# What are the analogues of $F_{\mu\nu}$ in General Relativity?

In electromagnetism, the measurable gauge-invariant quantities are the electric and magnetic fields or the six independent components of the field strength tensor $$F_{\mu\nu}$$. What are the analogues of $$F_{\mu\nu}$$ in General Relativity? I have a feeling that $$g_{\mu\nu}$$ is not the analogue of $$F_{\mu\nu}$$ but I am not sure. Any help?

Warning: this answer takes the perspective of the second table, rather than the first, in a question @knzhou linked.

We need to first explain what's analogous to $$A_\mu$$. The Christoffel symbols are. We can then ask what's analogous to $$F_{\mu\nu}$$; the Riemann tensor is. I suspect someone else's answer will provide a rationale for why that differs from what I'll say here, because I will look at it, perhaps unfortunately, from a perspective outside of GR.

Classical electromagnetism and general relativity are both gauge theories; the latter's equivalent of local $$U(1)$$ transformations is general coordinate transformations. In both cases, a very limited symmetry that preserves the partial derivatives of those tensors it preserves is expanded to something for which the preserved notion of a derivative is partial-plus-extra. One need only compare the gauge covariant derivative $$\partial_\mu\phi+qA_\mu\phi$$ (or in a non-Abelian Yang-Mills theory, $$\partial_\mu\phi_a+qf_a^{\:bc}A_{\mu b}\phi_c$$) with the Riemannian connection (aka "covariant derivative") $$\partial_\mu V^\nu+\Gamma_{\mu\rho}^\nu V^\rho$$ to appreciate the analogy.

We'll respectively denote the linear operators used above here as $$D_\mu,\,\nabla_\mu$$. This gives us commutators: $$F_{\mu\nu}\propto[D_\mu,\,D_\nu]$$, while $$[\nabla_\mu,\,\nabla_\nu]V_\rho=R_{\mu\nu\rho\sigma}V^\sigma$$. (The latter lacks a derivative of the vector field, because GR has zero torsion.)

In the above analogy, what's actually analogous to $$g_{\mu\nu}$$ is the above scalar field $$\phi$$.

• Isn't $A$ analogous to the Levi-Civita connection? Commented Sep 11, 2021 at 19:36
• @Filippo if you write $\nabla = \partial + A$, the symbol $\nabla$ is the "Levi-Civita connection" and $A$ is the "Christoffel". Am I missing something? (or maybe it's a matter of semantics?). See this: physics.stackexchange.com/a/1950/226902 Commented Feb 17, 2022 at 12:44
• @Quillo If $A$ denotes a covariant derivative, then it is in some sense analogous to the Levi-Civita connection, since it is a covariant derivative, too, right? Commented Feb 17, 2022 at 13:28
• @Filippo the covariant derivative is typically written as $\nabla$ or $D$, see physics.stackexchange.com/q/618415/226902 physics.stackexchange.com/q/466911/226902 . Probably your doubt is addressed here: physics.stackexchange.com/a/8152/226902 Commented Feb 17, 2022 at 13:33
• @Quillo Thank you for the links! Regarding the last link: I'd say that a connection defines a covariant derivative, not the other way around. Commented Feb 17, 2022 at 13:59

From the perspective of Principal Lie-group (G) bundle, the curvature 2-form $$\textbf{F}=\textbf{dA}+\frac{1}{2}[\textbf{A,A}]$$corresponding to Lie group $$G=SU(N)$$ is essentially the Yang-Mills field $$F_{\mu\nu}$$. For $$G=GL(n;C)$$ or $$GL(n;R)$$, the components of curvature 2-form are the Riemann tensor components $$R_{\alpha\beta\gamma\delta}$$.

The propagating d.o.f. for gravitational field is contained in Weyl curvature components $$C_{\alpha\beta\gamma\delta}$$ (traceless part of Riemann tensor) and hence from physics point of view, the analogue for $$F_{ab}$$ in GR can be taken as the Weyl curvature $$C_{abcd}$$. There are certain similarities b/w Maxwell tensor $$F_{ab}$$ and Weyl tensor $$C_{abcd}$$:

Both $$F_{ab}$$ and $$C_{abcd}$$ are traceless satisfying the source free equations $$\nabla ^aF_{ab}=0$$ (Maxwell) and $$\nabla ^aC_{abcd}=0$$ (Einstein). In language of representation theory, one can decompose $$F_{ab}$$ as the $$(1,0)\oplus (0,1)$$ irreducible representation of $$su(2)_L\times su(2)_R$$. This can be expressed using the symmetric spinor $$\phi_{AB}$$: $$F_{ab} =F^{+}_{ab}+F^{-}_{ab} \leftrightarrow \phi_{AB}\epsilon_{A'B'}+\bar{\phi}_{A'B'}\epsilon_{AB}$$Similarly one can decompose Weyl tensor as the $$D(2,0)$$ and $$D(0,2)$$ irreducible representation of $$SL(2,C)$$: $$C_{abcd}=C^{+}_{abcd}+C^{-}_{abcd}\leftrightarrow \Psi_{ABCD}\epsilon_{A'B'}\epsilon_{C'D'}+c.c.$$where $$\Psi_{ABCD}$$ (symmetric) is the gravitational spinor. This kind of decomposition is not possible for Riemann tensor. One can go on further to define analogue of other physical quantities such as:

1. Bel-Robinson tensor $$T_{abcd}=\Psi_{ABCD}\bar{\Psi}_{A'B'C'D'}$$ as the gravitational analogue for free electromagnetic stress energy tensor: $$T_{ab}=\frac{1}{2\pi}\phi_{AB}\bar{\phi}_{A'B'}$$ , both of them satisfying the Rainich's condition and "conservation law".

2. Electric and Magnetic parts of $$C_{abcd}$$ can be defined wrt some unit timelike vector $$u^a$$: $$E_{ab}=C_{abcd}u^cu^d$$ (electric part) and $$H_{ab}=\frac{1}{2}\eta_{ade}{C^{de}}_{bc}u^c$$ (magnetic part). Note, the similarity with Electric and magnetic vectors in Maxwell theory: $$E_a=F_{ab}u^b$$ , $$H_a=\frac{1}{2}\eta_{abc}F^{bc}$$.

This answer says roughly the same things as @JG's answer, but phrased slightly differently.

The analogy between the Riemann tensor $$R$$ and the electromagentic field strength tensor $$F$$ is not so apparent when we dress them up with indices $$F_{\mu\nu}$$ and $$R^\rho_{\,\,\mu\nu\sigma}$$. However, things become more apparent if we look slightly more abstractly.

