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I was trying to solve the problem 8.5 from the textbook An Introduction to General Relativity by Hughston & Tod. Given a Riemannian manifold and a vector field $V^a$ such that $\mathcal{L}_Vg_{ij}=2\phi g_{ij}$ for some scalar $\phi$, the reader is asked to prove that $$ \mathcal{L}_V \Gamma^c_{ab}=\phi_a \delta^c_b + \phi_b\delta^c_a - g_{ab}\phi^c, $$ where ${\Gamma^c_a}_b$ is the connection on the manifold, and $\phi_a \equiv \nabla_a\phi$ (and $\mathcal{L}_V$ is the Lie derivative).

However, I was only able to show that $$ \mathcal{L}_V \Gamma^c_{ab}=\phi_a \delta^c_b + \phi_b\delta^c_a - g_{ab}\phi^c - {V^c}_{,a}{_{,b}} \,\,\,\,, $$ and I see no reason why the last term here should vanish (commas denote partial derivatives). I double-checked the proof twice, beginning from scratch.

Could anyone please check this and confirm that there is a mistake in the book (or that I miss something here ;) ).


To give a hint for intermediate steps, I used the metric-connection relation and applied the Leibniz rule, $$ \mathcal{L}_V \Gamma^c_{ab}=\mathcal{L}_V\{ \tfrac{1}{2}g^{cd}(g_{bd,a}+g_{ad,b}-g_{ab,d}) \}= \\ =\tfrac{1}{2}\{\mathcal{L}_V g^{cd}\}(g_{bd,a}+g_{ad,b}-g_{ab,d}) + \tfrac{1}{2}g^{cd}\mathcal{L}_V (g_{bd,a}+g_{ad,b}-g_{ab,d}). $$ Then I deduced and employed the following relations: $$ \mathcal{L}_Vg^{ij}=-2\phi g^{ij} \,\,, $$ $$ \mathcal{L}_V g_{ab,c}=(\mathcal{L}_V g_{ab})_{,c}-g_{ib}{V^i}_{,a,c}-g_{ai}{V^i}_{,b,c}\,\,, $$ $$ (\mathcal{L}_V g_{ab})_{,c}=2\phi g_{ab,c}+2g_{ab}\phi_c $$ (the latter due to the problem statement). (UPDATE: the second formula is incorrect, see below).


UPDATE

So, I found the mistake in my calculations. When writing out $\mathcal{L}_V g_{ab,c}$, I naively used the standard formula for the Lie derivative of covariant tensors. But $g_{ab,c}$ is not a tensorial object, so I had to first deduce the formula of the Lie derivative for this object from scratch. The result is: $$ \mathcal{L}_V g_{ab,c} = V^i g_{ab,ci} + g_{ib,c}{V^i}_{,a} + g_{ai,c}{V^i}_{,b} + g_{ab,i}{V^i}_{,c}+ g_{ib}{V^i}_{,a,c}+g_{ai}{V^i}_{,b,c}. $$ As might be seen, this differs from the formula for the Lie derivative of a covariant tensor of valence 3: there are two extra terms here (the last ones). This formula happens to be equal to the partial derivative of the Lie derivative of the metric, so it turns out that partial derivative commutes with Lie derivative (at least for the metric tensor, but I think in deducing this result I never referred to $g_{ab}$ being actually a metric tensor, so the result should hold for any tensor $T_{ab}$): $$ \mathcal{L}_V g_{ab,c} = (\mathcal{L}_V g_{ab})_{,c}. $$

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  • $\begingroup$ The expression you got cannot be correct. The first 3 terms for $L_V\Gamma$ are all tensorial, but the last term is not (second partials of vector field components), so the entire expression is not tensorial. However the Lie derivative of a linear connection is a type (1,2) tensor field, so the result MUST be tensorial. $\endgroup$ – Bence Racskó Sep 2 '17 at 10:06
  • $\begingroup$ Why is the Lie derivative of a connection a tensor, if the connection itself is not? I understand that difference between two connections is a tensor, but when we write out the Lie derivative for $\Gamma$, we do not get just difference between connections. $\endgroup$ – Maximko Sep 2 '17 at 11:32
  • $\begingroup$ Because $L_X\Gamma=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left((\phi_\epsilon)^*\Gamma-\Gamma\right)$, where $\phi_\epsilon$ is the one-parameter group of diffeomorpfisms generated by $X$ and $(\phi_\epsilon)^*\Gamma$ is the "pullback" of the connection, which is definable as $(\phi^*\nabla)_XY=\phi^{-1}_*(\nabla_{\phi_*X}(\phi_*Y))$. The pullback of a connection is a connection itself so $\phi_\epsilon^*\Gamma-\Gamma$ is the difference between two connections. $\endgroup$ – Bence Racskó Sep 2 '17 at 15:07
  • $\begingroup$ Ok, thanks, now I see. And I see where I have mistaken. Writing out the Lie derivative $\mathcal{L}_V g_{ab,c}$, I naively applied the formula which is valid for tensorial objects, but $g_{ab,c}$ is not a tensor. I had to derive the formula for the Lie derivative of this object from scratch, and it picks extra terms with second partial derivatives, which will probably cancel out the term $-{V^c}_{,a,b}$ in my result (I still need to check that). $\endgroup$ – Maximko Sep 2 '17 at 18:29
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The equation of Hughston & Tod is correct. One way to check this is to use the conformal transformation. In fact, $V^a$ is a conformal Killing vector. This conformal Killing vector induces a diffeomorphism $\psi_\epsilon$ such that

$$g'_{ab}=\psi_\epsilon^*g_{ab}=(1+2\phi\epsilon)g_{ab}$$ Here, $0<\epsilon\ll1$. From this, one can read off the conformal factor $\Omega=\sqrt{1+2\phi\epsilon}\approx1+\phi\epsilon$. Under conformal transformation, the connection varies in the following way,

$$\delta\Gamma^c{}_{ab}=2\delta^c_{(a}\nabla_{b)}\ln\Omega-g_{ab}\nabla^c\ln\Omega$$

Therefore, the Lie derivative of the connection is

$$\mathscr L_V\Gamma^c{}_{ab}=\lim_{\epsilon\rightarrow0}\frac{\delta\Gamma^c{}_{ab}}{\epsilon}$$

You can easily check that Hughston & Tod give the correct formula.

For the conformal transformation, please check Appendix D in Wald's General Relativity.

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