Relationship between classical $q$-deformed General Relativity and the cosmological constant

The frame-connection formulation of pure General Relativity in 4 dimensions is given by the action $$S_{4d}[e, \omega] = \frac{1}{2 \kappa} \int \varepsilon_{IJKL} e^I \wedge e^J \wedge F^{KL},$$ where $$e^I = e^I_{\mu} dx^{\mu}$$ is the frame field, $$\omega^I_{\;J} = \omega^I_{\;J\;\mu} dx^{\mu}$$ is the spin connection and $$F^I_{\;J} = d\omega^I_{\;J} + \omega^{IK} \wedge \omega_{KJ}$$ is its curvature tensor. In my notation, $$I, J, \dots$$ are internal flat space indices and $$\mu, \nu, \dots$$ are coordinate indices on the space-time manifold, though I use differential forms whenever possible for bookkeeping.

The analogous action in 3 dimensions takes an even simpler form $$S_{3d}[e, \omega] = \frac{1}{2 \kappa} \int \varepsilon_{IJK} e^I \wedge F^{JK}.$$ This is a special case of the topological $$BF$$ theory action for the group $$SL(2,\mathbb{R})$$, the non-compact form of $$SU(2)$$. In Euclidean signature, the group is just the compact $$SU(2)$$.

A well known toy model of quantum gravity in $$3d$$ is the Ponzano-Regge model, which is also the result of applying the covariant Loop Quantum Gravity programme in Euclidean signature in $$3d$$.

The Ponzano-Regge model has a peculiar "twin" called the Turaev-Viro model defined for $$q \in \mathbb{C}$$ a root of unity. The definition mirrors the definition of the Ponzano-Regge model that makes use of the representation theory of $$SU(2)$$, except it employs the representation theory of the corresponding $$q$$-deformed Hopf algebra $$SU_q (2)$$.

It has many interesting features, like finiteness of its amplitudes (in fact, Turaev-Viro is frequently used to regularize the formal infinite Ponzano-Regge amplitudes), triangulation independence (which essentially means Turaev-Viro is a TQFT, even though it uses triangulations in its definition). But probably the most unexpected consequence is that Turaev-Viro turns out to give $$3d$$ General Relativity with a non-zero cosmological constant $$\Lambda$$ that depends on $$q$$, and vanishes in the $$q \rightarrow 1$$ limit when Turaev-Viro becomes Ponzano-Regge.

This relationship between $$q$$-deformed Lie groups (aka quantum groups, but I try to avoid this terminology, because physicists have a very different notion of "quantum" which is not to be confused) and the cosmological constant fascinates me.

I am curious if it also exists on a purely classical level. To be more precise, my hypothesis is that a similar action with the gauge group $$SL(2,\mathbb{R})$$ or $$SU(2)$$ replaced by the corresponding $$q$$-deformed Hopf algebra, whatever that means, somehow automatically contains a cosmological constant term without it being explicitly present in the theory.

Similarly, one may also hope that a similar mysterious relationship between $$q$$-deformed groups and the cosmological constant exists in $$4d$$.

I'm looking for references that pursued this direction.

• Have you seen Fairbairn & Meusburger, 2010 or Han, 2011? These papers seem to be along the lines of your reasoning. Jul 11, 2020 at 14:12
• @A.V.S. thanks – I am familiar with Han, 2011 (and I just had a glance over F&M, 2010). This question is about whether this correspondence exists for the classical General Relativity with $SO(3,1)$ replaced with the corresponding $q$-deformed Hopf algebra. My understanding is that both references you linked deal with spinfoam models, which are quantum. Jul 11, 2020 at 16:42
• Yes, c.c. there is recovered in classical limit from quantum theory. But if you put deformation on the classical level there would be no equivalence between deformed theory and EH action gravity with c.c., those would be two different theories, yes? Or are you just expecting an (A)dS vacuum as a solution? Jul 11, 2020 at 19:40
• @A.V.S. I expected there to be some sort of a connection, if not equivalence, between the classical theories Jul 11, 2020 at 19:49

In 4D case, if you q-deform the Lorentz group (or more precisely $$Spin(1,3)$$) $$SO(1,3)$$ to arrive at the de Sitter group $$SO_q(1,3) = SO(1,4),$$ then the curvature $$F^I_{\;J} = d\omega^I_{\;J} + \omega^{IK} \wedge \omega_{KJ}$$ will be q-deformed to (given that the q-deformed Lorentz connection has two parts $$\omega^I_{\;J} + qe^I$$) $$F^I_{\;Jq} = (d\omega^I_{\;J} + \omega^{IK} \wedge \omega_{KJ}) + q^2(e^I\wedge e_J).$$ And the Lagrangian $$S_{4d}[e, \omega] = \frac{1}{2 \kappa} \int \varepsilon_{IJKL} e^I \wedge e^J \wedge F^{KL}$$ will be q-deformed to $$S_{4dq}[e, \omega] = \frac{1}{2 \kappa} \int \varepsilon_{IJKL} e^I \wedge e^J \wedge F^{KL}_q.$$ Therefore the $$q^2$$ part will contribute a Lagrangian term $$\frac{q^2}{2 \kappa} \int \varepsilon_{IJKL} e^I \wedge e^J \wedge e^K \wedge e^L,$$ which amounts to a cosmological constant.
BTW, there is also a torsion part of the q-deformed curvature too $$F^I_{q} = q(de^I + \omega^{IJ} \wedge e_J),$$ which will not contribute to the q-deformed Lagrangian.
• Nice! How does $e$ make it into the $q$-deformed curvature though? Jul 13, 2020 at 18:49
• The dS (q-deformed Lorentz) connection has two parts $\omega^I_{\;J} + q e^I$. Jul 13, 2020 at 18:56
• I think I understand the intuition. $\varepsilon_{IJKL} e^K \wedge e^L$ is canonically conjugate to $\omega_{IJ}$, so it doesn’t commute with it and adding it to $\omega$ will make it a quantum group connection. Is that what you meant? Jul 14, 2020 at 2:25
• @Prof.Legolasov, actually it's adding the non-commuting $e^K$ (rather than $\varepsilon_{IJKL} e^K \wedge e^L$) to $\omega_{IJ}$, since the q-deformed connection is required to be a 1-form ( $\varepsilon_{IJKL} e^K \wedge e^L$ is 2-form). You can rewrite the q-deformed connection $\omega^I_{\;J} + qe^K=\omega^I_{\;J} + qe^K_{\;4} = \omega^a_{q\;b}$, with $a,b = 0, 1, 2, 3, 4$. Jul 14, 2020 at 13:49