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Dept of Mathematical Sciences, 
Portland State University, 
Portland, OR, 97207.
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Minimally separating, homology, geodesics, Riemannian manifolds, length spaces, Brillouin, connectedness.
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\begin{document}
\title{MEDIATRICES AND CONNECTIVITY}
\author{J. J. P. Veerman
\thanks{e-mail: veerman@pdx.edu}\\
Department of Mathematical Sciences, \\Portland State University,
Portland, OR 97207.} \maketitle

\begin{abstract}
If $(X,d)$ is a connected metric space and $a$ and $b$ are points
in $X$, then the locus $L$ of the points $x$ where
$d(x,a)-d(x,b)=0$ is called a mediatrix. For instance if $X$ is a
geodesic space then the geodesics emanating from $a$ and $b$
(starting at the same time, and travelling with unit speed) are
said to focus at $L$.

In an earlier paper (\cite{VPRS}), Brillouin Spaces were defined.
These are spaces in which mediatrices have desirable properties.
Most importantly: they are minimally separating. This means that
for every proper subset $L'$ of $L$ in $X$, $X-L$ is disconnected,
but $X-L'$ is connected.

The purpose of this note is twofold. First, we give a very simple
characterization of Brillouin Spaces, which shows that, for
example,  compact, connected Riemannian manifolds are Brillouin
Spaces. Second, we give a description in terms of homology of
these mediatrices. This leads to a complete topological
classification of mediatrices in 2-dimensional, compact,
connected, Riemannian manifolds.\\
KEYWORDS: Minimally separating, homology, geodesics, Riemannian
manifolds, length spaces.
\end{abstract}

\vskip .3in

\section{Introduction}
\label{introduction}
\setcounter{figure}{0}
\setcounter{equation}{0}

\vskip.2in

Suppose that $x_1(t)$ and $x_2(t)$ are two distinct solutions of a
second order differential equation with $x_1(0) = x_2(0)$ and
there is some $T\ne 0$ so that $x_1(T) = x_2(T)$. Then the
trajectories $x_1$ and $x_2$ are said to focus at time $T$.  One
can ask how the number of trajectories which focus varies with the
endpoint $x(T)$--- this gives rise to the concept of a focal
decomposition. This concept (originally called sigma
decomposition) was first proposed in \cite{Pe1}. In \cite{PT},
several explicit examples were worked out, in particular the
fundamental example of the pendulum $\ddot{x} = -\sin x$.

Later, in \cite{KP}, the idea of focal decomposition was
approached in the context of geodesics of a Riemannian manifold
$M$ (in addition to a reformulation of the main theorem of
\cite{PT}).  Here, one takes a basepoint $x_0$ in the manifold
$M$: two geodesics $\gamma_1$ and $\gamma_2$ focus at some point
$y\in M$ if $\gamma_1(T) = y = \gamma_2(T)$.  This gives rise to a
decomposition of the tangent space of $M$ at $x$ into regions
where the same number of geodesics focus. In this setting there is
also a connection with number theory (see \cite{Pe2}).

The flavor of these papers, and especially their connections with
various other branches of mathematics, seemed to invite a separate
treatment of this circle of ideas. We now present some of the
results (\cite{VPRS}) that this gave rise to. In this note we will
not go into details about the relation these results bear to the
original problem. (However, see \cite{VPRS} for a discussion of
this.)

Let $S$ be a discrete set in a path-connected, metric space $X$
with distance function $d(.,.)$. Assume further that the function
$d(x,.)$ is proper, so that closed balls of finite radius are
compact (see \cite{Ca} where it proved that such a space is
locally compact and complete). We will call such a space a proper
(path-connected, metric) space. By a discrete set $S =
\{x_i\}_{i\in I}$ we mean that any compact subset of $X$ contains
at most finitely many points of $S$. Note that if $\liminf_{a,b\in
S} d(a,b) > 0$, then $S$ is discrete. Choose a preferred point
$x_0$ (the origin) in $S$. The most important notion in
\cite{VPRS} is the formal definition of the $n$-th Brillouin zone
$B_n(x_0)$ with base point $x_0$. We use $N_r(x)$ for an open ball
of radius $r$ and centered in $x$ and $C_r(x)$ for the boundary of
the closed ball (the circumference). The following notion is
discussed in greater detail in \cite{VPRS} and several figures are
included there).

\begin{defn}
\label{defbrillouin} Let $x\in X$, let $n$ be a positive integer,
$n \le \#(S)$, and let $r=d(x,x_0)$. Then define the sets
$b_n(x_0)$ and $B_n(x_0)$ as follows.
\begin{itemize}
\item $x \in b_n(x_0)  \iff
  \#\left( N_r(x) \cap S \right) = n-1$   \quad and \quad
  $C_r(x) \cap S = \{x_0\}$ (one point only).
\item $x \in B_n(x_0) \iff
  \#\left( N_r(x) \cap S \right) = m$   \quad and \quad
  $\#\left( C_r(x) \cap S \right) = \ell \geq 1$,  \quad \mbox{where
  $l,m \in \Z^+$} with \mbox{$m +1 \leq n \leq m +\ell$}.
\end{itemize}
\end{defn}

\noindent {\bf Remark:} The notion of Brillouin Zone was
introduced by Brillouin in the 1930's, and plays an important role
in solid state physics (see \cite{Br} or \cite{AM}).

