Is there a duality between (2+1)D gravity and Chern-Simons Theory? Or they merely have related features? If so, of which type and why?


2 Answers 2


Consider the local Lorentz vector $\omega^a$ constructed from the spin connection via the Hodge dual: $\omega^a=\frac12\epsilon^{abc}e_c\omega^c{}_{ab}$, where $e_a=\eta_{ab}e^b{}_\mu\mathrm dx^\mu$. Essentially since the triad and this spin vector are on the same footing, we can construct a Chern-Simon formulation of 3D gravity. In the following we work with a negative cosmological constant for simplicity, but the generalisation to arbitrary cosmological constant merely involves a slightly different algebra of generators (e.g. $\mathfrak{iso}(2, 1)$ for $\Lambda=0$).

Introduce the connections transforming in the adjoint representation of the Lorentz group $\mathrm{SO}(2,1)$ : $$ \mathfrak{so}(2,1) : [J_a,J_b]=\epsilon_{ab}{}^cJ_c \\A^a_\pm=\omega^a\pm\frac{e^a}{\ell} \\\mathbf A_\pm=A^a_\pm J_a $$ where $\ell$ is the cosmological length scale. Additionally, the matrix representation of the Hodge dual of the Ricci tensor is $$ \mathbf R=R^aJ_a=\mathrm d\omega+\omega\wedge\omega $$ Finally, since we are trying to construct an alternative formulation of 2+1D gravity, we can form a scalar action by integrating the Chern-Simons form of $\mathbf A$ over a 3-manifold: $$ I_\mathrm{CS}[\mathbf A]=\mathbf A+\mathrm d\mathbf A +\frac23\mathbf{A\wedge A\wedge A} $$ Via a routine computation, $$ \mathrm{Tr}(I[\mathbf A_+]-I[\mathbf A_-])=\frac2\ell\mathrm{Tr}\left[2e\wedge\mathbf R+\frac{2}{3\ell^2}e\wedge e\wedge e-\mathrm d(\omega\wedge e)\right] $$

Tracing over all the generators via the Killing form definition $\mathrm{tr}(J_aJ_b)\sim\kappa(J_a,J_b)\equiv\eta_{ab}$, one ends up with the following: $$ \mathrm{Tr}(I[\mathbf A_+]-I[\mathbf A_-])=\frac2\ell\left(\frac12\sqrt{-g}R+\frac1{\ell^2}\sqrt{-g}\right)\mathrm d^3 x-\frac2\ell\mathrm d(\omega^a\wedge e_a) $$

Thus the following is equivalent to the Einstein-Hilbert action, up to an irrelevant total derivative term: $$ S_\mathrm{EH}\approx\frac{\ell}{16\pi G}\int_{\mathcal M}\mathrm{Tr}(I[\mathbf A_+]-I[\mathbf A_-]) $$

This is a concrete manifestation of gauge-gravity duality in 3 dimensions, relating a pure gravitational theory to a Chern-Simons gauge theory. A nice heuristic way I like to motivate this is that the symmetry algebra obeyed by the Killing vectors of $\mathrm{AdS}_3$ is $\mathfrak{so}(2, 2)$, which shows clean chiral bisplitting as $\mathfrak{so}(2, 2)\cong\mathfrak{so}(2, 1)\oplus\mathfrak{so}(2, 1)$, and these factors correspond to $I[\mathbf A_+]$ and $I[\mathbf A_-]$. As it turns out, the connections are flat on-shell.

One may compute, for instance, infinitesimal surface charges, phase space for $\mathrm{AdS}_3$, asymptotic symmetries (and hence the integrated surface charge) in either formalism, and see that they match classically. So we can attack the problem of 3D gravity through two angles, and we find that e.g. computing the conserved charges is easier in the CS formulation, while computing the asymptotic symmetry algebra is easier in gravity proper.

However, as Alan Garbarz remarks, this classical equivalence is only due to the invertibility of the dreibein, which may not be the case in the quantum theory if one starts probing beyond the saddle points. A naïve perturbative analysis would appear that the quantum theory is immediately valid, but the question of its validity still lingers due to non-perturbative effects. Classically and perturbatively, however, this duality is exact.

Witten (https://arxiv.org/abs/0706.3359) gives more evidence for the non-perturbative breaking of this duality, including:

  • Diffeomorphisms in 3D gravity are translated into gauge transformation of the connections. However, this only works for gauge transformations in the component connected to the identity, and so is not satisfied for a class of diffeomorphisms which may become important in the quantum theory
  • There is no a priori reason to sum over topologies in the Chern-Simons formulation

While these two features can be put in by hand, it raises issues of naturalness. Nonetheless, the Chern-Simons - 3D Gravity duality is very powerful and often serves as a foundation for exactly solvable models of 3D gravity.

  • $\begingroup$ Thanks a lot Nihar I will read it carefully! $\endgroup$ Jul 3, 2021 at 18:06

Let me make a few remarks after the answer by Nihar Karve.

  1. The relation between both actions is exactly that and nothing more. Of course, one can say that the equations of motion coincide, but this is more subtle, because one wants invertible vielbiens. See below.

  2. Einstein's equations imply that the Ricci curvature is constant, namely the space of solutions are 3D hyperbolic manifolds modulo diffeomorphisms. However, Chern-Simons' equations say that the 3D connection is flat, in other words the space of solutions is made of flat SO(2,2) principal bundles modulo gauge transformations. Both spaces are way different, in the latter you could consider (if topology permits) the zero connection, but this gives a non-invertible metric. There is also a similar Hamiltonian way of seeing this.

  3. I didn't mention boundary conditions so far, but they are actually a crucial ingredient for defining the space of solutions. In AdS3 gravity we have a somewhat canonical set of asymptotic boundary conditions, the so called Brown-Henneaux boundary conditions. In Chern-Simons you don't have any a priori reason to impose these boundary conditions, and actually they do not seem very natural, but of course you should if you want to make contact with gravity.

  4. Finally, and mainly because of point 2, the quantum theories may be different. There are many interesting works trying to make sense of Chern-Simons as a 3D quantum gravitY. Most of the times perturbatively around saddle points, where the theories should coincide. Just to name a more or less recent and relevant one, see Maloney-Witten's paper

  • $\begingroup$ Thanks a lot Alan!!!! I’ll read both answers and come back here. $\endgroup$ Jul 3, 2021 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.