• $$F$$ can be thought of as a $$2$$-form on the spacetime manifold, and it satisfies $$dF=0$$ (the exterior derivative). Or in components, for all $$a,b,c$$, $$\frac{\partial F_{bc}}{\partial x^a}+\frac{\partial F_{ca}}{\partial x^{b}}+\frac{\partial F_{ab}}{\partial x^{c}}=0$$. This equation encodes $$\nabla \cdot B=0$$ and $$\nabla \times E=-\frac{\partial B}{\partial t}$$

• Similarly, on any vector bundle with a linear connection $$(E,\pi,M, \nabla)$$, the curvature $$R$$ of the connection is an $$\text{End}(E)$$-valued $$2$$-form on $$M$$, i.e it is a smooth vector bundle morphism $$R:\bigwedge^2(TM)\to \text{End}(E)$$. Roughly speaking, this says to each point $$x\in M$$, and each pair of vectors $$h_x,k_x\in T_xM$$, we consider the plane/bivector $$h_x\wedge k_x$$, and to such a bivector, we have an endomorphism $$R(h_x\wedge k_x)\in \text{End}(E_x)$$. I explain the intuition more in this MSE answer. Now, one can also prove that $$d_{\nabla}R=0$$; i.e the exterior covariant derivative of the curvature vanishes. In components, this says $$\nabla_a(R_{bc})+\nabla_b(R_{ca})+\nabla_c(R_{ab})=0$$, which is none other than the differential Bianchi identity. In the case where $$E=TM$$ is the tangent bundle (a very common special case), then the description of $$R$$ as an $$\text{End}(TM)$$-valued $$2$$-form on $$M$$ is equivalent to saying the curvature is a $$(1,3)$$-tensor field on $$M$$.

So, both $$F,R$$ are morally speaking the same type of object (a $$2$$-form, the only difference is one is scalar-valued, the other is endomorphism-valued), and both satisfy a form of Bianchi's identity ($$dF=0$$ vs $$d_{\nabla}R=0$$). This is also why you may hear $$F$$ being referred to as a curvature.

I think it's worth saying that there is, in a way, not a great 1-1 analogue here:

the Maxwell action is proportional to $$F^{ab}F_{ab}$$, while, while JG is right that in a lot of ways, the "analogue" to choose for GR is $$R_{abcd}$$, the Hilbert action is proportional to $$g^{ab}R_{ab}$$ not $$R^{ab}R_{ab}$$. The coupling to matter happens via a $$g^{ab}$$ term, too, and not a $$\Gamma$$ term. General covariance just makes nailing down the "real" degrees of freedom of GR much more complex than it is for Electromagnetism, and there are just a lot of terms to worry about.

• It gets even murkier when you seek an analogoue for the source term $A_\mu j^\mu$; the answer then ends up contingent on what other physics we embed in GR.
– J.G.
Commented Sep 10, 2021 at 20:24
• Also, of note here, in the weak-field limit (and the "right coordinates"), the "classical potential" is $|g_{tt}| - 1$ Commented Sep 11, 2021 at 22:21

I will add a different point of view by using tetrad formalism. As @J.G. said, the Christoffel symbols are roughly the analogs of the gauge connection. In fact, the tetrad formalism has a more accurate analog: the spin connection $$\omega$$. This spin connection is $$\mathfrak{spin}(1,3)$$ valued, just like a gauge connection is valued in some gauge group (for example, in QCD one has a $$\mathfrak{su}(3)$$-valued gauge connection). So one has the following connection: $$$$\omega \in \Omega^1_{\mathfrak{spin}(1,3)}(\mathcal{M}) \equiv\Gamma(\mathfrak{spin}(1,3))\otimes_{\Omega^0(\mathcal{M})}\Omega^1(\mathcal{M})\simeq \Gamma(\mathfrak{spin}(1,3)\otimes \wedge^1 T^\ast \mathcal{M})$$$$ Where $$\Omega^i(\mathcal{M})\equiv \Gamma(\wedge^i T^\ast \mathcal{M})$$. The curvature of the spin connection is then obviously defined as: $$$$\Omega^2_{\mathfrak{spin}(1,3)}(\mathcal{M}) \ni \Omega\equiv d_\omega\omega \stackrel{!}{=}d\omega+[\omega \stackrel{\wedge}{,} \omega]_{\mathfrak{spin}(1,3)}$$$$ Adding a torsion-free condition $$d_\omega e=0$$, where $$e$$ is the tetrad, one can uniquely define $$\omega$$ in terms of $$e$$. Concretely, $$\Omega$$ defined by $$e$$ is nothing but the analog of the Riemann tensor. But I think that it is more clear why $$\Omega$$ is analogous to $$F$$ in the usual gauge theory framework, using the tetrad formalism. One has the following analogies: $$$$\begin{array}{ccc} \omega=\omega_\mu^{IJ}\sigma_{IJ}dx^\mu &\longleftrightarrow &A=A_\mu^a \tau^a dx^\mu \\ \Omega= \Omega_{\mu \nu}^{IJ}\sigma_{IJ}dx^\mu \wedge dx^\nu &\longleftrightarrow &F=F_{\mu \nu}^a \tau^a dx^\mu \wedge dx^\nu \end{array}$$$$ To go back to the usual formalism in GR, usually, the magnetic part and the electric part of the curvature of space-time are defined by the use of the Weyl tensor, just like @KP99 said. One can ask for a theory where the analogy is stronger: can one has a Lagrangian density defined as the contraction of the Weyl tensor with itself? The answer is Yes, it is the conformal gravity, but it is not strictly equivalent to the Einstein-Hilbert action.

The actual correspondence with gravity fields is with gauge fields, not electromagnetic fields, because of the non-linearity of the equations. So, I will lay out the basics of both in detail.

The fields involved in gravitational dynamics, when formulated in the setting of a Riemann-Cartan geometry, comprise a $$GA(4)$$ gauge field. The dynamics (and the defining constraint for the Riemann-Cartan geometry) reduces the $$GL(4)$$ part of the field to a $$SO(3,1)$$ gauge field, thereby effectively reducing the whole field to an $$ISO(3,1)$$ gauge field. However, its dynamics do not comprise any analogue of Yang-Mills dynamics, since the Lagrangian is only first order in the field strength and (in 3+1 dimensions) second order in the potentials, as you'll see below.

The Electromagnetic Field As A Gauge Field
The electric potential $$φ$$, magnetic potential $$𝐀 = \left(A_x, A_y, A_z\right)$$, electric field strength $$𝐄 = \left(E_x, E_y, E_z\right)$$ and magnetic induction $$𝐁 = \left(B^x, B^y, B^z\right)$$ can be combined into geometric objects $$A = 𝐀·d𝐫 - φ dt, \hspace 1em F = 𝐁·d𝐒 + 𝐄·d𝐫dt,$$ called the potential one-form $$A$$ and field strength two-form $$F$$; where I'm using the notation $$d𝐫 = (dx, dy, dz), \hspace 1em d𝐒 = (dy dz, dz dx, dx dy), \hspace 1em dV = dx dy dz,$$ and treating the products of the $$dx$$'s as anti-commuting, i.e. Grassmann algebra $$dx^μ dx^ν = -dx^ν dx^μ$$, for brevity, instead of denoting them by a "wedge" product (i.e. $$dx^μ ∧ dx^ν$$).

