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Hi there. This is regarding a proof from Wald that involves showing that for an axisymmetric, static space - time, the 2 - planes orthogonal to the (commuting) time - like and closed space - like killing vector fields \xi ^{a},\psi ^{a}, respectively, are integrable in the sense of Frobenius' theorem. One of the end of chapter problems is to show from direct calculation that \mathcal{L}_{\psi }\omega _{a} = 0 where \omega _{a} = \epsilon _{abcd}\xi ^{b}\triangledown^{c}\xi ^{d} is the twist of \xi ^{a} ,\mathcal{L}_{\psi } is the lie derivative along the flow of \psi ^{a}, and \epsilon _{abcd} is the levi - civita symbol. I was hoping someone could check my work to see if it is all legal.
We have that \mathcal{L}_{\psi}\omega _{a} = \mathcal{L}_{\psi}(\epsilon _{abcd}\xi ^{b}\triangledown^{c}\xi ^{d}) = \epsilon _{abcd}\mathcal{L}_{\psi}(\xi ^{b}\triangledown^{c}\xi ^{d}) + \xi ^{b}\triangledown^{c}\xi ^{d}\mathcal{L}_{\psi}\epsilon _{abcd}. Starting with the first expression, we have \mathcal{L}_{\psi}(\xi ^{b}\triangledown^{c}\xi ^{d}) = \triangledown ^{c}\xi ^{d}\mathcal{L}_{\psi}\xi ^{b} + \xi ^{b}\mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = \xi ^{b}\mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) where I have used the fact that [\xi ,\psi ] = 0.
\mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = (\psi ^{e}\triangledown _{e}\triangledown ^{c}\xi ^{d} - \triangledown ^{c}\xi ^{d}\triangledown _{e}\psi ^{c} - \triangledown ^{c}\xi ^{e}\triangledown _{e}\psi ^{d}). We have that \psi ^{e}\triangledown _{e}\triangledown ^{c}\xi ^{d} = R^{dc}{}_{ef}\psi ^{e}\psi ^{f} = R^{dc}{}_{[ef]}\psi ^{(e}\psi ^{f)} = 0 therefore \mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = - \triangledown ^{c}\xi ^{d}\triangledown _{e}\psi ^{c} - \triangledown ^{c}\xi ^{e}\triangledown _{e}\psi ^{d} = \triangledown ^{d}\xi ^{e}\triangledown _{e}\psi ^{c} - \triangledown ^{d}\psi ^{e}\triangledown _{e}\xi ^{c}. Now, \mathcal{L}_{\psi}\xi ^{c} = 0 \Rightarrow \xi ^{e}\triangledown _{e}\psi ^{c} = \psi^{e}\triangledown _{e}\xi ^{c}\Rightarrow \triangledown ^{d}( \xi ^{e}\triangledown _{e}\psi ^{c}) = \triangledown ^{d}(\psi^{e}\triangledown _{e}\xi ^{c}) so we can conclude that \triangledown ^{d}\xi ^{e}\triangledown _{e}\psi ^{c} - \triangledown ^{d}\psi ^{e}\triangledown _{e}\xi ^{c} = \psi ^{e}\triangledown ^{d}\triangledown _{e}\xi ^{c} - \xi ^{e}\triangledown ^{d}\triangledown _{e}\psi ^{c} = \psi ^{e}\xi ^{f}(R^{d}{}_{ef}{}^{c} + R^{d}{}_{f}{}^{c}{}_{e}) = -\psi ^{e}\xi ^{f}R^{dc}{}_{ef} = 0, where I have used the Bianchi identity, thus \mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = 0.
Now for the second expression, first I'll prove the following claim: let \mathbf{T} be a rank 2 tensor then \sum_{k}\epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = \epsilon _{a_1...a_n}T^{j}_j. To do this, first let k\in \left \{ 1,...,n \right \} be fixed but arbitrary. Note that \forall j\neq a_k, \epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = 0 because we would have a repeated index and by definition of the levi - civita symbol, a component with two or more repeated indices vanishes. Therefore, \epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = \epsilon _{a_1...a_{k-1}a_ka_{k+1}...a_{n}}T^{a_k}_{a_k} =\epsilon _{a_1...a_{n}}T^{a_k}_{a_k}. Because k was arbitrary, \sum_{k}\epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = \epsilon _{a_1...a_n}\sum_{k}T^{a_k}_{a_k} =\epsilon _{a_1...a_n}T^{j}_j. Using this, \mathcal{L}_{\psi}\epsilon _{abcd} = \psi ^{e}\triangledown _{e}\epsilon _{abcd} + \sum_{k}\epsilon _{a..e..d}\triangledown _{a_k}\psi ^{e} = \epsilon _{abcd}\triangledown _{e}\psi ^{e} = 0 where the facts that the covariant derivative of the levi - civita symbol vanishes and that killing fields are divergence free have been used. This finally gives us \mathcal{L}_{\psi}\omega _{a} = 0.
