\documentstyle[11pt]{article}
\title{The Thin Sandwich Conjecture \\[8mm]
  Ph.D. Thesis
}
\author{Gyula Fodor\thanks{This work has been done with the
   supervising of Professor Robert Bartnik, [1]} \\[8mm]
  H.A.S. C.R.I.P. \\
  Research Institute for \\
  Particle and Nuclear Physics \\
  H-1525 Budapest P.O.B. 49 \\
  Hungary
}
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\begin{document}
 
\maketitle
 
\begin{abstract}
 
In this paper we give a proof of the Thin Sandwich Conjecture of general
relativity for a restrictive but open set of initial data
$(g_{ab},\gd ab)$. In this case we show
the existence of unique solutions for the Thin Sandwich Equations and the
ellipticity of these equations. As a prominent application we examine the
Robertson-Walker spacetimes.
 
\end{abstract}
 
 
\section{Introduction}
 
 The  Thick  and  Thin  Sandwich  Conjectures  were  formulated by
Baierlein, Sharp and  Wheeler\cite{BaShWh}, and discussed  in more
detail  by   Wheeler\cite{Whe1}\cite{Whe2}\cite{MTW}.   From   the
viewpoint adopted in these papers, a spacetime is considered as  a
``geometrodynamical  history''  of  three-geometries.   One   such
three-geometry is characterized by a positive definite metric, (or
by a diffeomorphism equivalence class of them,) that is by a point
in superspace.  This picture is  also the basis of Feynman's  path
integral  approach  towards  quantum  gravity.  However, given two
points in superspace,  there is no  unique curve connecting  them,
which belongs to some  solution of the Einstein's  equations.  The
two  points  correspond  to  two  hypersurfaces  in  the embedding
spacetime, and if  we choose different  slicing between these  two
surfaces, we  get different  connecting paths  in superspace.  The
Sandwich Conjecture claims that this  is the only reason for  this
nonuniqueness.
 
\vskip9pt\noindent{\bf Sandwich Conjecture:}
{\em Given two three dimensional geometries, $g^{(1)}_{ab}$ and $g^{(2)}_{ab}$,
there exists a spacetime satisfying the vacuum Einstein's equations and
containing two Cauchy hypersurfaces with the intrinsic geometry $g^{(1)}_{ab}$ and
$g^{(2)}_{ab}$. This embedding spacetime is unique.}
 
\vskip9pt\noindent Contrary to the classical mechanical and
electrodynamical analogue of this conjecture, we do not have to specify the
time difference between the two configuration. The time difference between two
hypersurfaces is not even a globally and invariantly defined concept. On the
other hand the proper time interval along any timelike curve is determined by
the geometry of the spacetime, which is assumed to be fixed by the two spacelike
geometries. This would mean that in general relativity two 3-geometries contain
all the information about space and time. Or in other words, three-geometry is
the carrier of information about time. Essentially nothing is known about the
validity of the Sandwich Conjecture. One would expect that the above formulation
is excessively optimistic, and that serious restrictions would be needed on
$g^{(1)}_{ab}$ and $g^{(2)}_{ab}$ before the existence of a connecting spacetime
could be asserted. But the nature of such restrictions and the mathematical
methods by which the results could be demonstrated remain conjectural at best.
 
In view of the complexity of the problem, it is natural to investigate a
simplified version first, where the two hypersurfaces are infinitesimally close
to each other.
 
\vskip9pt\noindent{\bf Thin Sandwich Conjecture:}
{\em Suppose that a positive definite metric $g_{ab}$ and a symmetric tensor
field $\gd ab$ are given on a three-dimensional manifold \M. Then there exists a
spacetime satisfying the vacuum Einstein's equations, such that on a Cauchy
hypersurface of this spacetime the induced metric is $g_{ab}$, and the time
parameter derivative along a time flow vector field of the spacelike part of
the metric is $\gd ab$. This spacetime is unique on the Cauchy development of
the initial hypersurface.}
 
\vskip9pt\noindent
We note that here $\gd ab$ is just a notation for a tensor field independent
of $g_{ab}$, the dot does not denote a time derivative.
The Thin Sandwich Conjecture in this form is generally not
true. It is possible to specify $g_{ab}$ and $\gd ab$ such that no embedding
spacetime exists, and there are cases when the imbedding spacetime is not
unique. However we will prove that the conjecture is true for a large open
subset of initial data.
 
Suppose that the induced spacelike metric $g_{ab}$ and the extrinsic curvature
$K_{ab}$ is given on a three dimensional hypersurface, satisfying the
gravitational constraint equations. To simplify the situation, we will assume
that the hypersurface is compact and without boundary. Possibly this restriction
may be dropped after future investigation. Any spacelike slice of a spacetime
satisfying Einstein's equations generates such $g_{ab}$ and $K_{ab}$. On the
other hand, this data determine a unique spacetime on the Cauchy
development\cite{ChoBru}\cite{ChBrGe}.
We can choose a time flow vector field which is defined at least on the initial
hypersurface. This vector field can be specified by its hypersurface tangent and
orthogonal parts, which are called the shift vector field and the lapse function.
Calculating the time parameter derivative of the spacelike metric along this
vector field, we get a symmetric tensor field $\frac d{dt}g_{ab}\equiv\gd ab$
on the initial
hypersurface. Assuming that $(g_{ab},\gd ab)$ is generated by a solution
$(g_{ab},K_{ab})$ of the constraints, we will prove that the Thin Sandwich
Conjecture is true on an open neighbourhood of this $(g_{ab},\gd ab)$ if the
following two conditions hold:
 
\vskip9pt\noindent{\bf Condition 1a:}
{\em
\bb
2\e-R>0,
\ee
where $R$ is the scalar curvature belonging to the induced
three-metric $g_{ab}$ and $\e$ is the energy density of the (apriori prescribed)
matter fields on the initial hypersurface.}
 
\vskip9pt\noindent{\bf Condition 2a:}
{\em There is no choice of time flow vector field on the spacetime induced by
$(g_{ab},K_{ab})$, for which
\bb
\frac d{dt}g_{ab}\equiv\gd ab=0
\ee
on the initial hypersurface.}
 
\vskip9pt\noindent We will see that both of these conditions can be stated in
the following equivalent forms:
 
\vskip9pt\noindent{\bf Condition 1b:}
{\em The trace of the extrinsic curvature $K_{ab}$ dominates the
trace-free part $\bk_{ab}=K_{ab}-\frac13g_{ab}K^c_c$,
\bb
\frac23(K^a_a)^2>\bk_{ab}\bk^{ab}.
\ee}
 
\vskip9pt\noindent{\bf Condition 2b:}
{\em The equation
\bb
M_{(a|b)}=\mu K_{ab}
\ee
for the vector field $M^a$ and scalar field $\mu$ on the initial hypersurface
has only the trivial solution $M^a=0$ and $\mu=0$.}
 
The Thin Sandwich Conjecture can be viewed in another way, not just as the
`thin' version of the Sandwich Conjecture. It can be viewed as a
parametrisation of the constraint manifold. The constraint manifold
$\C\subset\left\{(g_{ab},K_{ab})\right\}$ consists of those metric and
extrinsic curvature function pairs $(g_{ab},K_{ab})$ on a three dimensional
manifold, which satisfies the gravitational constraint equations. Given
$(g_{ab},\gd ab)$, supposing that the Thin Sandwich Conjecture is true, they
determine a unique pair $(g_{ab},K_{ab})$, satisfying the constraints.
However, it is only a parametrisation modulo time flow vector field changes,
since different time derivatives $\gd ab$ can belong to the same pair
$(g_{ab},K_{ab})$, as one changes the time flow vector field.  It has been
shown by Arms, Fisher, Marsden and
Moncrief\cite{FisMar}\cite{AMM}, that the constraint manifold
$\C$ really is a submanifold of the space of all possible configurations
$(g_{ab},K_{ab})$, except in the cases when the imbedding spacetime admits a
Killing vector field. At these nongeneric $(g_{ab},K_{ab})$ pairs, $\C$ has a
quadratic singularity. This shows that the Thin Sandwich Conjecture may also
fail in these cases, and indeed, Condition 2 is false if the imbedding
spacetime admits Killing vectors.
 
\vskip9pt In this paragraph we outline the structure of the paper.
In Section \ref{secive} we derive the equations of the
Thin Sandwich problem and a variational principle leading to them. This was
originally done by Baierlein, Sharp and Wheeler\cite{BaShWh}, and extended by
Wheeler\cite{Whe1} to include matter fields. The variational principle was used
by Belasco and Ohanian\cite{BelOha} to prove a uniqueness theorem under
Conditions 1 and 2. We present the theorem and its proof in Section
\ref{secuni}. We note that there is an error in the proof of the second part of
the uniqueness theorem of \cite{BelOha}, and the result claimed there is easily
seen to be incorrect. In Section \ref{secexi}, using the
Implicit Function theorem we prove the existence and uniqueness of solutions of
the equations of the Thin Sandwich problem in a neighbourhood of a $(g_{ab},\gd
ab)$ generated by a solution $(g_{ab},K_{ab})$ of the constraint equations,
under the same Conditions 1 and 2 as in the uniqueness theorem. Under somewhat
less restrictive conditions we also prove the ellipticity of the linearised
Thin Sandwich Equations. They constitute an elliptic system, if and only if the
momentum tensor $\pi_{ab}=K_{ab}-g_{ab}K^c_c$ is positive or negative definite.
We will show that this is the case if Condition 1 holds. In section \ref{seccon}
we examine the Conditions 1 and 2 of existence and uniqueness. We will show that
Condition 2 fails to hold if the spacetime generated by $(g_{ab},K_{ab})$ admits
a vector field which is Killing at least on the initial hypersurface. The
difficulty that arises is very similar to that occurring in the
analysis of linearisation stability of the Einstein constraint
equations\cite{AMM}, in a neighbourhood of an initial data for
which the induced spacetime admits a Killing vector field.
Probably
one will be able to prove existence in the future even if Condition 2 is
false, and the nonuniqueness in this case may due only to different choices of
hypersurfaces. Condition 1 may seem too restrictive at first, but in section
\ref{secexa} we will see that it is  satisfied by most of the homogeneous
isotropic cosmology solutions. Following Belasco and Ohanian\cite{BelOha}, we
present a class of $(g_{ab},\gd ab)$ for which no embedding spacetime exists
with time flow vector field pointing everywhere to the same `future' side of
the hypersurface. This does not contradict our existence result, since we
prove existence only in a neighbourhood of a known solution. Finally, in
section \ref{secoth} we examine a Thin Sandwich formulation of
general relativity with different variables, introduced by
Komar\cite{Kom2}, and discuss the possibility of proving results
similar to ours for these equations.
 
We give a summary of some other important previous results.
Komar\cite{Kom0} has derived the vacuum Thin Sandwich Equations using canonical
methods. Bergmann\cite{Ber0} has calculated the linearised
Thin Sandwich Equations, but he did not examine the conditions for their
ellipticity. Christodoulou and Francaviglia\cite{ChrFra} have given an analysis
of the Thin Sandwich Equations in the case when the lapse function takes the
constant 1 value on the whole manifold. Naturally enough they found
nonexistence of
solutions under this unnatural condition. However their analysis shows
nonuniqueness if the three-dimensional initial hypersurface has flat regions.
Pereira\cite{Per} has given a Cauchy-Kowalewski type proof for the local
existence of unique analytic solutions of the Thin Sandwich Equations in a
small open set of the three dimensional initial hypersurface. These solutions
belong to analytic initial data, most of them are given on a two dimensional
surface in the initial hypersurface, and one is given on a line in this surface.
The initial data are the lapse and shift functions and their various partial
derivatives. The analysis of Pereira requires implicit conditions on the
initial data, and the approach is not natural either from geometrodynamical or
from partial differential equations viewpoint. Teitelboim\cite{Tei} has
presented a discussion about the quantum
mechanical origin of the Thin Sandwich Conjecture, including the case where the
two sides of the sandwich intersects each other.
 
