ$, resp. $XZZXYY$), then the labels of the corners of the faces incident to $v$ show that at least one of the neighbours of $v$ does not have label $XZXYYY$ (resp. $XYXZZZ$, resp. $XZZXYY$), and hence has degree at least 7.
\end{proof}

We are now in a position to be able to establish the existence of a suitable isoperimetric function.

\begin{lemma}\label{lem:curvaturenonpositivecase}
Let $n\geq 2$, $0\leq m,k<n$, $m\neq k, k\neq 0$ and set $A=k,B=k-m$ and let $\Gamma$ be the star graph of $P_n(x_0x_mx_k^{-1})$. Suppose that none of $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$, hold and that exactly one of the congruences ($\alpha.j$),($\beta+.j$),($\beta-.j$),($\gamma-$) of Table~\ref{tab:nonposshortcycles} holds ($j\in \{0,1\}$). Then $G_n(x_0x_mx_k^{-1})$ has a linear isoperimetric function.
\end{lemma}

\begin{proof}
Let $N\in \mathbb{N}$ and let $W$ be a freely reduced word in the free group $F_n$ of length at most $N$ that represents the identity of $G$ and let $D$ be a reduced van Kampen diagram whose boundary is a simple closed curve with label $W$. We let $I$ denote the set of interior vertices of $D$, $B$ denote the set of boundary vertices of $D$, and $F$ denote the set of faces of $D$. Then $\mathrm{Area}(W)\leq |F|$. Writing $\pi$ to denote $180$, we define the curvature of a face $f$ by $\kappa(f)=-\pi + (\mathrm{sum~of~angles~in}~f)$, the curvature of an interior vertex $v$ by $\kappa (v)=2\pi - (\mathrm{sum~of~angles~at}~v)$, and the curvature of a boundary vertex $\hat{v}$ by $\kappa(\hat{v})=\pi - (\mathrm{sum~of~angles~at}~\hat{v})$. Again it follows from the Gauss-Bonnet Theorem that equation~(\ref{eq:GaussBonnet}) holds.

Since none of the congruences $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$ hold, every interior vertex of $D$ is of degree at least~6 and since exactly one of the congruences ($\alpha.j$),($\beta+.j$),($\beta-.j$),($\gamma-$) hold, then label of an interior vertex of degree 6 is the corresponding label given in Table~\ref{tab:nonposshortcycles}.

We assign angles to the corners of faces in $D$ as follows. If $v$ is a boundary vertex assign 47 to every corner at $v$; if $v$ is an interior vertex of degree at least $7$ then assign 52 to every corner at $v$. Assume now that $v$ is an interior vertex of degree $6$ and consider a face $f$ with vertices $v$ and $u,w$: if $u,w$ are interior of degree 6 then assign 59 to the corner of $f$ at $v$; otherwise assign $63.5$ to the corner of $f$ at $v$.
 If a face $f$ contains a boundary vertex then $\kappa(f)\leq -\pi +47 +2(63.5)=-6$; if a face $f$ contains only interior vertices of degree 6 then $\kappa(f)=-\pi+3(59)=-3$; if a face $f$ contains only interior vertices of degree at least 7 then $\kappa(f)= -\pi +3(52)=-24$; if a face $f$ contains an interior vertex of degree 6 and two interior vertices of degree at least 7 then $\kappa(f)=-\pi+63.5+2(52)=-12.5$; if a face $f$ contains two vertices of degree 6 and one of degree at least 7 then $\kappa(f)=-\pi+2(63.5)+52=-1$. Therefore $\kappa(f)\leq -1$ for all $f\in F$.

We now turn to curvature of the vertices. If $v$ is an interior vertex of degree at least 7 then $\kappa(v)\leq 2\pi -7(52)=-4$; if $v$ is an interior vertex of degree 6 that has a neighbour that is either interior of degree at least 7 or is a boundary vertex then $\kappa(v)\leq 2\pi -4(59)-2(63.5)=-3$.

