text:
\section{Introduction}

Strongly shortcut graphs were introduced in earlier work of this
author \cite{Hoda:2018b} as graphs satisfying a weak notion of
nonpositive curvature.  They were shown to unify a broad family of
graphs of interest in geometric group theory and metric graph theory,
including hyperbolic graphs, $1$-skeleta of finite dimensional
$\CAT(0)$ cube complexes and standard Cayley graphs of finitely
generated Coxeter groups \cite{Hoda:2018b}.  Strongly shortcut groups
are defined as those groups admitting a proper and cocompact action on
a strongly shortcut graph \cite{Hoda:2018b}.  They are finitely
presented and have polynomial isoperimetric functions and so have
decidable word problem \cite{Hoda:2018b}.

A graph $\Gamma$ is \newterm{strongly shortcut} if, for some $K > 1$,
there is a bound on the lengths of cycles $\alpha \colon S \to \Gamma$
for which
$d_X\bigl(\alpha(p), \alpha(\bar p)\bigr) \ge \frac{1}{K} \cdot
\frac{|S|}{2}$ for every antipodal pair of points $p, \bar p \in S$.
By a theorem of this author, a graph $\Gamma$ is \newterm{strongly
  shortcut} if and only if, for some $K > 1$, there is a bound on the
lengths of the $K$-bilipschitz embedded cycles of $\Gamma$
\cite{Hoda:2018b}.

In this paper, we introduce the following generalization of this
notion to rough geodesic metric spaces: an $R$-rough geodesic metric
space $X$ is \newterm{strongly shortcut} if, for some $K > 1$, there
is a bound on the lengths of Riemannian circles $S$ for which there
exists an $R$-rough $1$-Lipschitz map $\alpha \colon S \to X$ that
satisfies
$d_X\bigl(\alpha(p), \alpha(\bar p)\bigr) \ge \frac{1}{K} \cdot
\frac{|S|}{2}$ for every antipodal pair of points $p, \bar p \in S$.
We give several characterizations of the strong shortcut property, we
show that a group acting metrically properly and coboundedly on a
strongly shortcut rough geodesic metric space is a strongly shortcut
group and we prove a few other results that may be of independent
interest in metric geometry and geometric group theory.  The results
of this paper are applied in several upcoming papers of the present
author and his coauthors \cite{hoda_krishna:relhyp:2020,
  haettel_hoda_petyt:hhshelly:2020, hoda_przytycki:heisenberg:2020}.

Below is a summary of our main results.

\begin{mainthm}[\Corref{sshortcut_equiv}]
  \mainthmlabel{sshortcut_equiv} Let $X$ be an $R$-rough geodesic
  metric space.  The following conditions are equivalent.
  \begin{enumerate}
  \item $X$ is strongly shortcut.
  \item There exists an $L > 1$ such that there is a bound on the
    lengths of the $(L,4R)$-quasi-isometric embeddings of Riemannian
    circles in $X$.
  \item There exists an $L > 1$ such that for every $C \ge 0$ there is
    a bound on the lengths of the $(L,C)$-quasi-isometric embeddings
    of Riemannian circles in $X$.
  \item \itmlabel{mainthm_ngons_bounded} For some $L > 1$ and some
    $n \in \N$, there is a bound on the $\lambda > 0$ for which there
    exists an $L$-bilipschitz embedding of $\lambda S_n^0$ in $X$,
    where $S_n^0$ is the vertex set of the cycle graph $S_n$ of length
    $n$ and $\lambda S_n^0$ is $S_n^0$ with the metric scaled by
    $\lambda$.
  \item \itmlabel{mainthm_no_ascone_circle} No asymptotic cone of $X$
    contains an isometric copy of the Riemannian circle of unit
    length.
  \end{enumerate}
\end{mainthm}

Conditions \pitmref{mainthm_ngons_bounded} and
\pitmref{mainthm_no_ascone_circle} of \Mainthmref{sshortcut_equiv}
also make sense for general metric spaces and we prove that they are
equivalent for general metric spaces.  (See
\Corref{general_metric_equiv}.)  Thus one may consider condition
\pitmref{mainthm_ngons_bounded} of \Mainthmref{sshortcut_equiv} as the
definition of the strong shortcut property for general metric spaces.

\begin{mainthm}[\Corref{graph_equiv}, \Corref{sshortcut_equiv}]
  Let $\Gamma$ be a graph.  Then $\Gamma$ is strongly shortcut as a
  graph if and only if $\Gamma$ is strongly shortcut as a geodesic
  metric space.
\end{mainthm}

\begin{mainthm}[\Corref{ss_group}]
  \mainthmlabel{ss_group} Let $G$ be a group.  The following
  conditions are equivalent
  \begin{enumerate}
  \item $G$ is strongly shortcut.
  \item $G$ acts metrically properly and coboundedly on a strongly
    shortcut rough geodesic metric space.
  \item $G$ has a finite generating set $S$ for which the Cayley graph
    of $(G,S)$ is strongly shortcut.
  \end{enumerate}
\end{mainthm}

Asymptotically $\CAT(0)$ spaces and groups were first introduced and
studied by Kar \cite{Kar:2011}.  A metric space $X$ is
\newterm{asymptotically $\CAT(0)$} if every asymptotic cone of $X$ is
$\CAT(0)$.  A group is \newterm{asymptotically $\CAT(0)$} if it acts
properly and cocompactly on an asymptotically $\CAT(0)$ proper
geodesic metric space.  Examples of asymptotically $\CAT(0)$ spaces
include $\CAT(0)$ spaces, Gromov-hyperbolic spaces and
$\widetilde{\SL(2,\R)}$ with the Sasaki metric \cite{Kar:2011}.

\begin{mainthm}[\Thmref{as_cat0_space}, \Thmref{as_cat0_group}]
  Asymptotically $\CAT(0)$ rough geodesic metric spaces are strongly
  shortcut.  Consequently, asymptotically $\CAT(0)$ groups are
  strongly shortcut.
\end{mainthm}

\begin{mainthm}[\Corref{as_cones_ss}]
  Let $X$ be an $R$-rough geodesic metric space.  If $X$ is strongly
  shortcut then every asymptotic cone of $X$ is strongly shortcut.
\end{mainthm}

\begin{mainthm}[\Corref{rough_approx_inv}]
  \mainthmlabel{mainthm_rough_approx_inv} Let $X$ be a strongly
  shortcut rough geodesic metric space.  Then there exists an
  $L_X > 1$ such that whenever $Y$ is a rough geodesic metric space
  and $C > 0$ and $f \colon Y \to X$ is an $(L_X,C)$-quasi-isometry up
  to scaling, then $Y$ is also strongly shortcut.  In particular, the
  strong shortcut property is a rough similarity invariant of rough
  geodesic metric spaces.
\end{mainthm}

In fact, \Mainthmref{mainthm_rough_approx_inv} holds for general
metric spaces with condition \pitmref{mainthm_ngons_bounded} of
\Mainthmref{sshortcut_equiv} in place of the strong shortcut property.
(See \Propref{not_approx_ngon_invariant}.)

We call the following result the Circle Tightening Lemma.  It is
essential in the proof of \Mainthmref{sshortcut_equiv}.  We believe it
may be of independent interest.  Note that in the statement of the
Circle Tightening Lemma, the rough geodesicity constant $R$ may be
equal to $0$ in which case the result is about $1$-Lipschitz maps and
$L$-bilipschitz maps in a geodesic metric space.

\begin{mainthm}[Circle Tightening Lemma, \Lemref{circle_tightening}]
  Let $N > 1$, let $L > 1$, let $K > 1$ be small enough (depending on
  $N$ and $L$), let $R \ge 0$, let $M > 0$ be large enough (depending
  on $N$, $L$, $K$ and $R$) and let $C \ge 4R$.

  Let $\alpha \colon S \to X$ be an $R$-rough $1$-Lipschitz map from a
  Riemannian circle $S$ to an $R$-rough geodesic metric space $X$.
  If
  \[ d_X\bigl(\alpha(p), \alpha(\bar p)\bigr) \ge \frac{1}{K} d_S(p,
    \bar p) \] for every antipodal pair $p, \bar p \in S$ and
  $|S| > M$ then there exists a countable collection $\{Q_i\}_i$ of
  closed segments in $S$ such that the following conditions hold.
  \begin{enumerate}
  \item $\sum_i|Q_i| < \frac{|S|}{N}$
  \item For each $i$ there is a function
    $\gamma_i \colon \bar Q_i \to X$ from an interval $\bar Q_i$ such
    that $\gamma_i$ has the same endpoints as $\alpha|_{Q_i}$ and
    $|\bar Q_i| < |Q_i|$.
  \item Replacing each $Q_i \to X$ in $\alpha$ with $\bar Q_i \to X$
    results in an $(L,C)$-quasi-isometric embedding of a circle that
    is also $R$-rough $1$-Lipschitz.
  \end{enumerate}
  Moreover, if $C > 0$ then $\{Q_i\}_i$ is a finite collection.
\end{mainthm}

We call the following refinement of the Milnor-Schwarz Lemma the Fine
Milnor-Schwarz Lemma.  It is used in the proof of
\Mainthmref{ss_group}.  It essentially says that if a group $G$ acts
metrically properly and coboundedly on a rough geodesic space $X$
then, up to scaling, the group $G$ has word metrics that are
quasi-isometric to $X$ with multiplicative constant arbitrarily close
to $1$.  We believe it may be of independent interest.

\begin{mainthm}[Fine Milnor Schwarz Lemma, \Lemref{fine_ms}, \Rmkref{metric_proper_finite}]
  Let $(X,d)$ be a rough geodesic metric space.  Let $G$ be a group
  acting metrically properly and coboundedly on $X$ by isometries.
  Fix $x_0 \in X$.  For $t > 0$ let $S_t$ be the finite set defined by
  \[ S_t = \bigl\{ g \in G \sth d(x_0,gx_0) \le t \bigr\} \] and
  consider the word metric $d_{S_t}$ defined by $S_t$.  (For those $t$
  where $S_t$ does not generate $G$, we allow $d_{S_t}$ to take the
  value $\infty$).  Let $K_t$ be the infimum of all $K > 1$ for which
  \begin{align*}
    (G,td_{S_t}) &\to X \\
    g &\mapsto g \cdot x_0
  \end{align*}
  is a $(K, C_K)$-quasi-isometry for some $C_K \ge 0 $.  Then
  $K_t \to 1$ as $t \to \infty$.
\end{mainthm}

\subsection{Structure of the paper}

In \Secref{definitions} we introduce basic notions that will be used
throughout the paper.  In \Secref{characterizations} prove various
characterizations of the strong shortcut property and prove that it is
a rough similarity invariant.  In \Secref{circle_tightening} we state
and prove the Circle Tightening Lemma.  In \Secref{fine_ms} we state
and prove the Fine Milnor-Schwarz Lemma.  In \Secref{as_cat0}, we
apply the results of the previous sections to prove that
asymptotically $\CAT(0)$ groups are strongly shortcut.

\subsection{Acknowledgements}

This work was partially supported by the ERC grant GroIsRan.

\section{Basic notions and definitions}
\seclabel{definitions}

Let $X$ and $Y$ be metric spaces, let $S$ be a set and let $R$ be a
positive real.  A function $f \colon S \to Y$ is $R$-roughly onto if
every $y \in Y$ is at distance at most $R$ from some point in
$f(X) \subset Y$.  An \defterm{$R$-rough isometric embedding} from $X$
to $Y$ is a function $f \colon X \to Y$ such that
\[ d(x_1,x_2) - R \le d\bigl(f(x_1),f(x_2)\bigr) \le d(x_1,x_2) + R \]
for all $x_1,x_2 \in X$.  An $R$-rough isometric embedding
$f \colon X \to Y$ is an \defterm{$R$-rough isometry} if it is roughly
onto.

An \defterm{$R$-rough geodesic} in $X$ from $x_1$ to $x_2$ is an
$R$-rough isometric embedding $f$ from the interval
$\bigl[0,\ell\bigr] \subset \R$ to $X$ with $\ell = d(x_1,x_2)$ such
that $f(0) = x_1$ and $f(\ell) = x_2$.  A metric space $(X,d)$ is
\defterm{$R$-rough geodesic} if every pair of points in $X$ is joined
by an $R$-rough geodesic.

Let $X$ and $Y$ be metric spaces, let $R \ge 0$ and let $K \ge 1$.  An
\defterm{$R$-rough $K$-Lipschitz map} from $Y$ to $X$ is a function
$\alpha \colon Y \to X$ such that
\[ d\bigl(\alpha(p), \alpha(q)\bigr) \le K d(p, q) + R \] for all
$p,q \in Y$.  An \defterm{$R$-path} in $X$ is an $R$-rough
$1$-Lipschitz map $\alpha \colon P \to X$ from an interval
$P \subset \R$.  An \defterm{$R$-circle} in $X$ is an $R$-rough
$1$-Lipschitz map $\alpha \colon S \to X$ from a Riemannian circle
$S$.  We use the notation $|F|$ to denote the length of $F$, where $F$
is an interval, a Riemannian circle or a finite union of closed
segments in an interval or in a Riemannian circle.

\begin{rmk}
  \rmklabel{rpath_concatenation} The concatenation of two $R$-paths
  need not be an $R$-path.  However, if $\alpha_1 \colon P_1 \to X$
  and $\alpha_2 \colon P_2 \to X$ are a pair of concatenatable
  $R$-paths and $\gamma \colon [0,R] \to X$ is the constant path at
  the point of concatenation then the concatenation
  $\alpha_1\gamma\alpha_2$ is an $R$-path.
\end{rmk}

An $R$-circle $\alpha \colon S \to X$ is \defterm{$\frac{1}{K}$-almost
  isometric}, for some $K > 1$, if
\[ d\bigl(\alpha(p), \alpha(\bar p)\bigr) \ge \frac{1}{K} \cdot
  \frac{|S|}{2} \] for every antipodal pair of points
$p, \bar p \in S$.

\begin{defn}
  \defnlabel{sshortcut} An $R$-rough geodesic metric space $X$ is
  \defterm{strongly shortcut} if, for some $K > 1$, there is a bound
  on the lengths of the $\frac{1}{K}$-almost isometric $R$-cycles of
  $X$.
\end{defn}

\begin{rmk}
  By \Corref{sshortcut_equiv}, the apparent dependence on $R$ in
  \Defnref{sshortcut} is not essential.  That is, if $X$ is an
  $R$-rough geodesic metric space and $R' > R$ then $X$ is strongly
  shortcut if and only if it is strongly shortcut when viewed as an
  $R'$-rough geodesic metric space.
\end{rmk}

We view graphs as geodesic metric spaces with each edge isometric to a
unit interval.  For a graph $\Gamma$, we use the notation $\Gamma^0$
to denote the vertex set of $\Gamma$ with its subspace metric.  The
\defterm{cycle graph} $S_n$ of length $n$ is the graph isometric to a
Riemannian circle of length $n$.  A \defterm{cycle} in a graph
$\Gamma$ is a combinatorial map $S_n \to \Gamma$ from some cycle graph
$S_n$ to $\Gamma$.  A \defterm{path graph} is a graph isometric to a
real interval.  A \defterm{combinatorial path} in a graph $\Gamma$ is
a combinatorial map $P \to \Gamma$ from a path graph $P$.

Note that if $\alpha \colon S_n \to \Gamma$ is a cycle in a graph
$\Gamma$ then $\alpha$ is a $1$-Lipschitz map from a Riemannian circle
to a geodesic metric space or, in the language we have established
above, an $R$-circle in an $R$-rough geodesic metric space where
$R = 0$.

\begin{defn}
  A graph $\Gamma$ is \defterm{strongly shortcut as a graph} if, for
  some $K > 1$, there is a bound on the lengths of the
  $\frac{1}{K}$-almost isometric cycles of $\Gamma$.
\end{defn}

\begin{rmk}
  \rmklabel{ss_asgraph_asmetric} By \Corref{graph_equiv},
  \Corref{sshortcut_equiv} and \Corref{rough_approx_inv}, the
  following conditions are equivalent for a graph $\Gamma$.
  \begin{enumerate}
  \item $\Gamma$ is strongly shortcut as a graph.
  \item $\Gamma$ is strongly shortcut as a geodesic metric space.
  \item $\Gamma^0$ is strongly shortcut as a rough geodesic metric
    space.
  \end{enumerate}
\end{rmk}

If $X$ is a metric space and $\lambda > 0$ then we write $\lambda X$
to denote the metric space obtained from $X$ by scaling the metric by
$\lambda$.

