%@@EDD2  37   518   497  70 AUTOR:<unregistered> CONT:
\input epsf  %@@

\def\ni{\ \\ \noindent}  %@@
\def\eg{{\it e.g.}}  %@@
\def\etc{{\it etc.}}  %@@
\def\ie{{\it i.e.}}  %@@
\def\12{{\textstyle\frac 12}}  %@@
\def\t#1{ {\textstyle{#1}} }  %@@

\documentstyle{article}  %@@
\begin{document}  %@@
\setlength{\textheight}{8in}  %@@

\title{\bf Fun With Curvature Tensors}  %@@
\author{\large Roberto B. Salgado}  %@@
\maketitle  %@@
\abstract{We collect some facts about curvature tensors.}  %@@

\section{The Riemann Curvature Tensor}  %@@
Let $M$ be a manifold (of dimension $n \geq 2$), and let $g_{ab}$  %@@
be a pseudo-Riemannian metric tensor of arbitrary signature. Let
$\nabla_a$ be the  %@@
unique torsion-free derivative operator compatible with the metric  %@@
$g_{ab}$.\footnote{Recall that a torsion-free derivative operator
satisfies $\nabla_{[a}\nabla_{b]}\varphi=0$ for all scalar fields
$\varphi$. Recall  %@@
that a derivative operator that is compatible with the metric
``annihilates the metric'', \ie, $\nabla_{a}g_{bc}=0$.}  %@@

The Riemann Curvature Tensor is a type-(1,3) tensor field
$R_{abc}{}^d$  %@@
defined\footnote{using the conventions of Wald.} by  %@@
$$R_{abc}{}^d\omega_d=2\nabla_{[a}\nabla_{b]}\omega_c,$$  %@@
for all covectors $\omega_a$.  %@@
\ni  %@@
The tensor satisfies three algebraic symmetries:  %@@
\begin{itemize}  %@@
      \item $R_{abc}{}^d=R_{[ab]c}{}^d$ (by definition)  %@@
      \item $R_{[abc]}{}^d=0$ (the Algebraic Bianchi Identity  %@@
            ${\bf dd\omega}=0$)  %@@
      \item $R_{abcd}=R_{ab[cd]}$  %@@
      (follows from metric-compatibility)\footnote{  %@@
      where $R_{abcd}=g_{de}R_{abc}{}^e$.}  %@@
\end{itemize}  %@@
Note that these three algebraic identities imply that
$R_{abcd}=R_{cdab}$.  %@@
\ni  %@@
This tensor also satifies one differential symmetry:  %@@
\begin{itemize}  %@@
\item $\nabla_{[a}R_{bc]d}{}^e=0$ (the Differential Bianchi Identity)  %@@
\end{itemize}  %@@

\ni  %@@
There are several ways one can view the Riemann Curvature Tensor.  %@@
\begin{itemize}  %@@
      \item As a type-(1,3) tensor, it can be viewed as a linear map  %@@
$R:V\otimes V\otimes V\otimes V^* \to {\bf R}$, mapping three vectors  %@@
of the tangent space (say, $X^a$, $Y^b$, $Z^c$) and one covector of  %@@
cotangent space (say, $\omega_d$) to a real number (say, $k$). In
symbols, $R_{abc}{}^dX^a Y^b Z^c \omega_d=k$.  %@@
      \item As a type (1,3) tensor, it can also be viewed as a linear  %@@
map $R:V\otimes V \to Lin(V,V)$, mapping two vectors of the tangent
space (say, $X^a$, $Y^b$) into the vector space of ``linear
transformations of the tangent space [into itself]''---in other
words, a ``transformation matrix''.  %@@
In symbols, $(R_{abc}{}^d X^a Y^b)Z^c=T^d$  %@@
or $[R(X,Y)]_c{}^d Z^c=T^d$ or $R(X,Y)Z=T$.  %@@
      \item The previous interpretation lends itself to  %@@
a geometric interpretation. The Riemann Curvature Tensor linearly
maps a vector $Z^c$ that is being parallel-transported around an
infinitesimal loop, with tangent vectors $X^a$ and $Y^b$ at the
starting point, into another vector $T^d$. If $Z^c$ and $T^d$
coincide for all choices of $X$, $Y$, $Z$, then the space is said to
be {\em flat} at that point.  %@@
      \item As a type-(0,4) tensor, it can be viewed as  %@@
a symmetric bilinear map $R:(\Lambda^2 V)\otimes(\Lambda^2 V) \to
{\bf R}$, mapping a pair of Grassman- or oriented-planes [\ie,
bivectors] to the reals. In symbols,
$R_{abcd}\omega^{ab}\sigma^{cd}=k$.  %@@

\end{itemize}  %@@

\section{Induced Tensors}  %@@

The Riemann Curvature Tensor has an intuitive geometrical
interpretation, as was discussed in the last section.  %@@
Though lacking as clear a geometrical interpretation,  %@@
other tensors measuring ``curvature'' can be obtained from  %@@
the Riemann Curvature Tensor. For example,  %@@

\smallskip  %@@
\ni Some algebraically-obtained tensors  %@@
\begin{itemize}  %@@
\item the Ricci Tensor $R_{ac}=R_{abc}{}^b$, which is obtained by  %@@
            contraction. The choice contracting with the second  %@@
            lower-index is a convention.  %@@
            Because of the symmetries of the Riemann  %@@
            tensor Curvature Tensor, the other contractions yield  %@@
            $R_{abc}{}^a=-R_{bc}$ and $R_{abc}{}^c=0$.  %@@
\item the Ricci Scalar $R=R_{ac}g^{ac}$, which is obtained by  %@@
            transvection with the metric.  %@@
      \item the Weyl [Conformal] Curvature Tensor  %@@
            $$C_{abc}{}^d=  %@@
                 R_{abc}{}^d  %@@
                 - \frac{2!}{n-2}g^{de}\left(\left(g_{a[c}R_{e]b}-  %@@
                                            g_{b[c}R_{e]a}\right)  %@@
                            - \frac{1}{n-1}Rg_{a[c}g_{e]b}\right),$$  %@@
            which is obtained by subtracting out from the  %@@
            Riemann Curvature tensor all of its traces.  %@@
            In other words, the Weyl tensor is the traceless part  %@@
            of the Riemann Curvature tensor.  (We will derive this  %@@
            in a later section.)  %@@
            Remarkably, it turns out this tensor is also invariant  %@@
            under conformal-transformations of the metric, \ie,  %@@
            the Weyl tensor is unchanged if the metric  %@@
            $g_{ab}$ is transformed to  %@@
            $\hat{g}_{ab}=\Omega^2 g_{ab}$, where  %@@
            $\Omega$ is a smooth scalar field on $M$.  %@@
            (We will also derive this in a later section.) \\  %@@
            \hspace*{4ex}  %@@
            In this form, a metric is required to write down  %@@
            the expression for the Weyl tensor. However, since  %@@
            the Weyl tensor is conformally-invariant, it really only  %@@
            depends on the conformal-part of the metric. Is there a  %@@
            manifestly conformally-invariant way to express the Weyl  %@@
            (conformal) Curvature tensor?  %@@
\end{itemize}  %@@

\pagebreak  %@@
\smallskip  %@@
\ni Some differentially-obtained tensors  %@@
\begin{itemize}  %@@
       \item the Einstein tensor $G_{ab}=R_{ab}-\frac{1}{2}Rg_{ab}$,  %@@
which is obtained by contracting the differential Bianchi-identity  %@@
then tranvecting with metric tensor:  %@@
\begin{eqnarray*}  %@@
      0 &=& \nabla_{[a}R_{bc]d}{}^a\\  %@@
        &=& \t{\frac 13} ( \nabla_{a}R_{bcd}{}^a +  %@@
                \nabla_{b}R_{cad}{}^a + \nabla_{c}R_{abd}{}^a )\\  %@@
        &=& \t{\frac 13} ( \nabla_{a}R_{bcd}{}^a +  %@@
                \nabla_{b}R_{cd} - \nabla_{c}R_{bd} )\\  %@@
      0 &=& g^{bd}\left(\nabla_{a}R_{bcd}{}^a +  %@@
                \nabla_{b}R_{cd} - \nabla_{c}R_{bd}\right)\\  %@@
        &=& \nabla_{a}R_{c}{}^a +  %@@
                \nabla_{b}R_{c}{}^b - \nabla_{c}R\\  %@@
        &=& \nabla_{a}(2R_{c}{}^a - Rg_{c}{}^a)\\  %@@
      0 &=& \nabla^{a}(R_{ca}-{\textstyle \frac{1}{2}}Rg_{ca})  %@@
\end{eqnarray*}  %@@
The expression in parentheses is the Einstein tensor  %@@
$G_{ab}=R_{ab}-\frac{1}{2}Rg_{ab}$.  This calculation shows  %@@
that this tensor is divergence-free, and that the coefficient  %@@
$\frac{1}{2}$ is dimensionally-independent.\\  %@@
\hspace*{4ex}  %@@
Of course, the importance of this tensor for physics is that  %@@
this tensor appears in the Einstein equation:  %@@
$$G_{ab}=8\pi T_{ab},$$  %@@
where $T_{ab}$ is the stress-energy tensor for the matter.\\  %@@
\hspace*{4ex}  %@@
I would like to take this opportunity to derive an often-seen  %@@
variant of this equation.  %@@
\begin{eqnarray*}  %@@
       R_{ab}-\12 Rg_{ab} &=& 8\pi T_{ab} \\  %@@
       R-\12 Rn           &=& 8\pi T      \\  %@@
       \left(\frac{2-n}{2}\right)R           &=& 8\pi T      \\  %@@
       R       &=& 8\pi \left(\frac{2}{2-n}\right) T.         \\  %@@
\end{eqnarray*}  %@@
Thus,  %@@
\begin{eqnarray*}  %@@
R_{ab}-{\textstyle\frac{1}{2}}\left(8\pi \frac{2}{2-n}T\right)g_{ab}  %@@
                                         &=& 8\pi T_{ab}     \\  %@@
R_{ab} &=& 8\pi \left(T_{ab}-{\textstyle\frac{1}{2}}  %@@
                           \left(\frac{2}{n-2}\right)Tg_{ab}\right).  %@@
\end{eqnarray*}  %@@
For ordinary spacetime, with $n=4$, this reduces to  %@@
$$R_{ab} =8\pi \left(T_{ab}-{\textstyle\frac{1}{2}}Tg_{ab}\right).$$  %@@
It is only a coincidence for the $n=4$ case that the Ricci tensor  %@@
and the stress-energy tensor appear symmetrically.  %@@
\end{itemize}  %@@





