A) If by gauge theory one means the Yang-Mills type theory with its compact symmetry group, then claim that GR can be formulated as a Yang-Mills theory is just meaningless, because the symmetry group of GR is non-compact. However, the “limited similarities” between the dynamical variables of the two theories (Claim 1) and, in particular, the tetrad formalism of GR make general relativity very “similar” to a gauge theory. In fact, while the 4-dimensional GR action integral is definitely not a gauge theory action (Claim 2), the (1+2)-dimensional GR, without cosmological constant, is “equivalent” to a gauge theory with gauge group \mathcal{P}(1,2) = T(3) \rtimes SO(1,2) and a Chern-Simons action integral (Claim 3). Even in this case, by “equivalence” one means on-shell equivalence between the gauge group \mathcal{P}(1,2) and the group of diffeomorphisms, the symmetry group of the E-H action \int d^{3}x \sqrt{|g|} R. Adding a cosmological constant \Lambda to the 3-dimensional GR will only change the gauge group from the Poincare’ group \mathcal{P}(1,2) to the de Sitter SO(1,3) or anti-de Sitter SO(2,2) group, depending on the sign of \Lambda. The point is the following: the invariant metric (inner product) on the Lie algebra in question needs to be non-degenerate so that the action integral contains kinetic terms for each component of the gauge field. In particular, a Chern-Simons action exists for the gauge group \mathcal{P}(1,n-1) if and only if n = 3: For a general n, a Lorentz invariant bilinear expression in the generators would have to be of the form C = c_{1} P^{a}P_{a} + c_{2} J^{ab}J_{ab}, for some constants c_{1} and c_{2}. However, [C , P_{b}] = 0 forces us to set c_{2} = 0 and with it disappear our hope in constructing a non-degenerate bilinear form on the Lie algebra. So, for a general n, there will be no Chern-Simons 3-form \mbox{tr} \left(\mathbb{A} \wedge \mbox{d} \mathbb{A} \right) = \mbox{tr} \left( X_{a}X_{b}\right) A^{a} \wedge \mbox{d} A^{b} for \mathcal{P}(1,n-1). For n = 3 we are lucky because, in this case we can take C = \frac{1}{2} \epsilon_{abc}P^{a}J^{bc} \equiv P^{a}J_{a}. It is easy to see that [C , P] = [C , J] = 0 (i.e., Poincare invariant) as well as non-degenerate. Thus we are led to define the following invariant inner product on the Lie algebra \mathfrak{p}(1,2): \mbox{tr} \left( P_{a} J_{b}\right) = \eta_{ab} \ , \ \ \ \mbox{tr} \left( P_{a}P_{b}\right) = \mbox{tr} \left( J_{a}J_{b}\right) = 0 . \ \ \ \ (A.0) With this inner product, a Chern-Simons action for the gauge group \mathcal{P}(1,2) will exists and, as we shall see below, coincide with Einstein-Hilbert action on (1+2)-dimensional space-time.
As usual, since I don’t like to use ambiguous language in my posts, I will try to mathematically clarify the three claims that I made above, at least in a sketchy way.