There is an obvious objection to the sandwich philosophy which is also a
motivation for the Thin Sandwich formalism rather than the `Thick' Sandwich
formalism.
There are problems even with the classical mechanical and electrodynamical
analogue of the `Thick' Sandwich Conjecture. One can look at the motion of a
classical harmonic oscillator by specifying the coordinates at two moments. The
problem with the oscillator arises when the time difference is a multiple of
the half period. If it is exactly half period, and the two coordinates are not
negative of each other, then no solution exists for the time development, but
if they are negative of each other, then the solution is not unique. One can
avoid this trouble by looking only at time differences smaller than the half
period.
However, in electrodynamics there are arbitrarily high frequency
modes. In the electromagnetic version of the Thin Sandwich
Conjecture, the vector potential (or the magnetic field) is given
at two moments. In a coordinate system $x^\mu=(t,x^1,x^2,x^3)$,
let the components of the vector potential be $A^0=0$,
$A^1=\sin (nt)\sin (nx^2)$, $A^2=A^3=0$, for some $n>0$ integer.
This satisfies the
$\p^\mu A_\mu=0$ Lorentz condition, and the $\p^\nu\p_\nu A^\mu=0$
Maxwell's equation, so it can be considered as a standing wave
solution on a $2\pi$ sized 3-torus. The vector potential vanishes
on the whole space if $t=0$ and $t=2\pi/n$. This shows, that
contrary to the other cases, if the time difference is $2\pi/n$,
then it is not enough to give the vector potential on the two
sides of the sandwich to fix the electromagnetic field inside.
To overcome this problem it is natural to turn to the limit
where the time difference goes to zero. This means the
specification of the vector potential (or
alternatively the magnetic field) and its time derivative at a
moment. The analogue of
this in general relativity is to give the spacelike metric and its time
derivative on a hypersurface, to determine the structure of the spacetime. Thus
we arrive at the Thin Sandwich Conjecture.
 
We will use Greek indices for spacetime tensors. These indices will be raised
and lowered with the spacetime metric $\ti g_{\al\be}$. We denote the future
directed unit normal vector field of the spacelike hypersurfaces by $n^\al$,
$n^\al n_\al=-1$, and the metric orthogonal projection tensor into the
hypersurfaces by $h^\al_\be=\de^\al_\be+n^\al n_\be$. Tensors orthogonal in all
the indices to $n^\al$ and $n_\al$ are in one-one correspondence with the
tensors in the three dimensional hypersurfaces. The easiest to see this in the
natural coordinate system, where the $x^0$ coordinate is adapted to the
hypersurfaces. There $n_1=n_2=n_3=0$, and a vector field $v^\al$ is orthogonal
to $n_\al$ if and only if $v^0=0$. This is not true for an $n^\al$ orthogonal
one-form $u_\al$, but the value of $u_0$ is unimportant if we only contract
with $n_\al$ orthogonal tensors. In equations that contain only objects
defined in the three dimensional hypersurfaces we will use Latin indices.
These equations can also be interpreted as equations for four dimensional
spacetime tensors orthogonal in all indices to $n^\al$ and $n_\al$. Latin
indices will be raised and lowered by the spacelike hypersurface metric
$g_{ab}$, which corresponds to the projection of the spacetime metric
$g_{\al\be}=h^\g_\al h^\de_\be \ti g_{\g\de}$.
 
 
 
\section{The Thin Sandwich Equations}\label{secive}
 
 
 
In this section we derive the Thin Sandwich Equations from the gravitational
constraints. We also present a variational principle leading to them. The
induced metric and the extrinsic curvature as initial data given on a
hypersurface completely determine the spacetime metric in the Cauchy
development. Four of the ten Einstein's equations, the hypersurface orthogonal
components, do not contain any time derivative of these tensors, therefore they
are regarded as constraints on the initial data. The energy and momentum
constraints are
\bb \frac12\left[R+\left(K^a_a\right)^2-K^{ab}K_{ab}\right]=\e \label{enco}
\ee
and
\bb
K^{ab}_{\ \ |b}-K^{b|a}_b=S^a,\label{imco}
\ee
where $|$ and $R$ denote respectively the covariant derivative and the scalar
curvature associated to the metric $g_{ab}$ of the spacelike hypersurface.
$K_{ab}$ is the extrinsic curvature, $\e$ is the energy density and $S^a$ is the
density of the energy flow. The energy density and the energy flow are related
to the hypersurface orthogonal components of the stress-energy tensor
$T_{\al\be}$ by
\bb
\e=8\pi n_\al n_\be T^{\al\be},\ \ \ \ \
S^\al=8\pi h^\al_\be n_\g T^{\be\g}.
\ee
In the following we suppose that $\e$ and $S^a$ is specified
apriori on the initial hypersurface, as is the case for example with vacuum or
with perfect fluids in a Robertson-Walker cosmological model.
 
We specify a time flow vector field by giving its hypersurface parallel and
orthogonal parts. The hypersurface parallel part $N^a$ is called shift vector,
and the orthogonal part $N$ is called lapse function. This means that in
adapted coordinates $(t,x^a)$ the spacetime metric has the form
\bb
ds^2=-N^2dt^2+g_{ab}(dx^a+N^adt)(dx^b+N^bdt).
\ee
The derivative $\frac d{dt}g_{ab}\equiv\gd ab$ of the spacelike part of the
spacetime metric along this time flow vector field can be written in terms of
the extrinsic curvature,
\bb
\dot g_{ab}=2NK_{ab}+2N_{(a|b)}\ . \label{gdot}
\ee
Expressing the extrinsic curvature
\bb
K_{ab}=\frac1N\left(\frac12\gd ab-N_{(a|b)}\right), \label{excur}
\ee
we can substitute into the energy and momentum constraints,
\bb
N^2(2\e-R)=\left(\frac12g^{ab}\gd ab-N^a_{\ |a}\right)^2
-\left(\frac12\gd ab-N_{(a|b)}\right)\left(\frac12g^{ac}g^{bd}\gd cd-
N^{(a|b)}\right), \label{loenco}
\ee
and
\bb
\left[\frac1N\left(\frac12g^{ac}g^{bd}\gd cd-N^{(a|b)}-\frac12g^{ab}g^{cd}
\gd cd+g^{ab}N^c_{\ |c}\right)\right]_{|b}=S^a. \label{loimco}
\ee
If the three-metric $g_{ab}$ and instead of the extrinsic curvature the time
derivative $\gd ab$ of the three-metric is given as initial data together with
$\e$ and $S^a$, then equations (\ref{loenco}) and (\ref{loimco}) are four
differential equations for the lapse and shift, called the {\em Thin Sandwich
Equations}. The basic form of the Thin Sandwich Conjecture is
 
\vskip9pt\noindent{\bf Thin Sandwich Conjecture 1:}
{\em Suppose that a positive definite metric $g_{ab}$, a symmetric tensor
field $\gd ab$, a non negative energy density $\e$ and an energy flow density
$S^a$ is given on a three-dimensional manifold \M. Then the Thin Sandwich
Equations (\ref{loenco}) and (\ref{loimco}) have a unique solution for $N^a$
and $N$.}
 
\vskip9pt\noindent $N^a$ and $N$ determine the extrinsic curvature $K_{ab}$ of
the spacelike hypersurface by equation (\ref{excur}). If the matter fields
obey some physically reasonable conditions, (certainly vacuum is ok),
$(g_{ab},K_{ab},\e,S^a)$ and possibly some other initial data for the matter
fields specify a spacetime containing the hypersurface, which is unique in the
Cauchy development\cite{ChoBru}\cite{ChBrGe}.
We can write the conjecture in a form which is equivalent at
least in the vacuum case,
 
\vskip9pt\noindent{\bf Thin Sandwich Conjecture 2:}
{\em Suppose that a positive definite metric $g_{ab}$, a symmetric tensor
field $\gd ab$, a non negative energy density $\e$, an energy flow density
$S^a$ and appropriate initial data for the matter fields are given on a
three-dimensional manifold \M. Then there exists a spacetime satisfying the
Einstein's and the matter field equations, such that on a Cauchy hypersurface
of this spacetime the induced metric is $g_{ab}$, and the time parameter
derivative along a time flow vector field  of the spacelike part of the metric
is $\gd ab$. This spacetime is unique on the Cauchy development of the initial
hypersurface.}
 
\vskip9pt
It is clear that there exist data $(g_{ab},\gd ab,\e,S^a,N^a,N)$ satisfying the
Thin Sandwich Equations. Any choice of spacelike hypersurface in a known
solution of Einstein's equations generates $(g_{ab},K_{ab},\e,S^a)$ satisfying
the constraint equations. Any further choice of shift $N^a$ and positive lapse
$N$ defines the time derivative $\gd ab$ of the spacelike metric in the
direction fixed by them. Obviously $(g_{ab},\gd ab,\e,S^a,N^a,N)$ satisfies the
Thin Sandwich Equations. However, this does not answer the Thin Sandwich
Conjecture, which starts at $(g_{ab},\gd ab,\e,S^a)$ and seeks $N^a$ and $N$.
 
To reduce the size of the equations, we introduce as a notation the tensor
\bb
\g_{ab}=\g_{ab}(g_{ab},\gd ab,N^a)
=\frac12\gd ab-N_{(a|b)}\ , \label{gamma}
\ee
and the scalar
\bb
\gc=\gc(g_{ab},\gd ab,N^a)
=\left(\g^a_a\right)^2-\g^{ab}\g_{ab}\ .\label{gammas}
\ee
The Thin Sandwich Equations can be written in the compact form
\bb
N^2(2\e-R)=\gc  \label{shenco}
\ee
and
\bb
\left[\frac1N\left(\g^{ab}-g^{ab}\g^c_c\right)\right]_{|b}=S^a. \label{shimco}
\ee
 
The second derivative of the lapse function is not present in the Thin Sandwich
Equations. This makes it difficult to analyse the elliptic or hyperbolic
character of the linearisation of these equations. It is possible to overcome
this difficulty if
\begin{assu}\label{assuer}
The three dimensional metric $g_{ab}$ and the energy density $\e$ is given such
that
\bb
2\e-R\not=0
\ee
at all points of the initial three dimensional manifold \M.
\end{assu}
If we know apriori that the lapse function $N$ is strictly
positive, then we can use the scalar Thin Sandwich Equation
(\ref{shenco}) to eliminate the lapse $N$ from the vectorial Thin
Sandwich Equations (\ref{shimco}),
\bb
N=\sqrt{\frac\gc{2\e-R}}\ \ ,   \label{lapse}
\ee
thereby obtain a nonlinear system of equations for the shift vector $N^a$
alone,
\bb
\left[\sqrt{\frac{2\e-R}\gc}\left(\g^{ab}
-g^{ab}\g^e_e\right)\right]_{|b}=S^a. \label{cive}
\ee
However, Assumption \ref{assuer} does not apriori assure that
$N>0$. If $N$ and $N^a$ satisfies the Thin Sandwich Equations,
then equation (\ref{shenco}) implies that the square root in
(\ref{lapse}) is well defined, but in (\ref{cive}) we have to
assume apriori that $\gc>0$.
 
If we substitute $\g_{ab}$ and $\gc$ from equation (\ref{gamma}) and
(\ref{gammas}), we get the detailed form
\bb
\left\{
\frac
{\frac12g^{ac}g^{bd}\gd cd-N^{(a|b)}-\frac12g^{ab}g^{cd}
\gd cd+g^{ab}N^c_{\ |c}}
{\sqrt{\frac1{2\e-R}\left[\left(\frac12g^{ef}\gd ef-N^e_{\ |e}\right)^2
-\left(\frac12\gd ef-N_{(e|f)}\right)\left(\frac12g^{eh}g^{fi}\gd hi-
N^{(e|f)}\right)\right]}}
\right\}_{|b}=S^a.\label{civel}
\ee
This is a nonlinear system of differential equations for the shift $N^a$, called
the {\em Reduced Thin Sandwich Equations}. If Assumption \ref{assuer} holds,
and we know apriori that the lapse $N$ is positive, then the
Reduced Thin Sandwich Equations are equivalent to the Thin Sandwich Equations
(\ref{loenco}) and (\ref{loimco}) . The cost of this simplification, besides the
more complicated algebraic structure of the new equations, is that we cannot
consider cases where $2\e-R$ or $N$ changes sign in our three dimensional
manifold. The real advantage is that if these two conditions and a simple
further condition hold, then the linearised
Reduced Thin Sandwich Equations are elliptic, as we shall see later. The
condition $N>0$ on the solution is equivalent to the assumption that in the
resulting spacetime the time flow vector field points out to the
same side on the whole initial hypersurface, it cannot be
tangent to this surface anywhere.
 