By Lemma~\ref{lem:nonposalldegree6} every interior vertex of degree 6 has a neighbour on the boundary or a neighbour that is interior of degree at least 7. Then $\kappa(v)\leq -3$ for all interior vertices $v$ so~(\ref{eq:GaussBonnet}) implies

\begin{alignat*}{1}
  2\pi &= \sum_{v\in I} \kappa(v) + \sum_{\hat{v}\in B} \kappa (\hat{v}) +\sum_{f\in F} \kappa(f)\\
       &\leq \sum_{v\in I} (-3) + \sum_{\hat{v}\in B} (\pi -\mathrm{sum~of~vertices~at}~\hat{v}) + \sum_{f\in F} (-1)\\
       &= -3|I| + \sum_{\hat{v}\in B} (\pi -\mathrm{sum~of~vertices~at}~\hat{v}) -|F|\\
       &\leq |B|\pi - \sum_{\hat{v}\in B} (\mathrm{sum~of~vertices~at}~\hat{v}) -|F|
\end{alignat*}

so
\[ \sum_{\hat{v}\in B} (\mathrm{sum~of~vertices~at}~\hat{v}) \leq (|B|-2)\pi -|F|.\]

On the other hand, the corner angle at any boundary vertex is $47$ so the sum of angles over the boundary vertices is bounded below by $47|B|$. Therefore $47|B|\leq (|B|-2)\pi -|F|$ so $|F|\leq 133|B|-360$. But $\mathrm{Area}(W)\leq |F|$ and $|B|\leq N$ so $\mathrm{Area}(W) \leq 133N-360$, so $f(N)=133N-360$ is a linear isoperimetric function, as required.
\end{proof}

We now have all the ingredients to prove Theorem~\ref{thm:nonpositive}.

\subsection{Proof of Theorem~\ref{thm:nonpositive}}\label{sec:proofofnonpositive}

Suppose that $n\geq 2$, $0\leq m,k<n$, $m\neq k$, $k\neq 0$, $(n,m,k)=1$ and that the cyclic presentation $P_n(x_0x_mx_k^{-1})$ satisfies $T(6)$.

Then Lemma~\ref{lem:nonposshortcycles} implies that none of the congruences $(\rho.j)$, $(\sigma+.j)$, $(\sigma-.j)$, $(\tau+.j)$, $(\tau-.j)$ hold. If $(\gamma+)$ holds then $n=8$ or $n\geq 12$ and $G$ is not hyperbolic by Lemmas~\ref{lem:2(A-B)=0groups},\ref{lem:H(n,n/2+2)nothyp}, so suppose that $(\gamma+)$ does not hold.

If $\Gamma$ has no cycle of length less than 7, then $P_n(x_0x_mx_k^{-1})$ satisfies C(3)-T(7), so $G_n(x_0x_mx_k^{-1})$ is hyperbolic by~\cite[Corollary~4.1]{GerstenShortI}, so we may assume that $\Gamma$ has a cycle of length~6 so, by Lemma~\ref{lem:nonposshortcycles}, at least one of the congruences
$(\alpha.j)$, $(\beta+.j)$, $(\beta-.j)$, $(\gamma-)$ hold ($j\in \{0,1\}$). Suppose that more than one of them hold. Then $G$ is one of the groups in the conclusion of Lemma~\ref{lem:nonposmorethanonelength6NewVersion}. When $n=8$ the group $G\cong G_8(x_0x_4x_1^{-1})=H(8,4)$ or $G\cong G_8(x_0x_6x_1^{-1})=H(8,6)$, which are non-hyperbolic by Corollary~\ref{cor:G8nonposnothyp} and Lemma~\ref{lem:H(n,n/2+2)nothyp}, respectively. In the remaining cases $G$ is hyperbolic by Example~\ref{ex:hypgroupswithn=9,10,18}.

