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\begin{document}

\title[Nonpositive immersions, sectional curvature, subgroup 
properties]{Nonpositive immersions, sectional curvature,
and subgroup properties}

\author{Daniel T. Wise}
\address{Department of Mathematics, McGill University, 
Montreal, Quebec, CA H3A 2K6, Canada}
\email{wise@math.mcgill.ca}
\thanks{Research supported by grants from FCAR and NSERC}
\subjclass[2000]{Primary
20F05, 
20F67, 
57M07, 
57M20} 

\keywords{Coherent groups, nonpositive curvature, one-relator groups}

\date{October 21, 2002}

\commby{Walter Neumann}

\begin{abstract}
This announcement describes research concerning local quasiconvexity,
coherence, compact cores, and local indicability
for fundamental groups of certain $2$-complexes.
\end{abstract}

\maketitle

\section*{Introduction}
Gromov's word-hyperbolic groups have had a profound impact on the theory
of infinite groups.
For some purposes, however, word-hyperbolicity is far too general
a property to work with. Perhaps the single most important
motivating example of a word-hyperbolic group is the fundamental
group of a genus~2 surface. And yet, general word-hyperbolic
groups fail to satisfy one of the most basic properties of
a surface group. Namely, every finitely generated subgroup of a
surface group is itself a surface group, but not every finitely
generated subgroup of a word-hyperbolic group is
word-hyperbolic or even finitely presented \cite{Rips82}.

A group is {\em coherent} if each of its finitely generated subgroups
is finitely presented.
A significant early result on coherent groups
proven independently by Scott and Shalen and applying ideas of Swarup,
is that the fundamental group of any $3$-manifold is coherent \cite{Scott73}.
There has recently been a resurgence of interest in the coherence property.
Most notably Feighn-Handel proved that free-by-cyclic groups are coherent
\cite{FeighnHandelCoherence}, and McCammond-Wise proved that
groups satisfying a certain `perimeter' condition are coherent 
\cite{McCammondWiseCoherence}.
The purpose of this announcement is to describe a variety of further results
concerning coherence and related properties.
In many ways, this is the beginning of a theory offering
a more faithful generalization of surface groups
than the theory of word-hyperbolic groups.

Section~\ref{sec:nonpositive} describes my
 results from \cite{WiseNonPositiveCoherence02},
where the {\em nonpositive immersion property} for $2$-complexes is introduced.
The fundamental group of a $2$-complex with nonpositive immersions
is coherent and locally indicable.
 This is the strongest result on coherent groups to date.
It subsumes the Feighn-Handel result on coherence of ascending HNN extensions
\cite{FeighnHandelCoherence}, it subsumes the coherence of finitely generated
3-manifold groups with cohomological dimension~2, and it appears likely to
 subsume the perimeter coherence results given in \cite{McCammondWiseCoherence}.

Section~\ref{sec:sectional} describes results from
\cite{WiseSectional02}, which introduces a new form of sectional curvature
for 2-complexes with angles at the corners of their 2-cells.
The 2-complexes with nonpositive sectional curvature
have the nonpositive immersion property, and thus have coherent
fundamental groups.
The compact $2$-complexes with negative sectional curvature
have the compact core property for covering spaces with finitely generated fundamental
group, and this facilitates further understanding of their subgroups.
We emphasize that the sectional curvature notion is a ``conformal'' theory
requiring angles but not requiring a metric.
However, when the 2-complex has negative sectional curvature
arising from a genuine metric of nonpositive curvature
one can prove that the fundamental group is actually locally
 quasiconvex. These results are parallel to the
local quasiconvexity theorem proven for certain small-cancellation groups
 in \cite{McCammondWiseCoherence}.

Finally, Section~\ref{sec:onerelator} describes significant progress
made in \cite{WisePositiveRelatorCoherence02} towards Baumslag's
long-standing conjecture on the coherence of one-relator groups.
I propose a graph-theoretical conjecture that would imply
the coherence of all one-relator groups.