\section{Special Curvature Tensors}  %@@


In this section, we derive the expression for two special curvature
tensors related to the Riemann Curvature tensor.  %@@
These tensors arise as concommitants of the Riemann tensor and the  %@@
metric which are preserved under certain transformations.  %@@
Since the Riemann tensor will generally not be preserved by  %@@
the transformation, we will need to compute the change.  Since  %@@
curvature tensors are derived from a choice of
derivative-operator\footnote  %@@
{One need only specify a derivative-operator $\nabla_a$ in order to
define a curvature tensor. The specification of a metric is not
necessary. However, specifying a metric offers a distinguished
derivative-operator, namely, the unique torsion-free
derivative-operator compatible with that metric.},  %@@
we can compute the change in the curvature tensor by computing  %@@
the change in the derivative-operator due to the transformation.  %@@

\subsection{Some Useful Formulae}  %@@

Recall that the difference of two derivative operators applied  %@@
to a vector can be represented by a tensor field $C^a{}_{mn}$  %@@
applied to that vector:  %@@
$$(\hat\nabla_m -\nabla_m)\xi^a=C^a{}_{mn}\xi^n.$$  %@@
With a little algebra, one can compute the transformation  %@@
of the Riemann tensor, the Ricci tensor, and the scalar curvature:  %@@
\begin{eqnarray*}  %@@
     \t{\frac{1}{2!}} \hat R_{abc}{}^d\omega_d  %@@
     &=& \hat\nabla_{[a}\hat\nabla_{b]}\omega_c      \\  %@@
     &=&  %@@
         \nabla_{[a}\hat\nabla_{b]}\omega_c  %@@
                      - C^m{}_{[ab]}\hat\nabla_m\omega_c  %@@
                      -C^m{}_{c[a}(\hat\nabla_{b]}\omega_m)\\  %@@
     &=& \nabla_{[a}\hat\nabla_{b]}\omega_c  %@@
                      -C^m{}_{c[a}(\hat\nabla_{b]}\omega_m)\\  %@@
     &=& \nabla_{[a}(\nabla_{b]}\omega_c-C^n{}_{b]c}\omega_n)  %@@
      -C^m{}_{c[a}(\nabla_{b]}\omega_m-C^n{}_{b]m}\omega_n)\\  %@@
     &=& \t{\frac{1}{2!}} R_{abc}{}^d\omega_d  %@@
         -\nabla_{[a}(C^n{}_{b]c}\omega_n)\\  %@@
     & &\qquad -C^m{}_{c[a}\nabla_{b]}\omega_m  %@@
         +C^m{}_{c[a}C^n{}_{b]m}\omega_n\\  %@@
     &=& \t{\frac{1}{2!}} R_{abc}{}^d\omega_d  %@@
         -(\omega_n\nabla_{[a}C^n{}_{b]c}  %@@
                 + C^n{}_{[b|c|}\nabla_{a]}\omega_n )\\  %@@
     & &\qquad -C^m{}_{c[a}\nabla_{b]}\omega_m  %@@
         +C^m{}_{c[a}C^n{}_{b]m}\omega_n\\  %@@
     &=& \t{\frac{1}{2!}} R_{abc}{}^d\omega_d  %@@
         -\omega_n\nabla_{[a}C^n{}_{b]c}  %@@
                 - C^m{}_{c[b}\nabla_{a]}\omega_m \\  %@@
     & &\qquad -C^m{}_{c[a}\nabla_{b]}\omega_m  %@@
         +C^m{}_{c[a}C^n{}_{b]m}\omega_n\\  %@@
     &=& \left(\t{\frac{1}{2!}} R_{abc}{}^d-\nabla_{[a}C^d{}_{b]c}+  %@@
              C^m{}_{c[a}C^d{}_{b]m}\right)\omega_d \\  %@@
\end{eqnarray*}  %@@
Hence,  %@@
$$\hat R_{abc}{}^d = R_{abc}{}^d-\t{2!}\nabla_{[a}C^d{}_{b]c}+  %@@
              \t{2!}C^m{}_{c[a}C^d{}_{b]m}$$  %@@
and, therefore,  %@@
$$\hat R_{ac} = \hat R_{anc}{}^n=R_{ac}-\t{2!}\nabla_{[a}C^n{}_{n]c}+
\t{2!}C^m{}_{c[a}C^n{}_{n]m}  %@@
$$  %@@

If the transformation arises from a change in the metric tensor,  %@@
we will require that the derivative-operator transform so as to  %@@
preserve metric-compatibility, \ie, $\hat\nabla_a \hat g_{bc}=0$.  %@@
It is straightforward to obtain the resulting expression for the
tensor field $C^m{}_{ab}$:  %@@
\begin{eqnarray*}  %@@
      0&=&\hat\nabla_a\hat g_{bc}\\  %@@
       &=&\nabla_a \hat g_{bc} - C^m{}_{ab}\hat g_{mc}  %@@
                               - C^m{}_{ac}\hat g_{bm}  %@@
\end{eqnarray*}  %@@
Thus,  %@@
$$ \nabla_a \hat g_{bc} = C^m{}_{ab}\hat g_{mc} + C^m{}_{ac}\hat
g_{bm}$$  %@@
and, by index-substitution,  %@@
$$ \nabla_b \hat g_{ca} = C^m{}_{bc}\hat g_{ma} + C^m{}_{ba}\hat
g_{cm}$$  %@@
$$ \nabla_c \hat g_{ab} = C^m{}_{ca}\hat g_{mb} + C^m{}_{cb}\hat
g_{am}.$$  %@@
Now, using the combination  %@@
 $\mbox{``abc''}+\mbox{``bca''}-\mbox{``cab''}$,  %@@
\begin{eqnarray*}  %@@
\nabla_a \hat g_{bc} + \nabla_b \hat g_{ca} - \nabla_c \hat g_{ab}  %@@
&=& C^m{}_{ab}\hat g_{mc}+C^m{}_{ac}\hat g_{bm}  %@@
   +C^m{}_{bc}\hat g_{ma}+C^m{}_{ba}\hat g_{cm}\\  %@@
& &\qquad -C^m{}_{ca}\hat g_{mb}-C^m{}_{cb}\hat g_{am}\\  %@@
&=& (C^m{}_{ab}+C^m{}_{ba})\hat g_{mc}  %@@
   +(C^m{}_{ac}-C^m{}_{ca})\hat g_{mb}\\  %@@
& &\qquad +(C^m{}_{bc}-C^m{}_{cb})\hat g_{ma}\\  %@@
&=& 2C^m{}_{ab}\hat g_{mc}  %@@
\end{eqnarray*}  %@@
Thus,  %@@
\begin{eqnarray*}  %@@
      C^m{}_{ab}  %@@
&=&\12 2 \delta^m{}_n C^n{}_{ab}\\  %@@
&=&\12 2 \hat \delta^m{}_n C^n{}_{ab}\\  %@@
&=&\12 2 \hat g^{mc} \hat g_{cn} C^n{}_{ab}\\  %@@
&=&\12 \hat g^{mc}(2C^n{}_{ab}\hat g_{nc})\\  %@@
&=&\12 \hat g^{mc} (\nabla_a \hat g_{bc} + \nabla_b \hat g_{ca} -
\nabla_c \hat g_{ab})  %@@
\end{eqnarray*}  %@@

Let us now use these formulae to obtain two special  %@@
curvature tensors: the Weyl conformal-curvature  %@@
tensor $C_{abc}{}^d$ and the Weyl projective-curvature tensor  %@@
$P_{abc}{}^d$.  %@@

\subsection{The Weyl Conformal Curvature Tensor}  %@@

Under a conformal-transformation, the metric tensor is smoothly
transformed according to  %@@
$$\hat g_{ab}=\Omega^2 g_{ab}=e^{2\phi}g_{ab},$$  %@@
where $\Omega^2$ is the conformal factor, a positive scalar field on
$M$. For convenience, the conformal factor $\Omega^2$ is often
written as $e^{2\phi}$, where $\phi$ is a scalar field on $M$.  %@@
The exponential function guarantees the positivity of the conformal
factor so as to avoid a degenerate transformation. In addition, this
and the factor of $2$ will simplify the formulas in the calculations.  %@@

