Covariant derivative of the vielbein determinant

Question

The vielbein postulate says that

$$\nabla_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu-\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma=0.$$

$\nabla$ is the coordinate covariant derivative, and $\Gamma^\sigma_{\mu\nu}$ is connection defined as $$\Gamma^\sigma_{\mu\nu}=\tilde{\Gamma}^\sigma_{\mu\nu}+K^\sigma_{\mu\nu},$$

where $\tilde{\Gamma}^\sigma_{\mu\nu}$ is the Christoffel symbol and $K^\sigma_{\mu\nu}$ is the contorsion.

The Lorentz covariant derivative however is defined as $$D_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu,$$ and the vielbein postulate then implies that $$D_\mu e_v^{\,a}=\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma.$$

Now, I want to take the Lorentz covariant deriavtive of the vielbein determinant, namely $$D_\mu\left(\sqrt{|g|}\right),$$ which we can take in the local coordinates to be $$D_\mu\left(\sqrt{-\text{det}(e_\alpha^{\,\,a}e_\beta^{\,\,b}\eta_{ab})}\right)=D_\mu\sqrt{e^2}=D_\mu|e|.$$

I am stuck here though as I'm unsure how one exactly takes this derivative. I would assume it should become a partial derivative as this quantity is just a scalar. Also, if one goes back a few steps and considers the relationship between the Christoffel symbols and the metric, we could just write this as

$$D_\mu\left(\sqrt{|g|}\right)=\partial_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\,\tilde{\Gamma}^\sigma_{\sigma\mu}.$$

Is this correct, or is there a better way to express this derivative?

EDIT:

Just in response to the answer given by @Immanuel (to avoid a long comment section), I will try to make clear my confusion with their answer.

The Lorentz covariant derivative is $$D_\mu V^a=\partial_\mu V^a+\omega_{\mu\,\,b}^{\,\,a} V^b,$$ where the spin connection $\omega$ looks like $$\omega_{\mu\,\,b}^{\,\,a}=\tilde{\omega}_{\mu\,\,b}^{\,\,a}+K_{\mu\,\,b}^{\,\,a}.$$

So I suppose my first point of confusion is why you have a $\Gamma$ instead of an $\omega$ in your definition of the Lorentz covariant derivative of a density.

If my definition is correct, shouldn't the Lorentz covariant derivative of a density be

$$D_\mu\left(\sqrt{|g|}\right)=\partial_\mu\left(\sqrt{|g|}\right)+\left(\tilde{\omega}_{\mu\,\,a}^{\,\,a}+K_{\mu\,\,a}^{\,\,a}\right)\sqrt{|g|},$$ which becomes $$D_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\tilde{\Gamma}_{\mu\lambda}^{\lambda}+\left(\tilde{\omega}_{\mu\,\,a}^{\,\,a}+K_{\mu\,\,a}^{\,\,a}\right)\sqrt{|g|}.$$

If $\omega_{\mu\,\,a}^{\,\,a}=0$, then we would have $$D_\mu\left(\sqrt{|g|}\right)=\sqrt{|g|}\tilde{\Gamma}_{\mu\lambda}^\lambda,$$

so I hope that clears up some of my confusion.

Answer 1

$\sqrt{|g|}$ is not a scalar. It is a density. Therefore, the covaraint derivative acts differently on it, i.e., you will get an extra term. But it turns out that w.r.t. Lorentz covariant derivative one has only one term.

Let $\rho$ be a density. Then, we know that $$ D_\mu \rho = (\partial_\mu - \omega^a_{\mu a}) \rho. $$ For the similar case of coordinate covariant derivative it is easy to show this ssing the definition of parallel transport as definition of covariant derivative.

However, we know that metricity implies $\omega^a_{\mu a} = 0$. Hence, we have $ D_\mu \rho = \partial_\mu \rho$.

Therefore, in this frame ("Lorentz frame") we find \begin{align} D_\mu \left( \sqrt{|g|} \right) = \partial_\mu \sqrt{|g|} = \sqrt{|g|} \tilde{\Gamma}^\nu_{\mu \nu}. \end{align} However, this is not true in natural coordinates. In that case we have $\nabla_\mu \sqrt{|g|} = 0$.