The Maxwell equations are then rendered as: $$𝐁 = ∇×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t} \hspace 1em ⇒ \hspace 1em dA = F, \\ ∇·𝐁 = 0, \hspace 1em ∇×𝐄 + \frac{∂𝐁}{∂t} = 𝟎 \hspace 1em ⇒ \hspace 1em dF = 0,$$ where $$∇ = (∂/∂x, ∂/∂y, ∂/∂z)$$, with the second set of equations being derivable from the first set.

Written in component form, the respective fields are: $$A = A_μ dx^μ, \hspace 1em F = ½ F_{μν} dx^μ dx^ν,$$ (in here, and the following, I'm using the summation convention), with the following reductions corresponding to the coordinates $$\left(x^0, x^1, x^2, x^3\right) = (t, x, y, z)$$: $$A_0 = -φ, \hspace 1em \left(A_1, A_2, A_3\right) = 𝐀, \hspace 1em \left(F_{23}, F_{31}, F_{12}\right) = 𝐁, \hspace 1em \left(F_{10}, F_{20}, F_{30}\right) = 𝐄.$$ In component form, we have: $$F_{μν} = ∂_μ A_ν - ∂_ν A_μ, \hspace 1em ∂_μ F_{νρ} + ∂_ν F_{ρμ} + ∂_ρ F_{μν} = 0.$$ Since $$F_{μν} = -F_{νμ}$$, there are only 6 independent components to $$F$$, which correspond to those listed above.

The dynamics are given by an action principle with an action integral of the form $$S = \int L$$, where $$L = 𝔏 d^4x$$ is the Lagrangian 4-form, expressible in terms of a Lagrangian density $$𝔏(A,F)$$ that is assumed to be a function of $$A$$ and $$F$$ (and possibly other fields, as well). Here I'm using the notation $$d^4 x = dx^0 dx^1 dx^2 dx^3$$. The Lagrangian is used to generate a set of constitutive relations between the fields (which may be likened to, as "kinematic" variables) and the corresponding dynamic fields (the "dynamic" variables): $$𝔊^{μν} = -\frac{∂𝔏}{∂F_{μν}}, \hspace 1em 𝔍^μ = \frac{∂𝔏}{∂A_μ},$$ comprising, respectively, the response fields, which have the property $$𝔊^{μν} = -𝔊^{νμ}$$ and sources $$𝔍^μ$$. These are both, actually, tensor densities, not tensors.

The Euler-Lagrange equations that arise from the variational principle, and the continuity equation that arises from these, as a consequence, are respectively: $$∂_ν 𝔊^{μν} = 𝔍^μ, \hspace 1em ∂_μ 𝔍^μ = 0.$$ The Maxwell equations $$∇·𝐃 = ρ, \hspace 1em ∇×𝐇 - \frac{∂𝐃}{∂t} = 𝐉, \hspace 1em ∇·𝐉 + \frac{∂ρ}{∂t} = 0,$$ can be recovered from this, provided we identify $$𝐃 = \left(𝔊^{01}, 𝔊^{02}, 𝔊^{03}\right), \hspace 1em 𝐇 = \left(𝔊^{23}, 𝔊^{31}, 𝔊^{12}\right), \hspace 1em ρ = 𝔍^0, \hspace 1em 𝐉 = \left(𝔍^1, 𝔍^2, 𝔍^3\right)$$ and that recovers the electric displacement $$𝐃 = \left(D^x, D^y, D^z\right)$$, magnetic field strength $$𝐇 = \left(H_x, H_y, H_z\right)$$, charge density $$ρ$$ and current density $$𝐉 = \left(J^x, J^y, J^z\right)$$.

The constitutive relation generated by the Lagrangian density, expressed in terms of them, are: $$𝐃 = \frac{∂𝔏}{∂𝐄}, \hspace 1em 𝐇 = -\frac{∂𝔏}{∂𝐁}, \hspace 1em 𝐉 = \frac{∂𝔏}{∂𝐀}, \hspace 1em ρ = -\frac{∂𝔏}{∂φ}.$$

Expressed as differential forms, response fields and sources can be written as $$G = ½ 𝔊^{μν} ∂_ν ˩ ∂_μ ˩ d^4 x, \hspace 1em J = 𝔍^μ ∂_μ ˩ d^4 x,$$ where $$(⋯)˩(⋯)$$ denotes the contraction operator, a bi-linear operator defined recursively on differential forms by $$∂_μ ˩ \left(dx^ν ⋯\right) = δ_μ^ν (⋯) - dx^ν \left(∂_μ ˩ (⋯)\right), \hspace 1em ∂_μ ˩ 1 = 0.$$ For example, $$∂_0 ˩ d^4x = dx dy dz = dV, \\ \hspace 1em \left(∂_1 ˩ d^4x, ∂_2 ˩ d^4x, ∂_3 ˩ d^4x\right) = \left(-dt dy dz, dt dx dz, -dt dx dy\right) = -dt d𝐒 = -d𝐒 dt.$$ Written in terms of the Maxwell fields, we have $$G = 𝐃·d𝐒 - 𝐇·d𝐫dt, \hspace 1em J = ρdV - 𝐉·d𝐒dt,$$ and the Euler-Lagrange equation and continuity equation, respectively, become $$dG = J, \hspace 1em dJ = 0.$$

The variational can be carried out entirely in algebraic form, with the differentials of the Lagrangian density expressed via $$ΔL = ΔA J - ΔF G$$. Then applying the condition $$Δ(dA) = d(ΔA)$$, we have $$ΔF G = Δ(dA) G = d(ΔA) G = d(ΔA G) + ΔA dG,$$ noting that the Leibnitz rule for odd-order differential forms has a minus sign, e.g.: $$d(A(⋯)) = dA (⋯) - A d(⋯)$$. Thus, $$ΔL = ΔA (J - dG) + d(-ΔA G),$$ and under the action integral, the last term $$d(-ΔA G)$$ is integrated to a boundary term, which drops out of the analysis, leaving behind only the Euler-Lagrange equation $$J = dG$$, upon application of the variational principle.

The Maxwell-Lorentz Action
What type of dynamics you have entirely hinges on what the constitutive relations are and, in turn, on what the Lagrangian density is. That's where all the physical content of the relevant theory lies. The fields and equations, themselves, are little more than place-holders that provide a general framework for all formulations of dynamics that cast with the players just laid out here.

