# Spin connection in the tetradic Palatini-formalism of general relativity

$$\DeclareMathOperator\SO{SO}$$I am trying to understand the tetradic Palatini-formalism of general relativity from a mathematical point of view. I am graduate student and quite new to mathematical gauge theory and principal bundles in general yet, therefore I need some help. To be precise, I have the following question:

First of all, let us fix some notation and terminology. Let $$(\mathcal{M},g)$$ be a $$4$$-dimensional Lorentzian manifold of signature $$(+,-,-,-)$$ and let $$\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M})$$ denote the bundle of orthonormal (co-)frames on $$(\mathcal{M},g)$$. It can be shown that this bundle is in fact a principal $$\SO(1,3)$$-bundle, where $$\SO(1,3)$$ denotes the Lorentz group, as usual.

Now let us consider the associated fibre bundle \begin{align*}E:=\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M})\times_{\rho}\mathbb{M}^{4},\end{align*} where $$\rho:\SO(1,3)\to\mathrm{Aut}(\mathbb{M}^{4})$$ denotes the fundermental representation of the Lorentz group and where $$\mathbb{M}^{4}:=\mathbb{R}^{4}$$ is the Minkowski space in four dimension with signature $$(+,-,-,-)$$.

Now as far as I know, the spin connection is usually defined to be a connection 1-form in the orthonormal frame bundle, i.e. $$\omega\in \Omega^{1}(\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M}),\mathfrak{so}(1,3))$$. Choosing a local gauge, i.e. a local section $$s:U\subset\mathcal{M}\to\mathcal{F}_{\mathrm{Ort}}(T\mathcal{M})$$, we can define a local gauge field $$A\in\Omega^{1}(U,\mathfrak{so}(1,3))$$ on $$\mathcal{M}$$ via $$A:=s^{\ast}\omega$$.

So far so good. Now many texts mention (e.g. arXiv:gr-qc/9410018) that we can also view the connection $$1$$-form as a $$\bigwedge^{2}E$$-valued 2 form on $$\mathcal{M}$$, i.e. an element of $$\Omega^{1}(U,\bigwedge^{2}E)$$ where $$U\subset\mathcal{M}$$ is an open subset. This is needed, because in the end the first-order tetradic Palatini action for general relativity is defined via

$$S[e,\omega]:=\int_{\mathcal{M}}\operatorname{tr}(e\wedge e\wedge F)$$ where $$F$$ denotes the corresponding curvature $$2$$-form and where $$\operatorname{tr}$$ has to be understood as some kind of "internal volume form", which is a section of $$\bigwedge^{4}E$$ and maps a $$\bigwedge^{4}E$$-valued form into an ordinary (real-valued) form. The field $$e$$ above is a bundle isomorphism of the form $$e:T\mathcal{M}\to E$$, called the "cotetrad", which can also be viewed as an element of $$\Omega^{1}(\mathcal{M},E)$$.

Can anyone explain me how we can view the connection $$1$$-form as a $$\bigwedge^{2}E$$-form as explained above? Of course, if something I have explained above is totally wrong, I am happy about every error pointed out.

• For a finite dimensional inner product space $(V,\eta)$, $\bigwedge^2 V \cong_\eta \mathfrak{so}(\eta)$. Probably you already know this. If not, maybe you just need the formula for the isomorphism? Jun 2, 2021 at 20:41
• Ah okay, I see. No I didn't know that...Thanks a lot! Jun 2, 2021 at 22:17

For a finite dimensional inner product space $$(V,\eta)$$, $$\bigwedge^2 V \cong_\eta \mathfrak{so}(\eta) \subset \operatorname{End}(V) \cong V\otimes V^* \cong_\eta V\otimes V$$. The antisymmetry condition appears when expanding the identity $$\eta(e^{tA}v,e^{t A}u) = \eta(v,u)$$ to first order in $$t$$, to get $$\eta(Av,u)+\eta(v,Au)=0$$.