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% Structure of Inifnite Locally-Finite 2-connected Graphs
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% by Carl Droms, Brigitte Servatius and Herman Servatius
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(1995), \#R17 \hfill}
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\begin{document}
\begin{center}
{\Large\bf The Structure of Locally Finite Two-Connected Graphs}\\[5mm]
{\bf Carl Droms\footnote{Supported by a James Madison
University Faculty
Summer Research grant.},
Brigitte Servatius} and
{\bf Herman Servatius}\\[8mm]
Submitted: May 15, 1995; Accepted: September 4, 1995.\\[12mm]
\end{center}
\date{{\normalsize Submitted: May 15, 1995; Accepted: September 4, 1995.}}
\begin{abstract}
We expand on Tutte's theory of $3$-blocks for
$2$-connected graphs, generalizing it to apply to infinite,
locally finite graphs, and giving necessary and sufficient
conditions for a labeled tree to be the $3$-block tree of a
$2$-connected graph.
\end{abstract}
\medskip
\noindent {\small\em Mathematics Subject Classification:} 05C40, 05C38, and 05C05.
\noindent {\small\em Key Words and Phrases:}
Tutte connectivity, $3$-block, $3$-block tree
\medskip
\section{Introduction}
Connectivity properties
of graphs
are among the basic aspects of graph theory.
Every graph is the disjoint union of its connected components,
and every connected graph is
the edge disjoint union of its maximal $2$-connected subgraphs,
encoded in the block-cutpoint tree.
A canonical decomposition for finite $2$-connected graphs was given by
Tutte~\cite{Tutte1} in the form of the $3$-block tree,
and generalized to matroids by
Cunningham and Edmonds~\cite{CunninghamEdmonds}.
Such decompositions are important tools in inductive arguments and
constructions.
Hopcroft and Tarjan~\cite{HopcroftTarjan} gave an important
algorithm for computing
the $3$-block tree of a graph in $O(V + E)$ time,
which is comparable to the
complexity of computing other non-canonical decompositions, say the ear
decomposition, and is also applicable to matroids.
Effective decompositon schemes for graphs of connectivity $3$ and higher
have been given, but none are canonical, and
in Section~\ref{HigherSect} we argue that none will be forthcoming.
The uniqueness of Tutte's construction may be exploited to study the
symmetry properties of graphs with low connectivity,
\cite{SS95} and~\cite{DSS95a}, particularly in the case of planar
graphs~\cite{DSS95b}.
In this paper we will examine the interpretation of
Tutte's decomposition and extend the theory to infinite graphs.
\section{$n$-Connectivity}
We are concerned with the structure of locally finite graphs of
connectivity less than $4$, allowing graphs to have loops and
multiple edges.
If a graph $G$ has at least $3$ vertices and is not
a triangle, then $G$ is defined to be {\em $n$-connected}
if $G$ has girth
at least $n$ and
any two vertices of $G$ are joined by $n$ internally disjoint paths.
To avoid uninteresting special cases,
the connectivities of the small graphs in Figure~\ref{DumbyFig} are
said to be infinite.
\begin{figure}[htb]
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\caption{Infinitely connected graphs.\label{DumbyFig}}
\protect
\end{figure}
This definition of $n$-connectivity,
also known as {\em Tutte connectivity,}
was introduced in~\cite{Tutte2},
see also~\cite{Oxley}, and differs for simple graphs from the usual
definition of $n$-connectivity only for $n > 3$, with the exception of the
special cases in Figure~\ref{DumbyFig}, and has the advantage of being
generalizable to matroids,~\cite{Tutte3},
and being invariant under dualization.
We call a graph on two vertices connected by parallel edges a {\em
multilink}. Note, that with the above definition, multilinks on more than
three edges are only 2--connected.
For any graph $G$ , there is an equivalence
relation defined on the edges of $G$ by setting $e \similar e'$ if
$e$ and $e'$ lie on a common cycle, or $e=e'$. The
equivalence classes are either single edges or
induce maximal $2$-connected subgraphs of
$G$, and are called the {\em blocks} (or {\em $2$-blocks})
of $G$. The {\em block--cutpoint tree} is the graph defined
on the union of the set of blocks and
the set of cutpoints, with a cutpoint adjacent to each of the
blocks to which it belongs.
It is obvious that this graph is a tree.
In general, one might hope that any $k$--connected graph
decomposes as the union of subgraphs which are either
$k+1$--connected or have at most $k$ vertices. However, this
is not the case, since a $2$-connected graph
may have no non-trivial $3$-connected subgraphs at all,
as in
Figure~\ref{AmalgFig}.
In the interest of analogy, therefore,
it is preferable to regard a block--cutpoint tree as an
encoding of the instructions for assembling a $1$-separable graph
from simpler pieces using the operation of vertex--union.
Let $A$ and $B$ be graphs and suppose there are functions
$f_A:e \rightarrow A$ and
$f_B:e \rightarrow B$, where $e$ is a graph consisting of
two vertices $1$ and $2$ joined by an edge which we will also
call $e$.
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\caption{Edge amalgamation.\label{AmalgFig}}
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Define the {\em edge amalgam of $A$ and $B$ over $f$}---denoted
$A \join{f} B$---to be the
graph obtained from the disjoint union of $A$ and $B$ by
identifying vertex $f_A(1)$ with $f_B(1)$, vertex $f_A(2)$ with
$f_B(2)$, and erasing the edges $f_A(e)$ and $f_B(e)$,
see Figure~\ref{AmalgFig}.