\noindent {\bf Remark:} In geometry $b_1(x_0)$ coincides with the
Dirichlet domain associated with $x_0$. In computational geometry,
they are sometimes called Voronoi cells (\cite{PS}. There also is
an intimate connection between this boundary and the cut-locus in
Riemannian geometry (see \cite{doC} and \cite{Kl}).

In the second part of the definition, if $m=n-1$ and $\ell=1$,
then $x \in b_n(x_0)$. So $b_n(x_0) \subseteq B_n(x_0)$.  Note
also that $b_n(x_0)$ is open and that $B_n(x_0)$ is closed.
Finally, observe that for fixed $x_0$ the sets $b_n(x_0)$ are
disjoint, but the sets $B_n(x_0)$ are not: their ``boundaries''
overlap. Here by ``boundaries'' we really mean the complement of
$b_n(x_0)$ in $B_n(x_0)$. These consist of subsets of mediatrices:

\begin{defn}
\label{mediatrix} For $a$ and $b$ distinct points in $S$, the
associated mediatrix (sometimes called equidistant set or
bisector) $L_{ab}$ is the set given by: \bsenn L_{ab} = \{x\in X\;
|\; d(x,a)-d(x,b)=0\} \quad . \esenn
\end{defn}

The fact that $B_n(x_0)-b_n(x_0)$ consists entirely of pieces of
mediatrices does not imply that $B_n(x_0)$ is the closure of
$b_n(x_0)$. And in fact, this is not a very ``natural'' property.
However, a slightly weaker result, Theorem \ref{VPRS2}, is true in
many common spaces. To formulate it, we need some definitions
first.

\begin{defn}
\label{metr-cons} The space $X$ is called metrically consistent
if, for all $x$ in $X$, all $R > r >0$ in $\R$ with $r$
sufficiently small, and for each $a\in C_R(x)$, there is a $z \in
C_r(x)$ satisfying
             $N_{d(z,a)}(z) \subseteq N_R(x)$
and $C_{d(z,a)}(z) \cap C_R(x) = \{a\}$.
\end{defn}

\begin{defn}
\label{min-sepa} A proper subset $L$ of a connected set $X$ is
separating if $X-L$ consists of more than one component. A set $L$
in $X$ is minimally separating if it is separating and no proper
subset of $L$ is separating.
\end{defn}

\begin{defn}
\label{conditions} A proper, path connected metric space $X$ is
Brillouin if it is metrically consistent and if for all $a$, $b$
in $X$, the mediatrices $L_{ab}$ are minimally separating sets.
\end{defn}

For Brillouin spaces we have the following result.

\begin{theo} (See \cite{VPRS}.) If $X$ is Brillouin , then $b_n(x_i)$
equals the interior of $B_n(x_i)$ (itself a closed set) and
$B_n(x_0)$ is contained in the closure of $b_1(x_0)\cup \cdots
b_n(x_0)$. \label{VPRS2}
\end{theo}

In section 2 we give a very simple example (example
\ref{flat-cylinder-example}) where $B_3(x_0)$ consists of a set
with no interior (and therefore is not the closure of $b_3(x_0)$).
While it is possible to give general conditions on $(X,d)$ for
which $B_n(x_0)$ is the closure of $b_n(x_0)$, we would have to
exclude many common spaces (we will not pursue this here). One of
the aims of this note is to prove that Brillouin spaces are
``natural'' in the sense that ``common'' spaces such as the
$n$-sphere, hyperbolic $n$-space, and $\R^n$ belong to that
category. All of these thus satisfy the conclusion of Theorem
\ref{VPRS2}. This will be done in section 2.

Clearly, mediatrices play an important role in this discussion.
The fact that they are minimally separating sets leads to a
topological characterization of them. Let us suppose that $X$ is a
smooth, compact, connected $n$-dimensional manifold, and $L\subset
X$ is a mediatrix. Denote by $i: L\rightarrow X$ the inclusion. In
section 3, we assume a ``triangulability'' condition and prove
that $i$ induces a homomorphism $i_{n-1}:
H_{n-1}(L;\Z_2)\rightarrow H_{n-1}(X;\Z_2)$ on the $n-1$-st
homology groups with coefficients $\Z_2$ with the property that
its Kernel has exactly one generator. Since for compact smooth
$n$-manifolds the homology is finitely generated, this means that
$H_{n-1}(L;\Z_2)$ has at most one generator more than
$H_{n-1}(X;\Z_2)$. While this seems abstract, in the case where
$X$ is a smooth, compact, connected $n$-dimensional manifold, one
can give a topological classification of mediatrices. This is done
in section 5.

