Derivatives in Poincare' gauge theory

I have been reading the lectures: http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartIII_Supergravity.pdf about Poincare' gauge theory. The Poincare' group is considered as semidirect product of $$O(3,1)\ltimes \mathbb{R}^4$$. So the author represents the action of the Poincare' group $$U(\Lambda,a) = id + \omega_{a, b} M^{a, b} + a_{b} P^b$$ as a $$5x5$$ matrix:

$$\left( \begin{array}{cc} \Lambda^a_b & a^b \\ 0 & 1 \end{array} \right) \left( \begin{array}{c} x^b \\ 1 \end{array}\right) = \left( \begin{array}{c} \Lambda^a_b x^b +a^b \\ 1\end{array}\right)$$

where $$\Lambda$$ represents general Lorentz transformation (including rotations) and $$a$$ translations. In order to make a gauge theory with it, it has to be considered as a local symmetry. In order to establish a gauge symmetry a covariant derivative is needed which is constructed by a connection $$A$$ according to Cartan ($$e^a = e^a_\mu dx^\mu$$ and $$\omega^a_b = \omega^a_{\mu b} dx^\mu$$):

$$A = \left( \begin{array}{cc} \omega^a_{\mu b} & e^a \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} \omega^a_{\mu b} dx^\mu & e^a_\mu dx^{\mu} \\ 0 & 0 \end{array} \right).$$

Then the curvature of this connection is calculated:

$$F = dA+ A\wedge A = \left( \begin{array}{cc} d\omega^a_b +\omega^a_c \wedge \omega^c_b & de^a + \omega^a_b \wedge e^b \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} R^a_b & T^a \\ 0 & 0 \end{array} \right)$$

where $$R^a_b$$ is the curvature form and $$T^a$$ the torsion form. Finally a covariant derivative $$D$$ is introduced for the translational gauge symmetry:

$$De^a: = de^a +\omega^a_b \wedge e^b = T^a.$$

The appearance of the new covariant derivative is actually rather confusing because another covariant (metric compatible) derivative respectively connection already exists:

$$\nabla e^a =-\omega^a_ b \wedge e^b$$

where $$\omega^a_b$$ is the spin connection (Actually in the mentioned paper above the introduction of the spin connection is without minus sign, but I think this is wrong. Anyway, at the end my question is more of conceptual nature, so I consider the sign is in this context as not so important). I try to precise my question. When we use $$\nabla$$ as covariant derivative, in which respect is it covariant, to local (general) Lorentz transformation according to the type of indices $$a$$ and $$b$$ are marking? Or is it covariant to local diffeomorphisms, considering $$\nabla$$ as

$$\nabla_\mu e^a = -\omega^a_{\mu b} e^b .$$

And finally in which respect is $$De^a$$ covariant, to local translations? And how can it be shown that $$De^a$$ is covariant to local translations (can I check it out by a little calculation)? Could also be a combined covariant derivative be constructed like for a tensor with 2 indices $$X^a_\mu$$ (one covariant derivative acts on index $$a$$, the other on index $$\mu$$)? On the other hand as far as I know locally diffeomorphisms can also be represented as translations depending on the local position, so could it be that both covariant derivatives could be (more or less) the same?