Mathematically, a conformal transformation is a pointwise rescaling
of the metric tensor.  %@@
A consequence of this transformation is that the sign of the
inner-product of two vectors is unchanged. Since  %@@
$\hat g_{ab}k^a l^b = \Omega^2 g_{ab} k^a l^b$  %@@
and $\Omega^2 > 0$, it follows that $\hat g_{ab} k^a l^b$  %@@
and $g_{ab} k^a l^b$ have the same sign. In particular, it follows
that null-vectors  %@@
remain null and, in the context of a Lorentzian signature,  %@@
spacelike-vectors remain spacelike and timelike-vectors remain  %@@
timelike. Thus, physically, a conformal transformation preserves the
light-cone structure.  %@@
In addition, for two non-null vectors, the  %@@
``[cosine of the] angle between between them'' is unchanged since:  %@@
\begin{eqnarray*}  %@@
      \frac{\hat g_{ab}k^a l^b}  %@@
      {\sqrt{\hat g_{cd}k^c k^d}\sqrt{\hat g_{ef}l^e l^f}}  %@@
      &=&  %@@
      \frac{\Omega^2 g_{ab}k^a l^b}  %@@
      {\sqrt{\Omega^2 g_{cd}k^c k^d}\sqrt{\Omega^2 g_{ef}l^e l^f}}\\  %@@
      &=&  %@@
      \frac{g_{ab}k^a l^b}  %@@
      {\sqrt{g_{cd}k^c k^d}\sqrt{g_{ef}l^e l^f}}  %@@
\end{eqnarray*}  %@@


Under a conformal transformation $\hat g_{ab}=\Omega^2 g_{ab}$,  %@@
since $\hat g_{ab}\hat g^{bc}=\hat\delta_a{}^c=\delta_a{}^c$,  %@@
it follows that $\hat g^{ab}=\Omega^{-2} g^{ab}$, and:  %@@
\begin{eqnarray*}  %@@
      C^m{}_{ab}  %@@
      &=& \12(\Omega^{-2}g^{mc})  %@@
      \left(\nabla_a(\Omega^2 g_{bc}) +  %@@
      \nabla_b(\Omega^2 g_{ca}) - \nabla_c(\Omega^2 g_{ab})\right)\\  %@@
      &=& \12\Omega^{-2}g^{mc}  %@@
      \left(2\Omega g_{bc}\nabla_a\Omega  %@@
      +2\Omega g_{ca}\nabla_b\Omega  %@@
      -2\Omega g_{ab}\nabla_c\Omega\right)\\  %@@
      &=& \Omega^{-1}g^{mc}  %@@
      \left(g_{bc}\nabla_a\Omega  %@@
      +g_{ca}\nabla_b\Omega  %@@
      -g_{ab}\nabla_c\Omega\right)\\  %@@
      &=& g^{mc}  %@@
      \left(g_{bc}\nabla_a(\ln \Omega)  %@@
      +g_{ca}\nabla_b(\ln \Omega)  %@@
      -g_{ab}\nabla_c(\ln \Omega) \right)  %@@
\end{eqnarray*}  %@@
where in the last step we used  %@@
$\Omega^{-1}\nabla_a\Omega=\nabla_a(\ln \Omega)$.  %@@

At this point, it is convenient to express this expression  %@@
in terms of $\phi$, using the relation $\Omega^2=e^{2\phi}$.  %@@
Thus,  %@@
\begin{eqnarray*}  %@@
      C^m{}_{ab}  %@@
      &=& g^{mc}  %@@
      \left(g_{bc}\nabla_a \phi  %@@
      +g_{ca}\nabla_b \phi  %@@
      -g_{ab}\nabla_c \phi \right)  %@@
\end{eqnarray*}  %@@

It is a straightforward but tedious computation to show that  %@@
$\hat C_{abc}{}^d=C_{abc}{}^d$ is invariant under conformal
transformations. This tensor is the Weyl conformal-curvature tensor.  %@@


\begin{eqnarray*}  %@@
\hat R_{abc}{}^d  %@@
&=& R_{abc}{}^d-\t{2!}\nabla_{[a}  %@@
    \left(  %@@
      g^{de}  %@@
      \left(g_{b]e}\nabla_c \phi +g_{|ec|}\nabla_{b]} \phi  %@@
      -g_{b]c}\nabla_e \phi \right)  %@@
    \right)\\  %@@
& & + \t{2!} g^{me}  %@@
      \left(g_{ec}\nabla_{[a} \phi +g_{e[a}\nabla_{|c|} \phi  %@@
      -g_{c[a}\nabla_{|e|} \phi \right)  %@@
      g^{de}  %@@
      \left(g_{b]e}\nabla_m \phi +g_{|em|}\nabla_{b]} \phi  %@@
      -g_{b]m}\nabla_e \phi \right)  %@@
\end{eqnarray*}  %@@


and, therefore,  %@@


$$\hat R_{ac} = \hat R_{anc}{}^n=R_{ac}-\t{2!}\nabla_{[a}C^n{}_{n]c}+
\t{2!}C^m{}_{c[a}C^n{}_{n]m}  %@@
$$  %@@



\subsection{The Weyl Projective Curvature Tensor}  %@@

Under a projective transformation, the derivative-operator changes  %@@
while preserving the geodesics, \ie, the geodesic vector fields
remain geodesic. Geometrically, this means that the acceleration
vector remains tangent to the integral curves of the vector field.  %@@
In other words, given $$\xi^m\nabla_m\xi^a=0,$$ we find  %@@
$$\xi^m\hat\nabla_m\xi^a=\alpha\xi^a,$$  %@@
where $\alpha$ is some function on the path.  %@@

We now compute $C^m{}_{ab}$.  %@@
\begin{eqnarray*}  %@@
      \alpha\xi^a  %@@
      &=&\xi^n(\hat\nabla_n \xi^a) \\  %@@
      &=&\xi^n(\nabla_n\xi^a-C^a{}_{nc}\xi^c)\\  %@@
      &=&0 - C^a{}_{nc}\xi^n\xi^c \\  %@@
      &=& -C^a{}_{nc}\xi^n\xi^c \\  %@@
\end{eqnarray*}  %@@
Forming the zero-bivector by antisymmetrizing with $\xi^d$,  %@@
\begin{eqnarray*}  %@@
       \alpha\xi^{[a}\xi^{d]}  %@@
          &=&-C^{[a}{}_{nc}\xi^{d]}\xi^n\xi^c \\  %@@
       0  &=&-C^{[a}{}_{nc}\delta^{d]}{}_r\xi^r\xi^n\xi^c \\  %@@
\end{eqnarray*}  %@@
we find that, since $\xi^a$ is arbitrary, the  %@@
[lower-index-]symmetric part  %@@
of the type-(2,3) tensor $C^{[a}{}_{nc}\delta^{d]}{}_r$ is zero.  %@@
Recalling the symmetry $C^a{}_{(nc)}=C^a{}_{nc}$,  %@@
and then contracting over $d$ and $r$, one finds  %@@
\begin{eqnarray*}  %@@
       0&=&\delta^r{}_d C^{[a}{}_{(nc}\delta^{d]}{}_{r)}\\  %@@
        &=&{\textstyle\frac {1}{2!\cdot 3}}  %@@
           \delta^r{}_d\left((C^a{}_{nc}\delta^d{}_r  %@@
                        -C^d{}_{nc}\delta^a{}_r)  %@@
                        +(C^a{}_{rn}\delta^d{}_{c}  %@@
                         -C^d{}_{rn}\delta^a{}_{c})  %@@
                        +(C^a{}_{cr}\delta^d{}_{n}  %@@
                         -C^d{}_{cr}\delta^a{}_{n})\right)\\  %@@
       0&=& (C^a{}_{nc}(n)-C^a{}_{nc})  %@@
           +(C^a{}_{cn}-C^d{}_{dn}\delta^a{}_c)  %@@
           +(C^a{}_{cn}-C^d{}_{cd}\delta^a{}_n)\\  %@@
        &=& C^a{}_{nc}(n+1)  %@@
           -C^d{}_{dn}\delta^a{}_c  %@@
           -C^d{}_{cd}\delta^a{}_n\\  %@@
        &=& C^a{}_{nc}(n+1)-2\delta^a{}_{(n}C^d{}_{c)d}  %@@
\end{eqnarray*}  %@@
This equation places a constraint on the form of $C^a{}_{bc}$.  %@@
Let us make the ansatz that there exists a covector field  %@@
$\phi_c=\frac{1}{n+1}C^d{}_{dc}$ on $M$. Thus, we have  %@@
$$  %@@
      C^a{}_{nc}=\phi_n\delta^a{}_c+\phi_c\delta^a{}_n  %@@
                =2\delta^a{}_{(n}\phi_{c)}  %@@
$$  %@@
Indeed, this definition is consistent:  %@@
\begin{eqnarray*}  %@@
      C^a{}_{ac}&=&\phi_a\delta^a{}_c+\phi_c (n)\\  %@@
                &=&(1+n)\phi_c  %@@
\end{eqnarray*}  %@@
Using $\alpha\xi^a = -C^a{}_{nc}\xi^n\xi^c$,  %@@
we can find the relation between $\alpha$ and $\phi_c$,  %@@
\begin{eqnarray*}  %@@
      \alpha\xi^a  %@@
      &=&-(\phi_n\delta^a{}_c+\phi_c\delta^a{}_n)\xi^n\xi^c \\  %@@
      &=&-2\phi_n\xi^n \xi^a  %@@
\end{eqnarray*}  %@@
Thus,  %@@
      $$\alpha=-2\phi_n\xi^n$$  %@@
which manifests the dependence of $\alpha$ and $\phi_a$  %@@
on the geodesic vector field $\xi^a$. Recall that  %@@
$\alpha$ is a function defined on the integral curve and  %@@
$\phi_a$ is a covector field defined on $M$.  %@@

