aainstitutetext: Departament de Matemàtiques and IOC, Universitat Politècnica
de Catalunya
EPSEVG, Av. V. Balaguer 1, E-08800 Vilanova i la Geltrú, Spainbbinstitutetext:
Universidad Internacional de Valencia, Carrer Pintor Sorolla, 21, E-46002
València, Spainccinstitutetext: Departament de Matemàtiques, Universitat
Politècnica de Catalunya,
Campus Diagonal Sud, Carrer de Pau Gargallo, 14, 08028 Barcelona,
Spainddinstitutetext: Departament de Física Quàntica i Astrofísica and
Institut de Ciències del Cosmos (ICCUB), Universitat de Barcelona, Martí i
Franquès 1, E-08028 Barcelona, Spain

# Particle realization of Bondi-Metzner-Sachs symmetry in $2+1$ space-time

Carles Batlle b,c Víctor Campello d Joaquim Gomis<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We construct a Lorentz invariant massive particle model in (2+1) space-time
with an enlarged set of symmetries which includes Bondi-Metzner-Sachs (BMS)
translations (supertranslations), using the non-linear realization framework.
The Hamiltonian formalism for the resulting Lagrangian is constructed, and the
infinite phase-space constraints and the set of gauge transformations are
analysed. We also compute the massless limit of the theory in phase-space.
After eliminating the gauge degrees of freedom, the physical reduced space is
left only with the degrees of freedom of a standard Poincaré particle but with
a residual set of symmetries that we prove to be BMS. A similar result for the
massless limit, including in this case superrotations, is pointed out.

##  1 Introduction

Bondi-Metzner-Sachs symmetry Bondi:1962px ; Sachs:1962wk , which originated as
an extension of Poincaré symmetry for asymptotically flat space-times, has
received a lot of renewed attention in the last 15 years. One of the interest
is to deduce Weinberg’s soft graviton theorems Weinberg:1965nx  as the Ward
identities of BMS supertranslations He:2014laa ; He:2014cra ;
Campiglia:2015qka ; Kapec:2015ena . An overview of recent developments can be
found in Strominger:2017zoo . For the relation of BMS symmetry with celestial
holography see, for example, Pasterski:2021rjz ; Raclariu:2021zjz

The BMS group, which is the semi-direct product of Lorentz and
supertraslations (including ordinary space-time translations) was extended in
Barnich:2009se ; Barnich:2011ct ; Barnich:2010eb  with the inclusion of
superrotations, obtaining the extended BMS group, given by the semidirect
product of Lorentz and superrotations and supertranslations.

A canonical realization of the BMS group was constructed using the Fourier
modes of a free massive Klein-Gordon (KG) field in $2+1$. In the case of a
massless KG field, one can construct an extended BMS symmetry which includes
superrotations Batlle:2017llu . This approach was pioneered in Longhi:1997zt .
In this context, BMS symmetries are not associated with asymptotically flat
space-times. Instead, they appear as a generalization of the components
$p^{\mu}$ of ordinary momenta, by noting that the supertranslations obey the
Beltrami differential equation of the hyperbolic space $H_{2+1}$ or
alternatively as the differential equation associated with one of the Lorentz
Casimirs, $C=J^{2}-\vec{K}^{2}$, where $J$ is the angular momentum and
$\vec{K}$ are the two-dimensional boosts. The massless case is obtained from a
suitable limit of the Beltrami equation or through the Lorentz Casimir $C$
(see references above and also Delmastro:2017erq ).

With the idea of further exploring BMS symmetry, we propose a massive particle
realization in $2+1$ dimensions, based on non-linear realizations of symmetry
algebras (see Gomis:2006xw ; Bergshoeff:2022eog  and references therein). The
particle Lagrangian is constructed in an infinite-dimensional space,
generalizing the Minkowski space, that we call BMS space. The infinite number
of coordinates are associated with the supertranslations, and we also use two
Goldstone coordinates associated with the two broken boost generators.

We want to construct a Lagrangian in terms of the pull-back of the
Maurer–Cartan forms subject to two conditions: it should have the lowest
possible number of derivatives and it should be invariant under the unbroken
$SO(2)$ subgroup of the Lorentz group in $2+1$, which is the symmetry that is
preserved by the presence of a massive spinless point particle. One can
therefore choose any component of the Maurer–Cartan form which is invariant
under rotations and we take the component $\Omega_{P_{0}}$ along the generator
of time translations $P_{0}$. In $2+1$ dimensions one can also consider the
term proportional to the generator of rotations, which can be used to
construct models of particles with spin (see, for example, PLYUSHCHAY199154 ;
Mezincescu_2010 ; Gomis_2013 ; Batlle:2014sca ), but in this paper we are only
interested in the spinless case. The Lagrangian is then the pull-back of
$\Omega_{P_{0}}$ to the world-line of the particle, up to a multiplicative
constant that makes the role of the mass. If one does not consider the super-
translation degrees of freedom, this method produces the Lagrangian of a
standard spinless massive particle in canonical form. More details of the
method can be found, for instance, in Section 5.1 of Bergshoeff:2022eog .

The Hamiltonian formalism for the resulting Lagrangian is constructed, and the
infinite phase-space first-class constraints and set of gauge transformations
are analysed. We also compute the massless limit of the model. We obtain the
physical reduced space after eliminating the gauge degrees of freedom by
introducing an infinite set of gauge fixing constraints. This space is left
only with the degrees of freedom of a standard Poincaré particle. Since in the
gauge fixing procedure the rigid symmetries are maintained we prove that the
ordinary relativistic massive Poincaré particle is invariant under a
realization of the BMS symmetry that we construct using the compensating gauge
transformations, which are necessary to preserve the gauge fixing under
supertranslations. In the massless case, we further find an infinite set of
superrotations that are symmetries of the massless relativistic particle. It
turns out that the infinite set of BMS coordinates associated with the
supertraslations are not physical because they can always be gauged away, and
therefore the model does not have the so-called soft BMS modes. These results
agree with the fact that the quadratic Casimir of BMS algebra coincides with
the quadratic Casimir of the Poincaré algebra.

A different approach to the definition of BMS particles in $2+1$ dimensions is
based on the coadjoint orbit approach Oblak:2015sea ; Campoleoni:2016vsh ;
Oblak:2016eij ; Ciambelli:2022cfr . However, as far as we know, no particle
action has been constructed in the literature using this approach. For the
relation among the non-linear realization framework and the coadjoint orbit
method, see for example Bergshoeff:2022eog .

The paper is organized as follows. Section 2 we derive the BMS particle
Lagrangian in $2+1$ space-time using the non-linear realization approach. The
canonical analysis of this Lagrangian is given in Section 3, including the
discussion of the constraints, the reduced phase space and the gauge
transformations induced by the first-class constraints. Section 4 presents the
generators that realize the Poincaré symmetry in BMS coordinates, and shows
that the theory is indeed invariant under them. The massless limit of the
theory is computed in Section 5, and it is seen that the set of symmetry
generators is extended to include superrotations. The gauge fixing of the
theory is presented in Section 6, and the physical degrees of freedom of the
BMS particle are determined. Our results are discussed in Section 7, and some
possible extensions and connections with other approaches are also considered.
Detailed proofs of some of the results are presented in Appendixes (A), (B)
and (C), and Appendix (D) contains a discussion of the quadratic Casimir in
BMS space.

##  2 Non-linear realization of the BMS algebra in $2+1$ dimensions

The extended BMS algebra in $2+1$ dimensions Barnich:2011ct  without central
extensions is given by

$\displaystyle[L_{n},P_{m}]$ $\displaystyle=$ $\displaystyle i(n-m)P_{n+m},$
$\displaystyle\left[P_{n},P_{m}\right]$ $\displaystyle=$ $\displaystyle 0,$
$\displaystyle\left[L_{n},L_{m}\right]$ $\displaystyle=$ $\displaystyle
i(n-m)L_{n+m},$ (1)

with $m,n\in\mathbb{Z}$. We are interested in the subalgebra $BMS_{3}$ formed
by the Lorentz generators $L_{0}$, $L_{\pm 1}$ and the supertranslations
$P_{n}$:

$\displaystyle\left[L_{1},L_{-1}\right]=2iL_{0},\quad\left[L_{1},L_{0}\right]=iL_{1},\quad\left[L_{-1},L_{0}\right]=-iL_{-1},$
(2)
$\displaystyle\left[L_{-1},P_{m}\right]=-i(m+1)P_{m-1},\quad\left[L_{0},P_{m}\right]=-imP_{m},\quad\left[L_{1},P_{m}\right]=-i(m-1)P_{m+1},$
(3) $\displaystyle\left[P_{n},P_{m}\right]=0.$ (4)

In order to construct a massive BMS particle we should consider the coset
$BMS_{3}/SO(2)$, locally given by

$g(\\{x\\},u,v)=g_{0}(\\{x\\})e^{iL_{-1}v}e^{iL_{1}u}=g_{0}(\\{x\\})U(u,v),$
(5)

with

$g_{0}(\\{x\\})=\prod_{n\in\mathbb{Z}}e^{iP_{n}x^{n}}.$ (6)

Here $U(u,v)$ is a boost transformation generated by $L_{1}$, $L_{-1}$ and
parametrized by the Goldstone coordinates $u,v$ and
$\\{x^{n}\\}_{n\in\mathbb{Z}}$ are the BMS coordinates (with $x^{0}$, $x^{\pm
1}$ related to ordinary $2+1$ space-time). The BMS coordinates are complex,
with $x^{n}$ and $x^{-n}$ complex conjugate of each other.

