\section{Introduction}
\input{introduction}


\section{Preliminaries}\label{sec:preliminaries}
\input{preliminaries}
\input{WG}
\input{cutpointdetection}
\input{normalization}
\section{Splittings}\label{sec:splitting}



\input{uncrossed}

\subsection{The Splitting Criterion}\label{sec:otalsplit}
\input{otalsplit}

\subsection{Proof of the Decomposition Theorem}\label{sec:proofofdecomp}
\input{splitting}



\section{Virtually Geometric Multiwords}\label{sec:vg}
\input{virtuallygeometric}
\subsection{Examples}
\input{examples}



\bibliographystyle{hyperamsplain}

\subsection{Whitehead Graphs}
The primary tool for understanding the topology of the decomposition
space associated to a line pattern is the generalized Whitehead graph of a
generating multiword. 
This machinery was developed in \cite{CasMac11}.
In this section we will recall the relevant definitions and results,
but see \cite{CasMac11} for details.

The Whitehead graph $\Wh(*)=\Wh_{\mathcal{B}}(*)\{w\}$ of a cyclically reduced word $w$ with respect to a basis
$\mathcal{B}$ of $F$ is a graph with $2n$ vertices labeled with the
elements of $\mathcal{B}$ and their inverses. 
One edge joins vertex $x$ to vertex $y$ for each occurrence of
$\inv{x}y$ in the word $w$ as a cyclic word. 

Similarly, the Whitehead graph of a multiword is obtained by adding
edges for each element of the multiword.
The number of edges is therefore equal to the sum of the word lengths
 with
respect to 
$\mathcal{B}$.

Whitehead's Algorithm \cite{Whi36} gives an algorithm for reducing the
number of edges of the graph by applying a certain collection of
Whitehead automorphisms. The complexity, the number of edges, can be
reduced monotonically down to some minimum. 
A particularly important observation is that if some Whitehead graph
$\Wh_\mathcal{B}(*)$ is connected and has a cut vertex, a vertex whose
removal disconnects the graph, then this Whitehead graph is not
minimal complexity.

An easy consequence of Whitehead's results is the following:

\begin{proposition} \label{proposition:Whitehead}
Given a multiword $\multimot$ in a free group $F$,
  the following are equivalent:
  \begin{enumerate}
  \item Some Whitehead graph for $\multimot$ is not connected.
\item Every minimal Whitehead graph for $\multimot$ is not connected.
\item $F$ splits freely rel $\multimot$.
  \end{enumerate}
  
\end{proposition}

The next three proclamations relate \fullref{proposition:Whitehead} to
the decomposition space. 
They come from work of Otal \cite[Prop 2.1]{Ota92}, and can also
be found in the thesis of Reiner Martin \cite[Theorem 49]{Mar95}.
\begin{lemma}{\cite[Lemma 4.3]{CasMac11}}\label{lemma:disconnected}
 If for some choice of basis $\Wh(*)$ is disconnected, then $\decomp$ is disconnected.
\end{lemma}

\begin{lemma}[{\cite[Lemma 4.4]{CasMac11}}]\label{lemma:basicconnectivity}
Suppose there exists a free basis $\mathcal{B}$ of $F$ such that
$\Wh_\mathcal{B}(*)$ is connected without cut vertices. 
Let
$\tree$ be the Cayley graph of $F$ corresponding to $\mathcal{B}$.
Pick any edge $e$ in $\tree$.
Let $*$ and $v$ be the endpoints of $e$.
Let $\hat{A}$ be the collection of points $\xi\in\partial \tree$ such that $v$ is on $[*,\xi]$.
The set $A=\qmap(\hat{A})$ is connected in $\decomp$.
\end{lemma}






Let $\{l_1,\dots, l_k\}\subset\lines$ be the set of lines that contain $e$.
For each $i$, the two endpoints $l_i^+$ and $l_i^-$ are identified in the decomposition space.
Thus: \[\qmap(\hat{A})\cap \qmap(\hat{A}^c)=\underset{i=1..k}{\bigcup}\qmap(l_i^+)\]

So, if $\Wh(*)$ is connected without cut vertices, then for any edge
$e$ in $\tree$ the boundaries at infinity of
the two connected components of $\tree\setminus e$ correspond to connected
sets in the decomposition space. Since $\Wh(*)$ is connected
there is also at least one line in $\lines$ crossing $e$, so these connected sets have a point in common.

\begin{corollary}\label{corollary:connected}
Suppose $\Wh(*)$ has no cut vertices. The
decomposition space is connected if and only if $\Wh(*)$ is connected.
Furthermore, if $\decomp$ is connected it is also locally connected.
\end{corollary}

Moreover, if none of the $\qmap(l_i^+)$ is a cut point of $\decomp$ then, in fact, \[\qmap(\hat{A})\setminus\underset{i=1..k}{\bigcup}\qmap(l_i^+)\] is a connected subset of $\decomp$.

\begin{assumption}
  Unless otherwise noted, throughout the paper we will assume that we
  are working with a multiword such that the decomposition space is
  connected, and that we have chosen a basis for the free group such that
  the corresponding Whitehead graph is connected without cut vertices.
Equivalently, $F$ does not split freely relative to the multiword.
\end{assumption}


There is no loss of generality from this assumption. 
If the decomposition space were not connected we would first pass to a
maximal free splitting of $F$ relative to the multiword and then deal
with the free factors separately. Once the decomposition space is
connected we can use Whitehead's algorithm to produce a basis such
that the Whitehead graph is connected without cut vertices.
For computational purpose it would probably be wise to choose a basis that gives a minimal complexity Whitehead graph,
but all of our arguments still work with only the ``connected without
cut vertices'' assumption.

The classical Whitehead graph generalizes: if $\lines$ is a line pattern in $F$ and $\X$ is a connected subset of $\closure{\tree}$ we
define the Whitehead graph $\Wh_\mathcal{B}(\X)\{\lines\}$ to be the
graph with vertices in bijection with connected components of
$\closure{\tree}\setminus \X$ and one edge joining $v$ and $v'$ for
each line of $\lines$ that has one endpoint in the component
corresponding to $v$ and the other in the component corresponding to $v'$.
We will omit $\mathcal{B}$ and $\lines$ when they are clear and just write  $\Wh(\X)$.
The notation $\Wh(*)$ for the classical Whitehead graph just means the Whitehead graph at a vertex $*$, and because the line pattern is equivariant we get the same graph for any vertex.


Let
$\X\subset\mathcal{Y}\subset\closure{\tree}$.
 Let $e$ be an edge of $\tree$ incident to
exactly one vertex of $\X$.

The edge $e$ corresponds to a vertex in $\Wh(\X)$. The graph
$\Wh(\X)\setminus e$
is obtained from $\Wh(\X)$ by deleting this vertex, but retaining the
incident edges as \emph{loose ends at} $e$.

If $v$ is a vertex of $\tree$ that is distance 1 from $\X$, then there
is a unique edge $e$ with one endpoint equal to $v$ and the other in
$\X$. Define $\Wh(\X)\setminus v=\Wh(\X)\setminus e$.

Similarly, $\Wh(\X)\setminus \mathcal{Y}$ is obtained from $\Wh(\X)$ by deleting each vertex
of $\Wh(\X)$ that corresponds to an edge in $\mathcal{Y}$. 
Visualizing Whitehead graphs in the tree, $\Wh(\X) \setminus \mathcal{Y}$ is the
portion of $\Wh(\mathcal{Y})$ that passes through the set $\X$. 

An example follows (\fullref{ex:whiteheadgraphs}), but see
\cite{CasMac11} for a more detailed account.

\begin{lemma}[{\cite[Lemma 4.9]{CasMac11}}]\label{lemma:connectedhull}
  Let $S$ be a nonempty, finite subset of $\decomp$ whose preimage in $\partial \tree$ is more than one point. Let $\hull$ be the convex hull of $\qmap^{-1}(S)$. 
There is a bijection between connected components of $\Wh(\hull)$ and connected components of $\decomp\setminus S$.
\end{lemma}

Let $S$ be a finite cut set whose preimage in $\partial \tree$ is a pair of points, so that $\hull$ is a line. 
The fact that $\Wh(*)$ is connected without cut vertices implies that every vertex of $\Wh(\hull)$ belongs to a component that limits to both boundary points of $\hull$.
It follows that if $\{x,y\}$ is a cut pair in $\decomp$, then for any
small connected neighborhood $N$ of $x$ in $\decomp$ the number of
components of $N\setminus x$ is equal to the number of components of
$\decomp\setminus\{x,y\}$, which is the number of components of
$\Wh(\hull)$.

It is easy to recognize whether a given element $g\in F$ gives rise to
a cut set $\qmap(\{\inv{g}^\infty,g^\infty\})\in\decomp$.
Just consider the quotient of $\Wh([\inv{g}^\infty,g^\infty])$ by the
$g$--action.
This can be accomplished by looking at the quotient of a finite
graph. 
If $g=ha$ for $a\in\mathcal{B}\cup\inv{\mathcal{B}}$, let $[*,g)$
denote the segment $[*,h]\subset [*,g]$.
Then $\Wh([*,g))\setminus [\inv{g}^\infty,g^\infty]$ is almost exactly
the portion of $\Wh([\inv{g}^\infty,g^\infty])$ over $[*,g)$.
The only discrepancy is that if $g\in\multimot$ then the $g$--line
through $*$ does not contribute and edge to
$\Wh([\inv{g}^\infty,g^\infty])$, but there is a loose edge in
$\Wh([*,g))\setminus [\inv{g}^\infty,g^\infty]$ that the $g$--action glues to
itself to form a closed loop containing no vertices.
We consider such a loop a trivial component of the quotient; throw it away, if it exists.
The number of other components is the number of complementary
components of $\qmap(\{\inv{g}^\infty,g^\infty\})$ in $\decomp$.

\begin{example}\label{ex:whiteheadgraphs}
  Consider the multiword $\multimot=\{ab, \inv{a}\inv{b}ab\}$ in
  $F=\left<a,b\right>$. In all of these figures the $a$--direction is
  right, the $\inv{a}$ direction is left, the $b$--direction is up,
  and the $\inv{b}$--direction is down.
  \begin{figure}[h]
\begin{minipage}[b]{0.48\textwidth}
 \centering
    \includegraphics[height=1.4in]{Pictures/whex1}
    \caption{$\Wh(*)$}
    \label{fig:whex1}
\end{minipage}
\hfill
\begin{minipage}[b]{0.48\textwidth}
    \centering
     \includegraphics[height=1.4in]{Pictures/whex2}
    \caption{$\Wh(*)$ (loose)}
    \label{fig:whex2}
\end{minipage}
  \end{figure}

\fullref{fig:whex1} shows the classical Whitehead graph $\Wh(*)$.
We think of this as a picture of the line pattern $\lines_{\multimot}$ in a small
neighborhood of the vertex $*\in\tree$ representing the identity
element of $F$. 
Note that it is connected without cut vertices.

\fullref{fig:whex2} shows the same Whitehead graph with ``loose
ends''. 
The vertices have been opened up to indicate the group
action.
In this figure, the blue edge
comes from the $ab$--line through $*$. 
The red edge comes
from the $ab$--line through $\inv{a}$.
The $a$--action takes the $ab$--line through $\inv{a}$ to the
$ab$--line through $*$.
In the figure the $a$--action corresponds to a right shift, and the
diagram is drawn so that a right shift would match up an end of the
red edge to an end of the blue edge.
Similarly, the $b$--action is a vertical shift, and should identify
the $ab$--line through $*$, the blue edge, with the $ab$--line through
$b$.
Since $\inv{a}$ and $b$ differ by a multiple of $ab$, the $ab$--line through $b$ is the same as the $ab$--line through
$\inv{a}$, so, again, this is the red edge, and the diagram is drawn so
that a vertical shift matches up an end of the
blue edge to an end of the red edge.
The four black edges come from the four distinct
$\inv{a}\inv{b}ab$--lines passing through $*$, and the edges are drawn
in such a way that the vertical and horizontal shifts match up edges
in the same way that the $b$--action and $a$--action, respectively,
act on the collection of $\inv{a}\inv{b}ab$--lines.
\begin{figure}[h]
  \centering
   \includegraphics[height=1.4in]{Pictures/whex3}
  \caption{Splicing}
  \label{fig:whex3}
\end{figure}

\fullref{fig:whex3} illustrates how multiple copies of this loose
ended Whitehead graph can
be combined using Manning's splicing construction \cite{Man09} to
build Whitehead graphs over larger regions of $\tree$.
Here two copies of $\Wh(*)$ splice together to make $\Wh([*,a])$.


\begin{figure}[h]
\begin{minipage}[b]{0.51\textwidth}
        \centering
         \includegraphics[height=1.4in]{Pictures/whex4}
        \caption{\newline\mbox{$\Wh([*,ab))\setminus [\inv{(ab)}^\infty,(ab)^\infty]$}\newline}
        \label{fig:whex4}
\end{minipage}
\hfill
\begin{minipage}[b]{0.47\textwidth}
 \centering
         \includegraphics[height=1.4in]{Pictures/whex5}
        \caption{nontrivial components in the quotient by $ab$--action}
        \label{fig:whex5}
\end{minipage}
      \end{figure}

\fullref{fig:whex4} shows $\Wh([*,ab))\setminus
[\inv{(ab)}^\infty,(ab)^\infty]$.
This is the portion of the Whitehead graph of
$\Wh([\inv{(ab)}^\infty,(ab)^\infty])$ that is visible on the interval
$[*,ab)=[*,a]$.
(This is not quite right, as the edge with no vertices is actually the
$ab$--line through $*$, which does not contribute an edge to  $\Wh([\inv{(ab)}^\infty,(ab)^\infty])$.)
This is a fundamental domain for the $ab$--action on
$[\inv{(ab)}^\infty,(ab)^\infty]$.
The quotient of $\Wh([*,a])\setminus
[\inv{(ab)}^\infty,(ab)^\infty]$ by the $(ab)$--action has two
non-trivial components, as seen in \fullref{fig:whex5}, so we conclude that
$\qmap(\{\inv{(ab)}^\infty,(ab)^\infty\})$ is a cut set in $\decomp$. 
Furthermore, since $ab\in\multimot$, it is a cut point.
\end{example}




\subsection{Cut Point and Cut Pair Detection and Approximation}
\begin{lemma}\label{lemma:cutpointsareinmultiword}
If $x\in\decomp$ is a cut point then the stabilizer of $x$ is
conjugate to $\left< g\right>$ for an element $g\in\multimot$.
\end{lemma}
\begin{proof}
  Using an argument similar to \fullref{lemma:connectedhull}, it is
  easy to see that $\qmap^{-1}(x)$ must be two points. The result follows.
\end{proof}


\begin{lemma}[cf {\cite[Lemma 4.12]{CasMac11}}]\label{lemma:periodic}
Let $\xi_0$ and $\xi_1$ be a pair of points in $\partial\tree$ such that
$\qmap(\{\xi_0,\xi_1\})$ is a cut pair in $\decomp$ and for each $i$ we have $\xi_i=\qmap^{-1}(\qmap(\xi_i))$. 
There exist elements $g$, $h\in F$ and
$a\in\mathcal{B}\cup\inv{\mathcal{B}}$ such that:
\begin{enumerate}
\item the oriented edges $[h,ha]$ and $[hg,hga]$ belong to $[\xi_0,\xi_1]\cap[h\inv{g}^\infty,hg^\infty]$,
\item components of $\Wh([ha,hg])\setminus \{h,hga\}$ that
 are in different components of $\Wh([\xi_0,\xi_1])$ are in different
 components of $\Wh([h\inv{g}^\infty,hg^\infty])$, and
\item for each line $l\in\lines$ crossing the edge $[h,ha]$, the lines
  $l$ and $hg\inv{h}l$ (which crosses $[hg, hga]$) belong to the same
  component of $\Wh([\xi_0,\xi_1])$.\label{item:preserves}
\end{enumerate}  
Moreover, the conjugacy class of $g$ contains an element whose length
is uniformly bounded in terms of the rank of $F$, $\multimot$, and $\mathcal{B}$.
\end{lemma}

In other words, the cut pair $\qmap(\{\xi_0,\xi_1\})$ can be
approximated by a rational cut set
$\qmap(\{h\inv{g}^\infty,hg^\infty\})$. 
Item (1) says that $[\xi_0,\xi_1]\cap [h\inv{g}^\infty,hg^\infty]$ contains a copy of the fundamental
domain for the $hg\inv{h}$ action on
$[h\inv{g}^\infty,hg^\infty]$.
Items (2) and (3) say that the $hg\inv{h}$ action behaves nicely with
respect to
connected components of $\decomp\setminus\qmap(\{\xi_0,\xi_1\})$, at
least if we look at the Whitehead graph on the overlap.
\begin{proof}
Assume $[\xi_0,\xi_1]$ is oriented from $\xi_0$ to $\xi_1$.

