I started studying the BMS symmetry in connection with the paper: http://arxiv.org/abs/1312.2229 and there are a few strange things I noticed.

First of all, from reading the original papers by Bondi, Metzner and Sachs, I know that the "BMS symmetry" is just an allowed subset of coordinate diffeomorphisms which leaves the asymptotic flatness of the space-time intact. However, when I read the paper above, the BMS symmetry is stated in form of a vector field eq. (2.10) and (2.14). Therefore, my first question:

How do I obtain from a given subset of coordinate diffeomorphisms (which is being considered a symmetry) a corresponding vector field a'la Killing vector?

Furthermore, later in the paper they go from the BMS vector fields to generators of BMS symmetry in eq. (3.3). They do not mention how to do it, so I'd like to know:

How do I go from vector fields characterizing a symmetry to an actual generator of the symmetry?

I realize that this is some pretty advanced stuff. I am grateful for any help or suggestion!

  • $\begingroup$ In the formula $(2.10)$, you have six infinitesimal symmetries generator (because there ia an asymptotic SL(2;C) symmetry). $\zeta^z$ has $6$ components, and so has $\zeta^a \partial_a$. Each component represent a symmetry generator, and the coefficients in front of the $\partial_z, \partial_r$, etc... represent the coordinates of a Killing vector. $\endgroup$
    – Trimok
    Aug 9, 2014 at 12:45
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    $\begingroup$ Trimok, yes, there are 6 different vector fields (for SL(2,C) plus infinitely many more for the supertranslations). However, those are not Killing vectors, since the transformation is not really a symmetry but only preserves the asymptotically flat shape for the metric. Furthermore, my question was rather, how to come up with these vector fields starting from given coordinate diffeomorphisms. Also, it seems that the generator (3.3) has to do with symplectic theory on the boundary, so I think to get it is a bit more tricky. $\endgroup$
    – Kagaratsch
    Aug 10, 2014 at 14:07
  • $\begingroup$ Yes, the six $\zeta^z$ are only asymptotic $SL(2,C)$ Killing vectors. Killing vectors represent infinitesimal generators of symmetries. For instance, a time-translation symmetry has an infinitesimal generator $\partial_0$, which corresponds to a Killing vector $\zeta^\mu = \delta^\mu_0$ $\endgroup$
    – Trimok
    Aug 11, 2014 at 8:31

1 Answer 1


It is easiest to directly derive the form of the vector fields from the boundary conditions for asymptotically flat spacetimes. See for example this paper


or this one


Infinitesimally diffeomorphisms act on the metric via Lie derivative $\mathcal{L}_{\xi}g_{\mu \nu}$. In the case of BMS, you require this transformation to respect the boundary conditions on $g_{\mu \nu}$, for example $g_{uu}\approx -1 +\mathcal{O}(r^{-1})$. So the vector field should satisfy $\nabla_u \xi_u =\mathcal{O}(r^{-1})$. You may set up such equations for each component of the metric and its corresponding boundary conditions. In each case you require the vector field not to touch the "leading" Minkowski part of the metric. This is a system of differential equations, which you may then solve.

As far as the generators go, this is sort of addressed in the second paper I linked to. Though as far as eqn 3.3 goes, you may heuristically think of this just as the ADM hamiltonian weighted by a function on the sphere which turns the uniform time translation into a supertranslation (the spaces Andy is studying are Christodoulou-Klainerman spaces for which $i^0$ is a non-singular point and so you may hope to match $\mathscr I^+_-$ to $\mathscr I^-_+$ through $i^0$ where the ADM hamiltonian is defined). Demonstrating that these charges generate the correct transformations quantum mechanically is actually very subtle, as is discussed here


Hope this helps.

  • $\begingroup$ In the paper arxiv.org/pdf/1106.0213v2.pdf in equations (2.18)-(2.22) the action of the asymptotic Killing vectors on the solution space is given. They say, these results "can be worked out to be" this, but they do not say how to do it in practice. Maybe you could tell me what kind of computation that is? (It does not seem to be just a simple Lie derivative, since the indices of for instance C_{AB} do not range over {u,r,A} but only over {A}). $\endgroup$
    – Kagaratsch
    Jan 22, 2015 at 19:41
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    $\begingroup$ To derive for example the transformation law for C_{ab}, compute the lie derivative of the full metric g_{mu nu}. This will tell you how each component of the metric transforms. Then specialize to the transformation of the {ab} component, and isolate the piece linear in r. This is how C_ab transforms. $\endgroup$
    – Dan
    Jan 22, 2015 at 20:44
  • $\begingroup$ Makes sense, thank you! Wonder if I may ask one more question: In the same paper in eq. (2.11) they give a bracket for the symmetry algebra. The first bit involving $\hat Y$ is just the straightforward commutator for the Lie algebra of conformal Killing vectors on 2-sphere, but the result for $\hat T$ features extra terms involving $\bar{D}_AY^A_iT_j$. Do you know where these extra terms come from and how to derive them? (Also, in eq. (2.16) they modify the bracket even more, I wonder how to check that this should read exactly like this and not any different?) $\endgroup$
    – Kagaratsch
    Jan 22, 2015 at 21:00
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    $\begingroup$ 2.11 should be derivable from the composition rule for a semidirect product. I don't know that much about 2.16 except that similiar things are done in Brown & Henneaux for example. $\endgroup$
    – Dan
    Jan 28, 2015 at 2:39

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