# FeaturedA Is Gravity A Gauge Theory

1. Jan 29, 2018

### Staff: Mentor

I have been reviewing GR lately because as a mentor I find myself now answering more of those questions. I learnt GR years ago from Wald and other sources, but since then have been exposed to the symmetries of the Standard Model. What struck me during this review is I now have a different perspective. I now suspect like all the other fundamental forces it is a gauge theory. Its gauge being invariance to coordinate transformations. Is this the right way of looking at it or is an I missing something? Exactly what gauge group would it be? Or is this a matter of exactly what one considers a gauge theory? I am not sure it can be formulated as a Yang-Mills theory (if so it would of course it would be one in the ordinary sense) - but came across the following I need to study:
https://cds.cern.ch/record/338848/files/9711054.pdf

Thanks
Bill

2. Jan 29, 2018

### Staff: Mentor

3. Jan 29, 2018

### Staff: Mentor

I don’t know if it is a gauge theory, but if it is then I don’t think that the gauge can be invariance to coordinate transformations. That comes from writing the theory in terms of tensors, but you can do that for the Standard Model too.

4. Jan 29, 2018

### Staff: Mentor

That's interesting because in linearised gravity its gauge invariance is exactly that - invariance to infinitesimal coordinate transformations - Chapter 7 - Ohanian Gravitation and Space Time. Kretchmann proved merely writing a theory in the form of tensors is vacuous - it says nothing - and was eventually agreed to by Einstein. The exact foundation of GR was that the content of laws had to be invariant to coordinate transformations - merely formulating a theory in the form of tensors does not guarantee that. The Standard Model symmetries are formulated in terms of SR tensors which I don't think are general coordinate transformations.

Have quickly scanned my posted link - they have a theory that generalizes GR - but isn't exactly GR.

I am groping in the dark here - not sure exactly whats going on.

Might need some high powered help from people like Urs or Sal (cant remember his full name).

Thanks
Bill

Last edited: Jan 29, 2018
5. Jan 29, 2018

### Staff: Mentor

Thanks.

Had a look at both - but the above looks more reasonable than the others (my posted paper and the other one you posted). After reading them I am now suspicious its not a Yang-Mills theory - that seems to lead to theories that are a bit strange - not that it is a criteria for being correct - but it doesn't 'excite' me much

The above paper is something that looks really interesting to study.

Thanks
Bill

6. Jan 29, 2018

### haushofer

https://www.physicsforums.com/insights/general-relativity-gauge-theory/

Short answer: GR can be seen as a gauge theory of the Poincaré algebra. It differs from Yang-Mills significantly, because (1) you put one gauge curvature to zero (that of the translations), and (2) the action is not quadratic in the remaining field strength (that of the Lorentz transformations). Because of the curvature constraint in (1) the local translations are just general coordinate transformations plus local Lorentz transformations, leaving you with the right types of transformations.

7. Jan 29, 2018

### Staff: Mentor

Great article - loved it. but I have a lot of reading to do. However I was gladdened when I read - People like to say that gravity is the result of locally gauging Lorentz symmetry. That was my gut feeling sense.

However a lot of work on my part understanding the detail.

Thanks
Bill

8. Jan 30, 2018

### haushofer

Yes, it is a bit formal. On top of that, it is quite different from GR and subtely different from Yang-Mills :P

The reason why it is a great approach however, is because it is easily extended to other symmetries, giving you all kinds of different theories of gravity. E.g., if you apply the same gauging procedure to the Bargmann algebra (which is the Galilei algebra + central extension), you obtain Newton-Cartan theory. In my PhD research I developed, in a similar way, Newton-Cartan theory for gravitating strings instead of point particles. And if you like supergravity (SUGRA): the D=4 minimal SUGRA can be obtained by gauging the super-Poincaré algebra.

9. Feb 1, 2018

### Auto-Didact

Enjoyed your Insight article. It immediately reminded me of the Bacry & Levy-Leblond classification of the 8 possible kinematical groups.

If I recall correctly, John Baez, and more prominently, Derek Wise did some nice work on MacDowell-Mansouri GR using Cartan geometry. Moreover, almost 25 years ago, i.e. prior to Wise's work, Roger Penrose used Newton-Cartan theory to describe Diosi-Penrose quantum state reduction; how difficult would it exactly be to extend his argument to (MacDowell-Mansouri) GR? I'm assuming that this has already been tried.