Let $n$ be the rank of the $F$. Let $x$ be the maximum valence of
$\Wh(*)$ (which depends on the choice of $\mathcal{B}$). Let $y$ be the $x$--th
Bell number, the number of distinct partitions of $x$ items into
nonempty subsets. Let
$z=1+(2n)^{y+2}$.
Along any segment $\X$ of $[\xi_0,\xi_1]$ of length $z$ there is some
$a\in\mathcal{B}\cup\inv{\mathcal{B}}$ such that there are
  at least $y+2$ many directed $a$--edges in the segment.
Fix the first of these, $e=[g_0,g_0a]$.

There are finitely many lines of $\lines$ that cross $e$.
Fix a numbering of them $1,\dots, k$.
Of course, $k\leq x$.
Partition them into subsets according to the component of
$\Wh([\xi_0,\xi_1])$ to which they belong.

Consider an element $g'\in F$ such that the oriented edge $g'e$ is in
$\X$.
There is a bijection $l\mapsto g'l$ between lines of $\lines$ crossing $e$ and
lines of $\lines$ crossing $g'e$, so the numbering of the lines
crossing $e$ can be pushed forward to a numbering of the lines
crossing $g'e$.
We can also partition the lines crossing $g'e$ according to the component of
$\Wh([\xi_0,\xi_1])$ to which they belong.

There are at least $y+1$ such $g'$, but at most $y$ distinct partitions, so for some of these $g'$ the partitions
are the same.
Thus, there exist $g_1$ and $g_2$ such that the oriented
edges $g_1e$ and $g_2e$ are edges of $[\xi_0,\xi_1]$ (with $g_2e$
between $g_1e$ and $\xi_1$) and for each line
$l\in\lines$ crossing $g_1e$, the corresponding line $g_2\inv{g}_1l$
crossing $g_2e$ is in the same component of $\Wh([\xi_0,\xi_1])$.

The desired elements are $h=g_1g_0$ and $g=\inv{g}_0\inv{g}_1g_2g_0$.

Moreover, $g$ is conjugate to an element $\inv{g}_1g_2$ of length at
most $z$.

\end{proof}
 
We can test an element $g$ to see if it gives a cut point or cut pair
in the decomposition space by counting non-trivial components of
$\Wh([*,g))\setminus [\inv{g}^\infty,g^\infty]$.
If we test all elements of length less
than the bound from \fullref{lemma:periodic} and find no cut points
or cut pairs, then there are no cut points or cut pairs, so
$\multimot$ is rigid. Thus:

\begin{corollary}
  There is an algorithm to decide if $(F,\multimot)$ is rigid.
\end{corollary}

The algorithm described is far from efficient, and we will not spend
any time in this paper verifying that a multiword is rigid. 
The interested reader is referred to \cite{CasMac11} for a more
detailed account. 
However, there is one prototypical family of examples of  rigid
multiwords worth mentioning:

\begin{proposition}[{\cite[Section 6.2]{CasMac11}}]\label{proposition:completeisrigid}
  If $\multimot$ and $\mathcal{B}$ are a multiword and basis of $F=F_n$ such that
  $\Wh_{\mathcal{B}}(*)\{\multimot\}$ is the complete graph on $2n$ vertices, the graph with exactly
  one edge joining each pair of distinct vertices, then $(F,\multimot)$ is rigid.
\end{proposition}

Conversely, we can decide if $(F,\multimot)$ is a QH--surface:

\begin{lemma}[{\cite[Theorem 2]{Ota92}, \cite[Theorem
    6.1]{CasMac11}}]\label{lemma:circles} Given a $\multimot$
  multiword in a free group $F$,
  the following are equivalent:
\begin{enumerate}
\item $(F,\multimot)$ is a QH--surface.
\item $\multimot$ consists of  the boundary words of a compact surface with boundary with fundamental group $F$.
\item Some Whitehead graph for $\multimot$ is a circle.
\item Every Whitehead graph for $\multimot$ with no cut vertex is a circle.
\item Every minimal Whitehead graph for $\multimot$ is a circle.
\item $\decomp$ is a circle. 
\item Every minimal cut set of $\decomp$ has size two.
\end{enumerate}
\end{lemma}




\endinput

\subsection{Examples}

\subsubsection{Baumslag's Word}
The first example is \emph{Baumslag's word} $w=\inv{a}^2\inv{b}\inv{a}ba\inv{b}ab$ in $F_2=\left<a,b\right>$.
In response to a question of Gordon and Wilton, Manning showed, by
enumerating subgroups and checking geometricity,  that
this word becomes geometric in an orientable handlebody with
fundamental group an index four subgroup of $F_2$.

\begin{comment}
This word generates a line pattern whose decomposition space is connected with no cut points.
From $\Wh(*)\setminus[a^{-\infty},a^\infty]$, it is apparent that $a^{-\infty}$ and $a^\infty$ give a cut pair in the decomposition space. 
This pair has only two complementary connected components, so it requires more work to see that it is not crossed by any other cut pair. 
If it were crossed we would be able to find a disconnected Whitehead graph of the form $\Wh([*,a^m])\setminus \{\inv{b},a^mb\}$ for some $m\in\mathbb{Z}$.
The reader may verify that all such Whitehead graphs are connected, so
the cut pair $\qmap(\{a^{-\infty},a^\infty\})$ is not crossed by any other
cut pair.
\end{comment}

The rJSJ\xspace for $F=\left<a,b\right>\cong\left<a,b,c\mid c=\inv{b}ab\right>$ 
is shown in \fullref{fig:longword}. 

\begin{figure}[h]
\labellist
\small
\pinlabel $\left<a,c\right>$  at -4 23
\pinlabel $\left<a\right>$ at 84 23
\pinlabel $a$ [r] at 2 11
\pinlabel $c$ [r] at 2 36
\pinlabel $a$ [l] at 80 11
\pinlabel $a$ [l] at 80 36
\endlabellist  
\centering
  \includegraphics{Pictures/baBA2}
  \caption{rJSJ\xspace-Decomposition of $\left<a,b\right>$ for $\inv{a}^2\inv{b}\inv{a}ba\inv{b}ab$ ($c=\inv{b}ab$)}
  \label{fig:longword}
\end{figure}

The word $w$ becomes $\inv{a}^2\inv{c}ac$ when rewritten in the rank two vertex
group, so the induced multiword is $\{\inv{a}^2\inv{c}ac,a,c\}$. 
One can check that this multiword is rigid, so this is the rJSJ\xspace. 
(Unfortunately the Whitehead graph contains two edges joining
$\inv{a}$ to $a$, so \fullref{proposition:completeisrigid} does not
apply here. Checking rigidity takes a little work.)

In \fullref{fig:heegaard} we have a reduced Whitehead graph/Heegaard
diagram for induced multiword that shows it is geometric.
\fullref{fig:nonorientablehandlebody} shows a (non-orientable) handlebody with embedded
multicurve representing $\{\inv{a}^2\inv{c}ac,a,c\}$.

\begin{figure}[h]
\begin{minipage}[b]{0.465\textwidth}
\labellist
\small
\pinlabel $a$ [c] at 30 102
\pinlabel $\inv{a}$ [c] at 30 28
\pinlabel $\inv{c}$ [c] at 147 102
\pinlabel $c$ [c] at 147 28
\tiny
\pinlabel $2$ at 34 111.5
\pinlabel $4$ at 40 106.5
\pinlabel $1$ at 40 97.5
\pinlabel $6$ at 34 91.5
\pinlabel $2$ at 34 39
\pinlabel $4$ at 40 33
\pinlabel $1$ at 40 24
\pinlabel $6$ at 34 18
\pinlabel $5$ at 140 110
\pinlabel $3$ at 136 101.5
\pinlabel $7$ at 140 94
\pinlabel $7$ at 140 37
\pinlabel $3$ at 136 28.5
\pinlabel $5$ at 140 20
\endlabellist
  \centering
  \includegraphics[height=1.6in]{Pictures/baumslagheegaard}
  \caption{Whitehead graph/ Heegaard diagram}
  \label{fig:heegaard}
\end{minipage}
\hfill
\begin{minipage}[b]{0.525\textwidth}
\labellist
\small
\pinlabel $a$ [b] at 14 122
\pinlabel $\inv{a}$ [t] at 14 1
\pinlabel $c$ [l] at 179 60
\endlabellist
  \centering
\includegraphics[height=1.6in]{Pictures/nonorientable}
  \caption{Corresponding non-orientable handlebody for $\{\inv{a}^2\inv{c}ac,a,c\}$}
  \label{fig:nonorientablehandlebody}
\end{minipage}
\end{figure}


The obstruction to geometricity of $w$ is that if we tried to build a graph
of spaces with this non-orientable handlebody as the vertex space we
would need to conjugate a curve running around an
orientable handle to one running around a non-orientable
handle ($a$ is conjugate to $c$).
A neighborhood of the $a$ curve is a M\"obius strip, while a
neighborhood of the $c$ curve is an annulus, so we can not achieve
this conjugation with a fattened annulus.

To correct this problem, pass to the index two subgroup:
\[G=\left<A=a^2, b, B=ab\inv{a}\right>\]

Baumslag's word is not in this subgroup, but its square is:
\begin{align*}
w^2&=(\inv{a}^2\inv{b}\inv{a}ba\inv{b}ab)^2\\
&=  \inv{a}^2\inv{b}\inv{a}ba\inv{b}ab   \inv{a}^2\inv{b}\inv{a}ba\inv{b}ab\\
&=\inv{a}^2\cdot\inv{b}\cdot\inv{a}^2\cdot ab\inv{a}\cdot
a^2\cdot\inv{b}\cdot ab   \inv{a}\cdot \inv{a}^2\cdot
a\inv{b}\inv{a}\cdot b\cdot a\inv{b}\inv{a}\cdot a^2\cdot b\\
&=\inv{A}\inv{b}\inv{A}BA\inv{b}B\inv{A}\inv{B}b\inv{B}Ab
\end{align*}


Apply the Whitehead automorphism that pushes $B$ through $b$.
This sends $B$ to $bB$ and $\inv{B}$ to $\inv{B}\inv{b}$, and fixes
$b$ and $A$.
The word becomes
$\inv{A}(\inv{b}\inv{A}b)BAB\inv{A}\inv{B}^2(\inv{b}Ab)$.

The splitting over $\left<A\right>$ is therefore $\left<A,b,B,C\mid
  C=\inv{b}Ab\right>$.
The induced multiword in the vertex group $\left<A,B,C\right>$ is
$\{A,C, \inv{A}\inv{C}BAB\inv{A}\inv{B}^2C\}$.
This multiword is rigid (this takes some work) and geometric in a non-orientable handlebody,
as seen in \fullref{fig:nonorientableindex2baumslag}.

\begin{figure}[h]
\labellist
\small
\pinlabel $A$ [b] at 78 190
\pinlabel $C$ [t] at 78 3
\pinlabel $\inv{B}$ [b] at 241 170
\pinlabel $B$ [t] at 241 20
\endlabellist
  \centering
 \includegraphics[]{Pictures/nonorientableindex2baumslag} 
  \caption{A non-orientable handlebody for $\{A,C, \inv{A}\inv{C}BAB\inv{A}\inv{B}^2C\}$}
  \label{fig:nonorientableindex2baumslag}
\end{figure}


Although the handlebody is non-orientable, this time we can build a 3--manifold
graph of spaces because we only need to conjugate an orientable handle to
an orientable handle.
Gluing on a fattened annulus conjugating $A$ to $C$ gives a
non-orientable handlebody with fundamental group isomorphic to $G$ for
which the image of
$w^2$ is geometric. 
One could pass further to a twofold cover of this handlebody to
find an index four subgroup of $F$ for which the multiword is
geometric in an orientable handlebody, if desired.


\subsubsection{Baumslag-Solitar Words}\label{sec:bswords}
Another interesting family of examples are given by the
Baumslag-Solitar words $w_{p,q}=\inv{a}^q \inv{b}a^pb$ in
$F_2=\left<a,b\right>$.
We will assume that $0< p\leq q$.
Gordon and Wilton \cite{GorWil09} have shown that $w_{p,q}$ is
virtually geometric when $p$ and $q$ are relatively prime.

The decomposition space associated to this word is connected without cut points.
The pair $\qmap(\{\inv{a}^{\infty},a^\infty\})$ is a cut pair.
\fullref{fig:Whas} shows the Whitehead graph:\[\Wh_{\{a,b\}}([*,a^6))\{\inv{a}^3
\inv{b}a^2b\}\setminus[a^{-\infty},a^\infty]\]
This is six copies of $\Wh(*)$ spliced together. 
Note there are $p=2$
components containing the vertices along the top of the figure, and
$q=3$ components containing the vertices along the bottom.
The $a$--action is a shift that exchanges the two components on the
top and cyclically permutes the three components along the bottom, so
the $a^6$--action fixes all five components.
 
In general there are $p+q$ connected components in the complement
of $\qmap(\{\inv{a}^{\infty},a^\infty\})$.
There are two orbits of components under the $a$--action, one of size
$p$ and one of size $q$.

The  case when $p=q=1$ is special; in this case the Whitehead graph is
a circle, which implies the decomposition space is a circle and the
word is geometric.

