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I'm calculating the Christoffel symbols of the second kind which is of course defined as multiplying the symbol of the first kind multiplied by the contravariant metric. I was thinking of how to make this a less "involved" calculation by simplifying in whatever way I can. The first thing that I wanted to try is distributing the contravariant metric, but this of course makes me wonder whether or not $$ g^{ks} \partial_{j} g_{is} = \partial_{j}(g^{ks}g_{is}) $$ is a true statement, where of course the rhs will reduce to zero since the Kronecker delta is independent of coordinates. The fact that this happens leads me to believe that this expression is not true, and an example calculation replacing the covariant metric with a vector led to the expression being false as well. My concern stems from wondering or not this will always be the case or not. My question is then are the operations of partial differentiation and index raising/lowering interchangeable? ( I'm leaning towards no. I've also searched here and in related places for an answer and haven't found what I'm looking for. )

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Certainly it is false in general. (Assuming that the Einstein summation convention is being used,) we have $$\partial_j (g^{ks} g_{is}) = \partial_j \delta^k {}_i = 0,$$ but $g^{ks} \partial_j g_{is}$ is not always zero. If it were, we wouldn't bother writing those terms in the usual coordinate formula for the Christoffel symbol.

In particular, the operations of differentiating with respect to a coordinate vector field in a chart and raising/lowering indices do not in general commute.

That said, suppose $\nabla$ is compatible with the metric, that is that, $\nabla_a g_{bc} = 0$; the Levi-Civita connection of the metric is one such choice. In this case, the product rule for covariant differentiation gives that differentiating with $\nabla$ does commute with raising/lowering indices: For example, given a tensor $T^b{}_{c_1 \cdots c_k}$, we have \begin{align} \nabla_a T^b {}_{c_1 \cdots c_k} &= \nabla_a (g^{bd} T_{d c_1 \cdots c_k}) \\ &= \underbrace{(\nabla_a g^{bd})}_0 T_{d c_1 \cdots c_k} + g^{bd} \nabla_a T_{d c_1 \cdots c_k} \\ &= g^{bd} \nabla_a T_{d c_1 \cdots c_k}. \end{align}

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  • $\begingroup$ Thank you for clearing that up! I was thinking this was the case for the latter reason ( why write out what vanishes in a definition ). I think I was mistaking the commutation between the metric and the covariant derivative as you've stated. $\endgroup$ – Doryan Miller Dec 19 '14 at 21:00
  • $\begingroup$ You're welcome, I'm glad you found it useful. Notation is partly to blame for this confusion, I think; the notation $T_{a_1 \cdots a_k, b}$ is used both for $\nabla_b T_{a_1 \cdots a_k}$ and $\partial_b T_{a_1 \cdots a_k}$. Sometimes one distinguishes these by using a comma for one type of derivative and a semicolon for the other, but this, needless to say, can still lead to confusion. I prefer to avoid this altogether by using commas for covariant derviatives and writing out coordinate partial derivatives (which aren't invariant) with $\partial_b$. $\endgroup$ – Travis Willse Dec 20 '14 at 3:36
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Allow me to abbreviate $\partial_k(g^{is}g_{sj})=(g^{is}g_{sj})_{,k}$. Then the identity that arise in the mentioned context is $${g^{is}}_{,\ k}\ g_{sj}=-g^{is}(g_{sj,k}).$$ Proof: Since $g^{is}g_{sj}={\delta^i}_j$ then $(g^{is}g_{sj})_{,k}=0$. But by Leibniz' rule $$(g^{is}g_{sj})_{,k}={g^{is}}_{,\ k}g_{sj}+g^{is}(g_{sj,k})=0.$$ Hence the claim.

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