# Insights 11d Gravity From Just the Torsion Constraint - Comments

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1. Sep 6, 2016

2. Sep 6, 2016

### haushofer

You mention the gauging of the Lorentz symmetries. In four dimensions you can obtain N=1 SUGRA by gauging the full super-Poincaré group, including local translations, LT's and supertransformations. In order to remove the local translations from your algebra you need to set the torsion (which is the field strength of translations) to zero. This is called a conventional constraint, because it enables one to solve for the spin connection. The same can be done in four dimensions for N=1 super-AdS, the Bargmann algebra, the N=2 super Bargmann algebra etc.

For the N=1, D=11 case this gauging does not work because the graviton-multiplet also contains the 3-form which is not a gauge field of the corresponding algebra (but maybe of some corresponding higher-spin algebra?), but maybe this adds something to your article.

3. Sep 6, 2016

### Urs Schreiber

Yes, the 3-form in 11d sugra is seen by the "supergravity Lie 3-algebra" extension of the 11d super-symmetry algebra, that I had talked about a bit in Part 2 "Homotopy Lie-n algebras in Supergravity" of the series. It was first recognized in its dual incarnation (as a "CIS" later also called "FDA" ) in
• Riccardo D'Auria, Pietro Fré, "Geometric Supergravity in D=11 and its hidden supergroup", Nuclear Physics B201 (1982) 101-140 (pdf)
where they use it to give an elegant re-derivation of the action of 11d SuGra. (It starts appearing with equation (3.15d) in that article).

But I haven't thought about whether this allows to do the analogous gauging procedure that you mention for 4d SuGra. It seems like it should. I'd have to think about it. That's a good point.

4. Sep 8, 2016

### haushofer

Afaik one can only obtain certain SUGRA's (the fields, transformations and curvatures) via gauging for special cases. E.g. for N=2, D=4 the graviton-multiplet contains the graviton, gravitino and vector field, but the SUSY-transformation rules of the vector field do not follow from the gauging (one can gauge the R-symmetry group, but that's a different cookie).

5. Sep 12, 2016

### haushofer

Thanks for those links. I just scanned through the paper by de Azcarraga,

https://arxiv.org/abs/hep-th/0406020

which indeed suggests that the 11D case can be obtained by a gauging procedure. However, this three-form is a composite field, so is this three-form equivalent to the original one? (regarding degrees of freedom, susy-transformations etc.)

It would be interesting to see if such an analysis would be possible for the D=4, N=2 case.

6. Sep 12, 2016

### Urs Schreiber

I once made some tentative notes on precisely this question here:

11d SuGra from higher to exceptional geometry
(web, https://dl.dropboxusercontent.com/u/12630719/HigherToExceptional.pdf [Broken]).

Last edited by a moderator: May 8, 2017
7. Sep 13, 2016

### haushofer

I'm not used to the mathematical treatment you're using in your notes, so it takes some effort for me to get to the punchline. Is the gauge treatment exactly equivalent to the usual D=11 sugra, or are there additional constraints on the three-form?

8. Sep 13, 2016

### Urs Schreiber

There are no additional constraints, on the contrary the "gauge treatment" overparameterizes the 3-forms. That's hidden in corollary 2.2 of the writeup, which observes that locally every 3-form arises from the "gauge treatment", in direct analogy to how generalized gemetry allows to parameterize 2-forms.

So there is more degrees of freedom in the "gauge treatment", not less. The punchline of my note is to observe that this additional freedom is encoded by sections of the "exceptional tangent bundle" as originally proposed by Hull: the bosonic body of that "hidden supersymmetry algebra" of D'Auria-Fre and followups is in fact the M-theoretic exceptional tangent bundle of 11d super-Minkowski spacetime. The note means to point out that the "gauge treatment" comes down to proposing that the 3-form is always to be pulled back via a section of this exceptional tangent bundle, via a canonical 3-form carried by it.

So the evident conjecture is that the "gauge treatment" is part of a solution to the problem as to the nature of the "hidden microscopic degrees of freedom" that are expected to be present in M-theory, given that they do appear as U-duality groups after high codimension compactifications.

9. Sep 13, 2016

### haushofer

I see. But here the link between the D=11 SUGRA and M-theory is clear. It's not clear at all if something similar can be done for other SUGRA's or what to expect. It's linked to the question whether every SUGRA can be obtained from some string theory by compactification and taking the low energy limit.

The reason why I bring this up is because a while ago I tried to construct the D=4, N=2 supergravity version of Newton-Cartan, which, as in the relativistic case, cannot be obtained directly from a gauging of the underlying superalgebra. For the D=3, N=2 theory however this was possible. It would simplify life greatly if every SUGRA theory could be obtained in such a way by extended the underlying superalgebra similar to your notes :P

10. Sep 13, 2016

### Urs Schreiber

I don't know, but it does remind me of the following: on the level of just the Green-Schwarz sigma models for the branes then the pair of dimensions 11/10 appears in a pattern with the pair of dimensions 4/3, in that the "brane scan" sees a membrane/string pair in either case, see for instance Duff's original diagram here. This might indicate that the situation with 11d sugra that we just discussed has its counterpart in 4d. This would be really interesting.

11. Sep 18, 2016

### EnigmaticField

Why is the vanishing torsion (the invariance of the infinitesimal translation of the vielbein field up to the contribution from a Lorentz gauge transformation) a manifestation of the principle of equivalence? I have long understood that the vanishing curvature means the parallel transport is path independent, but keep wondering what's the physical interpretation of the vanishing torsion. Now I see this article says the vanishing torsion means the principle of equivalence; but how to see this? Further, if the vanishing torsion really means the equivalence principle, then isn't it the case that the teleparallel theory violates the equivalence principle?

12. Sep 18, 2016

### Urs Schreiber

By Guillemin's theorem, which I mentioned, vanishing torsion is equivalently the property that the spacetime looks like flat Minkowski spacetime to first order around every point. This immediately implies that locally to first order, the free fall of any particle in a field of gravity is inertial motion in Minkowski spacetime.

One may think of this statement as a more conceptual version of the maybe more familiar computational statement that around every point there exist Riemann normal coordinates in terms of which the Levi-Civita connection (i.e. the unique torsion-free metric connection) vanishes at that point.

13. Sep 18, 2016

### Urs Schreiber

Let me come back to this. I still don't know the answer, but it did occur to me that there ought to be a formulation of $N=2$ $D=4$ sugra that more directly reflects the formulation of 11d sugra for which D'Auria-Fre and followups found the "secret gauging" that we talked about. And indeed it does, as also found by D'Auria and collaborators:

• Laura Andrianopoli, Riccardo D'Auria, Luca Sommovigo, "$D=4$, $N=2$ Supergravity in the Presence of Vector-Tensor Multiplets and the Role of higher $p$-forms in the Framework of Free Differential Algebras", Adv.Stud.Theor.Phys.1:561-596,2008 (arXiv:0710.3107)
Using this, they give a re-derivation of $N=2$, $D=4$ sugra (in the generality of massive twists, see also the introduction of Gunyadin-McReynolds-Zagerman 05) directly paralleling their earlier re-derivation of $D=11$ sugra. So this is the necessary first step for finding the 4d analog of that "secret gauging" that we discussed.

I have made a quick note with more detailed pointers to the references on the nLab at 4d supergravity Lie 2-algebra

14. Sep 19, 2016

### haushofer

It will take some time for me to read your reference, especially since I have my PhD defence this Friday, but just a conceptual question which pops up in my mind: what's the status of the equivalence principle then in SUGRA, where torsion is generated by a bi-spinor? If you locally try to make spacetime flat by moving inertially, you can still have spinor contributions to the geometry.

15. Sep 19, 2016

### Urs Schreiber

Yes, in supergravity there will always be non-vanishing super-component of the torsion. However, as discussed in the entry, for bosonic solutions in 11d supergravity it is the proper super-torsion that vanishes, and we are back to the situation that around every point to first order, spacetime looks like super-Minkowski spacetime. We still have the principle of equivalence then: setting the fermionic coordinates to zero, the situation is that of ordinary vanishing torsion, and the rheonomy principle then says that such ordinary bosonic solutions uniquely extend to superspace.

All my best wishes!

16. Sep 19, 2016

### haushofer

You mean because supersymmetric backgrounds necessarily have vanishing fermionic vev's?

Thanks, I'll let the forum know whether I survived :P

17. Sep 19, 2016

### Urs Schreiber

So I was referring to "backgrounds with vanishing fermionic vev's" if you wish, yes. Whether they are supersymmetric or not (have Killing spinors or not) is not relevant for the particular statement I made, but of course these are examples, yes.