# Why using quantum group SUq(2) instead of SU(2) makes sense?

1. ### marcus

24,115
Increasingly QG is being done with a q-deformed symmetry group: replacing SU(2) by
the quantum group SUq(2). What do you see as the intuitive basis for this?

One intuitive justification is that in a universe with a minimum measurable size and an a maximum distance to horizon there is an inherent uncertainty in determining angle. A minimal angular resolution---which SUq(2) nicely implements. The argument is simple and is worked out in this 2-page paper.

http://arxiv.org/abs/1105.1898
A note on the geometrical interpretation of quantum groups and non-commutative spaces in gravity
Eugenio Bianchi, Carlo Rovelli
(Submitted on 10 May 2011)
Quantum groups and non-commutative spaces have been repeatedly utilized in approaches to quantum gravity. They provide a mathematically elegant cut-off, often interpreted as related to the Planck-scale quantum uncertainty in position. We consider here a different geometrical interpretation of this cut-off, where the relevant non-commutative space is the space of directions around any spacetime point. The limitations in angular resolution expresses the finiteness of the angular size of a Planck-scale minimal surface at a maximum distance $1/\sqrt{\Lambda}$ related the cosmological constant Lambda.
This yields a simple geometrical interpretation for the relation between the quantum deformation parameter
$$q=e^{i \Lambda l_{Planck}^2}$$
and the cosmological constant, and resolves a difficulty of more conventional interpretations of the physical geometry described by quantum groups or fuzzy spaces.
Comments: 2 pages, 1 figure

The cosmo constant is a reciprocal area: a handle on it is the distance $1/\sqrt{\Lambda}$ which is easy to calculate by typing this into the google window:
c/(70.4 km/s per Mpc)/sqrt 3/sqrt 0.728

Last edited: May 11, 2011
2. ### atyy

9,932
A couple reasons before this:

1) The Ponzano-Regge spin foam model of 3D gravity was first regularized by going to quantum groups in the Turaev-Viro state sum model. The TV model seems to have positive cc compared to the Ponzano-Regge model.

2) As Physics Monkey pointed out in https://www.physicsforums.com/showpost.php?p=2576472&postcount=7, the Levin-Wen models (which are somehow related to the TV model) use quantum groups.

Spin foams have a long history of thinking of gravity as being a constrained BF theory, and so related to TQFTs, which the TV and Levin-Wen models are. Rovelli still believes his current model is some sort of TQFT http://arxiv.org/abs/1010.1939 . Hellmann's http://arxiv.org/abs/1105.1334 has very interesting comments on p8 about how triangulation independence, which is characteristic of state sum TQFTs, is supposed to be achieved.

Last edited: May 10, 2011
3. ### tom.stoer

5,489
I am still not convinced that this algebraic setup for the cc (via the deformation parameter q) makes sense b/c we know from the asymptotic safety program that the cc is subject to renromalization.

So there are two competeing ideas how to introduce the cc
- as something very special via the q-deformation
- as an ordinary coupling constant

I still do not see how these two approaches can be related.

4. ### atyy

9,932
But if, as you have said, LQG should get unification, then the current interpretations of the formalism as geometry should be wrong anyway.

5. ### mitchell porter

769
In his version of dS/CFT, David Lowe used the quantum deformation of the isometry group of de Sitter space, because he wanted a finite-dimensional Hilbert space, because the cosmological horizon of an observer in de Sitter space has a finite entropy.

6. ### tom.stoer

5,489
No, not really. The current formalism is not geometry (neither today nor in future), but geometry emerges in a certain semiclassical limit (already today). Using SUq(2) could be a way to let several aspects emerge simultaneously - geometry, particles, interactions, ...

This is a fascinating idea and I am really a fan - but I don't see how to this can be harmonized with renormalization.

7. ### marcus

24,115
I don't either, but perhaps it is possible that they might be related. This paper gives some suggestions of how that might work, on an intuitive level.
http://arxiv.org/abs/1105.1898

It interprets the deformation parameter as something that could run.

The group deformation q is a function of the finest angular resolution φ = Lmin/Lmax
the ratio of the two distance scales.

In the limit as φ --> 0, the deformation q --> 1
and the quantum group SUq(2) goes to the ordinary SU(2).

One might say the idea of the finest angular resolution is more intuitively meaningful of the two. Or else that the difference between them is trivial, since they are so closely related:

q = exp(i φ2)

I suppose it is obvious that φ can change as the universe evolves. It can change even without Lmin changing, if Lmax does. With any definition I can think of, the latter does change over time.

There is a distance called the cosmological event horizon CEH which is at present slowly increasing towards a limit currently estimated to be about 16 billion lightyears.

Roughly speaking the asymptotic value of the CEH is 1/√Λ, a distance you see being used in the paper. But more precisely using standard model cosmology the limit is √(3/Λ).
There is a rogue factor of √3 which gets in there.