Calculating the derivative of the quotient in the vectorial Thin
Sandwich Equation (\ref{shimco}), and multiplying by $N^2$, we get
\bb
N\left(\g^{ab}-g^{ab}\g^c_c\right)_{|b}
-N_{|b}\left(\g^{ab}-g^{ab}\g^c_c\right)=N^2S^a.
\ee
Those solutions of
this equation for which $N=0$ at some points of the
manifold \M\  may be problematic, because at those points the extrinsic curvature
is not defined by equation (\ref{excur}). If $K^{ab}$ cannot be defined by
continuous extension from a neighbourhood, then it is possible that no embedding
spacetime exists, or if it exists then it may not be unique.
 
The apriori condition $N>0$ is ultimately a restriction on the class of
admissible $\gd ab$. Clearly, given a solution $(g_{ab},K_{ab},\e,S^a)$ of the
constraints, using equation (\ref{gdot}), one can construct $\gd ab$ for which
$N=0$ somewhere.  Hence, for the Thin Sandwich Equations to be solvable (with
$N>0$), $\gd ab$ must be restricted in some form.
 
{\em Remark:} Calculating the derivative of the square root in equation
(\ref{cive}) we get
\bb
\frac12\frac1{\sqrt{2\e-R}}\left(\gc\right)^{-\frac32}\left[
2\gc(2\e-R)(\g^{ab}-g^{ab}\g^c_c)_{|b}+\gc(2\e-R)_{|b}
-(2\e-R){\gc}_{|b}\right]=S^a.
\ee
This shows that in the vacuum case $S^a=0$ the square root can be eliminated
from the Reduced Thin Sandwich Equations.
 
 
The Thin Sandwich Conjecture generally can only be true if suitable boundary
conditions are given. To avoid technical problems, in the rest of the paper we
suppose
\begin{assu}
The three dimensional manifold \M\  representing the initial hypersurface is
compact, connected and without boundary.
\end{assu}
In this case we do not have to specify boundary conditions. The results in this
paper are probably true in the noncompact case if suitable boundary conditions
are given, but this needs further investigation. The most probable next step
would be the assymptotically hyperboloidal spacetimes. The assymptotically
flat vacuum spacetimes are unfortunately precluded by Assumption \ref{assuer}.
 
The Reduced Thin Sandwich Equations can be derived from a variational
principle\cite{Whe1}.
\begin{lemma}\label{leula}
The Reduced Thin Sandwich Equations (\ref{cive}) are the Euler-Lagrange
equations belonging to the action
\bb
I=-2\ii\left\{\sqrt{\gc(2\e-R)}+S_aN^a\right\}, \label{civp}
\ee
where the integral is performed on the three dimensional initial manifold \M,
with respect to the natural volume element $dv$ belonging to the spacelike
metric $g_{ab}$.
\end{lemma}
This can be verified directly by variation with respect to $N^a$, or can be
obtained by the following argument.
 
As all the field equations, the constraints also can be obtained by extremizing
the action $I=I_G+I_M$, where $I_M$ is the action associated to the matter
fields, $I_G=\int d^4x\sqrt{-\ti g}\ti R$, $\ti g$ is the determinant of the
spacetime metric $\ti g_{\al\be}$, and $\ti R$ is the scalar curvature
belonging to $\ti g_{\al\be}$. After dropping boundary terms $I_G$ can be
written in terms of three dimensional quantities,
\bb
I_G=\int d^4x\sqrt gN\left[R+K_{ab}K^{ab}-(K^a_a)^2\right]
=\int d^4x\sqrt g\left[NR-\frac1N\gc\right].
\ee
The action does not contain time derivatives of $N$ and $N^a$, and there are
no second time derivatives of the other variables either, so by varying with
respect to $N$ and $N^a$ we will get constraints. The variation of $I_M$ with
respect to $N$ and $N^a$ is
\bb
\frac{\de I_M}{\de N}=
\frac{\de I_M}{\de \ti g^{\al\be}}\frac{\p \ti g^{\al\be}}{\p N}
\ , \ \ \ \ \ \ \ \ \ \
\frac{\de I_M}{\de N^\al}=
\frac{\de I_M}{\de \ti g^{\al\be}}
\frac{\p \ti g^{\al\be}}{\p N^\al}\ .
\ee
Using the definition of the stress-energy tensor
\bb
T_{\al\be}=-\frac1{8\pi}\frac{\de I_M}{\de \ti g^{\al\be}}\ ,
\ee
and the decomposition of the spacetime metric
\bb
\ti g^{\al\be}=\left(\begin{array}{cc}-\frac1{N^2}&\frac{N^b}{N^2}\\
\\
\frac{N^a}{N^2}&g^{ab}-\frac1{N^2}N^aN^b\end{array}\right),
\ee
we get
\bb
\frac{\de I_M}{\de N}=-16\pi\sqrt
gn^\al n^\be T_{\al\be}=-2\sqrt g\e\ ,
\ee
\bb
\frac{\de I_M}{\de N^\al}=-16\pi\sqrt gh^\be_\al n^\g T_{\be\g}
=-2\sqrt g S_\al\ .
\ee
We assume that $\e$ and $S^a$ are prescribed functions on the initial manifold
\M, hence are independent of $N$ and $N^a$, so if we are only interested in the
variations with respect to $N$ and $N^a$, we can write
\bb
I_M=\int d^4x\left(\frac{\de I_M}{\de N}N+\frac{\de I_M}{\de N^a}N^a\right)
=-2\int d^4x\sqrt g\left(\e N+S_aN^a\right).
\ee
Since there are no time derivatives of $N$ and $N^a$ in the  action, we
obtain the same equations at the variations by dropping the time integral and
integrating only in the hypersurface,
\bb
I=\iii\sqrt g\left[-(2\e-R)N-\frac1N\gc-2S_aN^a\right].
\ee
Easy to check that the variation with respect to $N$ and $N^a$ indeed gives the
Thin Sandwich Equations (\ref{loenco}) and (\ref{loimco}).
Using the form (\ref{lapse}) of the energy constraint to eliminate $N$, gives
the form (\ref{civp}) of the action, involving only the shift $N^a$.
 
 
\section{Uniqueness}\label{secuni}
 
 
In this section we present the proof of the uniqueness part of the
Thin Sandwich Conjecture, under the Assumptions 1a and 2b of the Introduction.
It is easy to see from the form (\ref{cive}) of the Reduced Thin Sandwich
Equations that if we have a solution $N^a$ with
\bb
\g_{ab}=\frac12\gd ab-N_{(a|b)}\ , \label{gamma2}
\ee
and if $\ti N^a$ is a vector field such that
\bb
\gt_{ab}=\frac12\gd ab-\ti N_{(a|b)}=\al\g_{ab} \label{gammat}
\ee
for some positive function $\al$ on the three-dimensional
manifold \M, then $\ti N^a$
also satisfies the Reduced Thin Sandwich Equations. Belasco and
Ohanian\cite{BelOha} have proved the converse if Condition 1a holds.
\begin{theo}
Suppose $2\e-R>0$. If $N^a$ and $\ti N^a$ are two solutions of the Reduced
Thin Sandwich Equations (\ref{civel}), there exists a function $\al>0$ such
that $\gt_{ab}=\al\g_{ab}$. Conversely, if $N^a$ is a solution and
$\ti N^a$ satisfies (\ref{gammat}), then $\ti N^a$ also satisfies
(\ref{civel}).
\end{theo}
{\em Remark:} If $2\e-R>0$, then solution of the Reduced Thin Sandwich
Equations means that $\gc>0$ apriori, and the lapse $N$ defined by
(\ref{lapse}) is positive.\hfill\break
\pr Let
\bb
\gtc=\left(\gt^a_a\right)^2-\gt_{ab}\gt^{ab}.
\ee
Since $N^a$ and $\ti N^a$ are supposed to be solutions of the Reduced
Thin Sandwich Equations (\ref{cive}),
\bb
\frac{2\e-R}\gc>0,\ \ \ \ \ \ \ \ \ \ \frac{2\e-R}\gtc>0.
\ee
The condition $2\e-R>0$ implies that
\bb
\gc>0, \ \ \ \ \ \ \ \ \ \ \ \ \ \gtc>0.
\ee
Since the metric is positive definite, $\g_{ab}\g^{ab}\geq0$, so $\g^a_a$, and
similarly $\gt^a_a$, cannot be zero at any point of the manifold \M. Using
their definition (\ref{gamma2}) and (\ref{gammat}),
\bb
\ii \left(\g^a_a-\gt^a_a\right)=
\ii \left(\ti N^a-N^a\right)_{|a}=0,
\ee
since our manifold \M is compact without boundary. Because of the conectedness
of the manifold this means that at some point $\g^a_a=\gt^a_a$, and since they
cannot change sign, it follows that $\g^a_a$ and $\gt^a_a$ have the same
sign at every point of the manifold \M.
 
We define
\bb
\G_{ab}(u)=u\g_{ab}+(1-u)\gt_{ab}\ ,  \label{Gamma}
\ee
and
\bb
\Gc(u)=\left(\G^a_a(u)\right)^2
-\G_{ab}(u)\G^{ab}(u)\ ,  \label{Gammac}
\ee
for $0\leq u\leq1$. Since we will use the square root of $\Gc(u)$ in the proof,
we first show that it is positive. Substituting (\ref{Gamma}) into
(\ref{Gammac}),
\bb
\Gc(u)=u^2\left[\gc+\gtc-2\ggc\right]
+2u\left[\ggc-\gtc\right]+\gtc,     \label{Gcu}
\ee
where we have introduced the notation
\bb
\ggc=\g^a_a\gt^b_b-\g_{ab}\gt^{ab}.
\ee
Straightforward to check, that since $\gc>0$ and $\gtc>0$,
$\Gc(u)$ has root in the $0\leq u\leq1$ interval if and only if
both of the following conditions hold,
\bb
\ggc<0,  \label{gyok1}
\ee
\bb
\left[\ggc\right]^2-\gc\gtc\geq0. \label{gyok2}
\ee
 
We fix a point on the manifold and consider $\G^a_b(u)$ there, denoting its
eigenvalues by $G_1(u)$, $G_2(u)$, $G_3(u)$. These can be calculated for
example in a local (coordinate) frame in which the metric is
Euclidean, $g_{ab}=\de_{ab}$. We can introduce a
three-dimensional space of eigenvalues, in which the above
eigenvalues are represented by the point with coordinates
$\left(G_1(u),G_2(u),G_3(u)\right)$. Easy to check that the
condition $\Gc(u)>0$ means that the eigenpoints lie inside a double cone whose
apex is the direction $(1,1,1)$, and which has the three coordinate axes lying
on its surface. The two halves of the cone are distinguished by the value of
$\G^a_a(u)=G_1(u)+G_2(u)+G_3(u)$: it is positive in one cone and negative in
the other. We have seen that $\g^a_a$ and $\gt^a_a$ are both negative or both
positive, so the eigenpoints of $\G^a_b(0)$ and $\G^a_b(1)$ lie in the same
cone. We can define another function $\Gb_{ab}(u)$ by substituting $-\gt_{ab}$
instead of $\gt_{ab}$ into the definition (\ref{Gamma}),
\bb
\Gb_{ab}(u)=u\g_{ab}+(1-u)(-\gt_{ab}),
\ee
and defining
\bb
\Gbc(u)=\left(\Gb^a_a(u)\right)^2-\Gb_{ab}(u)\Gb^{ab}(u)
\ee
in the same way as in (\ref{Gammac}). Now the eigenvalues of $\Gb^a_b(0)$
and $\Gb^a_b(1)$ lie in the opposite cones, therefore the curve of the
eigenvalues of $\Gb^a_b(u)$ for $0\leq u\leq1$ must intersect the surface of
the cones somewhere. This means that $\Gbc(u)$ has root in the
$0\leq u\leq1$ interval. Since $(-\gt)^*=\gtc>0$, the conditions of existence
of roots are quite similar to (\ref{gyok1}) and (\ref{gyok2}); we have only
changed the sign of $\gt_{ab}$. From the existence of the root it follows
that the following two inequalities hold:
\bb
\ggc>0,
\ee
\bb
\left[\ggc\right]^2-\gc\gtc\geq0. \label{gyok3}
\ee
Comparing the first with condition (\ref{gyok1}) we see that
$\Gc(u)$ cannot have root in the interval $0\leq u\leq1$. Since $\Gc(0)=\gc>0$,
it follows that $\Gc(u)>0$.
 
We can now define the following integral,
\bb
I(u)=-2\ii\left\{\sqrt{\Gc(u)(2\e-R)}+S_aN^a(u)\right\},
\ee
where
\bb
N^a(u)=uN^a+(1-u)\ti N^a,
\ee
for $0\leq u\leq1$. Since by Lemma \ref{leula} $N^a$ and $\ti N^a$ satisfies
the Euler-Lagrange equation (\ref{cive}) for the action (\ref{civp}), and
$\G_{ab}(0)=\g_{ab}$, $\G_{ab}(1)=\gt_{ab}$, obviously
\bb
I'(0)=I'(1)=0,
\ee
where the prime denotes derivative with respect to $u$. The second
derivative  of the integral $I(u)$ is
\bb
I''(u)=-2\ii\sqrt{(2\e-R)}\frac{d^2}{du^2}\sqrt{\Gc}\ .
\ee
Using the expression (\ref{Gcu}) for $\Gc(u)$ we get
\bb
I''(u)=2\ii\sqrt{(2\e-R)}(\Gc(u))^{-\frac32}
\left\{\left[\ggc\right]^2-\gc\gtc\right\}.
\ee
From (\ref{gyok3}) we see that $I''(u)\geq0$. This together with
$I'(0)=I'(1)=0$ means that $I''(u)$ must vanish on the interval $0\leq u\leq1$.
Since the integrand is non negative, it must be zero at every point of the
manifold \M, so we have
\bb
\left[\ggc\right]^2-\gc\gtc=0. \label{nulla}
\ee
We introduce the traceless tensor
\bb
q_{ab}=\g_{ab}-\frac{\g^c_c}{\gt^d_d}\gt_{ab}\ .
\ee
Expressing $\g_{ab}$ and substituting into the equation (\ref{nulla}), we get
\bb
\left[\ggc\right]^2-\gc\gtc
=\left(\gt_{ab}q^{ab}\right)^2+\gtc q_{ab}q^{ab}=0.
\ee
Because of the positive definiteness of the metric this can be zero only if
$q_{ab}=0$, that is
\bb
\g_{ab}=\frac{\g^c_c}{\gt^d_d}\gt_{ab}
=\al\gt_{ab}\ ,\ \ \ \ \ \ \ \ \
\al=\frac{\g^c_c}{\gt^d_d}>0.
\ee
Easy to check that if this condition holds then (\ref{nulla}) is true and
$I''(u)=0$. The proof is complete.\q
 
{\em Remark:} Belasco and Ohanian also claim that $\al$ should be $1$, but
there is a mistake in that part of the proof, because they forgot the cases
where the region where $\al\not=1$ consists of disjoint sets. We will present a
counterexample when examining Robertson-Walker spacetimes, which shows that the
result claimed there is incorrect.
 
If we introduce the notation $M^a=\ti N^a-N^a$ and $\mu=N(1-\al)$, we can write
the condition $\ti \g_{ab}=\al\g_{ab}$ into the form
\bb
M_{(a|b)}=\mu K_{ab}\ . \label{cond}
\ee
The  condition $\al>0$ means $\mu<N$. This is not particularly restrictive
because of the linearity of equation (\ref{cond}). We can reformulate the
previous theorem as
\begin{theo}\label{uni2}
Suppose that $2\e-R>0$, and there exists a solution $N^a$ of the
Reduced Thin Sandwich Equations (\ref{civel}). The solution is
unique if and only if the equation (\ref{cond}) has only the
trivial solution $M^a=0$, $\mu=0$. Furthermore, $M^a$ satisfies
(\ref{cond}) with $\mu<N$, if and only if $N^a+M^a$ is also a
solution of the Reduced Thin Sandwich Equations (\ref{civel}).
\end{theo}
 
 
\section{Existence}\label{secexi}
 
 
In this section we will apply the Implicit Function theorem to prove the
existence of solutions of the Reduced Thin Sandwich Equations
nearby a known
solution, under the Conditions 1a and 2b of the Introduction. Since we assume
that the original solution is a solution of the Reduced Thin Sandwich
Equations, it satisfies $\gc>0$ and $N>0$, and the solutions in a
neighbourhood will also satisfy these conditions. Let $H^n(T^i_j)$
denote the set of tensors of valence $(i,j)$ belonging to the Sobolev space
$H^n=W^{n,2}$ for $n\geq2$. We define the mapping
\bb
F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)=\left[\sqrt{\frac{2\e-R}\gc}
\left(\g^{ab}-g^{ab}\g^e_e\right)\right]_{|b}-S^a,
\ee
which assigns a vector field $F^a\in H^n(T^1_0)$ to any positive definite
symmetric tensor field $g_{ab}\in H^{n+2}(T^0_2)$, symmetric tensor field
$\gd ab\in H^{n+1}(T^0_2)$, scalar function $\e\in H^{n+1}$, vector field
$S^a\in H^n(T^1_0)$, and vector field $N^a\in H^{n+2}(T^1_0)$, where
$\g_{ab}=\frac12\gd ab-N_{(a|b)}$, and $\gc=(\g^a_a)^2-\g_{ab}\g^{ab}$.
If we assume $2\e-R>0$ and $\gc>0$, then
$F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$
is continuous on these function spaces. We will need the partial
derivative with respect to $N^a$, which is a linear map that
assigns the vector field
$D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)(M^b)\in H^n(T^1_0)$
to any vector field $M^a\in H^{n+2}(T^1_0)$,
\bbb
D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)(M^b)=\Biggl\{
\sqrt{\frac{2\e-R}\gc}\left(-M^{(a|b)}+g^{ab}M^e_{\ |e}\right)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\nonumber\\
-\frac12\sqrt{\frac{2\e-R}{(\gc)^3}}
\left[-2\g^c_cM^c_{\ |c}+2\g_{cd}M^{c|d}\right]
\left(\g^{ab}-g^{ab}\g^e_e\right)
\Biggr\}_{|b}\ .
\eee
Defining the non-density momentum tensor, as function of $g_{ab}$, $\gd ab$
and $N^a$,
\bb
\pi_{ab}=K_{ab}-g_{ab}K^c_c=\frac1N\left(\g_{ab}-g_{ab}\g^c_c\right),
\label{momen}
\ee
where now $N$ denotes the function defined by (\ref{lapse}), we can write the
partial derivative as
\bb
D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)(M^b)=\left(-\frac1NM^{(a|b)}
+\frac1Ng^{ab}M^c_{\ |c}-\frac1{N(2\e-R)}\pi^{ab}\pi_{cd}M^{c|d}
\right)_{|b}. \label{parder}
\ee
Easy to verify that
\begin{theo}\label{sadj}
The Reduced Thin Sandwich Operator
$D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$ is self adjoint with respect to
the $L^2$ inner product of the metric $g_{ab}$ on \M.
\end{theo}
\pr Let $M^a$ and $\bm^a$ be elements of $H^2(T^1_0)$. Then the $L^2$ inner
product
\bbb
\left(\bm^a,D_2F^a(M^b)\right)=\ii\left\{\bm_a\left(-\frac1NM^{(a|b)}
+\frac1Ng^{ab}M^c_{\ |c}-\frac1{N(2\e-R)}\pi^{ab}\pi_{cd}M^{c|d}
\right)_{|b}\right\}=\nonumber\\ \
\ii\left\{\left(-\frac1N\bm^{(a|b)}
+\frac1Ng^{ab}\bm^c_{\ |c}-\frac1{N(2\e-R)}\pi^{ab}\pi_{cd}\bm^{c|d}
\right)_{|b}M_a\right\}=\left(D_2F^a(\bm^b),M^a\right),
\eee
where we dropped divergence terms, since the three dimensional manifold
\M\  is compact and without boundary.\q
 
\begin{theo}
The differential operator $D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$ is
elliptic, if and only if $N\not=0$, $2\e-R\not=0$ and the momentum tensor
$\pi_{ab}$ belonging to $g_{ab}$, $\gd ab$ and $N^a$ is positive or negative
definite.
\end{theo}
\pr We have to look at the leading second order terms, which are
\bbb
\frac1NM^{[b,a]}_{\ \ \ \ \ ,b}-\frac1{N(2\e-R)}\pi^{ab}\pi_{cd}
M^{c,d}_{\ \ \ ,b}=\ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
\left(\frac1{2N}g^{ac}\p_b\p_c
-\frac1{2N}\de^a_bg^{cd}\p_c\p_d
-\frac1{N(2\e-R)}\pi^{ac}\pi^d_b\p_c\p_d\right)M^b. \label{sots}
\eee
Hence the leading symbol is
\bb
f^a_b(\xi)=-\frac1{2N}\xi^a\xi_b+\frac1{2N}\de^a_b\xi^c\xi_c
+\frac1{N(2\e-R)}\pi^a_c\pi^d_b\xi^c\xi_d\ .
\ee
The operator $D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$ is elliptic, if
the determinant of the leading symbol is not zero when $\xi^a\not=0$. It is
easier to calculate the determinant using an orthonormal frame, in which
$\xi^a$ has only one nonvanishing component, $\xi^a=(\xi,0,0)$. This frame is
not necessarily continuous, but we perform only algebraic operations at a
point. Then $g_{ab}=\de_{ab}$, and
\bb
\det\ f^a_b(\xi)=\det\left(\begin{array}{cc}\frac1{N(2\e-R)}\xi^2(\pi^1_1)^2&
\frac1{N(2\e-R)}\xi^2\pi^1_1\pi^1_b \\
\\ \frac1{N(2\e-R)}\xi^2\pi^a_1\pi^1_1 &
\frac1{2N}\de^a_b\xi^2+\frac1{N(2\e-R)}\xi^2\pi^a_1\pi^1_b \end{array}\right)
=\frac1{4N^3(2\e-R)}\xi^6(\pi^1_1)^2.
\ee
Thus, in general,
\bb
\det\ f^a_b(\xi)=\frac1{4N^3(2\e-R)}\xi^e\xi_e(\pi_{cd}\xi^c\xi^d)^2,
\ee
which is not zero for all $\xi^a\not=0$ if and only if $\pi_{ab}$ is
definite.\q
 
We can also look at the ellipticity of the linearised Reduced Thin Sandwich
Equations in the vicinity of a $(g_{ab},\gd ab,\e,S^a)$ for which there exists
a solution $N^a$. Since $N^a$ is a solution of the Reduced Thin Sandwich
Equations, $\gc>0$ and $N>0$. By equation (\ref{gdot}),
$(g_{ab},\gd ab,\e,S^a)$ can be generated by a solution
$(g_{ab},K_{ab},\e,S^a)$ of the constraint equations and arbitrary choice of
shift and positive lapse.
\begin{theo}\label{thel}
Suppose that given $g_{ab}$, $\gd ab$, $\e$ and $S^a$ such that $N^a$ is a
solution of the Reduced Thin Sandwich Equations (\ref{civel}),
$F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)=0$. Then for small $h_{ab}$,
$\dot h_{ab}$, $\ti\e$, $\ti S^a$ and $M^a$, in the linearised approximation,
the condition of $N^a+M^a$ being a solution with the metric $g_{ab}+h_{ab}$,
its time derivative $\gd ab+\dot h_{ab}$, energy density $\e+\ti\e$, and
density of the energy flow $S^a+\ti S^a$, gives a system of linear second
order partial differential equations for the small $M^a$,
\bb
F^a\left((g_{cd}+h_{ab},\gd cd+\dot h_{ab},\e+\ti\e,S^a+\ti S^a)
,N^c+M^a\right)=0.
\ee
This system is elliptic if and only if the momentum tensor $\pi_{ab}$ defined
by (\ref{momen}) and (\ref{excur}) is positive or negative definite.
\end{theo}
\pr If we are interested only in ellipticity, then we only have to look at
the leading second order terms, which are independent of the small $h_{ab}$,
$\dot h_{ab}$, $\ti\e$ and $\ti S^a$, and are obviously the same as in
(\ref{sots}), so the condition is the same as in the previous theorem.\q
 
\begin{lemma}\label{leel}
Suppose that $g_{ab}$, $K_{ab}$ and $\e$ satisfies the energy constraint
(\ref{enco}). If $2\e-R>0$, then $\pi_{ab}$ defined by (\ref{momen}) and
(\ref{excur}) is positive or negative definite.
\end{lemma}
\pr From the definition (\ref{momen}) of the momentum tensor,
$K^a_a=-\frac12\pi^a_a$ and $K_{ab}=\pi_{ab}-\frac12g_{ab}\pi^c_c$.
Substituting into the energy constraint
\bb
2\e-R=(K^a_a)^2-K^{ab}K_{ab}=
\frac12\left(\pi^a_a\right)^2-\pi^{ab}\pi_{ab}\ . \label{enf}
\ee
Hence it is enough to prove that if $\frac12(\pi^a_a)^2-\pi^{ab}\pi_{ab}>0$,
then $\pi_{ab}$ is positive or negative definite. It is the easiest to do this
using an orthonormal frame in which $\pi_{ab}$ is diagonal,
\bb
\pi_{ab}=\left(\begin{array}{ccc}\al&0&0\\0&\be&0\\0&0&\g\end{array}\right).
\ee
Since now $g_{ab}=\de_{ab}$,
\bbb
\frac12(\pi^a_a)^2-\pi^{ab}\pi_{ab}
=\frac12(\al+\be+\g)^2-\left(\al^2+\be^2+\g^2\right)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
=-\frac12\left(\al^2+\be^2+\g^2\right)+\al\be+\al\g+\be\g
=2\al\be-\frac12(\al+\be-\g)^2,
\eee
which can only be positive if $\al\be$ is positive, that is if $\al$ and $\be$
have the same sign. Similarly, $\al$ and $\g$ must have the same sign,
so $\pi_{ab}$ must be positive or negative definite.\q
 
{\em Remark:} This lemma is not true in the other direction, there exist
$\pi_{ab}$ which is positive (or negative) definite, but for which
$\frac12(\pi^a_a)^2-\pi^{ab}\pi_{ab}<0$. For example if $\al=\be=1$ and
$\g=6$ then $\frac12(\pi^a_a)^2-\pi^{ab}\pi_{ab}=-6$.
 
