- #1

- 27

- 0

I was reading somewhere (I think wikipedia on quantum geometry?) that Loop Quantum Gravity is "noncommutative"...but I'm trying to figure this out on my own (naturally, there are no citations for this claim!).

Now, my reasoning is that one would try to express various differential forms in Loop Quantum Gravity using "quantized" differential forms (see, e.g., Connes'

Now, my reasoning is that one would try to express various differential forms in Loop Quantum Gravity using "quantized" differential forms (see, e.g., Connes'

*ftp://ftp.alainconnes.org/noncommutative_differential_geometry.pdf[/URL][/i] Page 10). If anyone knows any good reference, or has any idea whether this would be the correct method for verifying the claim that Loop Quantum Gravity is noncommutative, let me know :wink:*

The various forms that come to mind are:

The Coframe 1-form: [tex]e^{I} = e_{\mu}^{I}dx^{\mu}[/tex]

The Connection 1-form: [tex] \omega^{I}_{J} = \omega_{\mu}^{I}_{J}dx^{\mu}[/tex]

and the Curvature 2-form [tex] \mathcal{R}^{I}_{J} = (1/2)*R_{\mu\nu}^{I}_{J}dx^{\mu}dx^{\nu} [/tex]

(Boy I hope that tex code worked, if not I'll just edit the post)

Would one simply find the "quantum" differentials [tex]dx^{\mu}=i[F,x^{\mu}] = i(Fx^{\mu} - x^{\mu}F)[/tex] and say "Shazam! I'm done! It's noncommutative, time for a beer!"? (I'm rather new to this noncommutative geometry idea, so I'm unsure if that would be enough to "demonstrate" in a hand-wavy manner that LQG is noncommutative.)The various forms that come to mind are:

The Coframe 1-form: [tex]e^{I} = e_{\mu}^{I}dx^{\mu}[/tex]

The Connection 1-form: [tex] \omega^{I}_{J} = \omega_{\mu}^{I}_{J}dx^{\mu}[/tex]

and the Curvature 2-form [tex] \mathcal{R}^{I}_{J} = (1/2)*R_{\mu\nu}^{I}_{J}dx^{\mu}dx^{\nu} [/tex]

(Boy I hope that tex code worked, if not I'll just edit the post)

Would one simply find the "quantum" differentials [tex]dx^{\mu}=i[F,x^{\mu}] = i(Fx^{\mu} - x^{\mu}F)[/tex] and say "Shazam! I'm done! It's noncommutative, time for a beer!"? (I'm rather new to this noncommutative geometry idea, so I'm unsure if that would be enough to "demonstrate" in a hand-wavy manner that LQG is noncommutative.)

Last edited by a moderator: