# Why is the covariant derivative of the metric tensor zero?

I've consulted several books for the explanation of why

$$\nabla _{\mu}g_{\alpha \beta} = 0,$$

and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu \beta}$

$$\Gamma ^{\lambda } _{\beta \mu} = \frac{1}{2} g^{\alpha \gamma}(\partial _{\mu}g_{\alpha \beta} + \partial _{\beta} g_{\alpha \mu} - \partial _{\alpha}g_{\beta \mu}).$$

But I'm getting nowhere. May be I've to go through the concepts of manifold much deeper.

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The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not vanish. But we specifically want a connection for which this condition is true because we want a parallel transport operation which preserves angles and lengths.

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Good answer. Some details are given in Wald section 3.1. We want the inner product $(v,w) = g_{ab} v^a w^b$ to remain constant under parallel transport along a curve with tangent $t^c$, which gives rise to the condition $t^c \nabla_c (g_{ab} v^a w^b) = 0.$ But (using parallel transport), this is the same as $t^c v^a w^b \nabla_c g_{ab} = 0$ and this should be true for all $v,w,t.$ –  Vibert Dec 30 '12 at 11:14
Thanks for the wonderful answer. I will try to go through the wald book. –  Aftnix Dec 30 '12 at 11:18
also note that the condition $\nabla g = 0$ is not enough to specify a unique connection - another condition (eg vanishing torsion) is necessary for that –  Christoph Dec 30 '12 at 11:27
@Christoph yes, that's important, I should have mentioned it. –  twistor59 Dec 30 '12 at 11:42

It can be show easily by the next reasoning. $$DA_{i} = g_{ik}DA^{k},$$ because $DA_{i}$ is a vector (according to the definition of covariant derivative). On the other hand, $$DA_{i} = D(g_{ik}A^{k}) = g_{ik}DA^{k} + A^{k}Dg_{ik}.$$ So, $$g_{ik}DA^{k} + A^{k}Dg_{ik} = g_{ik}DA^{k} \Rightarrow Dg_{ik} = 0.$$ So, it isn't a condition, it is a consequence of covariance derivative and metric tensor definition.

The relation between Christoffel's symbols and metric tensor derivations can be earned by cyclic permutation of indexes in the covariance derivative $g_{ik; l}$ expression, which is equal to zero.

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