# Exterior derivative identity in vacuum space-time

1. Jun 23, 2013

### PhizKid

I was reading a paper by Geroch and I was confused by the following: given a scalar field $\omega$ satisfying $\nabla_{a}\omega = \omega_{a} = \epsilon_{abcd}\xi^{b}\nabla^{c}\xi^{d}$ and the scalar $\lambda = \xi^{a}\xi_{a}$, where $\xi^{a}$ is a killing vector field, can someone prove to me why $\nabla_{[e}(2\lambda \nabla_{a}\xi_{b]} + \omega\epsilon_{ab]cd}\nabla^{c}\xi^{d}) = 0$ if we are in vacuum i.e. $R_{ab} = 0$? Thanks.

2. Jun 23, 2013

### WannabeNewton

First, it is easiest to show that $\nabla_{[e}(\epsilon_{ab]cd}\nabla^{c}\xi^{d}) = 0$. Note that this is equivalent to showing $\epsilon^{abef}\nabla_{e}(\epsilon_{abcd}\nabla^{c}\xi^{d}) = 0$ since this then implies $\epsilon_{ijkf}\epsilon^{abef}\nabla_{e}(\epsilon_{abcd}\nabla^{c}\xi^{d}) = -6\delta^{[a}_{i}\delta^{b}_{j}\delta^{e]}_{k}\nabla_{e}(\epsilon_{abcd}\nabla^{c}\xi^{d}) = -6\nabla_{[k}(\epsilon_{ij]cd}\nabla^{c}\xi^{d}) = 0$ which is what we want. We have that $\nabla_{e}(\epsilon_{abcd}\nabla^{c}\xi^{d}) = \nabla_{e}(\epsilon^{abef}\epsilon_{abcd}\nabla^{c}\xi^{d}) = -4\nabla_{e}(\delta^{[e}_{c}\delta^{f]}_{d}\nabla^{c}\xi^{d}) = -4\nabla_{e}\nabla^{e}\xi^{f} = 4R^{f}{}{}_{d}\xi^{d} = 0$ where I have used the fact that $\nabla^{c}\xi^{d} = \nabla^{[c}\xi^{d]}$ and that $R_{ab} = 0$ since we are in vacuum. Hence $\nabla_{[e}(\epsilon_{ab]cd}\nabla^{c}\xi^{d}) = 0$. Now for the annoying part .

Note that by the same argument above, in order to show that the exterior derivative of the 2-form $\alpha_{ab} = 2\lambda \nabla_{a}\xi_{b} + \omega \epsilon_{abcd}\nabla^{c}\xi^{d}$ vanishes, it suffices to show that $\epsilon^{abef}\nabla_{e}\alpha_{ab} = 0$. Expanding this out, and using the result from the previous paragraph, we get $\epsilon^{abef}\nabla_{e}\alpha_{ab} = 2\xi^{c}\xi_{c}\epsilon^{abef}\nabla_{e}\nabla_{a}\xi_{b} + 4\epsilon^{abef}\xi_{c}\nabla_{e}\xi^{c}\nabla_{a}\xi_{b} - 4\nabla^{e}\xi^{f}\epsilon_{eabc}\xi^{a}\nabla^{b}\xi^{c}$ where I have used the explicit form of the twist of $\xi^{a}$.

Now, note that $\epsilon^{abef}\nabla_{e}\nabla_{a}\xi_{b} = \epsilon^{ebaf}\nabla_{a}\nabla_{e}\xi_{b} = -\epsilon^{abef}\nabla_{a}\nabla_{e}\xi_{b}$ where in the first equality I renamed the indices and in the second equality I used the antisymmetry of the volume form. Hence $2\epsilon^{abef}\nabla_{e}\nabla_{a}\xi_{b} = \epsilon^{abef}R_{eab}{}{}^{d}\xi_{d} = \epsilon^{fbea}R^{d}{}{}_{bea}\xi_{d}$. Using the antisymmetry of the volume form and the Bianchi identity $R_{a[bcd]} = 0$, we have that $3\epsilon^{fbea}R^{d}{}{}_{bea}\xi_{d} = 3\epsilon^{fbea}R^{d}{}{}_{[bea]}\xi_{d} = 0$ thus $\epsilon^{abef}\nabla_{e}\nabla_{a}\xi_{b} = 0$.