The edge amalgam, also called 2--sum in~\cite{Oxley},
is analogous to the symmetric difference of sets, and
in $A \join{f} B$ we can regard the graph $B - f_B(e)$ as taking
the place of $f_A(e)$ in $A$, and vice versa.
Whenever possible, we will indicate the edge functions $f_A$ and
$f_B$ by simply labeling the edges $f_A(e)$ and $f_B(e)$ the same in
$A$ and $B$, in which case we write $A \join{e} B$ or just
$A \join{} B$.
It is clear that
$A$ and $B$ are $2$-connected if and only if
$A \join{f} B$ is $2$-connected.
The edge $f_A(e)$ is called an {\em amalgamated edge} of $A$.
If $X$ denotes the subgraph of $A$ obtained by erasing the
amalgamated edge, $X$ is also a subgraph of $A \join{f} B$ and
we write $A = \Bar{X}$.
Edge amalgamation is a commutative operation, so
$A \join{f} B \iso B \join{f} A$.
but it is not in general associative,
since $ (A \join{f} B) \join{g} C$ may be rewritten as
$ A \join{f} (B \join{g} C ) $
only if $g_{A+B}(e)$ belongs to $B$.
If both expressions are defined, they are equal, and
note that, since the amalgamated edges are erased, it must be
true that $f_B(e) \not = g_B(e)$. Since the associativity
is conditional, it is not in
general possible to simply ignore the parentheses in a long
expression, in particular whenever some term has more than two
amalgamated edges. A more convenient notation for the result of a
sequence
of edge amalgamations, therefore, is a
labeled tree in which the nodes are labeled with graphs, and
the edges are labeled with the two functions indicating which edges
are amalgamated in the endpoint graphs.
Necessarily, the
amalgamating edges at any node of this tree must be distinct.
We call such a labeled tree $\Tree$ an {\em edge amalgam tree}, and let
$G(\Tree)$ denote the graph obtained from the disjoint union of
the node labels by amalgamating along the edges determined by the
edge functions.
To avoid confusion between $G(\Tree)$ and $\Tree$,
we will hereafter refer to the vertices and edges of
an edge amalgam tree as {\em nodes} and {\em links}, and denote them with
Greek as opposed to Roman letters.
For a finite edge amalgam tree $\Tree$, it is clear that $G(\Tree)$
is $2$-connected if and only if the node graphs are $2$-connected,
and that
$G(\Tree)$ is locally finite if and only if the node
graphs are locally finite.
For infinite trees, however, neither is the case, as can be seen
from Figures~\ref{Rotten2Fig} and~\ref{Rotten1Fig}.
Accordingly, we shall give conditions under which an edge amalgam tree
defines a locally finite $2$-connected graph.
\section{The $3$-block tree}
A graph $G$ is said to be a
{\em $3$-block\/}
if it contains at least three edges and is either
a circuit, a finite multilink, or a simple, locally finite
$3$-connected graph.
Let $\Tree$ be a countable edge amalgam tree.
We call $\Tree$ a {\em $3$-block tree\/}
if the following conditions are satisfied:
\begin{enumerate}
\item If $\{\alpha,\beta\} \in E(\Tree)$ then
$G_\alpha$ and $G_\beta$ are not both circuits,
nor are they both multilinks.
\item If
$\eta = \{\alpha ,\beta\}$ is a link in $\Tree$ amalgamating
an edge $e$ in $G_\alpha$,
then
there is a finite subtree $\Tree' \leq \Tree$ with
$\alpha \in \Tree'$ and $\beta \not \in \Tree'$,
and a path in $G(\Tree')$ joining the endpoints of $e$
which is made up entirely of edges unamalgamated
in $G(\Tree)$.
\item For each vertex $v$ of $G_\alpha$,
$\alpha$ is contained in a finite subtree
$\Tree_{v} < \Tree$ such that
$\starr{v}$ in $G(\Tree_{v})$ consists entirely of
edges unamalgamated in $\Tree$.
\end{enumerate}
We impose condition~1 since circuits and multilinks would
otherwise have many inequivalent $3$-block trees.
Condition~2 avoids the situation of Figure~\ref{Rotten2Fig}
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in which each amalgamation increases the length of a circuit, resulting
in a graph with cut vertices.
Condition~3 insures local finiteness, disallowing amalgamations
such as in Figure~\ref{Rotten1Fig}.
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\put(130.00,40.00){\makebox(0,0)[cc]{$+ \cdots$}}
\put(102.33,40.00){\makebox(0,0)[cc]{+}}
\put(77.67,40.00){\makebox(0,0)[cc]{+}}
\put(52.67,40.00){\makebox(0,0)[cc]{+}}
\put(27.67,40.00){\makebox(0,0)[cc]{+}}
\put(115.00,10.00){\makebox(0,0)[cc]{$\cdots$}}
\put(15.00,10.00){\makebox(0,0)[cc]{$\cdots$}}
\end{picture}
\caption{\label{Rotten1Fig}}
\end{figure}
Note that we do not require $\Tree$ itself to be locally finite;
for instance,
one may amalgamate a triangle to each of the edges of
an infinite locally finite $3$-connected graph (effectively
subdividing each edge). In order to describe this graph as $G(\Tree$), $\Tree$
must be an infinite star.