\section{Characterizing the strong shortcut property}
\seclabel{characterizations}

In this section we will give various characterizations of the strong
shortcut property.

\begin{lem}
  \lemlabel{global_to_local} Let $\alpha \colon S \to X$ be a
  $\frac{1}{K}$-almost isometric $R$-circle in a metric space $X$.
  Then
  \[ d\bigl(\alpha(p),\alpha(q)\bigr) \ge d(p,q) - \frac{K-1}{K}
    \cdot \frac{|S|}{2} - 2R \] for all $p,q \in S$.
\end{lem}
\begin{proof}
  Let $p',q' \in S \setminus \{p,q\}$ be antipodal and suppose that a
  geodesic segment of $S$ visits $p'$, $p$, $q$, and $q'$, in that
  order.  Then
  \begin{align*}
    \frac{1}{K} \cdot \frac{|S|}{2}
    &\le d\bigl(\alpha(p'),\alpha(q')\bigr) \\
    &\le d\bigl(\alpha(p'),\alpha(p)\bigr)
      + d\bigl(\alpha(p),\alpha(q)\bigr)
      + d\bigl(\alpha(q),\alpha(q')\bigr) \\
    &\le d(p',p) + R + d\bigl(\alpha(p),\alpha(q)\bigr) + d(q,q') + R \\
    &= d(p',p) + d(q,q') + d\bigl(\alpha(p),\alpha(q)\bigr)  + 2R \\
    &= \frac{|S|}{2} - d(p,q) + d\bigl(\alpha(p),\alpha(q)\bigr)  + 2R
  \end{align*}
  from which we can obtain the desired inequality.
\end{proof}

The following definition is very useful because it applies to general
metric spaces.

\begin{defn}
  A metric space $X$ \defterm{approximates $n$-gons} if, for every
  $K > 1$, and every $n \in \N$ there exist $K$-bilipschitz embeddings
  of $\lambda S_n^0$ in $X$ for arbitrarily large $\lambda > 0$.
\end{defn}

We will see that in the case of a graph or a rough geodesic metric
space \emph{non}approximation of $n$-gons is equivalent to the strong
shortcut property.  Thus it would make sense to define the strong
shortcut property for general metric spaces as nonapproximability of
$n$-gons.

\begin{defn}
  Let $X$ and $Y$ be metric spaces.  A function $f \colon Y \to X$ is
  a $(K,C)$-quasi-isometry \defterm{up to scaling} if there exists a
  $\lambda > 0$ such that $f$ is a $(K,C)$-quasi-isometry when viewed
  as a function from $\lambda Y$ to $X$.  A function
  $f \colon Y \to X$ is a \defterm{rough similarity} if, for some
  $C > 0$, the function $f$ is a $(1,C)$-quasi-isometry up to scaling.
  A property $\mathscr{P}$ of metric spaces is a \defterm{rough
    similarity invariant} if whenever $X$ satisfies $\mathscr{P}$ and
  $f \colon Y \to X$ is a rough similarity then $Y$ also satisfies
  $\mathscr{P}$.  A property $\mathscr{P}$ of metric spaces is a
  \defterm{rough approximability invariant} if, for any metric space
  $X$ satisfying $\mathscr{P}$, there exists an $L_X > 1$ such that
  whenever $C > 0$ and $f \colon Y \to X$ is an
  $(L_X,C)$-quasi-isometry up to scaling, then $Y$ also satisfies
  $\mathscr{P}$.
\end{defn}

\begin{prop}
  \proplabel{not_approx_ngon_invariant} Nonapproximability of $n$-gons
  is a rough approximability invariant of metric spaces.  In
  particular, nonapproximability of $n$-gons is a rough similarity
  invariant of metric spaces.
\end{prop}
\begin{proof}
  Let $X$ be a metric space that does not approximate $n$-gons.  Then
  is a $K > 1$ and an $n \in \N$ and a $\Lambda > 0$ such that any
  $K$-bilipschitz embedding of $\lambda S_n^0$ in $X$ satisfies
  $\lambda < \Lambda$.

  Let $Y$ be a metric space, let $t > 0$, let $L \in (1,K)$ and let
  $f \colon tY \to X$ be an $(L,C)$-quasi-isometry.  Let
  $K' \in \bigl(1, \frac{K}{L}\bigr)$ and let
  $\lambda' > \frac{CLK'}{t}$.  We will show that there is a bound on
  the $\lambda'$ for which there exists a $K'$-bilipschitz embedding
  $\alpha \colon \lambda' S_n^0 \to Y$.  Viewing such an $\alpha$ as
  map from $t\lambda' S_n^0$ to $t Y$ the composition
  $f \comp \alpha \colon t\lambda' S_n^0 \to X$ is an
  $(LK',C)$-quasi-isometric embedding.  But the minimum distance
  between distinct points in $t\lambda' S_n^0$ is $t\lambda'$ and so
  one can show that $f \comp \alpha$ is a
  $\frac{t\lambda' LK' + C}{t\lambda' - LK'C}$-bilipschitz embedding
  from $t\lambda' S_n^0$.  But
  $\frac{t\lambda' LK' + C}{t\lambda' - LK'C} \to LK' < K$ as
  $\lambda' \to \infty$ so there is a $\Lambda_0$ such that if
  $\lambda' \ge \Lambda_0$ then
  $\frac{t\lambda' LK' + C}{t\lambda' - LK'C} < K$.  So if we had
  $\lambda' \ge \Lambda' = \max\bigl\{\Lambda_0,
  \frac{\Lambda}{t}\bigr\}$ then $f \comp\alpha$ would be a
  $K$-bilipschitz embedding of $\lambda S_n^0$ in $X$ with
  $\lambda = t\lambda' \ge \Lambda$, which would be a contradiction.
  Thus $\Lambda'$ bounds the the $\lambda'$ for which there exists a
  $K'$-bilipschitz embedding $\alpha \colon \lambda' S_n^0 \to Y$, as
  required.
\end{proof}

\begin{thm}
  \thmlabel{graph_neg_equiv} Let $\Gamma$ be a graph.  Then the
  following conditions are equivalent.
  \begin{enumerate}
  \item \itmlabel{not_sshortcut_as_graph} $\Gamma$ is not strongly
    shortcut as a graph.
  \item \itmlabel{graph_ngons} $\Gamma$ approximates $n$-gons.
  \end{enumerate}
\end{thm}
\begin{proof}
  $\pitmref{not_sshortcut_as_graph} \Rightarrow \pitmref{graph_ngons}$
  Let $K' > 1$ and let $\alpha \colon S_{n'} \to \Gamma$ be a
  $\frac{1}{K'}$-almost isometric cycle.  Let $n \in \N$ and subdivide
  $S_{n'}$ into $n$ segments of equal length, ignoring the original
  graph structure on $S_{n'}$.  Let $Y \subset S_{n'}$ be the set of
  endpoints of the segments.  Then $Y$ is isometric to $\lambda S_{n}$
  for $\lambda = \frac{n'}{n}$.  Let $\alpha'$ be the composition of
  the inclusion $Y \hookrightarrow S_{n'}$ with $\alpha$.  Let
  $p, q \in Y$ be distinct.  Then $d(p,q) \ge \frac{|S_{n'}|}{n}$ and,
  by \Lemref{global_to_local},
  \begin{align*}
    d\bigl(\alpha'(p),\alpha'(q)\bigr)
    &\ge d(p,q) - \frac{K'-1}{K'} \cdot \frac{|S_{n'}|}{2} \\
    &\ge d(p,q) - \frac{K'-1}{K'} \cdot \frac{n d(p,q)}{2} \\
    &= \Bigl(1 - \frac{n(K'-1)}{2K'} \Bigr)d(p,q)
  \end{align*}
  but $d\bigl(\alpha'(p),\alpha'(q)\bigr) \le d(p,q)$ so, when $K'$ is
  small enough that $\frac{n(K'-1)}{2K'} < 1$, the map $\alpha'$ is
  $K$-bilipschitz for $K = \Bigl(1 - \frac{n(K'-1)}{2K'} \Bigr)^{-1}$.
  Thus, given an arbitrary $n \in \N$ and an $\alpha'$ as above with
  $K'$ small enough, we can obtain a $K$-bilipschitz embedding of
  $\lambda S_n$ in $X$ with
  $K = \Bigl(1 - \frac{n(K'-1)}{2K'} \Bigr)^{-1}$ and
  $\lambda = \frac{n'}{n}$.  Since $\Gamma$ is not strongly shortcut,
  there exist $\alpha$ as above with $K' > 1$ arbitrarily close to $1$
  and with $n'$ arbitrarily large.  But $K \to 1$ as $K' \to 1$ and
  $\lambda \to \infty$ as $n' \to \infty$ so we have $K$-bilipschitz
  embeddings of $\lambda S_n$ with $K$ arbitrarily close to $1$ and
  $\lambda$ arbitrarily large.

  $\pitmref{graph_ngons} \Rightarrow \pitmref{not_sshortcut_as_graph}$
  Let $n \in \N$, let $K > 1$, let $\lambda > K$ and let
  $\alpha \colon \lambda S_n^0 \to X$ be a $K$-bilipschitz embedding.
  There is a retraction $r \colon \Gamma \to \Gamma^0$ such that $r$
  is a $(1,1)$-quasi-isometry.  Then the composition $r \comp \alpha$
  is a $(K,1)$-quasi-isometric embedding.  But distinct points in
  $\lambda S_n^0$ are at distance at least $\lambda$ and, since
  $K < \lambda$, this implies that $r \comp \alpha$ is
  $L$-bilipschitz, where $L = \frac{K\lambda + 1}{\lambda - K}$.  View
  $S_n$ as the Cayley graph of $\Z/n\Z$ with generating set $\{1\}$
  and, for $i \in \Z/n\Z$, let $v_i$ be the vertex of $S_n$
  corresponding to $i$.  Then, for each $i$, we have
  $d\bigl(r \comp \alpha(v_i), r \comp \alpha(v_{i+1})\bigr) \le
  \lfloor L\lambda \rfloor$ so there is a combinatorial path
  $\gamma_i \colon P_i \to \Gamma$ of length
  $m_i \in \bigl\{ \lfloor L\lambda \rfloor-1, \lfloor L\lambda
  \rfloor \bigr\}$ from $r \comp \alpha(v_i)$ to
  $r \comp \alpha(v_{i+1})$.  For each $i$, identify the endpoint of
  $P_i$ with the initial point of $P_{i+1}$ to obtain a cycle
  $\gamma \colon S_m \to \Gamma$ with $m = \sum_{i=1}^n m_i$.  Then
  $\alpha$ factors through $\gamma$ via the embedding that sends $v_i$
  to the initial point of $P_i \subset S_m$.  So, viewing $S_n^0$ as a
  subset of $S_m^0$ via this embedding, we have
  $r \comp \alpha(v_i) = \gamma(v_i)$, for each $i$.  Let $x \in S_m$
  and let $v_i$ minimize $d(x,v_i)$.  Then
  $d(x,v_i) \le \frac{\lfloor L\lambda \rfloor}{2}$ and if $\bar x$ is
  the antipode of $x$ and
  $\bar i = i + \bigl\lfloor \frac{n}{2} \bigr\rfloor$ then
  $d(\bar x, v_{\bar i}) \le \lfloor L\lambda \rfloor + \frac{n}{2}$
  so we have the following computation.
  \begin{align*}
    d\bigl(\gamma(x), \gamma(\bar x)\bigr)
    &\ge d\bigl(\gamma(v_i), \gamma(v_{\bar i})\bigr)
      - d\bigl(\gamma(x), \gamma(v_i)\bigr)
      - d\bigl(\gamma(\bar x), \gamma(v_{\bar i})\bigr) \\
    &\ge d\bigl(\gamma(v_i), \gamma(v_{\bar i})\bigr)
      - d(x, v_i) - d(\bar x, v_{\bar i}) \\
    &\ge d\bigl(\gamma(v_i), \gamma(v_{\bar i})\bigr)
      - \frac{\lfloor L\lambda \rfloor}{2}
      - \lfloor L\lambda \rfloor - \frac{n}{2} \\
    &= d\bigl(r \comp \alpha(v_i), r \comp \alpha(v_{\bar i})\bigr)
      - \frac{3\lfloor L\lambda \rfloor}{2} - \frac{n}{2} \\
    &\ge \frac{1}{L} d_{\lambda S_n^0}\bigl(v_i, v_{\bar i})
      - \frac{3\lfloor L\lambda \rfloor}{2} - \frac{n}{2} \\
    &= \frac{\lambda}{L} \Bigl\lfloor \frac{n}{2} \Bigr\rfloor
      - \frac{3\lfloor L\lambda \rfloor}{2}
      - \frac{n}{2} \\
    &\ge \frac{\lambda}{L} \Bigl( \frac{n-1}{2} \Bigr)
      - \frac{3L\lambda}{2} - \frac{n}{2} \\
    &= \frac{1}{m}\Bigl( \frac{\lambda(n-1)}{L}
      - 3L\lambda - n \Bigr)\frac{|S_m|}{2}
  \end{align*}
  Since $m \le nL\lambda$, the above computation implies that
  \[ d\bigl(\gamma(x), \gamma(\bar x)\bigr) \ge \Bigl(
    \frac{n-1}{nL^2} - \frac{3}{n} - \frac{1}{L\lambda}
    \Bigr)\frac{|S_m|}{2} \] so, given $\alpha$ as above, we can
  obtain a $\frac{1}{K'}$-almost isometric cycle in $\Gamma$ of length
  $m$, where
  $\frac{1}{K'} = \bigl( \frac{n-1}{nL^2} - \frac{3}{n} -
  \frac{1}{L\lambda} \bigr)$.  We need only show there exist $\alpha$
  for which $\frac{1}{K'}$ is aritrarily close to $1$ and $m$ is
  arbitrarily large.  By hypothesis, there exist $\alpha$ for which
  $K = \frac{n+1}{n}$ and $\lambda > n$, for arbitrary $n \in \N$.
  But then, as $n \to \infty$, we have
  $m \ge n\bigl(\lfloor L\lambda \rfloor-1\bigr) \to \infty$ and
  $L = \frac{K\lambda + 1}{\lambda - K} \to K$ so
  $\frac{1}{K'} = \bigl( \frac{n-1}{nL^2} - \frac{3}{n} -
  \frac{1}{L\lambda} \bigr) \to 1$.
\end{proof}

\begin{cor}
  \corlabel{graph_equiv} Let $\Gamma$ be a graph.  Then the following
  conditions are equivalent.
  \begin{enumerate}
  \item \itmlabel{sshortcut_as_graph} $\Gamma$ is strongly shortcut as a
    graph.
  \item \itmlabel{graph_ngons_bounded} $\Gamma$ does not approximate
    $n$-gons.
  \end{enumerate}
\end{cor}

Let $X$ be a metric space.  Let $\mathscr{U}$ be a nonprincipal
ultrafilter on $\N$.  Let $(b^{(m)})_{m\in\N}$ be a sequence in $X$.
Let $(s^{(m)})_{m\in\N}$ be a sequence of positive reals such that
$s^{(m)} \to \infty$ as $m \to \infty$.  Consider the set
\[ \mathfrak{X}' = \Biggl\{(x_m)_{m\in\N} \sth
  \text{$\biggl(\frac{d(x_m,b^{(m)})}{s^{(m)}}\biggr)_{m\in\N}$ is
    bounded} \Biggr\} \] of \defterm{sequences in $X$ that are bounded
  with respect to} the \defterm{basepoint sequence} $(b^{(m)})_m$ and
the \defterm{scaling sequence} $(s^{(m)})_m$.  For
$(x_m)_m, (x'_m)_m \in \mathfrak{X}'$,
\[ \bar d\bigl((x_m)_m,(x'_m)_m\bigr) = \lim_{\mathscr{U}}
  \frac{d(x_m,x'_m)}{s^{(m)}} \] defines a pseudometric on
$\mathfrak{X}'$.  The \defterm{asymptotic cone} $\mathfrak{X}$ of $X$
with respect to the nonprincipal ultrafilter $\mathscr{U}$, the
\defterm{basepoint sequence} $(b^{(m)})_m$ and the \defterm{scaling
  sequence} $(s^{(m)})_m$ is the metric space obtained from
$\mathfrak{X}'$ and $\bar d$ by identifying $(x_m)_m$ and $(x'_m)_m$
whenever $\bar d\bigl((x_m)_m,(x'_m)_m\bigr) = 0$.