We have the corollary
\begin{coro}\label{nnee}
If $N^c$ is a solution of the Reduced Thin Sandwich Equations belonging to
the initial data $(g_{cd},\gd cd,\e,S^c)$, such that $2\e-R>0$ and $N>0$, then
the linearised Reduced Thin Sandwich Operator
$D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$ is elliptic.
\end{coro}
 
In order to apply the Implicit Function theorem, we need to show that
$D_2F^a\bigl((g_{cd},\gd cd,\e,S^c),\linebreak[4]N^c\bigr)$ is
an isomorphism. Since it is self adjoint, in the
elliptic case this is equivalent to the vanishing of its kernel.
\begin{lemma}\label{kernel}
Suppose that $N^c$ is a solution of the Reduced Thin Sandwich Equations
belonging to the initial data $(g_{cd},\gd cd,\e,S^c)$, such that $2\e-R>0$
and $N>0$. Then $M^a$ is an element of the kernel of the
operator $D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$, if and only
if there exists a scalar function $\mu$ such that $M^a$ and $\mu$ are solutions
of the equation
\bb
M_{(a|b)}=\mu K_{ab}\ .
\ee
\end{lemma}
\pr Suppose that $M^a$ is in the kernel of
$D_2F^a\left((g_{cd},\gd cd,\e,S^c),N^c\right)$. We have to look for the
nontrivial solutions of the following equation,
\bb
\left(-\frac1NM^{(a|b)}+\frac1Ng^{ab}M^c_{\ |c}-\frac1{N(2\e-R)}
\pi^{ab}\pi_{cd}M^{c|d}\right)_{|b}=0.
\ee
Multiplying with $M_a$, integrating on the three dimensional manifold \M, and
dropping divergence terms,
\bb
\ii\left[\frac1NM_{(a|b)}M^{a|b}-\frac1N\left(M^a_{\ |a}\right)^2
+\frac1{N(2\e-R)}\left(\pi^{ab}M_{a|b}\right)^2\right]=0.
\ee
Using equation (\ref{enf}), we get a form for the integral, which is more
homogeneous in the variable $\pi_{ab}$,
\bb
\ii\frac1{N(2\e-R)}\left\{\left[\frac12(\pi^c_c)^2-\pi^{cd}\pi_{cd}
\right]\left[M_{(a|b)}M^{a|b}-\left(M^a_{\ |a}\right)^2\right]
+\left(\pi^{ab}M_{a|b}\right)^2\right\}=0.
\ee
We can choose an orthonormal frame in which the extrinsic curvature
is diagonal. Let
\bb
g_{ab}=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right),
\ \ \ \ \
K_{ab}=\left(\begin{array}{ccc}\al&0&0\\0&\be&0\\0&0&\g\end{array}\right).
\ee
Then the momentum is also diagonal,
\bb
\pi_{ab}=\left(\begin{array}{ccc}-\be-\g&0&0\\0&-\al-\g&0\\0&0&-\al-\be
\end{array}\right).
\ee
From the energy constraint (\ref{enf}),
\bb
2\e-R=2(\al\be+\al\g+\be\g)\not=0.
\ee
We denote the components of $M_{(a|b)}$ by
\bb
M_{(a|b)}=\left(\begin{array}{ccc}a&d&e\\d&b&f\\e&f&c\end{array}\right).
\ee
It can be verified by a simple but lengthy calculation, that the expression in
the integral can be written in the form
\bbb\label{intpoz}
\left[\frac12(\pi^c_c)^2-\pi^{cd}\pi_{cd}\right]
\left[M_{(a|b)}M^{a|b}-\left(M^a_{\ |a}\right)^2\right]
+\left(\pi^{ab}M_{a|b}\right)^2=
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
=\left[\frac12(2\al+2\be+2\g)^2-(\be+\g)^2
-(\al+\g)^2-(\al+\be)^2\right]
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
\left[a^2+b^2+c^2+2d^2+2e^2+2f^2-(a+b+c)^2\right]
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
-\left[a(\be+\g)+b(\al+\g)+c(\al+\be)\right]^2=
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
  ={\al^2}{b^2} - 2a\al b\be + {a^2}{\be^2} + 2{\al^2}bc -
   2a\al\be c - 2\al b\be c + 2a{\be^2}c + {\al^2}{c^2}
\ \ \ \ \ \nonumber\\
   +2\al\be{c^2} + {\be^2}{c^2} - 2a\al b\g +
   2\al{b^2}\g + 2{a^2}\be\g - 2ab\be\g -
   2a\al c\g - 2\al bc\g
\nonumber\\
   - 2a\be c\g -
   2b\be c\g + {a^2}{{\g}^2} + 2ab{{\g}^2} +
   {b^2}{{\g}^2} + 4\al\be{d^2}+ 4\al\g{d^2} +
   4\be\g{d^2}
\ \ \ \nonumber\\
   + 4\al\be{e^2} + 4\al\g{e^2} +
   4\be\g{e^2} + 4\al\be{f^2} + 4\al\g{f^2} +
   4\be\g{f^2}=
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
=\frac1{(\be+\g)^2}\Biggl\{\left[(\be+\g)^2a+\g^2b+\be^2c
-(\al\be+\al\g+\be\g)(b+c)\right]^2
\ \ \ \ \ \ \ \nonumber\\
+4(\al\be+\al\g+\be\g)\left[(\be+\g)^2(d^2+e^2+f^2)
+(\g b-\be c)^2\right]\Biggr\}.\ \ \ \ \ \ \ \ \
\eee
If $2\e-R=2(\al\be+\al\g+\be\g)>0$, this expression cannot be negative, so it
must vanish on the whole manifold \M. Hence $d=e=f=0$, $\be c=\g b$ and
$(\be+\g)^2a+\g^2b+\be^2c-(\al\be+\al\g+\be\g)(b+c)=0$. Multiplying with
$\be$ and eliminating $\be c$,
\bb
0=\be\left[(\be+\g)^2a+\g^2b+\be^2c-(\al\be+\al\g+\be\g)(b+c)\right]
=(\be a-\al b)(\be+\g)^2.
\ee
Similarly, multiplying with $\g$ and eliminating $\g b$,
\bb
0=\g\left[(\be+\g)^2a+\g^2b+\be^2c-(\al\be+\al\g+\be\g)(b+c)\right]
=(\g a-\al c)(\be+\g)^2.
\ee
If $\be+\g=0$, then $2\e-R=-\be^2\leq0$. Hence, if $2\e-R>0$, then
$\be+\g\not=0$, and the expression (\ref{intpoz}) is zero if and only if
\bb
\al b=\be a\ ,\ \ \ \al c=\g a\ ,\ \ \
\be c=\g b\ ,\ \ \ d=e=f=0.
\ee
This means that in the case $2\e-R>0$ the integral can be zero, only if there
exists a function $\mu$ on the manifold, such that
\bb
M_{(a|b)}=\mu K_{ab}\ .
\ee
Substituting this into the form (\ref{parder}) of the partial derivative, we
see that if $M^a$ satisfies this equation, then it is really in the kernel.
The relationship between the solutions of this equation and spacetime Killing
vectors is described in the next section.\q
 
The main theorem of the paper is
\begin{theo}\label{exis}
Suppose that given $g_{ab}\in H^{n+2}(T^0_2)$,
$\dot g_{ab}\in H^{n+1}(T^0_2)$, $\e\in H^{n+1}$ and $S^a\in H^n(T^1_0)$,
such that $2\e-R>0$, where $R$ is the scalar curvature belonging to $g_{ab}$.
Suppose that $N^a\in H^{n+2}(T^1_0)$ is a solution of the Reduced
Thin Sandwich Equations with $N>0$, that is
$F^a\left((g_{cd},\dot g_{cd},\e,S^c),N^c\right)=0$. Let $K_{ab}$ be defined
by (\ref{excur}). If the equation $M_{(a|b)}=\mu K_{ab}$ has only the trivial
solution, then there exists a unique continuous map on an open neighbourhood of
$(g_{cd},\dot g_{cd},\e,S^c)$ in
$H^{n+2}(T^0_2)\tm H^{n+1}(T^0_2)\tm H^{n+1}\tm H^n(T^1_0)$,
which assigns $\bar N^a\in H^{n+2}(T^1_0)$ to
$(\bar g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c)$, such that
$\bar N^a$ is a solution of the Reduced Thin Sandwich Equations
with $(\bar g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c)$, that is
$F^a\left((\bar g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c),\bar N^c\right)=0$.
\end{theo}
In other words, the Thin Sandwich Equations are solvable in a neighbourhood of
this $(g_{cd},\linebreak[2]\dot g_{cd},\linebreak[2]\e,S^c)$, in
which still $2\e-R>0$. The solution is unique, and $N>0$.
 
\noindent\pr
We have proved that if $2\e-R>0$ and $M_{(a|b)}=\mu K_{ab}$ has only the
trivial solution, then $N^a$ is unique by Theorem \ref{uni2},
$D_2F^a\left((g_{cd},\dot g_{cd},\e,S^c),N^c\right)$ is elliptic by Corollary
\ref{nnee} , it is self adjoint by Theorem \ref{sadj}, and its kernel vanishes
by Lemma \ref{kernel}, so it is a linear isomorphism. The result follows by
direct application of the Implicit Function theorem\cite{Lang}.\q
 
There is a natural way to generate some
$(g_{ab},\dot g_{ab},\e,S^a)$ and a solution $N^a$ of the
Reduced Thin Sandwich Equations belonging to them.
Any choice of spacelike hypersurface in an arbitrary spacetime solution of
Einstein's equations induces a solution $(g_{ab},K_{ab},\e,S^a)$ of the
constraint equations. Then by equation (\ref{gdot}), an arbitrary choice of
shift vector $N^a$ and positive lapse function $N$ fixes the
time derivative $\dot g_{ab}$ of the spacelike metric along the
time flow vector field given by this shift and lapse. Obviously,
this $N^a$ is a solution of the reduced
Thin Sandwich Equations with $(g_{ab},\dot g_{ab},\e,S^a)$. In view of this we
can rewrite our main theorem into the more natural form
\begin{theo}\label{exis2}
Suppose that $g_{ab}\in H^{n+2}(T^0_2)$, $K_{ab}\in H^{n+1}(T^0_2)$,
$\e\in H^{n+1}$ and $S^a\in H^n(T^1_0)$ satisfies the constraint equations
(\ref{enco}) and (\ref{imco}), and $2\e-R>0$, where $R$ is the scalar
curvature belonging to $g_{ab}$. For any choice of shift vector
$N^a\in H^{n+2}(T^1_0)$ and positive lapse function
$N\in H^{n+1}$ we define the time derivative $\dot g_{ab}$ of
the spacelike metric by (\ref{gdot}). If
the equation $M_{(a|b)}=\mu K_{ab}$ has only the trivial solution, then there
exists a unique continuous map on an open neighbourhood of
$(g_{cd},\dot g_{cd},\e,S^c)$ in
$H^{n+2}(T^0_2)\tm H^{n+1}(T^0_2)\tm H^{n+1}\tm H^n(T^1_0)$, which assigns
$\bar N^a\in H^{n+2}(T^1_0)$ to
$(\bar g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c)$, such that $\bar N^a$ is a
solution of the Reduced Thin Sandwich Equations with
$(\bar g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c)$, that is
$F^a\left((\bar g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c),\bar N^c\right)=0$.
\end{theo}
 
 
\section{Conditions} \label{seccon}
 
 
If $N^a$ and $N\not=0$ are solutions of the Thin Sandwich
Equations for the initial values
$(g_{cd},\dot g_{cd},\e,\linebreak[4] S^c)$, then by equation
(\ref{excur}) these determine the extrinsic curvature $K_{ab}$,
such that $(g_{cd},K_{cd},\e,\linebreak[4] S^c)$ satisfies the
constraint equations. On the other hand, if we choose a solution
$(g_{cd},K_{cd},\e,S^c)$ of the constraints, then by equation
(\ref{gdot}) any choice of lapse $N$ and shift $N^a$ generates
$(g_{cd},\dot g_{cd},\e,\linebreak[4] S^c)$,
for which $N$ and $N^a$ are solutions of the Thin Sandwich Equations. If
$2\e-R\not=0$, and we are interested only in solutions with $N>0$, then the
Thin Sandwich Equations are equivalent to the Reduced Thin Sandwich
Equations.
 