(Claim 1) about the “limited similarities”:
Basically, one would like to set up a 1-to-1 correspondence (dictionary if you like) between gravitational variables and gauge theory variables and see if such a dictionary is complete. If for every GR variable one can find a gauge theory variable, one then concludes that GR is certainly equivalent to a gauge theory. Even though we have already spoiled the fun, let us see how far one can go with analogy. Recall that under a general coordinate transformation, the gauge potential \mathbb{A}_{\alpha}(x) = A_{\alpha}^{C}(x)T_{C}, which is a (matrix) field taking values in the Lie algebra of the gauge group, transforms as a co-vector up to a gauge transformation. That is, when x^{\alpha} \to \bar{x}^{\alpha}(x), the Yang-Mills connection transforms according to \bar{\mathbb{A}}_{\alpha} (\bar{x}) = \frac{\partial x^{\beta}}{\partial \bar{x}^{\alpha}} \left( U(x) \mathbb{A}_{\beta}(x) U^{-1}(x) + U(x) \partial_{\beta} U^{-1}(x)\right) \ . \ \ \ (A.1) This means that the space-time index \alpha carried by the gauge field matrix (\mathbb{A}_{\alpha})^{a}{}_{b}(x) transforms (as it should) by the inverse Jacobian matrix \frac{\partial x^{\beta}}{\partial \bar{x}^{\alpha}}, while the matrix (i.e., internal) indices (a,b) of (\mathbb{A}_{\alpha})^{a}{}_{b}(x) transform by the arbitrary gauge functions U^{a}{}_{b}(x). The corresponding suspect in GR is the Christoffel connection \Gamma^{\mu}_{\alpha \nu}(x) with its usual transformation law \bar{\Gamma}^{\mu}_{\alpha \nu}(\bar{x}) = \left( \frac{\partial \bar{x}^{\mu}}{\partial x^{\rho}} \Gamma^{\rho}_{\beta \sigma} (x) \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} + \frac{\partial \bar{x}^{\mu}}{\partial x^{\sigma}} \frac{\partial}{\partial x^{\beta}} ( \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} ) \right) \frac{\partial x^{\beta}}{\partial \bar{x}^{\alpha}} \ . \ \ \ (A.2) Now, for each \alpha = 0,1,2,3, let us define the following 4 \times 4 field matrices \Gamma^{\mu}_{\alpha \nu}(x) \equiv ( \Gamma_{\alpha})^{\mu}{}_{\nu} (x) \ . Let us also introduce the following 4 \times 4 matrix (and its inverse) \frac{\partial \bar{x}^{\nu}}{\partial x^{\rho}} = V^{\nu}{}_{\rho} (x) \ , \ \ \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} = (V^{-1})^{\sigma}{}_{\nu} (x) \ . With these definitions, equation (A.2) takes on the following matrix form \bar{\Gamma}_{\alpha}(\bar{x}) = \left( V(x) \Gamma_{\beta}(x) V^{-1}(x) + V(x) \partial_{\beta} V^{-1}(x) \right) \frac{\partial x^{\beta}}{ \partial \bar{x}^{\alpha}} \ . \ \ \ (A.3) Now, I will make the following false argument: Since Eq(A.1) (which is the transformation law of the gauge field matrix \mathbb{A}_{\alpha}(x)) is identical to Eq(A.3) (which is the transformation law of the Christoffel matrix field \Gamma_{\alpha}(x)) then \Gamma_{\alpha}(x) is the gauge field (i.e., connection) associated the GL(4 , \mathbb{R}) gauge group. In other words, GR is a gauge theory! Can you figure out why my argument is false? What is wrong with the saying that Eq(A.1) is identical to Eq(A.3)? So, \Gamma_{\alpha} \leftrightarrow \mathbb{A}_{\alpha} is the first entry in our Gravity-Gauge dictionary. Okay, let us populate the dictionary with more objects. For the Riemann tensor R^{\mu}{}_{\nu \alpha \beta} we define the anti-symmetric matrix R^{\mu}{}_{\nu \alpha \beta} (x) \equiv ( \mathbb{B}_{\alpha \beta})^{\mu}{}_{\nu}(x) \ . Thus, the definition of the Riemann tensor, in terms of the Christoffel connection, translates to the following matrix equation \mathbb{B}_{\alpha \beta} = \partial_{\alpha} \Gamma_{\beta} - \partial_{\beta}\Gamma_{\alpha} + [ \Gamma_{\alpha} , \Gamma_{\beta} ] \ . But this is exactly like the definition of the Yang-Mills field strength matrix \mathbb{F}_{\alpha \beta}(x) = F^{C}_{\alpha \beta}(x) T_{C} \mathbb{F}_{\alpha \beta}(x) = \partial_{\alpha} \mathbb{A}_{\beta} - \partial_{\beta}\mathbb{A}_{\alpha} + [ \mathbb{A}_{\alpha} , \mathbb{A}_{\beta} ] \ . And so we have R^{\mu}{}_{\nu \alpha \beta} \leftrightarrow \mathbb{F}_{\alpha \beta}. This seems easy and you can carry on adding more objects, for example a (1,1)-type GR tensor T^{\mu}{}_{\nu} corresponds to a Yang-Mills matrix M with values in the adjoint representation of the gauge group, and then the GR covariant derivative \nabla_{\alpha} translates according to \nabla_{\alpha} T^{\mu}{}_{\nu} \leftrightarrow D_{\alpha} M = \partial_{\alpha} M + [ \mathbb{A}_{\alpha} , M] \ . However, our dictionary ends when we try to contract the space-time indices (\alpha , \beta , ...) with the would be “gauge indices” (\mu , \nu , ...) (why is that?). This means that the Ricci tensor R_{\nu \beta} = R^{\mu}{}_{\nu \mu \beta} and the scalar curvature R = g^{\nu \beta} R_{\nu \beta} do not correspond to well-defined objects in gauge theories. Thus, the Einstein-Hilbert action \int d^{4}x \sqrt{|g|} R does not exist and we conclude that in 4 dimensions GR is not equivalent to a gauge theory as I claimed in (Claim 2) above. We will reach the same conclusion about (Claim 2) in the tetrad formalism below.