The discussion in section 3 depends upon the assumption that the
$n$-manifold is triangulated and that the subset $L$ is
triangulated in such a way that both triangulations have a common
refinement. We prove in section 4 that if $X$ is a smooth
$2$-dimensional manifold, and $L\subset X$ is a mediatrix, then
$L$ is the topological image of a finite graph (a finite number of
edges glued together in a finite number of vertices). Thus $L$
admits a natural triangulation --- namely the one inherited from
the graph. We defer the proof that these triangulations have a
common refinement to forthcoming work.

For completeness we mention that the two basic properties from
which ultimately the physical and mathematical interest of
Brillouin zones is derived.

\begin{theo} (See \cite{VPRS}.) If $X$ is a proper, path connected metric
space these Brillouin zones tile the underlying space $X$ in two
ways:
    \bsenn
        \cup_i B_i(x_n) = X   \quad\logand\quad
  b_i(x_n) \cap b_j(x_n) = \emptyset \quad \logif i\neq j.
\esenn \bsenn
  \cup_i B_{n}(x_i) = X  \quad\logand\quad
  b_{n}(x_i) \cap b_{n}(x_j) = \emptyset \quad\logif i\neq j.
\esenn \label{VPRS1}
\end{theo}

In fact, it follows that if $S$ is the image of $x_0$ under a
discrete group of isometries then the Brillouin zones are
fundamental set for the group (see \cite{VPRS}). Indeed, in the
case of a lattice in $\R^2$, both results in this Theorem were
proved by Bieberbach (see \cite{Bi}).

\vskip.2in \noindent {\bf Acknowledgements:} I am grateful to
Charles Pugh and Rudy Beyl and for many illuminating conversations
that were instrumental in writing up this note. I would also like
to thank John Milnor for pointing out the example in section 5 and
Dusa McDuff for pointing out an redundancy in our original
assumptions.


\vskip .4in

\section{Brillouin Spaces}
\label{brillouin}
\setcounter{figure}{0}
\setcounter{equation}{0}

\vskip.2in

The aim of this section is to prove that a large class of metric
spaces has the Brillouin property. That class is described in the
next theorem and includes the connected Riemannian manifolds. All
of these therefore obey the conclusion of Theorem \ref{VPRS2}.

\begin{theo}
\label{main1} Let $(X,d)$ be a path connected metric space
satisfying the following conditions
\begin{itemize}
\item[1:]
Closed balls are compact.
\item[2:]
Between any two points $a$ and $b$ there is always a shortest path whose
length equals $d(a,b)$.
\item[3:]
If two shortest paths, one from $a$ to $b$, and the other from $a$ to $c\neq b$,
have a common segment, then one of the paths is a subset of the other.
\end{itemize}
Then $(X,d)$ is Brillouin.
\end{theo}

Suppose $\gamma$ in $X$ is a (connected) subset of a shortest path
$c$, is longer than $\gamma'$, a path with the same endpoints.
Then $c$ is longer than the concatenation of $\gamma'$ and
$c-\gamma$ (contradicting the assumption that $c$ is shortest).
Therefore, we have

\begin{lem} Let X be a space satisfying the above requirements.
Any path in $X$ that is a subset of a shortest path, is itself a
shortest path. \label{dusa}
\end{lem}

\begin{exam}\rm
\label{Y} The figure Y with the usual shortest path distance does
not satisfy the last of these requirements. It also is not
Brillouin, as can be seen by choosing two points $a$ and $b$, one
in each of the upper branches and at the same ``height''. The
mediatrix consists of the entire lower branch of the figure.
\end{exam}

The central object of study are the mediatrices $L_{ab}$
associated with two points $a$ and $b$ in $X$. We will also denote
this set by $L$ when no confusion is possible. We will also use
the following conventions. Let $(X,d)$ be a metric space
satisfying the requirements of Theorem \ref{main1} and let $a$ and
$b$ be two distinct points in $X$.
    \benn
    L_+ &=& \{x \in X \,|\, d(a,x) -  d(b,x) > 0 \} \quad .\\
    L_- &=& \{x \in X \,|\, d(a,x) -  d(b,x) < 0 \} \quad .
    \eenn
We will speak of the mediatrix associated to $a$ and $b$. For ease
of notation, $a$ and $b$ will now be fixed. The function $f(x) =
d(a,x) - d(b,x)$ is a Lipschitz function (with Lipschitz constant
2) from $X$ to $\R$. For a curve $\gamma$, we denote by
$\stackrel{\circ}{\gamma}$ the set $\gamma - \{\gamma(0) \cup
\gamma(1)\}$.

\begin{lem}
\label{lem1} Suppose $d(a,x) \leq d(b,x)$, and let $\gamma$ be the
minimizing path connecting the points $x$ and $a$. Then
$\stackrel{\circ}{\gamma}\subset L_-$ (the component containing
$a$).
\end{lem}