Note also that, while the edges of $G(\Tree)$ are
partitioned according to the $3$-blocks to which they belong,
all edges of a particular $3$-block may be amalgamated, in which case
the $3$-block will not correspond to any collection of edges of
$G(\Tree)$.
\begin{lemma} If $\Tree'$ is a finite subtree of $\Tree$, then $G(\Tree')$
is homeomorphic to a subgraph of $G(\Tree)$.
\end{lemma}
\begin{proof} Every edge in $G(\Tree')$ which is not in $G(\Tree)$ is
amalgamated, so
replace each with the path of unamalgamated edges
in $G(\Tree) - G(\Tree')$ which is
guaranteed to exist by condition~2.
\end{proof}
\begin{lemma}
For any $3$-block tree $\Tree$,
$G(\Tree)$ is locally finite and $2$-connected.
\end{lemma}
\begin{proof}
$G(\Tree)$ is locally finite by condition 3.
Clearly, any {\em finite\/} $3$-block tree
represents a $2$-connected graph. Let $v$ and $w$ be distinct vertices
of $G(\Tree)$, and suppose they correspond to vertices $v'\in G_\alpha$
and $w'\in G_\beta$. Let $\Tree'$ be any finite subtree of $\Tree$
containing $\alpha$ and $\beta$. Then $G(\Tree')$ is $2$-connected, so there
are two internally disjoint paths in $G(\Tree')$ from $v'$ to $w'$. Since
$G(\Tree')$
is homeomorphic to a subgraph of $G(\Tree)$, there are two such paths
joining $v$ and $w$ in $G(\Tree)$, as well.
\end{proof}
It is our aim to prove
\begin{theorem}\label{MainTh}
Any locally finite $2$-connected graph $G$ corresponds to a
unique $3$-block tree $\Tree$.
\end{theorem}
We will prove Theorem~\ref{MainTh} in two steps. In the next section, we will
show how to
construct a particular $3$-block tree for $G$, and in the one following,
we show that this tree is unique.
\section{Existence of $3$-block trees.}
Given a pair of vertices $\{a,b\}$ in a graph $G$,
there is an equivalence relation
defined on the edges of $G$ by setting $e$ and $e'$ equivalent if
there is a path in $G$ containing both $e$ and $e'$, and which has
neither $a$ nor $b$ as an internal vertex.
The subgraphs of $G$ which carry the
equivalence classes are called the {\em \bridges} of $G$ with
respect to $\{a,b\}$. Two distinct bridges clearly can intersect only
at $a$ and $b$, and if $G$ is $2$-connected, their intersection
is exactly $\{a,b\}$. $G$ may thus be regarded as the union
of its bridges along $\{a,b\}$.
However, these bridges may not be themselves $2$-connected.
If a bridge with respect to $\{a,b\}$ in a $2$-connected graph
$G$ consists of more than a single edge,
then adding a new
edge connecting $a$ and $b$ to that bridge does give a
$2$-connected graph, and we call these graphs the
{\em \branches} of $G$ with respect to $\{a,b\}$.
The {\em deletion of $e$ from $G$}, denoted $G-e$, is the graph
obtained from $G$ by deleting the edge $e$, but not its
endpoints. The {\em contraction of $e$ in $G$}, denoted
$G \contract e$, is obtained from $G-e$ by identifying the endpoints
of $e$.
Let $G$ be a locally finite $2$-connected graph. Choose an arbitrary
edge $e$ of $G$, and let $a$ and $b$ be its endpoints.
We will construct a $3$-block tree $\Tree_e$ with $G(\Tree_e)=G$.
We call $e$
{\em undeletable} if $G - e$ is $1$-separable, {\em incontractible} if $G
\contract e$ is $1$-separable, and {\em ordinary} otherwise.
An edge $e$ cannot be both undeletable and incontractible. since the first
requires that every cycle containing the endpoints of $e$ also contain $e$,
while the second insures two vertices $u$ and $v$ so that every path from
$u$ to $v$ passes through one of the endpoints of $e$, so neither
the two disjoint
paths from $u$ to $v$ could contain $e$.
Note that
if $\Tree$ is a $3$-block tree, then condition 3, and the fact that
edge amalgamation does not destroy cutpoints,
guarantees that if the edge
$e\in G(\Tree)$ belongs to the $3$-block $G_\alpha$, then its
type---undeletable, incontractible or ordinary---in $G(\Tree)$
is the same as its type in $G_\alpha$.
We begin the construction of $\Tree_e$ by defining $\Tree_0$ to consist of a
single vertex $\alpha_0$ labeled
``$G$,'' with distinguished edge $e$. Then $\Tree_0$ is an edge amalgam tree
(though {\em not\/} a $3$-block tree) and $G=G(\Tree_0)$.
To define $\Tree_1$, we perform what we will call a {\em simple expansion
of the vertex $\alpha_0$ at the edge $e$;} that is, we will replace
$\alpha_0$ with a star which is an edge--amalgam tree for $G_{v_0}$ (though,
again, not a $3$-block tree.)
First, we define the graph $B_e$---the $3$-block of $G$ containing $e$.
This will be the label of the central node of $\Tree_1$. We consider three
cases, depending on the type of the edge $e$.
\begin{itemize}\renewcommand{\labelitemi}{\hspace{\parindent}}
\item \underline{\bf $e$ undeletable.}
In this case, $G-e$ is not $2$-connected, and its block--cutpoint tree is a
non-trivial path $A_1, v_1, A_2,\ldots,v_{k-1}, A_k$,
where the vertices $A_i$ correspond to blocks of $G-e$ and the
$v_i$ are the cut vertices of $G-e$.
Set $v_0$ = $a$ and $v_{k} = b$. Let $B_e$ be the $(k+1)$--cycle with vertices
$v_0,v_1,\ldots,v_k$, and let $\Bar{e_i}$ denote the edge of $B_e$ joining
$v_i$ and $v_{i+1}$ (subscripts modulo $n$.) If some $A_i$ consists of more than
a single edge, we define $\Bar{A_i}$ to be the graph obtained from $A_i$ by
adding an edge $\Bar{e_i}$ joining $v_i$ and $v_{i+1}$.
$G$ is then obtained by amalgamating each $\Bar{A_i}$
to $B_e$ along $\Bar{e_i}$, identifying any unamalgamated edges $\Bar{e_i}$
with the corresponding edge in $G$ (see Figure~\ref{UndeletableFig}.)
\begin{figure}[hbt]
\centering
\unitlength=0.80mm
\begin{picture}(101.00,40.00)
\put(5.00,15.00){\line(2,1){20.00}}
\put(25.00,25.00){\line(-4,1){20.00}}
\put(5.00,30.00){\line(0,-1){15.00}}
\put(5.00,15.00){\line(1,2){5.00}}
\put(10.00,25.00){\line(-1,1){5.00}}
\put(10.00,25.00){\line(1,0){15.00}}
\put(30.00,5.00){\line(2,3){10.00}}
\put(40.00,20.00){\line(-3,1){15.00}}
\put(5.00,0.00){\line(0,1){15.00}}
\put(5.00,0.00){\circle{2.00}}
\put(5.00,15.00){\circle{2.00}}
\put(5.00,30.00){\circle{2.00}}
\put(10.00,25.00){\circle{2.00}}
\put(25.00,25.00){\circle*{2.50}}
\put(30.00,5.00){\circle*{2.50}}
\put(40.00,20.00){\circle{2.00}}
\put(25.00,25.00){\line(1,-2){5.00}}
\put(30.00,15.00){\line(0,-1){10.00}}
\put(30.00,15.00){\line(2,1){10.00}}
\put(30.00,15.00){\circle{2.00}}
\put(60.00,15.00){\line(2,1){20.00}}
\put(60.00,0.00){\line(0,1){15.00}}
\put(60.00,0.00){\circle{2.00}}
\put(60.00,15.00){\circle{2.00}}
\put(80.00,25.00){\circle{2.00}}
\put(85.00,5.00){\circle{2.00}}
\put(60.00,0.00){\line(5,1){25.00}}
\put(85.00,5.00){\line(-1,4){5.00}}
\put(55.00,20.00){\line(2,1){20.00}}
\put(75.00,30.00){\line(-4,1){20.00}}
\put(55.00,35.00){\line(0,-1){15.00}}
\put(55.00,20.00){\line(1,2){5.00}}
\put(60.00,30.00){\line(-1,1){5.00}}
\put(60.00,30.00){\line(1,0){15.00}}
\put(55.00,20.00){\circle{2.00}}
\put(55.00,35.00){\circle{2.00}}
\put(60.00,30.00){\circle{2.00}}
\put(75.00,30.00){\circle{2.00}}
\bezier{104}(55.00,20.00)(70.00,20.00)(75.00,30.00)
\put(90.00,10.00){\line(2,3){10.00}}
\put(100.00,25.00){\line(-3,1){15.00}}
\put(85.00,30.00){\circle{2.00}}
\put(90.00,10.00){\circle{2.00}}
\put(100.00,25.00){\circle{2.00}}
\put(85.00,30.00){\line(1,-2){5.00}}
\put(90.00,20.00){\line(0,-1){10.00}}
\put(90.00,20.00){\line(2,1){10.00}}
\put(90.00,20.00){\circle{2.00}}
\put(90.00,10.00){\line(-1,4){5.00}}
\put(5.00,0.00){\line(5,1){25.00}}
\put(2.50,8.00){\makebox(0,0)[cc]{$e$}}
\put(52.50,8.00){\makebox(0,0)[cc]{$e$}}
\put(45.00,10.00){\makebox(0,0)[cc]{$\longrightarrow$}}
\end{picture}
\caption{Simple expansion at an undeletable edge.\label{UndeletableFig}}
\end{figure}
{\bf Remark 1.} If $A_i$ is not a single edge then it is $2$-connected
(since it is a block of $G-e$), so the edge $\Bar{e_i}$ is not undeletable in
$\Bar{A_i}$.
\item \underline{\bf $e$ incontractible.}
Let $A_1, \ldots, A_k$ denote the bridges of the endpoints of $e$; there are
only finitely many of them, since $G$ is locally finite. Then
$k\geq 3$ and one of the $A_i$ is
the edge $e$ itself. If $A_i$ consists of more than a single edge, let
$\Bar{A_i}$ denote the graph $A_i$ with a new edge $\Bar{e_i}$ joining
$a$ and $b$.
If $A_i$ is a single edge, set $\Bar{e_i} = A_i$.
Let $B_e$ denote the multilink consisting of $e$ together with all the
edges
``$\Bar{e_i}$.''