The Maurer-Cartan form associated to $g$ is

$\displaystyle\Omega(g)$
$\displaystyle=-ig^{-1}\mbox{d}g=-iU^{-1}g_{0}^{-1}\mbox{d}g_{0}\,U-iU^{-1}\mbox{d}U$
(7)
$\displaystyle=\sum_{n\in\mathbb{Z}}\mbox{d}x^{n}\,U^{-1}P_{n}U-iU^{-1}\mbox{d}U.$
(8)

In the spirit of obtaining spinless particle actions, see for example
Gomis:2006xw ; Bergshoeff:2022eog , we are only interested in the terms
$\Omega_{P_{0}}$ of $\Omega$ proportional to $P_{0}$, which can only come from
$U^{-1}P_{n}U$. The detailed computation is presented in Appendix A, and the
result is

$\Omega_{P_{0}}=\mbox{d}x^{0}(1-2uv)+\sum_{n=1}^{\infty}\mbox{d}x^{n}(-1)^{n}v^{n}(n+1-2uv)+\mbox{d}x^{-1}2u+\sum_{n=2}^{\infty}\mbox{d}x^{-n}u^{n}\frac{n+1-2uv}{(1-uv)^{n}}.$
(9)

Following the standard procedure, we integrate the pullback of
$\Omega_{P_{0}}$ to the world-line of the particle

$\displaystyle
S[\\{x\\},u,v]=-\mu\int\mbox{d}\tau(\dot{x}^{0}(1-2uv)-2\dot{x}^{1}v(1-uv)+2\dot{x}^{-1}u$
(10) $\displaystyle+$
$\displaystyle\sum_{n=2}^{\infty}\dot{x}^{n}(-1)^{n}v^{n}(n+1-2uv)+\sum_{n=2}^{\infty}\dot{x}^{-n}u^{n}\frac{n+1-2uv}{(1-uv)^{n}})$
$\displaystyle=$
$\displaystyle-\mu\int\\!\\!\mbox{d}\tau\\!\left(\sum_{n=0}^{\infty}\dot{x}^{n}(-1)^{n}v^{n}(n+1-2uv)+\sum_{n=1}^{\infty}\dot{x}^{-n}u^{n}\frac{n+1-2uv}{(1-uv)^{n}}\right),$

and define the BMS particle Lagrangian

${\cal
L}=-\mu\left(\sum_{n=0}^{\infty}\dot{x}^{n}(-1)^{n}v^{n}(n+1-2uv)+\sum_{n=1}^{\infty}\dot{x}^{-n}u^{n}\frac{n+1-2uv}{(1-uv)^{n}}\right).$
(11)

The contribution to (10) of the ordinary space-time coordinates, i.e. $x^{0}$,
$x^{\pm 1}$, is

$S_{0}=-\mu\int\mbox{d}\tau\left(\dot{x}^{0}(1-2uv)+2\dot{x}^{1}(-v+v^{2}u)+2\dot{x}^{-1}u\right)\equiv\int\mbox{d}\tau{\cal
L}_{0}.$ (12)

This action corresponds to an ordinary spinless relativistic particle in flat
(2+1) Minkowski space-time, as can be seen by computing the momenta

$\displaystyle p_{0}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{0}}{\partial\dot{x}^{0}}=-\mu(1-2uv),$ (13) $\displaystyle p_{1}$
$\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{0}}{\partial\dot{x}^{1}}=-2\mu(-v+v^{2}u),$ (14) $\displaystyle p_{-1}$
$\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{0}}{\partial\dot{x}^{-1}}=-2\mu u,$ (15)

and checking the mass-shell condition

$-p_{0}^{2}+p_{1}p_{-1}=-\mu^{2}.$ (16)

Actually, if one computes the EOM given by (12) for the boost variables,

$\displaystyle-v\dot{x}^{0}+v^{2}\dot{x}^{1}+\dot{x}^{-1}$ $\displaystyle=$
$\displaystyle 0,$ (17)
$\displaystyle-u\dot{x}^{0}-\dot{x}^{1}+2uv\dot{x}^{1}$ $\displaystyle=$
$\displaystyle 0,$ (18)

solves them for $u$, $v$,

$u=\frac{\dot{x}^{1}}{\pm\sqrt{(\dot{x}^{0})^{2}-4\dot{x}^{1}\dot{x}^{-1}}},\quad
v=\frac{\dot{x}^{0}\pm\sqrt{(\dot{x}^{0})^{2}-4\dot{x}^{1}\dot{x}^{-1}}}{2\dot{x}^{1}},$
(19)

makes the change of space variables from the complex ones $x^{\pm 1}$ to the
real ones $x_{1}$, $x_{2}$, given by

$x^{\pm 1}=\frac{1}{2}(x_{1}\pm ix_{2}),$ (20)

and substitutes the resulting expressions for $u$, $v$ in (12), one gets the
ordinary space-time action with Lagrangian

${\cal
L}_{0}^{*}=\mp\mu\sqrt{(\dot{x}^{0})^{2}-(\dot{x}_{1})^{2}-(\dot{x}_{2})^{2}}.$
(21)

##  3 Canonical analysis of the BMS particle action

In order to understand the structure of the BMS Lagrangian (11) we perform the
Hamiltonian analysis. The momenta are given by

$\displaystyle\pi_{u}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}}{\partial\dot{u}}=0,$ (22) $\displaystyle\pi_{v}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal L}}{\partial\dot{v}}=0,$ (23) $\displaystyle
p_{n}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}}{\partial\dot{x}^{n}}=-\mu(-1)^{n}v^{n}(n+1-2uv),\ \ n=0,1,2,\ldots,$ (24)
$\displaystyle\bar{p}_{n}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}}{\partial\dot{x}^{-n}}=-\mu u^{n}\frac{n+1-2uv}{(1-uv)^{n}},\ \
n=1,2,\ldots.$ (25)

Notice that, since the $x^{-n}$ are complex conjugates of the $x^{n}$, then
the fact that ${\cal L}$ is real implies that $p_{n}$ and $\bar{p}_{n}$ are
also complex conjugates of each other.

From the expressions of the momenta one gets the set of primary constraints
$\pi_{u}=0$, $\pi_{v}=0$ and $\phi_{n}=0$, $\bar{\phi}_{n}=0$, with

$\displaystyle\phi_{n}$ $\displaystyle:=$ $\displaystyle
p_{n}+\mu(-1)^{n}v^{n}(n+1-2uv),\ \ n=0,1,2,\ldots$ (26)
$\displaystyle\bar{\phi}_{n}$ $\displaystyle:=$ $\displaystyle\bar{p}_{n}+\mu
u^{n}\frac{n+1-2uv}{(1-uv)^{n}},\ \ n=1,2,\ldots$ (27)

The non-zero Poisson brackets between these constraints are

$\displaystyle\\{\phi_{n},\pi_{u}\\}$ $\displaystyle=$
$\displaystyle-2\mu(-1)^{n}v^{n+1},\ n=0,1,2,\ldots$ (28)
$\displaystyle\\{\phi_{n},\pi_{v}\\}$ $\displaystyle=$
$\displaystyle\mu(-1)^{n}v^{n-1}(n(n+1)-2(n+1)uv),\ n=0,1,2,\ldots$ (29)
$\displaystyle\\{\bar{\phi}_{n},\pi_{u}\\}$ $\displaystyle=$
$\displaystyle\frac{\mu u^{n-1}}{(1-uv)^{n+1}}(2u^{2}v^{2}-2nuv+n^{2}-2uv+n),\
n=1,2,\ldots$ (30) $\displaystyle\\{\bar{\phi}_{n},\pi_{v}\\}$
$\displaystyle=$ $\displaystyle\frac{\mu
u^{n+1}}{(1-uv)^{n+1}}(n-1)(-2uv+n+2),\ n=1,2,\ldots$ (31)

Since the Lagrangian is homogeneous of degree one in the velocities, the
canonical Hamiltonian is identically zero and one must consider the Dirac
Hamiltonian

$H_{D}=\lambda_{u}\pi_{u}+\lambda_{v}\pi_{v}+\sum_{n=0}^{\infty}\lambda_{n}\phi_{n}+\sum_{n=1}^{\infty}\bar{\lambda}_{n}\bar{\phi}_{n},$
(32)

where the $\lambda$ are arbitrary functions.

If we order the constraints as
$(\pi_{u},\pi_{v},\phi_{1},\bar{\phi}_{1},\phi_{2},\bar{\phi}_{2},\ldots)$,
the infinite-dimensional matrix of Poisson brackets between them has the form

$M_{\infty}=\left(\begin{array}[]{ccccc}0&A_{1}&A_{2}&A_{3}&\cdots\\\
-A_{1}^{T}&0&0&0&\cdots\\\ -A_{2}^{T}&0&0&0&\cdots\\\
-A_{3}^{T}&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right),$ (33)

where all the entries are $2\times 2$ blocks and

$A_{i}=\left(\begin{array}[]{cc}\\{\pi_{u},\phi_{i}\\}&\\{\pi_{u},\bar{\phi}_{i}\\}\\\
\\{\pi_{v},\phi_{i}\\}&\\{\pi_{v},\bar{\phi}_{i}\\}\\\
\end{array}\right),\quad i=1,2,3,\ldots$ (34)

Since, for instance,

$A_{1}=2\mu\left(\begin{array}[]{cc}-v^{2}&-1\\\ 1-2uv&0\end{array}\right),$
(35)

has non-zero determinant, provided that $1-2uv\neq 0$, it turns out that all
the (infinite dimensional) columns to the right of $A_{1}$ can be expressed as
linear combinations of the columns which contain $A_{1}$. Hence, considering
also the first two columns of $M_{\infty}$, one can show that $M_{\infty}$ has
a rank equal to $4$. This means that, at most, we can select $4$ second-class
constraints, including necessarily $\pi_{u}$ and $\pi_{v}$, plus another two
which allow us to eliminate $u$ and $v$ in terms of two of the momenta $p_{i}$
and $\bar{p}_{i}$. Also, notice that the number of first class constraints is
infinite.

Notice that, although $u$, $v$ can be eliminated from any two of the $\phi$,
$\bar{\phi}$, it is convenient to select $\phi_{1}$, $\bar{\phi}_{1}$, as it
was done for the case of the pure Poincaré particle. The four selected
constraints

$\pi_{u},\ \ \pi_{v},\ \ \phi_{1},\ \ \bar{\phi}_{1}$ (36)

are second class, with the Poisson bracket matrix

$M=\left(\begin{array}[]{cc}0&A_{1}\\\ -A_{1}^{T}&0\\\
\end{array}\right)=2\mu\begin{pmatrix}0&0&-v^{2}&-1\\\ 0&0&1-2uv&0\\\
v^{2}&-1+2uv&0&0\\\ 1&0&0&0\end{pmatrix},$ (37)

which has determinant $16\mu^{4}(1-2uv)^{2}$ and inverse

$M^{-1}=\frac{1}{2\mu}\begin{pmatrix}0&0&0&1\\\
0&0&-\frac{1}{1-2uv}&\frac{v^{2}}{1-2uv}\\\ 0&\frac{1}{1-2uv}&0&0\\\
-1&-\frac{v^{2}}{1-2uv}&0&0\end{pmatrix}.$ (38)

From this, one can define the Dirac bracket

$\\{A,B\\}_{D}=\\{A,B\\}-\\{A,\Psi_{i}\\}M^{-1}_{ij}\\{\Psi_{j},B\\},$ (39)

where $\Psi_{j}\in\\{\pi_{u},\pi_{v},\phi_{1},\bar{\phi}_{1}\\}$.

If one demands the stability of the primary second-class constraints, i.e.