Otherwise, the number of complementary components is $p+q>2$, so 
$\qmap(\{\inv{a}^\infty, a^\infty\})$ is an uncrossed cut pair.
The rJSJ\xspace is shown in \fullref{fig:baBA2}.


\begin{figure}[h]
  \centering
\includegraphics[width=\textwidth]{Pictures/Wh}
  \caption{$\Wh_{\{a,b\}}([*,a^6))\{\inv{a}^3 \inv{b}a^2b\}\setminus[a^{-\infty},a^\infty]$}
  \label{fig:Whas}
\end{figure}


\begin{figure}[h]
\labellist
\small
\pinlabel $\left<\inv{b}a^pb,a^q\right>$ [c] at -4 23
\pinlabel $\left<a\right>$ [c] at 84 23
\pinlabel $a^q$ [r] at 2 11
\pinlabel $\inv{b}a^pb$ [r] at 2 36
\pinlabel $a^q$ [l] at 80 11
\pinlabel $a^p$ [l] at 80 37
\endlabellist  
\centering
  \includegraphics{Pictures/baBA2}
  \caption{rJSJ\xspace-Decomposition of $\left<a,b\right>$ for $\inv{a}^q\inv{b}a^pb$}
  \label{fig:baBA2}
\end{figure}

The rank two vertex group is $\left<A=a^q, C=\inv{b}a^pb\right>$, and
the induced multiword in this vertex group is $\{A,C,\inv{A}C\}$. 
The Whitehead graph for this multiword is a circle, which implies the vertex
decomposition space is a circle and the induced multiword is geometric.
Thus, \fullref{theorem:planarequalsvg} says $w_{p,q}$ is at least virtually geometric.

The cyclic vertex group has edge inclusions of degrees $p$ and $q$. 

If $p=1$ and $q=2$ we can make this geometric by using a solid Klein
bottle for the cyclic vertex space.
(We saw the non-cyclic vertex space for this example back in \fullref{ex:bs21}.)

If $p=q$ the word is also geometric, because two disjoint degree $p$
curves fit into the boundary of a solid torus.
However, an additional degree one curve does not fit.
This example is notable because the word $w_{p,p}$ is geometric, but
the augmented multiword $\{w_{p,p}, a\}$ is only virtually geometric.
Augmenting the multiword does not change virtual geometricity, but it
may change geometricity.

In all other cases, the word $w_{p,q}$ is not geometric.
Let $m$ be the least common multiple of $p$ and $q$.
It suffices to pass to the index $m$ subgroup:
\[G=\left<A, B_0, B_1,\dots, B_{m-1}\mid A=a^{m}, B_i=a^ib\inv{a}^i\right>\]
Since $A=a^m$ is the smallest power of $a$ in this subgroup, there is
still only one orbit of uncrossed cut pair in the decomposition space.
The $A$--action fixes each of the $p+q$
complementary components of $\qmap(\{\inv{A}^\infty,A^\infty\})$.
Therefore, the rJSJ\xspace has a
single cyclic vertex group with all edge inclusions of degree one, and all
non-cyclic vertex groups of the rJSJ\xspace have a circle for the decomposition space
of the induced multiword.




\endinput

\section{Introduction}

Let $F=F_n$ be a free group of finite rank $n>1$, and let
$\multimot=\{w_1,\dots, w_k\}\subset F$ be a \emph{multiword}, a finite list of
nontrivial words. 
The goal of this paper is to find splittings of $F$ \emph{relative to
  \multimot} (rel $\multimot$), splittings of $F$
as a free product with amalgamation over cyclic subgroups in such a
way that each word of $\multimot$ is conjugate into some factor.

First consider the problem in a topological setting. 
Consider the
surface in \fullref{fig:surface}. 
The labeled curves are generators of
the fundamental group $F_5=\left<a,b,c,d,e\right>$.

\begin{figure}[h]
\begin{minipage}[b]{0.48\textwidth}
  \centering
\labellist
\small
\pinlabel $a$ at 66 15
\pinlabel $c$ at 53 15
\pinlabel $b$ at 75 56
\pinlabel $d$ at 44 56
\pinlabel $e$ at 60 63
\endlabellist
  \includegraphics[width=.95\textwidth]{Pictures/introsurfacegens}
  \caption{A surface with chosen generators of the fundamental group}
  \label{fig:surface}
\end{minipage}
\hfill
\begin{minipage}[b]{0.48\textwidth}
  \centering
  \includegraphics[width=.95\textwidth]{Pictures/introsurface}
  \caption{A multi-curve realizing $\multimot=\{a, c, d, e, \inv{a}c,
dc\inv{d}\inv{c}, dc\inv{d}\inv{c}ab\inv{a}\inv{b}\inv{e}\}$}
  \label{fig:surfacedecomp}
\end{minipage}
\end{figure}





Consider the multiword $\multimot=\{a, c, d, e, \inv{a}c,
dc\inv{d}\inv{c}, dc\inv{d}\inv{c}ab\inv{a}\inv{b}\inv{e}\}$.
\fullref{fig:surfacedecomp} shows a \emph{multicurve} realizing
$\multimot$.
This is a collection of curves freely
homotopic to curves representing the words of $\multimot$.


Splittings of surfaces as free products with cyclic amalgamation come
from simple closed curves on the surface. Splitting relative to a
multiword requires us to find a simple closed curve disjoint from the
curves representing the words of the multiword.

In this example, two of the curves are the boundary curves of the surface. 
Take a small regular neighborhood of the remaining curves. 
There are two components; one is an annulus, and the other is more
complicated.
The complement consists of two discs, an annulus, and a sphere with
four boundary components. 


Cutting off discs does not give us any information, so consider the
union of the discs with the more complicated surface. This subsurface is a torus
with two boundary components, and it is \emph{filled} by the
multicurve; every essential simple closed curve in the subsurface intersects
one of the curves of the multicurve.
This subsurface does not split relative to the multiword.
This filled subsurface is an example of what we will call a \emph{rigid} vertex.


We also have two parallel annuli, one a neighborhood of a curve and
one in the complement.
Take their union to get an annulus
with a curve of the multicurve as its core.
This core curve gives us a splitting of $F$ relative to the multiword.

The remaining complementary component is a sphere with four boundary
components. It is \emph{empty}, in the sense that there are no curves
of the multicurve in the interior of this subsurface. 
It is possible to split this subsurface relative to the multiword, but
not in a canonical way. 
Every simple closed curve would give us a splitting, but for each
simple closed curve there is another simple closed curve intersecting
it.
The splittings from intersecting curves would be incompatible, so
rather than make an arbitrary choice we keep the empty subsurface and
remember that it splits in incompatible ways.


So far, we have the splitting from the annulus and splittings from the
empty subsurface. There is one remaining. We can split along the
boundary curve of the filled subsurface where it meets the empty
subsurface.
To make the picture symmetric, insert another annulus at the interface
of the filled subsurface and the empty subsurface, as in
\fullref{fig:surfacedecomp2}.
\fullref{fig:surfacegog} shows the corresponding graph of groups
decomposition of $F$ rel $\multimot$.
It has two cyclic vertex groups corresponding to the two annuli and
two rank 3 free vertex groups corresponding to the two higher
complexity subsurfaces.
The edge groups are all cyclic.
The labels at the ends of each edge indicate the image of a generator
of the edge group in the adjacent vertex group.




\begin{figure}[h]
\begin{minipage}[b]{0.48\textwidth}
 \centering
  \includegraphics[width=.95\textwidth]{Pictures/introsurfacedecomp}
  \caption{Decomposition of a surface.}
  \label{fig:surfacedecomp2}
\end{minipage}
\hfill
\begin{minipage}[b]{0.48\textwidth}
  \centering
\labellist
\pinlabel $\left<a\right>$ at 33 -1
\pinlabel $\left<dc\inv{d}\inv{c}a\right>$ at 33 40
\pinlabel $\left<dc\inv{d}\inv{c}a,\,ba\inv{b},\,e\right>$ at 60 20
\pinlabel $\left<a,c,d\right>$ at 0 20
\pinlabel $dc\inv{d}\inv{c}a$ at 0 34
\pinlabel $a$ at 1 10
\pinlabel $a$ at 22 -1
\pinlabel $a$ at 44 -1
\pinlabel $ba\inv{b}$ at 68 10
\pinlabel $dc\inv{d}\inv{c}a$ at 68 34
\pinlabel $dc\inv{d}\inv{c}a$ at 11 41
\pinlabel $dc\inv{d}\inv{c}a$ at 55 41
\endlabellist
\includegraphics[width=.75\textwidth]{Pictures/introsplit}
\caption{Corresponding graph of groups}
\label{fig:surfacegog}
\end{minipage}
\end{figure}

In summary, we have a decomposition of the original surface into
subsurfaces. 
The decomposition is bipartite: there is a collection of
annuli and a collection of more complicated subsurfaces, and each
subsurface is adjacent only to members of the opposite collection. 
Among the more complicated subsurfaces there are those that are filled
and those that are empty. 
Every splitting of the total surface relative to the multicurve comes from
either the core curve of one of the annuli or a simple closed curve in
the interior of one of the empty surface pieces. 

We call this decomposition the \emph{relative
  JSJ--Decomposition (rJSJ)}. 
It is a decomposition that encodes all possible cyclic splittings
relative to the given multiword. 
The main result of this paper is that given a multiword in a free
group there is a canonical group-theoretic relative
JSJ--Decomposition.
In general it will not be the case that the free group corresponds to
the fundamental group of a surface with boundary and the pieces of the
rJSJ correspond to subsurfaces.
Nonetheless, the statement will be completely analogous to the simple
example above: 

\begin{decompositiontheorem}[\fullref{theorem:JSJ}]
\hypertarget{decompthmtarget}{
Let $\multimot$ be a multiword in a free group $F$ such that $F$
  does not split freely relative to $\multimot$. There exists a
  canonical relative JSJ--dceomposition (rJSJ), a graph of groups
  decomposition $\Gamma$ of $F$ relative to $\multimot$ with cyclic
  edge groups,
  satisfying the following conditions:
  \begin{enumerate}[(1)]
\item $\Gamma$ satisfies certain normalization hypotheses.
\item $\Gamma$ is universal. If $F$ splits over a cyclic subgroup relative to $\multimot$ then the
cyclic subgroup is conjugate into one of the vertex groups.
\item $\Gamma$ is maximal. It can not be refined and still satisfy
  these conditions.
  \end{enumerate}
Moreover, the rJSJ is characterized by splitting $F$ over the stabilizers of cut points
  and uncrossed cut pairs in a certain topological space $\decomp_{\multimot}$.
There are three mutually exclusive possibilities:
\begin{enumerate}[(a)]
\item $(F,\multimot)$ is rigid. $\decomp_{\multimot}$ has no cut points or cut
  pairs. The rJSJ is
  trivial, a single vertex stabilized by $F$.
\item $(F,\multimot)$ is a QH--surface. $\decomp_{\multimot}$ is a circle. The rJSJ is
  trivial, a single vertex stabilized by $F$.\label{item:introqhsurface}
\item The rJSJ is nontrivial. For every non-cyclic vertex group $G$ the pair $(G, \indg)$ is either
  rigid or a QH--surface.\label{item:intronicepieces}
\end{enumerate}
 Consequently,  if $F$ splits over a cyclic subgroup relative to $\multimot$ then the
cyclic subgroup is conjugate into one of the cyclic vertices or
one of the QH--surface vertices of the rJSJ.}
\end{decompositiontheorem}
We will say that $(F,\multimot)$ is \emph{rigid} if the corresponding
decomposition space $\decomp_{\multimot}$ is connected without
cut points or cut pairs. 
The fact that such a line pattern is quasi-isometrically
rigid is the main result of \cite{CasMac11}.
Rigidity implies that this vertex does not split relative to the
multiword (\fullref{corollary:rigiddontsplit}).

Statement (\ref{item:introqhsurface}) means that $F$ is the
fundamental group of a surface with boundary components and
$\multimot$ represents the boundary curves.

In the setting of the example, statement (\ref{item:intronicepieces})
means that $G$ is the fundamental group of a subsurface, and $\indg$
is the multiword in $G$ consisting of the words of $\multimot$
conjugate into $G$ as well as any boundary curves of the subsurface
that are not boundary curves of the total surface.

\begin{remarks}
In general one does not have a canonical JSJ--decomposition of a
  group.
Rather, there is a deformation space of JSJ--decompositions \cite{GuiLev09}.
The normalization hypotheses of condition (\ref{item:bipartite}) pick out a particular point of the deformation space.

It is usual in the various JSJ theories to call the vertex
  groups ``rigid'' and ``flexible''.
 The term ``quadratically hanging surface'' is from Sela's JSJ theory
\cite{Sel97}. 
In more general versions of JSJ theory the flexible vertex groups are
often certain fiber bundles over quadratically hanging 2--orbifolds
\cite{GuiLev09}.
Since we are working in free groups we will not need this generality;
QH--surfaces are all that will occur as flexible vertices.

 A ``rigid'' vertex is usually defined to be one that does not
  split further. This leads to the annoying possibility that ``rigid''
  and ``flexible'' may not be mutually exclusive. The example
  that is relevant in this paper is the sphere with three boundary
  components. We will use a definition of ``rigid'' that excludes this
  example, so ``rigid'' and ``QH-surface'' will be mutually exclusive.
\end{remarks}

In the surface example above there is a subtle point.
One splitting came from a boundary curve of the filled subsurface. 
This curve was not a member of the multicurve, but, in a sense, it might
just as well have been; the rJSJ\xspace would not have been any
different if it had been included in the first place. 
Characterizing such ``missing'' curves in the case of the free group
is not so trivial. 
In fact, it is our main task.
The rest of the theorem follows rather easily.

\subsection{Line Patterns and Decomposition Spaces}
Let $\mathcal{B}=\{b_1,\dots, b_n\}$ be a free basis for $F$, and let
$\multimot$ be a multiword in $F$. 
Think of the words of $\multimot$ as closed paths in the rose with $n$
edges labeled by the $b_i$.
The universal cover of the rose is a tree $\tree$ that is a Cayley
graph of $F$.
Define the \emph{line pattern} $\lines_{\multimot}$ generated by $\multimot$ to be the collection
of coarse equivalence classes of lifts of the closed paths in the rose
coming from $\multimot$.
Each individual lift picks out two points on the boundary at infinity
of $\tree$. The convex hull of these two points is a geodesic of
$\tree$ coarsely equivalent to the lifted curve.
These are the ``lines'' that make up the line pattern.

Changing a multiword by conjugating an element or replacing an element
with a root or a power
does not change the associated line pattern.
In other words, we are really interested in finite collections of
conjugacy classes of maximal cyclic subgroups.
We will implicitly treat multiwords $\multimot$ and $\multimot'$ as
equivalent if the words in $\multimot$ belong to the same collection of conjugacy classes of
maximal cyclic subgroups as the words of $\multimot'$. 
We will also say $g\in\multimot$ if the conjugacy class of the
maximal cyclic subgroup containing $g$ is in $\multimot$.
With this notion of equivalence, a line pattern determines a generating
multiword.

There is a topological space, the \emph{decomposition space},
$\decomp$ (or $\decomp_\lines$ or $\decomp_{\multimot}$ if we want to be
explicit) associated to a line
pattern.
It is obtained as a quotient of the boundary at infinity of
the tree $\tree$ by identifying the two endpoints of each line $l\in\lines$.

Any quasi-isometry of $F$ that coarsely preserves the line pattern
extends to a homeomorphism of the decomposition space. 
In particular, $F$ acts on $\decomp$ by homeomorphisms, and $\decomp$
is independent of the choice of basis of $F$. 
We will use the topology of the decomposition space to construct a
simplicial tree with an $F$--action.
This will give us the rJSJ.