$G$ is again the amalgam of $B_e$ with the graphs $\Bar{A_i}$ along the edges
$\Bar{e_i}$ (see Figure~\ref{IncontractibleFig}.)
\begin{figure}[hbt]
\centering
\unitlength=0.80mm
\begin{picture}(121.00,35.00)
\put(50.00,15.00){\makebox(0,0)[cc]{$\longrightarrow$}}
\put(20.00,0.00){\line(0,1){30.00}}
\put(20.00,30.00){\line(-2,-3){10.00}}
\put(10.00,15.00){\line(-1,0){10.00}}
\put(0.00,15.00){\line(4,3){20.00}}
\put(20.00,0.00){\line(-2,3){10.00}}
\put(0.00,15.00){\line(4,-3){20.00}}
\put(20.00,0.00){\line(4,3){20.00}}
\put(40.00,15.00){\line(-4,3){20.00}}
\put(20.00,30.00){\line(2,-3){10.00}}
\put(30.00,15.00){\line(1,0){10.00}}
\put(30.00,15.00){\line(-2,-3){10.00}}
\put(0.00,15.00){\circle*{2.00}}
\put(10.00,15.00){\circle*{2.00}}
\put(20.00,30.00){\circle*{2.00}}
\put(20.00,0.00){\circle*{2.00}}
\put(30.00,15.00){\circle*{2.00}}
\put(40.00,15.00){\circle*{2.00}}
\put(80.00,0.00){\line(0,1){30.00}}
\put(80.00,30.00){\line(-2,-3){10.00}}
\put(70.00,15.00){\line(-1,0){10.00}}
\put(60.00,15.00){\line(4,3){20.00}}
\put(80.00,0.00){\line(-2,3){10.00}}
\put(60.00,15.00){\line(4,-3){20.00}}
\put(60.00,15.00){\circle*{2.00}}
\put(70.00,15.00){\circle*{2.00}}
\put(80.00,30.00){\circle*{2.00}}
\put(80.00,0.00){\circle*{2.00}}
\put(90.00,0.00){\line(0,1){30.00}}
\put(90.00,30.00){\circle*{2.00}}
\put(90.00,0.00){\circle*{2.00}}
\put(100.00,0.00){\line(0,1){30.00}}
\put(100.00,0.00){\line(4,3){20.00}}
\put(120.00,15.00){\line(-4,3){20.00}}
\put(100.00,30.00){\line(2,-3){10.00}}
\put(110.00,15.00){\line(1,0){10.00}}
\put(110.00,15.00){\line(-2,-3){10.00}}
\put(100.00,30.00){\circle*{2.00}}
\put(100.00,0.00){\circle*{2.00}}
\put(110.00,15.00){\circle*{2.00}}
\put(120.00,15.00){\circle*{2.00}}
\bezier{144}(90.00,0.00)(100.00,15.00)(90.00,30.00)
\bezier{144}(90.00,0.00)(80.00,15.00)(90.00,30.00)
\put(22.00,15.00){\makebox(0,0)[cc]{$e$}}
\put(92.00,15.00){\makebox(0,0)[cc]{$e$}}
\end{picture}
\caption{Simple expansion at an incontractible edge.
\label{IncontractibleFig}}
\end{figure}
{\bf Remark 2.} Since $A_i$ has a single bridge with respect to
$\{a,b\}$, $\Bar{e_i}$ is not incontractible in $\Bar{A_i}$.
\item \underline{\bf $e$ ordinary.}
We first define a partial order on the collection of all subgraphs
$Y_i\subseteq G$ for which $G = \Bar{X_i}\join{}\Bar{Y_i}$ for some $X_i$
properly
containing $e$. Specifically, we set $Y_j \closerthan Y_i$ if
$Y_i = \Bar{Z}\join{} \Bar{Y_j}$ for some subgraph $Z$.
Let $\{A_i\}$ denote the collection of maximal elements with respect to this
partial order, and let $a_i$ and $b_i$ be the vertices of attachment of $A_i$.
\begin{lemma}\label{IntersectLem}
Two maximal elements $A_i$ and $A_j$ intersect in at most
one common vertex of attachment.
\end{lemma}
\begin{proof}
Let $G=\Bar{X_i}\join{}\Bar{A_i}=\Bar{X_j}\join{}\Bar{A_j}$.
If $A_i$ and $A_j$ had {\em both\/} vertices of attachment in common,
then we could write $G=\Bar{X_i\intersect X_j}\join{}\Bar{A_i\union A_j}$,
and $A_i\union A_j=\Bar{A_i}\join{}\Bar{A_j}$, contrary to the maximality
of $A_i$ and $A_j$.
Suppose $A_i$ and $A_j$ have an interior vertex $v$ in
common. Then there are two internally disjoint paths joining
$v$ and $e$, each passing from $A_i$ to $X_i$ and also from $A_j$ to
$X_j$. By the above, not both $a_i$ and $b_i$ can belong to $A_i$ nor to
$A_j$. Choose notation so
that the paths are labeled as in Figure~\ref{PathFig}.