$\dot{\Psi}_{i}=\\{\Psi_{i},H_{D}\\}\stackrel{{\scriptstyle{\cal M}}}{{=}}0,$
(40)

where ${\cal M}$ is the submanifold defined by the constraints $\Psi_{i}$, one
can determine the values of the four arbitrary functions $\lambda_{u}$,
$\lambda_{v}$, $\lambda_{1}$, $\bar{\lambda}_{1}$. The result is that
$\lambda_{u}=\lambda_{v}=0$, while $\lambda_{1}$ and $\bar{\lambda}_{1}$ are
quite involved functions of all the other $\lambda$. However, in the reduced
space ${\cal M}$ we can set $\phi_{1}=\bar{\phi}_{1}=0$, and the reduced Dirac
Hamiltonian is

$H_{\cal
M}=\lambda_{0}\phi_{0}+\sum_{n=2}^{\infty}(\lambda_{n}\phi_{n}+\bar{\lambda}_{n}\bar{\phi_{n}}),$
(41)

where all the constraints are assumed to be computed on ${\cal M}$. Using
$\phi_{1}=0$ and $\bar{\phi}_{1}=0$ to effectively eliminate $u$ and $v$ in
terms of $p_{1}$ and $\bar{p}_{1}$ one has

$\displaystyle u$ $\displaystyle=$ $\displaystyle-\frac{1}{2\mu}\bar{p}_{1},$
(42) $\displaystyle v$ $\displaystyle=$
$\displaystyle-\frac{\mu}{\bar{p}_{1}}\pm\frac{1}{\bar{p}_{1}}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}},$
(43)

from which it also follows that

$uv=\frac{1}{2}\mp\frac{1}{2\mu}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}.$ (44)

Then, on the reduced space,

$\phi_{0}=p_{0}\pm\sqrt{\mu^{2}+p_{1}\bar{p}_{1}},$ (45)

and

$\displaystyle\phi_{n}$ $\displaystyle=$ $\displaystyle
p_{n}+\mu(-1)^{n}\left(-\frac{\mu}{\bar{p}_{1}}\pm\frac{1}{\bar{p}_{1}}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}\right)^{n}\left(n\pm\frac{1}{\mu}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}\right),$
(46) $\displaystyle\bar{\phi}_{n}$ $\displaystyle=$
$\displaystyle\bar{p}_{n}+\mu(-1)^{n}\bar{p}_{1}^{n}\frac{n\pm\frac{1}{\mu}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}}{\left(\mu\pm\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}\right)^{n}}$
(47) $\displaystyle=$
$\displaystyle\bar{p}_{n}+\mu(-1)^{n}\left(\frac{\mu}{\bar{p}_{1}}\pm\frac{1}{\bar{p}_{1}}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}\right)^{-n}\left(n\pm\frac{1}{\mu}\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}\right),$

for $n\geq 2$ (as a check, these expressions become identities for $n=1$). The
case $n=0$ of (45) can also be included in either (46) or (47). Notice that
the mass-shell condition can be recovered by taking the square of $\phi_{0}$.
This is inherent to the nonlinear realization formalism, since the momenta
appear always linearly in the canonical constraints (see Gomis_2013  for a
discussion of this mechanism in the case of $p$-branes).

The constraints can also be written as

$\displaystyle\phi_{n}$
$\displaystyle=p_{n}+\frac{\mu}{\bar{p}_{1}^{n}}f_{n}^{\pm}(p_{1}\bar{p}_{1}),$
(48) $\displaystyle\bar{\phi}_{n}$
$\displaystyle=\bar{p}_{n}+\mu\bar{p}_{1}^{n}g_{n}^{\pm}(p_{1}\bar{p}_{1}),$
(49)

for $n=0,2,3,\ldots$, where

$\displaystyle f_{n}^{\pm}(x)$
$\displaystyle=\left(\mu\mp\sqrt{\mu^{2}+x}\right)^{n}\left(n\pm\frac{1}{\mu}\sqrt{\mu^{2}+x}\right),$
(50) $\displaystyle g_{n}^{\pm}(x)$
$\displaystyle=\left(-\mu\mp\sqrt{\mu^{2}+x}\right)^{-n}\left(n\pm\frac{1}{\mu}\sqrt{\mu^{2}+x}\right),$
(51)

which satisfy, for $n\geq 1$,

$\displaystyle\frac{\mbox{d}}{\mbox{d}x}f_{n}^{\pm}(x)$
$\displaystyle=\mp\frac{n+1}{2\sqrt{\mu^{2}+x}}f_{n-1}^{\pm}(x),$ (52)
$\displaystyle\frac{\mbox{d}}{\mbox{d}x}g_{n}^{\pm}(x)$
$\displaystyle=\pm\frac{n-1}{2\sqrt{\mu^{2}+x}}g_{n+1}^{\pm}(x),$ (53)

and also the second-order recurrence relation

$\displaystyle(n-1)f_{n+1}^{\pm}(x)\pm
2n\sqrt{\mu^{2}+x}f_{n}^{\pm}(x)+(n+1)xf_{n-1}^{\pm}(x)=0,\quad n\geq 1,$ (54)
$\displaystyle(n+1)g_{n-1}^{\pm}(x)\pm
2n\sqrt{\mu^{2}+x}g_{n}^{\pm}(x)+(n-1)xg_{n+1}^{\pm}(x)=0,\quad n\geq 1.$ (55)

The signs $\pm$ which appear in the above expressions correspond to different
sheets of the constraint manifold in reduced space, parametrized by $p_{1}$
and $\bar{p}_{1}$. Notice that constraints $\phi_{n},\bar{\phi}_{n}$ are first
class. Therefore we expect that the physical degrees of freedom of the BMS
particle will be finite-dimensional.

In the reduced phase space, with the boost variables $u$, $v$ eliminated in
terms of $p_{1}$, $\bar{p}_{1}$, the symmetry between the momenta
corresponding to coordinates with positive and negative index is restored, and
also that the constraint $\bar{\phi}_{n}$ is the complex conjugate of
$\phi_{n}$, thereby justifying the notation. It will also be convenient to
introduce the functions $P_{n}$, $\bar{P}_{n}$ of the variables $P_{1}$,
$\bar{P}_{1}$ defined by

$\displaystyle P_{n}$
$\displaystyle=-\frac{\mu}{\bar{p}_{1}^{n}}f_{n}^{\pm}(p_{1}\bar{p}_{1}),$
(56) $\displaystyle\bar{P}_{n}$
$\displaystyle=-\mu\bar{p}_{1}^{n}g_{n}^{\pm}(p_{1}\bar{p}_{1}),$ (57)

and, in particular,

$P_{0}=\mp\sqrt{\mu^{2}+p_{1}\bar{p}_{1}},$ (58)

in terms of which the constraints are $\phi_{n}=p_{n}-P_{n}$,
$\bar{\phi}_{n}=\bar{p}_{n}-\bar{P_{n}}$, $\phi_{0}=p_{0}-P_{0}$. Notice also
that $P_{1}=p_{1}$, $\bar{P}_{1}=\bar{p}_{1}$.

Due to the fact that $M^{-1}$ does not have contributions in the lower-right
square, the Dirac brackets of the variables $x^{n}$, $x^{-n}$, $p_{n}$,
$\bar{p}_{n}$ do not change with respect to the Poisson ones,

$\displaystyle\\{x^{n},p_{m}\\}_{D}=\\{x^{n},p_{m}\\}=\delta^{n}_{m},$ (59)
$\displaystyle\\{x^{-n},\bar{p}_{m}\\}_{D}=\\{x^{-n},\bar{p}_{m}\\}=\delta^{n}_{m}.$
(60)

First of all, the $x^{n}$, $x^{-n}$, $p_{n}$, $\bar{p}_{n}$ have zero Poisson
brackets with all the $4$ second-class constraints for $n\geq 2$, so only the
cases for $n=1$ need to be discussed. Since $p_{1}$, $\bar{p}_{1}$ have also
zero brackets with all the constraints, all brackets involving either of them
remain also unchanged. Finally,

$\\{x^{1},x^{-1}\\}_{D}=0-\\{x^{1},\Psi_{i}\\}M_{ij}^{-1}\\{\Psi_{j},x^{-1}\\},$

but the only non-zero brackets of the $x^{\pm 1}$ are with $\phi_{1}$ and
$\bar{\phi}_{1}$, and this selects the lower-right square of $M^{-1}$, which
is identically zero. The non-trivial Dirac brackets are those involving $u$,
$v$, $\pi_{u}$, $\pi_{v}$ and $x^{\pm 1}$:

$\displaystyle\\{u,x^{1}\\}_{D}=0,\quad\\{v,x^{1}\\}_{D}=-\frac{1}{2\mu}\frac{1}{1-2uv},$
(61)
$\displaystyle\\{u,x^{-1}\\}_{D}=\frac{1}{2\mu},\quad\\{v,x^{-1}\\}_{D}=\frac{1}{2\mu}\frac{v^{2}}{1-2uv},$
(62)
$\displaystyle\\{\pi_{u},u\\}_{D}=\\{\pi_{u},v\\}_{D}=\\{\pi_{v},u\\}_{D}=\\{\pi_{v},v\\}_{D}=0,$
(63)
$\displaystyle\\{\pi_{u},x^{1}\\}_{D}=\\{\pi_{u},x^{-1}\\}_{D}=\\{\pi_{v},x^{1}\\}_{D}=\\{\pi_{v},x^{-1}\\}_{D}=0.$
(64)

Summing up, in the reduced phase space one has coordinates

$\\{x^{n},\ x^{-m},\ p_{n},\ \bar{p}_{m}\\}_{\stackrel{{\scriptstyle
n=0,1,2,\ldots,}}{{m=1,2,\ldots}}}$ (65)

with Hamiltonian (41), where the constraints are given by (45), (46), (47),
and with ordinary brackets. The first class constraints
$\phi_{n},\bar{\phi}_{n}$ will generate infinite gauge transformations given
by the canonical generator

$G=\epsilon(\tau)\phi_{0}+\sum_{m\geq
2}\left(\alpha_{m}(\tau)\phi_{m}+\beta_{m}(\tau)\bar{\phi}_{m}\right).$ (66)

For instance, the re-parametrization associated to
$\phi_{0}=p_{0}\pm\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}$ acts only on the standard
space-time coordinates, and is given by

$\displaystyle\delta x^{0}$ $\displaystyle=$
$\displaystyle\epsilon(\tau)\\{x^{0},\phi_{0}\\}_{D}=\epsilon(\tau),$ (67)
$\displaystyle\delta x^{1}$ $\displaystyle=$
$\displaystyle\epsilon(\tau)\\{x^{1},\phi_{0}\\}_{D}=\pm\epsilon(\tau)\frac{\bar{p}_{1}}{2\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}}=-\epsilon\frac{\bar{p}_{1}}{2P_{0}},$
(68) $\displaystyle\delta x^{-1}$ $\displaystyle=$
$\displaystyle\epsilon(\tau)\\{x^{-1},\phi_{0}\\}_{D}=\pm\epsilon(\tau)\frac{{p}_{1}}{2\sqrt{\mu^{2}+p_{1}\bar{p}_{1}}}=-\epsilon\frac{{p}_{1}}{2P_{0}},$
(69)

where the function $P_{0}$ of $p_{1},\bar{p}_{1}$ has been used to rewrite the
final expression. Furthermore, $\delta p_{0}=\delta p_{1}=\delta\bar{p}_{1}=0$
and hence also $\delta u=\delta v=0$.