Note that \Thmref{general_metric_neg_equiv},
\Corref{general_metric_equiv} and \Corref{as_cones_general_ss} apply to
general metric spaces and not just rough geodesic metric spaces.

\begin{thm}
  \thmlabel{general_metric_neg_equiv} Let $X$ be a metric space.  Then
  the following conditions are equivalent.
  \begin{enumerate}
  \item \itmlabel{general_asymptotic_circle} There is an asymptotic
    cone of $X$ that contains an isometric copy of the Riemannian
    circle of unit length.
  \item \itmlabel{general_ngons} $X$ approximates $n$-gons.
  \end{enumerate}
\end{thm}
\begin{proof}
  $\pitmref{general_asymptotic_circle} \Rightarrow
  \pitmref{general_ngons}$ Suppose $S \subset \mathcal{X}$ is a
  subspace isometric to the Riemannian circle of unit length in the
  \defterm{asymptotic cone} $\mathcal{X}$ of $X$ with respect to a
  nonprincipal ultrafilter $\mathscr{U}$, a \defterm{basepoint
    sequence} $(b^{(m)})_m$ and a \defterm{scaling sequence}
  $(s^{(m)})_m$.  Take any $n \in \N$, any $K > 1$ and any
  $\Lambda > 0$.  We will construct a $K$-bilipschitz map
  $\alpha \colon \lambda S_n^0 \to X$ with $\lambda \ge \Lambda$.
  Subdivide $S$ into $n$ segments of equal length and let $S^0$ denote
  the set of endpoints of the segments.  For each $\epsilon > 0$ and
  each $p,q \in S^0$ reprented by $(p_m)_m$ and $(q_m)_m$, there is an
  $A_{\epsilon}^{p,q} \in \mathscr{U}$ such that
  \[ d(p,q) - \epsilon \le \frac{d(p_m,q_m)}{s^{(m)}} \le d(p,q) +
    \epsilon \] for all $m \in A_{\epsilon}^{p,q}$.  There are
  finitely many pairs $p,q \in S^0$ so
  $A_{\epsilon} = \bigcap_{p,q} A_{\epsilon}^{p,q} \in \mathscr{U}$.
  Then, for any distinct $p,q \in S^0$,
  \[ \Bigl(1 - \frac{\epsilon}{d(p,q)}\Bigr)s^{(m)}d(p,q) \le
    d(p_m,q_m) \le \Bigl(1 +
    \frac{\epsilon}{d(p,q)}\Bigr)s^{(m)}d(p,q) \] for all
  $m \in A_{\epsilon}$.  But $d(p,q) \ge \frac{1}{n}$ and so
  \[ (1 - n\epsilon)s^{(m)}d(p,q) \le d(p_m,q_m) \le (1 +
    n\epsilon)s^{(m)}d(p,q) \] for all $m \in A_{\epsilon}$.  So if
  $n\epsilon < 1$ then, for $m \in A_{\epsilon}$, the map
  \begin{align*}
    \alpha_m \colon s^{(m)} S^0 &\to X \\
    p &\mapsto p_m
  \end{align*}
  is bilipschitz with bilipschitz constant
  $\max\bigl\{1 + n\epsilon, \frac{1}{1-n\epsilon} \bigr\} =
  \frac{1}{1-n\epsilon}$.  The space $s^{(m)} S^0$ is isometric to
  $\frac{s^{(m)}}{n} S_n^0$ so if we chose $\epsilon$ small enough so
  that $n\epsilon < 1$ and $\frac{1}{1-n\epsilon} < K$ and we take
  $m \in A_{\epsilon}$ large enough that
  $\frac{s^{(m)}}{n} \ge \Lambda$ then we can take
  $\alpha = \alpha_m$.

  $\pitmref{general_ngons} \Rightarrow
  \pitmref{general_asymptotic_circle}$ For $m \in \N$, there exists a
  $\frac{m+1}{m}$-bilipschitz map
  $\alpha_m \colon \lambda_m S_{2^m}^0 \to X$ with $\lambda_m \ge m$.
  Metrize the group
  $\frac{1}{2^m}\Z = \bigl\{\frac{k}{2^m} \sth k \in \Z\bigr\} \subset
  \R$ with the subspace metric and metrize the quotient group
  $\frac{1}{2^m}\Z/\Z$ with the quotient metric.  Then
  $\frac{1}{2^m} S_{2^m}^0$ is isometric to $\frac{1}{2^m}\Z/\Z$.  Via
  this isometry we identify the vertex set $S_{2^m}^0$ with the
  elements of $\frac{1}{2^m}\Z/\Z$.  Thus we view
  $\frac{1}{2^m} S_{2^m}^0$ as a metric subspace of the Riemannian
  circle of unit length $S = \R/\Z$.  By this identification, the
  union $S_{\mathbb{D}} = \bigcup_{m \in \N} S_{2^m}^0 \subset S$ is
  the \defterm{dyadic circle} $\Z\bigl[\frac{1}{2}\bigr]/\Z$.  The
  dyadic circle $S_{\mathbb{D}}$ is dense in $S$.  Thus, since
  asymptotic cones are complete metric spaces
  \cite[Proposition~10.70]{Drutu:2018}, it will suffice to
  isometrically embed $S_{\mathbb{D}}$ into an asymptotic cone of $X$.

  View $\alpha_m$ as an $\frac{m+1}{m}$-bilipschitz map from
  $\frac{1}{2^m} S_{2^m}^0$ to $\frac{1}{\lambda_m 2^m}X$.  Set
  $b^{(m)} = \alpha_m(0)$ and set $s^{(m)} = \lambda_m 2^m$.  Every
  nonzero element of $S_{\mathbb{D}}$ can be uniquely represented as
  $\frac{k}{2^{\ell}}$ with $k$ odd and satisfying
  $0 \le k < 2^{\ell}$.  For any such representation
  $\frac{k}{2^{\ell}}$ and any $m \in \N$, set
  \[ x_{\frac{k}{2^{\ell}}}^{(m)} =
    \begin{cases}
      b^{(m)} &\text{if $m < \ell$} \\
      \alpha_m\bigl(\frac{k}{2^{\ell}}\bigr) &\text{if $m \ge \ell$}
    \end{cases} \] and set $x_0^{(m)} = b^{(m)}$.  Then, for any non
  principal ultrafilter $\mathscr{U}$, the expression
  $p \mapsto (x_p^{(m)})_m$ defines an isometric embedding of
  $S_{\mathbb{D}}$ into the asymptotic cone $\mathcal{X}$ of $X$ with
  respect to $\mathscr{U}$, the basepoint sequence $(b^{(m)})_m$ and
  the scaling sequence $S^{(m)}$.  Indeed, for every
  $p,q \in S_{\mathbb{D}}$,
  \[ \frac{d(x_p^{(m)},x_q^{(m)})}{s^{(m)}} =
    \frac{d\bigl(\alpha_m(p),\alpha_m(q)\bigr)}{\lambda_m 2^m} \le
    \frac{\frac{m+1}{m} \cdot \lambda_m d_{S_{2^m}}(p,q)}{\lambda_m
      2^m} = \frac{m+1}{m} \cdot d_{S_{\mathbb{D}}}(p,q) \]
  and
  \[ \frac{d(x_p^{(m)},x_q^{(m)})}{s^{(m)}} =
    \frac{d\bigl(\alpha_m(p),\alpha_m(q)\bigr)}{\lambda_m 2^m} \ge
    \frac{\frac{m}{m+1} \cdot \lambda_m d_{S_{2^m}}(p,q)}{\lambda_m
      2^m} = \frac{m}{m+1} \cdot d_{S_{\mathbb{D}}}(p,q) \] whenever
  $m$ is large enough.
\end{proof}

\begin{cor}
  \corlabel{general_metric_equiv} Let $X$ be a metric space.  Then the
  following conditions are equivalent.
  \begin{enumerate}
  \item \itmlabel{general_no_asymptotic_circle} No asymptotic cone of $X$
    contains an isometric copy of the Riemannian circle of unit
    length.
  \item \itmlabel{general_ngons_bounded} $X$ does not approximate $n$-gons.
  \end{enumerate}
\end{cor}

\begin{cor}
  \corlabel{as_cones_general_ss} Let $X$ be a metric space and let
  $\mathfrak{X}$ be an asymptotic cone of $X$.  Suppose that $X$ does
  not approximate $n$-gons.  Then $\mathfrak{X}$ does not approximate
  $n$-gons.
\end{cor}
\begin{proof}
  By \Corref{general_metric_equiv}, it suffices to show that any
  asymptotic cone $\mathfrak{X}'$ of $\mathfrak{X}$ does not contain
  an isometric copy of a Riemannian circle of unit length.  But
  $\mathfrak{X}'$ is isometric to an asymptotic cone of $X$
  \cite[Corollary~10.80]{Drutu:2018} so does not contain an isometric
  copy of a Riemannian circle of unit length by
  \Corref{general_metric_equiv}.
\end{proof}

\begin{thm}
  \thmlabel{not_sshortcut_equiv} Let $X$ be an $R$-rough geodesic
  metric space.  The following conditions are equivalent.
  \begin{enumerate}
  \item \itmlabel{not_sshortcut} $X$ is not strongly shortcut.
  \item \itmlabel{k_2r_quasi_circles} For every $L > 1$ there exist
    $(L,4R)$-quasi-isometric embeddings of arbitrarily long
    Riemannian circles in $X$.
  \item \itmlabel{quasi_circles} For every $L > 1$ there is a
    $C \ge 0$ such that there exist $(L,C)$-quasi-isometric embeddings
    of arbitrarily long Riemannian circles in $X$.
  \item \itmlabel{ngons} $X$ approximates $n$-gons.
  \item \itmlabel{asymptotic_circle} There is an asymptotic cone of
    $X$ that contains an isometric copy of the Riemannian circle of
    unit length.
  \end{enumerate}
\end{thm}
\begin{proof}
  Conditions \pitmref{ngons} and \pitmref{asymptotic_circle} are
  equivalent for general metric spaces, by
  \Thmref{general_metric_neg_equiv}.  So it will suffice to prove the
  equivalence of conditions \pitmref{not_sshortcut},
  \pitmref{k_2r_quasi_circles}, \pitmref{quasi_circles} and
  \pitmref{ngons}.

  $\pitmref{not_sshortcut} \Rightarrow \pitmref{k_2r_quasi_circles}$
  Let $N = 2$, let $L > 1$ be arbitrary, let $K > 1$ be small enough
  to satisfy \Lemref{circle_tightening} and let
  $\alpha \colon S \to X$ be a $\frac{1}{K}$-almost isometric
  $R$-circle in $X$ with $|S|$ arbitrarily larger than the $M$ from
  \Lemref{circle_tightening}.  Then the limit $R$-circle
  $\alpha_{\infty} \colon S_{\infty} \to X$ given by
  \Lemref{circle_tightening} is an $(L,4R)$-quasi-isometric embedding
  of a Riemannian circle of length at least $\frac{|S|}{2}$.

  $\pitmref{k_2r_quasi_circles} \Rightarrow
  \pitmref{quasi_circles}$ This is immediate.

  $\pitmref{quasi_circles} \Rightarrow \pitmref{ngons}$ Let
  $\alpha \colon S \to X$ be $(L,C)$-quasi-isometric embedding of a
  Riemannian circle.  Let $n \in \N$, subdivide $S$ into $n$ segments
  of equal length and let $Y$ be the set of endpoints of the segments.
  Then $Y$ is isometric to $\frac{|S|}{n} S_n$ and
  \[\Bigl(\frac{1}{L} - \frac{nC}{|S|} \Bigr)d(p,q)
    \le d\bigl(\alpha(p),\alpha(q)\bigr) \le \Bigl(L +
    \frac{nC}{|S|}\Bigr)d(p,q) \] for distinct $p,q \in Y$.  By
  hypothesis, there exist arbitrarily long $\alpha$ with $L$
  arbitrarily close to $1$ and so $\alpha|_Y$ is a $K$-bilipschitz
  embedding of $\lambda S_n$ for $\lambda = \frac{|S|}{n}$ arbitrarily
  large and
  $K = \max\Bigl\{\bigl(\frac{1}{L} - \frac{nC}{|S|}\bigr)^{-1}, L +
  \frac{nC}{|S|}\Bigr\}$ arbitrarily close to $1$.

  $\pitmref{ngons} \Rightarrow \pitmref{not_sshortcut}$ Let
  $n \in \N$, let $L > 1$, let $\lambda > 0$ and let
  $\alpha \colon \lambda S_n^0 \to X$ be an $L$-bilipschitz embedding.
  View $S_n$ as the Cayley graph of $\Z/n\Z$ with generating set
  $\{1\}$ and, for $i \in \Z/n\Z$, let $v_i$ be the vertex of $S_n$
  corresponding to $i$.  Then, for each $i$, we have
  $d\bigl(\alpha(v_i), \alpha(v_{i+1})\bigr) \le L\lambda$ so, by
  scaling an $R$-rough geodesic, there is an $R$-path
  $\gamma'_i \colon P_i \to X$ of length $|P_i| = L\lambda$ from
  $\alpha(v_i)$ to $\alpha(v_{i+1})$.  Let $\gamma_i$ be the
  concatenation $c_i\gamma'_i$ where $c_i\colon [0,R] \to X$ is the
  constant path of length $R$ at $\alpha(v_i)$.  For each $i$,
  identify the endpoint of $P_i$ with the initial point of $P_{i+1}$
  to obtain an $R$-circle $\gamma \colon S \to X$ with
  $|S| = n(L\lambda + R)$.  Then $\alpha$ factors through $\gamma$ via
  the embedding that sends $v_i$ to the initial point of
  $P_i \subset S$.  So, viewing $S_n^0$ as a subset of $S$ via this
  embedding, we have $\alpha(v_i) = \gamma(v_i)$, for each $i$.  Let
  $x \in S$ and let $v_i$ minimize $d(x,v_i)$.  Then
  $d(x,v_i) \le \frac{L\lambda + R}{2}$ and if $\bar x$ is the
  antipode of $x$ and
  $\bar i = i + \bigl\lfloor \frac{n}{2} \bigr\rfloor$ then
  $d(\bar x, v_{\bar i}) \le L\lambda + R$ so we have the following
  computation.
  \begin{align*}
    d\bigl(\gamma(x), \gamma(\bar x)\bigr)
    &\ge d\bigl(\gamma(v_i), \gamma(v_{\bar i})\bigr)
      - d\bigl(\gamma(x), \gamma(v_i)\bigr)
      - d\bigl(\gamma(\bar x), \gamma(v_{\bar i})\bigr) \\
    &\ge d\bigl(\gamma(v_i), \gamma(v_{\bar i})\bigr)
      - d(x, v_i) - R - d(\bar x, v_{\bar i}) - R \\
    &\ge d\bigl(\gamma(v_i), \gamma(v_{\bar i})\bigr)
      - \frac{L\lambda + R}{2}
      - (L\lambda+R) - 2R \\
    &= d\bigl(\alpha(v_i), \alpha(v_{\bar i})\bigr)
      - \frac{3L\lambda + 7R}{2} \\
    &\ge \frac{1}{L} d_{\lambda S_n^0}\bigl(v_i, v_{\bar i})
      - \frac{3L\lambda + 7R}{2} \\
    &= \frac{\lambda}{L} \Bigl\lfloor \frac{n}{2} \Bigr\rfloor
      - \frac{3L\lambda + 7R}{2} \\
    &\ge \frac{\lambda}{L} \Bigl( \frac{n-1}{2} \Bigr)
      - \frac{3L\lambda + 7R}{2} \\
    &= \frac{1}{|S|}\Bigl( \frac{\lambda(n-1)}{L}
      - 3L\lambda - 7R \Bigr)\frac{|S|}{2} \\
    &= \frac{1}{n(L\lambda + R)}\Bigl( \frac{\lambda(n-1)}{L}
      - 3L\lambda - 7R \Bigr)\frac{|S|}{2} \\
    &= \biggl( \frac{\lambda n-\lambda}{L^2\lambda n + nLR}
      - \frac{3L\lambda}{L\lambda n + nR}
      - \frac{7R}{L\lambda n + nR} \biggr)\frac{|S|}{2}
  \end{align*}
  So, given $\alpha$ as above, we can obtain a $\frac{1}{K}$-almost
  isometric $R$-circle in $X$ of length $|S| = n(L\lambda + R)$, where
  $\frac{1}{K} = \Bigl( \frac{\lambda n-\lambda}{L^2\lambda n + nLR}
  - \frac{3L\lambda}{L\lambda n + nR} - \frac{7R}{L\lambda n + nR}
  \Bigr)$.  We need only show there exist $\alpha$ for which
  $\frac{1}{K}$ is aritrarily close to $1$ and $|S|$ is arbitrarily
  large.  By hypothesis, there exist $\alpha$ for which
  $L = \frac{n+1}{n}$ and $\lambda > n$, for arbitrary $n \in \N$.
  But then, as $n \to \infty$, we have
  $|S| = n(L\lambda + R) \to \infty$ and
  $\frac{1}{K} = \Bigl( \frac{\lambda n-\lambda}{L^2\lambda n + nLR}
  - \frac{3L\lambda}{L\lambda n + nR} - \frac{7R}{L\lambda n + nR}
  \Bigr) \to 1$.
\end{proof}

\begin{cor}
  \corlabel{sshortcut_equiv} Let $X$ be an $R$-rough geodesic metric
  space.  The following conditions are equivalent.
  \begin{enumerate}
  \item \itmlabel{sshortcut} $X$ is strongly shortcut.
  \item \itmlabel{k_2r_quasi_circles_bounded} There exists an $L > 1$
    such that there is a bound on the lengths of the
    $(L,4R)$-quasi-isometric embeddings of Riemannian circles in $X$.
  \item \itmlabel{quasi_circles_bounded} There exists an $L > 1$ such
    that, for every $C \ge 0$, there is a bound on the lengths of the
    $(L,C)$-quasi-isometric embeddings of Riemannian circles in $X$.
  \item \itmlabel{ngons_bounded} $X$ does not approximate $n$-gons.
  \item \itmlabel{no_asymptotic_circle} No asymptotic cone of $X$
    contains an isometric copy of the Riemannian circle of unit
    length.
  \end{enumerate}
\end{cor}

\begin{cor}
  \corlabel{as_cones_ss} Let $X$ be a metric space.  If $X$ is
  strongly shortcut then every asymptotic cone of $X$ is strongly
  shortcut.
\end{cor}
\begin{proof}
  Follows immediately from \Corref{sshortcut_equiv} and
  \Corref{as_cones_general_ss}.
\end{proof}

\begin{cor}
  \corlabel{rough_approx_inv} The strong shortcut property is a rough
  approximability invariant of rough geodesic metric spaces.  In
  particular, the strong shortcut property is a rough similarity
  invariant of rough geodesic metric spaces.
\end{cor}
\begin{proof}
  Follows immediately from \Corref{sshortcut_equiv} and
  \Propref{not_approx_ngon_invariant}.