It follows that  %@@
\begin{eqnarray*}  %@@
      \hat R_{abc}{}^d  %@@
      &=& R_{abc}{}^d-2\nabla_{[a}C^d{}_{b]c}+  %@@
                      2C^m{}_{c[a}C^d{}_{b]m} \\  %@@
      &=& R_{abc}{}^d  %@@
          -2\nabla_{[a}(\delta^d_{b]}\phi_c + \phi_{b]}\delta^d_c)  %@@
           +2(\delta^m_c\phi_{[a}+ \phi_c\delta^m_{[a})  %@@
            (\delta^d_{b]}\phi_m +\phi_{b]}\delta^d_m)\\  %@@
      &=& R_{abc}{}^d  %@@
          -2(\delta^d_{[b}\nabla_{a]}\phi_c +  %@@
             \delta^d_c\nabla_{[a} \phi_{b]})\\  %@@
          & &\qquad +2(\delta^m_c\phi_{[a}\delta^d_{b]}\phi_m +  %@@
             \delta^m_c\phi_{[a}\phi_{b]}\delta^d_m +  %@@
             \phi_c\delta^m_{[a}\delta^d_{b]}\phi_m +  %@@
             \phi_c\delta^m_{[a}\phi_{b]}\delta^d_m)\\  %@@
      &=& R_{abc}{}^d  %@@
          -2(\delta^d_{[b}\nabla_{a]}\phi_c +  %@@
             \delta^d_c\nabla_{[a} \phi_{b]})\\  %@@
          & &\qquad +2(\delta^m_c\phi_{[a}\delta^d_{b]}\phi_m +  %@@
             0+  %@@
             \phi_c\delta^d_{[b}\phi_{a]} +  %@@
             \phi_c\delta^d_{[a}\phi_{b]})\\  %@@
      &=& R_{abc}{}^d  %@@
          -2(\delta^d_{[b}\nabla_{a]}\phi_c +  %@@
             \delta^d_c\nabla_{[a} \phi_{b]})  %@@
           +2\delta^d_{[b}\phi_{a]}\phi_c\\  %@@
\end{eqnarray*}  %@@
and, therefore,  %@@
\begin{eqnarray*}  %@@
      \hat R_{ac}  %@@
      &=& R_{ac}  %@@
          -2(\delta^b_{[b}\nabla_{a]}\phi_c +  %@@
             \delta^b_c\nabla_{[a} \phi_{b]})  %@@
           +2\delta^b_{[b}\phi_{a]}\phi_c\\  %@@
      &=& R_{ac}  %@@
          -(\delta^b_{b}\nabla_{a}\phi_c  %@@
            -\delta^b_{a}\nabla_{b}\phi_c  %@@
            +\delta^b_c\nabla_{a} \phi_{b}  %@@
            -\delta^b_c\nabla_{b} \phi_{a})  %@@
           +(\delta^b_{b}\phi_{a}-\delta^b_{a}\phi_{b})\phi_c\\  %@@
      &=& R_{ac}  %@@
          -(n\nabla_{a}\phi_c  %@@
            -\nabla_{c} \phi_{a})  %@@
           +(n\phi_{a}-\phi_{a})\phi_c  %@@
\end{eqnarray*}  %@@
Note that the symmetry of the Ricci Tensor is not necessarily  %@@
preserved by a general projective transformation.  %@@
Thus, in order to preserve its symmetry, we must consider a
restricted class of projective-transformations, \ie, we must  %@@
place conditions on the form of the covector $\phi_a$ which  %@@
characterizes the projective transformation.  %@@

By demanding that the symmetry of the Ricci Tensor be preserved,  %@@
we find that  %@@
\begin{eqnarray*}  %@@
      \hat R_{[ac]}  %@@
      &=&R_{[ac]}-(n\nabla_{[a}\phi_{c]}-\nabla_{[c} \phi_{a]})  %@@
                    +(n\phi_{[a}-\phi_{[a})\phi_{c]} \\  %@@
      0&=&0-(n+1)\nabla_{[a}\phi_{c]}+0\\  %@@
\end{eqnarray*}  %@@
In other words, we require that $\phi_c$ satisfy
$\nabla_{[a}\phi_{c]}=0$. This condition is equivalent  %@@
to requiring that $\phi_c=\nabla_c \psi$, \ie, this covector  %@@
is the torsion-free gradient of some scalar field $\psi$.  %@@
Indeed, the symmetry of the Ricci Tensor follows  %@@
generally from the torsion-free condition on the  %@@
derivative-operator.  Thus, we have  %@@
\begin{eqnarray*}  %@@
         \hat R_{ac}  %@@
         &=& R_{ac}-(n\nabla_{a}\phi_c  %@@
                     -\nabla_{a} \phi_{c})  %@@
                   +(n\phi_{a}-\phi_{a})\phi_c\\  %@@
         &=& R_{ac}  %@@
             -(n-1)(\nabla_{(a}\phi_{c)}-\phi_{a}\phi_c)\\  %@@
         &=& R_{ac}-(n-1)P_{ac}  %@@
\end{eqnarray*}  %@@
where $P_{ac} \equiv \nabla_{(a}\phi_{c)}-\phi_{a}\phi_c  %@@
=-\frac{1}{n-1}(\hat R_{ac} -R_{ac})$.  %@@

So,  %@@
\begin{eqnarray*}  %@@
      \hat R_{abc}{}^d  %@@
      &=& R_{abc}{}^d  %@@
          -2(\delta^d_{[b}\nabla_{a]}\phi_c + 0)  %@@
           +2\delta^d_{[b}\phi_{a]}\phi_c\\  %@@
      &=& R_{abc}{}^d  %@@
           -2\delta^d_{[b}(\nabla_{a]}\phi_c-\phi_{a]}\phi_c)\\  %@@
      &=& R_{abc}{}^d  %@@
          -2\delta^d_{[b}P_{a]c}\\  %@@
      &=& R_{abc}{}^d  %@@
          +2\frac{1}{n-1}\delta^d_{[b}(\hat R_{a]c} -R_{a]c})\\  %@@
      \hat R_{abc}{}^d -\frac{2}{n-1}\delta^d_{[b}\hat R_{a]c}  %@@
      &=&  %@@
           R_{abc}{}^d -\frac{2}{n-1}\delta^d_{[b}R_{a]c}\\  %@@
      \hat P_{abc}{}^d  %@@
      &=& P_{abc}{}^d  %@@
\end{eqnarray*}  %@@
Thus, we find that  %@@
$P_{abc}{}^d=R_{abc}{}^d-\frac{2}{n-1}R_{c[a}\delta_{b]}{}^d$  %@@
is invariant under a restricted projective transformation.  %@@
This is the Weyl Projective Curvature tensor.  %@@

Note that this tensor is also traceless:  %@@
\begin{eqnarray*}  %@@
      P_{abc}{}^b  %@@
      &=&R_{abc}{}^b -\frac{1}{n-1}  %@@
          (R_{ca}\delta_b{}^b - R_{cb}\delta_a{}^b)\\  %@@
      &=&R_{ac}-\frac{1}{n-1}  %@@
          (R_{ca}(n) - R_{ca})\\  %@@
      &=&0  %@@
\end{eqnarray*}  %@@





\section{Decompositions of the Riemann Curvature Tensor}  %@@

\subsection{Generalized Curvature Tensors}  %@@

It is useful to define the notion of a  %@@
{\em generalized curvature tensor}.  %@@
A {\bf generalized curvature tensor} is an element $Z_{abc}{}^d$  %@@
of the space of type-(1,3) tensor fields on $M$ satisfying the
following algebraic relations  %@@
\begin{itemize}  %@@
      \item $Z_{[ab]c}{}^d=Z_{abc}{}^d$  %@@
      \item $Z_{[abc]}{}^d=0$  %@@
      \item $Z_{ab[cd]}=Z_{abcd}$  %@@
\end{itemize}  %@@
In addition, if $Z_{abc}{}^d$ also satisfies the differential  %@@
relation  %@@
\begin{itemize}  %@@
      \item $\nabla_{[a}Z_{bc]d}{}^e=0$,  %@@
\end{itemize}  %@@
then we say that $Z_{abc}{}^d$ is {\bf proper}.  %@@