In order to check the action of this symmetry on the Lagrangian ${\cal L}$ it
suffices to consider the action on ${\cal L}_{0}$, since the re-
parametrization acts only on $x^{0},x^{\pm 1}$. It is convenient to re-define
the arbitrary function $\epsilon(\tau)$ as $\epsilon(\tau)/(2P_{0})$, so that

$\delta x^{0}=2P_{0}\epsilon(\tau),\ \delta x^{1}=-\epsilon(\tau)\bar{p}_{1},\
\delta x^{-1}=-\epsilon(\tau){p}_{1},$ (70)

and also to write the Lagrangian in terms of the momenta
$p_{1},\bar{p}_{1}$,111If one uses the equations of motion for
$p_{1},\bar{p}_{1}$ to eliminate these non-dynamical variables, the result is
the standard Poincaré Lagrangian in the form (21).

${\cal L}_{0}=\dot{x}^{0}P_{0}+\dot{x}^{1}p_{1}+\dot{x}^{-1}\bar{p}_{1}.$ (71)

One has then, taking into account that the $p_{1}$, $\bar{p}_{1}$ do not
transform under this symmetry,

$\displaystyle\delta{\cal L}_{0}$
$\displaystyle=2\frac{\mbox{d}}{\mbox{d}\tau}(\epsilon
P_{0})P_{0}-\frac{\mbox{d}}{\mbox{d}\tau}(\epsilon\bar{p}_{1})p_{1}-\frac{\mbox{d}}{\mbox{d}\tau}(\epsilon
p_{1})\bar{p}_{1}$ (72)
$\displaystyle=\epsilon(2P_{0}\dot{P}_{0}-\frac{\mbox{d}}{\mbox{d}\tau}(p_{1}\bar{p}_{1}))+\dot{\epsilon}(2P_{0}^{2}-2p_{1}\bar{p}_{1})=\epsilon\frac{\mbox{d}}{\mbox{d}\tau}(P_{0}^{2}-p_{1}\bar{p}_{1})+2\dot{\epsilon}(P_{0}^{2}-p_{1}\bar{p}_{1})$
(73) $\displaystyle=\frac{\mbox{d}}{\mbox{d}\tau}(2\mu^{2}\epsilon),$ (74)

where $P_{0}^{2}-p_{1}\bar{p}_{1}=\mu^{2}$ has been used.

Similarly, one can study the transformation of the Lagrangian (11) under the
transformation generated by $\phi_{m}$ for $m\geq 2$. We write the Lagrangian
(11) in the notationally convenient form

${\cal
L}=\sum_{n=0}^{\infty}\dot{x}^{n}P_{n}+\sum_{n=1}^{\infty}\dot{x}^{-n}P_{-n},$
(75)

where $P_{-n}=\bar{P}_{n}$ and all of them are functions of $p_{1}$,
$p_{-1}=\bar{p}_{1}$ (or of $u$ and $v$).

As shown in Appendix B, if $G=\alpha_{m}(\tau)\phi_{m}$ one has

$\delta x^{n}=\\{x^{n},G\\}_{D}=\begin{cases}\alpha_{m}&\text{if $n=m$},\\\
-\alpha_{m}(m+1)\frac{P_{m-1}}{2P_{0}}&\text{if $n=1$},\\\
\alpha_{m}(m-1)\frac{P_{m+1}}{2P_{0}}&\text{if $n=-1$},\\\
0&\text{otherwise},\end{cases}$ (76)

and then

$\displaystyle\delta{\cal
L}=\frac{\mbox{d}}{\mbox{d}\tau}\left(\epsilon_{m}(2P_{0}P_{m}-(m+1)p_{1}P_{m-1}+(m-1)\bar{p}_{1}P_{m+1})\right),$
(77)

where the parameter of the transformation has been written as
$\alpha_{m}=2P_{0}\epsilon_{m}$. Notice that the function inside the total
derivative depends on $p_{1}$, $\bar{p}_{1}$ and hence, through (14), (15), on
the original Lagrangian variables $u,v$. Similarly, for
$G=\beta_{m}\bar{\phi}_{m}$, one has

$\delta x^{n}=\\{x^{n},G\\}_{D}=\begin{cases}\beta_{m}&\text{if $n=-m$},\\\
\beta_{m}(m-1)\frac{\bar{P}_{m+1}}{2P_{0}}&\text{if $n=1$},\\\
-\beta_{m}(m+1)\frac{\bar{P}_{m-1}}{2P_{0}}&\text{if $n=-1$},\\\
0&\text{otherwise},\end{cases}$ (78)

and then,

$\displaystyle\delta{\cal
L}=\frac{\mbox{d}}{\mbox{d}\tau}\left(\epsilon_{m}(2P_{0}\bar{P}_{m}-(m+1)\bar{p}_{1}\bar{P}_{m-1}+(m-1){p}_{1}\bar{P}_{m+1})\right),$
(79)

with $\beta_{m}=2P_{0}\epsilon_{m}$.

Notice that, in fact, these generators do not yield real transformations of
${\cal L}$, since they do not respect the fact that $x^{n}$ and $x^{-n}$ are
complex conjugates, but this can be easily solved by working with the real
generators

$\displaystyle
G_{m}=\epsilon_{m}(\phi_{m}+\bar{\phi}_{m}),\quad\epsilon_{m}^{*}=\epsilon_{m},$
(80)
$\displaystyle\bar{G}_{m}=\epsilon_{m}(\phi_{m}-\bar{\phi}_{m}),\quad\epsilon_{m}^{*}=-\epsilon_{m},$
(81)

for $m=2,3,\ldots$, and using (77) and (79) to obtain the corresponding
variations of the Lagrangian.

##  4 Realization of Lorentz symmetry in BMS space

From now on we will use the notation $p_{-n}=\bar{p}_{n}$ in order to obtain
more compact expressions. The generators of the Lorentz group in physical
$2+1$ space-time with coordinates $x^{0}$, $x_{1}$, $x_{2}$ and corresponding
canonical momenta $p_{0}$, $P_{1}$, $P_{2}$ are given by
$K_{0}=x_{1}P_{2}-x_{2}P_{1}$ for rotations and $K_{i}=x^{0}P_{i}+x_{i}p_{0}$,
$i=1,2$, for boosts. The relation with the $x^{\pm 1}$ coordinates is given by
$x^{\pm 1}=\frac{1}{2}(x_{1}\pm ix_{2})$, which induces for the momenta the
relation $p_{\pm 1}=P_{1}\mp iP_{2}$. Defining $J=-iK_{0}$, $K_{\pm}=K_{1}\mp
iK_{2}$, one has, in terms of the coordinates $x^{0}$, $x^{\pm 1}$ and their
associated canonical momenta $p_{0}$, $p_{\pm 1}$,

$\displaystyle J$ $\displaystyle=x^{1}p_{1}-x^{-1}p_{-1},$ (82) $\displaystyle
K_{+}$ $\displaystyle=x^{0}p_{1}+2p_{0}x^{-1},$ (83) $\displaystyle K_{-}$
$\displaystyle=x^{0}p_{-1}+2p_{0}x^{1},$ (84)

which obey the Lorentz $SO(2,1)$ algebra $\\{K_{+},K_{-}\\}=2J$,
$\\{J,K_{+}\\}=K_{+}$, $\\{J,K_{-}\\}=-K_{-}$. These generators can be
extended to the BMS space with coordinates $x^{n},p_{n}$, $n\in\mathbb{Z}$ by
defining

$\displaystyle J$ $\displaystyle=\sum_{n=-\infty}^{+\infty}nx^{n}p_{n}$
$\displaystyle=\cdots-2x^{-2}p_{-2}\boxed{-x^{-1}p_{-1}+0\cdot
x^{0}p_{0}+x^{1}p_{1}}+2x^{2}p_{2}+\cdots,$ (85) $\displaystyle K_{+}$
$\displaystyle=\sum_{n=-\infty}^{+\infty}(1-n)x^{n}p_{n+1}$
$\displaystyle=\cdots+3x^{-2}p_{-1}\boxed{+2x^{-1}p_{0}+x^{0}p_{1}}+0\cdot
x^{1}p_{2}-x^{2}p_{3}+\cdots,$ (86) $\displaystyle K_{-}$
$\displaystyle=\sum_{n=-\infty}^{+\infty}(1+n)x^{n}p_{n-1}$
$\displaystyle=\cdots-x^{-2}p_{-3}+0\cdot
x^{-1}p_{-2}\boxed{+x^{0}p_{-1}+2x^{1}p_{0}}+3x^{2}p_{1}+\cdots,$ (87)

Notice that, under complex conjugation, $(K_{+})^{*}=K_{-}$, and that
$J^{*}=-J$, as it must be according to their definition from the real
generators $K_{0}$, $K_{1}$ and $K_{2}$.

This set of generators, together with the supertranslation generators

${\cal P}_{n}=p_{n},$ (88)

provides a realization of BMS (Lorentz + supertranslations) in phase space,
with $\\{x^{n},p_{m}\\}=\delta^{n}_{m}$. Indeed, one has

$\displaystyle\\{K_{+},K_{-}\\}=2J,\quad\\{J,K_{+}\\}=K_{+},\quad\\{J,K_{-}\\}=-K_{-},$
(89) $\displaystyle\\{K_{+},{\cal P}_{n}\\}=(1-n){\cal
P}_{n+1},\quad\\{K_{-},{\cal P}_{n}\\}=(1+n){\cal P}_{n-1},$ (90)
$\displaystyle\\{J,{\cal P}_{n}\\}=n{\cal P}_{n},\quad\\{{\cal P}_{n},{\cal
P}_{m}\\}=0,$ (91)

with $n,m\in\mathbb{Z}$. The connection with the abstract algebra (2-4) is
made by means of the identifications $K_{+}\mapsto-iL_{1}$, $K_{-}\mapsto
iL_{-1}$, $J\mapsto iL_{0}$, ${\cal P}_{n}\mapsto P_{n}$.

The fact that the extended generators satisfy the correct algebra is not
enough to state that we have a realization of BMS symmetry. Indeed, one must
prove that the generators $K_{+}$, $K_{-}$, $J$, and ${\cal P}_{n}$ are
conserved charges of the system. In this case, one must prove that the
generators have weakly zero Poisson brackets with all the first-class
constraints $\phi_{0}$, $\phi_{n}$, $\bar{\phi}_{n}$, $n=2,3,\ldots$,
appearing in the reduced Hamiltonian (41). Here, weakly zero means zero up to
the constraints, and we will denote this by $\simeq 0$.

