$\DeclareMathOperator\SO{SO}$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.

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    $\begingroup$ 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? $\endgroup$ Jun 2, 2021 at 20:41
  • $\begingroup$ Ah okay, I see. No I didn't know that...Thanks a lot! $\endgroup$ Jun 2, 2021 at 22:17

1 Answer 1


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 $\eta(Av,u)+\eta(v,Au)=0$.


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