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\controldates{25-JAN-2005,25-JAN-2005,25-JAN-2005,25-JAN-2005}
 
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\issueinfo{11}{02}{}{2005}
\dateposted{January 28, 2005}
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\PII{S 1079-6762(05)00142-3}
\copyrightinfo{2005}{American Mathematical Society}
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\begin{document}

\title[A combinatorial curvature flow]{A combinatorial curvature flow for compact 3-manifolds with boundary}

\author{Feng Luo}
\address{Department of Mathematics, Rutgers University, Piscataway, NJ 07059}
\email{fluo@math.rutgers.edu}

\subjclass[2000]{Primary 53C44, 52A55}

\commby{Tobias Colding}

\date{May 14, 2004}

\begin{abstract} We introduce a combinatorial curvature flow for piecewise
constant curvature metrics on compact triangulated 3-manifolds  with boundary
consisting of surfaces of negative Euler characteristic. The flow tends to find
the complete hyperbolic metric with totally geodesic boundary on a manifold.
Some of the basic properties of the combinatorial flow are established.  The
most important one is that the evolution of the combinatorial curvature
satisfies a combinatorial heat  equation. It implies that the total curvature
decreases along the flow. The local convergence of the flow to the hyperbolic
metric is also established if the triangulation is isotopic to a totally
geodesic triangulation.  \end{abstract}

\maketitle

\section{Introduction}\label{sec1}

\subsection{}   
The purpose of this paper is to construct  a combinatorial
curvature flow which is a 3-dimensional  analog of the flows considered in
\cite{CL}. In \cite{CL},  we introduced a 2-dimensional combinatorial curvature
flow for triangulated surfaces of non-positive Euler characteristic.  It is
shown that for any initial choice of PL metric of circle packing type, the flow
exists for all times and converges exponentially fast to the
Andreev-Koeb-Thurston's circle packing metrics. In the 3-dimensional case, it
is shown that the curvature evolution equation of the combinatorial curvature
flow satisfies a combinatorial heat equation. Furthermore, the flow tends to
find the complete hyperbolic metric of totally geodesic boundary on the
3-manifold.

Following Hamilton \cite{Ha}, a curvature flow should deform the metrics in
such a fashion that, as the metric evolves, the curvature evolves according
to some heat type equation. It indicates infinitesimally, the metrics tend to
improve themselves. In the case of combinatorial curvature flow, instead of
using the space of all Riemannian metrics,  we consider  the space of all
piecewise constant curvature metrics supported in a fixed triangulation. The
goal is to produce a flow, or a vector field, in the space so that the
combinatorial curvature evolves according to a combinatorial Laplace equation. 
The most general form of a combinatorial Laplacian is defined as follows. Given 
finite graph with vertices labelled by $\{1,2,\dots, n\}$, a {\it combinatorial
Laplace operator} is a negative semidefinite $n \times n $ matrix
$A=[a_{ij}]_{n \times n}$ so that it acts on the space of all functions defined 
on vertices by the linear transformation $A$. To be more precise, suppose
$x=[x_1, \dots, x_n]^t$ is a function (representated as a column vector) whose
value at the $i$th vertex is $x_i$; then the combinatorial Laplace operator
sends  $x$ to $Ax$. In the construction of the combinatorial curvature flow, we
are guided by the principle that the curvature evolves according to some
combinatorial Laplacian operator.  Finding the correct combinatorial Laplacian,
or more precisely, the  negative semidefinite matrix, is the main ingredient
of the paper.  The combinatorial Laplacian operator used in this paper is
obtained as the Hessian of the hyperbolic volume function with dihedral angles
as variables. That the hyperbolic volume function for a hyperideal
tetrahedron is strictly  concave in terms of the dihedral angles was first
observed by J. Schlenker \cite{Sc}. The author thanks Igor Rivin for bringing 
him this
earlier work of Schlenker. In fact, for strictly hyperideal polytopes of the
same combinatorial type, the hyperbolic volume, viewed as a function
of the dihedral angles, is smooth and strictly concave (\cite{Sc}).