Several facts about the relationship between relative splittings and
the topology of the decomposition space are already known:
\begin{enumerate}
\item $F$ splits freely rel $\multimot$ if and only if
  $\decomp$ is not connected. (\fullref{corollary:connected})
\item Any cut point of $\decomp$ is stabilized by a
  conjugate of some word $w$ of $\multimot$, and $F$ splits over
  $\left<w\right>$ rel
  $\multimot$. (\fullref{lemma:cutpointsareinmultiword} and
  \fullref{proposition:otalsplit})
\item If $F$ splits over $\left<g\right>$ rel $\multimot$ then the
  image of $(g^\infty, \inv{g}^\infty)$ in $\decomp$ is either a cut
  point or a cut pair. It is a cut point if and only if
  $g\in\multimot$. (\fullref{lemma:splitimpliescut} and \fullref{lemma:cutpointsareinmultiword})
\item $(F,\multimot)$ is a QH--surface if and only if $\decomp$ is a
  circle. (\fullref{lemma:circles})
\end{enumerate}

These observations can be attributed to Otal \cite{Ota92}. 
He proves in addition that if the decomposition space is planar and each element
of the multiword is full width, then the presence of cut pairs implies
the existence of a splitting rel $\multimot$.

In \fullref{theorem:splitting} we remove the additional hypotheses:
If $(F,\multimot)$ is not rigid then either $F$ splits rel $\multimot$
or $(F,\multimot)$ is a sphere with three boundary components.

To do this we characterize the ``missing'' curves mentioned earlier.
In a circle every pair of points is a cut pair, and for each cut pair
there is another whose two points lie in opposite complementary
components of the first, and vice versa.
We say that such cut pairs are \emph{crossing}. 

In the decomposition space it may be that there are cut pairs that are
not crossed by any other cut pairs. We call these \emph{uncrossed}.
We will show that every cut pair in the decomposition space either is
uncrossed or belongs to a circle, in a certain sense.
\fullref{proposition:periodic} shows that an uncrossed cut pair is rational, it is
stabilized by a cyclic subgroup $\left<g\right><F$. 
Once we know this it is not hard to see Otal's splitting argument
still applies, so $F$ splits over $\left<g\right>$ rel $\multimot$.
This subgroup $\left<g\right>$ is the analogue of the ``missing'' boundary curve of
the filled subsurface.

We build a simplicial tree from the collection of cut points and
uncrossed cut pairs of $\decomp$.
The $F$ action on $\decomp$ gives us a simplicial cocompact $F$ action
on this tree. 
The quotient graph of groups is the rJSJ. 
The result is canonical, since the tree is determined by the topology
of $\decomp$.

\subsection{Virtually Geometric Multiwords}

In \fullref{sec:vg} we will apply the \hyperlink{decompthmtarget}{Relative JSJ-Decomposition Theorem}\xspace to characterize
virtually geometric multiwords.

A multiword in $F$ is said to be \emph{geometric} if it can be represented by
an embeded multicurve in the boundary of a handlebody with fundamental
group $F$. A multiword is \emph{virtually geometric} if it becomes geometric upon
lifting to a finite index subgroup of $F$.

Otal's main result in \cite{Ota92}, suitable reinterpreted,
is that in the case that the rJSJ\xspace is trivial the multiword
is geometric if and only if the decomposition space is planar.
Furthermore, planarity of the decomposition space can be deduced from the
Whitehead graph of the multiword.

The \hyperlink{decompthmtarget}{Relative JSJ-Decomposition Theorem}\xspace provides a reduction of the virtual geometricity question:

\begin{vgtheorem}[\fullref{theorem:planarequalsvg}]
\hypertarget{vgthmtarget}{
  For a multiword in a free group, the following are equivalent:
\begin{enumerate}
\item The multiword is virtually geometric.
\item The decomposition space is planar.
\item For every non-cyclic vertex group of the rJSJ\xspace, the induced multiword is geometric.
\end{enumerate}
Thus, virtually geometric multiwords are exactly those that are built
from geometric pieces.}
\end{vgtheorem}

\subsection{Relation to Other People's Work}
\subsubsection{Boundaries of Hyperbolic Groups}
It follows from Stallings's Theorem \cite{Sta68} that the boundary of a hyperbolic group is disconnected if and only if the
group splits over a finite group.

Bowditch \cite{Bow98b}  shows that if the boundary is connected, and
if the group is not a ``virtual semitriangle group'', then splittings
over virtually cyclic groups come from the cut pair structure of the boundary.

The results of this paper say that the analogous statements are true
for relative splittings of free groups with the decomposition space replacing the
boundary of the hyperbolic group.

\fullref{proposition:circle} is a generalization of a result of
Bowditch that determines when collections of cut pairs form a circle.
Among the technical differences are the facts that for hyperbolic
boundaries there are no global cut points and every local cut point belongs
to a cut pair.
Neither of these are true for our decomposition spaces.


\subsubsection{The Relatively Hyperbolic Perspective}
In the construction of the decomposition space we take an element $w$ of
the free group and identify the two endpoints of its axis. 
Essentially we have declared that a hyperbolic element now acts
parabolically.
One can make this precise by saying that the free group is hyperbolic
relative to the cyclic groups generated by elements of the multiword. 
It can be shown that the decomposition space is precisely Bowditch's boundary of a relatively
hyperbolic group \cite{Bow11}.
Bowditch \cite{Bow01} extends the ideas used in splitting hyperbolic
groups according to the topology of the boundary to show that the cut point
structure of the relatively hyperbolic boundary gives a canonical
``peripheral splitting'', a splitting of the relatively hyperbolic
group over the parabolic subgroups such that parabolic subgroups are
conjugate into the factors. 


Bowditch's results
specialized to free groups hyperbolic relative to a multiword give the
same graph of groups as Otal's argument
applied to the cut points of the decomposition space. 
However, this does not capture all relative splittings.
It does not
recognize the ``missing curve'' in the introduction, for example.


\subsubsection{Cactus Trees}
Since this paper was first written, a paper of Papasoglu and Swenson
\cite{PapSwe11} about the ``cactus tree'' of a metric space has
appeared. 
They prove a very general result about encoding the minimal cut sets
of a continuum in a tree, again, generalizing Bowditch's result for
boundaries of hyperbolic groups.
In our setting their theorem says that if the decomposition space
is connected with no cut points then the set of cut pairs can be
encoded in a simplicial tree.
An alternate approach to producing the rJSJ would then be to apply
results of Otal or Bowditch to produce a splitting from the cut points
of the decomposition space, and then apply Papasoglu and Swenson's
result to produce splittings of the vertex groups from their cut
pairs, refining the original decomposition.

We would still have to do the work of determining the properties of
the rJSJ\xspace. 
Also, philosophically, the uncrossed cut pairs are ``just as good as''
cut points. 
The missing curve in the surface example really ought to
have been there, and the same splitting argument works just as well
when they are included.




\subsection{Plan of the Paper}
In \fullref{sec:preliminaries} we fix definitions and recall some
results about Whitehead graphs and decomposition spaces. 
This machinery was developed by the author and Macura in a previous paper
\cite{CasMac11}. The reader is referred to that paper for proofs and
more details.

The \hyperlink{decompthmtarget}{Relative JSJ-Decomposition Theorem}\xspace is proven in \fullref{sec:splitting}.
In \fullref{sec:crossed} we generalize a result of Bowditch that
tells us when the decomposition space has cut points or uncrossed cut
pairs. 
In \fullref{sec:uncrossed} we show that cut points and
uncrossed cut pairs have cyclic stabilizers and that there are
finitely many orbits of them.
In \fullref{sec:otalsplit} we
generalize a result of Otal that shows that we can use any equivariant collection
of cut points and uncrossed cut pairs in the decomposition space to
produce a graph of groups decomposition of $F$ rel $\multimot$. 
In \fullref{sec:refine} we examine the structure of the decomposition
space in comparison to a graph of groups splitting and determine when
a graph of groups decomposition can be refined. 
Finally, in \fullref{sec:proofofdecomp} we put these pieces together
to prove the \hyperlink{decompthmtarget}{Relative JSJ-Decomposition Theorem}\xspace.

\fullref{sec:vg} applies the \hyperlink{decompthmtarget}{Relative JSJ-Decomposition Theorem}\xspace to prove the \hyperlink{vgthmtarget}{Characterization of Virtual Geometricity}\xspace.


\bigskip

I would like to thank Jason Manning for bringing Otal's paper to my
attention and for helpful comments on an earlier version of this paper.


\endinput


\subsection{Normalization} \label{sec:normalization}
Suppose $\Gamma$ is a graph of groups decomposition of $F$ with cyclic edge groups. 
It will be useful to normalize $\Gamma$
without changing the non-cyclic vertex groups.

First subdivide each edge incident to a non-cyclic vertex so that the edge map into each non-cyclic vertex maps onto a maximal cyclic subgroup in the vertex.
Now fold all the edges together that map into a common conjugacy class
of maximal cyclic subgroup in a given non-cyclic vertex (changing
stable letters, if necessary). 
Finally, since $F$ is free, every edge groups maps onto a maximal cyclic subgroup in one of its vertex groups. 
Thus, any interior edges, edges adjacent to two cyclic vertices, can be collapsed.

We are left with a new graph of groups decomposition of $F$ that is bipartite: vertex groups are either maximal cyclic subgroups or are non-cyclic. 
Cyclic vertex groups are adjacent only to non-cyclic vertex groups, and vice versa. 
Furthermore, each non-cyclic vertex has at most one edge mapping into any conjugacy class of maximal cyclic subgroup, and each edge map maps an edge group onto a maximal cyclic subgroup containing its image in the vertex group. 

\fullref{fig:normalization} shows an example of this process.
In the diagram all edges have cyclic stabilizers.
Unlabeled vertices have cyclic stabilizers. 
Assume a generator has been fixed for each of the cyclic vertex and
edge groups.
Edge labels indicate the image of the generator of the edge stabilizer
into the incident vertex group. 
Labels of $1$ are omitted.

  \begin{figure}[h]
    \centering
\labellist
\small
\pinlabel $\left<a,b\right>$ [c] at 0 97
\pinlabel $\left<x,y,z\right>$ [c] at 106 97
\pinlabel $\left<a,b\right>$ [c] at 172 97
\pinlabel $\left<x,y,z\right>$ [c] at 279 97
\pinlabel $\left<a,b\right>$ [c] at 345 97
\pinlabel $\left<x,y,z\right>$ [c] at 455 97
\pinlabel $\left<a,b\right>$ [c] at 85 31
\pinlabel $\left<x,y,z\right>$ [c] at 211 31
\pinlabel $\left<a,b\right>$ [c] at 269 31
\pinlabel $\left<x,y,z\right>$ [c] at 397 31
\tiny
\pinlabel $ab\inv{a}$ at -5 112
\pinlabel $b^2$  at -1 80
\pinlabel $b$ at 172 112
\pinlabel $b$  at 172 81
\pinlabel $b$  at 360 102
\pinlabel $b$  at 101 35
\pinlabel $b$  at 290 35
\pinlabel $x$ at 78 96
\pinlabel $y$ at 110 112
\pinlabel $z$ at 110 81
\pinlabel $x$ at 250 96
\pinlabel $y$ at 282 112
\pinlabel $z$ at 282 81
\pinlabel $x$ at 423 96
\pinlabel $y$ at 457 112
\pinlabel $z$ at 457 81
\pinlabel $x$ at 181 32
\pinlabel $y$ at 200 48
\pinlabel $z$ at 200 16
\pinlabel $x$ at 372 35
\pinlabel $y$ at 385 47
\pinlabel $z$ at 385 16
\pinlabel $2$ at 29 128
\pinlabel $3$ at 41 131
\pinlabel $2$ at 85 128
\pinlabel $5$ at 41 62
\pinlabel $3$ at 83 63
\pinlabel $2$ at 201 128
\pinlabel $3$ at 213 131
\pinlabel $2$ at 257 128
\pinlabel $5$ at 213 62
\pinlabel $3$ at 255 63
\pinlabel $2$ at 192 69
\pinlabel $2$ at 375 92
\pinlabel $3$ at 390 131
\pinlabel $2$ at 378 127
\pinlabel $2$ at 433 128
\pinlabel $5$ at 390 62
\pinlabel $3$ at 431 63
\pinlabel $2$ at 162 65
\pinlabel $6$ at 175 62
\pinlabel $3$ at 175 0
\pinlabel $10$ at 119 25
\pinlabel $60$ at 321 17
\pinlabel $20$ at 338 25
\pinlabel $2$ at 320 35
\pinlabel $6$ at 324 43
\endlabellist
    \includegraphics[width=.9\textwidth]{Pictures/normalization}
    \caption{Normalization}
    \label{fig:normalization}
  \end{figure}



\endinput


\subsection{Refining Splittings}\label{sec:refine}
\fullref{theorem:splitting} tells us when we can split $F$ rel
$\multimot$. 
In this subsection we determine when a splitting can be refined.



\begin{lemma}\label{lemma:projection}
 Let $\Gamma$ be a graph of groups decomposition of $F$ rel
 $\multimot$ produced by \fullref{proposition:multipleotal}. 
Let $\mathcal{D}(\Gamma)$ be the bipartite tree constructed in the
proof, which is the Bass-Serre tree of the graph of groups decomposition.
There is a natural
 $F$--equivariant surjective quasi-map $\pi_\Gamma\colon\thinspace \decomp \to \closure{\mathcal{D}(\Gamma)}$.
\end{lemma}
\begin{proof}
  If $x\in S_i$ then define $\pi_\Gamma(x)$ to be the corresponding
  Type 1 vertex. 

Otherwise, define a partial order relative to $x$ on the set
  $\{S_i\}_{i\in I}$ by $S_i<S_j$ if $x$ and $S_j$ are in different components
  of $S_i$. 
There may or may not be minimal elements with respect to this partial
order.
First suppose there is a minimal element $S_i$.
The point $x$ is in one of the complementary components of $S_i$. 
Associated to this complementary component is a Type 2 vertex
consisting of the other cut sets in the collection that are in the
component and $S_i$--minimal.
These cut sets are also $x$--minimal, since if $S_j$ separates $x$
from $S_k$ it would also separate $S_i$ from $S_k$.
Define $\pi_\Gamma(x)$ to be the closed star of this Type 2 vertex.

Note that if $x=\qmap(\{\inv{w}^\infty,w^\infty\})$ then,
as in the proof of \fullref{proposition:otalsplit}, there do exist
minimal elements, because if $S_i$ separates $x$ from $S_j$ it is
necessary that the axis of $g_i$ intersects the finite geodesic
segment joining the axis of $g_j$ to the convex hull of $\{\inv{w}^\infty,w^\infty\}$.
By construction, there are only finitely many such axes.

Now suppose there are no minimal elements for the $x$--order.
Let $\xi=\qmap^{-1}(x)\in\partial\tree$.
There is a
sequence $S_{i_1}>S_{i_2}>\cdots$.
Define $\pi_\Gamma(x)$ to be the boundary point in
$\partial\mathcal{D}(\Gamma)$ to which the $\pi_\Gamma$ images of the
$S_{i_j}$ converge.
We show this is well defined.
Passing to a subsequence, we may assume that the axes of the
corresponding $g_{i_j}$ are disjoint in $\tree$.
For each $S_{i_j}$ there is an edge $e_{i_j}\in\tree$ that is adjacent
to the axis of $g_{i_j}$ and separates the axis from $\xi$.
Furthermore, for $k>j$ the axis of $g_{i_k}$ is on the $\xi$--side of $e_{i_j}$.
Thus, the edges $e_{i_j}$ are distinct edges on a geodesic ray in $\tree$ converging
to $\xi$. 
Now $\pi_\Gamma(x)$ is well defined since any other
geodesic ray converging to $\xi$ eventually coincides with this one,
so that the tails of any two sequences $S_{i_1}>S_{i_2}>\cdots$ and $S'_{i_1}>S'_{i_2}>\cdots$
can be merged into a single ordered sequence.