Suppose that $(g_{cd},\dot g_{cd},\e,S^c)$ is given together with a solution
$N^a$ and $N>0$ of the Thin Sandwich Equations, or equivalently,
$(g_{cd},\dot g_{cd},\e,S^c)$ can be obtained from a solution
$(g_{cd},K_{cd},\e,\hfill\break S^c)$ of the constraint equations by choosing
$N>0$ and $N^a$. Theorem \ref{exis2} means that there exists an open
neighbourhood of $(g_{cd},\dot g_{cd},\e,S^c)$ in which for any $(\bar
g_{cd},\dot{\bar g}_{cd},\bar\e,\bar S^c)$ the form 1 of the Thin Sandwich
Conjecture is true if the following two conditions hold:
 
\vskip9pt\noindent{\bf Condition 1a:}
{\em
\bb
2\e-R>0,
\ee
where $R$ is the scalar curvature belonging to the three-metric $g_{ab}$.}
 
\vskip9pt\noindent{\bf Condition 2b:}
{\em The equation
\bb
M_{(a|b)}=\mu K_{ab} \label{condi}
\ee
for the vector field $M^a$ and scalar field $\mu$ on the initial hypersurface
has only the trivial solution $M^a=0$ and $\mu=0$.}
 
\vskip9pt Introducing the trace-free part of the extrinsic curvature,
$\bk_{ab}=K_{ab}-\frac13g_{ab}K^c_c$, the energy constraint (\ref{enco})
becomes
\bb
2\e-R=(K^a_a)^2-K_{ab}K^{ab}=\frac23(K^a_a)^2-\bk_{ab}\bk^{ab}.
\ee
Hence
condition 1a is equivalent to
 
\vskip9pt\noindent{\bf Condition 1b:}
{\em The trace of the extrinsic curvature $K_{ab}$ dominates the
trace-free part
\bb
\bk_{ab}=K_{ab}-\frac13g_{ab}K^c_c
\ee
in the following way:
\bb
\frac23(K^a_a)^2>\bk_{ab}\bk^{ab}.
\ee}
 
\vskip9pt\noindent Obviously the same is true for the shift tensor $\g_{ab}$. An
advantage of the $2\e-R>0$ form of the condition is that it is expressed only in
terms of the initial data in the hypersurfaces, while the exterior curvature is
only known after solving the Thin Sandwich Equations for the lapse and shift.
The extrinsic curvature form is useful to examine the condition of when various
slicings of a spacetime induce appropriate initial data for the Thin Sandwich
Equations.
 
We can interpret (\ref{gdot}) as an equation determining the time derivative of
the spacelike part of the metric in the direction of the time flow vector field
specified by the shift $N^a$ and lapse $N$. Comparing with equation
(\ref{condi}) we get the following equivalent form of Condition 2b:
 
\vskip9pt\noindent{\bf Condition 2a:}
{\em There is no choice of time flow vector field on the spacetime induced by
$(g_{ab},K_{ab},\e,S^a)$, for which
\bb
\gd ab=0
\ee
on the whole initial hypersurface. (Now $\gd ab=\frac d{dt}g_{ab}$
is the derivative in the direction of the time flow vector
field.)}
 
\vskip9pt In view of this the following theorem is not surprising.
\begin{theo}\label{prkil}
Suppose $\ti M^\al$ is a spacetime vector field. Denote its
projection into
a spacelike hypersurface by $M^\al=h^\al_\be\ti M^\be$ and its hypersurface
orthogonal part by $\mu=\ti M^\al n_\al$, where $n^\al$ is the unit normal of
the hypersurface and $h^\al_\be=\de^\al_\be+n^\al n_\be$ is the metric
orthogonal projection into the hypersurface. Then the following are equivalent:
\begin{enumerate}
\item $M_{(a|b)}=\mu K_{ab}$ in the hypersurface
\item $\ti M^\al$ satisfies the hypersurface projected part of the Killing
equations on the hypersurface.
\end{enumerate}
\end {theo}
\pr We can express $\ti M^\al$ by its components,
\bb
\ti M^\al=\de^\al_\be\ti M^\be=(h^\al_\be-n^\al n_\be)
\ti M^\be=M^\al-\mu n^\al.
\ee
Denoting the spacetime covariant derivative related
to $\ti g_{ab}$ by $\n_\al$, the projected Killing equation is
\bb
0=h^\g_\al h^\de_\be\n_{(\g}\ti M_{\de)}
=h^\g_\al h^\de_\be\n_{(\g}M_{\de)}
-h^\g_\al h^\de_\be\n_{(\g}\left(n_{\de)}\mu\right).
\ee
Using the relation
$M_{\al|\be}=h^\g_\al h^\de_\be\n_\de M_\g$ between the two covariant
derivatives and the definition of the extrinsic curvature
$K_{\al\be}=h^\g_\al h^\de_\be\n_\g n_\de$, we get
\bb
0=M_{(\al|\be)}-\mu K_{ab}\ ,
\ee
which is exactly the condition (\ref{condi}).\q
 
As a special case we have
\begin{coro}\label{killing}
If $\ti M^\al$ is a spacetime Killing vector field, its projection into a
hypersurface with extrinsic curvature $K_{ab}$ satisfies the equation
$M_{(a|b)}=\mu K_{ab}$ with some function $\mu$.
\end{coro}
This means that Condition 2b  is false if the spacetime generated by
$(g_{ab},K_{ab},\e,S^a)$ admits a Killing vector, or admits a vector field
which satisfies at least the spacelike part of the Killing equations on the
initial hypersurface. Let $(g_{ab},\gd ab,\e,S^a)$ be generated by such
$(g_{ab},K_{ab},\e,S^a)$ and some choice of $N^a$ and $N$. Then by Theorem
\ref{uni2}, $N^a$ and $N$ is not the unique solution of the Thin Sandwich
Equations belonging to this $(g_{ab},\gd ab,\e,S^a)$. Furthermore, our
existence proof does not work either in the neighbourhood of such
$(g_{ab},\gd ab,\e,S^a)$. It is possible that existence can be proved even in
these cases, and the different lapse and shift solutions generate the same
spacetime with different foliations, but this needs further investigation.
 
An important question is whether the converse of Corollary \ref{killing} is
true in the following sense. Suppose that $M^a$ satisfies equation
$M_{(a|b)}=\mu K_{ab}$ on the initial hypersurface. Is there a Killing vector
$\ti M^\al$ on the induced spacetime such that $M^a$ is the projection of
$\ti M^\al$ on the initial hypersurface? In view of Theorem \ref{prkil} this
appears to be false, generally one can only find a vector field which satisfies
only the spacelike part of the Killing equations.
 
However there is a large class of initial data which satisfies Condition 2b.
\begin{theo}\label{negdef}
If the Ricci tensor belonging to the hypersurface metric $g_{ab}$
is negative definite,
the equation $M_{(a|b)}=\mu K_{ab}$ has only the trivial solution $M^a=0$,
$\mu=0$.
\end{theo}
\pr The proof is very similar to the well known argument that a compact
manifold with negative definite Ricci tensor has no nontrivial Killing
vectors. Our sign conventions for the Riemann and Ricci tensors are
$M_{c|ba}-M_{c|ab}=R_{abc}^{\ \ \ d}M_d$, and $R_{ab}=R_{acb}^{\ \ \ c}$. We
have
\bb
M_{b|a}^{\ \ \ |b}-M_{b\ \ |a}^{\ |b}=R_{ab}M^b.
\ee
Differentiating condition (\ref{condi}), we get
\bb
M_{a|b}^{\ \ \ b}+M_{b|a}^{\ \ \ b}=2(\mu K_{ab})^{|b}.
\ee
Comparing these and using that $M^a_{\ |a}=\mu K^a_a$,
\bb
M_{a|b}^{\ \ \ b}+R_{ab}M^b-2(\mu K_{ab})^{|b}+(\mu K^b_b)_{|a}=0.
\ee
We multiply with $M^a$ and integrate on the three dimensional manifold \M. We
can drop divergence terms, since our
manifold does not have boundary and it is compact, so all integrals are finite.
Substituting for $M_{(a|b)}$ from (\ref{condi}), and then using the form
(\ref{enco}) of the energy constraint, we get
\bbb
0=\ii M^a\left[M_{a|b}^{\ \ \ b}+R_{ab}M^b-2(\mu K_{ab})^{|b}
+(\mu K^b_b)_{|a}\right]
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
=\ii\left[-M^{a|b}M_{a|b}+M^aM^bR_{ab}+2\mu M^{a|b}K_{ab}
-\mu M^a_{\ |a}K^b_b\right]
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
=\ii\left[-M^{[a|b]}M_{[a|b]}-\mu^2K^{ab}K_{ab}+M^aM^bR_{ab}
+2\mu^2K^{ab}K_{ab}-\mu^2\left(K^a_a\right)^2\right]
\nonumber\\
=\ii\left[-M^{[a|b]}M_{[a|b]}+M^aM^bR_{ab}-\mu^2(2\e-R)\right].
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\eee
Since the Ricci curvature is negative definite, the second term is negative if
$M^a\not=0$. As the energy density $\e$ must be non negative, the last
term cannot be positive either. Since the metric is Riemannian, the first term
is non positive, so the only solution is $M^a=0$, $\mu=0$.\q
 
The proof also shows that
\begin{theo}
If the initial hypersurface has a flat intrinsic geometry and $\e\not=0$, then
the equation $M_{(a|b)}=\mu K_{ab}$ has only the Killing vector solutions with
$\mu=0$.
\end{theo}
 
