The covariant derivative may be defined as:

$$\nabla_{a}=e_{a}^{~\mu}(x)\partial_{\mu}+\frac{1}{2}e_{a}^{~\mu}(x)\omega_{\mu bc}(x)M^{bc}\tag{1}$$

where $e_{a}^{~\mu}(x)$ is the vielbein, $\omega_{\mu bc}(x)=-\omega_{\mu cb}(x)$ is the spin connection, $M_{bc}=-M_{cb}$ are the Lorentz generators (in an arbitrary representation), Latin indices are Lorentz (flat) indices and greek indices are world (curved) indices.

In order for the covariant derivative of Lorentz tensor fields to transform covariantly, we require the spin connection to transform inhomogeneously (and hence non-covariantly). For concreteness I will consider the covariant derivative of (the components of) a Lorentz one-form, $V_a(x)$ . If, under a general coordinate transformation (GCT) and a Local Lorentz transformation (LLT), we want the following to hold true, $$\nabla'_aV'_{b}(x')=\Lambda_{a}^{~c}\Lambda_{b}^{~d}\nabla_cV_{d}(x),\tag{2}$$ i.e. if it is to transform covariantly, then we require that the spin connection transforms as $$\omega'_{\mu ab}(x')=\frac{\partial x^{\nu}}{\partial x'^{\mu}}\Lambda_{a}^{~c}\Lambda_{b}^{~d}\omega_{\nu cd}(x)+\Lambda^{c}_{~~a}\partial'_\mu\Lambda_{bc}.\tag{3}$$ I have just finished deriving the origin of the inhomogeneous term in (3) but I had to rely on two things which I am not able to justify.

First, I had to assume that the Lorentz generators are invariant under a LLT, i.e I assumed that $$M'_{ab}=\Lambda_a^{~c}\Lambda_{b}^{~d}M_{cd}=M_{ab}.$$ I am not too surprised by this fact, but i'm not sure how to justify it?

Secondly, I also assumed that the inhomogeneous term in (3) is anti-symmetric in $a$ and $b$. This must be the case, otherwise the spin connection would not retain it's antisymmetry when it undergoes a transformation. But I would also like prove that it is the case, so I tried the following: \begin{align} \Lambda^{c}_{~~a}\partial'_\mu\Lambda_{bc}&=\partial'_{\mu}\big(\Lambda^{c}_{~~a}\Lambda_{bc}\big)-\Lambda_{bc}\partial'_{\mu}\Lambda^{c}_{~~a}\\ &=\partial'_{\mu}\big(\eta_{ab})-\Lambda_{bc}\partial'_{\mu}\Lambda^{c}_{~~a}\\ &=-\Lambda_{bc}\partial'_{\mu}\Lambda^{c}_{~~a}\\ &=-\Lambda_{b}^{~c}\partial'_{\mu}\Lambda_{ca}. \end{align} This is almost it, but not quite, since I need to show that $\Lambda^{c}_{~~a}\partial'_\mu\Lambda_{bc}=-\Lambda^{c}_{~~b}\partial'_\mu\Lambda_{ac}$. How may I proceed?

P.S. I use the mostly plus convention for my Minkovski metric.


1 Answer 1


To answer my first query, it is not true that:


Such a statement doesn't quite make (physical) sense to begin with, since the Lorentz generators are non-dynamical abstract basis elements of the Lorentz algebra. Therefore, one should be careful when considering the transformation of the covariant derivative under a GCT and LLT. More specifically, it transforms as $$\nabla_a\mapsto\nabla'_a=e_{a}'^{~\mu}(x')\partial'_{\mu}+\frac{1}{2}e_{a}^{'~\mu}(x')\omega'_{\mu bc}(x')M^{bc}$$ and not $$\nabla_a\mapsto\nabla'_a=e_{a}'^{~\mu}(x')\partial'_{\mu}+\frac{1}{2}e_{a}^{'~\mu}(x')\omega'_{\mu bc}(x')M'^{bc}.$$ This is where my problem arose.

To answer my second query, the transformation law (3) is wrong. It should be $$\omega'_{\mu ab}(x')=\frac{\partial x^{\nu}}{\partial x'^{\mu}}\Lambda_{a}^{~c}\Lambda_{b}^{~d}\omega_{\nu cd}(x)+\Lambda^{~c}_{a}\partial'_\mu\Lambda_{bc}.$$ Then it follows that the counter term is antisymmetric: $$\Lambda^{~c}_{a}\partial'_\mu\Lambda_{bc} =\partial'_{\mu}\big(\Lambda^{~c}_{a}\Lambda_{bc}\big)-\Lambda_{bc}\partial'_{\mu}\Lambda^{~c}_{a}=-\Lambda_{b}^{~c}\partial'_{\mu}\Lambda_{ac}.$$


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