Is $\partial_k g_{ij}=\partial_j g_{ik} $?

Question

I Found the following identity: $\partial_k g_{ij}=\partial_j g_{ik} $ ? or at least that is what it looks that it is being used. g is a riemannian metric by the way.

Why is it true? .I know the metric is symmetric , but this doesn't make sense to me.

Answer 1

What you wrote is not equal. What they wrote is simply because the summation is over both $k$ and $j$, so they have split up the sum and relabelled indices.

More explicitly, if $b^{kj}$ is symmetric, then \begin{align} 2a_{jk}b^{jk}&=a_{jk}b^{jk}+a_{jk}b^{jk}\tag{obvious}\\ &=a_{jk}b^{jk}+a_{kj}b^{kj}\tag{relabel dummy indices}\\ &=a_{jk}b^{jk}+a_{kj}b^{jk}\tag{$b$ symmetric}\\ &=(a_{jk}+a_{kj})b^{jk}. \end{align} In your case, apply this with $b^{kj}=\dot{\gamma}^k\dot{\gamma}^j$ and $a_{jk}=g_{ij,k}$ ($i$ is fixed).

Answer 2

No, $g_{ij,k}$ and $g_{ik,j}$ are in general not equal.

What is used here is that if $s^{ij}$ is a symmetric tensor, and $t_{\alpha ij}$ any tensor, then $$ t_{\alpha ij} s^{ij} = t_{\alpha ji} s^{ji} = t_{\alpha ji} s^{ij}. $$