Since the constraints do not depend on the coordinates, this condition is
trivially satisfied by the generators of supertranslations ${\cal P}_{n}$. For
$J$ one has, using that $\\{p_{n},J\\}=-np_{n}$, $\\{p_{-n},J\\}=np_{-n}$,
$\\{p_{1}p_{-1},J\\}=0$,

$\displaystyle\\{\phi_{n},J\\}$ $\displaystyle=\\{p_{n}+\mu
p_{-1}^{-n}f_{n}^{\pm}(p_{1}p_{-1}),J\\}=-np_{n}+\mu
f_{n}^{\pm}(p_{1}p_{-1})\\{p_{-1}^{-n},J\\}$ $\displaystyle=-np_{n}-n\mu
f_{n}^{\pm}(p_{1}p_{-1})p_{-1}^{-n-1}\\{p_{-1},J\\}=-np_{n}-n\mu
f_{n}^{\pm}(p_{1}p_{-1})p_{-1}^{-n-1}p_{-1}$ $\displaystyle=-n\phi_{n}\simeq
0,$ $\displaystyle\\{\bar{\phi}_{n},J\\}$ $\displaystyle=\\{p_{-n}+\mu
p_{-1}^{n}g_{n}^{\pm}(p_{1}p_{-1}),J\\}=np_{-n}+\mu
g_{n}^{\pm}(p_{1}p_{-1})\\{p_{-1}^{n},J\\}$ $\displaystyle=np_{-n}+n\mu
g_{n}^{\pm}(p_{1}p_{-1})p_{-1}^{n-1}\\{p_{-1},J\\}=np_{n}+n\mu
g_{n}^{\pm}(p_{1}p_{-1})p_{-1}^{n-1}p_{1}$
$\displaystyle=n\bar{\phi}_{n}\simeq 0.$

For $K_{+}$, using that $\\{p_{n},K_{+}\\}=-(1-n)p_{n+1}$,
$\\{p_{-n},K_{+}\\}=-(1+n)p_{-n+1}$, and in particular that
$\\{p_{1},K_{+}\\}=0$, $\\{p_{-1},K_{+}\\}=-2p_{0}$,

$\displaystyle\\{\phi_{n},K_{+}\\}$ $\displaystyle=\\{p_{n}+\mu
p_{-1}^{-n}f_{n}^{\pm}(p_{1}p_{-1}),K_{+}\\}$ $\displaystyle=-(1-n)p_{n+1}+\mu
f_{n}^{\pm}(p_{1}p_{-1})\\{p_{-1}^{-n},K_{+}\\}+\mu
p_{-1}^{-n}\\{f_{n}^{\pm}(p_{1}p_{-1}),K_{+}\\}$
$\displaystyle=-(1-n)p_{n+1}+\mu
f_{n}^{\pm}(p_{1}p_{-1})(-np_{-1}^{-n-1}(-2p_{0}))+\mu
p_{-1}^{-n}(f_{n}^{\pm})^{\prime}(p_{1}p_{-1})p_{1}(-2p_{0})$
$\displaystyle\stackrel{{\scriptstyle(\ref{derf})}}{{=}}-(1-n)p_{n+1}+2\mu
np_{0}p_{-1}^{-n-1}f_{n}^{\pm}(p_{1}p_{-1})\pm\mu(n+1)p_{0}p_{1}p_{-1}^{-n}\frac{f_{n-1}^{\pm}(p_{1}p_{-1})}{\sqrt{\mu^{2}+p_{1}p_{-1}}}$
$\displaystyle\stackrel{{\scriptstyle\phi_{0}}}{{\simeq}}-(1-n)p_{n+1}\mp 2\mu
n\sqrt{\mu^{2}+p_{1}p_{-1}}p_{-1}^{-n-1}f_{n}^{\pm}(p_{1}p_{-1})-\mu(n+1)p_{1}p_{-1}^{-n}f_{n-1}^{\pm}(p_{1}p_{-1})$
$\displaystyle\stackrel{{\scriptstyle\phi_{n+1}}}{{\simeq}}(1-n)\mu
p_{-1}^{-n-1}f_{n+1}(p_{1}p_{-1})$ $\displaystyle\quad\quad\quad\mp 2\mu
n\sqrt{\mu^{2}+p_{1}p_{-1}}p_{-1}^{-n-1}f_{n}^{\pm}(p_{1}p_{-1})-\mu(n+1)p_{1}p_{-1}^{-n}f_{n-1}^{\pm}(p_{1}p_{-1})$
$\displaystyle=-\mu p_{-1}^{-n-1}\left((n-1)f_{n+1}(x)\pm
2n\sqrt{\mu^{2}+x}f_{n}^{\pm}(x)+(n+1)xf_{n-1}^{\pm}(x)\right)_{x=p_{1}p_{-1}}\stackrel{{\scriptstyle(\ref{rec3f})}}{{=}}0,$

and

$\displaystyle\\{{\bar{\phi}}_{n},K_{+}\\}$ $\displaystyle=\\{p_{-n}+\mu
p_{-1}^{n}g_{n}^{\pm}(p_{1}p_{-1}),K_{+}\\}$ $\displaystyle=-(1+n)p_{-n+1}+\mu
g_{n}^{\pm}(p_{1}p_{-1})\\{p_{-1}^{n},K_{+}\\}+\mu
p_{-1}^{n}\\{g_{n}^{\pm}(p_{1}p_{-1}),K_{+}\\}$
$\displaystyle=-(1+n)p_{-n+1}+n\mu
g_{n}^{\pm}(p_{1}p_{-1})p_{-1}^{n-1}(-2p_{0})+\mu
p_{-1}^{n}(g_{n}^{\pm})^{\prime}(p_{1}p_{-1})p_{1}(-2p_{0})$
$\displaystyle\stackrel{{\scriptstyle(\ref{derg}),\phi_{0}}}{{\simeq}}-(1+n)p_{-n+1}\mp
2\mu
n\sqrt{\mu^{2}+p_{1}p_{-1}}p_{-1}^{n-1}g_{n}^{\pm}(p_{1}p_{-1})+\mu(n-1)p_{1}p_{-1}^{n}g_{n+1}^{\pm}(p_{1}p_{-1})$
$\displaystyle\stackrel{{\scriptstyle\phi_{n-1}}}{{\simeq}}(1+n)\mu
p_{-1}^{n-1}g_{n-1}(p_{1}p_{-1})$ $\displaystyle\quad\quad\quad\mp 2\mu
n\sqrt{\mu^{2}+p_{1}p_{-1}}p_{-1}^{n-1}g_{n}^{\pm}(p_{1}p_{-1})+\mu(n-1)p_{1}p_{-1}^{n}g_{n+1}^{\pm}(p_{1}p_{-1})$
$\displaystyle=\mu p_{-1}^{n-1}\left((n+1)g_{n-1}(x)\pm
2n\sqrt{\mu^{2}+x}g_{n}^{\pm}(x)+(n-1)xg_{n+1}^{\pm}(x)\right)_{x=p_{1}p_{-1}}\stackrel{{\scriptstyle(\ref{rec3g})}}{{=}}0.$

Similarly, one can show that $\\{\phi_{n},K_{-}\\}\simeq 0$,
$\\{{\bar{\phi}}_{n},K_{-}\\}\simeq 0$. This can be done by direct calculation
as above or using that the Poisson bracket structure is real and then

$\displaystyle\\{\phi_{n},K_{-}\\}^{*}=\\{\bar{\phi}_{n},K_{+}\\}\simeq 0,$
(92)
$\displaystyle\\{{\bar{\phi}}_{n},K_{-}\\}^{*}=\\{{\phi}_{n},K_{+}\\}\simeq
0.$ (93)

One concludes then that the extended Poincaré generators are conserved charges
for our system. A discussion of the Casimirs of the Lorentz and Poincaré
groups in BMS space is presented in Appendix D, where, in particular, it is
shown that the only quadratic Casimir of the BMS group is the standard one
Poincaré Casimir, $C_{2}=p_{0}^{2}-p_{1}p_{-1}$, see appendix D.

One might be tempted to generalize the Lorentz generators to include
superrotations (or rather superboosts) by replacing the “1” which appear in
(86) and $(\ref{K-})$ with arbitrary positive integers $m$,

$\displaystyle K_{+}^{m}$
$\displaystyle=\sum_{n=-\infty}^{+\infty}(m-n)x^{n}p_{n+m},$ (94)
$\displaystyle K_{-}^{m}$
$\displaystyle=\sum_{n=-\infty}^{+\infty}(m+n)x^{n}p_{n-m},$ (95)

for $m=1,2,\ldots$ By appropriate identifications, these generators, together
with $J$ and the ${\cal P}_{n}$, provide a representation of the extended BMS
algebra in terms of Poisson brackets. However, the extended generators
obtained for $m=2,3,\ldots$ do not commute with all the first class
constraints of our system, and hence are not conserved quantities. To be more
precise, the constraints $\phi_{n}$, $n=2,3,\ldots$, are weakly invariant only
under $K_{+}^{m}$, while the $\bar{\phi}_{n}$ are invariant only under
$K_{-}^{m}$. We will see in Section 5 that the massless limit of our theory is
fully invariant under these generalized transformations.

##  5 Massless limit

Since the Lagrangian (11) is proportional to $\mu$ one cannot take the
massless limit directly in configuration space. This is not a problem in phase
space, since the system is in this case entirely defined by the set of
constraints, which have a non-trivial limit when $\mu\to 0$. Indeed,
performing this limit in (48) and its complex conjugate (49) one gets the
constraints

$\displaystyle\varphi_{n}$ $\displaystyle=p_{n}\pm(\mp
1)^{n}p_{-1}^{-n}(\sqrt{p_{1}p_{-1}})^{n+1},\quad n=0,1,2,\ldots,$ (96)
$\displaystyle\bar{\varphi}_{n}$ $\displaystyle=p_{-n}\pm(\mp
1)^{n}p_{-1}^{n}(\sqrt{p_{1}p_{-1}})^{-n+1},\quad n=1,2,\ldots,.$ (97)

Notice that $\varphi_{0}=p_{0}\pm\sqrt{p_{1}p_{-1}}$, and that $\varphi_{1}$
and $\bar{\varphi}_{1}$ are trivial, as in the massive case.

As shown in Appendix C, one has that

$\\{\varphi_{n},K_{+}^{m}\\}\simeq 0,\ \\{\varphi_{n},K_{-}^{m}\\}\simeq 0,\
\\{\bar{\varphi}_{n},K_{+}^{m}\\}\simeq 0,\
\\{\bar{\varphi}_{n},K_{-}^{m}\\}\simeq 0,$ (98)

where the superrotation generators $K_{\pm}^{m}$, $m=0,1,2,\ldots$ are defined
as in (94) and (95), and where the weak equality is now over the manifold
defined by $\varphi_{n}=0$, $\bar{\varphi}_{n}=0$. Thus, $K_{\pm}^{m}$ are
conserved quantities in the massless limit theory.

