# Interpretation of covariant derivative of metric tensor being zero, specific problem on sphere

A question about the covariant derivative of the metric tensor being zero, example: sphere.

I understand, that the metric tensor of a (unit-)sphere is calculated via the outer product of the base vectors. Doing that, we get the tensor $$g_{rr}=1$$, $$g_{\vartheta\vartheta}=\sin^2\vartheta$$ (can show it if necessary, if I ever remember how to get that LaTeX tool invoked here)

I also understand, that the covariant derivative of the metric tensor equals zero (can prove it mathematically), and I read, that the intuitive interpretation of that is, that the metric does not change when we move on the sphere.

But the metric above definitely changes, since it contains $$\sin^2\vartheta$$ -- except if we move on a geodesic curve. But I thought, that a non-changing metric means it does not change in whatever direction we move.
Am I wrong there?

• To use mathjax, wrap the relevant text with dollar signs (i.e., $x^2$ yields $x^2$). Two consecutive dollar signs yields a centered equation. See Help::notation for more details Commented Jun 14, 2023 at 17:43
• In short: the components of the metric may change depending on the coordinate system you are using, but this is not physical. The fact that $\nabla_\mu g_{\rho\sigma}=0$ is the tensorial formulation of the metric being constant which then holds for all coordinate systems. When you look at the metric in $\mathbb R^2$ using polar coordinates its components are changing as well but we know that this just an effect of the coordinates being used. The metric is flat. Commented Jun 14, 2023 at 18:04
• Thanks, Kurt, I added a follow-up question, but it ended up at the bottom (please, scroll all the way down) Commented Jun 14, 2023 at 19:03

Before answering your question, I want to clear out some misconceptions. The value of $$\nabla_{\mu}g_{\rho \sigma}$$ depends on your choice of connection. The reason why in curved spaces $$\nabla_{\mu} \neq \partial_{\mu}$$ is because the basis which we use to expand the vectors change from point to point. Note that this is not what we observe in flat Euclidean space. The basis vectors are the same everywhere. Mathematically speaking, the basis vectors (and therefore the vectors themselves) live in different tangent spaces. This is a fancy way of saying that not only the value of the component changes, your basis vectors are changing as you move around in your curved space.
This is where the connection comes to play. The connection tells you how the basis vectors change from point to point. Then, you use this to define the full covariant derivative which tells the change in the value of the vector $$\textbf{and}$$ the basis components. This is why it is misleading to say that the metric in question changes as we vary $$\theta$$ and therefore it is not constant. You are forgetting to account for the change in the basis! If you also account for the change in basis (via christoffel symbols), you can see that the metric won't change.
Now I also want to address the claim I made at the beginning where I said that the covariant derivative of the metric depends on the connection you choose. This will contain some mathematical jargon so you can skip to the end where I give the physical usefulness of the connection used in General Relativity. You can read more about it from Harvey Reall's lecture note on General Relativity. Generally, covariant derivatives are defined before we even talk about connection. These objects are defined such that they obey some fundamental properties (such as linearity, Leibniz Rule etc.). This definition does not single out a preferred connection (i.e an object that tells you how the basis vectors change). However, it turns out (this can be proved easily) that there exists a unique symmetric connection such that $$\nabla g = 0$$. This is called the Levi-civita connection and the main connection we use in General Relativity. The fact that it is covariantly conserved implies that the metric doesn't change depending on where we are in the manifold and therefore assign distances quite nicely everywhere in the space. Furthermore, Levi-Civita connection arises naturally when we consider Lagrangians which is a nice sign as Lagrangians have been a very good framework to write our theories in.
• Follow-follow up: or is it not possible to find such a system of base vectors which create a metric that does not depend on $\vartheta$ because I am moving from one tangential plane to the next while moving? In what coordinates on a sphere can I see the metric being constant? Or am I totally off? Commented Jun 14, 2023 at 19:14