# Question from Schutz's

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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?

Thanks.

This post has been migrated from (A51.SE)
asked Mar 10, 2012

## 1 Answer

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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.

This post has been migrated from (A51.SE)
answered Mar 10, 2012 by (40 points)

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