\begin{figure}[hbt]
\centering
\unitlength=0.70mm
\begin{picture}(50.00,45.00)
\put(5.00,25.00){\line(0,-1){10.00}}
\put(5.00,15.00){\line(1,-1){10.00}}
\put(15.00,5.00){\line(1,0){15.00}}
\put(30.00,5.00){\line(1,1){15.00}}
\put(45.00,20.00){\line(-1,1){15.00}}
\put(30.00,35.00){\line(-1,0){15.00}}
\put(15.00,35.00){\line(-1,-1){10.00}}
\put(5.00,15.00){\circle*{2.00}}
\put(5.00,25.00){\circle*{2.00}}
\put(15.00,35.00){\circle*{2.00}}
\put(30.00,35.00){\circle*{2.00}}
\put(45.00,20.00){\circle*{2.00}}
\put(30.00,5.00){\circle*{2.00}}
\put(15.00,5.00){\circle*{2.00}}
\put(0.00,20.00){\makebox(0,0)[cc]{$e$}}
\put(10.00,40.00){\makebox(0,0)[cc]{$a_i$}}
\put(10.00,0.00){\makebox(0,0)[cc]{$a_j$}}
\put(35.00,40.00){\makebox(0,0)[cc]{$b_j$}}
\put(35.00,0.00){\makebox(0,0)[cc]{$b_i$}}
\put(50.00,20.00){\makebox(0,0)[cc]{$v$}}
\end{picture}
\caption{\label{PathFig}}
\end{figure}
Thus $b_j \in A_i$ and $b_i \in A_j$, and every path from
any internal vertex of $A_i$ to $e$ which does not pass
through $a_i$ must pass through $b_i$, which is in $A_j$.
Thus, the path must continue through $a_j$, since if it
continued through $b_j$, it would still be within $A_i$.
But then $\{a_i,a_j\}$ separates
$A_i \cup A_j$ from $e$, violating the maximality of $A_i$,
so $A_i$ and $A_j$ have no interior vertex in common.
\end{proof}
In this case, we form $B_e$ by replacing each $A_i$ with a new
edge $\Bar{e_i}$ joining $a_i$ and $b_i$, and we form $\Bar{A_i}$ by
adding the edge $\Bar{e_i}$ to $A_i$. It follows from the maximality of
the $A_i$ that $B_e$ is simple (since otherwise a multilink could be split off
of $B_e$) and that $B_e$ is $3$-connected (since if there were a two-cutset,
one of its \bridges\ could be split off of $B_e$.) Further, $B_e$ is locally
finite since no vertex of $B_e$ has larger valence in $B_e$ than it has in $G$.
Once again, $G$ is the amalgam of $B_e$
with the $\{\Bar{A_i}\}$ (see Figure~\ref{OrdinaryFig}.)
\begin{figure}[hbt]
\centering
\unitlength=0.80mm
\begin{picture}(101.00,36.00)
\put(5.00,25.00){\line(0,-1){20.00}}
\put(5.00,25.00){\line(3,1){15.00}}
\put(20.00,30.00){\line(1,-1){15.00}}
\put(20.00,0.00){\line(-3,1){15.00}}
\put(5.00,5.00){\line(1,1){10.00}}
\put(15.00,15.00){\line(-1,1){10.00}}
\put(15.00,15.00){\line(1,0){20.00}}
\put(35.00,15.00){\line(-1,3){5.00}}
\put(30.00,30.00){\line(-1,0){10.00}}
\put(20.00,0.00){\line(3,1){15.00}}
\put(35.00,5.00){\line(0,1){10.00}}
\put(35.00,15.00){\line(-3,-1){15.00}}
\put(20.00,10.00){\line(0,-1){10.00}}
\put(20.00,10.00){\line(3,-1){15.00}}
\put(2.50,15.00){\makebox(0,0)[cc]{$e$}}
% \put(5.00,5.00){\vector(0,1){10.00}}
\put(45.00,15.00){\makebox(0,0)[cc]{$\longrightarrow$}}
\put(5.00,25.00){\circle*{2.00}}
\put(5.00,5.00){\circle*{2.00}}
\put(15.00,15.00){\circle*{2.00}}
\put(20.00,0.00){\circle*{2.00}}
\put(20.00,10.00){\circle*{2.00}}
\put(35.00,5.00){\circle*{2.00}}
\put(35.00,15.00){\circle*{2.00}}
\put(30.00,30.00){\circle*{2.00}}
\put(20.00,30.00){\circle*{2.00}}
\put(55.00,25.00){\line(0,-1){20.00}}
\put(60.00,30.00){\line(3,1){15.00}}
\put(75.00,35.00){\line(1,-1){15.00}}
\put(85.00,-5.00){\line(-3,1){15.00}}
\put(55.00,5.00){\line(1,1){10.00}}
\put(65.00,15.00){\line(-1,1){10.00}}
\put(65.00,15.00){\line(1,0){20.00}}
\put(90.00,20.00){\line(-1,3){5.00}}
\put(85.00,35.00){\line(-1,0){10.00}}
\put(85.00,-5.00){\line(3,1){15.00}}
\put(100.00,0.00){\line(0,1){10.00}}
\put(100.00,10.00){\line(-3,-1){15.00}}
\put(85.00,5.00){\line(0,-1){10.00}}
\put(85.00,5.00){\line(3,-1){15.00}}
\put(52.50,15.00){\makebox(0,0)[cc]{$e$}}
% \put(55.00,5.00){\vector(0,1){10.00}}
\put(55.00,25.00){\circle*{2.00}}
\put(55.00,5.00){\circle*{2.00}}
\put(65.00,15.00){\circle*{2.00}}
\put(85.00,-5.00){\circle*{2.00}}
\put(85.00,5.00){\circle*{2.00}}
\put(100.00,0.00){\circle*{2.00}}
\put(100.00,10.00){\circle*{2.00}}
\put(85.00,35.00){\circle*{2.00}}
\put(75.00,35.00){\circle*{2.00}}
\put(90.00,20.00){\circle*{2.00}}
\put(75.00,35.00){\circle*{2.00}}
\put(85.00,15.00){\circle*{2.00}}
\put(60.00,30.00){\circle*{2.00}}
\put(70.00,0.00){\circle*{2.00}}
\put(60.00,30.00){\line(3,-1){30.00}}
\put(85.00,15.00){\line(-3,1){30.00}}
\put(55.00,5.00){\line(3,1){30.00}}
\bezier{144}(70.00,0.00)(75.00,10.00)(100.00,10.00)
\end{picture}
\caption{Simple expansion at an ordinary edge.\label{OrdinaryFig}}
\end{figure}
\end{itemize}
$\Tree_1$ is a star whose central vertex $\beta$ is labeled
``$B_e$,'' and whose pendant vertices $\{\alpha_i\}$ are labeled with the
various $\Bar{A_i}$. Given a link $\eta=\{\beta,\alpha_i\}$
in $\Tree_1$, define the
edges $\eta_\beta$ and
$\eta_{\alpha_i}$ to be the edges $\Bar{e_i}$ in $B_e$ and $\Bar{A_i}$,
respectively, with the obvious orientations. Clearly, $G=G(\Tree_1)$.