B) The tetrad formalism of GR, (Claim 2) and (Claim 3):
Since GR can be, equivalently, formulated in terms of the tetrad field and spin connection, let me say few words about the geometrical meaning and the functioning of the tetrad field and the spin connection. This is important because it is this tetrad formalism that led many people to try to interpret GR as a gauge theory of the Poincare’ group \mathcal{P}(1,3). The idea is to interpret the tetrad field e^{a}{}_{\alpha}(x) as the gauge field associated with translation, while the spin connection \omega_{\alpha}{}^{a}{}_{b}(x) is taken to be the gauge field associated with the Lorentz group SO(1,3). So, what are these fields and in what vector spaces do their indices take values in? We say that space-time of dimension n can be modeled by a smooth n-manifold M if and only if M admits metric of Lorentzian signature. So, we introduce an abstract vector bundle \mathcal{V}^{n} with structure group SO(1,n-1). This means that \mathcal{V}^{n} is equipped with a Minkowskian metric \eta_{ab} = \mbox{diag}(1,-1,-1, \cdots , -1) and a volume form \epsilon_{a_{1} a_{2} \cdots a_{n}}. We assume that \mathcal{V}^{n} has the same topological structure as that of the tangent bundle TM so that isomorphisms exist between \mathcal{V}^{n} and TM. At any p \in M, the “tetrad” e_{\mu}{}^{a} provides a choice of (vector space) isomorphism e_{\mu}{}^{a}(p) : M_{p}T \to \mathcal{V}^{n}, i.e., a \mathcal{V}^{n}-valued 1-form on M e^{a} (x) = e_{\mu}{}^{a}(x) \mbox{d}x^{\mu}. I think it is clear to you that I am using (a,b,c, \cdots ) as Lorentz indices (i.e., they define the geometrical objects on \mathcal{V}^{n}), and (\mu , \nu \cdots ) as tangent space indices (i.e., they define geometrical object on the spacetime manifold M). The metric \eta and the volume form \epsilon on \mathcal{V}^{n} together with isomorphism e_{\mu}{}^{a} between MT and \mathcal{V}^{n} give a metric g_{\mu\nu} = e_{\mu}{}^{a} \ e_{\nu}{}^{b} \ \eta_{ab} \ , on M having the same signature as \eta_{ab} and a volume form \sqrt{|g|} \epsilon_{\mu_{1} \mu_{2} \cdots \mu_{n}} = e_{\mu_{1}}{}^{a_{1}} \ e_{\mu_{2}}{}^{a_{2}} \ \cdots \ e_{\mu_{n}}{}^{a_{n}} \ \epsilon_{a_{1} a_{2} \cdots a_{n}} \ . \ \ \ (B.1) The spin connection \omega, which is an \mathfrak{so}(1,n-1)-valued connection on \mathcal{V}^{n}, can be regarded as a 1-form on M with values in the Lie algebra of SO(1,n-1) \omega^{a}{}_{b}(x) = \omega_{\mu}{}^{a}{}_{b}(x) \ \mbox{d}x^{\mu} \ . \ \ \ \ \ \ (B.2) Now, instead of the fields (g_{\mu\nu} , \Gamma^{\rho}_{\mu\nu}), GR can be (equivalently) described in terms of the fields (e_{\mu}{}^{a} , \omega_{\mu}{}^{a}{}_{b}). The curvature tensor is defined by R_{\alpha \beta}{}^{ab}( \omega ) = e_{\mu}{}^{a} \ e_{\nu}{}^{b} R_{\alpha \beta}{}^{\mu\nu}( \Gamma ) = \partial_{[ \alpha} \omega_{ \beta ]}{}^{ab} + [ \omega_{\alpha} , \omega_{\beta} ]^{ab} \ , or simply as a 2-form on M with values in \wedge^{2}\mathcal{V}^{n}: \mathcal{R}^{a}{}_{b} \equiv \frac{1}{2} R_{\alpha \beta}{}^{a}{}_{b} \ \mbox{d}x^{\alpha} \wedge \mbox{d}x^{\beta} = \left( \mbox{d} \omega + \omega \wedge \omega \right)^{a}{}_{b} \ . \ \ (B.3)
