# 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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This is only meant to supplement the first answer.

If we think physically, then we live in one particular (pseudo-)Riemannian world. In this world, there is only one metric tensor (up to scalar) and it can pretty much be measured. If I found it here, and if an alien measured it, and we compared our answers, they would be scalar multiples of each other (choice of Parisian metre stick for me, choice of Imperial foot for the alien, or, vice versa..). There is precisely one connection, and it can be calculated from the metric.

So I quarrel with the word used by @twistor59, «chosen». There is no choice. Given a metric, the connection is determined. I agree with the rest of the answer, but would like to see the word «chosen» replaced by «given». I would rather say,

given a metric, the connection is determined by the metric.

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Given a metric, the Levi-Civita connection is determined by the metric. We often find it convenient to choose the Levi-Civita connection over other possible choices. –  Timothy Wofford Apr 10 '14 at 8:10

The calculation is quite easy and you can find it here:

http://www.physicspages.com/2014/01/02/covariant-derivative-of-the-metric-tensor/

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## protected by Qmechanic♦Jun 10 at 10:07

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