\subsection{}  
The basic building blocks for a 2-dimensional flow  are
hyperbolic and Euclidean triangles in \cite{CL}. In the
3-dimensional case, the basic building blocks are the {\it strictly hyperideal
tetrahedra} discovered by Bao and Bonahon \cite{BB}.  See also
Frigerio and Petronio \cite{FP}.

Given an ideal triangulation of a compact 3-manifold with boundary consisting
of surfaces of negative Euler characteristic, we replace each (truncated)
tetrahedron with a strictly hyperideal tetrahedron by assigning the edge
lengths. The isometric gluing of these tetrahedra gives a hyperbolic cone
metric on the 3-manifold. The {\it combinatorial curvature} of the cone metric
at an edge is  $2\pi$ less the sum of dihedral angles at the edge. The
combinatorial curvature flow that we propose is the following system of
ordinary differential equations: 
\begin{equation}\label{eq11} d x_i/dt = K_i,
\end{equation}  
where $x_i$ is the length of the $i$th edge and $K_i$
is the combinatorial curvature of the cone metric $(x_1, \dots, x_n)$ at the
$i$th edge. The equation \eqref{eq11} captures the essential features of the
2-dimensional combinatorial Ricci flow in \cite{CL}. The most important of all
is that the combinatorial curvature evolves according to a combinatorial heat
equation. Thus the corresponding maximum principle applies. The flow has the
tendency of finding the complete hyperbolic metric of totally geodesic boundary
on the manifold. By analyzing the singularity formations in equation
\eqref{eq11}, it is conceivable that  one could give a new proof of Thurston's
geometrization theorem for these manifolds using \eqref{eq11}.  Furthermore,
the flow \eqref{eq11} will be a useful tool to find algorithmically  the
complete hyperbolic metric.

\subsection{}   
Suppose $M$ is a compact 3-manifold whose boundary is
non-empty and is a union of surfaces of negative Euler characteristic.  Let
$C(M)$ be the compact 3-manifold obtained by coning off each boundary component
of $M$ to a point. In particular, if $M$ has $k$ boundary components, then
there are exactly $k$ cone points $\{v_1, \dots, v_k\}$ in $C(M)$ so that $C(M)
-\{v_1, \dots, v_k\}$ is homeomorphic to $M - \partial M$. An {\it ideal
triangulation}  (or truncated triangulation) $T$ of $M$ is a triangulation
$\Cal T$ of  $C(M)$ such that the vertices of the triangulation are exactly the
cone points $\{v_1, \dots, v_k\}$. By  Moise \cite{Mo}, every
compact 3-manifold can be ideally triangulated. We  identify $M$ with a subset
of $C(M)$ as follows. Take $st(v_1, \dots, v_k)$ to be the open star of the
vertices $\{v_1, \dots, v_k\}$ in the second barycentric subdivision of the
triangulation. Then we take $M = C(M) - st(v_1, \dots, v_k)$. By abuse
of the language,   the {\it edges, triangles} and {\it tetrahedra} in $M$ in the
ideal triangulation $T$ are defined to be the intersection $a \cap M$, where $a$
is an $i$-dimensional simplex in the triangulation $\Cal T$ of $C(M)$ for
$i=1,2,3$ respectively. Note that the boundary $\partial M$ of $M$ has the
induced triangulation from $\Cal T$. The 1-simplex and 2-simplex in the
boundary triangulation of $\partial M$ are not called edges or triangles in
the ideal triangulation. 