Next, suppose $\Tree_n$ has been defined and
construct $\Tree_{n+1}$ by performing, for each pendant node $\alpha$
of $\Tree_n$ which is not a $3$-block,
a simple expansion of $\alpha$ at the edge
$\eta_\alpha$ (where $\eta$ is the unique edge of $\Tree_n$ incident to
$\alpha$).
Then $G=G(\Tree_{n+1})$,
and each nonpendant node of $\Tree_{n+1}$ is labeled with
a $3$-block.
Moreover, by remarks~1 and~2 above,
no two adjacent nodes of $\Tree_{n+1}$ are labeled with
either links or circuits.
Let $\Tree_e=\lim \Tree_n$.
Let $\Tree^0_n$ denote the edge amalgam tree obtained from
$\Tree_n$ by removing the pendant nodes which are not $3$-blocks
and let $G^0(\Tree_n)$ be the graph
obtained from $G(\Tree^0_n)$ by removing the edge $\eta_\alpha$ for each
pendant link $\eta\in\Tree_n$ with endpoint $\alpha\in\Tree^0_n$ (in effect,
we remove from $G(\Tree^0_n)$ all those edges which are amalgamated in
$G(\Tree_e)$).
It is clear
that the $G^0(\Tree_n)$ constitute an increasing sequence of subgraphs of
$G(\Tree_e)$ and that $\lim G^0(\Tree_n)=G(\Tree_e)$.
It is also clear that the $G^0(\Tree_n)$ are subgraphs of $G$; thus, to
prove that $G=G(\Tree_e)$, it suffices to show
\begin{lemma} Each finite subgraph of $G$ is contained in $G^0(\Tree_n)$ for
some $n$.
\end{lemma}
\begin{proof}
It suffices to show that each edge of $G$ belongs to one of the graphs
$G_\alpha$ for some node $\alpha\in\Tree$, for then we may choose $n$
so large that $\alpha$ is a node of $\Tree^0_n$.
Let $e'$ be any edge of $G$. If $e'\in B_e$, then $e'\in G^0(\Tree_1)$,
so suppose $e'\not\in B_e$. Let $d$ be the length of a shortest circuit in
$G$ containing both $e$ and $e'$. Let $\alpha$ be the node of $\Tree_1$
with $e'\in G_\alpha$, and let $e''=\eta_\alpha$, where $\eta$ is the link of
$\Tree_1$ connecting $\alpha$ to the central vertex.
Then the length of a shortest circuit in $G_\alpha$
containing both $e'$ and $e''$ is either $d$
(if $B_e$ is a multilink) or
strictly less than $d$ (otherwise).
Since two multilinks cannot be adjacent
in any $\Tree_n$,
it follows by induction that $e'$ belongs to $G_\gamma$ for some
node $\gamma\in\Tree^0_n$ for some $n$, and since $e'$ is not
amalgamated, it belongs to $G^0(\Tree_{n+1})$.
\end{proof}
So we have that $G = G(\Tree_e)$, and by the lemma, $\Tree_e$ satisfies
conditions~2 and~3, and so $\Tree_e$ is a $3$-block tree representing $G$.
\section{Invariance of $\Tree_e$}
\begin{proposition}
Let $G$ be a locally finite $2$-connected graph, and let
$e$ be an edge of $G$. If $\Tree$ is any $3$-block tree
for $G$, then $\Tree = \Tree_e$.
\end{proposition}
\begin{proof}
We will show that the $3$-block of $\Tree$ which
contains the edge $e$ is equal to $B_e$ in $\Tree_e$.
Once this is established, then the subgraphs $A_i$ are
determined, and an induction shows that, for all $k$,
the subtree of $\Tree$
consisting of all nodes a distance $k$ from the node
labeled $B_e$ is identical with the corresponding subtree of
$\Tree_e$, and the result follows.
Let $\alpha$ be the node of $\Tree$ whose graph
$G_\alpha$ contains $e$.
