The isometry group of global $AdS_{d+1}$ is well-known to be $SO(d,2)$. I have a suspicion that when the spacetime is asymptotically AdS, with dynamical gravity in the bulk, the symmetry group gets enhanced to the Weyl group. This would be analogous to how the Poincare group gets enhanced to the BMS symmetry group of asymptotically flat spacetime. I'd like to know if my suspicion is correct.

My rough reasoning is as follows. Let us write the asymptotically AdS metric in Fefferman-Graham form:

$$ \tag{1} ds^2 = \frac{l^2}{r^2}\left(dr^2 + (g^{(0)}_{ij}+ r g^{(1)}_{ij} + ... + r^d g^{(d)}_{ij}+...)dx^idx^j \right),$$

where the Einstein equations fix all other coefficients in terms of $g_{ij}^{(0)}$ and $g_{ij}^{(d)}$.

In the asymptotically flat case, the BMS symmetry group consists of transformations that leave the asymptotic form of the metric invariant. Likewise, in the AdS case there is a subgroup of diffeomorphisms (the so-called PBH transformations) which leave the metric in FG form but with new coefficients. In particular, they act on $g_{ij}^{(0)}$ as Weyl-recalings. Therefore, I'd like to conclude that gravity in asymptotically AdS has Weyl group symmetry.

My issue is that I don't truly know what "symmetry group" means in this context. Since I study QFT, I want to consider symmetries of the gravitational path integral viewed as a function of the boundary data $g_{ij}^{(0)}$ and $g_{ij}^{(d)}$. More concretely, to say that the theory has Weyl symmetry, I want the path integral to be invariant under Weyl-rescaling both the "background" $g_{ij}^{(0)}$ and the "dynamical field" $g_{ij}^{(d)}$. The problem is, I don't see how this type of symmetry is related to the notion of "leaving the asymptotic form of the metric invariant".

So, what is the appropriate notion of "symmetry group" in an asymptotically AdS theory of gravity, and why? And is it correct that this symmetry group is enhanced from $SO(d,2)$ to Weyl, just as Poincare is enhanced to BMS in the flat case?

  • $\begingroup$ You will probably find this paper useful: arxiv.org/abs/1905.00971. In short, this depends on the boundary conditions chosen at the timelike boundary of ${\rm AdS}$. Employing the standard Dirichlet boundary conditions (which are the ones chosen in AdS/CFT most of the time) there is no enhancement and the asymptotic symmetry group is just ${\rm SO}(d,2)$. You will find, though, that for a particular choice of boundary conditions you get something else. $\endgroup$
    – Gold
    Commented Nov 21, 2022 at 23:38
  • 1
    $\begingroup$ In any case, you might be interested in understanding the general definition of an asymptotic symmetry group. This review arxiv.org/abs/1801.07064 by Compère is a great resource at that. This paper arxiv.org/abs/2009.14334 is also great, they have a very complete review in section 2. The basic idea is that one must look to the Hamiltonian charge that generates the symmetry through the appropriate Poisson bracket. The asymptotic symmetry group will be the quotient of the allowed gauge transformations by the trivial ones, where the later are defined by a vanishing charge. $\endgroup$
    – Gold
    Commented Nov 21, 2022 at 23:42
  • $\begingroup$ +1 for the the review by Compere - it looks excellent. Having looked at the proper definition, I can see now that an asymptotic symmetry of AdS should leave all leading components invariant, so in particular $g_{ij}^{(0)}$ must be left invariant: it can't suffer a Weyl-rescaling. This is a stricter requirement than what I'd been thinking before, so it makes sense that the symmetry group doesn't get enhanced. $\endgroup$ Commented Nov 22, 2022 at 19:58
  • $\begingroup$ That said, in AdS/CFT, the conformal invariance of the CFT is supposed to be enhanced to full Weyl symmetry. So I'd still be very interested to know what this corresponds to in the bulk. $\endgroup$ Commented Nov 22, 2022 at 20:00

1 Answer 1


For any weakly asymptotically simple space-time $(\tilde{\mathcal{M}},\tilde{g})$, you can define asymptotic symmetry group as the group of conformal self-transformation of the boundary $\mathscr{I}=\partial\mathcal{M}$. If $\mathcal{M}$ is asymptotically flat, then this group is precisely the BMS-group (see Penrose1963, Tamburino and Winicour (1966)). If $\mathcal{M}$ is asymptotically $AdS_d$, then this group corresponds to the isometry of the line element (1) at infinity written in Fefferman-Graham form (see BMS/CFT correspondence, Kerr/CFT correspondence) (to derive this result, simply solve for equation (8.8) as considered in Tamburino and Winicour (1966) for the line element (1), considering scale factor $\Omega=\frac{1}{r}$).

This definition for asymptotic symmetry follows naturally from the condition that they should manifest as isometry transformation of the physical metric at $r\to\infty$. In Penrose's Conformal compactification approach, the idea of infinity ($r\to\infty$) is replaced by the three-dimensional surface $\mathscr{I}$ and therefore, the isometry transformation manifest as conformal transformation of $\mathscr{I}$. This is reminiscent of the fact that Poincare motions of Minkowski space-time become conformal motions for any conformally flat metric. Just like the Poincare group, which defines momentum and angular-momentum in flat space-time, the asymptotic symmetry group captures all information about momentum and angular momentum of a gravitating system in an asymptotically simple space-times (see T.Dray and M.Streubel 1984).

It should be noted that free z.r.m. spinor fields are invariant under the conformal transformation (conformal rescaling or conformal mapping) $g_{ab}\to \omega^2g_{ab}$:

$$\hat{\nabla}^{AA'}\hat{\phi}_{\underbrace{AB\cdots L}_n}=\omega^{-3}\underbrace{\nabla^{AA'}\phi_{AB\cdots L}}_{=0}=0$$ provided $\phi_{AB\cdots L}$ is a conformal density of weight -1 (in 3+1 dim space-times) (Refer to the end of section 5.7 in Spinors and spacetime. Vol-I). Here $a,b$ are abstract space-time indices whereas $A,B,C,...$ are abstract spinor indices. Under the action of asymptotic symmetry group, the metric on $\mathscr{I}$ transforms as $\hat{g}_{ab}=e^{2\partial_l\xi^l}g_{ab}$. Thus, any free z.r.m. spinor field $\phi_{AB...L}$ which transforms as $\phi_{AB...L}\to e^{-\partial_l\xi^l}\phi_{AB...L}$ will still satisfy mass-less field equations on $\mathscr{I}$. Under appropriate limit (Refer to section 6.8 in Spinors and spacetime. Vol-II), similar results hold for conformally coupled scalar fields as well.

  • $\begingroup$ I'd like to understand your second paragraph better. What exactly do you mean by the symmetries being "relevant". Do you mean that the S-matrix of a massless field theory coupled to asymptotically AdS gravity possesses those symmetries? Also, I don't see why a free massless field must have conformal symmetry. For example, a free scalar field is only has conformal symmetry if a curvature coupling is introduced, with exactly the right coefficient (i.e. "conformal coupling") $\endgroup$ Commented Nov 22, 2022 at 20:08
  • $\begingroup$ @nodumbquestions I have elaborated the second paragraph. I might be wrong, but as far as I know, the idea of BMS group as symmetry group for S-matrix was primarily motivated by Hawking's proposal that super-translation hairs on BH event horizon will carry all information about infalling particles. This is not a generic result and only applies when there is a matching condition b/w data on $\mathscr{I}^{\pm}$. I don't know if the same applies for asymptotically AdS gravity $\endgroup$
    – paul230_x
    Commented Nov 23, 2022 at 9:29

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