It is clear that the map is equivariant and onto
$\mathcal{D}(\Gamma)$.
Let $\eta$ be a point in $\partial\mathcal{D}(\Gamma)$.
Pick a sequence of Type 1 vertices converging to $\eta$, and let
$S_{i_1}, S_{i_2},\dots$ be the corresponding cut sets. 
By construction, for any $j<k<l$ we have that $S_{i_k}$ separates
$S_{i_j}$ from $S_{i_l}$.
As in the previous paragraph, from such a sequence we can pick a
sequence of edges that lie on a geodesic ray in $\tree$, converging to
a point $\xi\in\partial\tree$.
Then $\pi_\Gamma(\qmap(\xi))=\eta$.
\end{proof}



\begin{definition}
  Let $\Gamma$ be a graph of groups decompositions of $F$ rel
  $\multimot$ with cyclic edge stabilizers.
Define the \emph{augmented multiword}, denoted $\augw$, to be the
multiword in $F$ consisting of words of $\multimot$ together with
generators of each of the edge groups of $\Gamma$.
\end{definition}


Recall that we consider multiwords equivalent if they determine the
same set of conjugacy classes of maximal cyclic subgroups. 
The number of words in $\augw$ is not necessarily the number of words
of $\multimot$ plus the number of edges of $\Gamma$; we only need to
add a new word if an edge stabilizer is not contained in a conjugate
of $\left<w\right>$ for any $w\in\multimot$.

Adding new words to a multiword results in a smaller decomposition
space.

\begin{definition}
  Define the \emph{augmentation map}, $\augmap$, to be the quotient map:
  \[\augmap\colon\thinspace \decomp_{\multimot} \to \decomp_{\augw}\]
\end{definition}

Note that $\qmap_{\augw}=\augmap \circ \qmap_{\multimot}$.

\begin{definition}
  Let $G$ be a vertex stabilizer of a graph of groups
  decomposition $\Gamma$ of $F$ rel $\multimot$ with cyclic edge stabilizers.
If $G$ is non-cyclic, define the \emph{induced multiword} in $G$,
denoted $\indg$, to be the
multiword in $G$ consisting of any conjugates of elements of
$\multimot$, along with generators of the maximal cyclic subgroups
into which the incident edge groups include.
\end{definition}

The point here is that a splitting of $G$ rel the induced multiword
should give a refinement of the graph of groups decomposition.
This is why we add the incident edge groups, to ensure that a
splitting of $G$ will be compatible with the splitting we already have.
The following lemmas make this precise.

\begin{lemma}\label{lemma:embed}
 Let $G$ be a non-cyclic
  vertex group of $\Gamma$.
The decomposition space $\decomp_{\indg}$ of $G$ corresponding
to $\indg$ embeds naturally into the decomposition space
$\decomp_{\augw}$ of $F$ corresponding to $\augw$. 
\end{lemma}
\begin{proof}
The inclusion $\iota\colon\thinspace G\hookrightarrow F$ is a quasi-isometric embedding, so
it extends to a topological embedding $\partial \iota \colon\thinspace \partial G \hookrightarrow \partial F$.
The equivalence relation on $\partial G$ coming from $\indg$ is the
restriction to $\partial\iota(\partial G)$ of the equivalence relation on
$\partial F$ coming from $\augw$.
\end{proof}

\begin{theorem}[Structure Theorem for Decomposition Spaces]\label{theorem:structure}
The decomposition
space $\decomp_{\augw}$ is a compactified tree of decomposition spaces
of the vertex groups of $\Gamma$. 
\end{theorem}
\begin{proof}
 According to \fullref{lemma:projection} there is an $F$--equivariant surjective
 quasi-map
 $\pi_\Gamma\colon\thinspace\decomp_{\augw}\to\closure{\mathcal{D}(\Gamma)}$.
The preimage of a Type 1 vertex of $\mathcal{D}(\Gamma)$ is a cut
point in $\decomp_{\augw}$.
The preimage of the closed star of a Type 2 vertex stabilized by $G$
is the copy of $\decomp_{\indg}$ in $\decomp_{\augw}$.
\end{proof}

\begin{lemma}\label{lemma:connectedpieces}
Let $\Gamma$ be a graph of groups decomposition of $F$ rel $\multimot$
with cyclic edge groups. 
Let $G$ be a non-cyclic vertex of $\Gamma$.
The decomposition space $\decomp_{\indg}$ of $G$ with respect to $\indg$ is connected.
\end{lemma}
\begin{proof}
By \fullref{corollary:connected}, if $\decomp_{\indg}$ is not connected there is a free splitting of
  $G$ rel $\decomp_{\indg}$.
This gives a free splitting of $F$ rel $\augw$, which implies that
$\decomp_{\augw}$ is not connected. 
This is not possible since $\decomp_{\augw}$ is a quotient of
$\decomp_{\multimot}$, which, by our standing assumption, is connected.
\end{proof}


\begin{lemma}\label{lemma:rigidboundary}
Let $\Gamma$ be a graph of groups decomposition of $F$ rel $\multimot$
with cyclic edge groups. 
Let $G$ be a non-cyclic vertex of $\Gamma$ such that $\decomp_{\indg}$
is rigid.
Let $\left<g\right>$ be the stabilizer of an edge incident to $G$.
Then $S=\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is either a
cut point or uncrossed cut pair of $\decomp_{\multimot}$.
\end{lemma}
\begin{proof}
$S$ is a cut point if a root of a conjugate of $g$ is in $\multimot$
and is a cut pair otherwise.

If $\decomp_{\multimot}\setminus S$ has at least three components then $S$ is
uncrossed, so assume there are exactly two components.

If $S$ is crossed then there is another cut pair $R$ of $\decomp_{\multimot}$
whose points are in different components of $\decomp_{\multimot}\setminus S$.
Conversely, $R$ is crossed by $S$, so the cut pair $R$ also has exactly two complementary components, $A_0$ and $A_1$.

Let $\{S_i\}_{i\in I}$ be the collection of cut points and cut pairs
corresponding to the edges of $\Gamma$.
Let $J=\{j\mid S_j \text{ crosses or intersects } R\}$.
Then $\augmap(R\cup\{S_j\}_{j\in
  J})\cap \decomp_{\indg}$ is a cut pair in $\decomp_{\indg}$,
contradicting rigidity.
\end{proof}

We will use the following kind of collections several times, so we give them a name.
\begin{definition}
An \emph{uncrossed collection} in $\decomp_{\multimot}$ is a non-empty
union of orbits of cut points and uncrossed cut pairs.
\end{definition}

Note that an uncrossed collection satisfies the hypotheses of
\fullref{proposition:multipleotal}, since by
\fullref{proposition:boundedperiodicity} there are only finitely many orbits
of cut points or uncrossed cut pairs.
Thus, for any uncrossed collection there is a corresponding graph of
groups decomposition of $F$ rel $\multimot$.

\begin{lemma}[Universality of Uncrossed
  Splittings]\label{lemma:uncrossedsplittingsareuniversal}
Let $\{S_i\}_{i\in I}$ be an uncrossed collection.
Let $\Gamma$ be the corresponding graph of groups decomposition of $F$.
The stabilizer of a cut point or cut pair of $\decomp_{\multimot}$ is elliptic in $\Gamma$.
\end{lemma}
\begin{proof}
Let $S$ be a cut point or cut pair.
By considering the $F$--action on the convex hull of $\qmap^{-1}(S)$
it is clear that the stabilizer of $S$ is either trivial or a maximal
cyclic subgroup.
 If its stabilizer is trivial we
are done, so assume its stabilizer is $\left<g\right>$.

 Since $S$ has cyclic stabilizer $\pi_\Gamma(S)\subset\mathcal{D}(\Gamma)$.
We claim that $\pi_\Gamma(S)$ is either a single Type 1 vertex or is
the star of a Type 2 vertex. Then $g$ fixes the
appropriate vertex.

If $S$ is a cut point this is obvious, so suppose it is a cut pair and
that the image is not contained in the closed star of some Type 2
vertex. Then there is some Type 1 vertex between the two points of
$\pi_\Gamma(S)$.
This corresponds to a cut set $S_i$ separating the two points of $S$. 
But this would mean $S$ and $S_i$ cross, contradicting the hypothesis
that the $S_i$ are uncrossed.
\end{proof}

For the remainder of this section we assume the hypotheses and
notation of \fullref{lemma:uncrossedsplittingsareuniversal}:
\begin{enumerate}
\item $\{S_i\}_{i\in I}$ is an uncrossed collection.
\item $\Gamma$ is the corresponding graph of groups decomposition of $F$.
\end{enumerate}

\begin{lemma}\label{lemma:cutsarecutsinaug}
If $S$ is a cut point or cut pair of
  $\decomp_{\multimot}$ then $\augmap(S)$ is a cut point or cut pair of
  $\decomp_{\augw}$.
If $R$ is a cut point or cut pair of
  $\decomp_{\augw}$ then $\augmap^{-1}(R)$ is a cut point or cut pair of $\decomp_{\multimot}$.
\end{lemma}
\begin{proof}
 Consider an element $h$ of $\augw\setminus\multimot$. It
 generates the stabilizer of an $S_i$ that is an uncrossed cut pair
 in $\decomp_{\multimot}$.
Thus, both $\inv{h}^\infty$ and $h^\infty$ are in the same
complementary component of $S$.
Identifying them does not reduce
the number of components, so $\augmap(S)$ is a cut set in
$\decomp_{\augw}$.

Conversely, let $R$ be a cut point or cut pair of
  $\decomp_{\augw}$. 
It is clear that $\augmap^{-1}(R)$
is a cut set in $\decomp_{\multimot}$; we just need to show that it consists of at most two points. 

Suppose a point $r\in R$ has preimage $\augmap^{-1}(r)$ consisting of
two points. 
Then $r$ is stabilized by a conjugate of an element $h\in \augw\setminus\multimot$,
so $\augmap^{-1}(r)=S_i$ for some $i$. 
By the first part of the lemma, $\augmap(S_i)=\{r\}$ is a cut point,
so $R=\{r\}$, and $\augmap^{-1}(R)=S_i$ is a cut pair.
\end{proof}



\begin{lemma}\label{lemma:prerefinement}
Let $G$ be a non-cyclic
  vertex group of $\Gamma$.
Let $g$ be an element of $G$ such that 
$\qmap_{\indg}(\{\inv{g}^\infty,g^\infty\})$ is a cut point
or uncrossed cut pair in $\decomp_{\indg}$. 
Then $\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is a cut
point or uncrossed cut pair, respectively, of $\decomp_{\multimot}$ that is not
in the uncrossed collection.
\end{lemma}
\begin{proof}
There is a Type 2 vertex of $\mathcal{D}(\Gamma)$ corresponding to
$G$. 
To each adjacent Type 1 vertex there is an associated cut set
$S_j$. Let $J$ be the set of indices of these cut sets.
Each $S_j$ becomes a cut point in $\decomp_{\indg}$, and each complementary component of the image of
$\qmap_{\indg}(\{\inv{g}^\infty,g^\infty\})$ in $\decomp_{\indg}$ 
contains some of these $S_j$ points, so $\augmap^{-1}(
\qmap_{\indg}(\{\inv{g}^\infty,g^\infty\}))= \qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is a cut point or cut
pair in $\decomp_{\multimot}$ separating some of these $S_j$.
By definition, no two cut sets in $\{S_j\}_{j\in J}$ are separated by
  a cut set in $\{S_i\}_{i\in I}$, so
$\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is a cut set and is
not in the uncrossed collection.

Since $\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is not one of the $S_i$, its
cardinality is the same as that of the image in $\decomp_{\indg}$.
Thus, they are either both cut points or both cut pairs. 

It remains only to show that if $\qmap_{\indg}(\{\inv{g}^\infty,g^\infty\})$ is
an uncrossed cut pair then $\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is uncrossed.

Suppose $\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ is crossed
by some other cut pair $R$ of $\decomp_{\multimot}$. 
If $\pi_\Gamma(R)$ is not contained in the star of the $G$ vertex then
there is an $S_i$ separating a point of $R$ from 
$\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$.
For $R$ to cross $\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\})$ it
would have to cross $S_i$, but $S_i$ is uncrossed.
This is a contradiction, so $\pi_\Gamma(R)$ is contained in the star of the $G$ vertex.
This means that $\augmap(R)$ and $\augmap(\qmap_{\multimot}(\{\inv{g}^\infty,g^\infty\}))$ are
crossing cut pairs in $\decomp_{\augw}$.
Since these both project to the star of the $G$ vertex, their $\augmap$
images are in the copy of $\decomp_{\indg}$ embedded in
$\decomp_{\augw}$ as in \fullref{lemma:embed}.
Thus we get crossing cut pairs of $\decomp_{\indg}$, contrary to hypothesis.
\end{proof}

\begin{lemma}[Refinement Lemma]\label{lemma:refinement}
  Let $G$ be a non-cyclic vertex of $\Gamma$.
If $\decomp_{\indg}$ is neither a circle nor rigid then there is a
refinement $\Gamma'$ of $\Gamma$ obtained by splitting $G$ rel $\indg$.
The refinement $\Gamma'$ is a splitting over an uncrossed collection
containing $\{S_i\}_{i\in I}$.
\end{lemma}
\begin{proof}
By \fullref{lemma:connectedpieces}, $\decomp_{\indg}$ is connected. If
it is not a circle and not rigid, then by
  \fullref{mainlemma} there is an element $g\in G$ such that
  $\qmap_{\indg}(\{\inv{g}^\infty, g^\infty\})$ is a cut point or
  uncrossed cut pair.
By \fullref{lemma:prerefinement}, $\qmap_{\multimot}(\{\inv{g}^\infty, g^\infty\})$ is a cut point or
  uncrossed cut pair of $\decomp_{\multimot}$.
Thus, we can add the orbit of $\qmap_{\multimot}(\{\inv{g}^\infty,
g^\infty\})$ to the set $\{S_i\}_{i\in I}$ to get a larger uncrossed
collection, and hence a graph of groups decomposition $\Gamma'$
refining $\Gamma$.
\end{proof}

\endinput

\subsection{Free Groups, Line Patterns and Decomposition Spaces}
Let $F=F_n$ be a free group of rank $n>1$. For $g\in F$, let $\inv{g}$
denote inverse of $g$.

A nontrivial element $g\in F$ is \emph{indivisible} if is not a proper
power of another element. Equivalently, the cyclic subgroup
$\left<g\right>$ is maximal; it is not properly contained in a cyclic
subgroup of $F$.

The \emph{width} of $g$ is the rank of the smallest free
factor of $F$ containing $g$.
An element is \emph{full width} if its width is equal to the rank of $F$.
Full width is also known as \emph{diskbusting}, particularly in the
context of 3--manifolds.