All we're left with now is $\epsilon^{abef}\nabla_{e}\alpha_{ab} = 4\epsilon^{abef}\xi_{c}\nabla_{e}\xi^{c}\nabla_{a}\xi_{b} - 4 \epsilon_{eabc}\xi^{a}\nabla^{b}\xi^{c}\nabla^{e}\xi^{f}$. Using again the antisymmetry of the volume form and permuting the indices around, we can express this as $\epsilon^{abef}\nabla_{e}\alpha_{ab} = 8\xi_{c}\nabla^{a}\xi^{b}\epsilon_{abe}{}{}^{[c}\nabla^{f]}\xi^{e}$. As usual, in order to show that this vanishes all we have to show is that $\epsilon^{klcf}(\nabla^{a}\xi^{b}\epsilon_{abec}\nabla_{f}\xi^{e}) = 0$. This can be easily done as $\epsilon^{klcf}(\nabla^{a}\xi^{b}\epsilon_{abec}\nabla_{f}\xi^{e}) = \epsilon^{klfc}\epsilon_{abec}\nabla^{a}\xi^{b}\nabla^{e}\xi_{f} = -6\nabla^{[k}\xi^{l}\nabla^{f]}\xi_{f}$. Expanding this out, we have $-6\nabla^{[k}\xi^{l}\nabla^{f]}\xi_{f} = -2(\nabla^{k}\xi^{l}\nabla^{f}\xi_{f} - \nabla^{k}\xi^{f}\nabla^{l}\xi_{f} + \nabla^{l}\xi^{f}\nabla^{k}\xi_{f}) = - \nabla^{k}\xi^{f}\nabla^{l}\xi_{f}+\nabla^{l}\xi_{f}\nabla^{k}\xi^{f} = 0$ where I used the fact that $\nabla^{f}\xi_{f} = 0$ for killing vector fields.

So we finally have that $\epsilon^{abef}\nabla_{e}\alpha_{ab} = 0$ therefore $\nabla_{[e}\alpha_{ab]} = 0$ as desired.

3. Jun 23, 2013

### PhizKid

Thank you WannabeNewton, this was very clarifying

4. Jun 23, 2013

### WannabeNewton

Yep anytime. I should probably note that this is also problem 5 of chapter 7 in Wald's general relativity text, for future reference.

5. Jun 24, 2013

### PhizKid

I have another question to ask: in the paper Geroch then says that since $\nabla_{[e}(\epsilon_{ab]cd}\nabla^{c}\xi^{d}) = 0$, there exists an $\alpha_{a}$ such that $\nabla_{[a}\alpha_{b]} = \frac{1}{2}\epsilon_{abcd}\nabla^{c}\xi^{d}$ and that we can "add a gradient" to make it so that $\xi^{a}\alpha_{a} = \omega$ but why is any of this true?

6. Jun 24, 2013

### WannabeNewton

Well $\omega_{ab} = \epsilon_{abcd}\nabla^{c}\xi^{d}$ is just a 2-form so the statement $\nabla_{[e}\omega_{ab]} = \nabla_{[e}\{\epsilon_{ab]cd}\nabla^{c}\xi^{d}\} = 0$ is equivalent to saying that $d\boldsymbol{\omega} = 0$. By the Poincare lemma, this then implies that locally there exists a 1-form $\boldsymbol{\alpha}$ such that $\boldsymbol{\omega} = d\boldsymbol{\alpha}$ i.e. $\nabla_{[a}\alpha_{b]} = \frac{1}{2}\epsilon_{abcd}\nabla^{c}\xi^{d}$.

As for your second question, note that if we take an arbitrary smooth scalar field $\varphi$ and consider $\alpha_{b} + \nabla_{b}\varphi$ then $\nabla_{[e}\{\alpha_{b]} + \nabla_{b]}\varphi \} = \nabla_{[e}\alpha_{b]}$ since $\nabla_{a}$ is torsion free. So consider $\tilde{\alpha}_{b} = \alpha_{b} + \nabla_{b}\varphi$ such that $\xi^{b}\tilde{\alpha}_{b} = \omega = \xi^{b}\alpha_{b} + \xi^{b}\nabla_{b}\varphi$. Then all you have to do is, in principle, find a $\varphi$ solving $\omega = \xi^{b}\alpha_{b} + \xi^{b}\nabla_{b}\varphi$ and you can then just use $\tilde{\alpha}_{b}$ as the 1-form that solves $\nabla_{[a}\tilde{\alpha}_{b]} = \frac{1}{2}\epsilon_{abcd}\nabla^{c}\xi^{d}$.