Some examples of generalized curvature tensors:  %@@
\begin{itemize}  %@@
      \item Trivial: $I_{abc}{}^d=2 g_{c[a}g_{b]}{}^d$  %@@
      \item Riemann: $R_{abc}{}^d$ (as defined by  %@@
            $R_{abc}{}^d\omega_d=2\nabla_{[a}\nabla_{b]}\omega_c$)  %@@
      \item Kulkarni-Nomizu product of symmetric tensors $K_{ab}$  %@@
      and $N_{ab}$:  $$Q_{abc}{}^d=4 K_{[c[a}N_{b]e]}g^{ed}$$  %@@
      (The Trivial tensor defined above, of course, is a special  %@@
      case of this tensor.) It is easy to see that this tensor  %@@
      satisfies the above conditions:  %@@
      The first and the third follow immediately from the definition  %@@
      of $Q_{abc}{}^d$.  %@@
      The second follows from the symmetry of $K_{ca}$ (\ie,  %@@
      $K_{(ca)}=K_{ca}$ implies that $K_{[ca]}=0$).  Thus,  %@@
      $$Q_{[abc]}{}^d=K_{[ca}N_{b]}{}^d=0$$  %@@
      \item Weyl Conformal and Projective tensors:  %@@
      $C_{abc}{}^d$ and $P_{abc}{}^d$.  This follows since these  %@@
      tensors are linear combinations of the Riemann and  %@@
      Kulkarni-Nomizu tensors.  %@@
\end{itemize}  %@@


\subsection{The Traceless Part of the Riemann Curvature Tensor}  %@@

Let us now derive the expression for the Weyl Conformal Curvature  %@@
Tensor as ``the traceless part of the Riemann tensor''. For
convenience, we will work with $C_{abcd}=C_{abc}{}^eg_{ed}$  %@@
\begin{eqnarray*}  %@@
      C_{abcd}  %@@
      &=& R_{abcd} +\lambda(g_{a[c}L_{d]b} - g_{b[c}L_{d]a}) \\  %@@
\end{eqnarray*}  %@@
where $\lambda$ and $L_{ab}$ are to be determined.  %@@
$\lambda$ is a coefficient and $L_{ab}$ is symmetric tensor field
concommitant of $R_{ab}$, $g_{ab}$, and, therefore, $R$.  %@@
\begin{eqnarray*}  %@@
      C_{abcd}g^{bd}  %@@
      &=& R_{abcd}g^{bd} +\frac{\lambda}{2}g^{bd}\left(  %@@
                           (g_{ac}L_{db}-g_{ad}L_{cb})  %@@
                           - (g_{bc}L_{da}-g_{bd}L_{ca})\right)\\  %@@
    0 &=& R_{ac} +\frac{\lambda}{2}\left(  %@@
                           (g_{ac}L-L_{ca})  %@@
                           - (L_{ca}-nL_{ca})\right)\\  %@@
      &=& R_{ac} +\frac{\lambda}{2}\left(  %@@
                           (g_{ac}L-L_{ca})  %@@
                           - (1-n)L_{ca})\right)\\  %@@
      &=& R_{ac} +\frac{\lambda}{2}\left(  %@@
                           (g_{ac}L+(n-2)L_{ac}\right)\\  %@@
\end{eqnarray*}  %@@
So, by transvection with $g^{ac}$, we have
$0=R+\frac{\lambda}{2}\left(2(n-1)L\right)$, or equivalently,  %@@
$\lambda L = -\frac{1}{n-1}R$.  And, so,  %@@
$$0=R_{ac}-\frac{1}{2(n-1)}Rg_{ac}+  %@@
                \frac{\lambda}{2}(n-2)L_{ac}$$  %@@
Thus,  %@@
$$\lambda L_{ac}=-\frac{2}{n-2}  %@@
                 \left(R_{ac}-\frac{1}{2(n-1)}Rg_{ac}\right)$$  %@@
We identify $\lambda=-\frac{2}{n-2}$ and define the  %@@
``Curious Tensor''  %@@
$$L_{ac} \equiv R_{ac}-\frac{1}{2(n-1)}Rg_{ac}.$$  %@@

Thus, we obtain the following expression:  %@@
$$C_{abcd}=R_{abcd}-\frac{2}{n-2}(g_{a[c}L_{d]b} - g_{b[c}L_{d]a}).$$  %@@


\subsection{The Ricci Decomposition}  %@@

Thus, the Riemann Curvature Tensor can be decomposed into  %@@
$$R_{abcd}=C_{abcd}+\frac{2}{n-2}(g_{a[c}L_{d]b} - g_{b[c}L_{d]a}).$$  %@@

Upon substituting the definition  %@@
$L_{ac} \equiv R_{ac}-\frac{1}{2(n-1)}Rg_{ac}$,  %@@
we obtain another decomposition:  %@@
$$R_{abcd}=C_{abcd}+\frac{2}{n-2}  %@@
(g_{a[c}R_{d]b} - g_{b[c}R_{d]a})  %@@
                   -\frac{1}{(n-1)(n-2)}  %@@
R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a})  %@@
,$$  %@@
which can be written  %@@
\marginpar{\scriptsize Ricci decomposition}  %@@
\begin{eqnarray*}  %@@
R_{abcd}&=&C_{abcd}+\frac{2}{n-2}  %@@
                    (g_{a[c}R_{d]b} - g_{b[c}R_{d]a})  %@@
                   -\frac{2}{(n-1)(n-2)}Rg_{a[c}g_{d]b}  %@@
\end{eqnarray*}  %@@
since $2g_{[a[c}g_{d]b]}=g_{a[c}g_{d]b}-g_{b[c}g_{d]a}  %@@
                           =g_{a[c}g_{d]b}+g_{b[d}g_{c]a}  %@@
                           =g_{a[c}g_{d]b}+g_{a[c}g_{d]b}  %@@
                           =2g_{a[c}g_{d]b}$.  %@@

However, the $g_{[a[c}R_{d]b]}$-term is reducible (\ie, it can  %@@
be written as the sum of a traceless-part and a trace-part).  %@@

\subsection{The Irreducible Decomposition}  %@@

Define the traceless-Ricci tensor  %@@
$S_{db}\equiv R_{db}-\frac{1}{n}g_{db}R$.  %@@
For convenience, define  %@@
$L_{abcd} \equiv \frac{4}{n-2}g_{[a[c}L_{d]b]}$ so that  %@@
$$R_{abcd}=C_{abcd}+L_{abcd}.$$  %@@
Then, with a little algebra:  %@@
\begin{eqnarray*}  %@@
      L_{abcd}  %@@
      &=&\frac{2}{n-2}\left(  %@@
                  (g_{a[c}R_{d]b} - g_{b[c}R_{d]a})  %@@
                   -\frac{1}{2(n-1)}  %@@
                  R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a})\right)\\  %@@
      &=&\frac{2}{n-2}\left(  %@@
                  (g_{a[c}S_{d]b} - g_{b[c}S_{d]a})  %@@
                   +\left(\frac{1}{n}-\frac{1}{2(n-1)}\right)  %@@
                  R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a})\right)\\  %@@
      &=&\frac{2}{n-2}\left(  %@@
                  (g_{a[c}S_{d]b} - g_{b[c}S_{d]a})  %@@
                   +\frac{n-2}{2n(n-1)}  %@@
                  R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a})\right)\\  %@@
      &=&\frac{2}{n-2}(g_{a[c}S_{d]b} - g_{b[c}S_{d]a})  %@@
         +\frac{1}{n(n-1)} R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a})  %@@
\end{eqnarray*}  %@@
Occasionally, the following generalized curvature tensors are  %@@
defined:  %@@
\begin{eqnarray*}  %@@
      E_{abcd} &\equiv&  %@@
         \frac{2}{n-2}(g_{a[c}S_{d]b} - g_{b[c}S_{d]a})=  %@@
         \frac{4}{n-2}g_{[a[c}S_{d]b]}\\  %@@
      G_{abcd} &\equiv&  %@@
         \frac{1}{n(n-1)} R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a}) =  %@@
         \frac{2}{n(n-1)} R g_{a[c}g_{d]b}  %@@
\end{eqnarray*}  %@@
so,  $$L_{abcd}=E_{abcd}+G_{abcd}.$$  %@@
It is easy to check that  %@@
\begin{eqnarray*}  %@@
      E_{abcd}g^{bd}  %@@
      &=&\frac{2}{n-2}g^{bd}(g_{a[c}S_{d]b} - g_{b[c}S_{d]a})\\  %@@
      &=&\frac{2}{n-2}g^{bd}\12((g_{ac}S_{db}-g_{ad}S_{cb}) -  %@@
                             (g_{bc}S_{da}-g_{bd}S_{ca}) \\  %@@
      &=&\frac{1}{n-2}((g_{ac}S-S_{ac}) -  %@@
                             (S_{ca}-nS_{ca}) \\  %@@
      &=&\frac{1}{n-2}(0g_{ac}+(n-2)S_{ac})\\  %@@
      &=&S_{ac}  %@@
\end{eqnarray*}  %@@
so  %@@
\begin{eqnarray*}  %@@
      E_{abcd}g^{bd}g^{ac}  %@@
      &=&g^{ac}S_{ac}\\  %@@
      &=&0  %@@
\end{eqnarray*}  %@@
and  %@@
\begin{eqnarray*}  %@@
      G_{abcd}g^{bd}  %@@
      &=&\frac{1}{n(n-1)}g^{bd}R(g_{a[c}g_{d]b} - g_{b[c}g_{d]a})\\  %@@
      &=&\frac{1}{n(n-1)}Rg^{bd}\12((g_{ac}g_{db}-g_{ad}g_{cb}) -  %@@
                             (g_{bc}g_{da}-g_{bd}g_{ca})) \\  %@@
      &=&\frac{1}{2n(n-1)}R((g_{ac}n-g_{ca}) -  %@@
                             (g_{ca}-ng_{ca})) \\  %@@
      &=&\frac{1}{2n(n-1)}R(2(n-1)g_{ca})\\  %@@
      &=&\frac{1}{n}Rg_{ca}  %@@
\end{eqnarray*}  %@@
so  %@@
\begin{eqnarray*}  %@@
      G_{abcd}g^{bd}g^{ac}  %@@
      &=&g^{ac}\frac{1}{n}Rg_{ca}\\  %@@
      &=&\frac{1}{n}R(n)\\  %@@
      &=&R  %@@
\end{eqnarray*}  %@@
Thus, we obtain an irreducible decomposition:  %@@
\marginpar{\scriptsize irreducible decomposition}  %@@
\begin{eqnarray*}  %@@
R_{abcd}&=&C_{abcd}+E_{abcd}+G_{abcd}\\  %@@
        &=&C_{abcd}  %@@
        +\frac{2}{n-2}(g_{a[c}S_{d]b} - g_{b[c}S_{d]a})  %@@
        +\frac{2}{n(n-1)} R g_{a[c}g_{d]b}  %@@
\end{eqnarray*}  %@@