A \emph{basis} or \emph{free basis} or \emph{free generating set} of
$F$ is a generating set consisting of exactly $n$ elements
$\mathcal{B}=\{b_1,\dots, b_n\}$.

A subset $\{g_1,\dots, g_k\}$ is \emph{basic} is it is a subset of a
some basis. 
In particular, a single element $g$ is \emph{basic} if and only if it is indivisible and width one.
The term \emph{primitive} is also commonly used for our basic elements.

The \emph{degree} of a homomorphism from the integers into a free
group is the index of the image in the maximal cyclic subgroup containing
the image.

The Cayley graph of $F$ with respect to a basis $\mathcal{B}$ is a
tree $\tree$ whose vertices are in bijection with elements of $F$.
There is an edge from vertex $g$ to vertex $h$ if and only if there
exists a $b_i\in\mathcal{B}$ such that $gb_i=h$. 
We make this a metric space by assigning each edge length one; $F$
acts isometrically on $\tree$ by left multiplication.

The tree $\tree$ has
a boundary at infinity $\partial\tree$ that is homeomorphic to a Cantor set.
This compactifies the tree, $\closure{\tree}=\tree\cup \partial\tree$ is
a compact topological space whose topology on $\tree$ agrees with the
metric topology.

Two subsets of $\tree$ are \emph{coarsely equivalent}, written
$\stackrel{c}{=}$, if they have bounded Hausdorff distance.

For any two points $\xi, \xi'\in\closure{\tree}$ there exists a unique
geodesic $[\xi,\xi']$ joining them.

For a nontrivial $h\in F$, we denote by $gh^\infty$ the point in $\partial\tree$ that is the limit
of the sequence of vertices $(gh^i)$ in $\tree$.
The $w$--line through $g$ is the coarse equivalence class of the set
$\{gw^m\}_{m\in\mathbb{Z}}$.
This coarse equivalence class contains a unique geodesic $l$ with
endpoints on the boundary: $l^+=gw^\infty$ and $l^-=gw^{-\infty}$.
If $w$ is cyclically reduced then $g$ is actually a point on the line $l$.

Let $\lines_{\multimot}$ be the line pattern generated by a multiword $\multimot$.
The decomposition space $\decomp_{\multimot} =\decomp_{\lines_{\multimot}}$ is the quotient of
$\partial\tree$ obtained by identifying the two points $l^+$ and $l^-$
for each $l\in\lines_{\multimot}$. 
Let $\qmap_{\multimot} \colon\thinspace\partial\tree\to\decomp_{\multimot}$ be the
quotient map.
We will often drop the $\multimot$ subscript when it is clear that we
are working with a fixed multiword.
Note that a change of basis of $F$ gives equivariant homeomorphisms
of $\partial\tree$ and of $\decomp$, so we can talk about these spaces
without reference to a specific basis.
It also makes sense to define $\partial F=\partial\tree$.

The decomposition space is a perfect, metrizable, compact, Hausdorff topological space. 
When it is connected it is also locally connected and has topological
dimension one. 

By construction, distinct lines $l$ and $l'$ in $\lines$ never have a
common endpoint in $\partial\tree$.
Thus, the preimage of a point in the decomposition space is either a
single point in $\partial\tree$ or the pair of endpoints of a line in the pattern.

Any homeomorphism of $\partial\tree$ that preserves the collection of
pairs of endpoints of lines of $\lines$ descends to a homeomorphism of
the decomposition space. In particular, any quasi-isometry of $\tree$
that, up to coarse equivalence, preserves the line pattern will induce
a homeomorphism of the decomposition space.

We will primarily be interested in the case that the decomposition
space is connected. 
A \emph{cut set} is a set whose complement is not connected.
The cut set is \emph{minimal} if no proper subset is a cut set.
A \emph{cut point} is a point that is a cut set.
A two point cut set ought to be called a cut pair, but we
will be interested in minimal cut pairs, so the term
\emph{cut pair} will be reserved for a two point cut set, neither
point of
which is a cut point.

If $\{x_0,x_1\}$ and $\{y_0,y_1\}$ are cut pairs, we say $\{x_0,x_1\}$
\emph{crosses} $\{y_0,y_1\}$ if $x_0$ and $x_1$ are in different
complementary component of $\decomp\setminus\{y_0,y_1\}$.
Since we assume that no point of a cut pair is a cut point, crossing
is symmetric: $\{x_0,x_1\}$ crosses $\{y_0,y_1\}$ if and only if $\{y_0,y_1\}$
crosses $\{x_0,x_1\}$.
Thus, we can say $\{x_0,x_1\}$ and
$\{y_0,y_1\}$ \emph{cross}.

Note that a cut point can not be
crossed by a cut pair: the cut pair would not
disconnect the space unless one of its points was actually a cut
point, but the term ``cut pair'' is reserved for two point cut sets
that are minimal.
Similarly, a cut pair with more than two complementary connected components can
not be crossed by a cut pair.



\subsection{A Line Pattern Lifted to a Finite Index Subgroup}
Let $G$ be a finite index subgroup of $F$. 
Let $1=f_1,\dots, f_k$ be right coset representatives, so that
$F=\underset{i=1\dots k}{\coprod} Gf_i$. 
The map $\inv{\iota}\colon\thinspace F\to G$ that sends $gf_i$ to $g\in G$ is a
quasi-isometry. 
It is a coarse inverse to the inclusion $\iota\colon\thinspace
G\to F$.


Let $w_j$ be an element of a multiword $\multimot$.
For each $i$ and $j$ there exists a minimal positive integer $a=a(i,j)$ such that
$f_iw_j^{a}\inv{f}_i\in G$. Therefore, for any $h=gf_i$, we have:
\[\{hw_j^m\}_{m\in\mathbb{Z}}\stackrel{c}{=}\{hw_j^{am}\}_{m\in\mathbb{Z}}=\{gf_iw_j^{am}\}_{m\in\mathbb{Z}}\stackrel{c}{=}\{g(f_iw_j^a\inv{f}_i)^m\}_{m\in\mathbb{Z}}\subset G\]
Thus, if $h\in Gf_i$, the map $\inv{\iota}$ sends the $w_j$--line through $h$ in $F$ to a set of points
coarsely equivalent to the $(f_i w_j^a \inv{f}_i)$--line through
$\inv{\iota}(h)$ in $G$.

Consider the set of conjugacy classes in $G$ given by
$\{[f_iw_j^{a(i,j)}\inv{f}_i]\mid \text{ for all } i,j\}$. 
Pick a cyclically reduced representative from each distinct conjugacy
class.
This gives a multiword $\multimot'$ that generates a line pattern in
$G$ that is equivalent to the image of $\lines_{\multimot}$ under
$\inv{\iota}$.
We call this the \emph{line pattern $\lines_{\multimot}$ lifted to}
$G$, and call $\multimot'$ the \emph{multiword $\multimot$ lifted to} $G$.

In topological terms, think of $F$ as the fundamental group of a
surface (with boundary).
The multiword $\multimot$ can be represented by a multicurve on the
surface. 
Let $G$ be the fundamental group of a covering surface.
For each curve in the multicurve, take all of its lifts to the $G$ surface.
The collection of all of these is a multicurve representing $\multimot'$.

The lifted line pattern is independent of the choice of coset
representatives and choice of $\multimot'$, and
$\inv{\iota}$ takes each line of $\lines_{\multimot}$ to a set coarsely equivalent
to a line in the lifted pattern, and similarly for $\iota$. 
Therefore, $\inv{\iota}$ extends to a $G$--equivariant homeomorphism
$\partial F\to\partial G$ that preserves pairs of endpoints of lines in the
pattern.
This homeomorphism descends to a $G$--equivariant homeomorphism of
decomposition spaces.




\endinput


\subsection{Quasi-isometry invariance of the rJSJ}
 Let $(F,\multimot)$ and $(F',\multimot')$ be two free groups with
 multiwords. 
Suppose there is a quasi-isometry $\phi \colon\thinspace F \to F'$ that
bijectively matches up the line patterns $\lines_{\multimot}$ and
$\lines_{\multimot'}$ in the sense that there exists some constant $C$
such that for every line $l\in\lines_{\multimot}$ there is a line of
$\lines_{\multimot'}$ within Hausdorff distance $C$ of $\phi(l)$, and
vice versa. 
We say that $\phi$ 
\emph{preserves line patterns}.

Such a quasi-isometry gives a homeomorphism of the corresponding
decomposition spaces. 
In particular, it preserves the cut set structure.
Thus, the Bass-Serre trees $\mathcal{D}(\Gamma)$ and
$\mathcal{D}(\Gamma')$ for the respective rJSJ's of $\multimot$ and
$\multimot'$ are the same. 
The trees are isomorphic, and even the kinds of vertices ---
cut point, uncrossed cut pair, rigid, QH--surface --- are preserved.
There is no reason in general that this isomorphism should be
compatible with the group actions, but it will be, for instance, in
the case of lifting a multiword to a finite index subgroup of $F$.
Such a lift has rJSJ a finite sheeted covering graph of groups of the original rJSJ.


A result of Behrstock and Neumann \cite{BehNeu06} shows that there is
a line pattern preserving quasi-isometry between any two QH--surface multiwords.
In particular, the sphere with three boundary components is
quasi-isometric to the other QH--surfaces. 
This is part of the motivation for insisting that the sphere with
three boundary components is not rigid, even though it has no relative
splittings.
This example also shows that relative splittings need not be preserved
by pattern preserving quasi-isometries, only universal splittings need be.

\endinput

\subsection{Crossing Pairs and the Circle}\label{sec:crossed}
In this subsection we give criteria for the
decomposition space to be a circle.

In a circle, the only minimal cut sets are cut pairs, and every cut
pair is crossed by a cut pair.

 \begin{lemma}\label{lemma:3of4}
   Suppose $\{x_0,x_1\}$ and $\{y_0,y_1\}$ are crossing cut pairs in $\decomp$. 
Let  $A_0$ and $A_1$ be the two connected components of
$\decomp\setminus\{x_0,x_1\}$, and assume $y_0\in A_0$ and $y_1\in
A_1$.
Similarly, let  $B_0$ and $B_1$ be the connected components of
$\decomp\setminus\{y_0,y_1\}$, and assume $x_0\in A_0$ and $x_1\in
A_1$.
Then $\{x_0,y_0\}$ is a cut pair with connected components $C_0=A_0\cap
B_0$ and $C_1=A_1\cup B_1\cup\{x_1\}\cup\{y_1\}$.
 \end{lemma}
 \begin{proof}
Since $\{x_0,x_1\}$ and $\{y_0,y_1\}$ cross, $y_0$ is a
cut point of the connected set $A_0$.
Any neighborhood of $y_0$ contains a connected open set $N$ such that
$N\setminus y_0$ has exactly two connected components.  
Therefore, $A_0\setminus y_0$ has exactly two connected components,
which are $A_0\cap B_0$ and $A_0\cap B_1$.
Similarly, the sets $A_1\cap B_0$ and $A_1\cap B_1$ are connected.

It follows that $C_1$ is a connected set and $\closure{C}_1\setminus
   C_1=\{x_0,y_0\}$. 

Now:
\[\{x_0,y_0\}\subset\closure{C}_0\setminus C_0\subset
\{x_0,x_1,y_0,y_1\}\]

We are done if $x_1$ and $y_1$ are not limit points of $C_0$.
Suppose, without loss of generality, that $x_1$ is a limit point of $C_0=A_0\cap B_0$.
Since $x_1$ is also a limit point of $A_0\cap B_1$ and $A_1\cap B_1$, this implies that for any small neighborhood $N$ of $x_1$, the
complement $N\setminus x_1$ has at least three connected components,
which is a contradiction.
 \end{proof}


The following proposition is a generalization of a construction of Bowditch for boundaries
of hyperbolic groups \cite{Bow98b}.

 \begin{proposition}\label{proposition:circle}
 The decomposition space $\decomp$ is a circle if and only if all of the
 following conditions are satisfied:
 \begin{enumerate}
 \item $\decomp$ is connected.
\item $\decomp$ has no cut points.
\item $\decomp$ has cut pairs.
\item  Every cut pair in $\decomp$ is crossed by a cut pair.
\end{enumerate}
\end{proposition}
\begin{proof}
A circle satisfies these conditions, so one direction is clear.

Define a relation on $\decomp$ by $x\sim y$ if $x=y$
  or if $\{x,y\}$ is a cut pair.
  \begin{claim}\label{claim:transitive}
    $\sim$ is an equivalence relation.
  \end{claim}
  \begin{claimproof}
We must show transitivity.

Suppose $x$, $y$ and $z$ are distinct points with $x\sim y$ and $y\sim
z$, so that $\{x,y\}$ and $\{y,z\}$ are cut pairs.
By hypothesis, every cut pair is crossed by a cut pair. Therefore,
every cut pair has exactly two complementary connected components.
Let $A_0$ and $A_1$ be the connected components of
$\decomp\setminus\{x,y\}$. 
Let $B_0$ and $B_1$ be the connected components of
$\decomp\setminus\{y,z\}$.
Assume that $z\in A_0$ and $x\in B_0$.
Arguing as in the proof of \fullref{lemma:3of4},
$\decomp\setminus\{x,z\}$ has two connected components, $A_0\cap B_0$
and $A_1\cup B_1\cup \{y\}$.    
  \end{claimproof}

\begin{claim}
 Equivalence classes are closed. 
\end{claim}
\begin{claimproof}
If an equivalence class consists of a single point it is closed, so
suppose $[x]$ is not a single point.
  
Let $(y_i)\to y$ for $y_i\in [x]$.
Choose points $\xi\in \qmap^{-1}(y)$ and $\xi_i\in \qmap^{-1}(y_i)$.
After passing to a subsequence, $(\xi_i)\to\xi$ in $\partial\tree$.

Pick $\eta\in \qmap^{-1}(x)$. 
For each $i$, $\Wh([\eta,\xi_i])$ has two components. 

Now consider $[\eta,\xi]$. Number the vertices $v_j$ along this geodesic
with consecutive integers, increasing in the $\xi$ direction, where
$v_0$ may be any vertex on $[\eta,\xi]$. Since $(\xi_i)\to\xi$, for
every $j$ there is an $I_j$ such that for all $i>I_j$, we have $[\eta,v_{j+1}]\subset [\eta,\xi_i]$.
Therefore, $\Wh([\eta,v_j])\setminus
[\eta,\xi]=\Wh([\eta,v_j])\setminus[\eta,\xi_i]$.

Thus, for all $j$, the Whitehead graph $\Wh([\eta,v_j])\setminus
[\eta,\xi]$ has two components, which implies
$\qmap(\{\eta,\xi\})=\{x,y\}$ is a cut pair. Hence, $y\in[x]$, and $[x]$ is closed.
\end{claimproof}

\begin{claim}
All of $\decomp$ is in one equivalence class.
\end{claim}
\begin{claimproof}
We have assumed that a cut pair exists, so there is an equivalence
class $[x]$ consisting of more than one point.
Suppose that $[x]$ is not all of $\decomp$. 

Let $U$ be a connected component of $\decomp\setminus[x]$.
Since $\decomp$ is locally connected and $[x]$ is closed, $U$ is open
in $\decomp$. 
Since $\decomp$ is connected without cut points, $U$ has at least two
limit points in $[x]$.
Pick distinct points $y_0$ and $y_1$ in $\closure{U}\cap [x]$.
These points are  a cut pair, and $\decomp\setminus\{y_0,y_1\}$ has exactly two
connected components, $A_0$ and $A_1$.
Assume $U\subset A_0$.

