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Why using quantum group SUq(2) instead of SU(2) makes sense? 
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#1
May1011, 11:00 PM

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PF Gold
P: 23,274

Increasingly QG is being done with a qdeformed symmetry group: replacing SU(2) by
the quantum group SU_{q}(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 resolutionwhich SU_{q}(2) nicely implements. The argument is simple and is worked out in this 2page paper. http://arxiv.org/abs/1105.1898 A note on the geometrical interpretation of quantum groups and noncommutative spaces in gravity Eugenio Bianchi, Carlo Rovelli (Submitted on 10 May 2011) Quantum groups and noncommutative spaces have been repeatedly utilized in approaches to quantum gravity. They provide a mathematically elegant cutoff, often interpreted as related to the Planckscale quantum uncertainty in position. We consider here a different geometrical interpretation of this cutoff, where the relevant noncommutative space is the space of directions around any spacetime point. The limitations in angular resolution expresses the finiteness of the angular size of a Planckscale minimal surface at a maximum distance [itex]1/\sqrt{\Lambda}[/itex] related the cosmological constant Lambda. This yields a simple geometrical interpretation for the relation between the quantum deformation parameter [tex]q=e^{i \Lambda l_{Planck}^2} [/tex] 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 [itex]1/\sqrt{\Lambda}[/itex] which is easy to calculate by typing this into the google window: c/(70.4 km/s per Mpc)/sqrt 3/sqrt 0.728 


#2
May1011, 11:08 PM

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A couple reasons before this:
1) The PonzanoRegge spin foam model of 3D gravity was first regularized by going to quantum groups in the TuraevViro state sum model. The TV model seems to have positive cc compared to the PonzanoRegge model. 2) As Physics Monkey pointed out in http://www.physicsforums.com/showpos...72&postcount=7, the LevinWen 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 LevinWen 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. 


#3
May1111, 01:04 AM

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P: 5,464

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 qdeformation  as an ordinary coupling constant I still do not see how these two approaches can be related. 


#4
May1111, 01:08 AM

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



#5
May1111, 03:05 AM

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



#6
May1111, 05:12 AM

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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
May1111, 12:50 PM

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PF Gold
P: 23,274

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 = L_{min}/L_{max} the ratio of the two distance scales. In the limit as φ > 0, the deformation q > 1 and the quantum group SU_{q}(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 φ^{2}) I suppose it is obvious that φ can change as the universe evolves. It can change even without L_{min} changing, if L_{max} 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 L_{max} would change over time. That is if we take it to be the CEHwhich 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 40some 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 L_{min}) and in the CEH (related to L_{max}). 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. 


#8
May1111, 02:36 PM

Astronomy
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PF Gold
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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
May1111, 05:11 PM

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#10
May1111, 05:30 PM

P: 640

http://arxiv.org/abs/1105.1898
So, now, who can argue ... why this scale cannot be at 10^18m? 


#11
May1111, 06:07 PM

P: 343

It seems strange to me to have the quantity
[tex]l_{pl}^2 \Lambda = G \Lambda[/tex] turn up in an equation. Since in the action we always have the cosmological constant in the form [tex] \frac{\Lambda}{G}[/tex] This is the energy density that must couple to spacetime in order to form a dimensionless quantity. Or just think of the Einstein equations [tex]G_{\mu \nu} + \Lambda = 8 \pi G T_{\mu \nu} [/tex] One would have to multiply through by G to get [tex] G \Lambda[/tex] 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? 


#12
May1111, 06:11 PM

Astronomy
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PF Gold
P: 23,274

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


#13
May1111, 06:16 PM

Astronomy
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PF Gold
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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) 


#14
May1111, 06:39 PM

P: 343




#15
May1411, 08:08 AM

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


#16
May1411, 08:33 AM

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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
May1511, 04:08 PM

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P: 5,464

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


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