\subsection{For the $n=4$ case: Hodge-duality}  %@@

In the $n=4$ case, this decomposition reveals additional symmetries  %@@
under the Hodge-dual operations.  The left and right Hodge-duals of  %@@
a tensor $X_{abcd}=X_{[ab][cd]}$ are defined by  %@@
$${}^\sim X_{abcd}\equiv \frac{1}{2!}\epsilon_{abmn}X^{mn}{}_{cd}  %@@
\qquad  %@@
X_{abcd}^\sim \equiv \frac{1}{2!}X_{ab}{}^{mn}\epsilon_{cdmn}.$$  %@@
We wish to evaluate  %@@
$${}^\sim X_{ab}^\sim {}^{cd} = \left(\frac{1}{2!}\right)^2  %@@
                       \epsilon_{abmn}X^{mn}{}_{pq}\epsilon^{pqcd}.$$  %@@
for $C_{abcd}$, $E_{abcd}$, and $G_{abcd}$.  %@@
It will be convenient to use the following identities:  %@@
$$\epsilon_{abmn}\epsilon^{pqcd}  %@@
   =(-1)^\sigma\delta_{abmn}^{pqcd}  %@@
   =(-1)^\sigma (4!)\delta_{[a\vphantom{b}}^{p}  %@@
                    \delta_{b\vphantom{]}}^q  %@@
                    \delta_{m\vphantom{]}}^{c\vphantom{p}}  %@@
                    \delta_{n]}^{d\vphantom{p}},$$  %@@
where $\sigma$ is the number of negative signs in the metric tensor,  %@@
and  %@@
$$\delta_{abmn}^{pqcd}  %@@
       =\delta_{ab}^{pq}\delta_{mn\vphantom{b}}^{cd\vphantom{p}}  %@@
       +\delta_{mn\vphantom{b}}^{pq}\delta_{ab}^{cd\vphantom{p}}  %@@
       +(2!)^2\delta_{[m[a}^{pq}\delta_{b]n]}^{cd\vphantom{p}},  %@@
$$  %@@
where $\delta_{ab}^{pq}=(2!)\delta_{[a}^p\delta_{b]}^q  %@@
=(2!)\delta_{a\vphantom{b}}^{[p}\delta_{b}^{q]}$.  %@@

Thus,  %@@
\begin{eqnarray*}  %@@
{}^\sim X_{ab}^\sim {}^{cd}  %@@
&=&\left(\frac{1}{2!}\right)^2  %@@
         \epsilon_{abmn}X^{mn}{}_{pq}\epsilon^{pqcd}\\  %@@
&=&\left(\frac{1}{2!}\right)^2  %@@
         (-1)^\sigma\delta_{abmn}^{pqcd}X^{mn}{}_{pq}\\  %@@
&=&(-1)^\sigma \left(\frac{1}{2!}\right)^2 \left(  %@@
\delta_{ab}^{pq}\delta_{mn\vphantom{b}}^{cd\vphantom{p}}  %@@
       +\delta_{mn\vphantom{b}}^{pq}\delta_{ab}^{cd\vphantom{p}}  %@@
       +(2!)^2\delta_{[m[a}^{pq}\delta_{b]n]}^{cd\vphantom{p}}  %@@
       \right)  %@@
        X^{mn}{}_{pq}\\  %@@
&=&(-1)^\sigma \left(\frac{1}{2!}\right)^2\left(  %@@
       (2!)^2X^{cd}{}_{ab}  %@@
       +(2!) \delta_{ab}^{cd\vphantom{p}} X^{mn}{}_{mn}  %@@
       -(2!)^4 \delta_{[a}^{[c} X^{d]m}{}^{\vphantom{[]}}_{b]m}  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       X^{cd}{}_{ab}  %@@
       +\frac{1}{2!} \delta_{ab}^{cd\vphantom{p}} X^{mn}{}_{mn}  %@@
       -(2!)^2\delta_{[a}^{[c} X^{d]m}{}^{\vphantom{[]}}_{b]m}  %@@
       \right)  %@@
\end{eqnarray*}  %@@
where we have used  %@@
\begin{eqnarray*}  %@@
\delta^{pq}_{ab} X^{mn}{}_{pq}  %@@
&=& (2!\delta^{p}_{[a}\delta^{q}_{b]}) X^{mn}{}_{pq}\\  %@@
&=& 2! X^{mn}{}_{[ab]}\\  %@@
&=& 2! X^{mn}{}_{ab}  %@@
\end{eqnarray*}  %@@
and  %@@
\begin{eqnarray*}  %@@
\delta_{[m[a}^{pq}\delta_{b]n]}^{cd\vphantom{p}} X^{mn}{}_{pq}  %@@
&=&  %@@
2! X^{mn}{}_{[m[a} \delta_{b]n]}^{cd}=  %@@
2! X^{mn}{}_{m[a}\delta_{b]n}^{cd} =  %@@
2! X^{mn}{}_{m[a}(2!\delta_{b]}^{[c}\delta_{n\vphantom{[]}}^{d]})\\  %@@
&=&(2!)^2  X^{m[d}{}^{\vphantom{[]}}_{m[a} \delta_{b]}^{c]}=  %@@
(2!)^2 (-1)^2\delta_{[b}^{[c} X^{d]m}{}^{\vphantom{[]}}_{a]m}\\  %@@
&=&-(2!)^2 \delta_{[a}^{[c} X^{d]m}{}^{\vphantom{[]}}_{b]m}  %@@
\end{eqnarray*}  %@@

With this formula, we can easily compute:  %@@
\begin{eqnarray*}  %@@
{}^\sim C_{ab}^\sim {}^{cd}  %@@
&=&(-1)^\sigma \left(  %@@
       C^{cd}{}_{ab}       + 0       - 0  %@@
       \right)\\  %@@
&=&(-1)^\sigma C^{cd}{}_{ab}  %@@
\end{eqnarray*}  %@@
\begin{eqnarray*}  %@@
{}^\sim E_{ab}^\sim {}^{cd}  %@@
&=&(-1)^\sigma \left(  %@@
       E^{cd}{}_{ab}  %@@
       + 0  %@@
       -(2!)^2\delta_{[a}^{[c} E^{d]m}{}^{\vphantom{[]}}_{b]m}  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       E^{cd}{}_{ab}  %@@
       + 0  %@@
       -(2!)^2\delta_{[a}^{[c} S^{d]}{}^{\vphantom{[]}}_{b]}  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       E^{cd}{}_{ab}  %@@
       + 0  %@@
       -(2!)^2 \left(\frac{n-2}{4}E_{ab}{}^{cd}\right)  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       E^{cd}{}_{ab}  %@@
       + 0  %@@
       -(4-2)E^{cd}{}_{ab}  %@@
       \right) \qquad{\mbox{using $n=4$}}\\  %@@
&=&-\ (-1)^\sigma E^{cd}{}_{ab}  %@@
\end{eqnarray*}  %@@
\begin{eqnarray*}  %@@
{}^\sim G_{ab}^\sim {}^{cd}  %@@
&=&(-1)^\sigma \left(  %@@
       G^{cd}{}_{ab}  %@@
       +\frac{1}{2!} \delta_{ab}^{cd\vphantom{p}} G^{mn}{}_{mn}  %@@
       -(2!)^2\delta_{[a}^{[c} G^{d]m}{}^{\vphantom{[]}}_{b]m}  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       G^{cd}{}_{ab}  %@@
       +\frac{1}{2!} \delta_{ab}^{cd\vphantom{p}} R  %@@
       -(2!)^2\delta_{[a}^{[c}  %@@
       (\frac{1}{n}g^{d]}{}^{\vphantom{[]}}_{b]}R)  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       G^{cd}{}_{ab}  %@@
       +\frac{1}{2!} \delta_{ab}^{cd\vphantom{p}} R  %@@
       -(2!)^2 \left(\frac{1}{n}\right)  %@@
       \delta_{a\vphantom{b}}^{[c}  %@@
       \delta^{d]}_{b}R  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       G^{cd}{}_{ab}  %@@
       +\frac{1}{2!} \delta_{ab}^{cd\vphantom{p}} R  %@@
       -(2!)^2 \left(\frac{1}{n}\right)  %@@
       \left(\frac{1}{2!}\delta_{ab}^{cd}\right)R  %@@
       \right)\\  %@@
&=&(-1)^\sigma \left(  %@@
       G^{cd}{}_{ab}  %@@
       +\frac{1}{2!} \delta_{ab}^{cd\vphantom{p}} R  %@@
       -(2!)\left(\frac{1}{4}\right)  %@@
       \delta_{ab}^{cd}R  %@@
       \right)\qquad{\mbox{using $n=4$}} \\  %@@
&=&(-1)^\sigma G^{cd}{}_{ab}  %@@
\end{eqnarray*}  %@@