A similar argument holds for $\nabla_{[e}\{2\lambda \nabla_{a}\xi_{b]} + \omega\epsilon_{ab]cd}\nabla^{c}\xi^{d}\} = 0$.

7. Jun 25, 2013

### PhizKid

Thanks. One last question (hopefully): can you explain why the exterior derivative of the twist of the killing vector field identically vanishes in vacuum space-time because Geroch just mentions it, he doesn't prove it.

Last edited: Jun 25, 2013
8. Jun 25, 2013

### WannabeNewton

Sure! It is actually quite easy to show that $\nabla_{[a}\omega_{b]} = 0$. As usual we just have to show that $\epsilon^{abcd}\nabla_{c}\omega_{d} = \epsilon^{abcd}\epsilon_{defg}\nabla_{c}(\xi^{e}\nabla^{f}\xi^{g}) = 6 \nabla_{c}(\xi^{[b}\nabla^{c}\xi^{d]})$ vanishes. Expanding this out we find that $6\nabla_{c}(\xi^{[b}\nabla^{c}\xi^{d]}) = 2\nabla_{c}(\xi^{b}\nabla^{c}\xi^{d}- \xi^{c}\nabla^{b}\xi^{d} + \xi^{d}\nabla^{b}\xi^{c}) \\ = 2( \xi^{b}\nabla_{c}\nabla^{c}\xi^{d} - \xi^{c}\nabla_{c}\nabla^{b}\xi^{d} - \xi^{d}\nabla_{c}\nabla^{c}\xi^{b} + \nabla_{c}\xi^{b}\nabla^{c}\xi^{d} - \nabla_{c}\xi^{c}\nabla^{b}\xi^{d} + \nabla_{c}\xi^{d}\nabla^{b}\xi^{c}) \\= 2( \xi^{b}\nabla_{c}\nabla^{c}\xi^{d} - \xi^{c}\nabla_{c}\nabla^{b}\xi^{d} - \xi^{d}\nabla_{c}\nabla^{c}\xi^{b} + \nabla_{c}\xi^{b}\nabla^{c}\xi^{d} - \nabla^{c}\xi^{d}\nabla_{c}\xi^{b})\\ = 2(R^{d}{}{}_{e}\xi^{e}\xi^{b} - R^{b}{}{}_{e}\xi^{e}\xi^{d} - \xi^{c}\nabla_{c}\nabla^{b}\xi^{d}) = -2R^{db}{}{}_{ce}\xi^{c}\xi^{e} = 0$

where I used the fact that $R_{ab} = 0$ in vacuum space-time. Hence $\nabla_{[a}\omega_{b]} = 0$.

9. Jun 25, 2013

### PhizKid

Is it also true that $\mathcal{L}_{\psi}\omega_{a} = 0$ where $\psi^{a}$ is another killing vector field which is lie transported by $\xi^{a}$? I ask because geometrically it seems like it should: because one killing vector field is lie transported along the other and the metric tensor is lie transported along any killing vector field (hence intuitively the volume should be preserved along the flow as well), the twist should then be lie transported since it is constructed only out of those quantities.

10. Jun 25, 2013

### WannabeNewton

Yes that is true and your intuition is exactly correct; proving that $\mathcal{L}_{\psi}\omega_{a} = 0$ is also problem 1 of chapter 7 in Wald's general relativity textbook. What has to be shown is that $\mathcal{L}_{\psi}(\epsilon_{abcd}\xi^{b}\nabla^{c}\xi^{d}) = \xi^{b}\nabla^{c}\xi^{d}\mathcal{L}_{\psi}\epsilon_{abcd} + \xi^{c}\epsilon_{abcd}\mathcal{L}_{\psi}(\nabla^{c}\xi^{d}) = 0$ since $\mathcal{L}_{\psi}\xi^{a} = 0$.