B1) 4-dimensional Gravity and (Claim 2):
In this subsection we will put n = 4 and show that one cannot hope to interpret GR as a gauge theory. Using the above formalism we can now use the 1-form e^{a} and the 2-form \mathcal{R}^{cd} together with the volume form \epsilon_{abcd} on \mathcal{V}^{(1,3)} to construct an invariant (action) integral on M^{(1,3)} (the 4-form e \wedge e \wedge \mathcal{R} on M which takes values in \mathcal{V} \otimes \mathcal{V} \otimes \wedge^{2} \mathcal{V} and maps to \wedge^{4} \mathcal{V}^{(1,3)} can be used to form an invariant integral on M because section of \wedge^{4} \mathcal{V}^{(1,3)} is just a function) S(e , \omega ) = - \frac{1}{2} \int_{M^{(1,3)}} \ \epsilon_{abcd} \ e^{a} \wedge e^{b} \wedge \mathcal{R}^{cd} \ , \ \ \ \ \ \ (B.4) Indeed, this is nothing but the Einstein-Hilbert action written in terms of the fields ( e ,\omega ): To obtain eq(B.4) from the E-H action, write the Ricci scalar in the form \begin{align*}R & = \frac{1}{2} \left( \delta^{\mu}_{\alpha} \ \delta^{\nu}_{\beta} - \delta^{\mu}_{\beta} \ \delta^{\nu}_{\alpha}\right) R_{\mu \nu}{}^{\alpha \beta} \\ & = - \frac{1}{4} \epsilon^{\mu \nu \rho \sigma} \ \epsilon_{\alpha \beta \rho \sigma} \ R_{\mu \nu}{}^{\alpha \beta} \end{align*} Now multiply this with \sqrt{|g|} and use the 4-dimensional version of eq(B.1): \sqrt{|g|} \ \epsilon_{\alpha \beta \rho \sigma} = \epsilon_{abcd} \ e_{\alpha}{}^{c} \ e_{\beta}{}^{d} \ e_{\rho}{}^{a} \ e_{\sigma}{}^{b} \ , then integrate over M \begin{align*} \int d^{4}x \ \sqrt{|g|} \ R &= - \frac{1}{2} \int \left( d^{4}x \ \epsilon^{\mu \nu \rho \sigma}\right) \ \epsilon_{abcd} \ e_{\rho}{}^{a} \ e_{\sigma}{}^{b} \left( \frac{1}{2} e_{\alpha}{}^{c} \ e_{\beta}{}^{d} \ R_{\mu \nu}{}^{\alpha \beta}\right) \\ &= - \frac{1}{2} \int \ \epsilon_{abcd} \left( e_{\rho}{}^{a} \mbox{d}x^{\rho}\right) \wedge \left( e_{\sigma}{}^{a} \mbox{d}x^{\sigma}\right) \wedge \left( \frac{1}{2} R_{\mu\nu}{}^{cd} \ \mbox{d}x^{\mu} \wedge \mbox{d}x^{\nu}\right) \\ &= - \frac{1}{2} \int_{M^{(1,3)}} \ \epsilon_{abcd} \ e^{a} \wedge e^{b} \wedge \mathcal{R}^{cd} \ . \end{align*} So, on the 4-dimensional spacetime M^{(1,3)}, the E-H action is of the general form S_{EH}(e,\omega ) \sim \int_{M^{(1,3)}} \ \epsilon \ e \wedge e \wedge \left( \mbox{d} \omega + \omega^{2} \right) \ . Now if we interpret the fields (e , \omega ) as components of a gauge connection \mathbb{A}, then the above action will have the following general form S( \mathbb{A} ) \sim \int_{M^{(1,3)}} \ \mathbb{A} \wedge \mathbb{A} \wedge \left( \mbox{d} \mathbb{A} + \mathbb{A}^{2} \right) \ . But there is no such action in gauge theories and, therefore, 4-dimensional gravity is not a gauge theory.