For instance, the free field in a vacuum: $$𝐃 = ε_0 𝐄, \hspace 1em 𝐁 = μ_0 𝐇, \hspace 1em 𝐉 = 𝟎, \hspace 1em ρ = 0$$ arises from the Lagrangian density $$𝔏 = ½ \left(ε_0 E^2 - B^2/μ_0\right)$$. This is the Maxwell-Lorentz Lagrangian density; which, expressed in component form, is: $$𝔏 = -¼ ε_0 c \sqrt{|g|} g^{μρ} g^{νσ} F_{μν} F_{ρσ}.$$ The dynamics are dependent on a metric $$g_{μν}$$ and are expressed in terms of its determinant $$g$$ and inverse $$g^{μν}$$. The response field and source for this Lagrangian density are given by: $$𝔊^{μν} = ε_0 c \sqrt{|g|} g^{μρ} g^{νσ} F_{ρσ}, \hspace 1em 𝔍^μ = 0.$$

The Gauge Fields: "Kinematics"
Gauge fields have the same form as the electromagnetic fields, and the same players are involved, but there are now multiple copies of each type of field.

In place of the "kinematic" fields are $$φ^a$$, $$𝐀^a$$, $$𝐁^a$$, $$𝐄^a$$, which are indexed to a basis $$\left(Y_0, Y_1, ⋯, Y_{n-1}\right)$$ of an $$n$$-dimensional Lie algebra that is endowed with a bi-linear Lie bracket $$[⋯,⋯]$$ satisfying the identities $$[Y_a, Y_b] = f^c_{ab} Y_c, \hspace 1em [u,v] = -[v,u], \hspace 1em [u,[v,w]] + [v,[w,u]] + [w,[u,v]] = 0,$$ which entail the following relations for the structure coefficients $$f^c_{ab}$$: $$f^c_{ab} = -f^c_{ba}, \hspace 1em f^d_{ae} f^e_{bc} + f^d_{be} f^e_{ca} + f^d_{ce} f^e_{ab} = 0.$$

They can be written as the gauge potentials $$A^a_μ$$ and gauge field strengths $$F^c_{μν}$$. The corresponding geometric objects are, respectively, the potential one-form and field strength two-form: $$A = A^a_μ dx^μ Y_a, \hspace 1em F = ½ F^c_{μν} dx^μ dx^ν Y_c.$$

In addition to treating the products of the $$dx$$'s as anti-commuting, I'm also assuming the products of the $$dx$$'s to intersperse freely with the $$Y$$'s; e.g. $$dx^μ Y_a = Y_a dx^μ$$, rather than denoting them explicitly as tensor products (e.g. $$dx^μ ⊗ Y_a$$ or $$dx^μ ⊗ Y_a$$); and that the $$dx$$'s also intersperse freely with the dual basis $$y_a$$ introduced later below.

For convenience, we will also assume that the Lie algebra is embedded in an associative linear algebra $$u v - v u = [u, v]$$, i.e. $$Y_a Y_b - Y_b Y_a = \left[Y_a, Y_b\right] = f^c_{ab} Y_c.$$ This is already the case for any matrix representation of the Lie algebra.

Various combinations can be represented in various ways. \begin{align} A^a &= A^a_μ dx^μ, & F^c &= ½ F^c_{μν} dx^μ dx^ν, \\ A_μ &= A^a_μ Y_a, & F_{μν} &= F^c_{μν} Y_c. \end{align} Thus, the gauge fields are, component-wise, Lie-vector-valued fields each of whose Lie components $$A^a$$ and $$F^c$$ is a copy of a field analogous to the electromagnetic potentials and field strengths. Each the coordinate components $$A_μ$$ and $$F_{μν}$$ are, themselves, Lie-valued vectors.

With these conventions, we may write the following products - making use of the anti-symmetry of the wedge product: $$A^2 = A^a_μ A^b_ν dx^μ dx^ν Y_a Y_b = -A^b_ν A^a_μ dx^ν dx^μ Y_a Y_b = -A^a_μ A^b_ν dx^μ dx^ν Y_b Y_a,$$ thus $$A^2 = ½ A^a_μ A^b_ν dx^μ dx^ν \left(Y_a Y_b - Y_b Y_a\right) = ½ A^a_μ A^b_ν dx^μ dx^ν \left[Y_a, Y_b\right] = ½ f^c_{ab} A^a_μ A^b_ν dx^μ dx^ν Y_c,$$ which can be equivalently written in any number of other ways: $$A^2 = ½ f^c_{ab} A^a A^b Y_c = ½ \left[A_μ, A_ν\right] dx^μ dx^ν = ½ [A, A].$$ Similarly $$AF - FA = ½ f^c_{ab} A^a_μ F^b_{νρ} dx^μ dx^ν dx^ρ Y_c = f^c_{ab} A^a F^b Y_c = ½ \left[A_μ, F_{νρ}\right] dx^μ dx^ν dx^ρ = [A, F].$$

The definition of the field strength in terms of the potentials is modified with the inclusion of the extra terms arising from the structure coefficients by: $$F^c_{μν} = ∂_μ A^c_ν - ∂_ν A^c_μ + f^c_{ab} A^a_μ A^b_ν,$$ and can equivalently be written in any of the number of following ways: $$F^c = dA^c + ½ f^c_{ab} A^a A^b, \hspace 1em F_{μν} = ∂_μ A_ν - ∂_ν A_μ + ½ \left[A_μ, A_ν\right].$$

A direct consequence of this is the analogue of the "Bianchi identities": \begin{align} dF + AF - FA &= d\left(dA + A^2\right) + A\left(dA + A^2\right) - \left(dA + A^2\right)A \\ &= d^2 A + dA A - A dA + A dA + A A^2 - dA A - A^2 A \\ &= (d^2 A) + (A A^2 - A^2 A) \\ &= 0 \end{align} since $$d(A⋯) = dA (⋯) - A d(⋯)$$ (because $$A$$ is a differential form of odd-degree), $$d^2(⋯) = 0$$ and $$A A^2 = A^2 A$$.

The gravity field, too, will fit within this framework; not just the electromagnetic field or its generalization to "Yang-Mills" fields.