Suppose then that exactly one of the congruences $(\alpha.j)$, $(\beta+.j)$, $(\beta-.j)$, $(\gamma-)$ holds. Then $G$ has a linear isoperimetric function, and hence is hyperbolic, by Lemma~\ref{lem:curvaturenonpositivecase}. By~\cite[Corollary~11]{HowieWilliams} $G$ contains a non-abelian free subgroup so it is non-elementary hyperbolic.

\subsection*{Acknowledgements}

The first named author thanks Jim Howie for his hospitality during a research visit to the School of Mathematical and Computer Sciences, Heriot Watt University, in November 2019; that visit was central to the development of the non-hyperbolicity results in this paper. Both authors thank Jim Howie and William Bogley for insightful conversations relating to this work, and helpful comments on a draft of this article.

config: arxiv


TITLE: Is there a way to prove a boolean operator isn't universal?
QUESTION [3 upvotes]: In boolean algebra, I could prove an operator is universal by implementing a NAND or NOR gate with it. But is there a way to prove a boolean operator isn't universal?
I would like to know a general method that should work for every (or almost every) incomplete operator.
Right now, I want to prove incompleteness of this operator:
$$ T(w,x,y,z) = (\neg w \lor \neg x \lor \neg y) \oplus (xyz)$$
After simplification I could write it like one of the following:
$$(\neg(wx y)) \oplus (xyz) =wxyz \lor \neg w\neg x \lor \neg w\neg y \lor \neg w\neg z \lor \neg x\neg y \lor \neg x\neg z \lor \neg y\neg z$$
$\oplus$ is the eXclusive OR operator, $\lor$ is OR, and I'm using implicit AND for simple reading

REPLY [1 votes]: Following this relevant question which I asked on cs.stackexchange, a relatively general and easy way that works many times is to use the property that if an operator $T$ is universal, then necessarily $T(x,...,x)=\neg x$. In your case:
$$T(x,x,x,x)=(\neg x \vee \neg x\vee \neg x)\oplus(xxx)=\neg x\oplus x=1$$
And since this isn't $\neg x$, then $T$ necessarily isn't universal.


TITLE: Notation: subset of powerset containing sets of given cardinality
QUESTION [1 upvotes]: We are given a set $S$. Is there a generally accepted symbol to denote the set of all subsets with cardinality $\kappa$ of $S$? The notation that comes closest to what I want is the one I found on wikipedia:
https://en.wikipedia.org/wiki/Power_set#Subsets_of_limited_cardinality,
where they write:
``The set of subsets of $S$ of cardinality less than $\kappa$ is denoted by $\mathcal{P}_{\kappa}(S)$ or $\mathcal{P}_{<\kappa}(S)$. Similarly, the set of non-empty subsets of S might be denoted by  $\mathcal{P}_{\ge1}(S)$''
Is a notation like
$$\mathcal{P}_{=\kappa}(S)$$
generally accepted?

REPLY [1 votes]: I cannot say that I've seen that notation, but I am likely to understand it from context. A slightly more accepted notation is $[X]^\kappa$.
But the usual advice in this situation is to explicitly define your notation if you feel it might be non-standard, and sometimes even if it is standard it is worth reminding the reader what exactly you mean.


TITLE: Can we combine convolution and higher powers for locally maximising a function?
QUESTION [0 upvotes]: Can we somehow find local maximum function value (for strictly positive functions) using a convolution?
My idea is based on the result that $$ \lim_{p\to \infty}\left[\frac{1}{N}\sum_{k=1}^N {(a_k)} ^p\right]^{\frac 1 p} = \max(a_k)$$
Similarly we might be able to use the result that
$$ \min(a_k) = \max(a_k)-\max(\max(a_k)-a_k)$$
Could we perhaps use this for building envelope detection? It seems theoretically sound to me, but would it be practically feasible?

