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In q. 22 in page 141, I am asked to show that if

$$U^{\alpha}\nabla_{\alpha} V^{\beta} = W^{\beta},$$

then

$$U^{\alpha}\nabla_{\alpha}V_{\beta}=W_{\beta}.$$

Here's what I have done: $$V_{\beta}=g_{\beta \gamma} V^{\gamma},$$ so $$U^{\alpha} \nabla_{\alpha} (g_{\beta \gamma} V^{\gamma})=U^{\alpha}(\nabla_{\alpha} g_{\beta \gamma}) V^{\gamma} + g_{\beta \gamma} (U^{\alpha} \nabla_{\alpha} V^{\gamma}).$$

Now, I understand that the second term is $W_{\beta}$, but how come the first term vanishes?

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2 Answers 2

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The covariant derivative is metric compatible, so $\nabla_{\alpha} g_{\beta \gamma} = 0$. This is the condition that the inner product is preserved under parallel transport.

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I don't think you need metric compatibility to prove this although you can use it. There is a much simpler way with repeated use of (any) metric to lower the index.

$ U^\alpha\nabla_\alpha V^\beta=W^\beta $

$ \Rightarrow U^\alpha g^{\beta\gamma}\nabla_\alpha V_\gamma=g^{\beta\gamma}V_\gamma $

$ \Rightarrow U^\alpha g_{\mu\beta}g^{\beta\gamma}\nabla_\alpha V_\gamma=g_{\mu\beta}g^{\beta\gamma}V_\gamma $

$ \Rightarrow U^\alpha\delta_\mu^\gamma\nabla_\alpha V_\gamma=\delta_\mu^\gamma V_\gamma $

$ \Rightarrow U^\alpha\nabla_\alpha V_\mu=V_\mu $

there are three ways to do the first step, only one uses metric compatibility. More at https://www.general-relativity.net/2019/10/symmetries-and-killing-vectors.html

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