\noindent {\bf Proof:} Assume that the lemma is false. So there is
a shortest path $\gamma$ from $x\in (L\cup L^-)$ to $a\in L^-$
such that $\stackrel{\circ}{\gamma}\cap (L\cup L_+)$ contains a
point $y$. Then
    \bse
    d(a,x) \leq d(b,x) \quad \logand \quad d(a,y)\geq
    d(b,y) \quad .
    \label{eq=1}
    \ese
In the following, $\gamma$ is a minimizing path from $a$ to $x$.
\be
    d(a,x)&= d(a,y) + d(y,x) & \mbox{($y$ lies on $\gamma$)} \label{eq=2}\\
        &\geq d(b,y) + d(y,x) & \mbox{($y$ lies in the set $L\cup L_+$)} \label{eq=3}\\
        &\geq d(b,x)   & \mbox{(triangle inequality)} \quad .
    \label{eq=4}
    \ee
It thus follows that the first inequality in (\ref{eq=1}) and the
one in (\ref{eq=4}) must be equalities, and so:
    \bsenn
    d(a,x) = d(b,x) \quad .
    \esenn
It then follows that the inequality in (\ref{eq=3}) must also be
an equality. That implies that
    \bsenn
    d(a,y) = d(b,y) \quad .
    \esenn
Finally, the equality in (\ref{eq=4}) implies that
    \bsenn
    d(x,y)+d(b,y)=d(b,x) \quad .
    \esenn
Thus we have that the path $(x,y,b)$ is a shortest path. Since
this is by equation (\ref{eq=2}) also true for $(x,y,a)$, this
violates the last condition of Theorem \ref{main1}. \QED

\begin{cory}
Let $X$ have the properties in Theorem \ref{main1} and $L$ be a
mediatrix in $X$. Then $X-L$ has exactly 2 components, and each of
these is path connected. \label{rudy}
\end{cory}

\noindent
    {\bf Proof:}
Suppose there are at least 3 components. One of these contains
neither $a$ nor $b$. Let $\gamma$ be a shortest path from a point
$x$ in this component to $a$. Then $\gamma(0)$ and $\gamma(1)$ are
not in $L$ and its endpoint lie in distinct components. This
contradicts the previous lemma. To see that both components are
path connected, note that if $x$ and $y$ in $L_+$ then there are
paths from $x$ to $a$ and from $y$ to $a$ (and both do not
intersect $L$).
    \QED

For future reference we prove a slightly stronger version of Lemma
\ref{lem1} .

\begin{lem}
\label{lem2} Suppose $x$ and $y$ are (not necessarily distinct)
points in a mediatrix $L$. Let $\gamma$ be a minimizing path
connecting the points $x$ and $a$ or $b$ and $\eta$ a minimizing
path connecting $y$ to either $a$ or $b$. Then
$\stackrel{\circ}{\gamma}\cap \stackrel{\circ}{\eta}=\emptyset$.
\end{lem}

\noindent {\bf Proof:} First assume that $\gamma$ lands in $a$ and
$\eta$ lands in $b$ (or vice versa). Lemma \ref{lem1} implies that
$\stackrel{\circ}{\gamma}$ and $\stackrel{\circ}{\eta}$ lie in
distinct components of $X-L$. Hence their intersection is empty.

Now assume that $\gamma$ and $\eta$ land in the same point, say
$a$. Assume that the lemma is false. So there is a point $z\in
\stackrel{\circ}{\gamma}\cap\stackrel{\circ}{\eta}$. By Lemma
\ref{dusa}, the sub-path along $\gamma$ from $x$ to $z$, and the
one from $z$ to $a$ are shortest paths with respective lengths
$d(x,z)$ and $d(z,a)$. The path along $\eta$ from $z$ to $a$ is
also a shortest path of length $d(z,a)$. Thus from $x$ to $a$
there are two distinct shortest paths coinciding in a sub path,
contradicting Condition 3. \QED

\begin{cory}
\label{minsep}
The set $L$ is a closed, minimally separating set.
\end{cory}

\noindent {\bf Proof:} The $g(x)=d(a,x)$ is continuous and
therefore so is $f(x)=d(a,x)-d(b,x)$. Thus $L_-$, the set of $x$
such that $f(x) <0$, is open and non-empty (it contains the point
$a$). The same holds for $L_+$. This proves in the first place
that $X-L$ is not connected, and furthermore, that the complement
$L$ of $L_-\cup L_+$ is closed.

To prove minimality, suppose that for some $x\in L$, the set $L -
x$ also separates $X$ into two components, one containing $a$, the
other containing $b$. Then $x$ is in one of the components
relative to $L - x$. Consider the minimizing segment from $x$ to
$a$ and from $x$ to $b$. At least one of them must pass through
$L-x$, contradicting Lemma \ref{lem1}. \QED

To prove the main result of the section, it is now sufficient to prove the
following lemma.

\begin{lem}
\label{consistent}
A space $X$ as defined in Theorem \ref{main1} is metrically consistent.
\end{lem}