B2) 3-dimensional Gravity is a gauge theory (Claim 3):
Let us now repeat what we have done in (B1) for the (1+2)-dimensional spacetime M^{(1,2)}. I am sure you can follow all the steps in the derivation of the following E-H action on M^{(1,2)}:
<br />
\begin{align*}<br />
S_{EH}(e , \omega ) & = \frac{1}{2} \int \left(d^{3}x \ \epsilon^{\mu \nu \rho} \right) \ e_{\rho}{}^{a} \left(\epsilon_{abc} \ R_{\mu \nu}{}^{bc} \right) \\<br />
& = \int \left( e_{\rho}{}^{a} \mbox{d}x^{\rho} \right) \wedge \left( \frac{1}{2} \epsilon_{abc} \ R_{\mu \nu}{}^{bc} \mbox{d}x^{\mu} \wedge \mbox{d}x^{\nu} \right) \\<br />
& = \int_{M^{(1,2)}} \ e^{a} \wedge \left( \epsilon_{abc} \ \mathcal{R}^{bc} \right) \\<br />
& = 2 \int_{M^{(1,2)}} \ e^{a} \wedge \mathcal{R}_{a} \ \ \ \ \ \ \ \ \ \ (B.5) \\<br />
& = 2 \int_{M^{(1,2)}} \ e^{a} \wedge \left( \mbox{d} \omega_{a} + \frac{1}{2} \epsilon_{abc} \ \omega^{b} \wedge \omega^{c} \right) \ \ \ (B.5)<br />
\end{align*}<br />
where \omega_{a} = \frac{1}{2} \epsilon_{abc} \ \omega^{bc} =\frac{1}{2} \ \epsilon_{abc} \ \omega_{\mu}{}^{bc} \mbox{d}x^{\mu} \ , and \mathcal{R}_{a} = \frac{1}{2} \epsilon_{abc} \ \mathcal{R}^{bc} = \mbox{d} \omega_{a} + \frac{1}{2} \epsilon_{abc} \ \omega^{b} \wedge \omega^{c} \ . Now, if we interpret (e, \omega ) as components of gauge field matrix \mathbb{A}, then the E-H action of 3-dimensional gravity will be of the form S_{EH}( \mathbb{A}) \sim \int_{M^{(1,2)}} \mathbb{A} \wedge \left( \mbox{d}\mathbb{A} + \mathbb{A}^{2} \right) \ . But this looks very much like gauge theory action of Chern-Simons type. So, it is conceivable to interpret (1+2)-gravity as a gauge theory of the Poincare’ group \mathcal{P}(1,2) with a pure Chern-Simons action. Indeed, you can show that the action eq(B.5) is invariant under local SO(1,2) (Lorentz) transformations generated by the infinitesimal parameter \beta^{a}(x):
\delta e^{a} (x) = \epsilon^{abc} \ e_{b}(x) \beta_{c}(x) \ , \delta \omega^{a} (x) = \mbox{d} \beta^{a} + \epsilon^{abc} \ \omega_{b} (x) \beta_{c}(x) \ , as well as local translations generated by the infinitesimal T^{(3)}-parameter \alpha^{a}(x):
\delta e^{a} (x) = \mbox{d} \alpha^{a} + \epsilon^{abc} \ \omega_{b}(x) \ \alpha_{c}(x) \ , \ \ \delta \omega^{a} (x) = 0 \ . Moreover, when the fields (e^{a} , \omega^{a}) satisfy their equations of motion (i.e., on shell), one can show that the combination of transformations with parameters \alpha^{a}(x) = \chi \ e^{a}(x) and \beta^{a}(x) = \chi \ \omega^{a}(x) is equivalent to a M^{(1,2)}-diffeomorphism generated by the vector field \chi. It remains to prove that 3D-gravity action eq(B.5) is the Chern-Simons action associated with the gauge group \mathcal{P}(1,2). Below, I will present you with two