A linear combination of solutions of equation (\ref{condi}) is also a solution.
We prove
\begin{theo}\label{finite}
If the determinant of $K_{ab}$ is not zero, the space of the solutions of the
equation $M_{(a|b)}=\mu K_{ab}$ is finite dimensional.
\end{theo}
\pr We follow the proof of Taub\cite{Taub} for conformal Killing vectors.
Suppose now that we are looking for solutions on an $n$ dimensional manifold.
Substituting (\ref{condi}) into the definition of the Riemann tensor, we get
\bb
M_{c|ba}+M_{a|cb}=R^{\ \ \ d}_{abc}M_d+2(\mu K_{ac})_{|b}\ .
\ee
Adding and subtracting the same equation with permuted indices and using the
symmetry property $R^{\ \ \ \ \ d}_{[abc]}=0$ of the Riemann tensor, we have
\bb
M_{c|ba}=-R^{\ \ \ d}_{bca}M_d+(\mu K_{ac})_{|b}+(\mu K_{bc})_{|a}
-(\mu K_{ab})_{|c}\ . \label{Mcba}
\ee
By the definition of the curvature,
\bb
M_{c|bad}-M_{c|bda}=R^{\ \ \ e}_{dac}
M_{e|b}+R^{\ \ \ e}_{dab}M_{c|e}\ .
\ee
Substituting $M_{c|ba}$ and  $M_{c|bd}$ from (\ref{Mcba}), we get
\bbb
(\mu K_{ac})_{|bd}+(\mu K_{bc})_{|ad}+(\mu K_{db})_{|ca}-(\mu K_{ab})_{|cd}
-(\mu K_{dc})_{|ba}-(\mu K_{bc})_{|da}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber\\
+R^{\ \ \ e}_{bcd}M_{e|a}-R^{\ \ \ e}_{dac}M_{e|b}-R^{\ \ \ e}_{dab}M_{c|e}
-R^{\ \ \ e}_{bca}M_{e|d}+R^{\ \ \ e}_{bcd\ |a}M_e-R^{\ \ \ e}_{bca\ |d}M_e=0.
\eee
We wish to use this equation to express $\mu_{|ab}$. If the inverse of
$K_{ab}$ exists, that is the determinant is nonvanishing, then we can multiply
by $K^{-1cd}$, and the $\mu$ second derivative terms will be
\bb
(2-n)\mu_{|ab}-K_{ab}K^{-1cd}\mu_{|cd}.
\ee
After multiplying also by $K^{-1ab}$, we get an expression for
\hbox{$2(1-n)K^{-1cd}\mu_{|cd}$}. If $n>2$ we can substitute into
the previous equation to get an expression for $\mu_{|ab}$ in
terms of $\mu$, $\mu_{|a}$, $M_a$, $M_{a|b}$ and $g_{ab}$.
Introducing the condensed notation
\bb
\T_A=(\mu,\mu_{|a},M_a,M_{a|b}),\ \ \ \  A=1,\ldots,(n+1)^2,
\ee
the equation (\ref{Mcba}) and the expression for $\mu_{|ab}$,
together with
\bb
\frac{\p\mu}{\p x^a}=\mu_{|a}
\ee
and
\bb
\frac{\p M_a}{\p x^b}=M_{a|b}+\G^{\ \ c}_{ab}M_c,
\ee
where $\G^{\ \ c}_{ab}$ is the Christoffel symbol belonging to
$g_{ab}$, form a system of first order  partial differential
equations
\bb
\frac{\p\T_A}{\p x^a}=f_{aA}(x,\T),\ \ \ \ \ A=1,\ldots,(n+1)^2,\ \ \
a=1,\ldots,n,
\ee
where $f_{aA}(x,\T)$ are functions of the coordinates $x^b$ and
the variables $\T_A$. The $(n+1)^2$ components of $\T_A$ at a
point determines all the partial
derivatives of $\T_A$ at that point. However there are $n(n+1)/2$ algebraic
relations of the form (\ref{condi}) between the values $\T_A$, hence the
maximum number of independent parameters is $(n+1)(n+2)/2$. We proved that
the space of solutions of equation (\ref{condi}) is finite dimensional, and
maximum $(n+1)(n+2)/2=10$ parameters given at a point determine the solution
uniquely. It is interesting that the maximum number of independent Killing
vector fields in the embedding spacetime is also $(n+1)(n+2)/2$.\q
 
 
\section{Examples}\label{secexa}
 
 
If the Ricci tensor belonging to the spacelike metric $g_{ab}$
of a given spacetime
solution is negative definite, then $2\e-R>0$, and we proved in Theorem
\ref{negdef}, that $M_{(a|b)}=\mu K_{ab}$ has only the trivial solution. The
existence of unique solutions of the Thin Sandwich Equations in a
neighbourhood of the initial data follows by Theorem \ref{exis2}. Any known
spacetime which has a Cauchy surface with negative scalar curvature generates
such initial data. An obvious example with a compact Cauchy surface without
boundary is the spatially compactified hyperbolic Robertson-Walker
spacetime. Hence
\begin{coro}
The Thin Sandwich Equations are solvable with initial data
$(g_{ab},\gd ab,\e,S^a)$ in a neighbourhood of the initial data
$(g_{ab}^{RW},\dot g_{ab}^{RW},\e_{RW},S^a_{RW})$ generated by the standard
slicing of the spatially compactified hyperbolic Robertson-Walker
spacetime. The solution for $N^a$ and $N>0$ is unique.
\end{coro}
 
We look at the spherical Robertson-Walker solutions in detail.
The spacetime metric is
\bb
ds^2=-d\tau^2+a^2\left[d\psi^2+\sin^2\psi(d\vartheta^2
+\sin^2\vartheta d\varphi)\right],
\ee
where the coordinates are $x^\al=(\tau,\psi,\vartheta,\varphi)$,
and $a$ is a function of $\tau$. Since the shift is zero, the
extrinsic curvature is
$K_{ab}=\frac12\gd ab=\frac{\dot a}ag_{ab}$, where the dot
denotes derivative with respect to $\tau$. Equation (\ref{condi})
becomes the conformal Killing equation $M_{(a|b)}$. From the
energy constraint,
\bb
2\e-R=(K^a_a)^2-K_{ab}K^{ab}=6\frac{\dot a^2}{a^2}\geq0.
\ee
$2\e-R$ is zero, if
and only if $\dot a=0$. Then the universe is maximal in size,
turning from expansion to contraction. It follows by Theorem
\ref{thel} and Lemma
\ref{leel}, that except from the maximal sized case, the linearisation of the
Reduced Thin Sandwich Equations is elliptic at this solution. However, the
zero shift is not the only solution of the Thin Sandwich Equations induced on
the initial hypersurface by this spacetime. In view of Corollary \ref{killing},
equation (\ref{condi}) has non trivial solutions, the hypersurface projections
of the spacetime Killing vectors. This includes the Killing vectors in the
spacelike hypersurfaces, with $\mu=0$. Although $2\e-R>0$, contrary to the
assertion of Belasco and Ohanian\cite{BelOha}, there exist non Killing vector
solutions of this equation. For example the vector field
$M^a=(\sin\psi,0,0)$ is a
solution with $\mu=-\frac a{\dot a}\cos\psi$. $M^a_{\ |a}=-3\cos\psi$ is
positive on one half of the manifold, and negative on the other, but if
$M^a_{\ |a}=0$, then also $\mu=0$. In view of the second part of Theorem
\ref{uni2}, this example also shows that generally the solutions of the
Thin Sandwich Equations are not unique, and this nonuniqueness is not only
because of the possible isometries of the spacelike hypersurfaces generated by
the Killing vectors belonging to the three-metric in the initial data. This is
also contrary to one might expect from the analysis of Wheeler\cite{Whe2} of
the spherical cosmology solution in the small shift approximation.
 
We have seen in Theorem \ref{finite}, that the space of solutions of the
equation $M_{(a|b)}=\mu K_{ab}$ is finite dimensional and completely
determined by the values of $M^a$ and $\mu$ and their first derivatives at any
point. It has been shown by Arms, Fisher, Marsden and
Moncrief\cite{FisMar}\cite{AMM}, that those
$(g_{ab},K_{ab})$ which satisfies the constraint equations constitute a smooth
submanifold of the Hilbert space of all possible configurations
$(g_{ab},K_{ab})$, except in the nongeneric case when the spacetime induced by
$(g_{ab},K_{ab})$ admits Killing vectors. In these cases there are also
solutions of the constraints in any  neighbourhood of such $(g_{ab},K_{ab})$,
but instead of forming a submanifold, they have a quadratic singularity at the
point $(g_{ab},K_{ab})$ of the configuration space. These
considerations show, that probably there exists
$(g_{ab},K_{ab},\e,S^a)$ close to the Robertson-Walker initial
data, for which the equation $M_{(a|b)}=\mu K_{ab}$ has only the
trivial solution on the initial hypersurface, and $2\e-R$ is
still positive. By Theorem \ref{exis2}, the Thin Sandwich
Equations would have a unique solution in a neighbourhood of
$(g_{ab},\gd ab,\e,S^a)$ generated by this
$(g_{ab},K_{ab},\e,S^a)$ and any choice of shift and positive
lapse.
\begin{conj}
There exists a neighbourhood of the initial data
$(g_{ab}^{RW},\dot g_{ab}^{RW},\e_{RW},S^a_{RW})$ generated by
the standard slicing of the spherical Robertson-Walker spacetime,
in which those $(g_{ab},\gd ab,\e,S^a)$ for which the Thin
Sandwich Equations are uniquely solvable form a dense open
subset.
\end{conj}
 
It is possible to specify $g_{ab}$ and $\gd ab$ on the initial hypersurface,
such that no solution of the Thin Sandwich Equations (\ref{shenco}) and
(\ref{shimco}) exists with positive lapse. Suppose that $g_{ab}$ and $\e$ is
given such that $2\e-R>0$, on a compact connected three dimensional manifold
\M\   without boundary. Then from the energy constraint
(\ref{shenco}), if $N>0$ then
\bb
(\g^a_a)^2=N^2(2\e-R)+\g_{ab}\g^{ab}>0.
\ee
Using the expression (\ref{gamma}) for $\g_{ab}$ and dropping
boundary terms,
\bb
\ii\g^a_a=\frac12\ii g^{ab}\gd ab\ .
\ee
If $\gd ab$ is specified such that $\int_{\cal M}dv g^{ab}\gd ab=0$, then
$\g^a_a$ must vanish at some point of the manifold, which is impossible if
$N>0$, because of the condition $2\e-R>0$. This means that if $2\e-R>0$,
then
\bb
\ii g^{ab}\gd ab\not=0
\ee
is a necessary condition to the existence of solutions with
positive lapse of the Thin Sandwich Equations. Belasco and
Ohanian\cite{BelOha} has given two examples to this, the first is
$\gd ab=L_{(a|b)}$ for some vector field $L^a$, and the second is
$g^{ab}\gd ab=0$.
 
There is another counterexample given by Belasco and Ohanian\cite{BelOha},
which is rather a restriction on the admissible energy flow densities. Let our
three dimensional manifold \M\  be compact without boundary. Suppose that it
admits a Killing vector field $X^a$. The momentum constraint (\ref{imco})
tells us that the energy flow density is the divergence of the
momentum tensor,
\bb
S^a=\pi^{ab}_{\ \ |b}\ .
\ee
Because of the symmetry of the momentum tensor and the
$X_{(a|b)}=0$ Killing-equation,
\bb
\ii S^aX_a=\ii\pi^{ab}_{\ \ |b}X_a=-\ii\pi^{ab}X_{(a|b)}=0.
\ee
This shows that if the initial manifold \M\ is compact without boundary and
admits a Killing vector $X^a$, then
\bb
\ii S^aX_a=0
\ee
is a necessary condition on the energy flow density $S^a$ in order to have a
solution of the Thin Sandwich Equations.
 
 
 
\section{Other variables}\label{secoth}
 
 
In this section we investigate a different approach towards the Thin Sandwich
formulation of general relativity, introduced by Komar\cite{Kom2}.
The new metric variable is conformally related
to the spacelike metric. The conformal factor is the trace of
the extrinsic curvature. This formalism leads to elliptic
differential equations in more general cases than the usual
formalism, but because of the complicated structure of the
equations we cannot apply the Implicit Function theorem to get
existence and uniqueness results. Komar has introduced the new
variables using a canonical transformation\cite{Kom1}. It has been
shown by Bergmann\cite{Ber1}, that all canonical transformations
involving the extrinsic curvature lead to elliptic differential
equations under this less restrictive condition.
 
