Different types of covariant derivatives in Poincare' invariant differential geometry 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 $b$ ?
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? 
 A: I am not sure if this answers OP's question, but here's something.
I think the proper way to think about Poincaré gauge theory is to first imagine the Poincaré symmetry to be a purely internal one. The following can be formalized precisely using principal fibre bundles and Cartan geometry, but let us use a local formalism here.
The idea is that if we work initially on Minkowski spacetime, but using arbitrary $x^\mu$ local coordinates (not necessarily pseudo-euclidean ones), then we still have the pseudo-euclidean coordinates $\xi^a$. We can use the pseudo-euclidean coordinates to take the components of tensor fields, but we measure their position-dependence using the arbitrary coordinates $x^\mu$.
We try to think of $\xi^a$ as a single geometric object, which we call here a Cartan radius vector.
Since the pseudo-euclidean coordinates can be subjected to Poincaré-transformations, the transformation $$ \xi^{a^\prime}=(\Lambda^{-1})^{a^\prime}_{\ a}(\xi^a-a^a) $$ is a symmetry transformation (here $\Lambda$ is a constant Lorentz transformation and $a$ is a constant translation).
We have $\vartheta^a=\mathrm d\xi^a$ being an orthonormal covielbein, which is also holonomic, since it is an exterior derivative, the metric is given by $$ g=\eta_{ab}\mathrm d\xi^a\mathrm d\xi^b. $$
Under the Poincaré transformation above, the vielbein transforms as $$ \vartheta^{a^\prime}=(\Lambda^{-1})^{a^\prime}_{\ a}\vartheta^a, $$ so all is good. Tensor fields transform as expected, i. e. via Lorentz transformations, without the translational part $a^a$.
If we try to perform a local Poincaré-transformation as $$ \xi^{a^\prime}(x)=(\Lambda^{-1}(x))^{a^\prime}_{\ a}(\xi^a(x)-a^a(x)), $$ we'd have $\vartheta^{a^\prime}=\mathrm d\xi^{a^\prime}=\mathrm d(\Lambda^{-1})^{a^\prime}_{\ a}(\xi^a-a^a)+(\Lambda^{-1})^{a^\prime}_{\ a}(\mathrm d\xi^a-\mathrm da^a),$ which is clearly not a good frame transformation.
We can now introduce a covariant derivative $D_\mu$ which acts on anything transforming under the Poincaré group as follows. On any Lorentz-tensor, it acts the same way an ordinary linear connection does, i. e. $D_\mu V^a=\partial_\mu V^a+\omega^a_{\ b\mu}V^b$, but on the only non-linear object in our theory (the Cartan vector) it acts as $$ D_\mu\xi^a=\partial_\mu\xi^a+\omega^a_{\ b\mu}\xi^b+\theta^a_\mu. $$
As it turns out $D_\mu\xi^a$ transforms linearly as $D^\prime_\mu\xi^{a^\prime}=(\Lambda^{-1})^{a^\prime}_{\ a}D_\mu\xi^a$, provided that $A_\mu=\frac{1}{2}\omega^{ab}_\mu M_{ab}+\theta^a_\mu P_a$ transforms as a connection under the Poincaré group.
If we define $D\xi^a=D_\mu\xi^a\mathrm dx^\mu$, then we can define the vielbein as $$ \vartheta^a=D\xi^a$$ (provided that there is a constraint on the connection such that $D\xi^a$ will be linearly independent, this is needed to be enforced by hand!),  which will be orthonormal by definition, but will not be holonomic in general, as $D$ is not an actual exterior derivative.

