# Notes on gravity as a gauge theory

Gravity has often been called a gauge theory of the Poincaré or Lorentz group. Here, I develop general relativity in direct analogy to Yang-Mills theory, avoiding geometry entirely1. None of this is original, but I have tried to simplify the presentation compared to the literature, where the similarities and differences between the two theories are often unclear.

### Gauge fields and field strengths

The Poincaré algebra is:

$[P_a, P_b] = 0$

$[P_a, M_{bc}]=\eta_{ab}P_c - \eta_{ac} P_b$

$[M_{ab}, M_{cd}] = \eta_{ad}M_{bc}+\eta_{bc}M_{ad} - \eta_{bd}M_{ac}-\eta_{ac}M_{bd}$

Roman letters $a,b, \cdots$ are gauge indices, while Greek letters $\mu, \nu, \cdots$ are coordinate indices. We use “mathematician’s convention” for the generators where the $i$ is absorbed: $T_{math}=i T_{physics}$. We proceed just as in Yang-Mills theory, taking the Poincaré group as the gauge group. It has 10 generators: 4 translations $P_a$ and 6 rotations/boosts $M_{ab}$.

Introduce the covariant derivative:

$\displaystyle D_\mu = \partial_\mu - e_\mu^a P_a - \frac{1}{2}\omega^{ab}_\mu M_{ab}$

where $e_\mu^a$ (the vielbein) and $\omega^{ab}_\mu$ (the spin connection) are the gauge fields associated with translations and rotations, respectively. Note the units: $P_a$ has unit 1, so $e_\mu^a$ is unitless, while $M_{ab}$ is unitless, so $\omega^{ab}_\mu$ has unit 1. We can take $\omega^{ab}_\mu$ to be antisymmetric in $ab$ since $M_{ab}$ is antisymmetric. The field strengths are found in the usual way:

\begin{aligned} F_{\mu\nu}&=D_\mu D_\nu - D_\nu D_\mu \\ &= -C^a_{\mu\nu}P_a -\frac{1}{2}R^{ab}_{\mu\nu}M_{ab} \end{aligned}

where we have defined the field strengths $C^a_{\mu\nu}$ (the torsion) and $R^{ab}_{\mu\nu}$ (the curvature tensor).
We obtain:

$C^a_{\mu\nu}=\partial_\mu e^a_\nu - \partial_\nu e^a_\mu - \omega^{a}_{\mu b} e^b_\nu + \omega^{a}_{\nu b} e^b_\mu$

$R^{ab}_{\mu\nu}=\partial_\mu \omega_\nu^{ab} - \partial_\nu \omega_\mu^{ab}-\omega_\mu^{ac}\omega_{\nu c}^{\;\;\;\;b} + \omega_\nu^{ac}\omega_{\mu c}^{\;\;\;\;b}$

As usual, we raise and lower indices using $\eta_{ab}$ and $\eta^{ab}$.

General relativity is obtained by setting the torsion $C^a_{\mu\nu}=0$. Certainly, theories with torsion have been extensively considered, but we will not do so here. Experimental data have not ruled out theories involving both torsion and curvature. However, the bottom-up construction of the Lagrangian of an interacting massless spin-2 particle produces general relativity2.

This constraint allows us to solve for the spin connection in terms of the vielbein. After some calculation (e.g. listing out all possible terms and matching coefficients), the answer is:

$\displaystyle \omega_\mu^{ab}=\frac{1}{2}(e^{\rho b}\partial_\mu e_\rho^a-e^{\rho a}\partial_\mu e_\rho^b+ e^{\rho a} e^{\sigma b} \partial_\rho g_{\mu\sigma}-e^{\rho b}e^{\sigma a}\partial_\rho g_{\mu\sigma} )$

where $g_{\mu\nu}=e_\mu^a \eta_{ab} e_\nu^b$.

### Representations and Lagrangians

Just as in Yang-Mills theory, the Poincaré group here acts as an internal symmetry group. Fields transform as a finite-dimensional representation of the Lorentz algebra, and transform trivially under translations3: $P_a=0$. This has an important consequence for constructing Lagrangians. Recall that the gauge field $A_\mu(x)=A^a_\mu(x) T^a$ in Yang-Mills theory transforms as

$A_\mu\rightarrow U A_\mu U^{-1} + (\partial_\mu U) U^{-1}$

under a gauge transformation $U(x)$. The $(\partial_\mu U) U^{-1}$ is required to cancel out the $(\partial_\mu U)\phi(x)$ in the transformation of $\partial_\mu\phi(x)$. However, since $P_a=0$, this additional term is not needed. $e^a_\mu$ is already gauge-covariant and can be placed directly in the Lagrangian.