\section{Nonpositive immersions}\label{sec:nonpositive}
We shall work in the category of combinatorial $2$-complexes
and combinatorial maps. An {\em immersion} is a local injection.
\begin{defn}
A $2$-complex $X$ has {\em nonpositive immersions} if for every
immersion $Y\rightarrow X$ with $Y$ compact, either $\euler(Y)\leq
0$ or $\pi_1Y$ is trivial.
\end{defn}

Our main theorem is the following surprising coherence result:
\begin{thm}\label{thm:nonpositive coherence}
If $X$ has nonpositive immersions, then $\pi_1X$ is coherent.
\end{thm}

A group $G$ is {\em locally indicable} if every nontrivial
finitely generated subgroup of $G$ has an infinite cyclic
quotient. The best known classes of locally indicable groups are
torsion-free one-relator groups  \cite{Brodskii80,Brodskii84} and
fundamental groups of irreducible 3-manifolds with boundary
\cite{Howie82}.
The following plays a role in the proof of
Theorem~\ref{thm:nonpositive coherence}.
\begin{thm}\label{thm:locally Z}
If $X$ has nonpositive immersions, then $\pi_1X$ is
locally indicable.
\end{thm}

The following is proven through a simple  counting argument utilizing the
 graph of spaces decomposition:
\begin{thm}\label{thm:ascending}
Let $X$ be the standard $2$-complex of an ascending HNN extensions
of a free groups. Then $X$ has nonpositive immersions.
\end{thm}

The following is proven by combining an asphericity result
on $3$-manifold spines \cite{Huebschmann81,CorsonTrace2000}
with the fact that the Euler
characteristic of a closed $3$-manifold is zero:
\begin{thm}
Every compact irreducible 3-manifold with boundary has a spine
with nonpositive immersions.
\end{thm}

One of the main tools employed in the
 proofs of the theorems presented here are {\em towers}
 (see \cite{Howie87}).
Towers first appeared explicitly in Papakyriakopolous's proof of
Dehn's Lemma \cite{Papakyriakopoulos57}. However, the tower method
is implicit in Magnus's original solution of the word-problem and
Freiheitssatz for one-relator groups \cite{Magnus30,Magnus32}.

\begin{defn}[Tower]
A {\em tower} $T\rightarrow X$ is a map which can be expressed as
a composition
\[T=\hat X_n\rightarrow X_n\hookrightarrow \cdots
\hookrightarrow \hat X_2\rightarrow X_2
\hookrightarrow \hat X_1\rightarrow X_1 = X,\]
where each $\hat
X_i\rightarrow X_i$ is a covering map, and each
$X_{i+1}\hookrightarrow \hat X_i$ is an inclusion.
\end{defn}

We can now briefly sketch the proofs of our Theorems \ref{thm:locally Z}~and~\ref{thm:nonpositive coherence}.

\begin{proof}[Sketch of Theorem~\ref{thm:locally Z}]
Let $\hat X$ be a covering space, and suppose $\homology^1(\hat
X)=0$. Let $Y\rightarrow \hat X$ be a $\pi_1$-surjection with $Y$
compact and $\homology^1Y=0$.
 Let $Y\rightarrow T \rightarrow \hat X$
be a  maximal tower lift of $Y\rightarrow \hat X$. Then
$Y\rightarrow T$ is $\pi_1$-surjective, and hence
$\homology^1T=0$, so $\euler(T)\geq 1$, and so $\pi_1T$ is
trivial. But then $\pi_1\hat X$ is trivial, since $T\rightarrow
\hat X$ is $\pi_1$-surjective.
\end{proof}


\begin{proof}[Sketch of Theorem~\ref{thm:nonpositive coherence}]
Consider a finitely generated subgroup $H\subset \pi_1X$ presented
by
\[\langle a_1,\dots, a_r\mid
R_1, R_2, \dots \rangle.\]
 For each $i$ there is a subpresentation
$P_i = \langle a_1,\dots, a_r\mid R_1,\dots, R_i\rangle$, and  a
maximal tower lift of $P_i\rightarrow X$ to $P_i\rightarrow T_i$.
In this way we obtain a sequence $T_1\rightarrow T_2 \rightarrow
\cdots$. We choose $m$ such that $\euler(T_m)$ is maximal, and then
$T_m\rightarrow X$ is $\pi_1$-injective, and hence a
$\pi_1$-isomorphism to $H$. This final step employs Howie's
Freiheitssatz for nonsingular systems of equations over locally
indicable groups \cite{Howie81}.
\end{proof}