\end{proof}

\section{The Circle Tightening Lemma}
\seclabel{circle_tightening}

The Circle Tightening Lemma describes how one may perform surgery on
an almost isometric $R$-circle to obtain a quasi-isometrically
embedded $R$-circle, assuming the various constants are chosen
appropriately.  A version of this lemma first appeared implicitly in
the proof of a proposition in an earlier work of this author
\cite[Proposition~3.5]{Hoda:2018b} where it applied only to graphs.
Here we state and prove a generalization to (rough) geodesic metric
spaces.

\subsection{Tightening sequence for a Riemannian circle}
\sseclabel{tightening_riemannian_circle}

Let $S$ be a Riemannian circle.  A \defterm{tightening sequence} for
$S$ is a sequence of intervals and Riemannian circles $(P_i)_i$, a
sequence of Riemannian circles $(S_i)_i$ and sequences of maps
\[ \begin{tikzcd}
    S = S_0  & P_0 \ar[l,hook'] \ar[r] & S_1  & P_1 \ar[l,hook'] \ar[r] & S_2 & P_2 \ar[l,hook'] \ar[r] & \cdots
  \end{tikzcd} \] such that the following conditions are satisfied.
\begin{enumerate}
\item For each $i$, either
  \begin{enumerate}
  \item
    \begin{enumerate}
    \item $P_i \hookrightarrow S_i$ and $P_i \to S_{i+1}$ are
      continuous paths of unit speed, and
    \item $P_i \to S_{i+1}$ is injective on the interior $\pc{i}$ of
      $P_i$, and
    \item $|P_i| \ge \frac{|S|}{2}$, and
    \item $|S_{i+1}| < |S_i|$; or
    \end{enumerate}
  \item
    \begin{enumerate}
    \item $\pc{i} = P_i = S_i = S_{i+1}$, and
    \item $P_i \hookrightarrow S_i$ and $P_i \to S_{i+1}$ are the
      identity map on $S_i$.
    \end{enumerate}
  \end{enumerate}
\end{enumerate}
Then, for each $i$, the circle $S_{i+1}$ is obtained from $S_i$ by
replacing $Q_i = S \setminus \pc{i}$ with $\bar Q_i$, where either
$Q_i = \bar Q_i = \emptyset$ or $Q_i$ and $\bar Q_i$ are intervals
with $|\bar Q_i| < |Q_i|$.  So we also have a commutative diagram of
$1$-Lipschitz maps
\[ \begin{tikzcd}
    Q_0 \ar[r,two heads] \ar[dd,hook] & \bar Q_0 \ar[ddr,bend left=15,hook] & Q_1 \ar[r,two heads] \ar[dd,hook] & \bar Q_1 \ar[ddr,bend left=15,hook] & Q_2 \ar[r,two heads] \ar[dd,hook] & \bar Q_2 \ar[ddr,bend left=15,hook] & \cdots \\
    \\
    S=S_0 \ar[rr,bend left=45,two heads,"\pi_0"] & P_0 \ar[l,hook'] \ar[r] & S_1 \ar[rr,bend left=45,two heads,"\pi_1"] & P_1 \ar[l,hook'] \ar[r] & S_2  \ar[rr,bend left=45,two heads,"\pi_2"] & P_2 \ar[l,hook'] \ar[r] & \cdots
  \end{tikzcd} \] where each $Q_i \twoheadrightarrow \bar Q_i$ is
affine.  We call the $Q_i$ the \defterm{tightened segments} of the
tightening sequence.  We let $\pi^{(i)}$ denote the composition
$\pi_{i-1} \comp \pi_{i-2} \comp \cdots \comp \pi_0$.  A tightening
sequence is \defterm{eventually constant} if $S_i = S_{i+1}$ for all
large enough $i$.

Let $\pc{0,j}$ be the limit of the diagram
\[ \begin{tikzcd}
    & \pc{0} \ar[d,hook] \ar[dl,hook'] \ar[dr,hook] & & \pc{1} \ar[d,hook] \ar[dl,hook'] \ar[dr,hook] & \cdots & \pc{j-1} \ar[d,hook] \ar[dl,hook'] \ar[dr,hook] \\
    S=S_0 & P_0 \ar[l,hook'] \ar[r] & S_1 & P_1 \ar[l,hook'] \ar[r] & \cdots & P_{j-1} \ar[l,hook'] \ar[r] & S_j
  \end{tikzcd} \]
in the category of topological spaces and continuous maps.
Concretely, we have $\pc{0,0} = S_0$ and $\pc{0,1} = \pc{0}$ and
$\pc{0,j} = \pc{0,j-1} \cap \pc{j-1}$ where the intersection is
taken in $S_{j-1}$.  Thus we have the following commutative
diagram.
\[ \begin{tikzcd}
    & \pc{0,1} \ar[d,equal] && \pc{0,2} \ar[d,hook] \ar[ll,hook'] && \pc{0,3} \ar[d,hook] \ar[ll,hook'] & \cdots \ar[l,hook'] \\
    & \pc{0} \ar[d,hook] \ar[dl,hook'] \ar[dr,hook] & & \pc{1} \ar[d,hook] \ar[dl,hook'] \ar[dr,hook] & & \pc{2} \ar[d,hook] \ar[dl,hook'] \ar[dr,hook] & \cdots \\
    S=S_0 & P_0 \ar[l,hook'] \ar[r] & S_1 & P_1 \ar[l,hook'] \ar[r] & S_2 & P_2 \ar[l,hook'] \ar[r] & \cdots
  \end{tikzcd} \] We can think of $\pc{0,j}$ as the original points of
$S$ that are not replaced until at least step $j$ of the
``construction'' of the $S_i$, where the $j$th step of the
construction refers to the operation of replacing $Q_j$ with
$\bar Q_j$ in order to obtain $S_{j+1}$ from $S_j$.

We say that a tightening sequence is \defterm{disjoint up to $j$} if
$Q_i \subset \pc{0,i}$ for all $i < j$, where $\pc{0,i}$ is viewed as
a subspace of $S_i$ via the embedding $\pc{0,i} \hookrightarrow S_i$.
We say that a tightening sequence is \defterm{completely disjoint} if
it is disjoint up to $j$ for every $j$.

If a tightening sequence is disjoint up to $j$, for $i<j$, we have
$Q_i \sqcup \pc{0,i+1} = \pc{0,i} \hookrightarrow S$.  So, for
$i < j$, we may think of the $Q_i$ as disjoint subspaces of $S$ and
$S \setminus \pc{0,j} = \bigcup_{i=0}^{j-1}Q_i$ in $S$.  Since $S_j$
is obtained from $S$ by replacing $Q_i$ with $\bar Q_i$, for each
$i < j$, we see that the $\bar Q_i$, with $i < j$, embed disjointly in
$S_j$ and the complement in $S_j$ of $\pc{0,j}$ is
$\bigcup_{i=0}^{j-1}\bar Q_i$.

If a tightening sequence is completely disjoint then the $Q_i$ all
embed disjointly in $S$ and the complement of their union in $S$ is
$\pc{0,\infty} = \bigcap_{j=1}^{\infty} \pc{0,j}$.

\begin{lem}
  \lemlabel{lim_riemannian_circle} Consider a tightening sequence for
  a Riemannian circle $S$ with the same notation as above.  If the
  tightening sequence is completely disjoint and the sum
  $\sum_{i=0}^{\infty} |Q_i|$ of the tightened segment lengths is
  strictly less than $|S|$ then
  \begin{align*}
    \delta \colon S \times S &\to \R_{\ge 0} \\
    (x,y) &\mapsto \lim_{i \to \infty} d_{S_i}\bigl(\pi^{(i)}(x),
    \pi^{(i)}(y) \bigr)
  \end{align*} defines a pseudometric on $S$ such that
  the induced metric quotient $S_{\infty}$ of $(S,\delta)$,
  called the \defterm{limit Riemannian circle} of the tightening
  sequence, is a Riemannian circle of length
  $\lim_{i \to \infty} |S_i|$.
\end{lem}
\begin{proof}
  The $\pi^{(i)}$ are isomorphisms on fundamental group so, for each
  $i$, we have a commuting diagram \[
    \begin{tikzcd}
      \R \ar[d] \ar[r,"\tilde \pi_i"]  & \R \ar[d] \\
      S_i \ar[r, "\pi_i"] & S_{i+1}
    \end{tikzcd} \] where $\R \to S_i$ and $\R \to S_{i+1}$ are the
  quotient maps from $(\R,+)$ by the subgroups $|S_i|\Z$ and
  $|S_{i+1}|\Z$, respectively.  Then, since $\pi_i$ is $1$-Lipzchitz,
  so is $\tilde \pi_i$.  Without loss of generality, the map
  $\tilde \pi_i$ sends $0$ to $0$ and preserves order, in the sense
  that $s \le r$ implies $\tilde \pi_i(s) \le \tilde \pi_i(r)$.  Let
  $\tilde \pi^{(i)} \colon \R \to \R$ be the composition
  $\tilde \pi_0 \comp \tilde \pi_1 \comp \cdots \comp \tilde
  \pi_{i-1}$ so that the diagram \[ \begin{tikzcd}
      \R \ar[d] \ar[r,"\tilde \pi^{(i)}"]  & \R \ar[d] \\
      S \ar[r, "\pi^{(i)}"] & S_i
    \end{tikzcd} \] commutes and satisfies the same properties as the
  previous diagram.  Then for $r \in \R$, the sequence
  $\bigl(\pi^{(i)}(r)\bigr)_i$ is either nonnegative and nonincreasing
  or nonpositive and nondecreasing.  In either case the limit exists
  so we can define a limit function
  \begin{align*}
    \pi^{(\infty)} \colon \R &\to \R \\
    r &\mapsto  \lim_{i \to \infty} \pi^{(i)}(r)
  \end{align*}
  which is also $1$-Lipschitz, sends $0$ to $0$ and preserves order.

  Since
  $\pi^{(i)}\bigl(|S|\bigr) = |S_i| \ge |S| - \sum_{i=1}^{\infty}|Q_i|
  > 0$, we have the following.
  \[\pi^{(\infty)}\bigl(|S|\bigr) = \lim_{i \to \infty}|S_i| > 0 \]
  For $r \in \R$, we have
  $\pi^{(i)}\bigl(r + |S|\bigr) = \pi^{(i)}(r) + |S_i|$ so
  \[ \pi^{(\infty)}\bigl(r + |S|\bigr) = \pi^{(\infty)}(r) +
    \lim_{i\to\infty}|S_i| \] which implies that if $\R \to \bar S$ is
  the quotient map of $(\R,+)$ with kernel
  $\bigl(\lim_{i \to \infty}|S_i|\bigr)\Z$ then the map
  \begin{align*}
    \pi^{(\infty)} \colon S &\to S_{\infty} \\
    r + |S|\Z &\mapsto \pi^{(\infty)}(r) + \bigl(\lim_{i \to \infty}|S_i|\bigr)\Z
  \end{align*}
  is well defined and makes the diagram \[
    \begin{tikzcd}
      \R \ar[d] \ar[r,"\tilde \pi^{(\infty)}"]  & \R \ar[d] \\
      S \ar[r, "\pi^{(\infty)}"] & \bar S
    \end{tikzcd} \]
  commute.

  Then $|\bar S| = \lim_{i \to \infty}|S_i|$ and, for $r+|S|\Z$ and
  $s+|S|\Z$ in $S$, we have
  \begin{align*}
    &d_{\bar S}\Bigl(\pi^{(\infty)}\bigl(r+|S|\Z\bigr),
    \pi^{(\infty)}\bigl(s+|S|\Z\bigr)\Bigr) \\
    &= d_{\bar S}\Bigl(\pi^{(\infty)}(r) + |\bar S|\Z,
      \pi^{(\infty)}(s) + |\bar S|\Z\Bigr) \\
    &= \min_{k \in \Z}\Bigl|\pi^{(\infty)}(r) -
      \pi^{(\infty)}(s) + |\bar S|k \Bigr| \\
    &= \min_{k \in I}\Bigl|\pi^{(\infty)}(r) -
      \pi^{(\infty)}(s) + |\bar S|k \Bigr| \\
    &= \min_{k \in I}\lim_{i \to \infty}\Bigl|\pi^{(i)}(r) -
      \pi^{(i)}(s) + |\bar S_i|k \Bigr| \\
    &= \lim_{i \to \infty}\min_{k \in I}\Bigl|\pi^{(i)}(r) -
      \pi^{(i)}(s) + |\bar S_i|k \Bigr| \\
    &= \lim_{i \to \infty}\min_{k \in \Z}\Bigl|\pi^{(i)}(r) -
      \pi^{(i)}(s) + |\bar S_i|k \Bigr| \\
    &= \lim_{i \to \infty}d_{S_i}\Bigl(\pi^{(i)}(r) + |S_i|\Z,
      \pi^{(i)}(s) + |S_i|\Z\Bigr) \\
    &= \lim_{i \to \infty}d_{S_i}\Bigl(\pi^{(i)}\bigl(r+|S|\Z\bigr),
      \pi^{(i)}\bigl(s+|S|\Z\bigr)\Bigr)
  \end{align*}
  where
  $I = \biggl\{k \in \Z \sth |k| \le \Bigl\lceil\frac{|r-s|}{|\bar
    S|}\Bigr\rceil\biggr\}$.  So the pseudometric on $S$ pulled back
  from $\pi^{(\infty)}$ is $\delta$. Then, since $\pi^{(\infty)}$ is
  surjective, this implies that $\bar S$ is $S_{\infty}$, the induced
  metric quotient of $(S,\delta)$ and $\pi^{\infty}$ is the quotient
  map.
\end{proof}

\begin{rmk}
  If a completely disjoint tightening sequence of a Riemannian circle
  is eventually constant then, for large enough $i$, the limit
  Riemannian circle $S_{\infty}$ is isometric to the $i$th Riemannian
  circle of the sequence $S_i$.
\end{rmk}

