Integration trouble (integral over a 2-sphere)

[etotheipi]

There's an integral over a 2-sphere $S$ with unit normal $N^a$ within a hypersurface orthogonal to a Killing field $\xi^a$$$F = \int_S N^b (\xi^a / V) \nabla_a \xi_b dA = \frac{1}{2} \int_S N^{ab} \nabla_a \xi_b dA, \quad N^{ab} := 2V^{-1} \xi^{[a} N^{b]}$$which follows because the Killing equation is $\nabla_{a} \xi_b = \nabla_{[a} \xi_{b]}$ and we can also write $\xi^a N^b \nabla_{[a} \xi_{b]} = \xi^a N^b \delta^{[c}_{a} \delta^{d]}_b \nabla_c \xi_d = \xi^{[c} N^{d]} \nabla_c \xi_d$. The original integral is supposed to transform into$$F = \frac{-1}{2} \int_S \epsilon_{abcd} \nabla^c \xi^d$$but I don't see how yet. Can anyone provide a hint? Thanks.

 

[Twigg]

Did part of the last equation get lost to a typo? The final result is rank 2 (a and b are free) but the original integral is a scalar. Am I missing something?

 

[etotheipi]

As far as I can tell they're the same as in the book; the indices in this case are abstract, so I reckon the second should be understood as the integral of a 2-form over the submanifold.

 

[etotheipi]

After some helpful discussions with @Twigg, here's a possible idea: first we will use that $\nabla_a \xi_b = \nabla_{[a} \xi_{b]}$, and also use that the volume form $\epsilon_{ab}$ on the 2-sphere is totally antisymmetric, i.e. $\epsilon_{ab} = \epsilon_{[ab]}$,\begin{align*}F = \frac{1}{2} \int_S N^{ab} \nabla_a \xi_b \mathrm{d}A &= \frac{1}{2} N^{ab} \nabla_{[a} \xi_{b]} \epsilon_{cd} \\

&= \frac{1}{2} \int_S N_{[ab]} \nabla^a \xi^b \epsilon_{[cd]} \\

&= \frac{1}{2} \int_S \nabla^a \xi^b \delta^{[e}_a \delta^{f]}_b \delta^{[g}_c \delta^{h]}_d N_{ef} \epsilon_{gh}

\end{align*}However, since $\delta^{[e}_a \delta^{f]}_b \delta^{[g}_c \delta^{h]}_d = \frac{1}{4} \delta^{e}_a \delta^{f}_b \delta^{g}_c \delta^{h}_d = 6 \delta^{[e}_a \delta^{f}_b \delta^{g}_c \delta^{h]}_d$, this is simply\begin{align*}

F &= \frac{1}{2} \int_S \nabla^a \xi^b \cdot 6 N_{[ab} \epsilon_{cd]} \\

&= \frac{-1}{2} \int_S \nabla^a \xi^b \epsilon_{abcd} \\

\end{align*}where the last line follows because $\epsilon_{abcd} = -6N_{[ab} \epsilon_{cd]}$