REPLY [0 votes]: Yes, we can do this.
If we consider the function $$f(t)=\cases{\frac 1 {2N}\phantom{0},|t|<N\\0\phantom{\frac 1 {2N}},|t|\geq N}$$
We can with the convolution

If we are in continuous time:

$$(g*f)(t)=\int_{-\infty}^\infty g(\tau)\cdot f(t-\tau) d\tau$$

If we are in discrete time:
$$(g*f)(t)=\sum_{\tau=-\infty}^\infty g(\tau)\cdot f(t-\tau)$$

Define the non-linear operation $$O(g,p) = (g^p*f)(t)^{1/p}$$
Where exponent means multiplicative power.
For example: if $g(t) = \sin(t), O(g,4) = ((\sin(x)^4*f))(t)^{1/4}$
In practice we usually want much higher $p$ exponents, maybe 32 or 1024.
We now investigate the test-function
$$g(t) = \sin(60t^2)^2\cdot (\sin(16t)^2 + \cos(8t)^2)$$
The square is to get same level of minimum values: 0.
If we analyze $g(t)$, we can see that we have a chirp function modulated by a slower wave. This is to investigate behaviors for different frequencies.

We see that for higher frequencies we get reasonable envelope behavior. It is also possible to observe that as we change parameter $N$ above, which frequency range we can calculate maxima for are affected - but also the rectangular shape of local maxima will grow. We may need to smooth these out somehow. But how to do so is outside scope of question.


text:
\begin{definition}[Definition:Mach Number]
'''Mach number''' is a [[Definition:Unit of Measurement|unit of measurement]] of [[Definition:Speed|speed]] of a [[Definition:Body|body]] $B$ in a [[Definition:Fluid|fluid]] $F$.
It is defined as the [[Definition:Ratio|ratio]] of the [[Definition:Speed|speed]] of $B$ to the [[Definition:Speed|speed]] of [[Definition:Sound (Physics)|sound]] in $F$.
{{NamedforDef|Ernst Waldfried Josef Wenzel Mach|cat = Mach}}
\end{definition}

config: wiki


TITLE: How is $\Bbb{Z}_2\otimes \Bbb{Z}_3$ isomorphic to $\Bbb{Z}_6$?
QUESTION [1 upvotes]: I am currently reading the book  Group Theory in a Nutshell for Physicists by Anthony Zee and he wrote something about isomorphism I could not quite understand.

Now we come in for a bit of a surprise. Consider $\Bbb{Z}_2 \otimes \Bbb{Z}_3$ consisting of $(n, m)$, which we compose by $(n + n' \mod 2, m + m' \mod3)$. Again, we start with $(0, 0)$ and add $(1, 1)$ repeatedly:
$(0,0)\{+(1,1)\}\rightarrow (1,1) \{+(1,1)\}\rightarrow(0,2)\{+(1,1)\}\rightarrow(1,0)\{+(1,1)\}\rightarrow(0,1)\{+(1,1)\}\rightarrow(1,2)\{+(1,1)\}\rightarrow (0,0)$
We are back where we started! In the process, we cycled through all
six elements of $\Bbb{Z}_2 \otimes \Bbb{Z}_3$. We conclude that the
six elements $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)$ } describe
$\Bbb{Z}_6$

Further he writes-

Thus, $\Bbb Z_2 \otimes \Bbb Z_3$ and $\Bbb Z_6$ are isomorphic; they are literally the same group. Note that this
phenomenon, of a possible isomorphism between $\Bbb Z_p \otimes \Bbb Z_q$ and $\Bbb{Z}_{pq}$, does not require p
and q to be prime, only relatively prime. (Consider the example of $\Bbb Z_4 \otimes \Bbb Z_9$.)

And hence my question: How did the author establish the one-to-one mapping between the groups claimed to be isomorphic? Any help is appreciated.