Following Bao-Bonahon \cite{BB}, a  {\it strictly hyperideal tetrahedron}  in
the 3-dimen\-sional hyperbolic space is a compact convex polyhedron that 
is diffeomorphic to a truncated tetrahedron in the 3-dimensional Euclidean
space and its four hexagonal   faces are right-angled hyperbolic hexagons.
(Note that two compact subsets of $\mathbf R^n$ are diffeomorphic if there is a
diffeomorphism between two open neighborhoods of them sending one compact set
to the other.) See Figure 1. In \cite{BB}, Bao and Bonahon  give a
complete characterization of hyperideal convex polyhedra.  As a very special
case of their work,  they obtained a characterization of  strictly hyperideal
tetrahedra using dihedral angles. This characterization of a hyperideal simplex
was also obtained by Frigerio and Petronio \cite{FP}.  Recall that an {\it
edge} in a hyperideal tetrahedron is the intersection of two hexagonal faces. 
The works \cite{BB}  and  \cite{FP} show that the strictly hyperideal tetrahedra
are completely characterized  by their six dihedral angles at the six edges
such
that the sum of  three dihedral angles associated to edges adjacent to each
vertex is less than $\pi$. In particular, the space of all strictly hyperideal
tetrahedra forms an open convex polytope in $\mathbf R^6$ when parametrized by
the dihedral angles.


\begin{figure}
\includegraphics{era142el-fig-1}
\caption{The six edges of a hyperideal tetrahedron are the 
intersections of its hexagonal faces.}
\end{figure}

The main technical observation is the following. We were informed by I. Rivin
that this was first known to J. Schlenker in \cite{Sc}.

\begin{thm}[Schlenker]\label{thm1} The volume of a strictly
hyperideal tetrahedron is a strictly  concave   function of its dihedral
angles. \end{thm} 

The Hessian of the volume is a negative definite matrix by
Theorem \ref{thm1}. This provides a basis for constructing the combinatorial
Laplacian operator for the curvature evolution equation. 

Given an ideal triangulated 3-manifold $(M, T)$, let  $E$ be the set of edges
in the ideal triangulation and let $n$ be the number of edges in $E$.  An
assignment $x: E \to \mathbf R_{>0}$ is called a {\it hyperbolic cone metric
associated to the ideal triangulation $T$} if for each tetrahedron $t$ in $T$
with edges $e_1, \dots, e_6$, the six numbers $x_i = x(e_i)$ $(i=1,\dots, 6)$ 
are
the edge lengths of a strictly hyperideal tetrahedron in $\mathbf H^3$. The set
of all hyperbolic cone metrics associated to $T$ is denoted by $L(M, T)$, which
will be regarded as an open subset of $\mathbf R^n = \mathbf R^E$ by
measuring the edge lengths. The  combinatorial curvature of a cone metric $x
\in L(M,T)$ is the map $K: E \to \mathbf R$ sending an edge $e$ to the
combinatorial curvature of $x$ at the edge $e$. Again we  identify the set of
all combinatorial curvatures $\{ K \mid x \in L(M, T)\}$ with a subset of $\mathbf
R^n$. The combinatorial curvature flow is the vector field in $L(M, T)$ defined
by equation \eqref{eq11}, where $K_i=K_i(t)$ on the right-hand side is the
combinatorial curvature of the metric $x=(x_1,\dots,x_n)$ at time $t$ at the
$i$th edge. 
\begin{thm}\label{thm2}    
For any ideal
triangulated 3-manifold, under the combinatorial curvature flow \eqref{eq11},
the combinatorial curvature $K_i(t)$ evolves according to a combinatorial heat 
equation,  
\begin{equation}\label{eq12} dK_i(t)/dt = \sum_{j=1}^n a_{ij} K_j(t),
\end{equation}  
where the matrix $[a_{ij}]_{n \times n}$ is symmetric
negative definite. Furthermore, the combinatorial curvature flow is the
negative gradient flow of a locally strictly convex function. \end{thm}

\begin{cor}\label{cor3} For any ideal triangulated
3-manifold $(M,T)$, under the combinatorial curvature flow \eqref{eq11},
\begin{enumerate} 
\item[(a)] the total curvature $\sum_{i=1}^n K_i^2(t)$ is
strictly decreasing along the flow unless $K_i(t) =0$ for all $i$;

