2
$\begingroup$

Beyond the Standard Model and the General relativity invariant groups, recently we have met (again) the BSM groups of asymptotic symmetries given by the Bondi-Metzner-Sachs (BMS) or the extended BSM groups. As far as I understand, there are at least 3 BSM symmetries:

  1. Supertranslations. Meaning: angle-dependent directional time translations (angle with respect what?I suppose the whole flat spacetime but not sure) so I could understand it as the energy entering into a given direction, I think.
  2. Superrotations. Meaning: angle-dependent directional space translations. I believe it can be understood as the angular momentum entering from a given direction.
  3. SuperLorentz transformations. Meaning: angle-dependent directional boost. Interpretation: centre-of-mass-energy conservation.

Is there any other asymptotic symmetry out there? What is the physical interpretation of these new possible symmetries?

BONUS: are the 3 symmetries above contained in the Diff(S²) group? Can they be generalized to higher dimensions?

$\endgroup$
1
  • $\begingroup$ There are also large gauge transformations, which are local gauge transformations that approach angle dependent functions at the asymptotic limit. $\endgroup$
    – schris38
    Commented Jul 8, 2022 at 17:20

1 Answer 1

1
$\begingroup$

I believe the answer to this question demands one general definition of what an asymptotic symmetry is. In general, this can be answered employing the Covariant Phase Space formalism. References are https://arxiv.org/abs/1801.07064 and https://arxiv.org/abs/2009.14334. I'll provide a summary here. Suppose we have a theory in which we have fields $\Phi_i$ possibly including the metric tensor. Let $\mathbf{L}$ be its Lagrangian form. The Covariant Phase Space prescription tells us that starting from a generic variation of $\mathbf{L}$ it can be put in the form $$\delta{\mathbf{L}}=E[\Phi_i]\delta \Phi_i+d{\pmb\Theta}[\Phi_i,\delta\Phi_i]\tag{1}$$ which defines a pre-sympletic potential density ${\pmb \Theta}$. Then, taking one anti-symmetric variation we define the pre-sympletic density $${\pmb \omega}[\Phi_i,\delta_1\Phi_i,\delta_2\Phi_i]=\delta_1{\pmb\Theta}[\Phi_i,\delta_2\Phi_i]-\delta_2{\pmb\Theta}[\Phi_i,\delta_1\Phi_i].\tag{2}$$

From ${\pmb \omega}$ we may construct the pre-sympletic form ${\pmb \Omega}$

$${\pmb\Omega}[\Phi_i,\delta_1\Phi_i,\delta_2\Phi_i]=\int_\Sigma {\pmb \omega}[\Phi_i,\delta_1\Phi_i,\delta_2\Phi_i]\tag{3}.$$

Now suppose the theory under consideration has gauge/local symmetries. Let $\delta_1\Phi_i=\delta_\varepsilon\Phi_i$ be such a gauge transformation and $\delta_2\Phi_i=\delta\Phi_i$ a generic variation. From standard Hamiltonian mechanics, we would expect $${\pmb\Omega}[\Phi_i,\delta_\varepsilon\Phi_i,\delta\Phi_i]=-\delta Q_\varepsilon[\Phi_i]\tag{4}$$ where $Q_\varepsilon[\Phi_i]$ is the charge associated to such symmetry, so one is motivated to study (4) for such local symmetries. What one finds is that as a consequence of Noether's second theorem, valid for local symmetries, (4) is actually one codimension two integral over $\partial \Sigma$. One then finds the following possibilities:

  1. For some set of $\varepsilon$ that integral vanishes. This means from (4) that $\delta_\varepsilon$ is a degenerate vector field for ${\pmb \Omega}$ thereby making the phase space ill-defined. These are therefore not true symmetries, these are the gauge transformations which are mere redundancies and must be gauge-fixed.

  2. For some set of $\varepsilon$ one finds $Q_\varepsilon\neq 0$ as a codimension two integral over $\partial\Sigma$. These are the asymptotic symmetries.

  3. For some set of $\varepsilon$ one does not actually find (4). Instead of finding $-\delta Q_\varepsilon[\Phi_i]$ one finds and additional term which is not $\delta$ of something. This gives one non-integrable charge.

The subject of your question is (2). This is the definition of an asymptotic symmetry: a gauge transformation with non-trivial action at the boundary of spacetime, so that the charge computed from (4) is non-zero and therefore generates the symmetry in a classical phase space through the Poisson brackets defined by ${\pmb \Omega}$.

So it is important to understand that your question "is there any other asymptotic symmetry out there?" on the outset has the issue that this question is theory dependent. What is the Lagrangian? What are the boundary conditions? What is the background? All these things have impact on what asymptotic symmetries you will find.

Now some examples, in Einstein gravity in asymptotically flat spacetimes one finds the extended BMS group with supertranslations and superrotations. On the other hand, for Einstein gravity in asymptotically AdS$_4$ spacetimes with Dirichlet boundary conditions one finds instead the AdS isometry group while with Neumann boundary conditions the asymptotic symmetry group is empty. There is even the possibility of adopting a certain mixed boundary condition which gives you $\mathbb{R}\times{\cal A}$ where $\mathbb{R}$ are time translations and ${\cal A}$ is a group of two-dimensional area preserving diffeomorphisms, see https://arxiv.org/abs/1905.00971 for details. In asymptotically AdS$_3$ spacetimes one finds a Virasoro algebra. These are just Einstein gravity examples. One also may study asymptotic symmetries in gauge theories with generic gauge group, which you can found in https://arxiv.org/abs/2009.14334. There one finds for example Large Gauge Transformations.

Finally, about ${\rm Diff}(S^2)$ on asymptotically flat spacetimes and its relation to the extended BMS group. Lorentz transformations and their Virasoro extension to superrotations indeed comprise a subset of ${\rm Diff}(S^2)$. Supertranslations have a different nature because they are actually shifts along the generators of ${\cal I}^\pm$ shifting $u\to u+f(z,\bar z)$ and $v\to v+f(z,\bar z)$.

Regarding the full ${\rm Diff}(S^2)$ as an asymptotic symmetry, there are some authors who have argued in the past that the superrotations should be extended to the whole ${\rm Diff}(S^2)$, but there has been one recent argument against that constructed from the celestial holography point of view. See https://arxiv.org/abs/2208.13304 for the details of that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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