REPLY [1 votes]: Lemma: If a group $G$ has order $2p$ for prime $p>2$, then $G$ is isomorphic to either $\Bbb Z_{2p}$ or $D_p$.
Since $\Bbb Z_2\times \Bbb Z_3$ is abelian and $D_3$ is not, we must have that $\Bbb Z_2\times\Bbb Z_3$ is isomorphic to $\Bbb Z_6$.
Here $D_p$ is the dihedral group of order $2p$.


TITLE: On non-uniform convergence
QUESTION [0 upvotes]: In the book of MATHEMATICAL
METHODS FOR
PHYSICISTS by Arfken et al., (Example 1.2.1 Non uniform convergence):
The series of functions $S(x)=\sum_{n=0}^{\infty} \left(1-x\right)x^n$ is shown to not be uniformly convergent by providing the following explanation:
$$|1-(1-x^N)|<ε$$
"No matter what the values of N and a sufficiently small ε may be, there will be an x value
(close to 1) where this criterion is violated. The underlying problem is that x = 1 is the
convergence limit of the geometric series, and it is not possible to have a convergence rate
that is bounded independently of x in a range that includes x = 1."
Can anyone please explain why is it possible to have such an $x$ value which will always invalidate the inequality?

REPLY [1 votes]: HINT:
What happens if $x=1-1/N$?
Alternatively, take $\epsilon=\frac12$ and $x=\sqrt[N]{\frac{1}{2}}$.


TITLE: Possible values of $\frac{x_{1}}{x_{n} + x_{1} + x_{2}} + \frac{x_{2}}{x_{1} + x_{2} + x_{3}} + ... + \frac{x_{n}}{x_{n-1} + x_{n} + x_{1}}$
QUESTION [3 upvotes]: Let $n > 3$ be fixed integer. If $x_{1}, x_{2}, x_{3}, ..., x_{n}$ are from $\mathbb{R}^{+}$, what are the possible values for
$$ \frac{x_{1}}{x_{n} + x_{1} + x_{2}} + \frac{x_{2}}{x_{1} + x_{2} + x_{3}} + ... + \frac{x_{n}}{x_{n-1} + x_{n} + x_{1}} $$

