What does it mean for covariant derivative of metric tensor is zero in general relativity?

Question

Let $(M,g)$ be a Riemannian manifold and $\nabla$ be the Levi-Civita connection. I understand that the total covariant derivative $\nabla g$ of the metric tensor is zero. However, In general relativity's book, it is saying that the covariant derivative $\nabla_{\partial_k}g_{ij}$ is zero. It seem to me that $\nabla_{\partial_k}g_{ij}$ is just the function $\partial_{k}g_{ij}$ since $g_{ij}$ is a smooth function defined on the manifold. Does it mean that partial derivatives of metric components are zero or is there something wrong with my understanding? Many thanks to every answer!

Answer 1

The part you're missing is that although you originally define the Levi-Civita connection on vector fields, you naturally extend it to general tensors essentially via the product rule. Namely let $X,Y$ be vector fields, then a general $(0,2)$-tensor satisfies the following product rule $$\partial_Z(g(X,Y))=\nabla_Z (g(X,Y))=(\nabla_Z g)(X,Y)+g(\nabla_Z X,Y)+g(X,\nabla_ZY);$$ in other words, $\nabla_Z g$ is defined as the $(0,2)$-tensor to satisfy the above formula. The requirement of metric-compatibility of the Levi-Civita connection is exactly $\nabla_Z g\equiv 0$ for all vector fields $Z$. If you want to work with local coordinates, metric-compatibility tells us for $X=\partial_i,Y=\partial_j,Z=\partial_k$, $$\partial_k g_{ij}=g(\Gamma^l_{ki}\partial_l,\partial_j)+g(\partial_i,\Gamma^l_{kj}\partial_l)=\Gamma^l_{ki}g_{lj}+g_{il}\Gamma^l_{kj}.$$

Answer 2

As I mentioned in my comments, usually it is a good idea to stay away from indices and coordinates as much as possible. So I will try to do that.

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Start with a pseudo-Riemannian manifold $(\mathcal M,g)$. It was proved by Lev-Civita that there is a unique torsion-free affine connection which we denote $\nabla$ that preserves the metric, i.e $\nabla g=0$.

Why do we like this connection? Well, it makes computations much easier. For one, it is torsion free, which means that we don't need to worry about the order of the lower indices of our connection coefficients. This is already convenient. And, the other key point is that if we let $v,\omega$ be a vector and covector in $\mathrm T\mathcal M,\mathrm T^*\mathcal M$ respectively, if we define the sharp and flat of $\omega,v$ as $$\omega_{\sharp}=g^{-1}(\omega,\cdot) \\ v^\flat=g(v,\cdot)$$ The metric compatibility means that sharping and flatting commutes with covariant differentiation, that is $$\nabla^*(\omega_{\sharp})=(\nabla^*\omega)_{\sharp}\\\nabla(v^\flat)=(\nabla v)^{\flat} $$ I am using the notation $\nabla^*=\nabla_\sharp$ to denote basically the dual of the covariant derivative, and we have extended the notion of the musical isomorphism to tensor products in the obvious way.

More concretely, in terms of coordinates, this basically means that you can do things like take the divergence whichever way you please: $$\operatorname{div}u=(\nabla u)^k{}_k=\nabla_ku^k\\=g_{ik}\nabla^i(g^{jk}u_j) =g_{ik}\big(g^{jk}\nabla ^i u_j+u_j\underbrace{\nabla^ig^{jk}}_{=0}\big) \\=g^{jk}g_{ik}\nabla^iu_j \\ =\nabla^ku_k$$ So $\nabla \cdot u=\nabla^*\cdot u^{\flat}$.