\item[(b)] the equilibrium  points  of the combinatorial curvature flow
\eqref{eq11} are the complete hyperbolic metric with totally geodesic boundary;

\item[(c)] each  equilibrium  point is a local attractor of the flow.
\end{enumerate}
\end{cor}


Another consequence of the convexity is the following local rigidity result for
hyperbolic cone metrics without constrains on cone angles. Note that  by
\cite{HK},  hyperbolic cone metrics with cone angles at most $2\pi$ are 
locally rigid.   
\begin{thm}\label{thm4} For any ideal
triangulated 3-manifold $(M,T)$, the curvature map $\Pi: L(M, T) \to \mathbf
R^n$ sending a metric $x$ to its curvature $K(x)$ is a local diffeomorphism. In
particular, a hyperbolic cone metric associated to an ideal triangulation is
locally determined by its cone angles.  \end{thm}   
\subsection{} In the
rest of the paper, we prove the results. In the last section, we propose
several  questions related to the combinatorial curvature flow whose resolution
will lead to a new proof of Thurston's geometrization theorem for this class of
3-manifolds.  
\subsection{}\textbf{Acknowledgment.}  We thank Ben Chow for
conversations on the topic of Ricci flow and X.~S. Lin for  discussions. I thank
Igor Rivin for informing me about the earlier work of Schlenker \cite{Sc}.  
This work
was supported in  part by the NSF and a research grant from Rutgers University.

\section{Proofs}
In this section we prove theorems and corollaries stated in \S \ref{sec1}.

\subsection{}\textbf{Proof of Theorem \ref{thm1}.} Suppose $x_1,
\dots, x_6$ are the lengths of the six edges of a strictly hyperideal
tetrahedron so that the corresponding dihedral angles are $a_1, \dots, a_6$.
Let $a=(a_1, \dots, a_6)$, $x=(x_1, \dots, x_6)$, and let 
$V$ be the volume of the
strictly hyperideal tetrahedron. By the results of \cite{BB} and \cite{FP}, 
$x=x(a)$ is a function of $a$. Conversely, $a=a(x)$ is a function of $x$. This
is due to the following results on convex polyhedron. First of all, by the 
Cauchy
rigidity of a convex polytope, the isometry class of the convex polytope in the
hyperbolic 3-space is determined by the intrinsic geometry of the boundary
surface. In particular, the dihedral angles are determined by the induced
metric on the boundary surface. On the other hand,  the metric on the boundary
surface is determined by the  metrics on the four hexagonal faces since the
other four faces are triangles.  Finally, the metric on a right-angled
hyperbolic hexagon is determined by the three lengths of its three pairwise
non-adjacent edges. Thus $a=a(x)$. Obviously, these two functions $a=a(x)$ and
$x=x(a)$ are inverses of each other. Next we claim that both functions $a=a(x)$
and $x=x(a)$ are smooth (in fact real analytic functions). Indeed, to
express $x_i$ in terms of $a$, we use the cosine law for the triangular faces
of the strictly hyperideal tetrahedron. The cosine law expresses twelve edge
lengths $y_j$  of these four triangles in terms of $a$ analytically. Next, we
use the cosine law for the right-angled hexagon to express the length $x_i$
analytically in terms of the $y_j$'s. Thus we see that 
$x=x(a)$ is an analytic map.
Conversely, we use the cosine law for hexgonal faces to express $y_j$
analytically in terms of $x_i$'s. Then we use the cosine law for triangles to
express $a_i$ analytically in terms of $y_j$'s. Thus we see that both $a=a(x)$
and $x=x(a)$ are local diffeomorphisms. In particular, the Jacobi matrix
$[\partial x_i/\partial a_j]_{6 \times 6}$ is non-singular.