Attempt:
Start with $n=4$, we have
$$ S = \frac{x_{1}}{x_{4} + x_{1} + x_{2}} + \frac{x_{2}}{x_{1} + x_{2} + x_{3}} + \frac{x_{3}}{x_{2} + x_{3} + x_{4}} + \frac{x_{4}}{x_{3} + x_{4} + x_{1}}   $$
then extending this we get
$$ \frac{x_{1}}{x_{4} + x_{1} + x_{2}} + \frac{x_{2}}{x_{1} + x_{2} + x_{3}} + \frac{x_{3}}{x_{2} + x_{3} + x_{4}} + \frac{x_{4}}{x_{3} + x_{4} + x_{1}}   $$
$$ + \frac{x_{2}}{x_{4} + x_{1} + x_{2}} + \frac{x_{3}}{x_{1} + x_{2} + x_{3}} + \frac{x_{4}}{x_{2} + x_{3} + x_{4}} + \frac{x_{1}}{x_{3} + x_{4} + x_{1}}   $$
$$ + \frac{x_{4}}{x_{4} + x_{1} + x_{2}} + \frac{x_{1}}{x_{1} + x_{2} + x_{3}} + \frac{x_{2}}{x_{2} + x_{3} + x_{4}} + \frac{x_{3}}{x_{3} + x_{4} + x_{1}}   $$
$$ = 4 $$
Next
$$ \frac{x_{2} + x_{4}}{x_{4} + x_{1} + x_{2}}  \le \frac{x_{3} + x_{2} + x_{4}}{x_{4} + x_{1} + x_{2} + x_{3}}  $$
$$ \frac{x_{1} + x_{3}}{x_{1} + x_{2} + x_{3}}  \le \frac{x_{4} + x_{1} + x_{3}}{x_{4} + x_{1} + x_{2} + x_{3}}  $$
$$ \frac{x_{2} + x_{4}}{x_{2} + x_{3} + x_{4}}  \le \frac{x_{1} + x_{2} + x_{4}}{x_{4} + x_{1} + x_{2} + x_{3}}  $$
$$ \frac{ x_{3} + x_{1} }{x_{3} + x_{4} + x_{1}}  \le \frac{x_{3} + x_{2} + x_{1}}{x_{4} + x_{1} + x_{2} + x_{3}}  $$
So we can get
$$ \frac{x_{1}}{x_{4} + x_{1} + x_{2}} + \frac{x_{2}}{x_{1} + x_{2} + x_{3}} + \frac{x_{3}}{x_{2} + x_{3} + x_{4}} + \frac{x_{4}}{x_{3} + x_{4} + x_{1}} $$
$$ = 4 -  \left( \frac{x_{2} + x_{4}}{x_{4} + x_{1} + x_{2}} + \frac{x_{1} + x_{3}}{x_{1} + x_{2} + x_{3}} + \frac{x_{2} + x_{4}}{x_{2} + x_{3} + x_{4}} + \frac{ x_{3} + x_{1} }{x_{3} + x_{4} + x_{1}}  \right) \ge 1$$
Also using similar approach we get:
$$ \frac{x_{1}}{x_{4} + x_{1} + x_{2}} + \frac{x_{2}}{x_{1} + x_{2} + x_{3}} + \frac{x_{3}}{x_{2} + x_{3} + x_{4}} + \frac{x_{4}}{x_{3} + x_{4} + x_{1}} \le \frac{x_{1} + x_{3}}{ x_{3} + x_{4} + x_{1} + x_{2}} + \frac{x_{4} +x_{2}}{x_{4} + x_{1} + x_{2} + x_{3}} + \frac{x_{1} + x_{3}}{x_{1} + x_{2} + x_{3} + x_{4}} + \frac{x_{2} + x_{4}}{x_{2} + x_{3} + x_{4} + x_{1}} = 2  $$
so far we have  $1 \le S \le 2 $

REPLY [1 votes]: By the same way we obtain:
$$1=\sum_{cyc}\frac{x_1}{\sum\limits_{i=1}^nx_i}<\sum_{cyc}\frac{x_1}{x_n+x_1+x_2}.$$
Also, for even $n$ we obtain: $$\sum_{cyc}\frac{x_1}{x_n+x_1+x_2}<\left(\frac{x_1}{x_1+x_2}+\frac{x_2}{x_1+x_2}\right)+...+\left(\frac{x_{n-1}}{x_{n-1}+x_n}+\frac{x_n}{x_{n-1}+x_n}\right)=\frac{n}{2}.$$
For odd $n$ we can assume that $x_1+x_2+x_3$ is a minimal value of any our denominators, which gives $$\frac{x_1}{x_n+x_1+x_2}+\frac{x_2}{x_1+x_2+x_3}+\frac{x_3}{x_2+x_3+x_4}\leq$$
$$\leq\frac{x_1}{x_1+x_2+x_3}+\frac{x_2}{x_1+x_2+x_3}+\frac{x_3}{x_1+x_2+x_3}=1$$ and by the same way we obtain:
$$\sum_{cyc}\frac{x_1}{x_n+x_1+x_2}<1+\frac{n-3}{2}=\frac{n-1}{2},$$ which gives
$$\sum_{cyc}\frac{x_1}{x_n+x_1+x_2}<\left[\frac{n}{2}\right]$$
It's enough to understand that we got infimum and supremum.