methods for proving that statement. Both methods will require the Poincare’ algebra as well as an invariant inner product on it. In 3 dimensions it is convenient to work with the Lorentz generator J_{a} = \frac{1}{2} \epsilon_{abc} J^{bc} instead of J^{ab}. With this definition, the Lie algebra of \mathcal{P}(1,2) can be rewritten as [P_{a} , P_{b}] = 0 \ , \ \ [J_{a} , P_{b} ] = \epsilon_{abc}P^{c} \ ,[J_{a} , J_{b}] = \epsilon_{abc}J^{c} \ , and the relevant invariant inner product on the algebra is then \mbox{Tr}(J_{a}P_{b}) = \eta_{ab} \ , \ \ \ \mbox{Tr}(P_{a}P_{b}) = \mbox{Tr}(J_{a}J_{b}) = 0 \ .
Method 1: In this method, we will borrow the topological Chern-Simon action from ordinary Yang-Mills theory, express the Yang-Mills potential in terms of the fields ( e , \omega ) and use the Poincare’ algebra and its inner product to show that it is equivalent to the (1+2)-dimensional GR action integral eq(B.5). In an ordinary gauge theory with compact Lie group and algebra
[X_{a} , X_{b}] = C_{ab}{}^{c} X_{c}, \ \ \ a, b, c = 1, 2, \cdots r \ ,
the topological Chern-Simons action is given (up to a constant which I set equal to 1) by S_{CS}(A) = \int_{M^{(1,2)}} \mbox{tr} \left( \mathbb{A} \wedge \mbox{d} \mathbb{A} + \frac{2}{3} \mathbb{A} \wedge \mathbb{A} \wedge \mathbb{A}\right) \ , \ \ \ (B.6) where \mathbb{A} = \mathbb{A}_{\mu} \mbox{d}x^{\mu} = A_{\mu}^{a} X_{a} \mbox{d}x^{\mu} = A^{a} X_{a} \ . \ \ \ \ (B.7) I hope Eq(B.7) makes it clear to you that \mathbb{A} is a matrix-valued 1-form, \mathbb{A}_{\mu} is a matrix-valued gauge field, A_{\mu}^{a} are real numbers denoting the components of the gauge field matrix \mathbb{A}_{\mu} in the Lie algebra basis X_{a} and finally the A^{a}’s are a set of 1-forms representing the “components” of the 1-form matrix \mathbb{A} in the Lie algebra basis X_{a}. So, with this notation, we have \mathbb{A} \wedge \mathbb{A} = \frac{1}{2}[X_{a} , X_{b}] \ A^{a} \wedge A^{b} \ . Therefore \mbox{tr} \left(\mathbb{A} \wedge \mathbb{A} \wedge \mathbb{A}\right) = \frac{1}{2} \mbox{tr} \left( X_{a}[X_{b} , X_{c}] \right) A^{a} \wedge A^{b} \wedge A^{c} \ . \ \ \ (B.8b) And we also write \mbox{tr} \left( \mathbb{A} \wedge \mbox{d} \mathbb{A} \right) = \mbox{tr} \left( X_{a} X_{b} \right) \ A^{a} \wedge \mbox{d} A^{b} \ . \ \ \ \ (B.8a) We will now identify the generators X_{a}, \ a = 1, \cdots r with six Poincare’ generators (P_{a} , J_{a}), and the set of r one forms A^{a} with (e^{a} , \omega^{a}), the Poincare-valued one forms on M^{(1,2)}. In other words, we simply write \mathbb{A} (x) = e^{a}(x) \ P_{a} + \omega^{a}(x) \ J_{a} \ . \ \ \ \ \ (B.9) Now, using the inner product on the Poincare’ algebra, we see that the RHS of eq(B.8a) contains only two non-zero