\section{Sectional curvature for $2$-complexes}\label{sec:sectional}

We begin by describing a
combinatorial generalization of the Gauss-Bonnet theorem. This was
discovered in conjunction with a study of small-cancellation
theory \cite{McCammondWiseFanLadder}, but had actually
 been observed previously in \cite{BallmannBuyalo96}.
Previous versions of the Gauss-Bonnet theorem were concerned with
disc diagrams (i.e. singular surfaces) of one type or another
\cite{LS77,Sieradski84,GerstenReducible87,PrideHyperbolicComplexes88}.
We shall state the Combinatorial Gauss-Bonnet Theorem
after making the following definitions:

The $2$-complex $X$ is an {\em angled $2$-complex}
if a real number called an {\em angle} is assigned to each {\em corner}
of each $2$-cell of $X$. Equivalently, the angles are assigned to the edges of the links of $0$-cells
of $X$, and in an obvious sense, each $2$-cell and $0$-cell has a set of corners with
assigned angles.
The most natural angled $2$-complexes arise by identifying each $2$-cell with
a Euclidean polygon.
The metric, however, plays no role for many of the arguments,
and it is often useful to allow zero angles or even negative angles.
The curvature $\curvature(v)$ of a $0$-cell $v\in X^0$ is defined by
\begin{equation}\label{eq:vertex curvature}
\curvature(v) = 2\pi - \pi\euler(\link(v)\big) \ - \  \sum_{c\in \text{Corners}(v)} \angle c.
\end{equation}
For a $2$-cell $f$ with $|\boundary f|$ sides,
the curvature $\curvature(f)$ is defined by
\begin{equation}\label{eq:face curvature}
\curvature(f) = \big(|\boundary f|-2)\pi \ - \sum_{c\in \text{Corners}(f)} \angle c.
\end{equation}


\begin{thm}[Combinatorial Gauss-Bonnet Theorem]\label{thm:CGBT}
Let $X$ be an angled $2$-complex.
Then
\[
   2\pi\euler(X) \ = \
\sum_{v\in X^0} \curvature(v) \  +  \sum_{f \in \text{$2$-cells}(X)} \curvature(f).
\]
\end{thm}


Since formula~(\ref{eq:vertex curvature}) allows us
to measure the curvature of a $0$-cell in
an arbitrary angled $2$-complex, we are able to consider
the curvatures of arbitrary sections in $X$, and not just the planar sections.
This motivates the following:

\begin{defn}[Sectional curvature]
Let $X$ be an angled $2$-complex and let $x\in X^0$.
A {\em regular section} at $x$ is an immersion $(S,s)\rightarrow (X,x)$, where
$S$ is a $2$-complex, and $\link(s)$ is a connected
nonempty graph with no vertices of valence $\leq 1$.
We say $X$ has {\em nonpositive sectional curvature at $x$}
if for every regular section $(S,s)\rightarrow (X,x)$,
after pulling back the angles to corners of $S$,
the curvature $\kappa(s)$ is nonpositive.
We say that $X$ has {\em nonpositive sectional curvature}
if each $2$-cell of $X$ has nonpositive curvature
and $X$ has nonpositive sectional curvature at each $0$-cell.

The section $(S,s)\rightarrow (X,x)$ is a {\em planar section}
if $\link(s)$ is a cycle.
We say $X$ has {\em nonpositive planar sectional curvature} if
each $2$-cell has nonpositive curvature and
$\curvature(s)\leq 0$
for each $x\in X^0$ and each planar section $(S,s)\rightarrow (X,x)$.