TITLE: Is taking limits in this proof really necessary?
QUESTION [0 upvotes]: Consider the following proof:

￼Theorem. Let $A$ be a unital Banach algebra and $a$ an element of $A$ such that $\Vert a \Vert < 1$. Then $1 - a \in \operatorname{Inv}(A)$ and
  $$(1-a)^{-1} = \sum_{n=0}^\infty a^n.$$
  Proof. Since
  $$\sum_{n=0}^\infty \Vert a^n\Vert \leq \sum_{n=0}^\infty \Vert a \Vert^n = \left(1-\Vert a \Vert\right)^{-1} < + \infty,$$
  the series $\sum_{n=0}^\infty a^n$ is convergent, to $b$ say, in $A$, and since $$\left(1-a\right) (1+\cdots+a^n) = 1 - a^{n+1}$$
  converges to $\left(1-a\right) b = b \left(1-a\right)$ and to $1$ as $n \to \infty$, the element $b$ is the inverse of $1 - a$. $\square$

I am wondering about whether it is correct to argue as follows instead:
Since $\sum a^n$ is convergent the multiplication $a \sum_{n=0}^\infty a^n = \sum_{n=1}^\infty a^n$ and the subtraction $\sum_{n=0}^\infty a^n  - \sum_{n=1}^\infty a^n = 1$ are justified. Therefore $\left(1-a\right)b = 1$.

REPLY [1 votes]: Disclaimer: this is an answer to the above series of comments by Student. It was too long for a comment.
OK for "first" and "second". For "third", you shouldn't start by writing down $\frac1{1-a}$, since you precisely want to show that this exists. (By the way, the notation $\frac1{1-a}$ is usually "forbidden" in a vector-valued setting: it is generally agreed that one "must" write $(1-a)^{-1}$).
Likewise, you shouldn't write $\sum_0^\infty a_n$ as long as you don't know that the series is indeed convergent in $A$. So: start by proving that the series $\sum\Vert a^n\Vert$ is convergent (as you did). Then deduce that the series $\sum a^n$ is convergent in $A$ (because $A$ is a Banach algebra), and finally check as you did that $b=\sum_0^\infty a^n$ satisfies $(1-a)b=1=b(1-a)$, so that $1-a$ is indeed invertible with inverse $b$.
Final remark: part "second" is in fact never used.


TITLE: Basis of closed of ordered topology
QUESTION [0 upvotes]: I have a question about ordered topologies. Suppose $X = [0,1] \times [0,1]$ is ordered by a relation $\leq$. Then a basis of open sets is given by the intervals of the form $\langle a,b \rangle = \{ x : a < x < b \}$, and sets of the form $[(0,0), a\rangle$ and $\langle b, (1,1)]$. My question is the following: is the set consisting on all intervals of the form $[a,b]$ a basis for the closed sets of the topology? I am specially interested in the case of the lexicographic order. My intuition is that it indeed is a basis of closed but I am not sure I have not missed any corner case.

REPLY [1 votes]: You have to take set $\mathcal{C}$ of all finite unions of closed intervals $[a,b]$ to get a base for the closed sets: all complements of th  open base elements are of that form, e.g. $X\setminus \langle a,b\rangle = [(0,0), a] \cup [b, (1,1)]$ etc. and the complements of an open base form a closed base (and vice versa). Also, we need these finite unions, because intersections of closed intervals are going to yield order-convex sets only, and there are non-order convex closed sets too.


TITLE: Data - Monte Carlo Correction techninques in Particle physics
QUESTION [1 upvotes]: It is said that the data does not always match with the Monte Carlo simulations in particle physics.(I guess even in the Higgs to gamma gamma channel, the peak in real data was at about 127GeV and thus it was corrected) .Thus,I wished to know what are the different ways in which the differences in Data and MC simulations are corrected, if possible, for the case of jets like bjets.

REPLY [2 votes]: If certain distributions of variables don't agree, then you can reweight the MC sample to match data (typically either using a nice clean control sample or s-weighted signal). There are a variety of reweighting techniques e.g. histograms, kernel-density estimators, boosted decision trees.
If resolution doesn't agree, you can apply 'smearing', whereby a variable is shifted by some random amount on an event-by-event basis according to a Gaussian distribution (or similar).