10. Feb 2, 2018

### haushofer

Thanks! I'm not familiar with the Diosi-Penrose quantum state reduction, to be honest, but I also don't see why people would first try Newton-Cartan instead of ordinary Newtonian gravity. Newton-Cartan theory is more complicated than GR (although the solutions of course are simpler): you need degenerate geometry, two metrics, the connection contains an extra 2-form, the action principle is much more complicated, etc.

Also, the question arises what conclusions one can draw from Newton-Cartan theory. E.g., people have written papers about its quantization, but Newton-Cartan theory doesn't contain any gravitational waves/propagating degrees of freedom, so I don't see how the quantization of NC-theory gives one any insight into quantum gravity in general (relativity :P ).

11. Feb 2, 2018

### martinbn

I don't know what a gauge theory is, but my impression is that mathematically it is described by a principal bundle over space-time and a choice of a connection. All of the mathematical notions having physical interpretation such as: the connection form is the field potential, the curvature form is the field strength and so on. General relativity as usually presented looks different, but since the connection used is a linear connection it can be viewed as a connection on the $GL(4)$ bundle of linear frames. And since it is a pseudo-Riemannian it reduces to the bundle of orthogonal frames.

12. Feb 2, 2018

### Auto-Didact

DP quantum state reduction is a prediction of intrinsic mass-dependent wave-function collapse as an actual phenomenon in an as yet to be discovered non-linear extension of QM, of which QM is the limiting case with quantum state reduction mass $m_r \rightarrow 0$ with the actual value for $m_r$ being the Planck mass.

The actual argument used is quite subtle, namely not quantization of gravity (i.e. quantum gravity) but gravitization of QM, i.e. not reformulating GR respecting the principles of QM, but instead reformulating QM respecting the principles of GR i.e. in curved spacetime using the equivalence and general covariance principles.

Moreover, Penrose explicitly chooses NC because he needs curved spacetime and a mathematically tractable way of dealing with the prohibition of pointwise identification of space-times due to the principle of general covariance. NC doesn't exactly resolve this prohibition but it at least makes a preliminary investigation of the problem mathematically more tractable; the fact that NC is a Newtonian approximation isn't problematic at all since quantum state reduction is independent of $c$. Last but not least, there are papers arguing exactly the opposite i.e. that the NC framework does in fact shed light on the role of gravity in the measurement problem.

13. Feb 2, 2018

### Fra

I have always been partly symphatetic with Penrose direction of thinking of a deeper connection between QM and GR. As is also supported by recent ponderings of "ER=EPR" and "QM=GR" as argued by Susskind and others.

So there is no doubt in my mind that his intuition was onto something. But I think to develop the ideas and to understand exactly in what sense QM gravitates, one needs a serious revision of our constructing principles.

I think trying to analyse the constructing principles of GR as well as the SM, trying to find the "common denominator" is indeed a rational approach.

The first paragraph och haushofers nice insight articler describes the three steps on howto typically USE this gauge principle as a construction tool to build theories.

One of the things often overlooked thinkgs - probably because it fringes to philosophy - is how to interpret the "ontological status" of these gauge symmetries in the first place? And the real question is not how to just "interpret them" to make us feel better and the usual discussions that follow, the core question is which is the most fruitful WAY of interpreting them in order to arrive at a coherent construction of the theory of all interactions. The way of interpreting this that leads to progress is thus the "right" one.
(This was partly discussed from Wittens perspetive in https://www.physicsforums.com/threads/ed-witten-on-symmetry-and-emergence.927897/)

But in order to state things more clearly, for me at least it has always been clear that the physical essence of the gauge principles observer equivalence. This means that the laws of physics "seen" by any observer in our universe, must be consistent with the views of the other observers. Equivalence may be a better word than invariance, because invariance makes it sounds like the gauges are non-physical and simply in the mathematical realm which imo is wrong. To think that all observers are "equivalent" is much more SANE that to think that the theory is invariant with respect to the observer choice.

