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I'm currently trying to solve problem 1, Chapter 11 of Wald, General Relativity. The request is to derive from the condition $$ \tilde\nabla_a \tilde\nabla_b \Omega=0\text{ at }\mathscr I^+, $$ where $\Omega$ is the conformal factor defining the compactification of spacetime, the fact that in the chart $(u, \Omega, \theta, \phi)$, the behavior of $\tilde g_{uu}$, $\tilde g_{u\theta}$ and $\tilde g_{u\phi}$ is $O(\Omega^2)$ near $\mathscr I^+$. We also know that at $\mathscr I^+$ the metric is $$ ds^2=d\Omega du + d\theta^2 + \sin^2\theta d\phi^2. $$

My try: since $\Omega$ is a coordinate we have $\partial_a \Omega=\delta^\Omega_a$; thus, at points of $\mathscr I^+$, $$ 0=\tilde\nabla_a \tilde\nabla_b \Omega=\partial_a \delta^\Omega_u-\tilde\Gamma^c_{ab}\delta^\Omega_c=0-\tilde\Gamma^\Omega_{ab}\implies \tilde\Gamma^\Omega_{ab}=0. $$ Now, since by compatibility $0=\tilde\nabla_a\tilde g_{bc}=\partial_a\tilde g_{bc}-\tilde \Gamma^d_{ab}\tilde g_{dc}- \tilde \Gamma^d_{ac}\tilde g_{bd}$, we have (at $\mathscr I^+$). $$ \partial_\Omega \tilde g_{uu}=2\tilde\Gamma^d_{\Omega u}\tilde g_{du}= 2\tilde \Gamma^\Omega_{u\Omega}=0 $$ This is fine, since $\partial_\Omega\tilde g_{uu}=0$ at $\mathscr I^+$ implies by Taylor expansion in $\Omega$ $$ \tilde g_{uu}(\Omega)=\tilde g_{uu}(0)+0 + O(\Omega^2)=O(\Omega^2) $$ as requested. However, a similar trick does not work for $\tilde g_{u\theta}$ and $\tilde g_{u\phi}$, since $$ \partial_\Omega \tilde g_{u\theta}=\tilde\Gamma^d_{\Omega u}\tilde g_{d\theta} + \tilde\Gamma^d_{\Omega \theta}\tilde g_{du}= \tilde\Gamma^\theta_{\Omega u}\tilde g_{\theta\theta}+\tilde \Gamma^\Omega_{\Omega\theta}= \tilde\Gamma^\theta_{\Omega u} + 0; $$ which at first sight does not vanish, at $\mathscr I^+$. I also tried using the mixed version $\tilde\nabla_a \tilde\nabla^b \Omega=0$, but this avails nothing as we get $$ \partial_a\tilde g^{b\Omega}=-\tilde \Gamma^b_{au} $$ and one cannot extract further information from this as far as I get.

Please, lend a hand!

EDIT: Perhaps I have found a way out. The metric is $u$-independent at $\mathscr I^+$, since it has the form given in the second formula above for any $u$, provided $\Omega=0$. Hence $\partial_u\tilde g_{ab}=0$ and the compatibility condition yields $$ 0=\tilde\nabla_u \tilde g_{ab}=0-\tilde \Gamma^c_{ua}\tilde g_{cb} - \tilde \Gamma^c_{ub}\tilde g_{ca} \implies \tilde\Gamma^c_{ua}\tilde g_{cb}=-\tilde\Gamma^c_{ub}\tilde g_{ca}, \text{ at }\mathscr I^+. $$ Now, substituting in the above computation, $$ \partial_\Omega\tilde g_{u\theta}=\tilde\Gamma^c_{u\Omega}\tilde g_{c\theta}=-\tilde\Gamma^c_{u\theta}\tilde g_{c\Omega}. $$ Now, also $\partial_A \tilde g_{u\Omega}=0$ at $\mathscr I^+$, for any angle $A=\theta,\phi$ yielding $$ \tilde\Gamma^c_{Au}\tilde g_{c\Omega}+\tilde\Gamma^c_{A\Omega}\tilde g_{cu}=0 $$ and hence $$ \partial_\Omega\tilde g_{u\theta} = \tilde\Gamma^c_{\Omega\theta}\tilde g_{cu}= \tilde\Gamma^\Omega_{\Omega\theta}=0 $$ and similarly for $\phi$.

I am still a little perplexed though, since this strategy also applies to getting $\partial_\Omega\tilde g_{u\Omega}=0$, which is not written down by Wald.

EDIT 2: Silly me! Even if $\partial_\Omega\tilde g_{u\Omega}=0$ is true, but one cannot infer $\tilde g_{u\Omega}=O(\Omega^2)$ near $\mathscr I^+$ because $\tilde g_{u\Omega}(\Omega)=\tilde g_{u\Omega}(0)+O(\Omega^2)=1+O(\Omega^2)$.

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Perhaps the most handy way is to proceed as follows: up to $\tilde\Gamma^\Omega_{ab}=0$ everything is fine. Now, write down this identity at $\mathscr I^+$ to get, since $\tilde g^{\Omega a}=\delta^c_u$ and $\partial_u\tilde g_{ab}=0$, $$ \frac{1}{2}\left(\partial_a\tilde g_{ub}+\partial_b\tilde g_{ua}\right)=0. $$ Now, fix $a=\Omega$ and pick $b=u,\theta,\phi$; then $\partial_b \tilde g_{u\Omega}=0$ by inspection and we have $$ \partial_\Omega\tilde g_{ub}=0,\text {at }\mathscr I^+. $$ Then, by Taylor expansion in $\Omega$ $$ \tilde g_{ub}(\Omega)=\tilde g_{ub}(0)+O(\Omega^2)=O(\Omega^2) $$ for $b=u,\theta,\phi.$

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    $\begingroup$ For those reading this in the future, it might he helpful for you to explain how exactly this proves the claim, it's not immediately obvious. $\endgroup$
    – Ryan Unger
    Jun 7, 2016 at 13:45

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