TITLE: Are there 'higher order moments' in physics?
QUESTION [8 upvotes]: This may be a rather noob question but please let me clarify: I'm struggling to understand the use of the word 'moments' w.r.t., probability distributions. It seems after some research and poking around it seems to have been derived from physics when trying to solve/prove something about/related to binomial distribution and the method was called method of moments. I've asked the corresponding question here: https://stats.stackexchange.com/q/17595/4426
Now 'Pearson' (one of the very famous statisticians) comments:

We shall now proceed to find the first four moments of the system of
  rectangles round GN. If the inertia of each rectangle might be
  considered  as concentrated along its mid vertical, we should have for
  the sth moment round NG, writing d = c(1 + nq).

Here are some of the details of the proof (as in the above post):

Now Pearson talks about calculating the 'rth' moment and uses a derivative function to do so:

Question: I'm not aware of such a function from my knowledge of elementary physics. What kind of moments are being calculated here? How do you calculate 'higher order moments'? Is there any such thing?
Basically looking to clarify something in statistics but was historically alluded to physics and hence just want to get it ironed out :)
UPDATE: Intent of question: What I want to know is does the above derivation have anything at all to do with the concept of moments in physics and how is it related? Since the 'word' moment (and its intent) seem to be borrowed from physics when the author is making the derivation. I personally want to know if something like this does exist in the field of physics and how are these two derivations (and 'moments' related)

REPLY [1 votes]: We can say that higher order moments represent the shape of the distribution: the mean the second moment, represent how much is the scattering, where this can be seen by a point scatter. the third represents the skewness /asymmetry, the fourth how much is the flatness of the curve. Further moments continue this way but it is as far as I know inexpressible in language or you can consider the first moments of the square of the variable, the cube...
In general, these higher orders represent the details of the variable: Going to higher order, you represent the data in more and more detail view. This observation has practical implications: if you have a sample of finite amount of data, you can calculate the moments up to an order: computing the higher order is useless and error because you don't have enough data. Or  better stated, the variance of the moment considered in large.
I f you are  interested, my book(PhD thesis) "Construction of Random Signals from their Higher Order moments", Ismail Chamseddine, 1997.


TITLE: Is there a 'Noetherialization' Process for Rings (Can I take a ring and get a "similar" Noetherian one)
QUESTION [7 upvotes]: I had asked this question in one of the seminars I attend and they said I should ask it here. Let R be a commutative ring with unity
where $1 \neq 0$
and suppose that R is not Noetherian (i.e. all of the 3 equivalent conditions: max.c, acc, and finitely generated ideals - all fail concurrently).
Can we take this R through some sort of algorithim to obtain a similar ring R(N) which is Noetherian? I get that the usage of "similar" here is really vague but I am not sure how else to phrase it. Also I would like for this new ring R(N) to be commutative with unity, and having $1 \neq 0$ ideally (or at least left and right Noetherian if it is not commutative let's say). If there is really nothing of this sort, then my question becomes, what is the next best thing? Again that is vague but not sure how else to phrase it. If there is such a thing, then can we generalize it to modules over a commutative unital ring with $1 \neq 0$?
(I tried googling this, no luck there).

REPLY [7 votes]: Let $\newcommand{\CRing}{\mathrm{CRing}}\CRing$ be the category of commutative, unital rings and let $\CRing_N$ be the full subcategory of Noetherian rings. Usually something like a "Noetherialization" would mean an adjoint to the inclusion of categories, so we can ask whether it is a reflective subcategory (which means that the inclusion $\CRing_N\hookrightarrow\CRing$ has a left adjoint). This however fails since this functor does not create limits. For instance, a countable product of the ring $\mathbb Z$ is not Noetherian, so it cannot be reflected to the subcategory of Noetherian rings.
Considering colimits should similarly show that this category is not coreflective either, so inclusion doesn't admit a right adjoint either.