We say $X$ has {\em negative $[$planar$]$ sectional curvature}
if $\curvature(s)<0$ for each $[$planar$]$ section $(S,s)\rightarrow (X,x)$.
Note that we permit the $2$-cells of $X$ to have curvature $\leq 0$ in this case.
\end{defn}


\subsection{ The weight test}
The nonpositive planar sectional curvature property is
a variation on a usual theme in geometric group theory,
and so we shall briefly review and compare similar notions.
An angled $2$-complex $X$ satisfies the Gersten-Pride
 {\em nonpositive weight test}
if for each $x\in X^0$, every immersed cycle in $\link(x)$ has
angle-length~$\geq 2\pi$, where the {\em angle-length} is the sum of
the angles of the edges traversed by the cycle
 \cite{GerstenReducible87,PrideHyperbolicComplexes88}.
 Similarly, the {\em negative weight test}  requires that the
 angle-length be $>2\pi$.
Among the many interesting weight test consequences is that $\pi_1X$
is word-hyperbolic provided that $X$ is compact and satisfies the negative
weight test.
Note that if $X$ has no negative angles, then $X$ satisfies the
nonpositive [negative] weight test if and only if it has
nonpositive [negative] planar sectional curvature.

The foremost examples of $2$-complexes satisfying the weight test
are piecewise Euclidean $2$-complexes with a metric of nonpositive curvature.
These are formed by gluing together a set of Euclidean polygons along their sides,
so that the gluing maps are isometries, and such that the nonpositive weight test
is satisfied. A finiteness condition is also necessary,
and we refer the reader to \cite{Paulin91} for a lucid and concise introduction to
metric spaces of nonpositive curvature,
and to \cite{BridsonHaefliger} for a more recent extensive treatment.


Because their angles are nonnegative,
piecewise Euclidean $2$-complexes have metrics of nonpositive curvature
exactly when they have nonpositive planar sectional curvature.
However, nonpositive sectional curvature is considerably more sensitive.
For instance, let $B_n$ denote a bouquet of $n$ circles, let $X_n=B_n\times B_n$
be the piecewise Euclidean $2$-complex whose $2$-cells
are unit squares, and let $x$ denote the unique $0$-cell in $X_n$.
Since $\link(x)$
is a complete bipartite graph and each angle is $\frac\pi2$, it is
easy to see that $X_n$ satisfies the nonpositive weight test.
However, $\curvature(x)>0$ if $n>2$, so
while $X_n$ has nonpositive planar sectional curvature, it does not
have nonpositive sectional curvature.

\subsection{Consequences of nonpositive and negative sectional curvature}
We first observe that
nonpositive sectional curvature
implies the nonpositive immersion property
because of the following:

\begin{thm}
Let $X$ have nonpositive sectional curvature.
Let $Y\rightarrow X$ be an immersion where $Y$ is compact and connected.
Then either $\euler(Y)\leq 0$ or $Y$ deformation retracts to a single $0$-cell.
\end{thm}
\begin{proof}[Sketch]
We first repeatedly collapse along free faces
to deformation retract to a subcomplex $Z$ of $Y$
such that $Z$ has no free faces and $\euler(Z)=\euler(Y)$.
Now suppose that $Z$ does not consist of a single $0$-cell,
and pull back the angles of $X$ to the corners of $Z$.
For each $z\in Z^0$, either $\link(z)$ consists of two or more vertices,
and so $\curvature(z)\leq 0$
or $\link(z)$ contains an edge, and so
$\curvature(z)\leq 0$ because of the nonpositive sectional curvature hypothesis.
Therefore $\euler(Z)\leq 0$ by Theorem~\ref{thm:CGBT}.
\end{proof}


Our two main results about $2$-complexes with {\em negative} sectional
curvature are Theorem~\ref{thm:negative sectional lq} below
and the following analogue of the compact core theorem
for $3$-manifolds:

\begin{thm}[Compact Core]\label{thm:negative sectional core}
Let $X$ have negative sectional curvature. Let $\hat X\rightarrow
X$ be a covering space such that $\pi_1\hat X$ is finitely generated. Then
there is a compact subcomplex $C\subset \hat X$ such that
$C\rightarrow \hat X$ induces a $\pi_1$-isomorphism.
\end{thm}

One immediate algebraic consequence of
 Theorem~\ref{thm:negative sectional core}
is that if $X$ has negative sectional curvature, then $\pi_1X$ is coherent.
The proof of Theorem~\ref{thm:negative sectional core}
is substantially less complex than the proof of Theorem~\ref{thm:nonpositive coherence}.
It would be desirable to have a similar direct proof of coherence from the
hypothesis of nonpositive sectional curvature.
However, there are $2$-complexes described in \cite{WiseNoCore}
that do not have the compact core property, and it is shown in
\cite{WiseSectional02} that angles can be assigned so that
these $2$-complexes have nonpositive sectional curvature.
Nevertheless, some of those angles are negative,
so there is hope of generalizing Theorem~\ref{thm:negative sectional core}
to the case where $X$ has nonnegative or positive angles.