Now consider the volume $V=V(a)$ as a function of the dihedral angles $a$.  By
the Schlaefli formula, we have $\partial V/\partial a_i =-x_i/2$.  This implies
that $-\partial x_i/\partial a_j = -\partial x_j/\partial a_i$ for all $i, j$
due to the symmetry of the Hessian matrix.  Since the space of all strictly
hyperideal tetrahedra parametrized by dihedral angles is connected and the
matrix $[\partial x_i/\partial a_j]_{6 \times 6}$ is non-singular,  the
signature of the symmetric matrix $[\partial x_i/\partial a_j]_{6 \times 6}$ is
independent of the choice of the strictly hyperideal tetrahedra.  One checks
directly that if the strictly hyperideal tetrahedron is regular (i.e., all
$x_i$'s are the same and all $a_i$'s are the same), the Jacobian matrix is
positive definite.  Using the Schlaefli formula, this implies that the Hessian
of the volume function $V$ is locally strictly concave. Since the domain is
convex by \cite{BB}, we see that the volume is a strictly concave function of
the dihedral angles. \qed
\smallskip

We do not have a proof that the volume function defined on the space of all
strictly hyperideal tetrahedra parametrized by the dihedral angles can be
extended continuously to all hyperideal tetrahedra. This fact should be true
and may  follow from \cite{Us}. See also the related work
\cite{Lu}.

Since the inverse of a symmetric positive definite matrix is again symmetric
and positive definite, we conclude:
\enlargethispage*{100pt}

\begin{cor}\label{cor5}
For a strictly hyperideal  tetrahedron with dihedral angle $a_i$ and  length $x_i$ at the $i$th edge,
the matrix $[\partial a_i/\partial x_j]_{6 \times 6}$ is symmetric and positive definite.
In particular, the function $F = 2V + \sum_{i=1}^6 a_i x_i$ is a locally strictly convex function of the length variables
$(x_1, \dots, x_6)$.
\end{cor}

Indeed, it suffices to verify that the Hessian matrix $[\partial^2 F/\partial
x_i \partial x_j]$ is strictly positive definite. Now, by construction,
\begin{align*}
\partial F/\partial x_j &= 2 \partial V/\partial x_j + a_j + \sum_{i=1}^6 x_i
\partial a_i/\partial x_j\\ 
&= 2 \sum_{i=1}^6 (\partial V/\partial a_i) (\partial
a_i/\partial x_j) + a_j + \sum_{i=1}^6 x_i \partial a_i/\partial x_j.
\end{align*}
\pagebreak

\noindent By the Schlaefli formula, $\partial V/\partial a_i=-x_i/2$.  Thus 
\begin{equation}\label{eq21} \partial F/\partial x_j = -\sum_{i=1}^6 x_i
\partial a_i/\partial x_j + a_j +  \sum_{i=1}^6 x_i \partial a_i/\partial x_j =
a_j.  
\end{equation}  
As a consequence, we see that the Hessian of
$F$ with respect to $(x_1, \dots, x_6)$ is exactly $[\partial a_i/\partial
x_j]_{6 \times 6}$, which is known to be positive definite. This shows that the
function $F$ is a strictly locally convex function. Unfortunately, the space of
all strictly hyperideal simplices parametrized by the edge lengths is not
convex. Thus there is no global convexity for $F$. 