The superrotation operators obey the algebra

$\displaystyle\\{K_{+}^{m},K_{+}^{n}\\}$ $\displaystyle=(m-n)K_{+}^{m+n},$
(99) $\displaystyle\\{K_{+}^{m},K_{-}^{n}\\}$
$\displaystyle=\begin{cases}2mJ&\text{if $m=n$},\\\ -(m+n)K_{+}^{m-n}&\text{if
$m>n$},\\\ (m+n)K_{-}^{n-m}&\text{if $m<n$},\end{cases}$ (100)
$\displaystyle\\{K_{-}^{m},K_{-}^{n}\\}$ $\displaystyle=(m-n)K_{-}^{m+n},$
(101)

Furthermore, they act on the supertranslation generators as

$\displaystyle\\{K_{\pm}^{m},p_{n}\\}=(m\mp n)p_{n\pm m},\quad
m=1,2,\ldots,\quad n\in\mathbb{Z}.$ (102)

If we now define

$L_{m}=\begin{cases}-J&\text{if $m=0$},\\\ K_{+}^{m}&\text{if $m>0$},\\\
-K_{-}^{-m}&\text{if $m<0$},\end{cases}$ (103)

it turns out that $\\{L_{m},L_{n}\\}=(m-n)L_{m+n}$ and the extended BMS
algebra (1) is obtained in terms of Poisson brackets of the massless limit BMS
particle.

##  6 Gauge fixing

In Section 3 it has been shown that, after eliminating the degrees of freedom
$u,v$ and its corresponding canonical momenta by means of the primary second-
class constraints $\pi_{u}=0$, $\pi_{v}=0$, $\phi_{1}=0$ and
$\bar{\phi}_{1}=0$, the theory still contains an infinite number of primary
first-class constraints which generate gauge transformations and indicate the
presence of gauge degrees of freedom.

These gauge degrees can be eliminated by converting the first-class
constraints to second class, by introducing appropriate gauge fixing
conditions. Since all the constraints $\phi_{n}$ (resp. $\bar{\phi}_{n}$)
depend linearly on $p_{n}$ (resp. $p_{-n}$), a sensible choice is to introduce
the constraints

$\psi_{n}=x^{n},\quad n=\pm 2,\pm 3,\ldots,$ (104)

so that

$\\{\psi_{n},\phi_{m}\\}=\delta_{n,m},\quad n,m=\pm 2,\pm 3,\ldots,$ (105)

and define a gauge fixing as

$GF=\\{\psi_{m}=0\\}_{|m|\geq 2},$ (106)

which allows the consistent elimination of all the extra BMS degrees of
freedom in phase space,

$\displaystyle p_{\pm n}=-\frac{\mu}{p_{\mp
1}^{n}}f_{n}^{\pm}(p_{1}p_{-1}),\quad x^{n}=0,\quad x^{-n}=0,\quad
n=2,3,\ldots,$ (107)

with only the gauge symmetry associated with $\phi_{0}$ remaining. In this
way, the physical degrees of freedom of the theory in phase space are reduced
to $x^{0},p_{0},x^{\pm 1},p_{\pm 1}$, and it can be seen that the Dirac
brackets between these remaining variables are the standard Poisson brackets.

Notice that the gauge condition is not invariant under supertranslations and
hence one must introduce a compensating gauge transformation so that the total
variation of the gauge condition, computed on the gauge condition, is zero. If
we consider $|m|\geq 2$ and denote by $\delta_{ST}^{n}x^{m}$ the
supertranslation of $x^{m}$ generated by $p_{n}$, and by $\epsilon^{m}(\tau)$
the gauge transformation on $x^{m}$, generated by $\phi_{m}$, one has

$0=\left.\left(\epsilon^{m}(\tau)+\delta_{ST}^{n}x^{m}\right)\right|_{GF},$
(108)

and, since $\delta_{ST}^{n}x^{m}=\epsilon^{n}\delta_{n}^{m}=\epsilon^{m}$, the
compensating gauge transformation associated to the supertranslation along the
$m$ coordinate, $|m|\geq 2$, is just

$\epsilon^{m}(\tau)=-\epsilon^{m}.$ (109)

Since the generators of the gauge transformations $\phi_{m}$, $|m|\geq 2$,
contain the momenta $p_{1},p_{-1}$, it turns out that these compensating gauge
transformations induce a residual transformation on the remaining variables
$x^{\pm 1}$, given by

$\delta^{m}_{\text{res}}x^{\pm 1}=\\{x^{\pm
1},-\epsilon^{m}\phi_{m}\\},\quad|m|\geq 2.$ (110)

Using $\\{x^{1},\phi_{n}\\}=-(n+1)P_{n-1}/(2P_{0})$,
$\\{x^{-1},\phi_{n}\\}=(n-1)P_{n+1}/(2P_{0})$, $n=\pm 2,\pm 3,\ldots$, one
gets

$\displaystyle\delta^{m}_{\text{res}}x^{1}$
$\displaystyle=\epsilon^{m}(m+1)\frac{P_{m-1}}{2P_{0}},$ (111)
$\displaystyle\delta^{m}_{\text{res}}x^{-1}$
$\displaystyle=-\epsilon^{m}(m-1)\frac{P_{m+1}}{2P_{0}},$ (112)

where it should be reminded that the several $P_{n}$ appearing on the right-
hand sides are functions of $p_{1}$, $p_{-1}$. That these transformations are
a symmetry of the theory is proved at the end of Appendix B. Notice that for
$m=1$ and $m=-1$, although no compensating gauge transformation is needed, one
formally obtains the standard translations in $x^{1}$ and $x^{-1}$,
respectively.

These residual transformations on the physical variables $x^{\pm 1}$ provide,
together with the Lorentz transformations, a realization of BMS in the
physical reduced space, up to reparametrizations. The need for a
reparametrization follows from the fact that $x^{0}$ does not transform under
$\delta^{m}_{\text{res}}$ but, under a boost, transforms into $x^{1}$ or
$x^{-1}$. For instance, for $K_{+}$ one has
$\delta_{+}x^{0}=\\{x^{0},K_{+}\\}=2x^{-1}$ and then (we drop the constant
parameters $\epsilon^{m}$)

$[\delta_{+},\delta^{m}_{\text{res}}]x^{0}=(m-1)\frac{P_{m+1}}{P_{0}},$ (113)

while $\delta^{m+1}_{\text{res}}x^{0}$, which should appear on the right-hand
side in order to have the BMS algebra, is zero. However, since
$\\{x^{0},\phi_{0}\\}=1$, the right-hand side can be interpreted as a
reparametrization with parameter

$\epsilon_{+}^{m}=(m-1)\frac{P_{m+1}}{P_{0}},$ (114)

so that the commutator is indeed a vanishing BMS supertranslation on $x^{0}$
plus a reparametrization,

$[\delta_{+},\delta^{m}_{\text{res}}]x^{0}=(m-1)\cdot
0+\delta_{0}^{\epsilon_{+}^{m}}x^{0}.$ (115)

For this to be consistent, the same reparametrization should appear when one
considers the action of the transformations on $x^{1}$ and $x^{-1}$,

$\displaystyle[\delta_{+},\delta^{m}_{\text{res}}]x^{1}$
$\displaystyle=\frac{m+1}{2}\left(\frac{p_{1}P_{m-1}}{P_{0}^{2}}+m\frac{P_{m-2}}{P_{0}}\right),$
(116) $\displaystyle[\delta_{+},\delta^{m}_{\text{res}}]x^{-1}$
$\displaystyle=-\frac{m-1}{2}\left(\frac{p_{1}P_{m+1}}{P_{0}^{2}}+m\frac{P_{m+2}}{P_{0}}\right).$
(117)

Using that

$\displaystyle\delta^{m+1}_{\text{res}}x^{1}=(m+2)\frac{P_{m}}{2P_{0}},$
$\displaystyle\quad\delta^{m+1}_{\text{res}}x^{-1}=-m\frac{P_{m+2}}{2P_{0}},$
(118)

Notice this residual transformation is no longer a point transformation. Under
reparametrizations with parameter $\epsilon_{+}^{m}$ (114),

$\displaystyle\delta_{0}^{\epsilon_{+}^{m}}x^{1}$
$\displaystyle=\epsilon_{+}^{m}\\{x^{1},\phi_{0}\\}=-\frac{m-1}{2}\frac{p_{-1}P_{m+1}}{P_{0}^{2}},$
(119) $\displaystyle\delta_{0}^{\epsilon_{+}^{m}}x^{-1}$
$\displaystyle=\epsilon_{+}^{m}\\{x^{-1},\phi_{0}\\}=-\frac{m-1}{2}\frac{p_{1}P_{m+1}}{P_{0}^{2}},$
(120)

one can check that (116), (117) can be rewritten as

$\displaystyle[\delta_{+},\delta^{m}_{\text{res}}]x^{1}$
$\displaystyle=(m-1)\delta^{m+1}_{\text{res}}x^{1}+\delta_{0}^{\epsilon_{+}^{m}}x^{1},$
(121) $\displaystyle[\delta_{+},\delta^{m}_{\text{res}}]x^{-1}$
$\displaystyle=(m-1)\delta^{m+1}_{\text{res}}x^{-1}+\delta_{0}^{\epsilon_{+}^{m}}x^{-1},$
(122)

which, together with (115) and up to the reparametrization, yield the correct
term for the BMS algebra. Similarly, for $K_{-}$, the reparametrization
parameter is

$\epsilon_{-}^{m}=-(m+1)\frac{P_{m-1}}{P_{0}},$ (123)

while no reparametrization is necessary to close the commutators of (111,112)
with the transformation given by the rotation generator $J$.

This is an example of a fact previously reported in the literature (see
Pons:1994hy , eq. (3.11)), i.e. the closure of the algebra of rigid symmetries
with the help of gauge transformations. In our case, the rigid transformations
correspond to Lorentz and supertranslations in physical space, and the gauge
transformation is given by the reparametrization invariance associated with
the first-class constraint $\phi_{0}$, which has not been fixed. That the
reparametrizations might be needed to close the algebra can be also be
inferred from the fact that the constraints $\phi_{n}$, $|n|\geq 2$, are
weakly Lorentz invariant on the manifold defined by $\phi_{0}$ (see the proof
in Section 4) and that those $\phi_{n}$ are the generators of the residual
gauge transformations that give rise to the BMS symmetry in physical space.

Under a Lorentz transformation,

$\displaystyle\delta_{J}x^{n}=\\{x^{n},J\\}=nx^{n},$ (124)
$\displaystyle\delta_{+}x^{n}=\\{x^{n},K_{+}\\}=(2-n)x^{n-1},$ (125)
$\displaystyle\delta_{-}x^{n}=\\{x^{n},K_{-}\\}=(2+n)x^{n+1},$ (126)

and one has

$\displaystyle\left.\delta_{J}x^{n}\right|_{GF}=0,\quad\left.\delta_{+}x^{n}\right|_{GF}=0,\quad\left.\delta_{-}x^{n}\right|_{GF}=0,$
(127)

so that the gauge condition is preserved, without the need for a compensating
gauge transformation. Notice that the factors $(2-n)$ and $(2+n)$ play a
fundamental role for $n=2$ and $n=-2$, respectively.