The Gauge Fields: "Dynamics"
In place of the "dynamic" fields are $$ρ_a$$, $$𝐉_a$$, $$𝐃_a$$ and $$𝐇_a$$, which are Lie co-vector valued, indexed to the dual Lie algebra basis $$\left(y^0, y^1, ⋯, y^{n-1}\right)$$. The basis and dual basis are together endowed with a (bi-linear) inner-product operation, given by $$y^a·Y_b = δ^a_b = Y_b·y^a.$$

The dynamics is generated from a Lagrangian density $$𝔏\left(A^a_μ, F^c_{μν}\right)$$, that is a function of the potentials and field strengths and possibly other fields. The derivatives give rise to the corresponding response fields and sources: $$𝔊^{μν}_c = -\frac{∂𝔏}{∂F^c_{μν}}, \hspace 1em 𝔍^μ_a = \frac{∂𝔏}{∂A^a_μ},$$ and when written out as geometric objects, they take on the respective forms: $$G = ½ 𝔊^{μν}_c ∂_ν ˩ ∂_μ ˩ d^4 x y^c, \hspace 1em J = 𝔍^μ_a ∂_μ ˩ d^4 x y^a.$$

The different combinations of the response fields can be written as: \begin{align} G_c &= ½ 𝔊^{μν}_c ∂_ν ˩ ∂_μ ˩ d^4 x, & J_a &= 𝔍^μ_a ∂_μ ˩ d^4 x, \\ 𝔊^{μν} &= 𝔊^{μν}_c y^c, & 𝔍^μ &= 𝔍^μ_a y^a. \end{align}

With these definitions, the variational of the Lagrangian can be written, similarly as before, except for the inclusion of the Lie inner product, as $$ΔL = ΔA·J - ΔF·G,$$ now with the component form: $$Δ𝔏 = ΔA^a_μ 𝔍^μ_a - ½ ΔF^c_{μν} 𝔊^{μν}_c.$$

We will assume that the algebraic representation used for the $$Y$$'s has been extended to the dual basis with the following operations: $$y^c Y_a - Y_a y^c = f^c_{ab} y^b,$$ and assume that a (linear) "trace" operation can be defined that satisfies the properties $$Tr(u⋯) = Tr(⋯u), \hspace 1em Tr\left(y^a Y_b\right) = δ^a_b,$$ and rewrite the variational as: $$ΔL = Tr(ΔA J - ΔF G).$$ Note that for form-valued objects, the [anti-]commutativity properties are applied under the trace operation, e.g. $$Tr(uv) = -Tr(vu)$$ if $$u$$ and $$v$$ are both odd-order differential-form-valued. The fields $$A$$ and $$J$$ are odd-order, while $$F$$ and $$G$$ are even order.

Noting that $$ΔF = Δ\left(dA + A^2\right) = ΔdA + ΔA A + A ΔA = d(ΔA) + ΔA A + A ΔA,$$ this allows us to directly rewrite the term $$ΔF G$$ under the trace as \begin{align} Tr(ΔF G) &= Tr(d(ΔA) G + ΔA A G + A ΔA G) \\ &= Tr(d(ΔA G) + ΔA dG + ΔA A G - ΔA G A) \\ &= d(Tr(ΔA G)) + Tr(ΔA (dG + AG - GA)). \end{align} where we used $$d(ΔA ⋯) = d(ΔA)(⋯) - ΔA d(⋯)$$, since $$ΔA$$ is odd-order.

Thus, the variational for the action integral can be written as: $$ΔS = \int ΔL = \int d(Tr(-ΔA G)) + \int Tr(ΔA (J - (dG + AG - GA))).$$ The "boundary term" expressing the divergence of $$Tr(-ΔA G)$$ gives us the basis for the symplectic geometry associated with the field, including such things as Poisson brackets, but otherwise drops out from the analysis; while the remainder yields the Euler-Lagrange equations: $$dG + AG - GA = J,$$ and, from this, the continuity equation: \begin{align} dJ + AJ + JA &= d^2 G + dA G - A dG - dG A - G dA \\ &+ (A dG + A AG - A G A) + (dG A + AG A - GA A) \\ &= \left(d^2 G\right) + \left(dA + A^2\right) G - G \left(dA + A^2\right) \\ &= FG - GF. \end{align} In component form, they may be written as $$∂_ν 𝔊^{μν}_a + f^c_{ab} A^b_ν 𝔊^{μν}_c = 𝔍^μ_a, \hspace 1em ∂_μ 𝔍^μ_a + f^c_{ab} A^b_ν 𝔍^ν_c = ½ f^c_{ab} F^b_{μν} 𝔊^{μν}_c$$ or equivalently as \begin{align} ∂_ν 𝔊^{μν} + [A_ν, 𝔊^{μν}] &= 𝔍^μ, & ∂_μ 𝔍^μ + [A_μ, 𝔍^μ] &= ½ [F_{μν}, 𝔊^{μν}], \\ dG_a + f^c_{ab} A^b G_c &= J_a, & dJ_a + f^c_{ab} A^b J_c &= ½ f^c_{ab} F^b G_c. \end{align}

Yang-Mills Dynamics
The gauge analogue of the Maxwell-Lorentz are the Yang-Mills Lagrangian densities which all have the form $$𝔏 = -¼ k_{ab} \sqrt{|g|} g^{μρ} g^{νσ} F^a_{μν} F^b_{ρσ},$$ where the gauge group metric $$k_{ab}$$ is assumed to be adjoint-invariant, which entails that $$f_{abc}$$ becomes fully anti-symmetric when lowered with the metric: $$f_{abc} = k_{ad} f^d_{bc} = -k_{bd} f^d_{ac} = -f_{bac}$$.

The corresponding response fields and sources are: $$𝔊^{μν}_c = k_{cd} \sqrt{|g|} g^{μρ} g^{νσ} F^d_{ρσ}, \hspace 1em 𝔍^μ_a = 0.$$ This generalizes the Maxwell field, which has only one Lie basis $$\left(Y_0\right)$$, whose Lie indices are all tacit, and where the sole metric component is $$k = ε_0 c$$, where $$ε_0$$ is the vacuum permittivity and $$c$$ the vacuum speed of light.

For the Yang-Mills response fields, $$FG - GF = 0$$ is always true, as a direct consequence of the adjoint-invariance of the gauge group metric $$k_{ab}$$, and resulting anti-symmetry of $$f_{abc}$$, so the right-hand side of the continuity equation is 0: $$dJ + AJ + JA = 0$$.

Gravity: "Kinematics". The Cartan Fields
For gravity, framed in the setting of a Riemann-Cartan geometry, the main players are, respectively, the frame one-form $$θ^a = h^a_μ dx^μ$$, connection one-form $$ω^c_b = Γ^c_{μb} dx^μ$$, torsion two-form $$Θ = ½ T^c_{μν} dx^μ dx^ν$$ and curvature two-form $$Ω = ½ R^a_{bμν} dx^μ dx^ν$$. The indices $$a,b,c,⋯ = 0, 1, 2, 3$$ are associated with the frame one-forms and their dual vectors $$e_a = e^μ_a ∂_μ$$, with $$e^μ_a h^a_ν = δ^μ_ν$$ and $$h^a_μ e^μ_b = δ^a_b$$. So, the $$e$$ and $$h$$ matrices are inverses of one another.

The literature has the unfortunate convention of referring to the coefficients of the connection as "spin coefficients" while - when referred to a coordinate frame - "connection coefficients". This obscures the fact that they are the components of the same object that transforms as a tensor with respect to two of its indices, but affinely with respect to the third. The $$θ$$-frame components, given by $$ω^c_b = Γ^c_{ab} θ^a$$, are related to the coordinate-frame components $$ω^ρ_ν = Γ^ρ_{μν} dx^μ$$ by $$Γ^c_{ab} = e^μ_a \left(∂_μ e^ρ_b + Γ^ρ_{μν} e^ν_b\right) h^c_ρ.$$ The $$a/μ$$ and $$c/ρ$$ index positions transform linearly, while the $$b/ν$$ position transforms only affinely (which means: "linear + offset").

