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$$\nabla_{j} v^{i}~=~g^{ik}\nabla_{j}v_{k}.$$

  1. Does this equality involve an intermediate step, where I take the metric inside the derivative, and then use the fact that covariant derivative of the metric is zero?

  2. Or is this true just because the covariant derivative transforms as a tensor (independent of the fact that the covariant derivative of the metric is zero)?

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Your first idea is correct: $\nabla_jv^i=\nabla_jg^{ik}v_k=g^{ik}\nabla_jv_k$ –  Danu Mar 23 '14 at 12:46

1 Answer 1

First recall that we define a co-vector field $\eta$ from a vector field $v$ via the flat map $ \eta~=~v^{\flat}.$ In components we have $\eta_k = g_{ki} v^i.$ Equivalently, we can reconstruct the vector field $v$ via the sharp map $v ~=~\eta^{\sharp}.$ In components we have $v^i = (g^{-1})^{ik} \eta_k.$

Now let us return to OP's question. As Danu writes in a comment above it is OP first idea that is correct in this context. Strictly speaking, the covariant derivative $\nabla_j$ acts on tensors rather than tensor components. Let us therefore include some parentheses in the formula for clarity: $$(\nabla_{j} \eta)_k~=~(\nabla_{j}g)_{ki} v^i+ g_{ki}(\nabla_{j}v)^i~=~g_{ki}(\nabla_{j}v)^i.$$

The last equality holds if the connection $\nabla$ and the metric $g$ are compatible, i.e. $(\nabla_{j}g)_{ki}=0.$

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