\subsection{Tightening sequence for an \texorpdfstring{$R$}{R}-circle}

Let $R \ge 0$ and let $X$ be an $R$-rough geodesic metric space. Let
$\alpha \colon S \to X$ be an $R$-circle.  A \defterm{tightening
  sequence} for $\alpha$ is a sequence of intervals and Riemannian
circles $(P_i)_i$, a sequence of Riemannian circles $(S_i)_i$ and
sequences of maps as in the commutative diagram
\[ \begin{tikzcd}
    S \ar[r,equal] \ar[drrr,bend right=15,"\alpha"'] & S_0 \ar[drr,bend right=15,"\alpha_0"] & P_0 \ar[l,hook'] \ar[r] & S_1 \ar[d,"\alpha_1"'] & P_1 \ar[l,hook'] \ar[r] & S_2 \ar[dll,bend left=15,"\alpha_2"'] & P_2 \ar[l,hook'] \ar[r] & \ar[dllll,bend left=10,"\alpha_3"'] \cdots \\
    & & & X & & & & \cdots
  \end{tikzcd} \] such that each $\alpha_i$ is an $R$-circle and the
sequence of maps
\[ \begin{tikzcd} S = S_0 & P_0 \ar[l,hook'] \ar[r] & S_1 & P_1
    \ar[l,hook'] \ar[r] & S_2 & P_2 \ar[l,hook'] \ar[r] & \cdots
  \end{tikzcd} \] is a tightening sequence for $S$.  Then, by the
discussion of \Ssecref{tightening_riemannian_circle}, we have a
diagram \[ \begin{tikzcd}
    & Q_0 \ar[r,two heads] \ar[dd,hook] & \bar Q_0 \ar[ddr,bend left=15,hook] & Q_1 \ar[r,two heads] \ar[dd,hook] & \bar Q_1 \ar[ddr,bend left=15,hook] & Q_2 \ar[r,two heads] \ar[dd,hook] & \bar Q_2 \ar[ddr,bend left=15,hook] & \cdots \\
    \\
    S \ar[r,equal] \ar[drrr,bend right=15,"\alpha"'] & S_0 \ar[rr,bend left=45,two heads,"\pi_0"] \ar[drr,bend right=15,"\alpha_0"] & P_0 \ar[l,hook'] \ar[r] & S_1 \ar[d,"\alpha_1"'] \ar[rr,bend left=45,two heads,"\pi_1"] & P_1 \ar[l,hook'] \ar[r] & S_2 \ar[dll,bend left=15,"\alpha_2"'] \ar[rr,bend left=45,two heads,"\pi_2"] & P_2 \ar[l,hook'] \ar[r] & \ar[dllll,bend left=10,"\alpha_3"'] \cdots \\
    & & & X & & & & \cdots
  \end{tikzcd} \] where the bounded planar regions are commuting
triangles and squares.

\begin{lem}
  \lemlabel{lim_r_circle} Consider a tightening sequence for an
  $R$-circle $\alpha \colon S \to X$ in an $R$-rough geodesic metric
  space $X$, with the same notation as above.  If the tightening
  sequence is completely disjoint and the sum
  $\sum_{i=0}^{\infty} |Q_i|$ of the tightened segment lengths is
  strictly less than $|S|$ then
  \begin{align*}
    \alpha_{\infty} \colon S_{\infty} &\to X \\
    x &\mapsto \lim_{i \to \infty} \alpha_i \comp \pi^{(i)} \comp \sigma(x)
  \end{align*}
  defines an $R$-circle, called a \defterm{limit $R$-circle} of the
  tightening sequence, where $S_{\infty}$ is the limit Riemannian
  circle of the tightening sequence and
  $\sigma \colon S_{\infty} \to S$ is a section of the quotient map
  $S \to S_{\infty}$.
\end{lem}
\begin{proof}
  Since the tightening sequence is completely disjoint, we may think
  of the $Q_i$ as a collection of disjoint segments in $S$.  For
  $x \in X$ either $x \notin \bigcup_{i=0}^{\infty} Q_i$ and
  $\bigl(\alpha_i \comp \pi^{(i)}(x)\bigr)_{i=0}^{\infty}$ is a
  constant sequence or $x \in Q_j$ for some $j$ and the tail sequence
  $\bigl(\alpha_i \comp \pi^{(i)}(x)\bigr)_{i=j+1}^{\infty}$ is
  constant.  In either case, the limit
  $\lim_{i \to \infty} \alpha_i \comp \pi^{(i)}(x)$ exists so we have
  a function $\alpha_{\infty}' \colon S \to X$ given by
  $\alpha_{\infty}'(x) = \lim_{i \to \infty} \alpha_i \comp
  \pi^{(i)}(x)$.

  Let $S_{\infty}$ be the limit Riemannian circle given by
  \Lemref{lim_riemannian_circle}.  So $S_{\infty}$ is the induced
  metric quotient of $(S,\delta)$ where $\delta$ is the pseudometric
  give by
  $\delta(x,y) = \lim_{i \to \infty} d_{S_i}\bigl(\pi^{(i)}(x),
  \pi^{(i)}(y) \bigr)$.  Since each $\alpha_i$ is an $R$-circle, for
  $x,y \in S$,
  \[ d_X\bigl(\alpha_i\comp \pi^{(i)}(x), \alpha_i\comp
    \pi^{(i)}(y)\bigr) \le
    d_{S_i}\bigl(\pi^{(i)}(x),\pi^{(i)}(y)\bigr) + R \] for all $i$,
  and thus
  \[ d_X\bigl(\alpha_{\infty}'(x), \alpha_{\infty}'(y)\bigr) \le
    \delta(x,y) + R \] by taking limits as $i \to \infty$.

  Then, for $x,y \in S_{\infty}$
  \begin{align*}
    d_X\bigl(\alpha_{\infty}(x), \alpha_{\infty}(y)\bigr)
    &= d_X\Bigl(\alpha_{\infty}'\bigl(\sigma(x)\bigr),
      \alpha_{\infty}'\bigl(\sigma(y)\bigr)\Bigr) \\
    &\le \delta\bigl(\sigma(x), \sigma(y)\bigr) + R \\
    &= d_{S_{\infty}}(x, y) + R
  \end{align*}
  which completes the proof.
\end{proof}

\begin{rmk}
  If a completely disjoint tightening sequence for an $R$-circle
  $\alpha \colon S \to X$ is eventually constant then, for large
  enough $i$, the limit $R$-circle
  $\alpha_{\infty} \colon S_{\infty} \to X$ is isometric over $X$ to
  the $i$th $R$-circle of the sequence $\alpha_i$.  This means that
  there is an isometry $S_{\infty} \to S$ such that diagram
  \[ \begin{tikzcd}
      S_{\infty} \ar[r] \ar[dr,"\alpha_{\infty}"'] & S_i \ar[d,"\alpha_i"] \\
      & X
    \end{tikzcd} \]
  commutes.
\end{rmk}

We are ready now to state the Circle Tightening Lemma.

\begin{lem}[Circle Tightening Lemma]
  \lemlabel{circle_tightening} Let $N > 1$, let $L > 1$, let $K > 1$
  be small enough (depending on $N$ and $L$), let $R \ge 0$, let
  $M > 0$ be large enough (depending on $N$, $L$, $K$ and $R$) and let
  $C \ge 4R$.

  Let $\alpha \colon S \to X$ be an $R$-circle in an $R$-rough
  geodesic metric space.  If $\alpha$ is $\frac{1}{K}$-almost
  isometric and $|S| > M$ then $\alpha$ has a completely disjoint
  tightening sequence such that the total length
  $\sum_{i=0}^{\infty} |Q_i|$ of the tightened segments is at most
  $\frac{|S|}{N}$ and the limit $R$-circle $\alpha_{\infty}$ is an
  $(L,C)$-quasi-isometric embedding.  If, additionally, we have
  $C > 0$ then such a tightening sequence exists that is eventually
  constant.
\end{lem}

\Lemref{circle_tightening} is a consequence of claims
\claimref{greedy_tightening_disjoint}, \claimref{greedy_bounds},
\claimref{greedy_tightening_qi} and
\claimref{positive_c_eventually_constant} and the strict inequalities
of \Claimref{rational_functions} but to understand these claims we
need to first define greedy tightening sequences and prove some
properties about them.

\subsection{Greedy tightening sequences}
\sseclabel{greedy_tightening}

In order to prove the \Lemref{circle_tightening} we will need to
describe a tightening sequence that is constructed inductively by
greedily choosing segments to tighten.  Let $\alpha \colon S \to X$ be
an $R$-circle in an $R$-rough geodesic metric space $X$, with
$R \ge 0$.  Let $C \ge 4R$ and let $L > 1$.  We will inductively
define a tightening sequence for $\alpha$ with the same notation as in
the previous sections.  Suppose we have $\alpha_i \colon S_i \to X$.
If $\alpha_i$ is an $(L,C)$-quasi-isometric embedding then we extend
the sequence as follows.
\begin{enumerate}
\item We set $\pc{i} = P_i = S_i = S_{i+1}$ and
  $Q_i = \bar Q_i = \emptyset$.
\item We let $P_i \hookrightarrow S_i$ and $P_i \to S_{i+1}$ be the
  identity map on $S_i$.
\item We let $\alpha_{i+1} = \alpha_i$.
\end{enumerate}
Otherwise, the set
\[ J_i = \Bigl\{(p,q) \in S_i \times S_i \sth
  d_X\bigl(\alpha_i(p),\alpha_i(q)\bigr) < \frac{1}{L} d_{S_i}(p,q) -
  C \Bigr\} \] is nonempty and $d_{S_i}(p,q) \ge LC$ for any
$(p,q) \in J_i$ and so
$s_i = \sup \bigl\{d_{S_i}(p,q) \sth (p,q) \in J_i \bigr\} > 0$.  By
compactness of $S$, there is a sequence $(p_i^{(n)},q_i^{(n)})_n$ in
$J_i$ that converges to some $(p_i,q_i) \in S \times S$ with
$d_{S_i}(p_i,q_i) = s_i$ as $n \to \infty$.  Then
\begin{align*}
  &d_X\bigl(\alpha_i(p_i), \alpha(q_i)\bigr) \\
  &\le d_X\bigl(\alpha_i(p_i), \alpha(p_i^{(n)})\bigr)
    + d_X\bigl(\alpha_i(p_i^{(n)}), \alpha(q_i^{(n)})\bigr)
    + d_X\bigl(\alpha_i(q_i^{(n)}), \alpha(q_i)\bigr) \\
  &< d_{S_i}(p_i, p_i^{(n)}) + R
    + \frac{1}{L}d_{S_i}(p_i^{(n)}, q_i^{(n)}) - C
    + d_{S_i}(q_i^{(n)}, q_i) + R \\
  &\to \frac{1}{L}d_{S_i}(p_i, q_i) - C + 2R
\end{align*}
as $n \to \infty$.  So
\begin{equation}
  \eqnlabel{alph_qi_alph_pi_distance}
  d_X\bigl(\alpha_i(p_i), \alpha(q_i)\bigr)
  \le \frac{1}{L}d_{S_i}(p_i, q_i) - C + 2R
\end{equation}
and
\begin{equation}
  \eqnlabel{qi_size}
  \text{$d_{S_i}(p_i,q_i) > 0$ and $d_{S_i}(p_i,q_i) \ge L(C-2R)$}
\end{equation}
hold.  Let $Q_i$ be a geodesic segment of $S_i$ between $p_i$ and
$q_i$.  Let $\pc{i}$ be the complement of $Q_i$ and let $P_i$ be the
closure of $\pc{i}$.  Let $\gamma'_i \colon \bar Q'_i \to X$ be an
$R$-rough geodesic from $\alpha_i(p_i)$ to $\alpha_i(q_i)$.  For
$x \in X$, let $c_x\colon [0,R] \to X$ denote the constant path of
length $R$ at $x$.  Let $\gamma_i \colon \bar Q_i \to X$ be the
concatenation $c_{\alpha_i(p_i)}\gamma'_ic_{\alpha_i(q_i)}$.  We have
\[ 0 \le |\bar Q_i| - 2R = d_X\bigl(\alpha_i(p_i),\alpha_i(q_i)\bigr)
  \le \frac{1}{L} |Q_i| - C + 2R \] and so, since $C \ge 4R$,
\begin{equation}
  \eqnlabel{qi_scale}
  |\bar Q_i| \le \frac{1}{L} |Q_i|
\end{equation}
holds.  We obtain $\alpha_{i+1} \colon S_{i+1} \to X$ from
$\alpha_i|_{P_i}$ and $\gamma_i$ by identifying the corresponding
endpoints of $\bar Q_i$ with $p_i$ and $q_i$ in $P_i$.  Then, by
consideration of \Rmkref{rpath_concatenation}, the map $\alpha_{i+1}$
is an $R$-circle.
\begin{rmk}
  \rmklabel{endpoint_distances}
  The inequalities
  \[ d_X\bigl(\alpha_{i+1}(p_i), \alpha_{i+1}(x)\bigr) \le
    d_{S_{i+1}}(p_i,x) \] and
  \[ d_X\bigl(\alpha_{i+1}(q_i), \alpha_{i+1}(x)\bigr) \le
    d_{S_{i+1}}(q_i,x) \] hold for any $x \in \bar Q_i$.
\end{rmk}

This completes the description of our inductive construction.  Any
tightening sequence for $\alpha$ obtained in this way is called an
\defterm{$(L,C)$-greedy tightening sequence} for $\alpha$.  The
importance of this construction for us is evident from the following
claim.

\begin{claim}
  \claimlabel{greedy_tightening_qi} Let $L > 1$, let $K > 1$, let
  $C \ge 4R$ and consider an $(L,C)$-greedy tightening sequence for a
  $\frac{1}{K}$-almost isometric $R$-circle $\alpha \colon S \to X$ in
  an $R$-rough geodesic metric space $X$.  If the tightening sequence
  is completely disjoint and the sum $\sum_{i=0}^{\infty} |Q_i|$ of
  the tightened segment lengths is strictly less than $|S|$ then any
  limiting $R$-circle $\alpha_{\infty} \colon S_{\infty} \to X$ is an
  $(L,C)$-quasi-isometric embedding.
\end{claim}
\begin{proof}
  Let $S_{\infty}$ be the limit Riemannian circle given by
  \Lemref{lim_riemannian_circle}.  So $S_{\infty}$ is the induced
  metric quotient of $(S,\delta)$ where $\delta$ is the pseudometric
  given by
  $\delta(x,y) = \lim_{i \to \infty} d_{S_i}\bigl(\pi^{(i)}(x),
  \pi^{(i)}(y) \bigr)$.  Let $\alpha_{\infty} \colon S_{\infty} \to X$
  be a limit $R$-circle as in \Lemref{lim_r_circle}.  So
  $\alpha_{\infty}$ is an $R$-circle defined by
  $\alpha_{\infty}(x) = \lim_{i \to \infty} \alpha_i \comp \pi^{(i)}
  \comp \sigma(x)$, where $\sigma \colon S_{\infty} \to S$ is a
  section of the quotient map $S \to S_{\infty}$.