The torsion is given by: \begin{align} T^a_{μν} &= ∂_μ h^a_ν - ∂_ν h^a_μ + Γ^a_{μc} h^c_ν - Γ^a_{νc} h^c_μ \\ &= ∂_μ h^a_ν - ∂_ν h^a_μ + Γ^a_{μν} - Γ^a_{νμ} \end{align} This can be written equivalently as the First Cartan Structure Equation: $$dθ^a + ω^a_c θ^c = Θ^a.$$

The curvature is given by: \begin{align} R^a_{bμν} = ∂_μ Γ^a_{νb} - ∂_ν Γ^a_{μb} + Γ^a_{μc} Γ^c_{νb} - Γ^a_{νc} Γ^c_{μb}, \end{align} or equivalently as the Second Cartan Structure Equation $$dω^a_b + ω^a_c ω^c_b = Ω^a_b.$$

Following from this are the "Bianchi identities": $$dΘ^a + ω^a_c Θ^c = Ω^a_c θ^c, \hspace 1em dΩ^a_b + ω^a_c Ω^c_b = Ω^a_c ω^c_b.$$

Gravity: "Dynamics"
As with the gauge "dynamic" fields, corresponding to the frame one-forms $$θ^b$$ and connection one forms $$ω^b_a$$, respectively, are the "sources" $$J_b$$ and $$J_a^b$$, and to the torsion two-forms $$Θ^b$$ and curvature two-forms $$Ω^b_a$$, respectively, the "response" fields $$G_b$$ and $$G^a_b$$, such that the total variational in the Lagrangian 4-form $$L$$ can be written as: $$ΔL = Δθ^b J_b + Δω^b_a J^a_b - ΔΘ^b G_b - ΔΩ^b_a G^a_b.$$ Using similar algebra as before, with $$Δdθ^b = dΔθ^b$$ and $$Δdω^b_a = dΔω^b_a$$ and integrating by parts, we can reduce this to: $$ΔL = d\left(-Δθ^b G_b - Δω^b_a G^a_b\right) + Δθ^b \left(J_b - dG_b + G_c ω^c_b\right) + Δω^b_a \left(J^a_b - dG^a_b - ω^a_c G^c_b + G^a_c ω^c_b - θ^a G_b\right).$$ From this results the boundary term $$-Δθ^b G_b - Δω^b_a G^a_b$$ and, the following Euler-Lagrange equations: $$dG_b - G_c ω^c_b = J_b, \hspace 1em dG^a_b + ω^a_c G^c_b - G^a_c ω^c_b + θ^a G_b = J^a_b,$$ and, derived from these, the "continuity" equations: $$dJ_b - J_c ω^c_b = -G_c Ω^c_b, \hspace 1em dJ^a_b + ω^a_c J^c_b - J^a_c ω^c_b + θ^a J_b = Ω^a_c G^c_b - G^a_c Ω^c_b + Θ^a G_b.$$

The Cartan Fields As Potentials And Field Strengths
The frame and connection one-forms together form the analogue of the potential one-form: $$A = θ^a Y_a + ω^a_b Y_a^b.$$ The equation $$dA + A^2 = F$$ yields the torsion and curvature two-forms together as the field strength $$F = Θ^a Y_a + Ω^a_b Y_a^b,$$ provided one adopts the following Lie-brackets for the basis elements $$Y_a$$ and $$Y_a^b$$: $$\left[Y_a,Y_c\right] = 0, \hspace 1em \left[Y_a^b,Y_c\right] = +δ^b_c Y_a, \hspace 1em \left[Y_a,Y_c^d\right] = -δ^d_a Y_c, \hspace 1em \left[Y_a^b,Y_c^d\right] = +δ^b_c Y_a^d - δ^d_a Y_c^b.$$ Those are the Lie brackets for the General Affine Group $$GA(4)$$.

For the dynamic fields, we have the following: $$G = G_b y^b + G^a_b y^b_a, \hspace 1em J = J_b y^b + J^a_b y^b_a$$ with the dual basis $$y^b$$ and $$y^b_a$$ and following algebra: $$y^a·Y_c = δ^a_c = Y_c·y^a, \hspace 1em y^a·Y_c^d = 0 = Y_c^d·y^a, \hspace 1em y^a_b·Y_c = 0 = Y_c·y^a_b, \hspace 1em y^a_b·Y_c^d = δ^a_c δ^d_b = Y_c^d·y^a_b, \\ y^a Y_c - Y_c y^a = -y^a_c, \hspace 1em y^a Y_c^d - Y_c^d y^a = δ^a_c y^d, \hspace 1em y^a_b Y_c - Y_c y^a_b = 0, \hspace 1em y^a_b Y_c^d - Y_c^d y^a_b = δ^a_c y^d_b - δ^d_b y^a_c$$

Notice that there is no space-time metric involved anywhere yet. The metric doesn't come from here, but (ultimately) from the field dynamics and from the players involved in them.

Also: this applies more generally than to Riemann-Cartan geometries; which specialize from metric affine geometries by placing a constraint on the metric. The constraint embodies the "everything is locally Minkowski" condition and effectively reduces the $$GL(4)$$ part of $$A$$ and $$F$$ (i.e. the $$ω^a_b Y_a^b$$ and $$Ω^a_b Y_a^b$$ fields) into an $$SO(3,1)$$ gauge field and, thus, the overall $$GA(4)$$ field into an $$ISO(3,1)$$ gauge field.

A further constraint on the torsion $$Θ^a = 0$$ reduces Riemann-Cartan geometry to (pseudo-)Riemannian geometry. General Relativity is historically formulated on a Riemannian geometry, defined by the Einstein-Hilbert action. Its lifting to a Riemann-Cartan is given by the Palatini action, which forces zero torsion as a constraint. Removing the constraint results in the Einstein-Cartan action and the Einstein-Cartan formulation, which differs slightly from General Relativity, be it Einstein-Hilbert-based or Palatini-based.

Einstein-Hilbert and Einstein-Cartan Dynamics
The Lagrangian densities, corresponding to the source-free gravity field (i.e. the field outside of matter), are of the form $$L = k ε_{abcd} Ω^a_e 𝔥^{eb} θ^c θ^d,$$ for some constant $$k$$, where $$ε_{abcd}$$ is the completely anti-symmetric tensor density defined by $$ε_{0123} = 1$$ and the tensor density $$𝔥^{eb} = g^{eb} \sqrt{|g|}$$ is the inverse metric multiplied by the determinant of the metric. It is referred to frame components and can (without much loss of generality) be assumed to be constant with a determinant equal to 1, but we won't actually do that just yet.

