# Insights General Relativity as a Gauge Theory - Comments

Tags:
1. Oct 17, 2016

### haushofer

2. Oct 17, 2016

### Orodruin

Staff Emeritus

Just to nitpick: The gauge group corresponding to electromagnetism is U(1), not SU(1) (SU(1) is the trivial group and therefore not very exciting to build a Yang-Mills theory on). Also, the U(1) in the SM before spontaneous symmetry breaking is not the electromagnetic U(1) (the generator is hypercharge, not charge) and the SU(2) of the SM is not just related to weak interactions. Upon SSB, the U(1)xSU(2) symmetry of the SM is broken into the electromagnetic U(1), whose generator is a linear combination of the hypercharge U(1) and one of the SU(2) generators.

3. Oct 17, 2016

### haushofer

Thanx, sloppiness :P

4. Oct 17, 2016

### Greg Bernhardt

5. Oct 18, 2016

### vanhees71

Great article. It would be great if you could put the complete references at the end, e.g., for the first one

R. Utiyama, Invariant theoretical interpretation of interaction, Phys. Rev. 101, 1597 (1955)
http://dx.doi.org/10.1103/PhysRev.101.1597

Last edited by a moderator: Oct 18, 2016
6. Oct 18, 2016

### haushofer

7. Oct 21, 2016

### RockyMarciano

Interesting insight.
Maybe you could clarify more about the difference between GR gauge and Yang-Mills gauge. You have delved into how to keep GCTs while removing local translations by using a curvature constraint. One first question is that this curvature constraint comes about through the differential Bianchi identities(that reflect the dependencies that Noether's second theorem for local gauge theories talk about in her seminal 1918 article) in the absence of local curvature source, so it can be viewd as an analogous for instance to the Ward identities in QED that are obtained automatically by applying Noether's second theorem in the context of QFT, so wouldn't you say this is a common point between Yang-Mills and GR gauges rather than the difference?

My other worry that I'm not completely sure if you have addressed in the insight, is that this constraint, if I have understood correctly what you refer to by it- the constraint leading to vacuum field equations-, is not a general feature of GR though, it only applies to certain isolated central objects situations. But curvature is indeed a general feature of GR as are the differential Bianchi identities that make clear the impossibility of using flat coordinates globally in a curved spacetime. So is GR only a gauge theory for those special isolated in vacuum objects cases? But then how are GCTs together with the LLTs kept in the general case when the curvature constraint is not present?. The dependencies and the Bianchi identities are still there, how are the local translations removed in the general case?

8. Oct 22, 2016

### haushofer

I'm not sure what you mean, so I hope this answer satisfies. The Bianchi identities for both curvatures imply, with the constraint R(P)=0 and the Vielbein postulate, the usual Bianchi identies for the Riemann tensor. These identities continue to hold when we couple the theory to matter, so I don't see any problem here.
I think I don't really understand your statement " the curvature constraint only applies to certain isolated central objects situations"

9. Oct 22, 2016

### RockyMarciano

Maybe I'm misinterpreting what you mean by R(P)=0? I thought you meant by it a constraint that only applies in vacuum i.e. where Rab=0.

You start from Minkowski spacetime(you switch off gravity), then gauge the Poincare symmetries and obtain gauge fields $R_{\mu\nu}{}^A(P)$ and $R_{\mu\nu}{}^{AB}(M)$ and then set the former to zero, leaving only the latter after wich you state that you have recovered the full GR. But are you sure you have the full GR and not some subset like spherically symmetric vacuum GR?

The reason I ask is that the GCTs in full GR are inextricably related with the "local translations" because of the well known issue that the metrics in full GR are dynamical fields while in the Minkowski setting you start with they are not. This relation is formalized mathematically with the Bianchi identities that guarantee the existence of a divergenceless Einstein tensor .

Last edited: Oct 22, 2016
10. Oct 22, 2016

### RockyMarciano

I've managed to get a copy of the second referenced paper(Kibble) in order to get some more understanding and maybe get up to speed on the subject of the insights article and what I found at the end of its section 6 is that after the gauging procedure, that includes removing the local translations and writing them as linear combinations of GCTs and LLTs as mentioned in the insight, what is obtained is a gauge theory of gravitation that is not exactly GR but the variant known as Einstein-Cartan-Sciama-Kibble theory, that doesn't have symmetric connection or Ricci tensors because it has torsion that allows to couple with matter with spin. Only then you can set the torsion to zero and then you get not the full EFE but the vacuum field equations of GR as I thought.

11. Oct 23, 2016

### haushofer

Ah, I see. Yes, I haven't stressed it too much, but the idea is that the gauging procedure gives the vacuum equations. As far as I can see those are exactly equivalent to the vacuum equations of GR. After that you can couple matter to your theory.

It's been a while since I've read that Kibble paper, but I don't see how one can keep torsion while removing the local translations: the R(P) curvature is the torsion. So R(P)=0 puts the torsion to zero, and the Bianchi identities then give you the corresponding symmetries on the Riemann tensor.