  If $\alpha_{\infty}$ is not an $(L,C)$-quasi-isometric embedding
  then, since $R \le C$,
  \[ d_X\bigl(\alpha_{\infty}(x), \alpha_{\infty}(y)\bigr) <
    \frac{1}{L} d_{S_{\infty}}(x,y) - C \] for some
  $x,y \in S_{\infty}$, which then must be distinct.  But
  \[ d_{S_{\infty}}(x,y) = \lim_{i \to \infty}
    d_{S_i}\Bigl(\pi^{(i)}\bigl(\sigma(x)\bigr),
    \pi^{(i)}\bigl(\sigma(y)\bigr) \Bigr)
  \]
  and
  \[ d_X\bigl(\alpha_{\infty}(x), \alpha_{\infty}(y)\bigr) = \lim_{i
      \to \infty} d_X\Bigl(\alpha_i \comp \pi^{(i)}\bigl(
    \sigma(x)\bigr), \alpha_i \comp \pi^{(i)}\bigl(
    \sigma(y)\bigr)\Bigr) \] where, by complete disjointness,
  $\Bigl(\pi^{(i)}\bigl( \sigma(x)\bigr)\Bigr)_i$ and
  $\Bigl(\pi^{(i)}\bigl( \sigma(y)\bigr)\Bigr)_i$ are eventually
  constant.  So, for all large enough $j$,
  \begin{align*}
    &d_X\Bigl(\alpha_j \comp \pi^{(j)}\bigl(
    \sigma(x)\bigr), \alpha_j \comp \pi^{(j)}\bigl(
    \sigma(y)\bigr)\Bigr) \\
    &= \lim_{i
      \to \infty} d_X\Bigl(\alpha_i \comp \pi^{(i)}\bigl(
      \sigma(x)\bigr), \alpha_i \comp \pi^{(i)}\bigl(
      \sigma(y)\bigr)\Bigr)
  \end{align*}
  but also, for all large enough $j$,
  \[ \lim_{i \to \infty} d_X\Bigl(\alpha_i \comp \pi^{(i)}\bigl(
    \sigma(x)\bigr), \alpha_i \comp \pi^{(i)}\bigl(
    \sigma(y)\bigr)\Bigr) < \frac{1}{L}
    d_{S_j}\Bigl(\pi^{(j)}\bigl(\sigma(x)\bigr),
    \pi^{(j)}\bigl(\sigma(y)\bigr) \Bigr) - C \] so, for all large
  enough $j$,
  \[ d_X\Bigl(\alpha_j\bigl(\pi^{(j)} \comp \sigma(x)\bigr),
    \alpha_j\bigl(\pi^{(j)} \comp \sigma(y)\bigr)\Bigr) < \frac{1}{L}
    d_{S_j}\Bigl(\pi^{(j)}\comp \sigma(x), \pi^{(j)}\comp \sigma(y)
    \Bigr) - C \] which implies that
  $\bigl(\pi^{(j)}\comp \sigma(x), \pi^{(j)}\comp \sigma(y)\bigr) \in
  J_j$, for all large enough $j$.  But then, for all large enough $j$,
  \[ d_{S_j}\Bigl(\pi^{(j)}\comp \sigma(x), \pi^{(j)}\comp \sigma(y)
    \Bigr) \le s_j = |Q_j| \] with $\lim_{j \to \infty}|Q_j| = 0$ so
  \[ d_{S_{\infty}}(x,y) = \lim_{j \to \infty}
    d_{S_j}\Bigl(\pi^{(j)}\comp \sigma(x), \pi^{(j)}\comp \sigma(y)
    \Bigr) = 0 \] a contradiction.
\end{proof}

\subsection{Eventual constantness and greedy tightening sequences}

Consider a greedy tightening sequence with notation as in
\Ssecref{greedy_tightening}.  Note that if $S_i = S_{i+1}$ for some
$i$ then $S_i = S_{i+1} = S_{i+2} = \cdots$ so the tightening sequence
is eventually constant.  Moreover, for any $i$ for which
$S_i \neq S_{i+1}$, we have
\begin{align*}
  |S_i| - |S_{i+1}|
  &= |Q_i| - |\bar Q_i| \\
  &\ge |Q_i| - \frac{1}{L}|Q_i| \\
  &= \Bigl(1- \frac{1}{L} \Bigr)|Q_i| \\
  &\ge \Bigl(1- \frac{1}{L} \Bigr)L(C-2R) \\
  & = (L-1)(C-2R)
\end{align*}
by \peqnref{qi_size} and \peqnref{qi_scale}.  If $R > 0$ then, since
$C \ge 4R$, we have $C > 2R > 0$.  If $R = 0$ then $C > 2R$ is
equivalent to $C > 0$.  Thus, if $C > 0$ then
$|S_i| - |S_{i+1}| \ge (L-1)(C-2R) > 0$.  This implies the following
claim.
\begin{claim}
  \claimlabel{positive_c_eventually_constant} If $R \ge 0$, $L > 1$
  and $C > 0$ then any $(L,C)$-greedy tightening sequence for an
  $R$-circle in an $R$-rough geodesic space is eventually constant.
\end{claim}

\subsection{Disjointness and greedy tightening sequences}

Consider a greedy tightening sequence with notation as in
\Ssecref{greedy_tightening}.  Assume that $\alpha$ is
$\frac{1}{K}$-almost isometric and that the tightening sequence is
disjoint up to $j$.  Recall that, by the discussion in
\Ssecref{tightening_riemannian_circle}, we may think of the $Q_i$,
with $i < j$, as disjoint subspaces of $S$.

If $i < j$ and $Q_i \neq \emptyset$ then, by \Lemref{global_to_local}
and \peqnref{alph_qi_alph_pi_distance},
\[ |Q_i| - \frac{K-1}{K}\cdot\frac{|S|}{2} - 2R \le
  d_X\bigl(\alpha(p_i),\alpha(q_i)\bigr) \le \frac{1}{L} |Q_i| - C + 2R \]
but $C \ge 4R$ so
\[ |Q_i| - \frac{K-1}{K}\cdot\frac{|S|}{2} \le \frac{1}{L} |Q_i| \]
for all $i < j$.  Hence, we have established the following claim.
\begin{claim}
  \claimlabel{onebound} If an $(L,C)$-greedy tightening sequence for a
  $\frac{1}{K}$-almost isometric $R$-circle is disjoint up to $j$ then
  \[ |Q_i| \le \biggl(\frac{K -1}{K} \cdot
    \frac{L}{L-1}\biggr)\frac{|S|}{2} \] for any $i < j$, where $Q_i$
  is the $i$th replaced segment of the tightening sequence.
\end{claim}
By this claim, we can find a pair of points $p,q$ in the closure of
$S \setminus \bigl(\bigcup_{i=1}^{j-1}Q_i\bigr)$ at distance
$d_S(p,q) \ge \frac{|S|}{2} - \frac{K -1}{K} \cdot
\frac{L}{L-1}\cdot\frac{|S|}{4}$.  Let $A_1$ and $A_2$ be the two
segments of $S$ between $p$ and $q$.  If
$I_1 = \{i < j \sth Q_i \subseteq A_1 \}$ then, since $\alpha_j$ is an
$R$-circle,
\begin{align*}
  d_X\bigl(\alpha(p),\alpha(q)\bigr)
  &\le d_{S_j}(p,q) + R \\
  &\le |A_1| - \sum_{i \in I_1}|Q_i| + \sum_{i \in I_1}|\bar Q_i| + R \\
  &\le |A_1| - \sum_{i \in I_1}|Q_i| + \frac{1}{L}\sum_{i \in I_1}|Q_i| + R \\
  &= |A_1| - \frac{L - 1}{L}\sum_{i \in I_1}|Q_i| + R
\end{align*}
where the last inequality follows by \peqnref{qi_scale}.  The same
corresponding relations also hold for $A_2$ and
$I_2 = \{i < j \sth Q_i \subseteq A_2 \}$ and so, by
\Lemref{global_to_local},
\begin{align*}
  &|S| - \frac{L-1}{L}\sum_{i<j}|Q_i| + 2R \\
  &\ge 2 d_X\bigl(\alpha(p),\alpha(q)\bigr) \\
  &\ge 2d_S(p,q) - \frac{K-1}{K} \cdot |S| - 4R \\
  &\ge |S| - \Bigl(\frac{K -1}{K} \cdot
    \frac{L}{L-1}\Bigr)\frac{|S|}{2} - \frac{K-1}{K} \cdot |S| - 4R
\end{align*}
which establishes the following claim.
\begin{claim}
  \claimlabel{sumbound} If an $(L,C)$-greedy tightening sequence for a
  $\frac{1}{K}$-almost isometric $R$-circle is disjoint up to $j$ then
  \[ \sum_{i<j}|Q_i| \le \biggl(\frac{K-1}{K} \cdot
    \frac{L(3L-2)}{2(L-1)^2}\biggr)|S| + \frac{6LR}{L-1} \] where
  $Q_i$ is the $i$th replaced segment of the tightening sequence.
\end{claim}

\Claimref{sumbound} implies that
\[ \sum_{i<j}|Q_i| \le \biggl(\frac{K-1}{K} \cdot
  \frac{L(3L-2)}{2(L-1)^2} + \frac{6LR}{|S|(L-1)} \biggr)|S| \] so if
$\frac{K-1}{K} \cdot \frac{L(3L-2)}{2(L-1)^2} < \frac{1}{N}$ then if
$|S| > M$ for some $M$ depending only on $K$, $L$, $R$ and $N$ then
$\frac{K-1}{K} \cdot \frac{L(3L-2)}{2(L-1)^2} + \frac{6LR}{|S|(L-1)} <
\frac{1}{N}$ and so $\sum_{i<j}|Q_i| < \frac{|S|}{N}$.  Since
$\frac{K-1}{K} \cdot \frac{L(3L-2)}{2(L-1)^2} < \frac{1}{N}$ is
equivalent to $K < \frac{NL(3L-2)}{(3N-2)L^2 - (2N-4)L - 2}$, we have
established the following claim.

\begin{claim}
  \claimlabel{greedy_bounds} Let $N > 1$, let $L > 1$, let $R \ge 0$
  and let $K > 1$ satisfy
  $K < \frac{NL(3L-2)}{(3N-2)L^2 - (2N-4)L - 2}$.  Then there exists
  an $M > 0$ such that if an $(L,C)$-greedy tightening sequence for a
  $\frac{1}{K}$-almost isometric $R$-circle $\alpha \colon S \to X$ of
  length $|S| > M$ is disjoint up to $j$ then
  \[ \sum_{i<j}|Q_i| < \frac{|S|}{N} \] where the $Q_i$ are the
  tightened segments of the tightening sequence.
\end{claim}

The following claim about rational functions has a short and
elementary proof.  We will make use of it below.

\begin{claim}
  \claimlabel{rational_functions} The inequalities
  \[ 1 < \frac{L\bigl(9L^2 -3L -4\bigr)}{7L^3 + 3L^2 - 10L + 2} \] and
  \[ 1 < \frac{NL(3L-2)}{(3N-2)L^2 - (2N-4)L - 2} \] and
  \[ \frac{L\bigl(9L^2 -3L -4\bigr)}{7L^3 + 3L^2 - 10L + 2} \le
    \frac{L(5L-4)}{3L^2 - 2} \] and
  \[ \frac{L\bigl(9L^2 -3L -4\bigr)}{7L^3 + 3L^2 - 10L + 2} \le
    \frac{L(7L-6)}{5L^2-2L-2} \] hold for any $L > 1$ and $N > 1$.
\end{claim}

The next claim is essential in proving disjointness of greedy
tightening sequences.

\begin{claim}
  \claimlabel{greedy_tightening_almost_disjoint} Let $K > 1$, let
  $L > 1$ and let $R \ge 0$.  There exists an $M > 0$ such that if $X$
  is an $R$-rough geodesic metric space and $\alpha \colon S \to X$ is
  a $\frac{1}{K}$-almost isometric $R$-circle with $|S| > M$ and
  \[ K < \frac{L\bigl(9L^2 -3L -4\bigr)}{7L^3 + 3L^2 - 10L + 2} \] and
  $C \ge 4R$ then any $(L,C)$-greedy tightening sequence for $\alpha$
  that is disjoint up to $j$ satisfies the following statement.  With
  notation as above, if $(p,q) \in J_j$ and $Q$ is a geodesic segment
  from $p$ to $q$ in $S_j$ then $Q \subset \pc{0,j}$, where we view
  $\pc{0,j}$ as a subspace of $S_j$ via the embedding
  $\pc{0,j} \hookrightarrow S_j$.
\end{claim}
\begin{proof}
  First we will show that $Q$ is not contained in $\bar Q_i$ for any
  $i < j$.  Recall that $\bar Q_i$ is the concatenation
  $A \bar Q'_i B$ where $\alpha_j|_{\bar Q'_i} \colon \bar Q'_i \to X$
  is an $R$-rough geodesic and $\alpha_j$ is constant on $A$ and $B$,
  each of which is isometric to $[0,R]$.  By \peqnref{qi_size}, we
  have $|Q| \ge (C-2R)L \ge (4R - 2R)L \ge 2R$ and $|Q| > 0$ so we
  cannot have $Q \subseteq A$ or $Q \subseteq B$.  We also cannot
  have $\bar Q_i' \subseteq Q \subseteq \bar Q_i$ since then,
  \begin{align*}
    |\bar Q_i| - 2R
    &= d_X\bigl(\alpha_i(p_i), \alpha_i(q_i)\bigr) \\
    &= d_X\bigl(\alpha_j(p), \alpha_j(q)\bigr) \\
    &< \frac{1}{L}|Q| - C \\
    &\le \frac{1}{L}|\bar Q_i| - C \\
    &< |\bar Q_i| - C
  \end{align*}
  which contradicts $C \ge 4R$.  So, if $Q \subseteq \bar Q_i$ then
  some endpoint of $Q$ is contained in $\bar Q'_i$.  But this
  implies that $\alpha_j|_{Q} \colon Q \to X$ is a $2R$-rough
  geodesic and so, by \peqnref{alph_qi_alph_pi_distance} and
  \peqnref{qi_size},
  \begin{align*}
    |Q| - 2R
    &\le d_X\bigl(\alpha_j(p), \alpha_j(q)\bigr) \\
    &< \frac{1}{L}|Q| - C \\
    &< |Q| - C
  \end{align*}
  which, again, contradicts $C \ge 4R$.  Thus we see that $Q$ is not
  contained in $\bar Q_i$ for an $i < j$.  Hence $Q$ intersects
  $\pc{0,j}$ nontrivially in $S_j$.

  If $p$ is contained in the interior of $\bar Q_i$ for some $i < j$
  then let $A_p = Q_i$ and let $\bar A_p = \bar Q_i$.  Otherwise, let
  $A_p = \bar A_p = \{p\}$.  Define $A_q$ and $\bar A_q$ similarly for
  $q$.  A priori, it is possible that $\bar A_p = \bar A_q$.  Let
  $Q^{-}$ be obtained from $Q$ by subtracting the interiors of
  $\bar A_p$ and $\bar A_q$ and let $Q^{+} \to S_j$ extend
  $Q \hookrightarrow S_j$ so as to include a full copy of $\bar A_p$
  and a full copy of $\bar A_q$.  Let $Q_0^{-} \subset S$ be obtained
  from $Q^{-} \subset S_j$ by replacing any $\bar Q_i \subset Q^{-}$
  with $Q_i \subset S$, for $i < j$.  Let $Q_0^{+} \to S$ be obtained
  from $Q^{+} \to S_j$ by replacing any
  $\bar Q_i \hookrightarrow S_j$, where $i < j$, with
  $Q_i \hookrightarrow S$.  Let $\bar Q_0^{+} \to X$ be obtained from
  $Q^{+} \to X$ by replacing $Q \to X$ with $\bar Q \to X$.

  Let $p^{+}$ and $q^{+}$ be the images of the endpoints of $Q_0^{+}$ in
  $S$, with $p^{+}$ the endpoint corresponding to $p$ and $q^{+}$
  the endpoint corresponding to $q$.  Let $p^{-}$ and $q^{-}$ be the
  endpoints of $Q_0^{-}$ in $S$, with $p^{-}$ the endpoint corresponding
  to $p$ and $q^{-}$ the endpoint corresponding to $q$.  Then we
  have
  \begin{align*}
    &d_X\bigl(\alpha(p^{+}), \alpha(q^{+})\bigr) \\
    &\le d_X\bigl(\alpha(p^{+}), \alpha_j(p)\bigr)
      + d_X\bigl(\alpha_j(p), \alpha_j(q)\bigr)
      + d_X\bigl(\alpha_j(q), \alpha(q^{+})\bigr) \\
    &\le d_{\bar A_p}(p^{+}, p)
      + d_X\bigl(\alpha_j(p), \alpha_j(q)\bigr)
      + d_{\bar A_q}(q, q^{+}) \\
    &< d_{\bar A_p}(p^{+}, p)
      + \frac{1}{L}|Q| - C
      + d_{\bar A_q}(q, q^{+}) \\
    &= d_{\bar A_p}(p^{+}, p)
      + \frac{1}{L}\bigl(d_{\bar A_p}(p, p^{-}) + |Q^{-}|
      + d_{\bar A_q}(q^{-}, q) \bigr)
      + d_{\bar A_q}(q, q^{+}) - C \\
    &\le d_{\bar A_p}(p^{+}, p)
      + d_{\bar A_p}(p, p^{-}) + \frac{1}{L}|Q^{-}| + d_{\bar A_q}(q^{-}, q)
      + d_{\bar A_q}(q, q^{+}) - C \\
    &= |\bar A_p| + \frac{1}{L}|Q^{-}| + |\bar A_q| - C \\
    &\le |\bar A_p| + \frac{1}{L}|Q_0^{-}| + |\bar A_q| - C \\
    &\le \frac{1}{L}|A_p| + \frac{1}{L}|Q_0^{-}| + \frac{1}{L}|A_q| - C \\
    &= \frac{1}{L}|Q_0^{+}| - C
  \end{align*}
  where the second inequality follows from \Rmkref{endpoint_distances}
  and the last inequality follows from \peqnref{qi_scale}.  By
  assumption, $Q$ nontrivially intersects at least one $\bar Q_i$,
  with $i < j$.  Let $m$ be minimal such that $Q$ nontrivially
  intersects $\bar Q_m$.  Then, since $Q$ intersects $\pc{0,j}$
  nontrivially, the image of $Q_0^{+} \to S_m$ must strictly contain
  $Q_m$.

