Riemann Curvature Tensor Symmetries Proof I am trying to expand $$\varepsilon^{{abcd}} R_{{abcd}}$$ by using
four identities of the Riemann curvature tensor:
Symmetry
$$R_{{abcd}} = R_{{cdab}}$$ 
Antisymmetry first pair of indicies
$$R_{{abcd}} = - R_{{bacd}}$$ 
Antisymmetry last pair of indicies
$$R_{{abcd}} = - R_{{abdc}}$$ 
Cyclicity
$$R_{{abcd}} + R_{{adbc}} + R_{{acdb}} = 0$$ 
From what I understand, the terms should cancel out and I should end up with
is $$\varepsilon^{{abcd}} R_{{abcd}} = 0.$$ 
What I ended up with was
this mess:
$$\begin{array}{l}
  \varepsilon^{{abcd}} R_{{abcd}} = R_{\left[ {abcd} \right]} =
  \frac{1}{4!} \left( \underset{- R_{{dcab}}}{\underset{+
  R_{{cdab}}}{\underset{- R_{{abdc}}}{\underset{{\color{dark green}
  + R_{{badc}}}}{\underset{- {\color{red} {\color{black}
  R_{{bacd}}}}}{{\color{blue} R_{{abcd}}}}}}}} +
  \underset{{\color{magenta} - R_{{adbc}}}}{\underset{{\color{red} +
  R_{{cbad}}}}{\underset{- R_{{cbda}}}{\underset{+
  R_{{bcda}}}{{\color{magenta} {\color{black} R_{{dabc}}}}}}}} +
  \underset{- R_{{abdc}}}{\underset{+ R_{{dcba}}}{\underset{-
  R_{{cdba}}}{\underset{- R_{{dcab}}}{\underset{{\color{dark green}
  + R_{{badc}}}}{\underset{- R_{{bacd}}}{\underset{+
  R_{{abcd}}}{R_{{cdab}}}}}}}}} + \underset{+
  R_{{dabc}}}{\underset{- R_{{cbda}}}{\underset{{\color{red} +
  R_{{cbad}}}}{\underset{- {\color{blue} {\color{black}
  R_{{bcad}}}}}{{R_{{bcda}}}}}}} - \underset{-
  R_{{bdca}}}{\underset{{\color{blue} + R_{{acdb}}}}{\underset{+
  R_{{dbca}}}{\underset{- R_{{dbac}}}{\underset{+
  R_{{bdac}}}{R_{{acbd}}}}}}} - \underset{{\color{blue} +
  R_{{adbc}}}}{\underset{- R_{{bcda}}}{\underset{+ {\color{blue}
  {\color{black} R_{{bcad}}}}_{}}{\underset{{\color{red} -
  R_{{cbad}}}}{R_{{adcb}}}}}} - \underset{{\color{black} +
  R_{{abcd}}}}{\underset{- R_{{bacd}}}{\underset{{\color{dark green}
  + R_{{badc}}}}{R_{{abdc}}}}} - \underset{{\color{magenta} +
  R_{{abcd}}}}{\underset{- R_{{abdc}}}{\underset{{\color{red}
  {\color{dark green} + R_{{badc}}}}}{\underset{- {\color{red}
  {\color{black} R_{{bacd}}}}}{R_{{cdba}}}}}} \right)
\end{array}$$
where I can get rid of the blue or the purple terms using cyclicity, but I'm
stuck because I cant see how I can get all the terms to cancel. The main
problem seems to be is that the last term in the cyclicity identity $$\left(
R_{{acdb}} \right)$$ can only be acquired from the 5th term $$\left(
R_{{acbd}} \right)$$ in the expression i have. After I get rid of 6 terms
with cyclicity I was thinking I could get of what remains with some symmetry
relationship. Am I going down the wrong path here? Do I need another
relationship? Carroll in ``Introduction to General Relativity'' says in eq
3.83 that all I have to do is expand the expression for $$R_{\left[ {abcd}
\right]}$$ and mess with the indicies using the 4 identities to proove that it
reduces to zero. 
To increase visibility I posted the same question on http://www.physicsforums.com/showthread.php?p=4852429#post4852429, but haven't gotten any replies yet.
 A: I think that it is only necessary to use the cyclic identity. Contracting both sides with the Levi-Civita, we should have $$0 = (R_{abcd} + R_{adbc} + R_{acdb}) \varepsilon^{abcd} \tag{1}.$$
Let $S = R_{abcd}\varepsilon^{abcd}$. Then $R_{adbc}\varepsilon^{abcd} = -R_{adbc}\varepsilon^{acbd} = R_{adbc}\varepsilon^{adbc} = S$ where the last step is renaming the dummy indices. With the same argument the third term is also equal to $S$, so we (1) says that $3S = 0$.
A: First, by definition of $\varepsilon$
 $$\varepsilon^{{abcd}} R_{{abcd}} = \varepsilon^{{abcd}} R_{{a[bcd]}} =\text{any symmatrization of the indices of $R$} $$
But
\begin{eqnarray}R_{{a[bcd]}} &=&\frac{1}{3!}\bigg\{  R_{abcd}-R_{abdc}+R_{adbc}-R_{adcb}+R_{acdb} -R_{acbd}  \bigg\}\;,\\
&=& \frac 1 3 \bigg\{   R_{abcd}+R_{adbc} +R_{acdb}\bigg\}\;,\\
&=& 0\;.\texttt{(by the first Bianchi identity)}
\end{eqnarray}
So, 
$$\varepsilon^{{abcd}} R_{{abcd}} = \varepsilon^{{abcd}} R_{{a[bcd]}} =\varepsilon^{{abcd}}\times 0=0\;.\quad\quad\quad\blacksquare $$