\subsection{}\textbf{Proof of Theorem \ref{thm2}.} To prove Theorem \ref{thm2}, one
uses equation \eqref{eq11} and Corollary \ref{cor5}. To begin with, we assume
that there are $n$ edges labelled by $1,2, \dots, n$. The length of the $i$th
edge is $x_i$ and the combinatorial curvature at the $i$th edge is $K_i$. If
$r$ is a tetrahedron in the ideal triangulation, let $I(r)$ be the set of all
indices $i$ so that the $i$th edge is a codimension-1 face of $r$. If $i \in
I(r)$, we use $a_i^r = a_i^r(x)$ to denote the dihedral angle of the strictly
hyperideal simplex $r$ at the $i$th edge in the hyperbolic cone metric $x$.
Define $a_i^r=0$ if  $i \notin I(r)$. Let $T^{(3)}$ be the set of all
tetrahedra in the ideal triangulation. Then  the combinatorial curvature $K_i $
is given by $ 2\pi - \sum_{r \in T^{(3)}} a_i^r$ by definition.  In particular, 
$$ dK_i/dt = -\sum_{r \in T^{(3)}} da_i^r/dt. 
$$ 
On the other hand, by the
chain rule, we can express 
$$ da_i^r/dt = \sum_{j=1}^n (\partial a_i^r/\partial
x_j)( dx_j/dt) = \sum_{j=1}^n \partial a_i^r/\partial x_j K_j.
$$ 
Thus we have,
$$ dK_i/dt =  \sum_{r \in T^{(3)}}  \sum_{j=1}^n (-\partial a_i^r/\partial x_j 
K_j). 
$$ 
In particular, if we express $dK_i/dt = \sum_{j=1}^n a_{ij} K_j$, then
$a_{ij} = -\sum_{r \in T^{(3)}} \partial a_i^r/\partial x_j$. Since the
Jacobian matrix $[\partial a_i/\partial x_j]_{6 \times 6}$ is symmetric, we
have $\partial a_i^r/\partial x_j =\partial a_j^r/\partial x_i$. (Indeed, by
definition $\partial a_i^r /\partial x_j =0$ unless $i, j \in I(r)$. In the
latter case, it follows from Corollary \ref{cor5}.)  Thus $a_{ij} = a_{ji}$. To
finish the proof, we need to show that the matrix $[a_{ij}]_{n \times n}$ is
negative definite. To this end, take a vector $(u_1, \dots, u_n)$ in $\mathbf
R^n$, and consider the quadratic  expression 
\begin{equation}\label{eq22}
\sum_{i, j =1}^n u_i u_j a_{ij} = - \sum_{r \in T^{(3)}} \sum_{i,j=1}^n u_i u_j
\partial a_i^r/\partial x_j.
\end{equation}  
For any fixed tetrahedron
$r \in T^{(3)}$, let $x_i^r = x_i$ when $i \in I(r)$. Then 
$$\sum_{i,j=1}^n u_i u_j \partial a_i^r/\partial x_j =\sum_{i,j \in I(r)} u_i
u_j \partial a_i^r/ \partial x_j^r.$$  
By Corollary \ref{cor5}, the expression
$-\sum_{i,j \in I(r)} u_i u_j \partial a_i^r/ \partial x_j^r$ is less than  or
equal to zero, and is zero if and only if $u_i =0$ for all indices $i \in I(r)$.
This shows first of all that 
the expression \eqref{eq22} is less than or equal to
zero. Furthermore, if it is zero, then all $u_i=0$ since each edge is adjacent
to some tetrahedron.  We have thus established that fact that the matrix
$[a_{ij}]_{ n \times n}$ is symmetric and negative definite. 

To see that the flow \eqref{eq11} is the gradient flow, recall that $L(M, T)$
denotes the open subset of $\mathbf R^n$ consisting of the hyperbolic cone
metrics associated to the ideal triangulation. To be more precise, if $x=(x_1,
\dots, x_n) \in L(M, T)$, then $x_i$ is the length of the $i$th edge in the
cone metric. For $x \in L(M, T)$, let $V = V(x)$ be the volume of the cone
metric which is the sum of all hyperbolic volumes of its strictly hyperideal
tetrahedra. Now define a function $H: L(M, T) \to \mathbf R$ by $H(x) = 2 V(x)
-\sum_{i=1}^n x_i K_i$ where $K_i$ is the curvature of the metric $x$ at the
$i$th edge.   We can express the function $H$ as the sum
\begin{equation}\label{eq23} H = \sum_{r \in T^{(3)}}  \Big( 
2V_r( x) + \sum_{ i
\in I(r) } x_i a_i^r\Big) - 2\pi \sum_{i=1}^n x_i  
\end{equation}  
where
$V_r(x) $ is the volume of the tetrahedron $r$ in the metric. By \eqref{eq21}
in the proof of Corollary \ref{cor5},  we have $\partial (2V_r(x) + \sum_{ i
\in I(r)} x_i a_i^r) /\partial x_j = a_j^r$. This implies that $\partial
H/\partial x_j = \sum_{ r \in T^{(3)}} a_j^r - 2\pi = -K_j$. Furthermore, the
Hessian matrix $[h_{ij}]$ of $H$ is given by  $h_{ij} =-\partial K_i/\partial
x_j = \sum_{r \in T^{(3)}} \partial a_i^r/\partial x_j = -a_{ij}$ where
$a_{ij}$ is the quantity used in the above proof. By the argument above, we see
that the Hessian of $H$ is positive definite. Thus the function $H$ is strictly
locally convex in $L(M, T)$. \qed