\noindent {\bf Proof:} Let $R>r>0$ and suppose that $C_R(x)$
contains at least 2 points (otherwise we are done). Choose $a\in
C_R(x)$, let $\gamma$ be the minimizing path from $x$ to $a$.
Choose $z=\gamma \cap C_r(x)$ and let $y\neq a$ in $C_R(x)$. Now
we have the strict inequality \bsenn d(z,a)<d(z,y) \quad . \esenn
This can be seen as follows. By the triangle inequality: \benn
d(x,y) & \leq & d(x,z)+d(z,y) \\
\Rightarrow d(z,y) & \geq & d(x,y)-d(x,z) = R-r = d(z,a) \quad .
\eenn
    Suppose this were an actual equality. Then the segments from
$x$ to $z$ and then from $z$ to $y$ would form a minimizing path
from $x$ to $y$, contradicting Lemma \ref{dusa}. \QED

From the previous it would seem that for Brillouin spaces
$B_n(x_0)$ is the closure of $b_n(x_0)$. There are conditions
under which it does hold. We give a simple example to show that
this is not always the case. As the example shows these conditions
will have to exclude many ``natural'' spaces. We will not pursue
this here.

\begin{exam}\rm\label{flat-cylinder-example} $X$ is the flat cylinder obtained by identifying opposite sides
of the strip $\{z\in \C \such -1\le \Re{z} \le 1\}$ in the usual
way. We will denote points in the cylinder by a corresponding
complex number. Let $x_0=1$, $x_1=i$, and $x_2=-i$. Each of the
mediatrices $L_{0,2}$ and $L_{0,3}$ consists of a circle with a
Lipschitz singularity at the origin. It is easy to see that in
this example $B_3(x_0)$ consists of only the origin. Note also
that $B_3$ is contained in $B_2$ and $B_1$, and also in the
closures of $b_1$ and $b_2$.
\end{exam}

\vskip .4in

\section{Structure Theorems}
\label{structure}
\setcounter{figure}{0}
\setcounter{equation}{0}

The aim of this section is to give a topological classification of
mediatrices. We assume that $X$ is a compact smooth $n$-manifold.
Then $X$ admits a triangulation (see \cite{Mu1}). We also assume
that $(X,L)$ is a triangulated pair, which means that $X$ admits
some triangulation of which $L$ is a subcomplex. We will be using
results from \cite{Mu}. For the record, we mention that the class
of compact triangulated relative $n$-homology manifolds $(X,L)$
(in terms of which Munkres states his results, see \cite{Mu},
section 63) includes the pairs $(X,L)$ stated above. We also
follow Munkres' convention concerning $n$-cells.

\begin{cory}
\label{boundary} The set $L$ is the boundary of the open set
$L_-$. ($L$ is also the boundary of the open set $L_+$.)
\end{cory}

\noindent {\bf Proof:} It suffices to prove the statement for
$L_-$. From the proof of Corollary \ref{minsep}, we conclude that
$L_-$ is open and $L\cup L_-$ is closed. From Lemma \ref{lem1} it
follows that each point in $L$ is a limit point of $L_-$. \QED

\begin{prop}
Let $(X,A)$ be a compact triangulated relative $n$-homology
manifold such that $X-A$ is connected and $A$ contains no
$n$-cells. Then the inclusion homomorphism $i_{n-1} :
H_{n-1}(A;\Z_2)\rightarrow H_{n-1}(X;\Z_2)$ is a monomorphism.
\label{inclusion}
\end{prop}

\noindent {\bf Proof:} We know that $H^0(|X|-|A|;\Z_2) = \Z_2$
(\cite{Mu}, Theorem 42.1). By Lefschetz duality, we have that
$H^0(|X|-|A|;\Z_2) = H_n(X,A;\Z_2) = \Z_2$ (\cite{Mu}, Theorem
70.2). Because $X$ is an $n$-manifold, we have that $H_n(X;\Z_2) =
\Z_2$. Part of the relative homology sequence is :
    \benn
    \cdots\longrightarrow H_n(A;\Z_2) & \stackrel{i_n}{\longrightarrow}
    H_n(X;\Z_2) & \stackrel{\pi_n}{\longrightarrow} H_n(X,A;\Z_2) \\
    \stackrel{\partial_n}{\longrightarrow} & H_{n-1}(A;\Z_2)
    \stackrel{i_{n-1}}{\longrightarrow} & H_{n-1}(X;\Z_2) \longrightarrow \cdots
    \label{relsequence}
    \eenn
By assumption $H_n(A;\Z_2)=0$ (since $A$ contains no $n$-cells).
It then follows that $\pi_n$ is an isomorphism. Thus
$Ker(\partial_n)=H_n(X,A;\Z_2)$. Hence $Ker(i_{n-1})=0$ and thus
$i_{n-1}$ is a monomorphism. \QED

\begin{prop}
Let $(X,L)$ be a compact triangulated relative $n$-homology
manifold such that $X-L$ has 2 components and $L$ contains no
$n$-cells. Then the inclusion homomorphism $i_{n-1} :
H_{n-1}(L;\Z_2) \rightarrow H_{n-1}(X;\Z_2)$ is not a monomorphism
(its kernel is $\Z_2$). \label{inclusion2}
\end{prop}

