Deriving the Covariant Derivative of the Metric Tensor

Question

First off, I did look through some other questions:

Covariant Derivative of Metric Tensor

Why is the covariant derivative of the metric tensor zero?

https://math.stackexchange.com/q/2174588/

But they either give the covariant derivative as:

$\nabla_{\rho} g_{\mu\nu}=\frac{\partial g_{\mu\nu}}{\partial x^{\rho}} - g_{\mu\sigma}\Gamma^{\sigma}_{\nu\rho}-g_{\tau\nu}\Gamma^{\tau}_{\mu\rho}$

By basis of the connection being chosen so that the covariant derivative of the metric is zero.

However, is this relation not derived from the tensor product rule of the covariant derivative, such that one can find:

$\nabla_{\rho}( g_{\mu\nu} \vec{e}^{\mu} \otimes \vec{e}^{\nu})=(\frac{\partial g_{\mu\nu}}{\partial x^{\rho}} - g_{\mu\sigma}\Gamma^{\sigma}_{\nu\rho}-g_{\tau\nu}\Gamma^{\tau}_{\mu\rho})\vec{e}^{\mu} \otimes \vec{e}^{\nu}$?

It seems to relate to the third resource I've linked, since I don't think it makes sense to say the covariant derivative of a tensor, that is, the tensor components and the tensor product of basis vectors/covectors, equals the covariant derivative of the COMPONENTS times the tensor product of basis stuff, since the covariant derivative is supposed to describe curved space by affecting the basis stuff, which it does, so we can't factor it out as that viewpoint would imply.

In the third linked source, it said to view the first formula as a component of the covariant derivative, not the covariant derivative of the component, which I can get behind.

So, in summa, why is it said:

$\nabla_{\rho} g_{\mu\nu}=\frac{\partial g_{\mu\nu}}{\partial x^{\rho}} - g_{\mu\sigma}\Gamma^{\sigma}_{\nu\rho}-g_{\tau\nu}\Gamma^{\tau}_{\mu\rho}$

And not:

$\nabla_{\rho}( g_{\mu\nu} \vec{e}^{\mu} \otimes \vec{e}^{\nu})=(\frac{\partial g_{\mu\nu}}{\partial x^{\rho}} - g_{\mu\sigma}\Gamma^{\sigma}_{\nu\rho}-g_{\tau\nu}\Gamma^{\tau}_{\mu\rho})\vec{e}^{\mu} \otimes \vec{e}^{\nu}$?

And if it is said as the latter, then we say:

$\nabla_{\rho}( g_{\mu\nu} \vec{e}^{\mu} \otimes \vec{e}^{\nu})=0$

And not:

$\nabla_{\rho}g_{\mu\nu}=0$

Since it might be that instead:

$\nabla_{\rho}g_{\mu\nu}=\frac{\partial g_{\mu\nu}}{\partial x^{\rho}}$?

I suppose part of my motivation for this question is in the Einstein-Hilbert Action, where we use the virtue of covariant derivative of the metric being zero to factor it into a covariant derivative so that we can prove one equation is zero, so that the EFEs pop out.

Addendum Auctoris: Okay, I made a fe- a lot of index oopsies, but that should be taken care of now. Thanks for pointing that out.

Addendum II: It seems that I'm bad at LaTeX, or I forget things easily.

Answer 1

It's just notation. We take $\nabla_\rho g_{\mu\nu}$ to mean $(\nabla g)_{\rho\mu\nu}$, that is, the component of the covariant derivative of the metric tensor, because it's convenient. After all, we already have the notation $\partial_\rho g_{\mu\nu}$ for the derivative of the components.