If $e$ is undeletable, then $G_\alpha$ is a cycle
$C$, and the internal vertices of $C-e$ correspond to
cutpoints of $G-e$. Let $e'$ be an amalgamated
edge along this cycle. Then $e'=\eta_\alpha$ for some link $\eta$ with
endpoint $\alpha$. Let $\beta$ be the other endpoint of $\eta$. Let $\Tree'$
denote
the component of $\Tree-\eta$ containing $\beta$, and let $e''=\eta_\beta$.
Then $e''$ is not undeletable in
$G(\Tree')$, since otherwise, $G_\beta$ would be a cycle, as well.
Therefore, every cutpoint of $G-e$ is a
vertex of $C$, and $G_\alpha = C = B_e$.
If $e$ is incontractible, then $G_\alpha$ is a multilink, and each
component of $\Tree-\alpha$ corresponds to a union of
some of the bridges of the endpoints of $e$.
If some such component, say $\Tree'$, corresponds to more than one
branch, then its amalgamated edge $e'$ will be incontractible in
$G(\Tree')$, and so the $3$-block $G_\beta$ containing
it is a multilink. But $\beta$ is adjacent to $\alpha$ in $\Tree$, so this
is impossible.
Therefore, each component of $\Tree-\alpha$
corresponds to exactly one \branch, and so $G_\alpha = B_e$.
If $G_\alpha$ is simple, locally finite and $3$-connected,
then $e$ is ordinary in $G$. Thus, $B_e$ is also simple, locally
finite and $3$-connected. We must show that $B_e=G_\alpha$.
Let $\eta$ be a link in $\Tree$ joining $\alpha$ with, say, $\beta$, and
let $\Tree'$ be the component of $\Tree-\alpha$ containing $\beta$. Then it
is clear that the subgraph $G(\Tree')-\eta_\beta$ of $G(\Tree)$ is
maximal with respect to the partial order defined earlier, and the same
is true for every link of $\Tree$ incident with $\alpha$. Since these maximal
elements are uniquely determined once $e$ has been chosen, it follows that
$G_\alpha=B_e$.
\end{proof}
\section{Higher connectivity}
\label{HigherSect}
The decomposition of a $2$-connected graph
into its $3$-block
tree is distinguished from other decompositions of $2$-connected graphs
by the fact that the $3$-block tree
is uniquely defined. Thus any symmetry
exhibited by the graph $\Tree(G)$ will be reflected in the tree $\Tree$.
Specifically, any automorphism of $G(\Tree)$ induces an automorphism of
$\Tree$. This is also true in the case of the block-cutpoint tree,
however, analogous decompositions of graphs of higher connectivity,
see~\cite{OxleyWu}, are not unique.
\begin{theorem}
Let $G$ be a simple $1$-connected vertex transitive graph.
Then either $G$ is
$2$-connected or its block-cutpoint tree is infinite.
\end{theorem}
\begin{proof}
Let $\Tree$ be the block-cutpoint tree. If $\Tree$ is finite,
then it has a center, which is an edge or a vertex. The center
cannot be an edge, since the cutpoint corresponding to one of its
endpoints would be distinguished in $G$.
Similarly, the center cannot
be a node corresponding to a cutpoint, so the center is a node
corresponding to a block.
Moreover, the central block must contain every vertex of $G$, so
any other block can at most be a loop, and since $G$ is simple,
there is
only one block.
\end{proof}
A similar result is also true for $2$-connected graphs.
\begin{theorem}
Let $G$ be a simple $2$-connected vertex transitive graph.
Then either $G$ is
either a cycle, $3$-connected or its $3$-block tree is infinite.
\end{theorem}
\begin{proof}
Let $\Tree$ be the $3$-block tree. If $\Tree$ is finite,
then it has a center, which is an edge or a vertex. The center
cannot be a link of $\Tree$, since the two endpoints of the amalgamated
edge corresponding to it would be distinguished, and $G$ is not a
multilink.
Thus the center is a node, and that node cannot represent a
multilink, since, again,
its endpoints would be distinguished in $G$.
Every vertex of $G$ must be in the $3$-block of the
central node, hence every other $3$-block is a mulitlink. Since $G$ is
simple, $\Tree$ consists of exactly one $3$-block, hence $G$ is a cycle
or $3$-connected.
\end{proof}
The pattern of these two proofs indicates the ``meta-result'' that
there cannot be a canonical tree decomposition of $3$-connected
graphs in analogy with two and one connectivity,
since we know there exists
infinitely many finite $3$-connected vertex transitive graphs
which are not
$4$-connected, e.g. the Cayley graphs of finite groups with three
generators, and these graphs would have to be indecomposable.
\medskip
\noindent {\small\bf Authors' addresses:}
\noindent {\small
James Madison University,
Harrisonburg, VA 22807,
cdroms@cayuga.math.jmu.edu
}
\noindent {\small
Worcester Polytechnic Institute,
Worcester, MA 01609-2280,
bservat@math.wpi.edu
}
\noindent {\small
11 Hackfeld Rd,
Worcester MA 01609,
jhs@cayuga.math.jmu.edu
}
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\end{document}
James Madison University
Harrisonburg, VA 22807
cdroms@cayuga.math.jmu.edu
Worcester Polytechnic Institute
Worcester, MA 01609-2280
bservat@math.wpi.edu
11 Hackfeld Rd
Worcester MA 01609
jhs@cayuga.math.jmu.edu