In the massless case, where the Lorentz group can be extended so as to include
superrotations, one has, for $m=2,3,\ldots$,

$\displaystyle\delta_{+}^{m}x^{n}=\\{x^{n},K_{+}^{m}\\}=(2m-n)x^{n-m},$ (128)
$\displaystyle\delta_{-}^{m}x^{n}=\\{x^{n},K_{-}^{m}\\}=(2m+n)x^{n+m},$ (129)

and a compensating gauge transformation must be introduced for $|n|\geq 2$ for
the values of $m$ such that the right-hand side of (128) or (129) are not zero
when evaluated on the gauge fixing condition (106). As in the case of
supertranslations, this will generate a residual gauge transformation for
$x^{\pm 1}$, which should provide a realization of superrotations.

In any case, after eliminating the gauge degrees of freedom, the remaining
variables are just $x^{0}$, $x^{\pm 1}$ and their canonical momenta, together
with the first class constraint $\phi_{0}$. This describes a Poincaré particle
in $2+1$, with the corresponding reparametrization invariance, with no extra
degrees of freedom, and with a realization of the supertranslations, plus
superrotations in the massless case, provided by the residual gauge
transformations. Summing up, the physical degrees of freedom of the BMS
particle do not contain the BMS coordinates for $|n|>1$ and hence no soft BMS
modes are present.

##  7 Conclusions and outlook

We have constructed a non-linear realization of a massive particle Lagrangian
for the BMS symmetry algebra in $2+1$ space-time. This Lagrangian depends on
an infinite set of BMS coordinates, which include the standard $2+1$ Poincaré
ones, together with the Goldstone boost variables.

The canonical analysis of this Lagrangian reveals the existence of a finite
set of second-class constraints, which can be eliminated using the standard
Dirac bracket construction, together with an infinite set of first-class
constraints, which generate a corresponding infinite set of gauge
transformations.

The standard Lorentz generators in $2+1$ are extended so that they act on the
full set of BMS variables, and the theory is shown to be invariant under them.
These extended Lorentz generators can be further generalized to an infinite
set, the so-called superrotations, that obey the extended $\text{BMS}_{3}$
algebra, but it is only in the massless limit of the theory that they are
conserved quantities and thus represent a symmetry of the system.

Upon fixing all the gauge degrees of freedom of the theory, except for the
standard reparametrization, one obtains a theory whose physical content is
that of an ordinary relativistic Poincaré particle, with the standard
reparametrization invariance provided by the remaining first-class constraint.
However, this gauge fixing procedure results in residual gauge transformations
acting on the standard space coordinates $x^{0},x^{1}$, $x^{-1}$ which, modulo
reparametrizations, realize the BMS symmetry. Since the remaining first-class
constraint is the standard one, the field theory associated with this particle
model is that of a free Klein-Gordon field.

The interpretation of these transformations for $x^{1}$, $x^{-1}$, which
depend on the associated canonical momenta $p_{1}$, $p_{-1}$, is a subject for
further study. The appearance of the momenta in a non-polynomial form seems to
indicate that, in the field theory associated with this model, the field
should transform non-locally. In Batlle:2017llu ; Batlle:2022hwf  a
realization of BMS in terms of a free scalar field was constructed, with
transformations that were non-local in ordinary space. Further investigations
are needed to clarify whether the two approaches can be related.

A possible link with the nonlocal transformations for the massless scalar
field presented in Batlle:2022hwf  is as follows.

Since the standard field equations can be obtained from the particle mass-
shell condition following the world-line approach to field theory, one can
consider, for the massless BMS particle, the infinite tower of squares of the
constraints

$\varphi_{n}\bar{\varphi}_{n}\simeq p_{n}p_{-n}-p_{1}p_{-1},\ n=0,2,3,\ldots$
(130)

(the case $n=1$ is trivial). For $n=0$, $p_{0}^{2}-p_{1}p_{-1}=0$, this yields
the standard massless Klein-Gordon equation. Indeed, using the relation with
the real coordinates $p_{\pm 1}=P_{1}\mp iP_{2}$ and
$p_{0}=-i\partial_{x^{0}}$, $P_{j}=-i\partial_{x_{j}}$, $j=1,2$, one has

$(p_{0}^{2}-p_{1}p_{-1})\Psi=(-\partial^{2}_{x^{0}}+\nabla^{2})\Psi=0.$ (131)

For $n=2,3\ldots$, one can introduce similar real coordinates $x^{(n)}_{1,2}$
related to the complex BMS coordinates $x^{\pm n}$, $n\geq 2$, by

$x^{\pm n}=\frac{1}{2}(x^{(n)}_{1}\pm ix^{(n)}_{2}),\quad n=2,3,\ldots$ (132)

so that

$p_{{\pm n}}=P^{(n)}_{1}\mp P^{(n)}_{2}=-i\frac{\partial}{\partial
x^{(n)}_{1}}\mp\frac{\partial}{\partial x^{(n)}_{2}}.$ (133)

The imposition of all the conditions on an scalar field
$\Psi(x^{0},x_{1},x_{2},x^{(2)}_{1},x^{(2)}_{2},\ldots)$, depending on all of
the BMS coordinates, leads then to the infinite set of equations

$(p_{n}p_{-n}-p_{1}p_{-1})\Psi=(-\nabla_{n}^{2}+\nabla^{2})\Psi=0,\quad\nabla_{n}^{2}=\frac{\partial^{2}}{\partial
x^{(n)2}_{1}}+\frac{\partial^{2}}{\partial x^{(n)2}_{2}},\ n=2,3,\ldots$ (134)

together with (131).

Notice that $p_{n}p_{-n}-p_{1}p_{-1}=0$, $n=0,2,3,\ldots$, can be written in
the equivalent form

$p_{n}p_{-n}-p_{n+1}p_{-(n+1)}=0,\ n=0,1,2,3,\ldots,$ (135)

which, upon converted to a field equation, yields the standard Klein-Gordon
equation plus an infinite tower of equations,

$\nabla^{2}_{n+1}\Psi=\nabla^{2}_{n}\Psi,\quad n=1,2,\ldots,$ (136)

or, equivalently, the infinite set of Klein-Gordon equations

$\nabla^{2}_{n}\Psi=\partial_{x^{0}}^{2}\Psi,\quad n=1,2,\ldots$ (137)

Whether the Lie symmetries of this set of PDE can be related to the nonlocal
symmetries of an scalar field depending only on the ordinary Poincaré
variables $x^{0},x^{(1)}$, which were obtained in Batlle:2022hwf  in terms of
polyharmonic functions, is a subject worthy of further study.

In the approach taken in this paper, the massless limit has been obtained in
the Hamiltonian formalism as the theory defined by the massless limit of the
Hamiltonian constraints. An alternative approach, which we will examine in the
future, is to obtain the model of a massless particle in the nonlinear
realization approach.

The construction of a particle model exhibiting BMS symmetry presented in this
paper could be, in principle, repeated for $\text{BMS}_{4}$, using, for
instance, the stereographic parametrization of $\text{BMS}_{4}$ Barnich:2009se
; Barnich:2011ct . BMS structure constants for $\text{BMS}_{4}$, $BMS_{5}$ and
$BMS_{6}$ using generalized spherical harmonics parametrizations can also be
found in Delmastro:2017erq , although they are much more involved.

###### Acknowledgements.

We acknowledge interesting discussions with Luca Ciambelli, Miguel Campliglia,
Jaume Gomis, Marc Henneaux, Axel Kleinschmidt and Sabrina Pasterski. JG
acknowledges the hospitality of the Max Planck Albert Einstein Institute in
Golm and the Perimeter Institute in Waterloo where this work has been
completed. The work of CB is partially supported by Project MASHED
(TED2021-129927B-I00), funded by MCIN/AEI/10.13039/501100011033 and by the
European Union Next Generation EU/PRTR. JG has been supported in part by
PID2019-105614GB-C21 and PID2019- 105614GB-C21 and from the State Agency for
Research of the Spanish Ministry of Science and Innovation through the Unit of
Excellence Maria de Maeztu 2020-2023 award to the Institute of Cosmos Sciences
(CEX2019-000918-M).

##  Appendix A Computation of the term of the Maurer-Cartan form proportional
to the $P_{0}$ generator

Using

$e^{X}Ye^{-X}=e^{\operatorname{ad}_{X}}Y=Y+\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots$
(138)

one has

$\displaystyle\left[Y,e^{-X}\right]$ $\displaystyle=$ $\displaystyle
e^{-X}\left(\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots\right)$
(139) $\displaystyle\equiv$ $\displaystyle e^{-X}K(X,Y),$

where

$K(X,Y)=\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots,$
(140)

which is linear in its second argument. We will call $K(X,Y)$ the $K$-action
of $X$ on $Y$.

By repeated use of (139) one arrives at

$U^{-1}P_{n}U=P_{n}+K(-iL_{1}u,P_{n})+K(-iL_{-1}v,P_{n})+K(-iL_{1}u,K(-iL_{-1}v,P_{n})).$
(141)

The second and third terms in (141) are

$\displaystyle K(-iL_{1}u,P_{n})$ $\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-iu)^{l}[L_{1},[L_{1},\stackrel{{\scriptstyle
l)}}{{\ldots}},[L_{1},P_{n}]\ldots]]$ (142) $\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-iu)^{l}(-i)^{l}(n-1)n\ldots(n+l-2)P_{n+l}$
$\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-1)^{l}u^{l}(n-1)n\ldots(n+l-2)P_{n+l}.$

$\displaystyle K(-iL_{-1}v,P_{n})$ $\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-iv)^{l}[L_{-1},[L_{-1},\stackrel{{\scriptstyle
l)}}{{\ldots}},[L_{-1},P_{n}]\ldots]]$ (143) $\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-iv)^{l}(-i)^{l}(n+1)n\ldots(n-l+2)P_{n-l}$
$\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-1)^{l}v^{l}(n+1)n\ldots(n-l+2)P_{n-l}.$

The fourth term in (141) is

$\displaystyle K(-iL_{1}u,K(-iL_{-1}v,P_{n}))$ $\displaystyle=$ $\displaystyle
K(-iL_{1}u,\sum_{l=1}^{\infty}\frac{1}{l!}(-1)^{l}v^{l}(n+1)n\ldots(n-l+2)P_{n-l})$
$\displaystyle=$
$\displaystyle\sum_{l=1}^{\infty}\frac{1}{l!}(-1)^{l}v^{l}(n+1)n\ldots(n-l+2)K(-iL_{1}u,P_{n-l}),$

where we have used the linearity of $K$ in its second argument. Finally

$\displaystyle
K(-iL_{1}u,K(-iL_{-1}v,P_{n}))=\sum_{l=1}^{\infty}\frac{1}{l!}(-1)^{l}v^{l}(n+1)n\ldots(n-l+2)$
(144)
$\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!}(-1)^{k}u^{k}(n-l-1)(n-l)\ldots(n-l+k-2)P_{n-l+k}.$

As mentioned before, we are interested only in the $P_{0}$ terms in (141):

  * •

$P_{n}$. Contributes only for $n=0$, with $P_{0}$.