The simplest term is:

$\displaystyle \mathcal{S}_\Lambda = \frac{\Lambda}{4!} \int \epsilon_{abcd}e^a e^b e^c e^d$

where $e^a=e^a_\mu dx^\mu$ is a 1-form and $\epsilon_{abcd}$ is the totally antisymmetric symbol4. This is the cosmological constant. It is equivalent to the standard form $\Lambda\int d^4 x \sqrt{-g}$.

On the other hand, the spin connection $\omega^{ab}_\mu$ does show up in the gauge transformation, so we must use the field strength $R^{ab}_{\mu\nu}$ in the Lagrangian. The next simplest term is then:

$\displaystyle \mathcal{S}_{EH} = \frac{M_{Pl}^2}{3}\int \epsilon_{abcd} e^a e^b R^{cd}$

where $R^{cd}=R^{cd}_{\mu\nu}dx^\mu dx^\nu$ is a 2-form. This is the Einstein-Hilbert action. Unlike Yang-Mills, we are permitted a term that is only linear in the field strength $R^{cd}$.

### Coupling to matter fields

Flat-space Lagrangians contain terms with global Lorentz indices, such as $\partial_\mu \varphi$ and $A_\mu$. We would like these to transform under the local Lorentz group with indices $a, b, \cdots$. The only object that can switch between global and local indices is $e_\mu^a$, or its inverse, $e^\mu_a$. Thus, the general prescription for coupling a flat-space Lagrangian to gravity is:

1. Contract all tensors with $e_\mu^a$ or $e^\mu_a$.
2. Make flat-space invariants use local indices: $\eta_{\mu\nu}\rightarrow \eta_{ab}$, $\epsilon_{\mu\nu\rho\sigma}\rightarrow \epsilon_{abcd}$.
3. Use covariant derivatives: $\partial_\mu\rightarrow \partial_\mu-\frac{1}{2}\omega_\mu^{ab} M_{ab}$.

Note that this even works on the volume form $d^4 x$, producing the familiar invariant measure $d^4 x \sqrt{-g}$:

$\displaystyle d^4 x = \frac{1}{4!}\epsilon_{\mu\nu\rho\sigma} dx^\mu dx^\nu dx^\rho dx^\sigma \rightarrow \frac{1}{4!} \epsilon_{abcd} e^a_\mu e^b_\nu e^c_\rho e^d_\sigma dx^\mu dx^\nu dx^\rho dx^\sigma$

For example, a scalar field coupled to gravity has the action:

$\displaystyle \mathcal{S} = \frac{1}{2\cdot 4!}\int \epsilon_{bcdf} e^b e^c e^d e^f (e_a^\mu e^{\nu a} \partial_\mu\varphi\partial_\nu\varphi - m^2 \varphi^2)$

An advantage of the vielbein formalism is that spinors can be coupled to gravity. For Dirac spinors, the Dirac matrices should also be converted to local indices $\gamma^\mu\rightarrow \gamma^a$, since they satisfy the Clifford algebra $\{\gamma^a,\gamma^b\}=2\eta^{ab}$. The Lagrangian for a massless fermion becomes:

$\mathcal{L}=i\bar\Psi \gamma^a e_a^\mu (\partial_\mu-\omega^{bc}_\mu M_{bc})\Psi$

where

$\displaystyle M_{ab}=S_{ab}=\frac{1}{4}[\gamma_a,\gamma_b]$

1 This is ironic from a historical perspective, since Yang and Mills were inspired by general relativity. Of course, in physics, there are many ways to skin a cat.

2 Schwartz, Matthew. Quantum Field Theory and the Standard Model, Ch. 8.

3 Thus, you could say gravity is the gauge theory of the Lorentz group instead. However, we had to introduce the vielbein as part of the covariant derivative in order to get the correct theory. So there is a slight wrinkle in the analogy.

4 Unlike Yang-Mills theory, we cannot write the Lagrangian using the “abbreviated” fields $e=e^a_\mu P_a dx^\mu$. In fact, $e e$ vanishes due to $[P_a,P_b]=0$.