terms \begin{align*} \mbox{tr} \left( \mathbb{A} \wedge \mbox{d} \mathbb{A} \right) &= \mbox{tr}(P_{a}J_{b}) \ e^{a} \wedge \mbox{d} \omega^{b} + \mbox{tr}(J_{a}P_{b}) \ \omega^{a} \wedge \mbox{d} e^{b} \\ & = e^{a} \ \wedge \ \mbox{d} \omega_{a} + \omega_{a} \ \wedge \ \mbox{d} e^{a} \ . \end{align*} Also, the Poincare’ algebra and inner product reduce the RHS of eq(B.8b) to only three non-zero and equal terms \begin{align*} \mbox{RHS of (B.8b)} & = \frac{1}{2} \mbox{tr} \left( [P_{a} , J_{b}] J_{c} \right) e^{a} \wedge \omega^{b} \wedge \omega^{c} \\<br />
& + \frac{1}{2} \mbox{tr} \left( [J_{a} , P_{b}] J_{c} \right) \omega^{a} \wedge e^{b} \wedge \omega^{c} \\<br />
& + \frac{1}{2} \mbox{tr} \left( [J_{a} , J_{b}] P_{c} \right) \omega^{a} \wedge \omega^{b} \wedge e^{c} \\<br />
& = \frac{3}{2} \epsilon_{abc} \ e^{a} \wedge \omega^{b} \wedge \omega^{c} \ . \end{align*} Substituting these results in eq(B.6) we obtain S_{CS}(e, \omega) = \int_{M^{(1,2)}} \left( \omega_{a} \wedge \mbox{d} e^{a} + e^{a} \wedge \mbox{d} \omega_{a} + \epsilon_{abc} \ e^{a} \wedge \omega^{b} \wedge \omega^{c}\right) \ . And, finally we integrate the first term by part and ignore total derivative to obtain the Chern-Simons action associated with gauge group \mathcal{P}(1,2) = T(3) \rtimes SO(1,2) (integrating the second term instead of the first produces another story)S_{CS}(e , \omega ) = 2 \int_{M^{(1,2)}} \ e^{a} \wedge \left( \mbox{d} \omega_{a} + \frac{1}{2} \epsilon_{abc} \omega^{b} \wedge \omega^{c} \right) = 2 \int_{M^{(1,2)}} \ e^{a} \wedge \mathcal{R}_{a} \ . Comparing this with eq(B.5), we see that S_{EH} (e , \omega ) = S_{CS} (e , \omega ). Thus, in (1+2)-dimensional spacetime, GR is equivalent to a gauge theory with gauge group \mathcal{P}(1,2) and topological Chern-Simons action.
Method 2: In here we will follow the usual method for constructing Chern-Simons action from topological invariant integral. Recall that, under gauge transformation by U(x) = e^{- \Theta (x)} \ , \Theta = \theta^{a}(x) X_{a}, the gauge field matrix changes according to \mathbb{A}_{\mu} \to U \left( \partial_{\mu} + \mathbb{A}_{\mu} \right) U^{-1} \ . The infinitesimal version of this transformation is simply \delta \mathbb{A}_{\mu} = \partial_{\mu} \Theta + [\mathbb{A}_{\mu} , \Theta ] \equiv D_{\mu} \Theta \ . \ \ \ \ (B.10) Now, we take U(x) to be local Poincare transformation U( \alpha , \beta ) = e^{- ( \alpha^{a} (x) \ P_{a} + \beta^{a} (x) \ J_{a} )} \ , and write the gauge field in terms of the triad field e_{\mu}{}^{a} and spin connection \omega_{\mu}{}^{a} on M^{(1,2)}: \mathbb{A}_{\mu} = e_{\mu}{}^{a} (x) \ P_{a} + \omega_{\mu}{}^{a}(x) \ J_{a} \ . \ \ \ \ \ (B.11) Substituting these in eq(B.10) and using the Poincare