Question about Komar integral derivation in Wald I have a question about derivation 11.2.10 in Wald (page 289). Here is a screenshot of the relevant passage: 

I don't get the step $$-\frac{1}{4\pi}\int _{\Sigma}R^{d}{}{}_{f}\xi^{f}\epsilon_{deab} = \frac{1}{4\pi}\int _{\Sigma}R_{ab}n^{a}\xi^{b}dV.$$ Could someone explain to me how he gets that equality? Thank you very much in advance. 
 A: [EDIT](Precisions for the sign in the definition of the normal)
It is possible that  the full expression for the LHS term would be : 
$$-\frac{1}{4\pi}\int _{\Sigma}R^{d}{}{}_{f}\xi^{f}\epsilon_{deab} dx^e \wedge dx^a \wedge dx^b$$
If so, and taking in account that the normal $n_d$ to the surface $\Sigma$ is simply defined by :
$$n_d ~dV= -\epsilon_{deab} dx^e \wedge dx^a \wedge dx^b$$
For the explanation of the minus sign, we have to take in account that the $\Sigma$ surface is a space-like hypersurface, so the normal is a time-like vector, that is $n_d n^d = -1$ (With Wald conventions).
Suppose, for one moment, that we are in a flat metric, and that the $\Sigma$ surface is the ordinary 3-spatial volume. Following the choice of the author of " $n^d$ as the unit future pointing normal to $\Sigma$", this means $n^0 = 1$.
But because $n_d n^d = -1$, this means $n_0 = -1$. This is the justification of the minus sign.
So, We finally get :
$$ \frac{1}{4\pi}\int _{\Sigma}R^{d}{}{}_{f}\xi^{f} n_d ~dV =  \frac{1}{4\pi}\int _{\Sigma}R_{ab}n^a\xi^{b}  ~dV$$