\subsection{}\textbf{Proof of Corollary \ref{cor3}.} This corollary easily
follows from Theorem \ref{thm2} by standard tools of differential
equations.

To see (a), that $f(t) = \sum_{i=1}^n K_i(t)^2$ is strictly decreasing in $t$, 
let us take the derivative $df(t)/dt$. We find $f'(t) = 2 \sum_{i=1}^n K_i(t)
dK_i/dt =2 \sum_{i, j =1}^n a_{ij} K_i K_j < 0$ unless $K_i(t) = 0$ for all
$i$.

To see (b),  first of all, observe that 
at the equilibrium points of \eqref{eq11}, all
curvatures $K_i(t)=0$ for all $i$. Thus there are no singularities at the
edges. This shows that the cone metric is a smooth complete hyperbolic metric
with totally geodesic boundary. Since the matrix $[a_{ij}]$ in \eqref{eq12} is
negative definite, the equilibrium points are always a local attractor. Thus
(c) follows. \qed
\medskip

We do not know if the equilibrium point is unique.


\subsection{}\textbf{Proof of Theorem \ref{thm4}.} Consider the
smooth map $\Pi: L(M, T) \to \mathbf R^n$ sending a metric $x$ to its curvature
$\Pi(x)=(K_1, \dots, K_n)$. By theorem \ref{thm1}, this map is the same as the
gradient map $\nabla(-H): L(M, T) \to \mathbf R^n$. Since the function $-H$ is
shown to be strictly locally concave, the gradient map is a local
diffeomorphism. This ends the proof. \qed

\section{Some remarks and questions} 

This work and \cite{CL} are
motivated by the work of Richard Hamilton \cite{Ha} on the Ricci flow.  The 
strategy of  \cite{Ha} is to find a flow deforming metrics such that its
curvature evolves according to a heat-type equation. The main focus of study
then shifts from the evolution of the metrics to the evolution of its curvature
using the maximum principle for heat equations. As long as one has control of
the curvature evolution, one gets some control of the metric evolutions by
studying either the singularity formation or the long time convergence. Theorem
\ref{thm2} above seems to indicate that the combinatorial flow \eqref{eq11}
deforms the cone metrics in the ``right" direction. There remains the task of
understanding the singularity formations in \eqref{eq11} which corresponds to
the degeneration of the strictly hyperideal tetrahedra. This is being
investigated. Below are some thoughts on this topic.  The motivations come from
\cite{Th},  \cite{CV}, \cite{Ri2}, \cite{Le} and \cite{CL}.
\subsection{}  
To understand the singularity formation, we will focus our
attention on the function $H$ in  \eqref{eq23}.  Following \cite{CV} and
\cite{Ri2}, our goal is to find the (linear) conditions on the ideal
triangulation which will guarantee the existence of critical points for
$H$. The existence of the ideal triangulation satisfying the (linear) condition
will be related to the topology of the 3-manifold and will be resolved by
topological arguments.

