# In what sense is BMS a symmetry? (What is kept invariant?)

Recently I've started to read about BMS (Bondi-Metzner-Sachs), and I've encountered several statements such as the following (from [1]).

[I]t turned out that the asymptotic symmetry group at null infinity in four dimensions is not the Poincaré group, but an enhanced group where translations are replaced by supertranslations.

My question: In what sense is this BMS a symmetry? I know that Poincaré is a symmetry in the sense that the fundamental equations of motion of the fields are kept invariant (look the same) once all the objects in the equations are transformed by Poincaré. Is that the same sense in which BMS is a symmetry, when we consider only fields at null infinity?

• According to physics.stackexchange.com/q/130117/17609, "the "BMS symmetry" is just an allowed subset of coordinate diffeomorphisms which leaves the asymptotic flatness of the space-time intact." Not sure if that helps. Perhaps the remainder in the link does. Perhaps not. Commented Sep 14, 2015 at 19:11
• Actually, maybe the sense in which Poincare is a symmetry is just that the metric is unchanged under its action (its generators are Killing vector fields)
– Lior
Commented Sep 14, 2015 at 19:28

You can look back at the original paper in which Sachs first constructed the full BMS group. (He called it the "generalized Bondi-Metzner group", but we have since added his name to get to "Bondi-Metzner-Sachs group".) The idea is that the BMS group is an asymptotic symmetry of the metric in an asymptotically flat spacetime. Sachs quotes earlier results that derive the general asymptotic behavior of the metric $g_{ab}$ in such a spacetime. Full details are in the paper, but it's relevant to note that he uses coordinates $u, r, \theta, \phi$, where $u$ is retarded time and $r, \theta, \phi$ are (basically) the usual spherical coordinates. An infinitesimal coordinate transformation will change the metric by $\delta g_{ab}$. Sachs argues that this change must behave asymptotically according to \begin{align} \tag{1} \delta g_{tt} &= \mathcal{O}(r^{-1}), \\ \tag{2} \delta g_{tr} &= \mathcal{O}(r^{-2}),\\ \tag{3} \delta g_{tA} &= \mathcal{O}(1),\\ \tag{4} \delta g_{rr} &= 0,\\ \tag{5} \delta g_{rA} &= 0,\\ \tag{6} \delta g_{AB} &= \mathcal{O}(r),\\ \tag{7} \delta g_{AB} g^{AB} &= 0,\\ \end{align} where $A$ and $B$ can range over $\theta$ and $\phi$. Any smooth infinitesimal coordinate transformation is generated by a vector field $\xi^a$, and will generally alter the metric according to $$\delta g_{ab} = -\nabla_a \xi_b - \nabla_b \xi_a.$$ Sachs uses these expressions to explicitly find the generators of the BMS group.

So the answer to the question is that the asymptotic behavior of the metric is kept invariant, and that is done by ensuring that any of the BMS transformations only change the metric in a way that asymptotically obeys conditions (1) through (7).

Nice question. I'm studying the same topic right now so maybe we can discuss it to understand better the topic.

The BMS trasformations are asymptotical symmetries; They leave the asympotic metric invariant. This isn't an exactly isometry, and the Killing field associated are only asympotically killing. The fact is that BMS trasformation leave the Boundary Condition (BC) at the Null infinity $I^\pm$ invariant.

In the original work of Bondi et al. is proved that the change in the mass of a physical system far from $I^\pm$ can change only if the "Bondi News" function is non zero. In this case this means that a gravitational wave coming along a null curve from $I^-$ interact with the system and then arriving at $I^+$ brings with itself some "news".

Sachs proved that BMS has a conformal subgroup which, made it regular on the whole $S^2$, is isomorphic to the $SO(3,1)$. Indicated that group with $L$, there is another subgroup, which is a Normal Subgroup of BMS, indicated by $N$. This is the Supetraslation subgroup $N$ and $$BMS/N \sim L$$

Ashtekar proved that there is not conserved quantity associated with Supetranslations. (They all vanish).

Of course this is what I got, so maybe is wrong or not exact. I hope we can discuss about it.