Let $\{z_0,z_1\}$ be a cut pair crossing $\{y_0,y_1\}$ with
complementary connected components $B_0$ and $B_1$. 
Assume $z_0\in A_0$, $z_1\in A_1$, $y_0\in B_0$ and $y_1\in B_1$.

By \fullref{lemma:3of4}, $y_0$ and $y_1$ are in $[x]\subset
\decomp\setminus U$.
Thus, $U$ is contained in $B_\epsilon$,
where $\epsilon$ is either 0 or 1.
Since $U\subset A_0$, we have $U\subset A_0\cap B_\epsilon$.

Now, $\{y_\epsilon,z_0\}$ is a cut pair whose connected
components are $C_0=A_0\cap B_\epsilon$ and $C_1=A_1\cup
B_{1+\epsilon}\cup\{y_{1+\epsilon}\}\cup\{z_1\}$ (subscripts modulo 2).
However, $U$, and hence $C_0$, has $y_{1+\epsilon}\in C_1$ as a limit point,
which is a contradiction. Thus, $[x]=\decomp$.
\end{claimproof}

Since $[x]=\decomp$, every point of $\decomp$ is a member of a cut pair, and the only
minimal cut sets are of size two.
It then follows from
\fullref{lemma:circles} that $\decomp$ is a circle.
\end{proof}

\begin{corollary}\label{corollary:uncrossedexist}
  If $\decomp$ is connected, not rigid, and not a circle, then
  $\decomp$ contains cut points or uncrossed cut pairs.
\end{corollary}


\subsection{Uncrossed Cut Pairs}\label{sec:uncrossed}
A cut point in $\decomp$ is of the form
$\qmap(\{\inv{w}^\infty,w^\infty\})$ for some $w\in\multimot$.
Therefore, the preimage
$\qmap^{-1}(\qmap(\{\inv{w}^\infty,w^\infty\}))$ is the pair of points
$\{\inv{w}^\infty,w^\infty\}\subset\partial\tree$.
The convex hull of these two points is a line on which the element $w$
acts cocompactly.
We will show that uncrossed cut pairs have a similar form.

In general the preimage in $\partial\tree$ of a cut pair in $\decomp$
could have as many as four points, so the convex hull may not be a line.
Even if a cut pair has a preimage consisting of two points, it may not
have a group element acting cocompactly on the convex hull of these
two points.

An easy argument, \fullref{lemma:uncrossedaxis}, shows that the convex hull
of the preimage of an uncrossed cut pair is a line.

Recall that  \fullref{lemma:periodic} says
that for any cut pair such that the convex hull of the preimage is a line, we can find a
group element $g$ of uniformly bounded length whose axis approximates
the hull in a certain strong sense.
\fullref{proposition:periodic} shows that if the cut
pair is uncrossed then the axis of $g$
and the convex hull of the preimage coincide.


\begin{lemma}\label{lemma:uncrossedaxis}
  The preimage of an uncrossed cut pair is two points, so its convex
  hull is a line.
\end{lemma}
\begin{proof}
Suppose we have a cut pair $\{x_0,x_1\}\subset\decomp$ that is not crossed
by any other cut pair. 
  Suppose $\qmap^{-1}(x_0)=\{l^-,l^+\}$ for some
  $w$--line $l\in\lines$ through some vertex $h$ in $\tree$.
Acting on everything by $\inv{h}$ and renaming, we may assume that $l$ is the
$w$--line through $*$. 

By an argument similar to the proof of \fullref{lemma:3of4}, for some
$i$ and $j$, the pair $\{w^{-i}x_1,w^{j}x_1\}$ is a cut pair
crossing $\{x_0,x_1\}$, contrary to hypothesis.
\end{proof}

\begin{proposition}\label{proposition:periodic}
If $\{x_0,x_1\}$ is an uncrossed cut pair and $g$ and $h$ are the 
elements from \fullref{lemma:periodic} such that the axis of
$hg\inv{h}$ approximates the convex hull of $\qmap^{-1}(\{x_0,x_1\})$, then
$\qmap(h\inv{g}^\infty)=x_0$ and $\qmap(hg^\infty)=x_1$.
\end{proposition}
\begin{proof}
Assume $h$ is trivial by replacing $\{x_0,x_1\}$ with
$\{\inv{h}x_0,\inv{h}x_1\}$ and renaming.

Let $\xi_i=\qmap^{-1}(x_i)$, which is a single point by
\fullref{lemma:uncrossedaxis}. 

Suppose $g^\infty\neq \xi_1$. 
Consider the last vertex $v\in\tree$ in the positive $g$ direction
such that
$[\inv{g}^\infty,g^\infty]$ and $[\xi_0,\xi_1]$ overlap.
Let $e_o$ be the edge incident to $v$ contained in the overlap.
Let $e_g$ be the edge leading to $g^\infty$.
Let $e_\xi$ be the edge leading to $\xi_1$.

The Whitehead graph $\Wh(v)\setminus\{e_o,e_g,e_\xi\}$ has four groups
of components:
Group $A$ consisting of components that go out $e_o$ and $e_\xi$ but
not $e_g$.
Group $B$ consisting of components that go out $e_o$ and $e_g$ but
not $e_\xi$.
Group $C$ consisting of components that go out $e_g$ and $e_\xi$ but
not $e_o$.
Group $D$ consisting of components that go out all three.

We further divide $A$ into groups $A'$ and $A''$. 
Group $A''$ consists of those components that are contained in the
component of $\Wh([\xi_0,\xi_1])$ containing $g^\infty$. Group $A'$
consists of the others.

All of the components in groups $A''$, $B$, $C$, and $D$ belong to a single
component of $\Wh([\xi_0,\xi_1])$, the one containing $g^\infty$.
Since $\Wh([\xi_0,\xi_1])$ has at least two components, $A'$ is
non-empty.

On the other hand, all of the components in groups  $A$, $C$, and $D$ belong to a single
component of $\Wh([\inv{g}^\infty,g^\infty])$, the one containing
$\xi_1$.
Now, components of $\Wh(\text{overlap})$ that are in separate
$\Wh([\xi_0,\xi_1])$ components must remain in separate
$\Wh([\inv{g}^\infty,g^\infty])$ components.
Thus, anything in the $A$ or $D$ groups must belong to a common
$\Wh([\xi_0,\xi_1])$ component. 
However, the $A'$ group pieces do not belong to the $g^\infty$ component of
$\Wh([\xi_0,\xi_1])$, while the $A''$ and $D$ groups do.
Since the $A'$ group is non-empty, the $A''$ and $D$ groups must be
empty. Also, the components in the $A'$ group all belong to a common
$\Wh([\xi_0,\xi_1])$ component. 
Thus, the cut set $\{x_0,x_1\}$ has exactly two complementary
components, the one containing $g^\infty$, and the one not containing $g^\infty$.

The $C$ group can not be empty or the $e_o$ edge would give a cut
vertex in the Whitehead graph at $v$. Similarly, the $B$ group can not
be empty or the $e_\xi$ edge would give a cut vertex.

Everything in the $C$ group, as it continues towards $\xi_1$, must
remain separate from things in the $A'$ group, because the $C$ group
belongs in the $g^\infty$ component of $\Wh([\xi_0,\xi_1])$, while the
$A'$ group does not.
Similarly, pieces in the $C$ group remain separate from pieces in the
$B$ group as they continue towards $g^\infty$, since the $C$ group
belongs to the component of $\Wh([\inv{g}^\infty,g^\infty])$
containing $\xi_1$, while things in the $B$ group do not.
This shows that the Whitehead graph $\Wh([\xi_1,g^\infty])$ has at
least two components, because the pieces from $C$ and the pieces from
$A'\cup B$ must be in separate components. Thus,
$\qmap(\{\xi_1,g^\infty\})$ is a cut pair.

There exists some number $k$ such that $\inv{g}^kv\in\text{overlap}$,
and $\qmap(\{\inv{g}^k\xi_1,g^\infty\})$ is also a cut pair. Since the
$g$ action on the overlap preserves components of
$\Wh([\xi_0,\xi_1])$, the edge $\inv{g}^ke_\xi$ corresponds to a
vertex in the component of $\Wh([\xi_0,\xi_1])$ not containing
$g^\infty$.

This gives us a contradiction: $\qmap(\{\inv{g}^k\xi_1,g^\infty\})$ is
a cut pair with one point in the $g^\infty$ component of
$\decomp\setminus\{x_0,x_1\}$ and one point in another, so
$\{x_0,x_1\}$ is crossed.

Thus, $\xi_1=g^\infty$.
A similar argument shows $\xi_0=\inv{g}^\infty$.

\end{proof}



\begin{proposition}\label{proposition:boundedperiodicity}

There are only finitely many orbits of cut points and
uncrossed cut pairs.
\end{proposition}
\begin{proof}
For cut points this is clear, since by
\fullref{lemma:cutpointsareinmultiword} the stabilizer of a cut point
is generated by a conjugate of a word in $\multimot$.

By \fullref{proposition:periodic} there is a member of each orbit of
uncrossed cut pairs that is stabilized by an element of bounded length provided by
\fullref{lemma:periodic}. 
\end{proof}


\begin{proposition}\label{mainlemma}
If $\multimot$ is a multiword in $F$ such that the decomposition space is
connected and not rigid, then either:
\begin{enumerate}
\item the decomposition space is a circle, or \label{case:circle}
\item there is an indivisible element $g\in F$ such that
 $\qmap(\{\inv{g}^\infty,g^{\infty}\})$ is a cut set that is not crossed by
 any cut pair. \label{case:split}
\end{enumerate}
\end{proposition}
\begin{proof}
If we are not in case (\ref{case:circle}) then by
\fullref{corollary:uncrossedexist} there exist cut points or uncrossed
cut pairs. Take $g$ to be the generator of the stabilizer of such a
cut set. For a cut point one may assume $g$ is in the multiword. For
an uncrossed cut pair the existence of such a $g$ is furnished by
\fullref{lemma:uncrossedaxis}.
\end{proof}


\endinput

\subsection{Rigid Multiwords and Geometricity}{\label{sec:rigidgeometric}
\begin{lemma}[{\cite[Proposition 0]{Ota92}}]\label{lemma:geometricimpliesplanar}
 The decomposition space of a geometric multiword is planar.
\end{lemma}
\begin{proof}
By definition, a geometric  multiword can be realized by an embedded
multicurve on the surface of a handlebody. 
The multicurve lifts to a collection of disjoint curves on the
boundary surface of the universal cover of the handlebody. 
This universal cover is a thickened tree, and may be compactified by
including the Cantor set boundary of the tree. 
The resulting space is a 3--ball with a collection of disjoint arcs
in the bounding 2--sphere. 
By Moore's Decomposition Theorem \cite{Moo25}, the quotient of the 2--sphere obtained by collapsing each of the
curves to a point is still a 2--sphere.
The image of the Cantor
set in this quotient is exactly the decomposition space.
Thus, the decomposition space embedds into $S^2$.
\end{proof}

Passing to a finite index subgroup induces a homeomorphism of
decomposition spaces, so we also have:
\begin{corollary}\label{corollary:vgimpliesplanar}
 The decomposition space of a virtually geometric multiword is planar.
\end{corollary}

\begin{theorem}[{cf \cite[Theorem 1]{Ota92}}]\label{theorem:rigidvg}
Let $(F,\underline{w})$ be rigid. The following are equivalent:
\begin{enumerate}
\item The multiword $\underline{w}$ is geometric.\label{item:rigidgeometric}
\item The decomposition space $\decomp_{\underline{w}}$ is planar.\label{item:rigidplanar}
\item Every minimal Whitehead graph for $\underline{w}$ has an
  embedding in the plane with
  consistent cyclic orderings of edges incident to inverse vertices.\label{item:planarwh}
\end{enumerate}
\end{theorem}
\begin{proof}
 \fullref{lemma:geometricimpliesplanar} shows \mbox{(\ref{item:rigidgeometric}) implies (\ref{item:rigidplanar})}.

\mbox{(\ref{item:rigidplanar}) implies (\ref{item:planarwh})} is the
content of \cite[Lemma 4.4]{Ota92}.
The hypotheses for this lemma are that every element of the multiword
is indecomposable and that the decomposition space embeds
into $S^2$ in such a way that closures of the complementary regions intersect
pairwise in at most one point.
The first hypothesis is too strong. 
Indecomposability of each element of the multiword
is only used to prove that the decomposition space has no cut points.
The second hypothesis is satisfied if the decomposition space has no
cut pairs. Therefore, rigidity is a sufficient hypothesis.
 
\begin{comment}
\mbox{(\ref{item:rigidplanar}) implies (\ref{item:rigidgeometric})}
follows from:
\begin{reftheorem}[{\cite[Theorem 1]{Ota92}}]
  Let $\underline{w}$ be an acylindrical multiword in $F$ in which all elements are indecomposable and such that the corresponding decomposition space is planar; then $\underline{w}$ is geometric.
\end{reftheorem}

\emph{Acylindrical} means that $F$ does not split over $\mathbb{Z}$ relative to the line pattern. 
The hypothesis that all elements of the multiword are indecomposable is too strong. It is used only to show that the decomposition space is connected with no cut points. 
\end{comment}


\mbox{(\ref{item:planarwh}) implies
  (\ref{item:rigidgeometric})} by \fullref{proposition:gooddiagramimpliesgeometric}.


\end{proof}


\begin{corollary}\label{corollary:rigidnotgimpliesnotvg}
  A rigid multiword is virtually geometric if and only if it is geometric.
\end{corollary}



Manning has shown \cite{Man09} that the word $w=bbaaccabc$ in
$F_3=\left<a,b,c\right>$ is not virtually geometric.
This is proved by showing that the structure of the Whitehead graph of
this word implies that the Whitehead graph in any finite cover is
still non-planar.

Alternatively, using the methods of \cite{CasMac11}  it is possible to
show $(F,w)$ is rigid, so
non-planarity of the Whitehead graph immediately implies the word is not
virtually geometric.

Similarly, we know by \fullref{proposition:completeisrigid} that a
multiword in $F_n$ whose Whitehead graph is the complete graph on the
$2n$ vertices is
rigid.
Such a Whitehead graph is planar if and only if $n=2$. 
When $n=2$ the
valence is 3, and any such multiword is geometric.
For any $n>2$ such a multiword is not virtually geometric.

\subsection{Non-rigid Multiwords and Virtual Geometricity}
In this section we prove the \hyperlink{vgthmtarget}{Characterization of Virtual Geometricity}\xspace, that virtually geometric
multiwords are those that are made up of geometric pieces.

We first prove the theorem in the special case that the decomposition
space has no uncrossed cut pairs, see \fullref{theorem:vgeq}.
This is the case where there are no ``missing curves'', or
equivalently, where they have been added to the multiword.

