Wald's derivation of Komar formula: passage on Eq. (11.2.5)

Question

In Wald's General Relativity derivation of the Komar formula for the mass of a static, asymptotically flat spacetime there is the following passage:

Using Killing's equation $\nabla_a\xi_b = \nabla_{[a}\xi_{b]}$, we may rewrite equation (11.2.4) as $$F = \frac{1}{2} \int_S N^{ab}\nabla_a\xi_b \mathrm{d}A = - \frac{1}{2} \int_S \boldsymbol{\epsilon}_{abcd}\nabla^c\xi^d, \tag{11.2.5}$$ where, in the first line [sic], $N^{ab} = 2 V^{-1} \xi^{[a}N^{b]}$ is the normal "bi-vector" to $S$, and, in the second line [sic] $\boldsymbol{\epsilon}_{abcd}$ is the volume element on spacetime associated with the spacetime metric and the integrand is viewed as two-form, $\boldsymbol{\alpha}$, to be integrated over the two-dimensional submanifold $S$ (see appendix B). (The orientation of $\boldsymbol{\epsilon}_{abcd}$ is chosen so that $\boldsymbol{\epsilon}_{abcd} = - 6 N_{[ab}\boldsymbol{\epsilon}_{cd]}$, where $\boldsymbol{\epsilon}_{cd}$ is the volume element on $S$.)

I don't understand this second step (the second "line", as Wald calls it) at all. Why exactly is this manipulation possible? How does the spacetime orientation appears? Why is this choice of orientation possible? I tried working a bit with differential forms (of which I have a very limited knowledge of) but couldn't even write the contraction on the "first line" in differential form language.

Answer 1

Wald's Eq. (B.2.24) relates the volume element $\boldsymbol{\epsilon}$ on an $n$-manifold with the volume element $\boldsymbol{\tilde{\epsilon}}$ on a hypersurface with unit normal vector $n^a$ through $$\frac{1}{n}\boldsymbol{\epsilon}_{a_1 \ldots a_n} = n_{[a_1}\boldsymbol{\tilde\epsilon}_{a_2 \ldots a_{n-1}]}. \tag{B.2.24}$$

The normal vector is chosen to be outward pointing if it is spacelike, but inward pointing if it is timelike. Repeatedly using this formula we get the relation between $\boldsymbol{\epsilon}_{abcd}$, the volume element on the spacetime, with $\boldsymbol{\epsilon}_{ab}$, the volume element on the 2-sphere $S$ we are integrating over, to be $$\frac{1}{12} \boldsymbol{\epsilon}_{abcd} = -V^{-1} \xi_{[a}N_{b}\boldsymbol{\epsilon}_{cd]} = -\frac{1}{2} N_{[ab}\boldsymbol{\epsilon}_{cd]},$$ where the minus sign comes from the normal to the spatial hypersurface having to be inward pointing, and hence it is $-V^{-1}\xi^a$ rather than $V^{-1}\xi^a$ (recall $V^{-1}\xi^a$ is the unit-norm future-pointing vector along the Killing trajectories being considered in the static spacetime).

Remark: I'm not fully confident about this minus sign. Feel free to correct me on this.

Therefore, we finally get to $$\boldsymbol{\epsilon}_{abcd} = -6 N_{[ab}\boldsymbol{\epsilon}_{cd]}.$$

With some algebraic work and using a lot of antisymmetric properties of $N_{ab}$, $\boldsymbol{\epsilon}_{ab}$, and $\nabla_a\xi_b$ we can find that $$- \nabla^c \xi^d \boldsymbol{\epsilon}_{abcd} = N_{ab}\boldsymbol{\epsilon}_{cd}\nabla^c\xi^d + N_{cd}\boldsymbol{\epsilon}_{ab}\nabla^c\xi^d - 4 N_{c[a}\boldsymbol{\epsilon}_{b]d}\nabla^c\xi^d.$$

Since $N_{ab}$ is the normal bi-vector to $S$, once we integrate both sides of this equation over $S$ we get $$- \int_S \nabla^c \xi^d \boldsymbol{\epsilon}_{abcd} = \int_S N_{cd}\boldsymbol{\epsilon}_{ab}\nabla^c\xi^d = \int_S N_{cd}\nabla^c\xi^d \mathrm{d}A,$$ as desired.