  So, if $Q_0^{+} \to S_m$ was the inclusion of a geodesic segment, we
  would have $(p^{+},q^{+}) \in J_m$, which would contradict
  $d_{S_m}(p_m,q_m) = s_m$.  Thus $|Q_0^{+}| > \frac{|S_m|}{2}$.  But
  then
  \begin{align*}
    |Q_0^{+}|
    &> \frac{|S_m|}{2} \\
    &\ge \frac{|S|}{2} - \frac{1}{2}\sum_{i < m}|Q_i| \\
    &\ge \frac{|S|}{2} - \biggl(\frac{K-1}{K} \cdot
      \frac{L(3L-2)}{2(L-1)^2}\biggr)\frac{|S|}{2} - \frac{3LR}{L-1} \\
    &= \biggl(1 - \frac{K-1}{K} \cdot
      \frac{L(3L-2)}{2(L-1)^2} - \frac{6LR}{|S|(L-1)}\biggr)\frac{|S|}{2} \\
  \end{align*}
  by \Claimref{sumbound} while
  \[ |Q_0| \le \biggl(\frac{K -1}{K} \cdot
    \frac{L}{L-1}\biggr)\frac{|S|}{2} \] by \Claimref{onebound}.  So
  $|Q_0^{+}| \le |Q_0|$ would imply
  \[ 1 - \frac{K-1}{K} \cdot \frac{L(3L-2)}{2(L-1)^2} -
    \frac{6LR}{|S|(L-1)} < \frac{K -1}{K} \cdot \frac{L}{L-1} \]
  which is equivalent to the following inequality.
  \begin{equation}
    \eqnlabel{if_qp_le_q0}
    1 < \frac{K-1}{K} \cdot \frac{L(5L-4)}{2(L-1)^2} +
    \frac{6LR}{|S|(L-1)}
  \end{equation}
  By hypothesis and \Claimref{rational_functions}, we have
  $K < \frac{L(5L-4)}{3L^2 - 2}$ which is equivalent to
  $1 > \frac{K-1}{K} \cdot \frac{L(5L-4)}{2(L-1)^2}$ so if $|S| > M'$
  for some $M'$ depending only on $K$, $L$ and $R$ then we would have
  $1 > \frac{K-1}{K} \cdot \frac{L(5L-4)}{2(L-1)^2} +
  \frac{6LR}{|S|(L-1)}$ and this would contradict
  \peqnref{if_qp_le_q0}.  Hence, assuming $|S|$ is greater than this
  $M'$, we have $|Q_0^{+}| > |Q_0|$.

  Then if $Q_0^{+} \to S$ were the inclusion of a geodesic segment
  then we would have $(p^{+},q^{+}) \in J_0$ and this would contradict
  $d_S(p_0,q_0) = s_0$.  Thus $|Q_0^{+}| > \frac{|S|}{2}$.  On the
  other hand
  \begin{align*}
    &|Q_0^{+}| \\
    &= |Q_0^{-}| + |A_p| + |A_q| \\
    &\le |Q^{-}|  + \sum_{i<j} \bigl(|Q_i|-|\bar Q_i|\bigr) +
      |A_p| + |A_q| \\
    &\le \frac{|S_j|}{2} + \sum_{i<j} \bigl(|Q_i|-|\bar Q_i|\bigr) +
      |A_p| + |A_q| \\
    &= \frac{|S|}{2} +
      \frac{1}{2}\sum_{i<j} \bigl(|Q_i|-|\bar Q_i|\bigr) +
      |A_p| + |A_q| \\
    &\le \frac{|S|}{2} + \frac{1}{2}\sum_{i<j} |Q_i| + |A_p| + |A_q| \\
    &\le \frac{|S|}{2} +
      \biggl(\frac{K-1}{K} \cdot
      \frac{L(3L-2)}{4(L-1)^2}\biggr)|S|
      + \frac{3LR}{L-1}
      + \biggl(\frac{K -1}{K} \cdot \frac{L}{L-1}\biggr)|S| \\
    &= \frac{|S|}{2} +
      \frac{K -1}{K} \cdot \frac{L}{L-1} \cdot \biggl(
      \frac{3L-2}{4(L-1)} + 1\biggr)|S|
      + \frac{3LR}{L-1} \\
    &= \frac{|S|}{2} +
      \frac{K -1}{K} \cdot \frac{L}{L-1} \cdot
      \frac{7L- 6}{4(L-1)} \cdot |S|
      + \frac{3LR}{L-1} \\
    &= \biggl(\frac{1}{2} + \frac{K-1}{K} \cdot \frac{L(7L- 6)}{4(L-1)^2}
       + \frac{3LR}{|S|(L-1)}\biggr) |S|
  \end{align*}
  where the last inequality follows from \Claimref{onebound} and
  \Claimref{sumbound}.  By hypothesis and
  \Claimref{rational_functions}, we have
  $K < \frac{L(7L-6)}{5L^2-2L-2}$, which is equivalent to
  $\frac{1}{2} + \frac{K-1}{K} \cdot \frac{L(7L- 6)}{4(L-1)^2} < 1$.
  Thus, if $|S| > M''$ for some $M''$ depending only on $K$, $L$ and
  $R$ then
  $\frac{1}{2} + \frac{K-1}{K} \cdot \frac{L(7L- 6)}{4(L-1)^2} +
  \frac{3LR}{|S|(L-1)} < 1$ and so $|Q_0^{+}| < |S|$ so that $Q_0^{+}$
  embeds in $S$.  In this case, the endpoints $p^{+},q^{+}$ of
  $Q_0^{+}$ in $S$ are at distance
  \[ d_S(p^{+},q^{+}) \ge \biggl(\frac{1}{2} - \frac{K-1}{K} \cdot
    \frac{L(7L- 6)}{4(L-1)^2} - \frac{3LR}{|S|(L-1)}\biggr)
    |S| \]
  but we also have
  \begin{align*}
    &d_X\bigl(\alpha(p^{+}),\alpha(q^{+})\bigr) \\
    &< \frac{1}{L}|Q_0^{+}| - C \\
    &\le \frac{1}{L}
      \biggl(\frac{1}{2} + \frac{K-1}{K} \cdot \frac{L(7L- 6)}{4(L-1)^2}
      + \frac{3LR}{|S|(L-1)}\biggr) |S| - C
  \end{align*}
  which, by \Lemref{global_to_local} and $C \ge 4R$, implies
  \begin{align*}
    &\frac{1}{L}
      \biggl(\frac{1}{2} + \frac{K-1}{K} \cdot \frac{L(7L- 6)}{4(L-1)^2}
      + \frac{3LR}{|S|(L-1)}\biggr) |S| \\
    &> \biggl(\frac{1}{2} - \frac{K-1}{K} \cdot
      \frac{L(7L- 6)}{4(L-1)^2} - \frac{3LR}{|S|(L-1)}\biggr) |S|
      - \frac{K-1}{K} \cdot \frac{|S|}{2}
  \end{align*}
  which is equivalent to the following inequality.
  \begin{equation}
    \eqnlabel{qplus_and_almost_isometric}
    \frac{K-1}{K} \cdot
    \frac{9L^2 - 3L -4}{2(L-1)^2}
    + \frac{6R(L+1)}{|S|(L-1)} > \frac{L-1}{L}
  \end{equation}
  By hypothesis, we have
  $K < \frac{L\bigl(9L^2 -3L -4\bigr)}{7L^3 + 3L^2 - 10L + 2}$ which
  is equivalent to
  $\frac{K-1}{K} \cdot \frac{9L^2 - 3L -4}{2(L-1)^2} < \frac{L-1}{L}$
  so if $|S| > M'''$ for some $M'''$ depending only on $K$, $L$ and
  $R$ then we would have
  $\frac{K-1}{K} \cdot \frac{9L^2 - 3L -4}{2(L-1)^2} +
  \frac{6R(L+1)}{|S|(L-1)} < \frac{L-1}{L}$ which contradicts
  \peqnref{qplus_and_almost_isometric}.

  Therefore, if $|S| > M = \max\{M', M'', M'''\}$, which depends only
  on $K$, $L$ and $R$ then assuming the existence of a $j$ for which
  $Q \not\subset \pc{0,j}$ leads us to a contradiction.
\end{proof}

\begin{claim}
  \claimlabel{greedy_tightening_disjoint} Let $K > 1$, let $L > 1$ and
  let $R \ge 0$.  There exists an $M > 0$ such that if $X$ is an
  $R$-rough geodesic metric space and $\alpha \colon S \to X$ is a
  $\frac{1}{K}$-almost isometric $R$-circle with $|S| > M$ and
  \[ K < \frac{L\bigl(9L^2 - 3L - 4\bigr)}{7L^3 + 3L^2 - 10L + 2} \]
  and $C \ge 4R$ then any $(L,C)$-greedy tightening sequence for
  $\alpha$ is completely disjoint.
\end{claim}
\begin{proof}
  Consider an $(L,C)$-greedy tightening sequence for $\alpha$ with
  notation as above.  For the sake of finding a contradiction, suppose
  $j \ge 1$ is the least integer with $Q_j \not\subset \pc{0,j}$.  As
  above we view the $\bar Q_i$ with $i < j$ as disjoint segments of
  $S_j$ with $S_j \setminus \bigcup_{i=0}^{j-1}\bar Q_m = \pc{0,j}$.

  Since $Q_j \not\subset \pc{0,j}$, we have
  $\bar Q_m \cap Q_j \neq \emptyset$, for some $m < j$.  Recall that
  $(p_j,q_j)$ is the limit of a sequence $(p_j^{(n)},q_j^{(n)})_n$ in
  $J_j$.  For each $n$, let $Q_j^{(n)}$ be a geodesic segment between
  $p_j^{(n)}$ and $q_j^{(n)}$ in $S_j$.  By
  \Claimref{greedy_tightening_almost_disjoint}, we have
  $Q_j^{(n)} \subset \pc{0,j}$, so we have
  $\bar Q_m \cap Q_j \subset \{p_m,q_m\} \cap \{p_j,q_j\}$. Without
  loss of generality, we may assume $q_m = p_j$.

  Since $Q_j^{(n)} \subset \pc{0,j}$, we may think of the $Q_j^{(n)}$
  as segments of $S_m$, by the embedding
  $\pc{0,j} \hookrightarrow S_m$.  Each $Q_j^{(n)}$ is a geodesic
  segment in $S_m$ since the complementary segment of $Q_j^{(n)}$ in
  $S_m$ is even longer than the complementary segment of $Q_j^{(n)}$
  in $S_j$.  Thus $(p_j^{(n)},q_j^{(n)})_n$ is a sequence in $J_m$.
  For each $n$, let $Q_m^{(n)}$ be a segment between $p_m^{(n)}$ and
  $q_m^{(n)}$ such that $(Q_m^{(n)})_n$ converges to $Q_m$ in
  Hausdorff distance.  The circular orders on the triples
  $(p_m^{(n)},q_m^{(n)},q_j^{(n)})$ and
  $(p_m^{(n)},p_j^{(n)},q_j^{(n)})$ are eventually constant and equal.
  Let $(A^{(n)})_n$ be a sequence of segments in $S_m$ from
  $p_m^{(n)}$ to $q_j^{(n)}$ such that $A^{(n)}$ eventually contains
  $q_m^{(n)}$ or, equivalently, eventually contains $p_j^{(n)}$.

  Let $\epsilon > 0$ satisfy $\epsilon < \frac{|Q_j|}{3}$ and
  $\epsilon \le \frac{LR}{L+1}$.  Then, for $n$ large enough,
  \begin{align*}
    |A^{(n)}|
    &\ge |Q_m^{(n)}| + |Q_j^{(n)}| - d_{S_m}(q_m^{(n)}, p_j^{(n)}) \\
    &> |Q_m| - \epsilon + |Q_j| - \epsilon - \epsilon \\
    &= |Q_m| + |Q_j| - 3\epsilon \\
    &> |Q_m|
  \end{align*}
  and
  \begin{align*}
    &d_X\bigl(\alpha_m(p_m^{(n)}), \alpha_m(q_j^{(n)})\bigr) \\
    &\le d_X\bigl(\alpha_m(p_m^{(n)}), \alpha_m(q_m^{(n)})\bigr)
      + d_X\bigl(\alpha_m(q_m^{(n)}), \alpha_m(p_j^{(n)})\bigr) \\
      &\;\;\;\;\;\; + d_X\bigl(\alpha_m(p_j^{(n)}), \alpha_m(q_j^{(n)})\bigr) \\
    &< \frac{1}{L}d_{S_m}(p_m^{(n)}, q_m^{(n)}) - C
      + d_{S_m}(q_m^{(n)}, p_j^{(n)}) + R
      + \frac{1}{L}d_{S_m}(p_j^{(n)}, q_j^{(n)}) - C \\
    &< \frac{1}{L}d_{S_m}(p_m^{(n)}, q_m^{(n)}) - C
      + d_{S_m}(q_m^{(n)}, p_j^{(n)}) + R
      + \frac{1}{L}d_{S_m}(p_j^{(n)}, q_j^{(n)}) - C \\
    &< \frac{1}{L}|Q_m| + \epsilon - C
      + \epsilon + R
      + \frac{1}{L}|Q_j| + \epsilon - C \\
    &= \frac{1}{L}(|Q_m| + |Q_j|) - 2C + R + 3\epsilon \\
    &< \frac{1}{L}(|A^{(n)}| + 3\epsilon) - 2C + R + 3\epsilon \\
    &= \frac{1}{L}|A^{(n)}| - 2C + R + \frac{3(L+1)}{L}\cdot\epsilon \\
    &\le \frac{1}{L}|A^{(n)}| - C
  \end{align*}
  since $C \ge 4R$.  So, if $A^{(n)}$ is a geodesic segment for
  arbitrarily large $n$ then, for some $n$, we would have
  $(p_m^{(n)},q_j^{(n)}) \in J_m$ and
  $d_{S_m}(p_m^{(n)},q_j^{(n)}) > |Q_m| = s_m$, which is a
  contradiction.  Thus eventually
  $|Q_m^{(n)}| + |Q_j^{(n)}| = |A^{(n)}| > \frac{|S_m|}{2}$ and so
  $|Q_m| + |Q_j| \ge \frac{S_m}{2}$.

  Then, by \Claimref{onebound}, we have
  \[ \frac{S_m}{2} \le 2\biggl(\frac{K - 1}{K} \cdot
  \frac{L}{L-1}\biggr)\frac{|S|}{2} \] while
  \begin{align*}
    \frac{|S_m|}{2}
    &\ge \frac{|S|}{2} - \frac{1}{2}\sum_{i < m}|Q_i| \\
    &\ge \frac{|S|}{2} - \biggl(\frac{K-1}{K} \cdot
      \frac{L(3L-2)}{2(L-1)^2}\biggr)\frac{|S|}{2} - \frac{3LR}{L-1} \\
    &= \biggl(1 - \frac{K-1}{K} \cdot
      \frac{L(3L-2)}{2(L-1)^2} - \frac{6LR}{|S|(L-1)}\biggr)\frac{|S|}{2}
  \end{align*}
  by \Claimref{sumbound}.  Combining these we obtain
  \[ \frac{K-1}{K} \cdot \frac{L(7L-6)}{2(L-1)^2} +
    \frac{6LR}{|S|(L-1)} \ge 1 \] but, by hypothesis and
  \Claimref{rational_functions}, we have
  $K < \frac{L(7L-6)}{5L^2 - 2L - 2}$, which is equivalent to
  $\frac{K-1}{K} \cdot \frac{L(7L-6)}{2(L-1)^2} < 1$ so, if $|S|$ is
  large enough (depending only on $K$, $L$ and $R$) then we have a
  contradiction.
\end{proof}

\section{The Fine Milnor-Schwarz Lemma}
\seclabel{fine_ms}

The Fine Milnor-Schwarz Lemma is a refinement of the Milnor-Schwarz
Lemma that gives finer control on the multiplicative constant of the
quasi-isometry.  In this section we will state and prove this version
of the Milnor-Schwarz Lemma.