\fullref{theorem:vgeq} contains the interesting features of the proof
and avoids some technical complications.
Given \fullref{theorem:vgeq}, the proof of
\fullref{theorem:planarequalsvg} amounts to showing that adding in the
``missing curves'' does not change planarity of the decomposition space.


\begin{lemma}\label{theorem:vgeq}
Let $\multimot$ be a multiword in $F$ such that the decomposition
space $\decomp_{\multimot}$ is connected with no uncrossed cut pairs. Let $\Gamma$ be the rJSJ\xspace.
The following are equivalent:
\begin{enumerate}
\item The multiword $\multimot$ is virtually geometric.\label{item:vg}
\item The decomposition space $\decomp_{\multimot}$ is planar.\label{item:planar}
\item For every non-cyclic vertex group $G$ of the $\Gamma$, the
  induced multiword $\indg$ is geometric.\label{item:piecesgeometric}
\end{enumerate}
\end{lemma}
\begin{proof}
\fullref{corollary:vgimpliesplanar} shows \mbox{(\ref{item:vg})
  implies (\ref{item:planar})}.

Let $G$ be a non-cyclic vertex group of $\Gamma$.
Since there are no uncrossed cut pairs, $\multimot=\augw$, so
\fullref{lemma:embed} shows that the decomposition space
$\decomp_{\indg}$ embeds into $\decomp_{\multimot}$.
Thus,  if
$\decomp_{\indg}$ is non-planar then $\decomp_{\multimot}$ is non-planar
as well. 
Therefore, (\ref{item:planar}) implies that $\decomp_{\indg}$ is
planar.
If $(G,\decomp_{\indg})$ is a QH--surface then $\indg$ is geometric.
If $(G,\decomp_{\indg})$ is rigid then since it is planar 
\fullref{theorem:rigidvg} says it is geometric.
Thus, \mbox{(\ref{item:planar}) implies (\ref{item:piecesgeometric})}.

Now assume (\ref{item:piecesgeometric}).
From a graph of groups we may build a corresponding graph of spaces \cite{ScoWal79}.
For each vertex group choose a vertex space with fundamental group isomorphic
to the vertex group.
For each edge group choose a space with fundamental group isomorphic
to the edge group, and let the edge space be the product of that space
with the unit interval. 
Use the edge injections of the graph of groups to define attaching
maps of edge spaces to the corresponding vertex spaces. 
The resulting space will have fundamental group isomorphic to the
fundamental group of the graph of groups.

For each non-cyclic vertex group, the induced multiword is geometric,
so we can choose the vertex space to be a handlebody with an embedded
multicurve in the boundary representing the induced multiword.

For the edge spaces we could use annuli, but later we will want to
thicken them to make the resulting graph of spaces a 3--manifold.

For the moment we will also make a geometricity assumption on the
cyclic vertex groups. 
Suppose for a cyclic vertex group $\left<g\right>$ there are $k$ incident edges and
each edge injection is degree one. 
In this case we choose the vertex space to be a solid torus with $k+1$
disjoint curves on the boundary representing the element $g$ and $k$ attaching curves
to which we will glue a boundary curve of an annulus edge space. 

Another possibility is that the degrees of the edge injections are all
two except for possibly one of degree one. In this case we choose the
vertex space to be a solid Klein bottle, and again we have disjoint
curves on the boundary representing $g$ and the attaching curves.

Suppose one of these possibilities is true for every cyclic vertex group.

The resulting graph of spaces has fundamental group $F$ and has an embedded
multicurve representing $\multimot$ such that the multicurve is
disjoint from the edge spaces.
It  is not yet a 3--manifold with boundary; we need to
fatten the annuli. 
To see if this is possible, consider for each boundary component of
each annulus the tubular neighborhood of the attaching curve in the
boundary of the corresponding handlebody.
If for each annulus the two neighborhoods are either both annuli or both
M\"{o}bius strips then the annuli may be fattened to make the graph of
spaces a 3--manifold.
Now, a fattened annulus is composed of a 1--handle and a 2--handle, so this
does not explicitly give the resulting space a handlebody structure.
However, a graph of aspherical spaces is aspherical \cite{ScoWal79},
and a compact aspherical 3--manifold with free fundamental group is a
handlebody \cite{Hem76}, so $\multimot$ is geometric. 

Thus, assuming (\ref{item:piecesgeometric}), there are two possible obstructions to geometricity:
\begin{enumerate}
\item The degrees of the edge injections into some cyclic vertex group
  are not of one of the two forms described above.\label{ob:degree}
\item Some annulus can not be fattened because one boundary
  neighborhood is an annulus and the other is a M\"{o}bius strip. \label{ob:mobius}
\end{enumerate}

\begin{claim}
  These obstructions vanish in a finite index subgroup of $F$, so
  $\multimot$ is virtually geometric.
\end{claim}
\begin{claimproof}
There are finitely many elements $g_i\in\underline{w}$ such that
$\qmap(\{\inv{g}_i^\infty,g_i^\infty\})$ is a cut point in
$\decomp_{\underline{w}}$.

From the proof of \fullref{proposition:otalsplit}, an edge injection of degree greater
than one into a cyclic
vertex group $\left<g_i\right>$ occurs when the $g_i$--action permutes some
components of $\decomp_{\underline{w}}\setminus
\qmap(\{\inv{g}_i^\infty,g_i^\infty\})$.
There are only finitely many components, so there exists some minimal positive
power $a_i$ of $g_i$ such that the $g_i^{a_i}$--action fixes each
complementary component.

Additionally, if some edge incident to the $\left<g_i\right>$ vertex
attaches to a handlebody around a non-orientable handle, and if $a_i$ is odd, then consider $g_i^{2a_i}$.

Let $H_i$ be a finite index subgroup of $F$ in which $g_i^{a_i}$ (or $g_i^{2a_i}$) is basic.
Let $H$ be the finite index subgroup $\cap_iH_i$.
If we apply the \hyperlink{decompthmtarget}{Relative JSJ-Decomposition Theorem}\xspace to $H$ we get a graph of groups covering
the graph of groups decomposition for $F$. 
By construction, the smallest power of $g_i$ in $H$ is a multiple of
$g_i^{a_i}$, so all edge inclusions are degree one.
This takes care of obstruction (\ref{ob:degree}), and we can choose
all the cyclic vertex spaces to be solid tori.

Furthermore, we can take the vertex spaces to be handlebodies finitely
covering the original handlebodies. 
If some attaching curve in the original decomposition ran along a
M\"{o}bius strip then it runs along an even covering of the M\"{o}bius
strip in the covering handlebodies.
Thus, all attaching attaching curves have annulus neighborhoods, which
takes care of obstruction (\ref{ob:mobius}).
\end{claimproof}

Thus, \mbox{$(\ref{item:piecesgeometric})\implies(\ref{item:vg})$}.
\end{proof}


\begin{theorem}[\hyperlink{vgthmtarget}{Characterization of Virtual Geometricity}\xspace]\label{theorem:planarequalsvg}
The following are equivalent:
\begin{enumerate}
\item The multiword is virtually geometric.\label{item:vg2}
\item The decomposition space  is planar.\label{item:planar2}
\item For every non-cyclic vertex group of the rJSJ\xspace, the induced multiword is geometric.\label{item:piecesgeometric2}
\end{enumerate}
Thus, virtually geometric multiwords are exactly those that are built
from geometric pieces.
\end{theorem}


\begin{proof}

\fullref{corollary:vgimpliesplanar} shows \mbox{(\ref{item:vg})
  implies (\ref{item:planar})}.



\begin{claim}\label{planarclaim2}
If  $\decomp_{\multimot}$ is planar
 then for each non-cyclic vertex group $G$ of the rJSJ\xspace, the
 decomposition space
 $\decomp_{\indg}$ is planar.
\end{claim}

\begin{claimproof}
For each cut point $p\in\decomp_{\multimot}$ there is a unique component $\mathcal{C}_p$
of $\decomp_{\multimot}\setminus p$ such that the image of $\partial G$ is contained
in $\mathcal{C}_p\cup p$.
The intersection over all cut points of the sets $\mathcal{C}_p\cup p$
is a connected subset of $\decomp_{\multimot}$ containing the image of
$\partial G$.
Thus, we may assume that there are no cut points.

Now, let $\{g_1,\dots, g_k\}=\indg\setminus \multimot$, so that $\{\inv{g}_i^\infty, g_i^\infty\}$ gives an
uncrossed cut pair in $\decomp_{\multimot}$.
This means that $\decomp_{\indg}$ is a quotient of the image of $\partial G$ in $\decomp_{\multimot}$ obtained by identifying pairs of points
$\{h\inv{g}_i^\infty, hg_i^\infty\}$ for each $h\in G$ and $i=1,\dots,
k$.

Embed $\decomp_{\underline{w}}$ into $S^2$. We will show that there is
a monotone upper semi-continuous decomposition of $S^2$ whose
non-degenerate elements are arcs whose two endpoints are
$\{h\inv{g}_i^\infty, hg_i^\infty\}$ for some $h$ and $i$, and whose
interiors are disjoint from the image of $\partial G$ in $S^2$.
Moore's Decomposition Theorem \cite{Moo25} says that the quotient of
the sphere obtained by collapsing each of these arcs to a point is again
the sphere, and the image of $\partial G$ in this quotient is
$\decomp_{\indg}$.
Thus, $\decomp_{\indg}$ is planar.

A \emph{decomposition} of $S^2$ is just a way of writing $S^2$ as a
disjoint union of finite unions of compact continua. 
The \emph{non-degenerate} elements are the non-singletons.
The decomposition is monotone if each element is connected.
The collection is upper semi-continuous if for each element $A$
of the decomposition, and for each neighborhood $U$ of $A$, there
exists a neighborhood $V$ of $A$ such that any element of the
decomposition that meets $V$ is contained in $U$.
For the quotient to be the sphere we also require that the elements of the decomposition are
non-separating subsets.

Each uncrossed cut pair $\qmap_{\multimot}(\{h\inv{g}_i^\infty, hg_i^\infty\})$ has
finitely many complementary components in $\decomp_{\multimot}$, one
of which, $\mathcal{C}$,  contains the image of $\partial G$.
Choose a connected component of $S^2\setminus \decomp_{\multimot}$
that limits to $\qmap_{\multimot}(h\inv{g}_i^\infty)$ and $\qmap_{\multimot}(hg_i^\infty)$.
The boundary of this set is a Jordan curve passing through the two
points $\qmap_{\multimot}(h\inv{g}_i^\infty)$ and
$\qmap_{\multimot}(hg_i^\infty)$. 

Choose the arc connecting  $\qmap_{\multimot}(h\inv{g}_i^\infty)$ and $\qmap_{\multimot}(hg_i^\infty)$
in our decomposition 
to be the sub-arc of this curve that does not go through the component
of $\decomp_{\multimot}\setminus
\{\qmap_{\multimot}(h\inv{g}_i^\infty),\qmap_{\multimot}(hg_i^\infty)\}$ containing the image of
$\partial G$.

To satisfy the requirements of Moore's theorem, we must show that for
every arc $A$ and every neighborhood $U$ of $A$ there is a
neighborhood $V$ of $A$ such that if an arc $A'$ meets $V$
it is contained in $U$.

By construction, each interior point of each arc has a neighborhood
that is not entered by any other arc in the collection, but this is
not true for the endpoints of the arc.

Fix an arc $A$ and let $b$ be one of its endpoints.
Let $\hat{b}$ be the point of $\partial\tree$ that is the preimage
of this endpoint.

The identity element of $F$ gives a basepoint $*$ for the topology on
$\closure{\tree}$ in the sense that the basic open neighborhoods $N_r(\xi)$
of a
point $\xi\in\partial\tree$ are the subsets of $\closure{\tree}$
consisting of points $\eta$ such that the geodesic from $*$
to $\eta$ coincides with the geodesic from $*$ to $\xi$ for
at least distance $r$.
Let $R$ be the length of the longest subword common to two elements of
$\augw$ (as cyclic words).
In other words, $R$ is the maximum overlap of two different axes in
the line pattern.
If $l^+$ and $l^-$ are the two endpoints of a line $l$ in the line pattern, and
if one endpoint of another line $l'$
is in $N_{r+R}(l^+)$ for some $r>d_\tree(*,l)$, then the other
endpoint of $l'$ is in $N_r{l^+}$.
Furthermore, for every $r>d_\tree(*,l)$ it is also true that
$\qmap_{\multimot}^{-1}(\qmap_{\multimot}(N_{r+R}(l^+)))\subset
N_r(l^+)\cup\l^-$.

Thus, for any sufficiently large $r$ there
is an open neighborhood $U_r$ of $b$ in $\decomp_{\multimot}$
such that $\qmap_{\multimot}(N_{r+R}(\hat{b}))\subset U_r \subset
\qmap_{\multimot}(N_r(\hat{b}))$.

Let $U$ be any neighborhood of $A$. 
For each interior point $\gamma$ of $A$ there is a small
neighborhood $V_\gamma$
contained in $U$ that does not contain points of the other arcs. 
If $b$ is an endpoint of $A$ there is a small open
neighborhood of $b$ contained in $U$ of the form $U_r$ for some $r$, as in the
previous paragraph.
Let $V_b$ be a neighborhood of $b$ of the form $U_{r+2R}$.
Let $V$ be the union of the $U_a$ for all $a\in A$.
If any arc of the collection enters $V$ then it must enter
$V_b\subset U_{r+2R}$.
This implies that the entire arc is contained in
$\qmap_{\multimot}(N_{r+R}(\hat{b}))\subset U$.
\end{claimproof}

\fullref{planarclaim2}, \fullref{theorem:rigidvg} and the fact that
QH--surface multiwords are always geometric show that \mbox{(\ref{item:planar})
  implies (\ref{item:piecesgeometric2})}.


By \fullref{theorem:vgeq}, if the induced multiword in each non-cyclic vertex group of
the rJSJ\xspace is geometric in its vertex group then $\augw$ is virtually
geometric.
Clearly this implies $\multimot$ is virtually geometric;
it is a subset of $\augw$.
Just
take the graph of spaces
from \fullref{theorem:vgeq} and omit some of the curves running
around solid torus vertex spaces to get an embedded multicurve representing
$\multimot$.
Thus  \mbox{(\ref{item:piecesgeometric2})
  implies (\ref{item:vg})}.
\end{proof}

\begin{remark}
  An argument similar to that of \fullref{planarclaim2} may be used to
  see \fullref{theorem:structure} from a different viewpoint, at
  least for virtually geometric multiwords.
Let $\multimot$ be a multiword with connected, planar decomposition space.
Suppose there is an uncrossed cut pair with $k$ complementary
components. It is possible to embed a graph in $S^2$ with the two
points as vertices and $k$ edges such that the interiors of the edges
lie in the complement of the decomposition space, and such that
complementary components of the decomposition space lie in
complementary components of the sphere minus the graph. 
We can find a similar graph for all uncrossed cut pairs and cut
points, and make the collection upper semi-continuous.

In general, the quotient of the sphere obtained by collapsing all of these graphs
to points is a \emph{cactoid}, but in this case we can say more: it is
a tree of spheres whose tree structure mirrors that of the Bass-Serre
tree of the rJSJ\xspace.
The vertices of the Bass-Serre tree with non-cyclic stabilizers
correspond to spheres. The cyclic vertices correspond to points of
intersection of different spheres. The decomposition space of the
augmented multiword embeds into the tree of spheres in such a way that
the decomposition space of an induced multiword in a non-cyclic vertex
group of the rJSJ\xspace embeds into the appropriate sphere in the
tree of spheres, and the cut points of the decomposition space are
exactly points of intersection of multiple spheres.
\end{remark}



\endinput