The dynamics can actually be expressed in an entirely self-contained way in terms of this object. You don't actually need the metric, per se, just the $$𝔥$$ tensor density. You can define the metric $$g$$ in terms of the density. In $$n$$ dimensions their determinants are related by: $$|\det 𝔥| = |\det g|^{n/2 -1}$$, and for $$n = 4$$, $$|\det 𝔥| = |\det g|$$. Thus, $$g^{ab} = 𝔥^{ab}/\sqrt{|\det 𝔥|}$$; and $$g_{ab}$$ may be defined as the inverse of $$g^{ab}$$. Only in $$n = 2$$ dimensions does $$g$$ fail to be recovered from $$𝔥$$. But for $$n = 2$$, $$g$$ can be freely rescaled without affecting the Lagrangian density, and you can just set $$g = 𝔥$$, and treat tensor densities as ordinary tensors.

Riemann-Cartan geometries are metric-affine geometries whose metrics are constrained to have zero covariant-derivative; specifically with metric having signature $$(+,+,+,-)$$ (or $$(+,-,-,-)$$). For the tensor density $$𝔥$$, the covariant derivative is given by: $$ℌ^{ab} = d𝔥^{ab} + ω^a_c 𝔥^{cb} + ω^b_c 𝔥^{ac} - ω^c_c 𝔥^{ab}.$$ This object, in turn, satisfies an analogue of the "Bianchi identity": $$dℌ^{ab} + ω^a_c ℌ^{cb} + ω^b_c ℌ^{ac} - ω^c_c ℌ^{ab} = Ω^a_c 𝔥^{cb} + Ω^b_c 𝔥^{ac} - Ω^c_c 𝔥^{ab}.$$

If the constraint $$ℌ_{ab} = 0$$ is applied, then it follows that $$Ω^a_c 𝔥^{cb} + Ω^b_c 𝔥^{ac} - Ω^c_c 𝔥^{ab} = 0,$$ from which we may derive $$Ω^c_c = 0$$ and $$Ω^{ab} = -Ω^{ba}$$, where the index is raised with $$g$$: $$Ω^{ab} = Ω^a_c g^{cb}$$. If we were also to assume $$𝔥$$ to be constant and non-singular, then a similar result would apply to the connection coefficients: $$ω^c_c = 0$$ and $$ω^{ab} = -ω^{ba}$$, where $$ω^{ab} = ω^a_c g^{cb}$$. This reduction applies in $$n = 4$$ dimensions, and in any number dimensions except $$n = 2$$.

This reduces the number of independent 1-forms in the connection and curvature 16 to 6 and the corresponding symmetry group from $$GL(4)$$ to $$SO(3,1)$$, provided the metric has signature $$(+,+,+,-)$$. Otherwise, more generally, the reduction is to $$SO(g)$$.

The corresponding response fields are: \begin{align} G_c &= 0, & G^e_a &= -k ε_{abcd} 𝔥^{eb} θ^c θ^d, \\ J_c &= 2 k ε_{abcd} Ω^a_e 𝔥^{eb} θ^d, & J^e_a &= 0. \end{align} The Euler-Lagrange equations reduce to: $$0 = J_b, \hspace 1em dG^a_b + ω^a_c G^c_b - G^a_c ω^c_b = 0,$$ and the "continuity" equations to: $$0 = dJ_b - J_c ω^c_b = 0, \hspace 1em 0 = θ^a J_b = Ω^a_c G^c_b - G^a_c Ω^c_b.$$

These are the dynamics for the Einstein-Cartan action. The dynamics for the Palatini action are obtained from this by constraining the torsion to be 0: $$Θ^a = 0$$. However, this will already follow from the second set of Euler-Lagrange equations. It's only in the presence of matter sources that a difference between Palatini and Einstein-Cartan arises.

To aid in writing out the component forms, define $$ε = \frac{ε_{abcd} θ^a θ^b θ^c θ^d}{4!}, \hspace 1em ε_a = e_a ˩ ε = \frac{ε_{abcd} θ^b θ^c θ^d}{3!}, \\ ε_{ab} = e_b ˩ ε_a = \frac{ε_{abcd} θ^c θ^d}{2!}, \hspace 1em ε_{abc} = e_c ˩ ε_{ab} = \frac{ε_{abcd} θ^d}{1!}$$ and note that $$ε_{abcd} = e_d ˩ ε_{abc}$$. Products with the frame one-forms include: $$θ^a ε = 0, \hspace 1em θ^b ε_a = δ^b_a ε, \hspace 1em θ^c ε_{ab} = δ^c_b ε_a - δ^c_a ε_b, \\ θ^d ε_{abc} = δ^d_c ε_{ab} - δ^d_b ε_{ac} - δ^d_a ε_{bc}, \hspace 1em θ^e ε_{abcd} = δ^e_d ε_{abc} - δ^e_c ε_{abd} + δ^e_b ε_{acd} - δ^e_a ε_{bcd}.$$ It will also prove convenient to define $$δ^{ab}_{cd} = δ^a_c δ^b_d - δ^a_d δ^b_c$$, so we can write $$θ^a θ^b ε_{cd} = δ^{ab}_{cd} ε, \hspace 1em θ^a θ^b ε_{cde} = δ^{ab}_{de} ε_c - δ^{ab}_{ce} ε_d + δ^{ab}_{cd} ε_e.$$

The Lagrangian density and its derivatives can, thus be written as: $$L = 2k \sqrt{|g|} Ω^{ab} ε_{ab}, \hspace 1em G_c = 0, \hspace 1em G^e_a = -2k \sqrt{|g|} g^{eb} ε_{ab}, \hspace 1em J_c = 2 k \sqrt{|g|} Ω^{ab} ε_{abc}, \hspace 1em J^e_a = 0.$$

The Lagrangian 4-form, we get: $$Ω^{ab} ε_{ab} = ½ R^{ab}_{cd} θ^c θ^d ε_{ab} = ½ R^{ab}_{cd} δ^{cd}_{ab} = R^{ab}_{ab} = R^b_b = R,$$ where the Ricci tensor is defined by $$R_{ab} = R^c_{acb}$$ and curvature scalar by $$R = R^a_a = g^{ab} R_{ab}$$, thus: $$L = 2k \sqrt{|g|} R ε.$$

For the "source" $$J_c$$, we have: \begin{align} Ω^{ab} ε_{abc} &= ½ R^{ab}_{de} θ^d θ^e ε_{abc} \\ &= ½ R^{ab}_{de} \left(δ^{de}_{bc} ε_a - δ^{de}_{ac} ε_b + δ^{de}_{ab} ε_c\right) \\ &= \left(R^{ab}_{bc} ε_a - R^{ab}_{ac} ε_b + R^{ab}_{ab} ε_c\right) \\ &= \left(-R^a_c ε_a - R^b_c ε_b + R ε_c\right) \\ &= -2\left(R^a_c - ½ δ^a_c R\right) ε_a \end{align} with the tensor $$R^a_c - ½ δ^a_c R$$ being the Einstein tensor, normally written as $$G^a_c$$ (but I have a notation conflict here).

