Symmetries of double dual of Riemann curvature tensor The definition of the double dual of Riemann is as follows:
$$G^{\alpha\beta}{}{}_{\gamma\delta} = \frac{1}{2}\epsilon^{\alpha\beta\mu\nu}R^{\rho\sigma}{}{}_{\mu\nu}\frac{1}{2}\epsilon_{\rho\sigma\gamma\delta} $$
Now my question is how to prove that this tensor satisfy: $G^{\alpha}{}_{[\beta\gamma\delta]} = 0$
 A: For simplicity, I use the Latin letters, and I rewrite the double dual of Riemann tensor as
$$
G_{abcd} = \frac{1}{4} \epsilon_{ab}{}^{ef} \, R_{efgh} \, \epsilon_{cd}{}^{gh} .
$$
Now, we want to see whether $G_{a [bcd]} = 0$. To do so, we calculate
$$
G_{abcd} + G_{acdb} + G_{adbc} = \frac{1}{4}  \left( \epsilon_{ab}{}^{ef} \epsilon_{cd}{}^{gh} + \epsilon_{ac}{}^{ef} \epsilon_{db}{}^{gh} + \epsilon_{ad}{}^{ef} \epsilon_{bc}{}^{gh} \right)R_{efgh}
\,.
$$
Using the property of the Levi-Civita tensor; i.e., 
$$
\epsilon_{ab}{}^{ef} \epsilon_{cd}{}^{gh}
=
(-1)^q \delta^{efgh}_{abcd} \,,
$$
where $q$ is the number of negatives in the metric signature (in GR this is equal to one) and $\delta^{efgh}_{abcd} := 4! \delta^{efgh}_{[abcd]}$ is the generalised Kronecker delta, we obtain
\begin{equation}
 G_{abcd} + G_{acdb} + G_{adbc} =
-\frac{1}{4} \left( \delta^{efgh}_{abcd} + \delta^{efgh}_{acdb} + \delta^{efgh}_{a dbc} \right) R_{efgh} . \tag{1}
\end{equation}
The expression in the bracket is a cyclic (symmetric)  permutation within a totally anti-symmetric (the generalised Kronecker delta) tensor, i.e., $\delta^{efgh}_{[a(bcd)]}$. But, this vanishes as you may show this by explicit calculation!
Alternatively:
One could use the definition of a totally antisymmetric tensor given by the generalised Kronecker delta. For a tensor $T_{abcd}$ one has
$$
 T_{[abcd]} = \frac{1}{4!} \delta^{efgh}_{abcd} \, T_{efgh}.
$$
Then, the equation (1) becomes:
$$
G_{abcd} + G_{acdb} + G_{adbc}
=
- 3! \left( R_{[abcd]} + R_{[acdb]} + R_{[adbc]} \right).
$$
This obviously vanishes. To show the result one can expand the expressions and use the algebraic symmetries of Riemann tensor.