\noindent {\bf Proof:} The same as above but now the Lefschetz
duality implies that $H_n(X,L;\Z_2) = \Z_2\times \Z_2 $, because
Corollary \ref{rudy} implies that $|X|-|L|$ has exactly 2
components. \QED

The two propositions together imply that the inclusion
homomorphism of a minimally separating set must have a kernel
equal to $\Z_2$. More precisely:

\begin{theo}
Let $L$ be a minimally separating set in $X$ and suppose that
$(X,L)$ is a compact triangulated relative $n$-homology manifold.
Then the kernel of inclusion homomorphism $i_{n-1} :
H_{n-1}(L;\Z_2) \rightarrow H_{n-1}(X;\Z_2)$ has exactly one
generator. Thus $H_{n-1}(L;\Z_2)$ has at most one generator more
than $H_{n-1}(X;\Z_2)$.
\end{theo}

We wish to remark here that the above result restricted to $n=2$
states that the maximum number of ``circular'' cuts which do not
disconnect the the surface is equal to the number of generators of
the first homology group with coefficients $\Z_2$. For this reason
this number is sometimes called the ``connectivity number''. This
formulation -- and the result -- appears in the book by Seifert
and Threlfall \cite{ST} (section 41). However, their reasoning is
different.

\vskip .2 in

We shall prove, in the next section, that, under suitable
conditions, $L$ is actually a homeomorphic image of a finite
closed graph (a graph without boundary). For such objects,
$H_1(L;\Z_2)$ is a free group with $d$ generators over $\Z_2$,
where $d$ is determined as follows (as was pointed out to us by R.
Beyl). Suppose $L$ consists of several disjoint subgraphs $L_i$.
The set of generators is the union of the generators of all of the
$L_i$. To determine the number of generators in a connected graph,
we must find the maximal tree it contains. By homotopy invariance,
we may collapse this tree to a point. All remaining edges will
then be circles glued together in a point. They are the generators
of $H_1(L_i)$. For reference see \cite{HW} (section 2.5).

\begin{exam}\rm : A mediatrix in a Riemannian 2-torus must
be homeomorphic to a closed graph (see next section) whose first
homology group has at most 3 generators, for instance 3 circles
glued together in a point, or 2 circles glued together in 2
distinct points.
\end{exam}

\vskip .4 in

\section{Regularity Theorems}
\label{regularity}
\setcounter{figure}{0}
\setcounter{equation}{0}

\vskip.2in

We will assume that $X$ is a compact, connected 2-dimensional
manifold with a smooth Riemannian metric giving rise to a distance
$d$. In this section we establish that $L$ is the homeomorphic
image of a finite graph. Thus $L$ comes with a natural
triangulation, namely the one inherited from the graph. We claim
that the triangulation of the underlying manifold $X$ and that of
$L$ have a common refinement (or $(X,L)$ is a triangulated pair,
see previous section). We will take this issue up in a forthcoming
work.

Note that $X$ is complete as a metric space, and thus the
Hopf-Rinow Theorem implies that $X$ is geodesically complete and
furthermore satisfies Condition 2 in Theorem \ref{main1} (see for
example \cite{Kl}). The third condition in that theorem follows
again from existence and uniqueness of the solutions of
differential equations. This establishes that $X$ is a Brillouin
space.

In this section we will prove the following result:

\begin{theo}
\label{graph} If $X$ is a compact 2 dimensional manifold with a
smooth Riemannian metric, then any mediatrix $L$ is a homeomorphic
image of a finite graph.
\end{theo}

To prove this, we need some notation and intermediate results. The
exponential at a point $x$ in $X$ will be denoted by $\exp_x$.
Since $X$ is compact, $\rho_m$, the minimum of the radii of
injectivity, is greater than 0. Thus for $\rho<\rho_m$ the map
\bsenn \exp_x : N_\rho(0)\subset T_xX \rightarrow X \esenn is a
diffeomorphism onto its image. However, the map $\exp_x$ is smooth
on {\it any} bounded domain of $T_xX$. Suppose $L$ is the
mediatrix associated with the points $a$ and $b$ in $X$. Let $x\in
L$ and $\rho$ small enough -- as specified before. First, apply
$\exp_x^{-1}$, a smooth coordinate transformation, to
$\exp_x(N_\rho(0))\subset X$. In the disk thus obtained, we use
geodesic polar coordinates $r$ and $\theta$. Thus $(r,\theta)$ is
a point in $T_xX$ and $(tr,\theta)$ is a point on line through the
origin and $(r,\theta)$ in $T_xX$. Note that $x$ is mapped to $0$.
Define
    \benn
    \Theta_a &=& \{ \theta\in S^1\, |\, t\rightarrow\exp_x(tr,\theta) {\rm \; is \; a\;
    minimal\; geodesic\; from \; } x {\rm\; to\; } a\}\\
    \Theta_b &=& \{ \theta\in S^1\, |\, t\rightarrow\exp_x(tr,\theta) {\rm \; is \; a\;
    minimal\; geodesic\; from \; } x {\rm \; to\; } b\}
    \eenn
Define a wedge to be a subset of $N_\rho(0)$ with the property
that
    \benn
    W= & \{(r,\theta) \in N_\rho(0)\; |\; \theta \in [\theta_-,
    \theta_+]\; , \; \theta_- \in \Theta_a \; , \; \theta_+ \in
    \Theta_b & \\
    &\logand \Theta_a\cap (\theta_-,\theta_+) = \Theta_b \cap
    (\theta_-,\theta_+) =\emptyset \} &
    \eenn
or the same definition with $a$ and $b$ interchanged. The proof of
the main theorem of this section now proceeds in a number of steps
here given as lemmas.