algebra, we obtain the following local (Poincare) gauge transformations \delta e_{\mu}{}^{a} = \partial_{\mu} \alpha^{a} + \epsilon^{abc} \left( e_{\mu b} \beta_{c} + \omega_{\mu b} \ \alpha_{c}\right) \ , \ \ \ (B.12a)\delta \omega_{\mu}{}^{a} = \partial_{\mu} \beta^{a} + \epsilon^{abc} \ \omega_{\mu b} \ \beta_{c} \ . \ \ \ \ \ \ \ (B.12b) The matrix-valued field tensor is defined as usual \mathbb{F}_{\mu\nu} = [ D_{\mu} , D_{\nu}] = \partial_{[ \mu} \mathbb{A}_{\nu ]} + [ \mathbb{A}_{\mu} , \mathbb{A}_{\nu}] \ . Substituting eq(B.11) and using the Poincare algebra, we get \mathbb{F}_{\mu\nu} (x) = \mathcal{E}_{\mu\nu}{}^{a} (x) \ P_{a} + \mathcal{R}_{\mu\nu}{}^{a} (x) \ J_{a} \ , where \mathcal{E}_{\mu\nu}{}^{a} = \partial_{[\mu} e_{\nu ]}{}^{a} + \epsilon^{abc} \left( e_{\mu b} \ \omega_{\nu c} + \omega_{\mu b} \ e_{\nu c}\right) \ ,\mathcal{R}_{\mu\nu}{}^{a} = \partial_{[\mu} \omega_{\nu ]}{}^{a} + \epsilon^{abc} \ \omega_{\mu b} \ \omega_{\nu c} \ . Now, suppose we want to put this \mathcal{P}(1,2) gauge theory on a 4-dimensional manifold N^{4}. Then, on N^{4} there will be a topological invariant given by the integral \mathcal{J} = \frac{1}{2} \int_{N^{4}} d^{4}x \ \epsilon^{\mu\nu\rho\sigma} \ \mbox{tr} \left( \mathbb{F}_{\mu\nu} \mathbb{F}_{\rho\sigma}\right) \ . using the inner product on the algebra, we get \mathcal{J} = \int_{N^{4}} d^{4}x \ \epsilon^{\mu\nu\rho\sigma} \ \mathcal{E}_{\mu\nu a} \ \mathcal{R}_{\rho \sigma}{}^{a} \ . \ \ \ \ \ \ \ \ \ \ \ (B.13) Now, with extremely painful algebra we can show that the integrand in eq(B.13) is actually a total divergence, and we use the divergence theorem to obtain \mathcal{J} = \int_{N^{4}} d^{4}x \ \partial_{\sigma} \left( \epsilon^{\mu\nu\rho\sigma} \ \mathcal{S}_{\mu\nu\rho}\right) = \int_{\partial N^{4}} d \Sigma_{\sigma} \ \epsilon^{\mu\nu\rho\sigma} \ \mathcal{S}_{\mu\nu\rho} \ , where \mathcal{S}_{\mu\nu\rho} = e_{\rho}{}^{a} \left( \partial_{[ \mu} \omega_{\nu ] a} + \epsilon_{abc} \ \omega_{\mu}{}^{b} \ \omega_{\nu}{}^{c}\right) \ . Now, if we identify our (1+2)-spacetime M^{(1,2)} with the boundary \partial N^{4} we obtain \int_{M} d^{3}x \ \epsilon^{\mu\nu\rho} \ \mathcal{S}_{\mu\nu\rho}, as a gauge invariant integral on M^{(1,2)} and this, by definition, is the Chern-Simons action \begin{align*} S_{CS}(e , \omega ) & = 2 \int_{M^{(1,2)}} d^{3}x \ \epsilon^{\mu\nu\rho} \ e_{\rho}{}^{a} \left( \frac{1}{2} \partial_{[ \mu} \omega_{\nu ] a} + \frac{1}{2} \ \epsilon_{abc} \ \omega_{\mu}{}^{b} \ \omega_{\nu}{}^{c} \right) \\ & = 2 \int_{M^{(1,2)}} \ e^{a} \wedge \left( \mbox{d} \omega_{a} + \frac{1}{2} \ \epsilon_{abc} \ \omega^{b} \wedge \omega^{c}\right) \\ & = S_{EH} ( e , \omega ) \ . \end{align*}
I think that is enough for now.