Denote by $\mathcal{D}_m$ the Lorentz covariant derivative, $$\mathcal{D}_m=\partial_m-\frac{1}{2}\omega_m{}^{ab}M_{ab} \tag{1}$$ where indices $m,n,p,\dots$ are world indices, indices $a,b,c,\dots$ are tangent indices, $\omega_{m}{}^{ab}$ is the Lorentz connection and $M_{ab}$ are the Lorentz generators (in an arbitrary representation). I would like to compute the covariant derivative of the determinant of the vielbein, $e:=\text{det}(e_m{}^{a})$, in both the torsion and torsionless cases. Lets consider $D=3+1$ for concreteness.
In the torsionless case, it is known that the Lorentz covariant derivative $\mathcal{D}_m$ coincides with the regular covariant derivative $\nabla_m$ of general relativity (in tosion free case) involving the Christoffel symbols. Therefore, since we know that $\nabla_m\sqrt{-g}=0$ and $\sqrt{-g}=\sqrt{-\text{det}(g_{mn})}=$ $=\sqrt{-\text{det}(e_m{}^{a}e_n{}^{b}\eta_{ab})}=\sqrt{e^2}=|e|$, we can conclude that $\mathcal{D}_me=0$.
How would I go about computing $\mathcal{D}_me$ without using the knowledge $\nabla_m\sqrt{-g}=0$?
In the torsionless case, the spin connection could be expressed in terms of the vielbein via the torsion free condition, which would eliminate $\omega_{m}{}^{ab}$ in $(1)$ from the calculation in favour of the vielbein. Further more the torsion free condition is equivalent to the statement that the vielbein is covariantly constant, $\hat{\nabla}_me_{n}{}^{a}=0$ where $\hat{\nabla}_a=\nabla_m-\frac{1}{2}\omega_{m}{}^{bc}M_{bc}$, that might also come in handy. But these are just ideas of what might be useful in the calculation, i'm not sure how to put them to use. To start with, i'm not even sure how to interpret the expression $M_{bc}e$. Normally we would have $M_{bc}\phi=0$ for any scalar field $\phi$, but I suspect that this is not the case here (given that I anticipate $\mathcal{D}_me=0$ and $\partial_m e\neq 0$). I know that under an infinitesimal local Lorentz transformation parameterized by $K^{ab}=-K^{ba}$, $e$ is invariant, but I don't think I can apply that here.
Once I see how the torsionless case is done, I'll have a crack at the case with torsion myself.
Edit 1
Might someone confirm whether or not the following is valid: \begin{align} \frac{1}{2}\omega_n{}^{bc}M_{bc}\text{det}(e_{m}{}^{a})&=\frac{1}{2}\omega_n{}^{bc}M_{bc}\big(\frac{1}{4!}\varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}e_{m_1}{}^{a_1}e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\big)\\ &=\frac{4}{4!}\varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}\big(\frac{1}{2}\omega_{nbc}M^{bc}e_{m_1}{}^{a_1}\big)e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\\ &=\frac{1}{3!}\varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}\big(\omega_{nbc}\eta^{a_1b}e_{m_1}{}^{c}\big)e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\\ &=-\frac{1}{3!}\omega_{nc}{}^{a_1}\varepsilon_{a_1a_2a_3a_4}\big(\varepsilon^{m_1m_2m_3m_4}e_{m_1}{}^{c}e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\big)\\ &=-\frac{1}{3!}\omega_{nc}{}^{a_1}\varepsilon_{a_1a_2a_3a_4}\big(\text{det}(e_{m}^{a})\varepsilon^{ca_2a_3a_4}\big)\\ &=-\frac{1}{3!}\cdot 3!e\omega_{nc}{}^{a_1}\delta^{c}_{a_1}\\ &=0 \end{align} where in the last line I have used the antisymmetry $\omega_{n}{}^{bc}=-\omega_{n}{}^{cb}$. This would imply that $\mathcal{D}_me=\partial_me$ in both the torsion and tosionless case, which I am doubtful of (but coincidentally, it is exactly what I need for a separate calculation I am doing).
Edit 2
Similar to the above I can show that $\nabla_m e=0$, \begin{align} \nabla_m e &=\nabla_m\big(\frac{1}{4!}\varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}e_{m_1}{}^{a_1}e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\big)\\ &=\partial_m\big(\frac{1}{4!}\varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}e_{m_1}{}^{a_1}e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\big)~~~~~~~~~~~~~~~~~~~-4\big(\frac{1}{4!} \varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}(\Gamma^p_{mm_1}e_{p}{}^{a_1})e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\big)\\ &=\partial_me-\frac{1}{3!}\varepsilon^{m_1m_2m_3m_4}\Gamma^p_{mm_1} e\varepsilon_{pm_2m_3m_4}\\ &=\partial_me-e\Gamma^p_{mp} \\ &=\partial_me-e(\frac{1}{2}g^{-1}\partial_mg)\\ &=\partial_me-\partial_me\\ &=0 \end{align}
These two results (in edit 1 and 2) are consistent with the fact that the Vielbein is covariantly constant (true for torsionless case) since on the one hand we have (using $\hat{\nabla}_me_{n}{}^{a}=0$) \begin{align} \hat{\nabla}_me=\hat{\nabla}_m\big(\frac{1}{4!}\varepsilon^{m_1m_2m_3m_4}\varepsilon_{a_1a_2a_3a_4}e_{m_1}{}^{a_1}e_{m_2}{}^{a_2}e_{m_3}{}^{a_3}e_{m_4}{}^{a_4}\big)=0 \end{align} and the the other hand we have \begin{align} \hat{\nabla}_me=(\nabla_m-\frac{1}{2}\omega_m{}^{bc}M_{bc})e=0. \end{align} However there is an inconsistency with the claim that $\mathcal{D}_m$ coincides with $\nabla_m$ in the torsionless case (since according to the above we have $\mathcal{D}_me=\partial_me$ and $\nabla_me=0$). Does anyone see where the problem is?