Proceeding, we have that $\mathcal{L}_{\psi}(\nabla^{a}\xi^{b}) = \psi^{c}\nabla_{c}\nabla^{a}\xi^{b} - \nabla^{c}\xi^{b}\nabla_{c}\psi^{a} - \nabla^{a}\xi^{c}\nabla_{c}\psi^{b}\\ = -R^{abcd}\psi_{c}\xi_{d} +\nabla^{b}\xi^{c}\nabla_{c}\psi^{a} - \nabla^{c}\xi^{a}\nabla^{b}\psi_{c}$.

Now note that $\nabla^{b}\xi^{c}\nabla_{c}\psi^{a} - \nabla^{c}\xi^{a}\nabla^{b}\psi_{c}\\ = \nabla^{b}(\xi^{c}\nabla_{c}\psi^{a}) - \xi_{c}\nabla^{b}\nabla^{c}\psi^{a} - \nabla^{b}(\psi^{c}\nabla_{c}\xi^{a}) + \psi_{c}\nabla^{b}\nabla^{c}\xi^{a} \\= \nabla^{b}\mathcal{L}_{\xi}\psi^{a} + R^{cabd}\xi_{c}\psi_{d} - R^{dabc}\xi_{c}\psi_{d}$.

Since $\mathcal{L}_{\xi}\psi^{a}= 0$ everywhere in space-time, $\nabla^{b}\mathcal{L}_{\xi}\psi^{a}= 0$ everywhere in space-time as well.
Thus, $\mathcal{L}_{\psi}(\nabla^{a}\xi^{b}) = \xi_{c}\psi_{d}(R^{dcba}- R^{dbca}+R^{dacb} ) = 0$ where in that last line the first Bianchi identity was used.

I personally know of two potential ways to show that $\mathcal{L}_{\psi}\epsilon_{abcd} = 0$ although they both come down to the same final equation. The formulaic way would be to make use of the fact that the volume form $\epsilon_{abcd}$ is totally antisymmetric. Evaluating the lie derivative and using the fact that $\nabla_{e}\epsilon_{abcd} = 0$ (as well as renaming the indices to something more convenient), we have $\mathcal{L}_{\psi}\epsilon_{a_1a_2a_3a_4} = \epsilon_{ca_2a_3a_4}\nabla_{a_1}\psi^{c} + \epsilon_{a_1ca_3a_4}\nabla_{a_2}\psi^{c} + \epsilon_{a_1a_2ca_4}\nabla_{a_3}\psi^{c} + \epsilon_{a_1a_2a_3c}\nabla_{a_4}\psi^{c}$. Now note that for each term in the RHS, the only index that survives after the sum over $c$ is the one that is not equal to the other three free indices because the volume form is totally antisymmetric i.e. it vanishes when two or more indices are equal. Therefore, $\mathcal{L}_{\psi}\epsilon_{a_1a_2a_3a_4} = \epsilon_{a_1a_2a_3a_4}\nabla_{c}\psi^{c} = 0$ because $\nabla_{c}\psi^{c} = 0$ for a killing vector field. Thus $\mathcal{L}_{\psi}\omega_{a} = 0$.

As a side note, another way to show that $\mathcal{L}_{X}\epsilon_{abcd} = \epsilon_{abcd}\nabla_{c}X^{c}$ for any vector field $X^{a}$ is to take an infinitesimal 4-parallelepiped in space-time spanned by $X^{a}$ and three infinitesimal connecting vectors $W^{a},Y^{a},Z^{a}$ (infinitesimal connecting vectors are always lie transported along a congruence) and consider the change in the volume $\epsilon_{abcd}X^{a}W^{a}Y^{a}Z^{a}$ of this infinitesimal 4-parallelepiped along the flow of $X^{a}$. Essentially whats going is that the change in the volume of the 4-parallelepiped along the flow of $X^{a}$ is given by the divergence of $X^{a}$ multiplied by the volume form. Since killing vector fields has vanishing divergence, they preserve the volume as you would expect since killing vector fields lie transport the metric tensor. Many smooth manifolds texts will in fact define the divergence of a vector field in this way.