I wrote this Insight after having a discussion with Urs here,

https://www.physicsforums.com/insights/11d-gravity-just-torsion-constraint/

Maybe you find that interesting too :)

12. Oct 23, 2016

### RockyMarciano

Ok, that was my initial undestanding, because only with the R(P) i.e. the torsion set to zero one recovers the Riemannian Bianchi identities, but in vacuum only. Then you can couple matter with spin but you are again back to the non-Riemannian connection and tensors, right?
Oh, sure. Sorry if that part was confused in my post, certainly getting rid of the torsion and of the local translations is in this case equivalent. The Einstein-Cartan theory is what you get before setting R(P) to 0.
I'll take a look, thanks.

13. Oct 23, 2016

### dextercioby

PGT (or Poincare gauge theory) is essentially the creation of Friedrich Hehl and his co-workers based on the article by Kibble in the JMP. That, in turn, was based on the (I would call revolutionary) idea of R. Utyiama who, in turn, adapted the original idea of H. Weyl with the subsequent work by E. Cartan and the Russians Ivanenko and Fock.
There's a book treating PGT and other gravity theories at depth which was written by Milutin Blagojevic https://www.amazon.com/Gravitation-...1477257936&sr=1-2&keywords=Blagojevic+Gravity and, to me, totally compensates the (well written, however incomplete) treatment in Ramond's book on QFT or the one in Weinberg's book on GR.

Last edited by a moderator: May 8, 2017
14. Oct 24, 2016

### haushofer

If you mean with "non- Riemannian" "torsionfull", then not necessarily. As fas as I know, spin-1 does not introduce torsion. Fermions can introduce torsion, and one example of this can be seen by the gauging of the N=1 super-Poincaré algebra. This gives you the gravitino gauge field psi, and the $$\{Q,Q\}=P$$ adds an extra term in the (now super!)covariant P-curvature, which schematically becomes
$$\hat{R}^A(P) = d e^A - \omega^{AB} \wedge e^B - \bar{\psi} \gamma^A \psi$$
So setting this to zero now gives
$$d e^A - \omega^{AB} \wedge e^B = \bar{\psi} \gamma^A \psi$$
instead of zero. The left hand side is the antisymmetric part of the connection, $\Gamma^A_{[\mu\nu]}$ due to the Vielbein postulate.

I'm not sure to which extent the generality is of the statement "fermions add torsion".

15. Oct 24, 2016

### Ben Niehoff

My understanding is that, with or without SUSY, the contribution of fermions to the energy-momentum tensor has a spin piece which is not symmetric, and thus always gives you nonzero torsion (due to the non-symmetric Ricci tensor). But the torsion is always non-propagating; i.e., it is nonzero only where the fermion fields are nontrivial.

16. Oct 26, 2016

### haushofer

I see. The fermionic action for e.g. spin-1/2 contains a kinetic term

$$\bar{\psi} \gamma^{\mu} D_{\mu} \psi = \bar{\psi} \gamma^{\mu} \Big( \partial_{\mu} - \frac{1}{4}\gamma_{AB}\, \omega_{\mu}{}^{AB} \Bigr)\psi$$
so in the first order formulation the variation with respect to the spin connection picks up a term

$$- \frac{1}{4} \bar{\psi} \gamma^{\mu} \gamma_{AB} \psi$$

giving torsion. The same goes for spin-3/2 and the Rarita-Schwinger action. This corresponds in the N=1,D=4 case to the gauging-argument I gave earlier for the gravitino.

17. Oct 26, 2016

### RockyMarciano

So this brings me back to the second question in my post #7. Once we make effective the necessary constraining condition(getting rid of "local translations" by setting the torsion to zero and therefore restricting to vacuum) to gauge GR's symmetry (CGTs) according to the equivalence principle(LLTs) acting as local symmetry, how can we introduce a stress-energy tensor in such a gauge theory?

18. Oct 30, 2016

### haushofer

Hi RockyMarciano,

The way I understand the procedure, is that the gauging procedure gives you GR with vacuum equations only. This is because the only gauge fields are the dependent spin connection and the vielbein (metric). After that you couple the theory to matter, but these matter fields in general will not be gauge fields anymore of an extension of the Poincaré algebra. An exception to this is the N=1 super-Poincaré theory in D=4, in which you introduce a gravitino; this field can be treated like the gauge field of the supertransformations. So after coupling of the matter, the spin connection will be given by the first order formalism and for fermions will obtain extra terms compared to the gauging.

Spin-1/2 fields cannot be considered as gauge fields of some 'extra generators' of the Poincaré algebra afaik (they don't even carry a vector index). So they introduce torsion because of the first order formalism (they couple to the spin connection), not because they appear in the curvature of local translations.

As such the gauging procedure for me is a way of obtaining the vacuum equations, but as soon as one wants to couple matter, the result in general will not be obtainable by a gauging procedure directly. If this would be the case, constructing supergravity theories would be much easier! So if I couple fermions to GR, I don't consider the underlying algebra anymore, but use the first order formalism to derive the spin connection.

If you want to see how matter couplings can be derived systematically in supergravity theories, you can check the "Super conformal tensor calculus", in which conformal symmetries are used as a trick to couple matter multiplets to Poincaré-SUGRA.

19. Nov 1, 2016

### haushofer

20. Nov 1, 2016

### RockyMarciano

Yes, thanks, this is basically the gravitational gauge anomaly for even dimensional spacetimes in the field theoretic context.