We note that not every word-hyperbolic group is coherent.
The first counterexamples were small-cancellation groups found
 by Rips in \cite{Rips82}. Examples of compact $2$-complexes
 with metrics of negative curvature but with incoherent fundamental group
 were given in \cite{Wise98}.
In many respects, the fundamental groups of $2$-complexes
with negative sectional curvature
behave very much like those of genus~$2$ surfaces.
This is further supported by the following consequences of
Theorem~\ref{thm:negative sectional core}.

\begin{cor}[Deficiency]\label{cor:deficiency}
Let $X$ have nonpositive sectional curvature, and suppose each
$2$-cell has curvature $\leq -2\pi \epsilon <0$. Any subgroup of
$\pi_1X$ generated by $r$~elements has a presentation with at most
$\frac{r-1}{\epsilon}$ relators.
\end{cor}

In particular, Corollary~\ref{cor:deficiency} implies that all
sufficiently low rank subgroups are free.

\begin{cor}[Finitely many conjugacy classes]
Let $X$ be a compact angled $2$-complex with negative sectional curvature.
For each $r$, there are finitely many conjugacy classes of
noncyclic freely indecomposable subgroups of $\pi_1X$ generated by
$r$ elements.
\end{cor}


We now describe a more powerful result about $2$-complexes
with negative sectional curvature.
It has a metric conclusion and requires a stronger metric hypothesis.
Our actual result is more general but involves a technical
condition involving the existence of certain ``straight lines''
that we shall not state here.
\begin{thm}[Local Quasiconvexity]\label{thm:negative sectional lq}
Let $X$ be a compact $2$-complex with a piecewise Euclidean
metric of nonpositive curvature.

If $X$ has negative sectional curvature,
then $\pi_1X$ is locally quasiconvex.
\end{thm}

\subsection{Examples with nonpositive and negative sectional curvature}
The results described above suggest that the class
of spaces with negative sectional curvature is very limited.
However, this class coincides quite accurately with many of the
spaces that have traditionally occupied the central stage.
We shall now describe a variety of such examples.

By taking the dual $2$-complex to the ideal triangulation
obtained in  \cite{EpsteinPenner88}
we have the following:
\begin{thm}
Let $M$ be a finite volume cusped hyperbolic $3$-manifold.
Then $M$ has a spine with nonpositive sectional curvature.
\end{thm}

One of our most intriguing classes of examples relates
the usual nonpositive planar sectional curvature with
the most basic $3$-manifold property.

\begin{thm}
Let $X$ have nonpositive $[$negative$]$ planar sectional curvature.
Suppose that \ $\link(x)$ embeds in $S^2$ for all $x\in X$.
Then $X$ has nonpositive $[$negative$]$ sectional curvature.
\end{thm}

In fact, a more general version of the above theorem holds with
the assumption that  $\lowlink(x)$ be spherical, where
$\lowlink(x)$ is the subgraph of $\link(x)$ containing all
edges corresponding to corners with angle $<\pi$.
This leads to the following two well-known classes that fit into this framework.

The first yields a class of ascending HNN extensions of free groups
which are fundamental groups of $2$-complexes with nonpositive
sectional curvature. Note that despite Theorem~\ref{thm:ascending},
 it seems unlikely that every ascending HNN extension of a free group
arises as $\pi_1X$, where $X$ has nonpositive sectional curvature.
\begin{thm}
Let $B$ be a bouquet of circles, and $\phi \colon B\rightarrow B$
be an immersion. Let $X$ be the $2$-complex for the mapping torus
of $\phi$. Then $X$ has nonpositive sectional curvature.
\end{thm}