Let $(X,d)$ be a rough geodesic metric space.  Let $G$ be a group
acting coboundedly on $X$ by isometries.  Let
\[ S_{x_0,t} = \bigl\{ g \in G \sth d(x_0,gx_0) \le t \bigr\} \] for
$x_0 \in X$ and $t \in \R_{\ge 0}$.

\begin{rmk}
  \rmklabel{metric_proper_finite} If the action of $G$ is metrically
  proper then the $S_{x_0,t}$ are finite.
\end{rmk}

Let $\Gamma_{x_0,t}$ be the graph with vertex set $G$ and with an edge
of length $t$ between $g$ and $g'$ whenever $g' = gs$ for some
$s \in S_{x_0,t}$.  So when $S_{x_0,t}$ generates $G$, the graph
$\Gamma_{x_0,t}$ is the Cayley graph of $G$ for the generating set
$S_{x_0,t}$ with edges scaled by $t$.  Let $d_{x_0,t}$ be the graph
metric on $\Gamma_{x_0,t}$ where we set $d_{x_0,t}(g,h) = \infty$ when
$g$ and $h$ are in different components of $\Gamma_{x_0,t}$.  Then,
when $S_{x_0,t}$ generates $G$, the metric $d_{x_0,t}$ is the word
metric on $G$ for the generating set $S_{x_0,t}$ scaled by $t$.  Let
$f_{x_0,t} \colon (G,d_{x_0,t}) \to (X,d)$ be defined by
$f_{x_0,t}(g) = gx_0$.  Let $K_{x_0,t}$ be the infimum of all
$K \ge 1$ for which there exists some $C_K > 0$ such that $f_{x_0,t}$
is a $(K,C_K)$-quasi-isometry.  Note that if $f_{x_0,t}$ is not a
quasi-isometry (e.g. if $S_{x_0,t}$ does not generate $G$) then
$K_{x_0,t} = \infty$.

\begin{lem}[Fine Milnor-Schwarz Lemma]
  \lemlabel{fine_ms} Let $(X,d)$ be a rough geodesic metric space.
  Let $G$ be a group acting coboundedly on $X$ by isometries.  Let
  $x_0 \in X$ and let $K_{x_0,t}$ be defined as above for
  $t \in \R_{\ge 0}$.  Then $K_{x_0,t} \to 1$ as $t \to \infty$.
\end{lem}
\begin{proof}
  Let $g \in G$.  We will prove that
  $d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr) \le d_{x_0,t}(1,g)$.  If
  $d_{x_0,t}(1,g) = \infty$ there is nothing to show and so
  $d_{x_0,t}(1,g) = Mt$ for some $M \in \N_{\ge 0}$ and there is a
  combinatorial path defined by
  \[ 1 = g_0, g_1, g_2, \ldots, g_M = g \] in $\Gamma_{x_0,t}$.  The
  $g_{i-1}^{-1}g_i$ are contained in $S_{x_0,t}$ and so, by the
  triangle inequality, we have the following.
  \begin{align*}
    d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr)
    &= d(x_0,gx_0) \\
    &\le d(x_0,g_1x_0) + d(g_1x_0,g_2x_0) + \ldots + d(g_{M-1}x_0,gx_0) \\
    &= d(x_0,g_1x_0) + d(x_0,g_1^{-1}g_2x_0) + \ldots + d(x_0,g_{M-1}^{-1}gx_0) \\
    &\le Mt \\
    &= d_{x_0,t}(1,g)
  \end{align*}

  We now establish a lower bound on
  $D = d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr)$.  Since $G$ acts
  coboundedly on $X$, the orbit $Gx_0$ is a quasi-onto subspace of
  $X$.  Hence $Gx_0$ is roughly isometric to $X$ and so $Gx_0$ is also
  a rough geodesic metric space.  Let $R$ be the rough geodesicity
  constant of $Gx_0$.  Let $\alpha \colon [0,D] \to Gx_0$ be an
  $R$-rough geodesic from $f_{x_0,t}(1) = x_0$ to
  $f_{x_0,t}(g) = gx_0$.  Assume that $t > R$ and subdivide $[0,D]$
  into at most $\bigl\lceil\frac{D}{t-R}\bigr\rceil$ segments of
  length at most $t-R$.  Let $0 = a_0 < a_1 < a_2 < \cdots < a_M = D$
  be the endpoints of the segments and let $g_ix_0 = \alpha(a_i)$, for
  each $i$, with $g_0 = 1$ and $g_M = g$.  Then
  \begin{align*}
    d(x_0, g_i^{-1}g_{i+1}x_0)
    &= d(g_ix_0, g_{i+1}x_0) \\
    &= d\bigl(\alpha(a_i), \alpha(a_{i+1})\bigr) \\
    &\le |a_i - a_{i+1}| + R \\
    &\le t
  \end{align*}
  for each $i$.  Thus $g_i^{-1}g_{i+1} \in S_{x_0,t}$, for each $i$,
  and so
  \[ 1 = g_0, g_1, g_2, \ldots, g_{M-1}, g_M = g \] defines a
  combinatorial path in $\Gamma_{x_0,t}$.  Hence
  \begin{align*}
    d_{x_0,t}(1,g)
    &\le tM \\
    &\le t\Bigl\lceil\frac{D}{t-R}\Bigr\rceil \\
    &= t\biggl\lceil\frac{d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr)}{t-R}\biggr\rceil \\
    &\le t\biggl(\frac{d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr)}{t-R} + 1\biggr) \\
    &= \frac{t}{t-R}d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr) + t
  \end{align*}
  and so we have
  $\frac{t-R}{t}d_{x_0,t}(1,g) - (t-R) \le
  d\bigl(f_{x_0,t}(1),f_{x_0,t}(g)\bigr)$.

  For $g,g' \in G$,
  \begin{align*}
    d\bigl(f_{x_0,t}(g),f_{x_0,t}(g')\bigr)
    &= d(gx_0, g'x_0) \\
    &= d(x_0,g^{-1}g'x_0) \\
    &= d\bigl(f_{x_0,t}(1),f_{x_0,t}(g^{-1}g')\bigr)
  \end{align*}
  and $d_{x_0,t}(g,g') = d_{x_0,t}(1,g^{-1}g')$ so we have
  \[ \frac{t-R}{t}d_{x_0,t}(g,g') - (t-R) \le
    d\bigl(f_{x_0,t}(g),f_{x_0,t}(g')\bigr) \le d_{x_0,t}(g,g') \]
  hence $1 \le K_{x_0,t} \le \frac{t-R}{t}$ when $t > R$.  This
  implies that $K_{x_0,t} \to 1$ as $t \to \infty$.
\end{proof}

\begin{cor}[Milnor-Schwarz Lemma]
  If the $G$-action on $X$ metrically proper then $G$ is finitely
  generated and, for any finite generating set $S$, the map
  \begin{align*}
    f_s \colon (G,d_S) &\to X \\
    g &\mapsto gx_0
  \end{align*}
  is a quasi-isometry, where $d_S$ is the word metric for the
  generating set $S$.
\end{cor}
\begin{proof}
  By \Lemref{fine_ms}, if $t$ is large enough then
  $K_{x_0,t} < \infty$.  Thus $S_{x_0,t}$ generates $G$.  By
  \Rmkref{metric_proper_finite}, the generating set $S_{x_0,t}$ is
  finite so $G$ is finitely generated.  Since $K_{x_0,t} < \infty$ and
  since scaling the metric $d_S$ by a factor of $t$ preserves the
  quasi-isometry type, the map $f_S$ is a quasi-isometry for
  $S = S_{x_0,t}$.  But the identity map on $G$ is a bilipschitz
  equivalence from $(G, d_{S'})$ to $(G, d_{S''})$ where $S'$ and
  $S''$ are any two finite generating sets and $d_{S'}$ and $d_{S''}$
  are the corresponding word metrics.  Thus $f_S$ is a quasi-isometry
  for any generating set $S$.
\end{proof}

\begin{cor}
  Let $G$ be a group.  If $G$ acts metrically properly and coboundedly
  on a strongly shortcut rough geodesic metric space $X$ then $G$ has
  a finite generating set $S$ for which the Cayley graph of $(G,S)$ is
  strongly shortcut.  In particular, the group $G$ is strongly
  shortcut.
\end{cor}
\begin{proof}
  By \Corref{rough_approx_inv}, there exists an $L_X > 1$ such that
  whenever $C > 0$ and $Y$ is a rough geodesic metric space and
  $f \colon Y \to X$ is an $(L_X,C)$-quasi-isometry up to scaling,
  then $Y$ is strongly shortcut.  But, by \Lemref{fine_ms}, there is a
  Cayley graph $\Gamma$ of $G$, a $C > 0$ and an
  $(L_X,C)$-quasi-isometry up to scaling $f \colon \Gamma \to X$.  So
  $\Gamma$ is strongly shortcut as a rough geodesic metric space.
  Then, by \Rmkref{ss_asgraph_asmetric}, the Cayley graph $\Gamma$ is
  strongly shortcut as a graph.
\end{proof}

\begin{cor}
  \corlabel{ss_group} Let $G$ be a group.  The following conditions
  are equivalent
  \begin{enumerate}
  \item $G$ is strongly shortcut.
  \item $G$ acts metrically properly and coboundedly on a strongly
    shortcut rough geodesic metric space.
  \item $G$ has a finite generating set $S$ for which the Cayley graph
    of $(G,S)$ is strongly shortcut.
  \end{enumerate}
\end{cor}

\section{Asymptotically \texorpdfstring{$\CAT(0)$}{CAT(0)} spaces}
\seclabel{as_cat0}

In this section we will apply the characterizations of
\Secref{characterizations} to prove that asymptotically $\CAT(0)$
rough geodesic metric spaces are strongly shortcut.  By the results of
\Secref{fine_ms}, this will imply that asymptotically $\CAT(0)$ groups
are strongly shortcut.

Asymptotically $\CAT(0)$ spaces and groups were first introduced and
studied by Kar \cite{Kar:2011}.  A metric space $X$ is
\defterm{asymptotically $\CAT(0)$} if every asymptotic cone of $X$ is
$\CAT(0)$.  A group is \defterm{asymptotically $\CAT(0)$} if it acts
properly and cocompactly on an asymptotically $\CAT(0)$ proper
geodesic metric space.  (Note that the condition that the action be on
a proper metric space does not make this definition more restrictive
than the definition given by Kar since the definition of proper action
in Kar \cite{Kar:2011} seems to be that of Bridson and Haefilger
\cite[Section~I.8.2]{Bridson:1999} \cite[Page 77]{Kar:2011} and any
geodesic metric space admitting a cocompact action that is proper by
this more restricted definition is a proper metric space.)  For an
introduction to $\CAT(0)$ geodesic metric spaces, see Bridson and
Haefliger \cite{Bridson:1999}.

\begin{thm}
  \thmlabel{as_cat0_space} Asymptotically $\CAT(0)$ rough geodesic
  metric spaces are strongly shortcut.
\end{thm}
\begin{proof}
  By uniqueness of geodesics in $\CAT(0)$ geodesic metric spaces,
  there is no isometric copy of a Riemannian circle in the asymptotic
  cone of an asymptotically $\CAT(0)$ metric space $X$.  So, by
  \Corref{sshortcut_equiv}, any asymptotically $\CAT(0)$ rough
  geodesic metric space is strongly shortcut.
\end{proof}

\begin{thm}
  \thmlabel{as_cat0_group} Asymptotically $\CAT(0)$ groups are
  strongly shortcut.
\end{thm}
\begin{proof}
  A proper and cocompact action on a proper metric space is metrically
  proper and cobounded.  So, by \Thmref{as_cat0_space}, any asymptotically
  $\CAT(0)$ group $G$ acts metrically properly and cobounded on a
  strongly shortcut geodesic metric space.  Thus, by
  \Corref{ss_group}, the group $G$ is strongly shortcut.
\end{proof}

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


Annmarie Lockhart is the founding editor of vox poetica, an online literary salon dedicated to poetry, and Unbound Content, an independent poetry press. A lifelong Bergen County, New Jersey resident, she lives, writes, and works two miles from the hospital where she was born. You can read her words at fine journals online and in print.
Pushcart Prize nominee KJ Hannah Greenberg's lightly pert and somewhat exuberant layered writing can be found in North American, European, Oceanic, Middle Eastern, and Far Eastern venues, as well as under select budgies. Her creative efforts are devoted to lovers of slipstream fiction, to second chair oboe players, and to mothers who despair of fining the bottom of Mt. Laundry. Her books are available at Amazon.


text:
" www.prosveta.com" and "www.prosveta.fr" are publications of Éditions Prosveta SA, ZI Capitou, 1277 avenue Lachenaud, 83600 FREJUS.
The information contained in this site is subject to French law and is provided "as is". "without warranties of any kind, either expressed or implied. This information may contain technical inaccuracies or typographical errors. They are non-contractual and subject to change without notice. They do not engage the responsibility of Éditions Prosveta SA, which strives to keep them up to date.
All content, pages, scripts, logos, texts, images, photos and sounds on this site are the exclusive property of Éditions Prosveta SA Any production, reproduction or representation of this site, in whole or in part (texts, sounds or images), on any medium whatsoever is prohibited. Failure to comply with this prohibition constitutes an infringement which may incur the civil and criminal liability of the infringer.
It is strictly forbidden to use or reproduce the name Prosveta or its logo , alone or associated, in any capacity whatsoever and in particular for advertising purposes without the prior written consent of Éditions Prosveta SA
Éditions Prosveta SA are not responsible for the content of any other site to which you may have access via the site www.prosveta.fr.
It is expressly understood by the user of this site that in no event shall Éditions Prosveta SA be held liable for any damage whatsoever, direct or indirect, material or immaterial or special, resulting in particular from the consultation and/or use of this website or other sites linked to it, such as the use of textual, sound or visual information that could have been there collected and in particular from any financial or commercial damage, loss of programs or data in its information system or other.
RCS of Fréjus: 76 B 21
APE Code: 5811Z
EEC VAT: FR61784116071
EORI No.: GB079467264000
Capital: 792,734.89 Euros
Brand Prosveta © Patrimonial rights
The brand PROSVETA belongs to ÉDITIONS PROSVETA SA, 1277 Avenue Jean Lachenaud , ZI du Capitou, 83600 Fréjus, France, which has registered it in many countries.
The use of the PROSVETA name and trademark is granted to allow the publication under contract or the distribution of the work of Master Omraam Mikhaël Aïvanhov in many countries.
However, the various companies and associations that bear the name PROSVETA are independent. There is no filiation between them.
Éditions Prosveta SA is also the owner of the property rights to the work of Master Omraam Mikhaël Aïvanhov.
Publications outside Editions Prosveta SA
Editions Prosveta SA are the only are worldwide holders of the publishing rights to the work of Master Omraam Mikhaël Aïvanhov. Its publications are made with the agreement of the holders of the moral rights to this work.
They publish this work mainly in French and English. They grant translation and publishing or co-publishing rights for other languages. The publishers concerned are of course authorized to provide information in the usual context of their sites and their activities.
It should be noted that, moreover, many sites use the name of Omraam Mikhaël Aïvanhov to present, comment on, communicate on, or give advice relating or supposed to relate to his work. The Internet allows unprecedented instantaneousness and multiplicity of communication in which everyone can publish and express themselves almost without control, and moreover often safe from prosecution.
Les Éditions Prosveta SA make it known that they only recognize as Omraam Mikhaël Aïvanhov's work what they themselves have published under his name. Any other, even claiming to be a partnership: website, publication, comment, advice or consideration made using the name of Omraam Mikhaël Aïvanhov or referring to truncated or insufficiently referenced quotes cannot be attributed to Omraam Mikhaël Aïvanhov and to Éditions Prosveta SA and therefore engage the sole responsibility of their author.
In accordance with article 34 of law n°78-17 of January 6, 1978 relating to data processing, files and freedoms, you have the right to access, rectify and delete data concerning you. You can exercise this right by writing to: ÉDITIONS PROSVETA SA – ZA CAPITOU - CS 30012 – 83601 FREJUS CEDEX – international@prosveta.com Use of cookies The user is informed that during his visits to the site www.prosveta.fr. fr, a cookie may be installed in its browser. A cookie does not identify the user. In general, he records information relating to the navigation of his device (computer, smartphone, tablet) on the site (the pages consulted, the date and time of the consultation) which can be read during his subsequent visits. The user can oppose the registration of cookies, or be warned before accepting cookies, by configuring the preferences of his browser.
Publishing Director
Les Éditions Prosveta SA, represented by its Chief Executive Officer.
Editions Prosveta SA @ Email
Design and creation of the site
Kiwik - 5 rue Claude Lewy – 45100 Orléans - France - www.studio-kiwik.fr
Planet-Work - 231 rue Saint-Honoré, 75001 Paris - France - www.planet-work.com

redpajama_set_name: RedPajamaCommonCrawl