If we calculate the curvature, then we get $R^a_{\ b}$ from $\omega^a_{\ b}$, the ordinary Lorentz-curvature, and get $T^a$ from $\theta^a$, the ordinary Lorentz-torsion, but one calculated from $\theta^a$ rather than $\vartheta^a$. But as the OP may verify easily themselves, these get mixed with each other under Poincaré-transformations, the "curvature" and the "torsion" separate only if we restrict to the Lorentz subgroup. Likewise, the translational part of the curvature, $\theta^a$, is not actually a vielbein, as under the full Poincaré group, this also gets mixed with the Lorentz part $\omega^a_{\ b}$.
But, we can make a gauge transformation with $(\Lambda^{-1})^{a^\prime}_{\ a}=\delta^{a^\prime}_{a}$ and $a^a=\xi^a$, then $$ \xi^{a^\prime}=\delta^{a^\prime}_a(\xi^a-\xi^a)=0, $$ and in this gauge we have $\vartheta^a=\theta^a$ i. e. the translational part of the connection $\theta^a$ is actually the vielbein, and $R^a_{\ b}$ and $T^a$ are actually the curvature and torsion of the pair $(\omega^a_{\ b},\vartheta^a=\theta^a)$.
The gauge $\xi^a=0$ is preserved only by pure Lorentz transformations, so to enforce this gauge, the translational subgroup has to die, and the Poincaré group is broken to the Lorentz subgroup.

Conclusion/discussion:


*

*The Poincaré symmetry is considered to be a purely internal symmetry, and the tensors are purely internal objects that transform only under the Lorentz subgroup, i.e. under representations of the Poincarl group whose kernel includes the $\mathbb R^4$ translational subgroup.

*We have an affine object in our theory, the $\xi^a$ Cartan vector. The Cartan vector along with the connection relates the internal and external geometries via $\vartheta^a_\mu=D_\mu\xi^a$. As we know, a vielbein can be seen as a soldering form that relates internal indices to spacetime indices. This also induces the spacetime geometry via $g_{\mu\nu}=\eta_{ab}D_\mu\xi^a D_\nu\xi^b$ and this metric is not flat in general.

*On anything other than the Cartan vector, the Poincaré connection acts only by its Lorentz part. I guess this also answers OP's question, that if - say - a spacetime vector field $V^\mu$ is introduced via $V^\mu=e^\mu_aV^a$, where $e^\mu_a$ is the inverse of $\vartheta^a_\mu$, then a spacetime Linear connection is introduced via $\nabla_\mu V^\nu\equiv e^\nu_a D_\mu V^a$. This connection is compatible with the $g_{\mu\nu}=\eta_{ab}D_\mu\xi^a D_\nu\xi^b$ metric but can be torsionful.

*We can choose the $\xi^a=0$ "unitary gauge", the enforcing of which kills the translation subgroup, but it makes the Lorentz part $\omega^a_{\ b}$ of the Poincaré-connection into a genuine Lorentz-connection, and the translational part $\theta^a$ actually equivalent with the vielbein, and makes the translational part of the curvature into the actual torsion of $\vartheta^a=\theta^a$.

*This shows that the translation subgroup of the Poincaré group plays an almost completely passive role in our theory. No objects other than $\xi^a$ transform under translations, translations are purely internal and are not related to spacetime translations or diffeomorphisms at all, and in the end they end up being a broken subgroup. Their only role was to die to link the internal geometry to spacetime geometry, and provide a metric, which gives spacetime a Riemann-Cartan geometry.

*The "unitary gauge" $\xi^a$ is enforceable globally, actually. From a high-brow point of view, we have an overlying principal bundle $\bar\pi:\bar P\rightarrow M$ with the $\mathrm{ISO}(3,1)$ Poincaré group as the structure group, the Cartan vector $\xi$ can be seen as a global cross-section of the quotient bundle $\bar P/\mathrm{SO}(3,1)$, which has the structure of an affine bundle. Because $\bar P/\mathrm{SO}(3,1)$ has $\mathbb R^4$ as fibres, which is contractible, it always admits global sections, so Cartan vectors always exist. If this bundle admits a global section (and as stated, it does!), then by the usual theorem on principal bundles, the structure group is reducible from $\mathrm{ISO}(3,1)$ to $\mathrm{SO}(3,1)$, so there exists a system of transition functions/gauge transformation where the translations are all zero.