Adian's ``cycle-free'' presentations, which were studied in
\cite{Stallings87} and \cite{GerstenReducible87}, are
presentations of the form
\[\langle a_1, a_2,\dots \mid
U_1=V_1, \ U_2=V_2, \ \dots \rangle,\]
where $U_i$ and $V_i$ are positive words  for each $i$,
and the {\em ascending} and {\em descending links} are forests.
These are the largest subgraphs of the link containing only
the $a_i^{+1}$ and only the $a_i^{-1}$ vertices, respectively.
\begin{thm}
Let $X$ be the standard $2$-complex of an Adian presentation.
Then $X$ has nonpositive sectional curvature.
\end{thm}
In particular, it follows from Theorem~\ref{thm:locally Z}
that Adian groups are locally indicable and hence left-orderable by
\cite{BurnsHale72}, which answers a question raised in
\cite{Stallings87}. Following a suggestion of Stallings,
it was observed in \cite{GerstenReducible87} that Adian presentations
satisfy Sieradski's asphericity test \cite{Sieradski84},
which is equivalent to
the Gersten-Pride weight test where only $\pi$ and $0$ angles are allowed.
In fact, we prove that every $2$-complex satisfying Sieradski's weight test
has nonpositive sectional curvature!

Our final family of examples of $2$-complexes gives some further sense of
the scope of nonpositive sectional curvature.
We say $X$ is a $(p,q,r)$ complex provided each $2$-cell has $\geq p$
sides, each cycle in $\link(x)$ has length~$\geq q$ for all $x\in
X^0$, and each $1$-cell appears $\leq r$ times on the boundary of
$2$-cells.
\begin{thm}
Let $X$ be a $(p,q,r)$ complex with $p\geq \frac{q}{q-2}r$. Assign
an angle of $\frac{(p-2)}{p}\pi$ to each corner. Then $X$ has
nonpositive curvature, and $X$ has negative curvature if the
inequality is sharp.
\end{thm}




\section{one-relator groups}\label{sec:onerelator}
Gilbert Baumslag raised the following in \cite{BaumslagProblems74}:
\begin{conj}\label{conj:one-relator coherence}
Every group defined by a presentation with a single defining relation is coherent.
\end{conj}

Let $X$ be the standard $2$-complex of a one-relator presentation
$\langle a_1, a_2, \dots \mid W \rangle$,
and suppose that $W$ is freely and cyclically reduced.
If $W$ is a proper power, then let $\hat X$ be a based cover corresponding
to a torsion-free finite index subgroup \cite{FischerKarrassSolitar72},
and let $X'$ be an aspherical $\pi_1$-isomorphic subcomplex of $\hat X$
obtained by removing duplicate $2$-cells.
By applying Theorem~\ref{thm:nonpositive coherence}
we hope to prove Conjecture~\ref{conj:one-relator coherence} by showing that:
\begin{conj}\label{conj:onerelatornonpositive}
If $W$ is not a proper power, then $X$ has nonpositive immersions.

If $W$ is a proper power, then $X'$ has nonpositive immersions.
\end{conj}

This conjecture has a lovely formulation in terms of
certain paths in a {\em labeled oriented graph}
$\Gamma$ which is a connected graph whose edges are directed
and labeled by letters in some alphabet set $\{a_1, a_2, \dots \}$
and which satisfies the following condition:
No vertex has two incoming edges with the same label,
and no vertex has two outgoing edges with the same label.
A combinatorial path $P\rightarrow \Gamma$ starting at some vertex
$v$ determines a word in $\{a_1^{\pm1}, a_2^{\pm1}, \dots\}$.
Conversely, a word $W$ in $\{a_1^{\pm1}, a_2^{\pm1}, \dots \}$ determines
at most one path $P$ starting at a vertex $v$.
A $W$-cycle is a closed path corresponding to the word $W$.
If $W=V^n$, then we shall not distinguish between two $W$-cycles if their starting
vertices are connected by a path corresponding to a word $V^i$ for some $i$.
For instance, the graph in Figure~\ref{fig:Wcycles} has exactly
\begin{figure}\centering
\includegraphics[width=.8\textwidth]{era105el-fig-1.eps}
\caption{}
\label{fig:Wcycles}
\end{figure}
three distinct $a^2$-cycles, no $b$-cycles,  two $b^4$-cycles, three $b^{12}$-cycles,
 three $aaba^{-1}b^{-1}$-cycles,
and no $ab$-cycles.