Thus, if $R_{abcd}=C_{abcd}+E_{abcd}+G_{abcd}$,  %@@
then  %@@
$${}^\sim R_{abcd}^\sim =(-1)^\sigma (C_{abcd}-E_{abcd}+G_{abcd}).$$  %@@


\section{Maxwellian Gravity?}  %@@

Using the decomposition of the Riemann tensor, we see that the
Curious Tensor and the Weyl Conformal Curvature Tensor are  %@@
algebraically-independent. However, using the Differential Bianchi
Identity, we see that these tensors are differentially-dependent.  %@@
\begin{eqnarray*}  %@@
      0&=&\nabla_{[a}R_{bc]}{}^{de}\\  %@@
       &=&\nabla_{[a}C_{bc]}{}^{de}  %@@
          +\frac{4}{n-2}\nabla_{[a}(g_{b}{}^{[d}L^{e]}{}_{c]})\\  %@@
       &=&\nabla_{[a}C_{bc]}{}^{de}  %@@
          +\frac{4}{n-2}g_{[b}{}^{[d}\nabla_{a}L^{e]}{}_{c]}\\  %@@
\end{eqnarray*}  %@@
Tracing this form of the Differential-Bianchi Identity,  %@@
\begin{eqnarray*}  %@@
      0&=&\delta_e{}^a\left(  %@@
          \nabla_{[a}C_{bc]}{}^{de}  %@@
          +\frac{4}{n-2}g_{[b}{}^{[d}\nabla_{a}L^{e]}{}_{c]}  %@@
          \right)\\  %@@
       &=&\nabla_{[a}C_{bc]}{}^{da}  %@@
          +\frac{4}{n-2}g_{[b}{}^{[d}\nabla_{a}L^{a]}{}_{c]} \\  %@@
       &=&{\textstyle\frac{1}{3}}(\nabla_{a}C_{[bc]}{}^{da}  %@@
          +\nabla_{[b}C_{c]a}{}^{da}  %@@
          +\nabla_{[c}C_{|a|b]}{}^{da})  %@@
          +\frac{4}{n-2}g_{[b}{}^{[d}\nabla_{a}L^{a]}{}_{c]}  %@@
          \\  %@@
       &=&{\textstyle\frac{1}{3}}  %@@
         (\nabla_{a}C_{[bc]}{}^{da}+0+0)  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3}}  %@@
          (g_{[b}{}^{[d}\nabla_{|a|}L^{a]}{}_{c]}  %@@
          +g_{a}{}^{[d}\nabla_{[c}L^{a]}{}_{b]}  %@@
          +g_{[c}{}^{[d}\nabla_{b]}L^{a]}{}_{a})\\  %@@
       &=&{\textstyle\frac{1}{3}}  %@@
         (\nabla_{a}C_{[bc]}{}^{da})  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3}}  %@@
          (g_{[b}{}^{[d}\nabla_{|a|}L^{a]}{}_{c]}  %@@
          -g_{a}{}^{[a}\nabla_{[c}L^{d]}{}_{b]}  %@@
          -g_{[b}{}^{[d}\nabla_{c]}L^{a]}{}_{a})\\  %@@
       &=&{\textstyle\frac{1}{3}}  %@@
         (\nabla_{a}C_{[bc]}{}^{da})  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3}}  %@@
            (g_{[b}{}^{[d}  %@@
              (\nabla_{|a|}L^{a]}{}_{c]}-\nabla_{c]}L^{a]}{}_{a})  %@@
            -g_{a}{}^{[a}\nabla_{[c}L^{d]}{}_{b]})\\  %@@
       &=&{\textstyle\frac{1}{3}}  %@@
         (\nabla_{a}C_{[bc]}{}^{da})  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3\cdot 2! }}  %@@
           \left[g_{[b}{}^{d}  %@@
              (\nabla_{|a|}L^{a}{}_{c]}-\nabla_{c]}L^{a}{}_{a})  %@@
              \right.\\  %@@
         & &\qquad\qquad\qquad\qquad\left.  %@@
            -g_{[b}{}^{a}  %@@
              (\nabla_{|a|}L^{d}{}_{c]}-\nabla_{c]}L^{d}{}_{a})  %@@
            -( g_{a}{}^{a}\nabla_{[c}L^{d}{}_{b]}  %@@
              -g_{a}{}^{d}\nabla_{[c}L^{a}{}_{b]}) \right]\\  %@@
       &=&{\textstyle\frac{1}{3}}  %@@
         (\nabla_{a}C_{[bc]}{}^{da})  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3\cdot 2! }}  %@@
           \left[g_{[b}{}^{d}  %@@
              (\nabla_{|a|}L^{a}{}_{c]}-\nabla_{c]}L^{a}{}_{a})  %@@
              \right.\\  %@@
         & &\qquad\qquad\qquad\qquad\left.  %@@
              -(\nabla_{[b}L^{d}{}_{c]}-\nabla_{[c}L^{d}{}_{b]})  %@@
            -( n\nabla_{[c}L^{d}{}_{b]}  %@@
              -\nabla_{[c}L^{d}{}_{b]}) \right]\\  %@@
       &=&\frac{1}{3}\nabla_{a}C_{[bc]}{}^{da}  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3\cdot 2!}}\left(  %@@
          g_{[b}{}^{d}(\nabla_{|a|}L^{a}{}_{c]}  %@@
                      -\nabla_{c]}L^{a}{}_{a})  %@@
          -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
       &=&\frac{1}{3}\nabla_{a}C_{[bc]}{}^{da}  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3\cdot 2!}}\left(  %@@
          g_{[b}{}^{d}(\nabla_{|a|}L^{a}{}_{c]}  %@@
                      -\nabla_{|a|} (g^{a}{}_{c]}L^{e}{}_{e}))  %@@
          -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
       &=&\frac{1}{3}\nabla_{a}C_{[bc]}{}^{da}  %@@
          -\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{3\cdot 2!}}\left(  %@@
          \nabla_{a}(L^{a}{}_{[c}- Lg^{a}{}_{[c}) g_{b]}{}^{d}  %@@
          -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
\end{eqnarray*}  %@@
Thus,  %@@
\begin{eqnarray*}  %@@
       \nabla_{a}C_{[bc]}{}^{da}  %@@
       &=&\left(\frac{4}{n-2}\right)  %@@
          {\textstyle\frac{1}{2!}}\left(  %@@
          \nabla_{a}(L^{a}{}_{[c}- Lg^{a}{}_{[c}) g_{b]}{}^{d}  %@@
          -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
\end{eqnarray*}  %@@
Let us transvect this form of the Differential-Bianchi Identity  %@@
with $g_d{}^b$:  %@@
\begin{eqnarray*}  %@@
       g_d{}^b\nabla_{a}C_{[bc]}{}^{da}  %@@
       &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}g_d{}^b\left(  %@@
          \nabla_{a}(L^{a}{}_{[c}- Lg^{a}{}_{[c}) g_{b]}{}^{d}  %@@
          -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
       0&=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
          \nabla_{a}(L^{a}{}_{[c}- Lg^{a}{}_{[c})  %@@
                         g_{b]}{}^{d}g_d{}^b  %@@
          -(n-3)\nabla_{[b} L^{d}{}_{c]}g_d{}^b  %@@
          \right)\\  %@@
        &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
          \nabla_{a}(L^{a}{}_{[c}- Lg^{a}{}_{[c})  %@@
                         g_{b]}{}^{b}  %@@
          -(n-3)\nabla_{[b} L^{b}{}_{c]}  %@@
          \right)\\  %@@
        &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
          \frac{1}{2!}\nabla_{a}(  %@@
                       (L^{a}{}_{c}- Lg^{a}{}_{c})g_{b}{}^{b}  %@@
                       -(L^{a}{}_{b}- Lg^{a}{}_{b})g_{c}{}^{b}  %@@
                                )  %@@
          -(n-3)\nabla_{[b} L^{b}{}_{c]}  %@@
          \right)\\  %@@
        &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
          \frac{1}{2!}\nabla_{a}(  %@@
                       (L^{a}{}_{c}- Lg^{a}{}_{c})n  %@@
                       -(L^{a}{}_{c}- Lg^{a}{}_{c})  %@@
                                )  %@@
          -(n-3)\nabla_{[b} L^{b}{}_{c]}  %@@
          \right)\\  %@@
        &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
            \frac{1}{2!}(n-1)\nabla_{a}  %@@
                    (L^{b}{}_{c}g^a{}_{b}- L^b{}_b g^{a}{}_{c})  %@@
          -(n-3)\nabla_{[b} L^{b}{}_{c]}  %@@
          \right)\\  %@@
        &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
            (n-1)\nabla_{a}(L^{b}{}_{[c}g^a{}_{b]})  %@@
          -(n-3)\nabla_a g^a{}_{[b} L^{b}{}_{c]}  %@@
          \right)\\  %@@
        &=&\left(\frac{4}{n-2}\right)  %@@
            \nabla_{a}(L^{b}{}_{[c}g^a{}_{b]})\\  %@@
\end{eqnarray*}  %@@
We find that $(L_{ac}- Lg_{ac})=2!L^{b}{}_{[c}g^a{}_{b]}$ is an
identically divergence-free symmetric tensor field.  %@@