Suppose $(M, T)$ is an ideal triangulated compact 3-manifold such that  each
boundary component of $M$ has negative Euler characteristic. A pair $(e, t)$,
where $e$ is an edge and $t$ is a tetrahedron containing $e$, is called a {\it
corner} in $T$.  Following Rivin \cite{Ri1}, and 
Casson and Lackenby \cite{La1}, we
say that the triangulated manifold $(M, T)$  supports a {\it linear hyperbolic
structure} if one can assign to each corner of $T$ a positive number called the
{\it dihedral angle} so that (1) the sum of  dihedral angles of all  corners
adjacent to each fixed edge be $2\pi$, and (2) the sum of  dihedral angles of
every triple of corners $(e_1, t), (e_2, t), (e_3, t)$, where $e_1, e_2, e_3$
are adjacent to a fixed vertex, be less than $\pi$.  By \cite{BB},
given a linear hyperbolic structure on $(M, T)$, we can realize each individual
tetrahedron by a strictly hyperideal tetrahedron whose dihedral angles are the
assinged numbers so that the sum of the dihedral angles at each edge be $2\pi$.
It can be shown  that if $(M, T)$ supports a linear hyperbolic structure, then
the manifold $M$ is irreducible without incompressible tori. One would ask if
the converse is also true.  The work of Lackenby \cite{La2} gives some
evidences that the following may have a positive answer. See also 
\cite{KR}.

\begin{question} Suppose $M$ is a compact irreducible
3-manifold with incompressible boundary consisting of surfaces of negative
Euler characteristic. If $M$ contains no incompressible tori and annuli,  is
there any ideal triangulation of $M$ which supports a linear hyperbolic
structure? \end{question} 

The next question is the 3-dimensional analog of the
2-dimensional singularity formation analysis presented in \cite{CV}, 
\cite{Ri2} and \cite{Le}. 

\begin{question} Suppose $(M, T)$ is an ideal
triangulated 3-manifold which supports a  linear hyperbolic structure. Does 
$H$ have a local minimal point in the space $L(M,T)$ of all cone metrics
associated to $(M, T)$? \end{question} 

Positive resolutions of these two
questions  will produce a new proof of Thurston's geometrization theorem for
this class of 3-manifolds. 
\subsection{}   
Suppose $(M,T)$ supports a
linear hyperbolic structure. We define the volume  of a linear hyperbolic
structure to be the sum of the volumes of its strictly hyperideal tetrahedra. 
Let $LH(M,T)$ be the space of all linear hyperbolic structures on $(M,T)$. It
can be shown, using Lagrangian multipliers, that the volume function is
strictly concave on $LH(M,T)$ whose maximal point is exactly the complete
hyperbolic metric on $M$. The situation is the same as for
ideal triangulations of
compact 3-manifolds whose boundary consists of tori. In this case, one realizes
each tetrahedron by an ideal tetrahedron in the hyperbolic space. It can  be
shown that the complete hyperbolic metric of finite volume is exactly equal to
the maximal point of the volume function defined on the space of all linear
hyperbolic structures given in \cite{Ri1}. This was also observed by Rivin
\cite{Ri3}.

\subsection{}  
We remark that the moduli space of all strictly hyperideal
tetrahedra para\-me\-trized by their edge lengths $x_1, \dots, x_6$ is not a
convex subset of $\mathbf R^6$. This is the main reason that we have only local
convergence and local rigidity in Corollary \ref{cor3} and 
Theorem \ref{thm4}. However,
it is conceivable that Theorem \ref{thm4} may still be true globally. We do not
know yet if the space  $L(M, T)$ of all cone metrics associated to the ideal
triangulated manifold is homeomorphic to a Euclidean space. It is likely to be
the case. In fact, one would hope that there is a diffeomorphism $h: \mathbf
R_{>0} \to \mathbf R_{>0}$ so that if we parameterize the space of all strictly
hyperideal tetrahedra by $(t_1, \dots, t_6)=(h(x_1), \dots, h(x_6))$, then  the
space becomes convex in $t$-coordinate. Evidently, if this holds, it implies
that
the space $L(M, T)$ is convex in the $t$-coordinate.

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\end{document}