Thus $$J_c = -4k \sqrt{|g|} \left(R^a_c - ½ δ^a_c R\right) ε_c.$$

Without going through the algebra in detail, the Euler-Lagrange equation $$G^a_b$$ leads to: $$0 = dG^a_b + ω^a_c G^c_b - G^a_c ω^c_b = -2k ε_{abcd} 𝔥^{eb} Θ^c θ^d,$$ which reduces to $$0 = -k 𝔥^{eb} \left(T^c_{bc} ε_a - T^c_{ac} ε_b + T^c_{ab} ε_c\right) = -k \sqrt{|g|} g^{eb} \left(T^c_{ab} - T^d_{db} δ^c_a - T^d_{ad} δ^c_b\right) ε_c$$ Thus, $$T^c_{ab} = T^d_{db} δ^c_a + T^d_{ad} δ^c_b.$$ For geometries of dimension other than $$n = 2$$, this leads to $$T^d_{ad} = 0$$ and $$T^d_{db} = 0$$, and therefore to $$T^c_{ab} = 0$$. So, even in Riemann-Cartan geometry, the Einstein-Cartan Lagrangian leads to zero torsion outside of matter.

Effective Reduction To ISO(3,1) Gauge Fields
If frame fields are restricted to those which give rise to constant components for the metric, then we may assume that the components $$𝔥^{ab}$$ are all constant, and therefore also the components $$g^{ab}$$ and $$g_{ab}$$. Then the constraint on $$𝔥$$ gives rise to $$ω^{ab} = -ω^{ba}$$ and $$Ω^{ab} = -Ω^{ba}$$, where $$ω^{ab} = ω^a_c g^{cd}$$ and $$Ω^{ab} = Ω^a_c g^{cd}$$.

This can be inverted to the following: $$ω^a_c = ω^{ab} g_{bc}$$ and $$Ω^a_c = Ω^{ab} g_{bc}$$. Then the relevant parts of the potential $$A$$ and field strength $$F$$ can be rewritten as $$ω^a_c Y_a^c = ω^{ab} g_{bc} Y_a^c = ½ ω^{ab} M_{ab}, \hspace 1em Ω^a_c Y_a^c = Ω^{ab} g_{bc} Y_a^c = ½ Ω^{ab} M_{ab},$$ where $$M_{ab} = g_{bc} Y_a^c - g_{ac} Y_b^c,$$ giving rise to the Lie brackets $$\left[M_{ab}, M_{cd}\right] = g_{bc} M_{ad} - g_{bd} M_{ac} - g_{ac} M_{bd} + g_{ad} M_{bc}.$$ Those are the Lie brackets for the gauge group $$SO(g)$$; or, if $$g$$ has signature $$(+,+,+,-)$$ or $$(+,-,-,-)$$, $$SO(3,1)$$. Combined with the $$Y_c$$, $$\left[M_{ab}, Y_c\right] = g_{bc} Y_a - g_{ac} Y_b,$$ this yields the brackets for $$ISO(3,1)$$.

The corresponding potential and field strength are $$A = ½ ω^{ab} M_{ab} + θ^a Y_a, \hspace 1em F = ½ Ω^{ab} M_{ab} + Θ^a Y_a.$$

For the dynamics, the variational in the Lagrangian becomes: $$ΔL = Δθ^a P_a - ½ Δω^{ab} S_{ab} - ΔΘ^a p_a + ½ ΔΩ^{ab} s_{ab},$$ with the dynamic fields defined by: $$p_a = G_a, \hspace 1em s_{ab} = ½ \left(g_{ac} G^c_b - g_{bc} G^c_a\right), \hspace 1em P_a = J_a, \hspace 1em S_{ab} = ½ \left(g_{ac} J^c_b - g_{bc} J^c_a\right),$$ and, with integration by parts, reduces to: \begin{align} ΔL &= d\left(-Δθ^a p_a + ½ Δω^{ab} S_{ab}\right) \\ &+ Δθ^b \left(P_b - dp_b + ω^a_b p_a\right) \\ &+ ½ Δω^{ab} \left(ds_{ab} - ω^c_a s_{cb} - ω^c_b s_{ac} + θ_a p_b - θ_b p_a - S_{ab}\right) \end{align} giving rise to the boundary term $$-Δθ^a p_a + ½ Δω^{ab} S_{ab}$$ and the Euler-lagrange equations $$dp_b - ω^a_b p_a = P_b \hspace 1em ds_{ab} - ω^c_a s_{cb} - ω^c_b s_{ac} = J_{ab} ≡ S_{ab} + θ_a p_b - θ_a p_b.$$ A curved space-time version of the spin-orbit decomposition arises, with the "orbital" part of the 3-form $$J_{ab}$$ given by $$L_{ab} ≡ θ_b p_a - θ_a p_b$$, where $$θ_a = g_{ac} θ^c$$. The corresponding "continuity" equations are: $$dP_b - ω^a_b P_a = -Ω^a_b p_a, \hspace 1em dJ_{ab} - ω^c_a J_{cb} - ω^c_b J_{ac} = -Ω^c_a s_{cb} - Ω^c_b s_{ac}.$$

The fields $$P_b = 𝔓^μ_b ∂_μ ˩ d⁴x$$, $$S_{ab} = 𝔖^μ_{ab} ∂_μ ˩ d⁴x$$ and $$J_{ab} = 𝔍^μ_{ab} ∂_μ ˩ d⁴x$$ correspond, respectively, to the 3-currents for momentum, internal angular momentum (a.k.a. "spin") and total angular momentum. The components $$𝔓^μ_b$$ play the role of the canonical stress tensor density. In the continuity equation, expressed in terms of $$S_{ab}$$: $$dS_{ab} - ω^c_a S_{cb} - ω^c_b S_{ac} + θ_a P_b - θ_b P_a = -Ω^c_a s_{cb} - Ω^c_b s_{ac} + Θ_a p_b - Θ_b p_a,$$ the term $$θ_a P_b - θ_b P_a = \left(𝔓_{ab} - 𝔓_{ba}\right) ∂_μ ˩ d⁴x$$, where $$𝔓_{ab} = g_{bc} 𝔓^c_a = g_{bc} h^c_μ 𝔓^μ_a$$, links the anti-symmetric part of $$𝔓_{ab}$$ to the continuity equation for the spin 3-current $$S_{ab}$$.

In U(1) gauge theory magnetic and the electric field are gauge invariant quantities. In non-abelian gauge theory however, their analouges are not gauge invariant and therefore not measureable (except if the coupling goes to zero). The only things that are, are the tuneling rate, topological charge and the Lagrangian itself.