\subsection{Aside: An Identity for the Einstein Tensor}  %@@

In this section, we digress to show that $(L_{ac}- Lg_{ac})$ is (now,
not surprisingly) exactly the Einstein Tensor. Using  %@@
$L_{ab}\equiv R_{ab}-\frac{1}{2(n-1)}R g_{ab}$  %@@
and $L = \frac{n-2}{2(n-1)}R$  %@@
\begin{eqnarray*}  %@@
      L_{ac}- Lg_{ac}  %@@
      &=&\left(R_{ac}-\frac{1}{2(n-1)}R g_{ac}\right)  %@@
         -\left(\frac{n-2}{2(n-1)}R\right)g_{ac}\\  %@@
      &=&R_{ac}  %@@
         -\left(\frac{1}{2(n-1)}+\frac{n-2}{2(n-1)}\right)Rg_{ac}\\  %@@
      &=&R_{ac}  %@@
         -\left(\frac{1+n-2}{2(n-1)}\right)Rg_{ac}\\  %@@
      &=&R_{ac}-\12 Rg_{ac}\\  %@@
      &=&G_{ac}  %@@
\end{eqnarray*}  %@@

In the appendix, we will pursue this identity further.  %@@

\subsection{An Analogy with the Maxwell Equations}  %@@

Thus,  %@@

\begin{eqnarray*}  %@@
       \nabla_{a}C_{bc}{}^{da}  %@@
       &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
             \nabla_{a}(G^{a}{}_{[c}) g_{b]}{}^{d}  %@@
              -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
       &=&\left(\frac{4}{n-2}\right)  %@@
          \frac{1}{2!}\left(  %@@
             0  %@@
              -(n-3)\nabla_{[b}L^{d}{}_{c]}  %@@
          \right)\\  %@@
       &=&-\left(\frac{2(n-3)}{n-2}\right)  %@@
          \nabla_{[b}L^{d}{}_{c]}  %@@
\end{eqnarray*}  %@@

Let us rewrite this equation in a very suggestive manner:  %@@
\begin{eqnarray*}  %@@
       \nabla_{a}C_{bc}{}^{da}  %@@
       &=&-\left(\frac{2(n-3)}{n-2}\right)  %@@
          \nabla_{[b}L_{c]}{}^{d}\\  %@@
       &=&J_{bc}{}^{d}  %@@
\end{eqnarray*}  %@@
where we have defined the {\bf Schouten Tensor}  %@@
$J_{bc}{}^d\equiv-(\frac{2(n-3)}{n-2}) \nabla_{[b}L_{c]}^d$.  %@@
It is interesting that this equation has a form remarkably similar  %@@
to that of the Maxwell Equation  %@@
\begin{eqnarray*}  %@@
       \nabla_{a}F^{da}  %@@
       &=&J^{d}  %@@
\end{eqnarray*}  %@@
Moreover, since the Weyl Conformal Curvature Tensor is  %@@
traceless, the Schouten Tensor $J_{bc}{}^d$ is identically
divergence-free (\ie, $\nabla_d J_{bc}{}^d=0$). This suggests  %@@
that if we view the Weyl Tensor as a {\em field strength}, the
Schouten Tensor $J_{bc}{}^d$ can be interpreted as its conserved {\em
current density}.  %@@

Incidentally, if $J_{bc}{}^d=0$ then $\nabla_a C_{bc}{}^{da}=0$,  %@@
\ie, the manifold is {\bf conformally-affine}(?). If $n=3$,  %@@
then the Weyl Tensor vanishes identically.  However, this  %@@
does not imply that the manifold is necessarily conformally-flat.  %@@
If $\nabla_{[b}L_{c]}{}^d=0$,  %@@
then the symmetric tensor field $L_{cd}$ is called a  %@@
{\bf Codazzi Tensor}.  %@@


Let us introduce the decompositon:  %@@
$$g_{mn}=h_{mn}+\epsilon u_m u_n,$$  %@@
where $\epsilon\equiv (g_{ab} u^a u^b)^{-1}$ and  %@@
$\bar\epsilon\equiv \epsilon^{-1}$.  %@@

Let us compute  %@@
\begin{eqnarray*}  %@@
      h^{[p}_m h^{c]}_n  %@@
      &=& (g^{[p}_m - \bar\epsilon u_m u^{[p})  %@@
          (g^{c]}_n - \bar\epsilon u^{c]}u_n)\\  %@@
      &=& g^{[p}_m g^{c]}_n  %@@
          -\bar\epsilon (g^{[p}_m u^{c]}u_n+g^{[c}_n u^{p]} u_m)  %@@
          +\bar\epsilon^2 u_m u_n u^{[p}u^{c]}\\  %@@
      &=& \bar\epsilon (g^{[p}_m g^{c]}_n u^b u_b  %@@
          -g^{[p}_m u^{c]}u_n-g^{[c}_n u^{p]} u_m)+0\\  %@@
      &=& \bar\epsilon u^b u_d (g^{[p}_m g^{c]}_n g^d_b  %@@
          -g^{[p}_m g^{c]}_b g^d_n-g^{[c}_n g^{p]}_b g^d_m)\\  %@@
      &=& \bar\epsilon u^b u_d (g^{[p}_m g^{c]}_n g^d_b  %@@
          +g^{[c}_m g^{|d|}_n g^{p]}_b+g^d_m g^{[p}_n g^{c]}_b)\\  %@@
      &=& \bar\epsilon u^b u_d (3 g^{[p}_m g^{c}_n g^{d]}_b)\\  %@@
      &=& \bar\epsilon u^b u_d (3 \frac{1}{3!} \delta ^{pcd}_{mnb})\\  %@@
      &=& \bar\epsilon u^b u_d 3 \frac{1}{3!}  %@@
               (\frac{(n-k)!}{(n-3)!}  %@@
               \delta ^{pcd a_4\ldots a_k}_{mnba_4\ldots a_k})\\  %@@
      &=& \bar\epsilon u^b u_d 3 \frac{1}{3!}  %@@
               (\frac{(4-4)!}{(4-3)!}\delta ^{pcda}_{mnba})\\  %@@
      &=& \bar\epsilon u^b u_d 3 \frac{1}{3!}  %@@
          (\bar\sigma \epsilon^{pcda} \epsilon_{mnba})\\  %@@
      &=& \bar\epsilon \bar\sigma u^b u_d \frac{1}{2!}  %@@
          \epsilon^{pcda} \epsilon_{mnba}  %@@
\end{eqnarray*}  %@@




\section{Suggestive Thoughts}  %@@



\section{References}  %@@

Bach, R.  %@@

Weyl, H.  %@@

Thomas, T.Y.  %@@

Schouten, J.A.  %@@

Yano, K.  %@@

Eisenhart, L.P.  %@@

Simon, U.  %@@

Penrose, R.  %@@

Malament, D.  %@@

Besse, A.  %@@

Kulkarni, R.S.  %@@

Nomizu,  %@@

Hawking, S.W. / Ellis, G.F.R.  %@@

de Felice, / Clarke, C.J.S.  %@@

Geroch, R.P.  %@@

Wald, R.M.  %@@

Murcharda, N.  %@@

Hawking, S.W.  %@@

Ellis, G.F.R.  %@@

Ehlers, J. / Kundt,  %@@

Lanczos, C.  %@@

Szekeres, P.  %@@

Lesche, B. / Som, M.M.  %@@

Lichnerowicz, A.  %@@

Paschual-S\'anchez, J.F.  %@@

Roberts, M.R.  %@@

Dolan, P. / Kim, C.W.  %@@

Novello / Duarte de Oliveira  %@@

Bampi, F. / Caviglia, G.  %@@

Hammon, K.S. / Norris, L.K.  %@@

Thorpe, J.A.  %@@

Fiedler, B.  %@@

Cresswll, A. / Zimmerman, R.L.  %@@

Kadzan, J.L.  %@@

Oliker, V.I. / Simon, U.  %@@

Derdzi\'nski, A. / Chun-Li Shen  %@@


Greenberg, P.J.  %@@

Kramer, D. / Stephani, H. / Herlt, E. / MacCallum, M. / Schmutzer, E.  %@@

Stephani, H.  %@@

Fekete, A.E.  %@@

\end{document}  %@@