I would like to recode Haushofers steps in general terms

(1) Encode/Represent the laws of physics with respect to an observer where we know how to do this
(2) Now, try to define the manifold of all physically existing observers, and find transformations that transforms one observer to another one. Transform the theory
the a general observer and note the appearance of additional terms in the theory that was not there before.
(3) Introduce counterterms to compensate for the additional terms implied by (2). These counterterms are identified with a NEW interaction.

So what to we need to know here?
A) The laws of physics with respect to one observer. (We are free to choose the observer where the theory just "happens" to be simlpest)
B) We need to know the full manifold of physically existing observers, and we need to understand the transformation groups that relates them.

The full understanding of (B) requires complementing the lossy renormalisation flow with the reverse step. It also needs to be complemented with details of the actual way laws are encoded, and how the internal structure both mirrorors and constraints the laws. This is indeed a holographic view, but given the observer equivalence view, rather than observer invariance view, i am personally convinced that the laws are a result of negotiation, and thus a kind of attractor in theory space. When all observers can agree to disagree without killing off each other, we have a stable law that is consistently coded throughout the equivalence class.

When we have sorted this mess out, I think the explicit connection between how information gravitates will be clear as well.

If we dont get stuck on the surface of current theories, and step back and look at the constructing principles, such as the observer equivalence, I see great hope to find a coherent picture which embraces both views.

/Fredrik

14. Feb 3, 2018

### haushofer

Thanks, I've done a little reading also by myself.

It was a bit confusing perhaps of me to mention quantization of gravity. I'm well aware that Penrose's approach is not this. I merely wanted to emphasize the fact that one has to be carefull to use Newton-Cartan theory as a playground. But if it's about general covariance, I see why Penrose choose it.

15. Feb 3, 2018

### haushofer

Hoping to give an answer which replies to this post (the "ontological status"):

The way I see gauge "redundancy", is that it enables us to make certain other symmetries manifest.

E.g., take U(1) gauge theory. We use this gauge "redundancy" such that we can put the photon in a spin-1 representation of the Lorentz algebra. E.g., we make the symmetries of special relativity manifest. The same can be said of GR: we put the graviton in a spin-2 representation (Fierz-Pauli theory) and we need certain gauge symmetries to avoid ghosts. The special thing about GR however is that these gauge "redundancies" are transformations in spacetime instead of some internal space ("fiber") . More generally, we refer to this as the Stückelberg trick. I regard Newton-Cartan theory also as a very sophisticated Stückelberg trick. Sophisticated, because it makes Newtonian gravity general covariant in a very non-trivial way, namely via geometrization.

https://arxiv.org/abs/1105.3735

16. Feb 3, 2018

### Fra

i will be away for a day or so, so i will be back with more comments on this later.

But I agree this is an important difference, but with ontological status, i meant where/how the information about these equivalence classes are physically encoded. Here we face another issue, which indeed is related to the internal vs external point. In QFT we study small susystems, and the observer is dominant. In GR we have a cosmological theory where the environment dominates the observer. How can we make sense of this? in the holographic sense?

/Fredrik

17. Feb 14, 2018

### samalkhaiat

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 modelled 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 $M^{(1,2)}$:
\begin{align*} 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) \\ & = \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) \\ & = \int_{M^{(1,2)}} \ e^{a} \wedge \left( \epsilon_{abc} \ \mathcal{R}^{bc} \right) \\ & = 2 \int_{M^{(1,2)}} \ e^{a} \wedge \mathcal{R}_{a} \ \ \ \ \ \ \ \ \ \ (B.5) \\ & = 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) \end{align*}
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-dimentional 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} \\ & + \frac{1}{2} \mbox{tr} \left( [J_{a} , P_{b}] J_{c} \right) \omega^{a} \wedge e^{b} \wedge \omega^{c} \\ & + \frac{1}{2} \mbox{tr} \left( [J_{a} , J_{b}] P_{c} \right) \omega^{a} \wedge \omega^{b} \wedge e^{c} \\ & = \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.

Last edited: Feb 14, 2018
18. Feb 16, 2018

### Paul Colby

The book is dated but it might be worth a look "Group Theory and General Relativity" by Moshe Carmeli pub McGraw-Hill 1977. Much in there about the Lorentz group representations, Yang Mills, spinors SL(2,C) and so forth. I've enjoyed reading sections over the years.