\begin{lem}
The sets $\Theta_a$ and $\Theta_b$ are disjoint compact subsets of
the circle. \label{continuity}
\end{lem}

    \noindent {\bf Proof:}
    The disjointness is clear from the that $d(x,a)=d(x,b)$, so
    that if the geodesics are not disjoint, then $a=b$.

Let $\{c_i\}_{i\in \N}$ be a sequence of minimal geodesics between
$x_i \in X$ and $a\in X$ with initial velocities $v_i$ (and
$|v_i|=1$). Furthermore, suppose they have the property that
$\lim_{i}x_i=x$ and $\lim_{i}v_i=v$. By geodesic completeness, the
initial point $x$ with initial (unit) velocity $v$ defines a
geodesic.

Now let $t\rightarrow\phi(t,x,v)$ be the geodesic at time $t$,
starting at $x$ with initial velocity $v$. Then $\phi$ is the
solution of a second order differential equation with a smooth
right hand side. It follows that $\phi(t,x,v)$ is also smooth
(therefore continuous) (see \cite{Ar}, section 32). Thus since
$\lim_{i}\phi(t_i,x_i,v_i)= a$, we also get $\phi(t,x,v)=a$.
Finally, the continuity of the distance function implies that
$t=\lim_i t_i= \lim_i d(x_i,a)=d(x,a)$. And so the limiting
geodesic is minimal.

Therefore, fixing $x_i=x$, the set of directions at $x$ in which
minimal geodesics to $a$ leave, is closed.
    \QED

\begin{lem} There are finitely many wedges $W_i$.
\end{lem}

\noindent {\bf Proof:} If the number of wedges were infinite, then
there would be a sequence in $\Theta_a$ accumulating to a point in
$\Theta_b$ (and vice versa). This contradicts the previous lemma.
\QED

\begin{lem} Each wedge $W_i$ is separated by $L'$, and $L'$ does not
intersect the sides of the wedge.
\end{lem}

\noindent {\bf Proof:} Consider the connected component of
$\exp_x^{-1}(L)\cap N_\rho(0)$ which contains the point
$0=\exp_x^{-1}(x)$. For brevity we will call this set $L'$ from
now on. $L$ separates $X$ into a component containing $a$ and
another one containing $b$. Thus the geodesics parting in a
direction in $\Theta_a$ must be separated by $L$ from the
geodesics parting in a direction $\Theta_b$. Since $L$ does not
meet $0$, they must be separated by $L'$. On the other hand, by
Lemma \ref{lem2}, $L'$ cannot intersect these geodesics (except at
$x$).
    \QED

Let us call the set $W_i\cap L' = L_i$ a spoke based in $x$.

\begin{lem} For $r$ small enough, each spoke $L_i$ is diffeomorphic to
\\$\{(x,f(x))\;|\; x\in [0,r)\}$ in $\R^2$, for some Lipschitz
function $f$.
\end{lem}

\noindent {\bf Proof:} Given a spoke $L_i$ in a wedge
$W_i=\{(r,\theta)\; |\; r\in [0,\rho] \logand \theta\in [\theta_-,
\theta_+]\}$. We know that the spoke $L_i$ meets $0$. Let $y\in
L_i$ be distinct from $0$. Suppose $\theta_1$ is a direction in
which a minimal geodesic from $y$ to $a$ departs, and $\theta_2$
one in which a geodesic to $b$ departs. By Lemma \ref{lem2},
$\theta_1 \in [\theta_--\epsilon, \theta_++\epsilon ]$, for some
$\epsilon>0$, otherwise the geodesics will cross. The same holds
for $\theta_2$. Notice by the same lemma that $L_i$ cannot cross
these rays leaving from $y$. Thus $L_i$ preserves a cone field
$\{\theta_--\epsilon\} \cup \{\theta_++\epsilon\}$. Now perform a
linear coordinate change on $N_\rho(0)$ that maps
$\theta=\theta_--2\epsilon$ to $\theta=0$ and
$\theta=\theta_++2\epsilon$ to $\theta=\pi/2$. Then $L_i$ is a
figure in the positive quadrant of $\R^2$ leaving invariant the
cone field $\{\epsilon\}\cup \{\pi/2 - \epsilon\}$ which implies
the result.

