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?


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