Answer 2

For this derivation, we first need to calculate the partial derivative of the covarinat metric tensor (which can be expressed, as the dot product of two covariant basis vectors). \begin{align} \partial_{\omega}{g_{{\mu}{\nu}}} =\partial_{\omega}\langle{\boldsymbol{\varphi}_{\mu},\boldsymbol{\varphi}_{\nu}}\rangle= \langle{\partial_{\omega}\boldsymbol{\varphi}_{\mu},\boldsymbol{\varphi}_{\nu}}\rangle+ \langle{\boldsymbol{\varphi}_{\mu},\partial_{\omega}\boldsymbol{\varphi}_{\nu}}\rangle \end{align} By the definition of the covariant derivative, acting on a vector field: \begin{align} \nabla_{\omega}\mathbf{F}=\boldsymbol{\varphi}_{\mu}\nabla_{\omega}F^{\mu} \end{align} According to the product rule and the fact that any vector can be expressed as $\mathbf{F}=\boldsymbol{\varphi}_{\mu}F^{\mu}$ this implies that, the covariant derivative of the basis vectors $\nabla_{\omega}\boldsymbol{\varphi}_{\mu}$ is zero. Now, we can compute the partial derivative of the covariant basis vector. \begin{align} 0=\nabla_{\omega}\boldsymbol{\varphi}_{\mu} =\partial_{\omega}\boldsymbol{\varphi}_{\mu}- \Gamma^{\alpha}_{{\mu}{\omega}}{\,}\boldsymbol{\varphi}_{\alpha} \Longrightarrow \partial_{\omega}\boldsymbol{\varphi}_{\mu} =\Gamma^{\alpha}_{{\mu}{\omega}}{\,}\boldsymbol{\varphi}_{\alpha} \end{align} We can now write the partial derivative of the metric tensor as follows: \begin{align} \partial_{\omega}{g_{{\mu}{\nu}}} =\Gamma^{\alpha}_{{\mu}{\omega}}\langle{\boldsymbol{\varphi}_{\alpha},\boldsymbol{\varphi}_{\nu}}\rangle +\Gamma^{\alpha}_{{\omega}{\nu}}\langle{\boldsymbol{\varphi}_{\mu},\boldsymbol{\varphi}_{\alpha}}\rangle =g_{{\alpha}{\nu}}\Gamma^{\alpha}_{{\mu}{\omega}}+g_{{\mu}{\alpha}}\Gamma^{\alpha}_{{\omega}{\nu}} \end{align} But the covariant derivative of the metric tensor was: \begin{align} \nabla_{\omega}g_{{\mu}{\nu}}=\partial_{\omega}{g_{{\mu}{\nu}}}-g_{{\alpha}{\nu}}\Gamma^{\alpha}_{{\mu}{\omega}}-g_{{\mu}{\alpha}}\Gamma^{\alpha}_{{\omega}{\nu}}=0 \end{align} And therefore, it has to be zero. Q.E.D.

Answer 3

So, $g$ is a tensor field which in a chart with the appropriate base could be written as $$ g = g_{\mu \nu} \, \mathrm{d}x^{\mu} \otimes \mathrm{d}x^{\nu}.$$

Now, the covariant derivative has a Leibniz rule (it is indeed a derivative), when applied to $g$ acts as $$\nabla_{\rho}g = \nabla_{\rho}g_{\mu \nu} \, \mathrm{d}x^{\mu} \otimes \mathrm{d}x^{\nu} + g_{\mu \nu}\, (\nabla_{\rho}\mathrm{d}x^{\mu}) \otimes \mathrm{d}x^{\nu} + g_{\mu \nu}\, \mathrm{d}x^{\mu} \otimes (\nabla_{\rho} \mathrm{d}x^{\nu}).$$

Since by components $g_{\mu \nu} = g_{\mu \nu}(x)$ is a scalar function and $(\nabla_{\rho}\mathrm{d}x^{\mu}) = - \Gamma^{\mu}_{\,\rho \sigma} \mathrm{d}x^{\sigma},$ then you recover the expression you want. Have in mind that by definition the affine connection takes your tensor and a vector and maps them onto another tensor:

$$\nabla_{\mathrm{_X}} g = \mathrm{X}^{\rho} \, \nabla_{\rho} g,$$ (where you plug in the expression you found). I hope you can see now why your second equation is wrong.

Finally, for different reasons one demands that the metric is covariantly constant $$\nabla_{\mathrm{_X}} g = 0,$$ mainly to obtain a metric compatible connection (and a torsion-free metric), something that is very useful in general relativity.

Note that $\nabla_{\mathrm{_X}} g = 0,$ would lead to $$\nabla_{\rho}\,[ g(Y,Z)] = 0$$ when you parallel transport the vectors the along a geodesic with a metric compatible (Levi-Civita) connection.

Answer 4

One of your points is symbolic and pedagogical, and I like it a lot. We often write $g_{ij} $ or whatever and don't mean the component, we mean the whole tensor. So $\nabla _k g_{ij}$ doesn-t mean the kth component of the covariant derivative of the ijth component of the metric tensor, we use our common sense (which none of us has a lot of) and realise we mean the covariant derivative in the direction of the kth basis vector field of the entire metric tensor $G$, and then take the ijth component of that. (Or not, i.e., once again, pars pro toto, we mean the resulting tensor.)

But time and time again, on this website, intelligent students have misinterpreted a professor's formula to mean what you protest against, as if it meant the covariant derivative of the ijth component, which makes no sense.

So, bully for you. And now to business: you do not sufficiently emphasise that there exists a unique connexion which satisfies two properties: no torsion, and, the covariant derivative of the metric tensor is zero. Usually this all goes without saying: if someone mentions the metric, and then mentions covariant differentiation, they mean covariant differentiation with respect to that unique connexion.

There are certainly connexions which yield a non-zero covariant derivative of the metric tensor, but all the formulas for the covariant derivative you cite are for the unique compatible connexion defined so as to yield zero.