I really, really appreciate any and all help on finding mistakes in the solution and/or flaws in any arguments. Sorry for the long winded post. Cheers!
We have that \mathcal{L}_{\psi}\omega _{a} = \mathcal{L}_{\psi}(\epsilon _{abcd}\xi ^{b}\triangledown^{c}\xi ^{d}) = \epsilon _{abcd}\mathcal{L}_{\psi}(\xi ^{b}\triangledown^{c}\xi ^{d}) + \xi ^{b}\triangledown^{c}\xi ^{d}\mathcal{L}_{\psi}\epsilon _{abcd}. Starting with the first expression, we have \mathcal{L}_{\psi}(\xi ^{b}\triangledown^{c}\xi ^{d}) = \triangledown ^{c}\xi ^{d}\mathcal{L}_{\psi}\xi ^{b} + \xi ^{b}\mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = \xi ^{b}\mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) where I have used the fact that [\xi ,\psi ] = 0.
\mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = (\psi ^{e}\triangledown _{e}\triangledown ^{c}\xi ^{d} - \triangledown ^{c}\xi ^{d}\triangledown _{e}\psi ^{c} - \triangledown ^{c}\xi ^{e}\triangledown _{e}\psi ^{d}). We have that \psi ^{e}\triangledown _{e}\triangledown ^{c}\xi ^{d} = R^{dc}{}_{ef}\psi ^{e}\psi ^{f} = R^{dc}{}_{[ef]}\psi ^{(e}\psi ^{f)} = 0 therefore \mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = - \triangledown ^{c}\xi ^{d}\triangledown _{e}\psi ^{c} - \triangledown ^{c}\xi ^{e}\triangledown _{e}\psi ^{d} = \triangledown ^{d}\xi ^{e}\triangledown _{e}\psi ^{c} - \triangledown ^{d}\psi ^{e}\triangledown _{e}\xi ^{c}. Now, \mathcal{L}_{\psi}\xi ^{c} = 0 \Rightarrow \xi ^{e}\triangledown _{e}\psi ^{c} = \psi^{e}\triangledown _{e}\xi ^{c}\Rightarrow \triangledown ^{d}( \xi ^{e}\triangledown _{e}\psi ^{c}) = \triangledown ^{d}(\psi^{e}\triangledown _{e}\xi ^{c}) so we can conclude that \triangledown ^{d}\xi ^{e}\triangledown _{e}\psi ^{c} - \triangledown ^{d}\psi ^{e}\triangledown _{e}\xi ^{c} = \psi ^{e}\triangledown ^{d}\triangledown _{e}\xi ^{c} - \xi ^{e}\triangledown ^{d}\triangledown _{e}\psi ^{c} = \psi ^{e}\xi ^{f}(R^{d}{}_{ef}{}^{c} + R^{d}{}_{f}{}^{c}{}_{e}) = -\psi ^{e}\xi ^{f}R^{dc}{}_{ef} = 0, where I have used the Bianchi identity, thus \mathcal{L}_{\psi}(\triangledown ^{c}\xi ^{d}) = 0.
Now for the second expression, first I'll prove the following claim: let \mathbf{T} be a rank 2 tensor then \sum_{k}\epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = \epsilon _{a_1...a_n}T^{j}_j. To do this, first let k\in \left \{ 1,...,n \right \} be fixed but arbitrary. Note that \forall j\neq a_k, \epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = 0 because we would have a repeated index and by definition of the levi - civita symbol, a component with two or more repeated indices vanishes. Therefore, \epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = \epsilon _{a_1...a_{k-1}a_ka_{k+1}...a_{n}}T^{a_k}_{a_k} =\epsilon _{a_1...a_{n}}T^{a_k}_{a_k}. Because k was arbitrary, \sum_{k}\epsilon _{a_1...a_{k-1}ja_{k+1}...a_{n}}T^{j}_{a_k} = \epsilon _{a_1...a_n}\sum_{k}T^{a_k}_{a_k} =\epsilon _{a_1...a_n}T^{j}_j. Using this, \mathcal{L}_{\psi}\epsilon _{abcd} = \psi ^{e}\triangledown _{e}\epsilon _{abcd} + \sum_{k}\epsilon _{a..e..d}\triangledown _{a_k}\psi ^{e} = \epsilon _{abcd}\triangledown _{e}\psi ^{e} = 0 where the facts that the covariant derivative of the levi - civita symbol vanishes and that killing fields are divergence free have been used. This finally gives us \mathcal{L}_{\psi}\omega _{a} = 0.
I really, really appreciate any and all help on finding mistakes in the solution and/or flaws in any arguments. Sorry for the long winded post. Cheers!