\begin{conj}\label{conj:W cycles in graph}
Let $\Gamma$ be a labeled oriented graph.
Then the number of $W$-cycles in $\Gamma$ is at most
$\betti{\Gamma}$. In particular,
when $\Gamma$ is connected, the number of $W$-cycles
is bounded by $1+e-v$, where $e$ and $v$ are the numbers
of edges and vertices in $\Gamma$.
\end{conj}

Here are two additional algebraic formulations in terms of
subgroups of free groups.
\begin{conj}\label{conj:positive rank 1 hanna neumann}
Let $H$ be a subgroup of $F$,
and let $Z$ be a cyclic subgroup of $F$.
Then the number of distinct conjugates of $Z$
that intersect $H$ nontrivially is bounded by $\rank(H)$.
\end{conj}


Readers familiar with Walter Neumann's generalized Hanna Neumann conjecture
(see \cite{WNeumann89,Dicks94})
will note that Conjecture~\ref{conj:positive rank 1 hanna neumann}
bears intriguing similarity with a `rank~1 variant'.
Unfortunately, the Hanna Neumann conjecture carries no information
when one of the subgroups has rank~$1$, and the standard approach
towards the Hanna Neumann conjecture does not
immediately lead to a proof of Conjecture~\ref{conj:positive rank 1 hanna neumann}.
Nevertheless, hopefully some fruitful connection will
eventually be realized.

A subgroup $H\subset F$ is {\em isolated} if $g^n\in H-\{1_F\} $ implies that $g\in H$.
When $W$ is not a proper power,
Conjecture~\ref{conj:W cycles in graph} can also be interpreted as follows:
\begin{conj}\label{conj:multiple conjugate intersection}
Let $H$ be an isolated rank~$r$ subgroup of a free group $F$. Let $H_1,
\dots, H_{r+1}$ be distinct conjugates of $H$ in $F$. Then
$\bigcap_{i=1}^{r+1} H_i = \{1_F\}$.
\end{conj}


Thus far, we have been able to prove
 Conjecture~\ref{conj:onerelatornonpositive}
in several cases.
For example:

\begin{thm}\label{thm:torsion-free solvable}
Suppose the one-relator group $\langle a_1, a_2, \dots \mid W\rangle $
 is residually torsion-free solvable.
Then $X$ has nonpositive immersions.
\end{thm}



A {\em positive} one-relator group is a one-relator group
such that for some generator $a_i$, the relator $W$ contains
$a_i$ but does not contain $a_i^{-1}$.
Of course, the motivating case is when $W$ is a positive word
in the generators.
\begin{thm}
Let $X$ be the standard $2$-complex of a positive one-relator group.
Then $X$ $[$respectively $X']$ has nonpositive immersions.
\end{thm}

As for many results in the theory of one-relator groups,
the  proofs employ staggered $2$-complexes \cite{Howie87}, and the analogous
theorems actually hold for staggered $2$-complexes.



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\end{thebibliography}
\end{document}



\bibliographystyle{plain}

\bibliography{C:/papers/wise}


First COMMENT
\begin{com}
I think we don't need to assume that $Z$ is maximal, because
this is covered by "distinct conjugates".
This wouldn't work if we were conjugating in $H$ instead. Neat.
\end{com}

SECOND COMMENT:
\begin{com}
Check this reference. It isn't exactly where I remembered it.
\end{com}

THIRD COMMENT:
\begin{com}
For $1\leq i\leq n$ we can attach a $b^i$ arc to an $a^{2n}$ circle.
The result is a rank $r=n+1$ subgroup $H$ such that
$\cap_{i=1}^{2n} H^{a^i} =\langle a^{2n}\rangle$.

Perhaps the general statement should be that
one of the conjugators normalizes the intersection of conjugates!
 \end{com}

FOURTH COMMENT:
\begin{com}
The sequence of abelianizations that make the $2$-cells embed,
Gives a free abelian tower.
Maybe this extends to residually solvable. I don't know.
\end{com}