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\begin{document}
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\begin{center}
{\large\bf Riemann Curvature Tensor}
\end{center} \large
\bigskip

In the following we use the notation colon $:$ instead of semi-colon $;$
for the covariant derivative. Covariant derivatives do {\bf not commute}.
As in Rindler before Eq.~(10.55) on p.217 we define
\begin{eqnarray} \label{bracket}
  \left[\matrix{h\cr j}\right] = V^h_{:j} = V^h_{,j}+V^a\,\Gamma^h_{aj}\ .
\end{eqnarray}
Then we have
\begin{eqnarray} 
  V^h_{:jk} &=& \left[\matrix{h\cr j}\right]_{:k} = 
  \left[\matrix{h\cr j}\right]_{,k} + 
  \left[\matrix{b\cr j}\right]\,\Gamma^h_{bk} - 
  \left[\matrix{h\cr b}\right]\,\Gamma^b_{jk}\,,\\
  V^h_{:kj} &=& \left[\matrix{h\cr k}\right]_{:j} = 
  \left[\matrix{h\cr k}\right]_{,j} + 
  \left[\matrix{b\cr k}\right]\,\Gamma^h_{bj} - 
  \left[\matrix{h\cr b}\right]\,\Gamma^b_{kj}\,,
\end{eqnarray}
for the two orders of the covariant derivatives. Using the symmetry
$\Gamma^b_{jk}=\Gamma^b_{kj}$ the difference becomes
\begin{eqnarray} \nonumber
  V^h_{:jk}-V^h_{:kj} &=& \left[\matrix{h\cr j}\right]_{,k} + 
  \left[\matrix{b\cr j}\right]\,\Gamma^h_{bk} - 
  \left[\matrix{h\cr k}\right]_{,j} - 
  \left[\matrix{b\cr k}\right]\,\Gamma^h_{bj}\ .  
\end{eqnarray}
Comparing corresponding terms we get
\begin{eqnarray} \nonumber
  \left[\matrix{h\cr j}\right]_{,k} - 
  \left[\matrix{h\cr k}\right]_{,j} &=&
  V^h_{,jk} + V^a_{,k}\,\Gamma^h_{aj} + V^a\,\Gamma^h_{aj,k} 
  \\ \nonumber &-& 
  V^h_{,kj} - V^a_{,j}\,\Gamma^h_{ak} - V^a\,\Gamma^h_{ak,j} 
\end{eqnarray}
where the product rule has been applied. Using now the 
symmetry $V^h_{,jk}=V^h_{,kj}$ we get
\begin{eqnarray} \label{bracket1}
  \left[\matrix{h\cr j}\right]_{,k} - 
  \left[\matrix{h\cr k}\right]_{,j} &=&
     V^a_{,k}\,\Gamma^h_{aj} + V^a\,\Gamma^h_{aj,k} 
   - V^a_{,j}\,\Gamma^h_{ak} - V^a\,\Gamma^h_{ak,j}\,. 
\end{eqnarray}
Next, inserting the definitions of the brackets
\begin{eqnarray} \label{bracket2}
  \left[\matrix{b\cr j}\right]\,\Gamma^h_{bk} - 
  \left[\matrix{b\cr k}\right]\,\Gamma^h_{bj} &=&
  V^b_{,j}\,\Gamma^h_{bk} + V^a\,\Gamma^b_{aj}\,\Gamma^h_{bk} - 
  V^b_{,k}\,\Gamma^h_{bj} - V^a\,\Gamma^b_{ak}\,\Gamma^h_{bj} 
\end{eqnarray}
holds. Combining from (\ref{bracket1}) and (\ref{bracket2})
\begin{eqnarray} \nonumber
  V^a_{,k}\,\Gamma^h_{aj} + V^b_{,j}\,\Gamma^h_{bk} - 
  V^a_{,j}\,\Gamma^h_{ak} - V^b_{,k}\,\Gamma^h_{bj} = 0
\end{eqnarray}
is seen to be true due to symmetry under exchanging $j$ and $k$. 
Left over are from (\ref{bracket1}) $V^a\,\Gamma^h_{aj,k}-V^a\,
\Gamma^h_{ak,j}$ and from (\ref{bracket2}) $V^a\,\Gamma^b_{aj}\,
\Gamma^h_{bk}-V^a\, \Gamma^b_{ak}\,\Gamma^h_{bj}$. So, the result is
\begin{eqnarray} \nonumber
  V^h_{:jk}-V^h_{:kj} &=& V^a\,\Gamma^h_{aj,k}+V^a\,\Gamma^b_{aj}\,
  \Gamma^h_{bk}-V^a\,\Gamma^h_{ak,j}-V^a\,\Gamma^b_{ak}\,\Gamma^h_{bj}
  \\ &=& -V^a\,R^h_{~ajk}\,,
\end{eqnarray}
where
\begin{eqnarray} \label{Riemann}
  R^h_{~ijk} &=& \Gamma^h_{ij,k}+\Gamma^b_{ij}\,\Gamma^h_{bk}
               -\Gamma^h_{ik,j}-\Gamma^b_{ik}\,\Gamma^h_{bj}
\end{eqnarray}
defines the Riemann curvature tensor. These are 256 functions, which
are related by many symmetries (identities). See Rindler p.218/219.
\vfill
\end{document}