We define the momentum density tensor
\bb
p^{ab}=\s\pi^{ab}=\s\left(K^{ab}-K^c_cg^{ab}\right),
\ee
where
\bb
g=\det(g_{ab})
\ee
is the determinant of the spacelike metric. We denote the trace
of the momentum by
\bb
p=p^{ab}g_{ab}\ ,
\ee
and the covariant derivative associated to $g_{ab}$ by $D_a$. The
momentum and energy constraints in the vacuum case are
\bb
\k^a=-2\s D_b\left(\frac1\s p^{ab}\right)=0,\label{imcop}
\ee
and
\bb
\k=\frac1\s p^{ab}p_{ab}-\frac1{2\s}p^2-\s R=0.\label{encop}
\ee
The Hamiltonian is
\bb
H=\iii\left(N_a\k^a+N\k\right).\label{hami}
\ee
 
We introduce the new variables
\bb
\bg_{ab}=\frac{p^2}{4g}g_{ab}
\ee
and
\bb
\bp^{ab}=\frac{4g}{p^2}\left(p^{ab}-\frac12pg^{ab}\right).
\ee
These can be obtained by a canonical transformation with the
generator\cite{Kom1}
\bb
G=-\iii\ g\left(\bp^{ab}g_{ab}\right),
\ee
but we do not necessarily need to use this fact in the following
calculations. The determinant of the new metric is
\bb
\bg=\det(\bg_{ab})=\frac{p^6}{2^6g^2}\ \ ,
\ee
and the trace of the new canonically conjugate momentum tensor is
\bb
\bp=\bp^{ab}\bg_{ab}=-\frac12p\ .\label{tr1}
\ee
 
Since we assume both $g_{ab}$ and $\bg_{ab}$ to be positive definite,
\bb
g=\frac{\ap^3}\z\ \ .
\ee
The inverse relations to the old variables are
\bb
g_{ab}=\frac\ap\z\bg_{ab}\ ,
\ee
and
\bb
p^{ab}=\frac\z\ap\left(\bp^{ab}-\bp\bg^{ab}\right).
\ee
 
Denoting the covariant derivative belonging to the new metric $\bg_{ab}$ by
$\bd_a$, we have the following relation between the two kind of covariant
derivatives of a one-form $u_a$,
\bb
D_au_b=\bd_au_b-C^c_{ab}u_c\ ,\label{tr5}
\ee
where
\bb
C^c_{ab}=\frac\z{2\bp}\left(\de^c_a\bd_b\pg+\de^c_b\bd_a\pg
-\bg_{ab}\bg^{cd}\bd_d\pg\right).
\ee
The relation between the two curvature scalars, $R$ belonging to
$g_{ab}$, and $\br$ belonging to $\bg_{ab}$, is
\bb
R=\frac\z\ap\br-2\frac\bg{\bp^2}\bg^{ab}\bd_a\bd_b\frac\ap\z
+\frac32\frac{\bg^{\frac32}}
{\ap^3}\bg^{ab}\left(\bd_a\pg\right)\bd_b\pg\ .
\ee
Using these transformation formulas we get the following forms for
the constraints in terms of the new variables
\bb
\k^a=-2\frac\bg\ap\bd_b\left(\frac1\z\bp^{ab}\right)=0,
\ee
and
\bb
\k=\frac{\bg^{\frac14}}{\ap^{\frac32}}\left[\bp^{ab}\bp_{ab}-\bp^2-\bp^2\br
+2\z\bp\bg^{ab}\bd_a\bd_b\pg-\frac32\bg\bg^{ab}\left(\bd_a\pg\right)\bd_b\pg
\right]=0.\label{nec}
\ee
 
The time derivative of the new metric variable is
\bb
\gp=\left(\frac{p^2}{4g}g_{ab}\right)^.=\frac p{2g}g_{ab}\left(\pp^{cd}g_{cd}+
p^{cd}\gd cd\right)-\frac{p^2}{4g}
g_{ab}g^{cd}\gd cd+\frac{p^2}{4g}\gd ab\ .
\ee
Using the dynamical equations
\bb
\gd ab=\frac2\s N\left(p_{ab}-\frac12pg_{ab}\right)+2D_{(a}N_{b)},
\ee
\bbb
\pp^{ab}=-N\s\left(R^{ab}-\frac12Rg^{ab}\right)
+\frac1{2\s}Ng^{ab}\left(p^{cd}p_{cd}-\frac12p^2\right)\nonumber\\
\ \ -\frac2\s N\left(p^{ac}p^b_c-\frac12pp^{ab}\right)
+\s\left(D^aD^bN-g^{ab}D^cD_cN\right)\nonumber\\
+\s D_c\left(\frac1\s N^cp^{ab}\right)
-2p^{c(a}D_cN^{b)}
\ ,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\eee
and eliminating the curvature term using the energy constraint (\ref{encop}), we
get for the time derivative
\bb
\gp=\frac{p^2}{2g}D_{(a}N_{b)}+\frac{p^2}{2g^{\frac32}}Np_{ab}
+\frac{p^2}{g^{\frac32}}Ng_{ab}\left(p^{cd}
p_{cd}-\frac12p^2\right)+\frac p{2\s}g_{ab}N^cD_c\frac p\s
-\frac p\s g_{ab}D^cD_cN.
\ee
 
There are now two one-forms associated to the shift vector $N^a$,
the first is $N_a=g_{ab}N^b$, and the second is
\bb
\bn_a=\bg_{ab}N^b=\frac\z\ap N_a\ .
\ee
For ease of notation we also use $\bn^a=N^a$. In place of the lapse function $N$
we introduce a new variable,
\bb
\la=\frac{\bg^{\frac14}}{\ap^{\frac32}}\bp N\ \ .
\ee
Using the transformation rules (\ref{tr1}-\ref{tr5}), we can express the time
derivative of the new metric in terms of the new variables,
\bbb
\gp=2\bd_{(a}\bn_{b)}+2\bg_{ab}\bd^c\bd_c\la+3\frac\z\bp\bg_{ab}\bg^{cd}\left(
\bd_c\la\right)\bd_d\pg\nonumber\\
+\la\left[\frac2\bp\bp_{ab}-\frac2{p^2}\bg_{ab}\bp^{cd}\bp_{cd}
+\frac\z\bp\bg_{ab}\bd^c\bd_c\pg\right].\label{gbdot}
\eee
 
This is the formula where the differences between the new and the usual
formalisms are most apparent. To the derivation of equation (\ref{gbdot}) we
had to use the dynamical field equations, while the corresponding formula with
the usual variables
\bb
\gd ab=2D_{(a}N_{b)}+\frac2\s Np_{ab}-\frac1\s Ng_{ab}p^c_c
\ee
follows simply from the definition of the extrinsic curvature. The other major
difference is that $\bp_{ab}$ cannot be expressed from equation (\ref{gbdot})
as we shall see later.
 
  Using the fact that the new variables emerge from a canonical
transformation, there is another way of getting the formula (\ref{gbdot}) for
the time derivative of the new metric variable. The Hamiltonian (\ref{hami})
takes the form
\bbb
H=\iii\biggl\{\la\left[\frac1\bp\bp^{ab}\bp_{ab}-\bp-\bp\br
+2\z\bd^a\bd_a\pg-\frac3{2\bp}\bg\bg^{ab}\left(\bd_a\pg\right)\bd_b\pg
\right]\nonumber\\
-2\z\bn_a\bd_b\left(\frac1\z\bp^{ab}\right)\biggr\}.
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\eee
Since $\bp^{ab}$ is the canonically conjugate variable to $\bg_{ab}$, variation
of this Hamiltonian with respect to $\bp^{ab}$ gives the same result for $\gp$
as in (\ref{gbdot}), apart from terms vanishing by the energy constraint. We do
not have to consider the $\bp^{ab}$ dependence of $\la$, since at the variation
it would only yield an expression proportional to the energy constraint.
 
We introduce the notation
\bb
l_{ab}=\gp-2\bd_{(a}\bn_{b)}\ .\label{lab}
\ee
The trace of equation (\ref{gbdot}) is
\bb
l=\bg^{ab}l_{ab}=6\bd^a\bd_a\la+9\frac\z\bp\bg^{ab}\left(\bd_a\la\right)
\bd_b\pg+\la\left[2-\frac6{\bp^2}\bp^{ab}\bp_{ab}
+3\frac\z\bp\bd^a\bd_a\pg\right].\label{trgd}
\ee
The trace-free part of (\ref{gbdot}) gives an equation, from which the
trace-free part of $p^{ab}$ can be expressed in terms of the trace,
\bb
\bp^{ab}=\bp\left(\frac1{2\la}l^{ab}-\frac1{6\la}l\bg^{ab}
+\frac13\bg^{ab}\right).\label{pab}
\ee
However, in contrast with the formalism with the usual
variables, the trace of $\bp$ cannot be expressed algebraically
from the equation. Instead, we get one more differential
equation, and we have to treat $\bp$ as a fifth unknown function
together with the shift and lapse functions. Multiplying equation
(\ref{pab}) by itself,
\bb
\frac1{\bp^2}\bp^{ab}\bp_{ab}=\frac1{4\la^2}l^{ab}l_{ab}
-\frac1{12\la^2}l^2+\frac13\ .\label{bpbp}
\ee
Substituting this into the form (\ref{nec}) of the energy constraint, we can
rearrange to express the only secondary derivative term, the Laplacian of
$\bp/\z$,
\bb
\bd^a\bd_a\pg=\pg\left[\frac12\br+\frac13-\frac1{8\la^2}l^{ab}l_{ab}
+\frac{l^2}{24\la^2}+\frac34\frac\bg{\bp^2}\bg^{ab}\left(\bd_a\pg\right)
\bd_b\pg\right].\label{el1}
\ee
Substituting this and the expression (\ref{bpbp}) into the equation
(\ref{trgd}), we can express the Laplacian of $\la$,
\bb
\bd^a\bd_a\la=\frac l6-\frac\la6-\frac\la4\br+\frac5{16\la}l^{ab}l_{ab}
-\frac5{48\la}l^2-\frac32\frac\z\bp\bg^{ab}\left(\bd_a\la\right)\bd_b\pg
-\frac38\frac\bg{\bp^2}\bg^{ab}\left(\bd_a\pg\right)\bd_b\pg.\label{el2}
\ee
Substituting the form (\ref{pab}) of $\bp^{ab}$ into the momentum constraint
(\ref{imcop}), and using the formula
\bb
\bd^b\bd_a\bn_b-\bd_a\bd^b\bn_b=\br_{ab}\bn^b
\ee
for commuting the second derivatives, we can rearrange to express the second
derivative terms of $\bn^a$,
\bbb
\bd^b\bd_b\bn_a+\frac13\bd_a\bd_b\bn^b=\left(l_{ab}-\frac13l\bg_{ab}\right)
\left(\frac\z\bp\bd^b\pg-\frac1\la\bd^b\la\right)\nonumber\\
+\frac23\la\frac\z\bp\bd_a\pg+\bd^b\gp-\frac13\bg^{bc}\bd_a\dot{\bar g}_{bc}
-\br_{ab}\bn^b.\label{el3}
\eee
 
Suppose that $\bg_{ab}$ and $\gp$ is given on the initial three dimensional
manifold \M. Then after the substitution of $l_{ab}$ and $l=\bg^{ab}l_{ab}$ from
(\ref{lab}), the differential equations (\ref{el1}), (\ref{el2}) and
(\ref{el3}), form a quasilinear system of five partial differential equations
for the determination of the five variables
\bb
\left(\pg,\la,\bn^1,\bn^2,\bn^3\right).
\ee
All the second derivative terms are on the right hand side of these equations,
so the symbol of the leading second order terms is
\bb
f=\left(
\begin{array}{ccccc}
\xi^a\xi_a&0&0&0&0\\
0&\xi^a\xi_a&0&0&0\\
0&0&\xi^a\xi_a+\frac13\xi^1\xi_1&\frac13\xi^1\xi_2&\frac13\xi^1\xi_3\\
0&0&\frac13\xi^2\xi_1&\xi^a\xi_a+\frac13\xi^2\xi_2&\frac13\xi^2\xi_3\\
0&0&\frac13\xi^3\xi_1&\frac13\xi^3\xi_2&\xi^a\xi_a+\frac13\xi^3\xi_3
\end{array}
\right).
\ee
The determinant is
\bb
\det(f)=\frac43\left(\xi^a\xi_a\right)^5.
\ee
Because of the positive definiteness of the spacelike metric
this is not zero if $\xi^a\not=0$. We have proved that this system
of differential equations is elliptic, and the only restriction we
had to make is that $\bp$ cannot be zero anywhere on
the manifold \M.
 
As with the usual variables, we can define a mapping
\bb
F^A\left(\left(\bg_{ab},\gp\right),
\left(\pg,\la,\bn^a\right)\right),\ \ \
A=1\ldots5,
\ee
which maps to zero when the equations (\ref{el1}), (\ref{el2})
and (\ref{el3}) are satisfied. However the partial derivative operator with
respect to the variables $\left(\pg,\la,\bn^a\right)$ is not self adjoint. It
does not seem to exist any obvious way of determining when the kernel of the
partial derivative and the kernel of its adjoint vanishes. Probably, similarly
to the formalism with the usual variables, this condition is much more
restrictive than the condition of ellipticity.
 
 
 
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\end{document}