The wedges do not necessarily partition the disk. The remaining
pieces (if any) are wedge shaped and have the form
    \bsenn
    S=\{(r,\theta) \in N_\rho(0)\, |\, \theta \in [\theta_-,
    \theta_+]\; , \; \theta_+ \, ,\, \theta_- \in \Theta_a
    \logand \Theta_b\cap (\theta_-,\theta_+) = \emptyset \}
    \esenn
(or the same definition with $a$ and $b$ interchanged). Suppose
that S contained a spoke $\ell$ connected to $x$. Then at each
point $y$ of the spoke $\ell$ there would be a minimal geodesic
departing in the direction $\theta_1(y)$ to $a$. As before
$\theta_1(y) \in [\theta_--\epsilon, \theta_++\epsilon ]$. Since
this would be true for $y$ arbitrarily close to $x$, by the
reasoning in the proof of Lemma \ref{continuity}, there would have
to be a minimal geodesic from $x$ to $a$ departing in the
direction $\theta_1(x) \in [\theta_-, \theta_+ ]$. This
contradicts the definition of $S$. \QED

From this proof we conclude the following. Each of the wedges
contains exactly one spoke, and each spoke is homeomorphic to a
radial segment. The complement of the wedges does not contains any
points of $L'$. This implies that there are finitely many spokes
associates with $x$, and that if $x$ has $n$ spokes associated
with it, then there must be at least $n$ distinct minimal
geodesics connecting $x$ with $a$ or $b$ (namely the boundaries of
the wedges). we are now in a position to perform the last step of
the proof of the main theorem. Denote a point in $L$ with more
than 2 spokes associated with it as a vertex.

\begin{lem} $L$ contains a finite number of vertices.
\end{lem}

\noindent {\bf Proof:} By compactness, if the lemma is false then
there is an accumulation point $x_0$ of vertices $\{x_i\}$.
Transform again to the disk $N_\rho(0)$ by applying
$\exp_{x_0}^{-1}$. Possibly by thinning out the sequence
$\exp_{x_0}(x_i)$, we obtain a sequence $\{y_i\}$ that lies
entirely in one of the wedges associated with $x_0$ or in the
complement of the wedges. In the first case, the $x_i$ lie in a
single wedge associated with 0. For the same reason as in the
previous lemma, the directions $\theta$ in which the minimal
geodesics from $x_i$ to $a$ or $b$ leave, must be contained in the
wedge. Since each $x_i$ is a vertex, there must be at least 3
minimal geodesics leaving to $a$ or $b$ from each $y_i$. By
continuity (see the proof of Lemma \ref{continuity}), this is also
true for $0$. But this contradicts the assumption that the $x_i$
lie in a single wedge associated with 0. When $x_i$ lies in the
complement of the wedges, a similar contradiction can be derived,
because now there must be geodesics to both $a$ and $b$ parting
from $x_i$. \QED


\vskip .4 in

\section{Further Remarks}
\label{remarks}
\setcounter{figure}{0}
\setcounter{equation}{0}

\vskip.2in

The topological classification of mediatrices $L$ in a compact
Riemannian 2-manifold $X$ is now an easy exercise. One calculates
the rank of $H_1(X;\Z_2)$ and call this number $q_1$ (the first
connectivity number, see \cite{ST}). Then the mediatrix is
topological graph whose first homology group has rank less than or
equal to $q_1+1$ (the graph has less than $q_1+1$ independent
cycles). For the torus this means that the mediatrix consists of
three or less topological circles forming one, two, or three
connected components. Curiously, it appears that the case where
$H_1(L;\Z_2)$ has three generators cannot occur. The
non-orientable manifolds were not excluded. It is fun to draw
mediatrices of a projective plane. According to section 3, a
mediatrix of the projective plane has at most two cycles.

It is possible that one has more information. For instance a
mediatrix in an orientable, connected, smooth compact Riemannian
2-manifold of genus $g$ with no vertices consists of at most $g$
disjoint ``circles'' (this was pointed out to us by J. Milnor).
The reasoning is as follows: Let $X$ be a surface with $g$
handles. A ``circular cut'' either disconnects the surface or else
opens up one of the handles. Since vertices are forbidden, a new
cut cannot go through a previous cut. So fill the obtained holes
by disk, which increases the genus by 2. Keep going until you
disconnect the surface or until the genus is zero. In the last
case the Jordan Curve Theorem now implies that with the next cut
we separate the surface.

We end by giving an intuitive example of a mediatrix in a
3-dimensional manifold.

\begin{exam}
\label{exam3} \rm : In $\R^3$, we can obtain a variety of
two-dimensional shapes by choosing an appropriate metric. Start
with the flat metric and consider two concentric spheres of almost
the same radius. In the region between them (a ``fat'' sphere), we
make the metric very big, in such a way that it is harder to cross
the fat sphere than to travel the circumference of the outer
sphere. Let $a$ be a point inside the sphere, and $b$ a point
outside, but less than one radius away from the outer sphere.
Clearly, the set $L_{1,0}$ is a topological sphere. In a similar
way we can obtain a torus for $L_{1,0}$. If we vary the metric
smoothly to go from one to the other, at the transition, we will
have a branching at the center of the figure.
\end{exam}


\vskip .4in

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