So even if √Λ would remain constant, the Lmax would change over time.
That is if we take it to be the CEH---which is the bound on how close a galaxy must be if it can, today, send us a message which will eventually reach us. If a galaxy is closer than the cosmological event horizon then whatever events happen there today, we will eventually get the news. If it is farther than that, we will never get the news.

Most people here will know the CEH, but I put it in for completeness in case some do not.

Even with Λ constant, that CEH distance is slowly increasing towards estim. 16 billion lightyears. And CEH used to be much smaller. At the time the microwave background light was emitted it was several orders of magnitude smaller. I would guess something like 40-some million lightyears, instead of the present 15 billion or so.

I don't pretend to understand in what sense the cosmological constant Λ itself could run. None of this makes clear sense to me yet. But it seems to have possibilities. Regardless of what happens with Λ itself, there is clearly some play in the both the Planck length (related to Lmin) and in the CEH (related to Lmax).

So intuitively there must be some play in their ratio, the finest angular resolution φ.

The message of this short paper, in my view, is that we can think of the quantum group deformation simply as a measure of the finest angular resolution.

(Perhaps it even has something to do with the "relative locality" line of investigation currently being pursued. This puts emphasis on the individual observer's momentum space and therefore necessarily on determining the incoming angles of whatever the observer is detecting/measuring.)

If anyone is curious, the asymptotic value of the CEH can be gotten by putting this in google:
c/(70.4 km/s per Mpc)/sqrt 0.728 in light years
The numbers 70.4 and 0.728 are from the 7 year WMAP report.

Last edited: May 11, 2011
8. ### marcus

24,115
As you mentioned, we think from asymptotic safety that Lambda runs, so how does this fit with SUq(2)? How do the two approaches relate? I don't think there is any mismatch. If indeed Lambda runs then that must affect the maximum length scale Lmax. Then certainly the angle resolution bound must run and therefore also the deformation parameter q!

If we assume Lambda runs then also must q, because Lambda places limits on measurement (the available spherical harmonics etc, as explained in the paper) so there is no incompatibility. Or so I think.

9. ### tom.stoer

5,489
The inability to show the presence of an incompatibility does not automatically proof its absence.

10. ### jal

642
http://arxiv.org/abs/1105.1898
Okayyyy!

So, now, who can argue ... why this scale cannot be at 10^-18m?

11. ### Finbar

343
It seems strange to me to have the quantity

$$l_{pl}^2 \Lambda = G \Lambda$$

turn up in an equation. Since in the action we always have the cosmological constant in the form
$$\frac{\Lambda}{G}$$

This is the energy density that must couple to space-time in order to form a dimensionless quantity. Or just think of the Einstein equations
$$G_{\mu \nu} + \Lambda = 8 \pi G T_{\mu \nu}$$

One would have to multiply through by G to get $$G \Lambda$$ turn up but this is arbitrary to some extent. It seems to imply that the Planck length has been put in by hand and some stage in the construction.

Maybe someone knows at what stage and why we get this combination of the Planck length and the cosmological constant?

Last edited: May 11, 2011
12. ### marcus

24,115
That's right! I spoke carelessly there. I should have said, not that there is no incompatibility but that there was no incompatibility shown. AFAIK it not been proven that there is a natural compatibiity between letting the group be deformed a variable amount into SUq(2), on the one hand, and treating Lamba as a running coupling constant, on the other hand.

I have no idea how or if that would work out.

13. ### marcus

24,115
I think it is the natural way to get a dimensionless quantity involving Lambda.

Lambda, as I understand it, is always a reciprocal area. So to get a dimensionless quantity one must multiply it by an area. So one multiplies by the Planck area. (What else ?)

Just kidding, but you can see where it comes in already in equation (2)

Last edited: May 11, 2011
14. ### Finbar

343
So basically by choosing to measure all quantities in Planck units. Hmm thinking about it should come somehow when one computes the entropy of the cosmological horizon. Since the area of the cosmological horizon is the reciprocal of Lambda.

15. ### atyy

9,932
I don't understand how the EPRL/FK semiclassical limit is going to work in the q-deformed case. Is there still a J->infinity limit, or does J have a maximum value now? Or if q has to be taken to zero for the limit, then the cosmological constant disappears in the semiclassical limit?

Actually that sounds right, given your idea that the cc should arise naturally from some sort of renormalization flow.

16. ### atyy

9,932
Carfora, Maezuoli and Rasetti's review of quantum tetrahedra http://arxiv.org/abs/1001.4402 begins with a great comment inspired by "Nescit Labi Virtus (Virtue cannot fail)".

17. ### tom.stoer

5,489
The semicallsical limit is an interesting aspect; honestly, I can't see how a q = zero or a j = infinity limit could make sense.

It's not my idea that the cc could arise from renromalization. It's just an idea that the non-gaussian fix point in AS gives us a non-zero cc, but I do not see how to get a running cc (which means a running q!) for q-deformation.

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