  * •

$K(-iL_{1}u,P_{n})$. Only the terms with $l=-n$ yield a $P_{0}$. Since $l\geq
1$, this means that there is no contribution for $n\geq 0$, while for $n=-m<0$
one picks the term

$\frac{1}{m!}(-1)^{m}u^{m}(-m-1)(-m)\ldots(-2)P_{0}=(m+1)u^{m}P_{0}.$

  * •

$K(-iL_{-1}v,P_{n})$. The $P_{0}$ contribution is obtained now for $l=n$ and
since $l\geq 1$, there is only contribution if $n>0$, which is

$\frac{1}{n!}(-1)^{n}v^{n}(n+1)n\ldots 2\,P_{0}=(-1)^{n}(n+1)v^{n}P_{0}.$

  * •

$K(-iL_{1}u,K(-iL_{-1}v,P_{n}))$. This term has multiple $P_{0}$
contributions, given by $l-k=n$, subjected to $l\geq 1$, $k\geq 1$. For given
$l$ one picks the $k=l-n$ term in the $k$ series, but $k\geq 1$ implies that
$l$ must satisfy, besides $l\geq 1$, the constraint $l\geq 1+n$. If $n\leq 0$
this just means $l\geq 1$, but, for $n>0$, $l$ is restricted by $l\geq 1+n$.

Selecting $k=l-n$ in (144) and restricting the series over $l$ according to
the above discussion one has, after re-arranging terms and cancelling some
signs,

$\sum_{l=n+1}^{\infty}\frac{l-n+1}{l!}(-1)^{l}(n+1)n\ldots(n-l+2)v^{l}u^{l-n}P_{0}$
(145)

for $n>0$, and

$\sum_{l=1}^{\infty}\frac{m+l+1}{l!}(m-1)m\ldots(m+l-2)v^{l}u^{l+m}P_{0}$
(146)

for $n=-m\leq 0$.

Putting everything together, the coefficient of $P_{0}$ in (8) can be computed
as follows, collecting the contributions proportional to the different
$\mbox{d}x^{n}$.

For $n=0$, there is only contribution from the first and fourth terms in
(141),

$\mbox{d}x^{0}(1+\sum_{l=1}^{\infty}\frac{l+1}{l!}(0-1)(0)\ldots(0+l-2)v^{l}u^{l}$

Notice, however, that the above series finishes in fact after $l=1$, so one
gets

$\mbox{d}x^{0}(1-2uv).$ (147)

For $n>0$, only the third and fourth terms have a non-vanishing contribution,
given by

$\mbox{d}x^{n}\left((-1)^{n}(n+1)v^{n}+\sum_{l=n+1}^{\infty}\frac{l-n+1}{l!}(-1)^{l}(n+1)n\ldots(n-l+2)v^{l}u^{l-n}\right).$
(148)

Actually, the product in the coefficients of the series always contains a zero
except if $l=n+1$, and hence the above expression collapses to

$\mbox{d}x^{n}\left((-1)^{n}(n+1)v^{n}+2(-1)^{n+1}uv^{n+1}\right)=\mbox{d}x^{n}(-1)^{n}v^{n}(n+1-2uv).$
(149)

Finally, for $n=-m<0$, the contributions come from the second and fourth terms
and are given by

$\mbox{d}x^{-m}\left((m+1)u^{m}+\sum_{l=1}^{\infty}\frac{m+l+1}{l!}(m-1)m\ldots(m+l-2)v^{l}u^{l+m}\right).$
(150)

The series is identically zero for $m=1$, while for $m\geq 2$ it can be
rewritten as

$\mbox{d}x^{-m}\left((m+1)u^{m}+\sum_{l=1}^{\infty}\frac{m+l+1}{l!}\frac{(l+m-2)!}{(m-2)!}v^{l}u^{l+m}\right).$
(151)

This series can be summed (provided that $|uv|<1$) and, after adding the
$(m+1)u^{m}$ term, one gets

$\mbox{d}x^{-m}u^{m}\frac{m+1-2uv}{(1-uv)^{m}}.$ (152)

Adding all the terms, the coefficient of $P_{0}$ in the Maurer-Cartan form is

$\Omega_{P_{0}}=\mbox{d}x^{0}(1-2uv)+\sum_{n=1}^{\infty}\mbox{d}x^{n}(-1)^{n}v^{n}(n+1-2uv)+\mbox{d}x^{-1}2u+\sum_{n=2}^{\infty}\mbox{d}x^{-n}u^{n}\frac{n+1-2uv}{(1-uv)^{n}}.$
(153)

The fact that the $\dot{x}^{-n}$ contribution is much more complex than that
of $\dot{x}^{n}$, actually involving the series that has been mentioned, is
due to the form of the last term in (141), which in turn is a consequence of
the ordering that we have selected for the two exponentials in $U$. For $n\geq
2$, the $K$-action of $-ivL_{-1}$ on $P_{n}$ can only descend to $P_{-1}$
(since the Poincaré part is BMS invariant), and then there is only one term in
the $K$-action of $-iL_{1}u$ that returns to $P_{0}$. Instead, for $n\geq 2$,
$K(-ivL_{-1},P_{-n})$ produces terms $P_{k}$ for any $k=-3,-4,\ldots$, and
then, for each of them, there is a way to return to $P_{0}$ by the $K$-action
of $-iL_{1}u$.

##  Appendix B Quasi-invariance of the Lagrangian under gauge transformations

We consider first the full Lagrangian (11) and its variation under the full
set of gauge transformations given by the first class constraints $\phi_{m}$
and $\bar{\phi}_{m}$, $m\geq 2$. In order to compute the transformation of the
phase-space variables induced by $\phi_{m}$, $\bar{\phi}_{m}$ one needs

$\displaystyle\\{x^{1},\phi_{m}\\}$ $\displaystyle=\\{x^{1},p_{m}+\mu
p_{-1}^{-m}f_{m}^{\pm}(p_{1}p_{-1})\\}$ $\displaystyle=\mu
p_{-1}^{-m}(f_{m}^{\pm})^{\prime}(p_{1}p_{-1})p_{-1}=\mp\mu
p_{-1}^{-m+1}\frac{m+1}{2\sqrt{\mu^{2}+p_{1}p_{-1}}}f_{m-1}^{\pm}(p_{1}p_{-1})$
$\displaystyle=\mu
p_{-1}^{-m+1}\frac{m+1}{2P_{0}}f_{m-1}^{\pm}(p_{1}p_{-1})=-(m+1)\frac{P_{m-1}}{2P_{0}},$
(154) $\displaystyle\\{x^{1},\bar{\phi}_{m}\\}$
$\displaystyle=\\{x^{1},\bar{p}_{m}+\mu p_{-1}^{m}g_{m}^{\pm}(p_{1}p_{-1})\\}$
$\displaystyle=\mu p_{-1}^{m}(g_{m}^{\pm})^{\prime}(p_{1}p_{-1})p_{-1}=\pm\mu
p_{-1}^{m+1}\frac{m-1}{2\sqrt{\mu^{2}+p_{1}p_{-1}}}g_{m+1}^{\pm}(p_{1}p_{-1})$
$\displaystyle=-\mu
p_{-1}^{m+1}\frac{m-1}{2P_{0}}g_{m+1}^{\pm}(p_{1}p_{-1})=(m-1)\frac{P_{-m-1}}{2P_{0}}.$
(155)

From these two it also follows that

$\displaystyle\\{x^{-1},\phi_{m}\\}$
$\displaystyle=\\{x^{1},\bar{\phi}_{m}\\}^{*}=(m-1)\frac{P_{m+1}}{2P_{0}},$
(156) $\displaystyle\\{x^{-1},\bar{\phi}_{m}\\}$
$\displaystyle=\\{x^{1},{\phi}_{m}\\}^{*}=-(m+1)\frac{P_{-m+1}}{2P_{0}},$
(157)

which, together with $\\{x^{n},\phi_{m}\\}=\delta^{n}_{m}$,
$\\{x^{-n},\bar{\phi}_{m}\\}=\delta^{n}_{m}$,
$\\{P_{n},\phi_{m}\\}=\\{P_{-n},\phi_{m}\\}=\\{P_{n},\bar{\phi}_{m}\\}=\\{P_{-n},\bar{\phi}_{m}\\}=0$,
allow to compute the transformations of all the terms in the Lagrangian. For
instance, if $G=\alpha_{m}\phi_{m}=\epsilon_{m}2P_{0}\phi_{m}$, one has

$\displaystyle\delta{\cal L}$
$\displaystyle=\frac{\mbox{d}}{\mbox{d}\tau}(2P_{0}\epsilon_{m})P_{m}+\frac{\mbox{d}}{\mbox{d}\tau}(-(m+1)\epsilon_{m}P_{m-1})p_{1}+\frac{\mbox{d}}{\mbox{d}\tau}((m-1)\epsilon_{m}P_{m+1})p_{-1}$
$\displaystyle=\epsilon_{m}(2\dot{P}_{0}p_{m}-(m+1)\dot{P}_{m-1}p_{1}+(m-1)\dot{P}_{m+1}p_{-1})$
$\displaystyle\quad+\dot{\epsilon}_{m}(2P_{0}P_{m}-(m+1)p_{1}P_{m-1}+(m-1)P_{m+1}p_{-1}).$
(158)

This can be written as a total derivative, $\delta{\cal
L}=\frac{\mbox{d}}{\mbox{d}\tau}F$, provided that

$2P_{0}\dot{P}_{m}-(m+1)P_{m-1}\dot{p}_{1}+(m-1)P_{m+1}\dot{p}_{-1}=0.$ (159)

Using222We do not display the dependence of $f_{m}^{\pm}$ on $p_{1}p_{-1}$.

$\displaystyle\dot{P}_{m}$
$\displaystyle=\frac{\mbox{d}}{\mbox{d}\tau}\left(-\frac{\mu}{p_{-1}^{m}}f_{m}^{\pm}\right)=m\frac{\mu}{p_{-1}^{m+1}}\dot{p}_{-1}f_{m}^{\pm}-\frac{\mu}{p_{-1}^{m}}(f_{m}^{\pm})^{\prime}\cdot(\dot{p}_{1}p_{-1}+p_{1}\dot{p}_{-1})$
(160)
$\displaystyle=m\frac{\mu}{p_{-1}^{m+1}}\dot{p}_{-1}f_{m}^{\pm}-\frac{\mu}{p_{-1}^{m}}\left(\mp\frac{m+1}{2\sqrt{\mu^{2}+p_{1}p_{-1}}}f_{m-1}^{\pm}\right)(\dot{p}_{1}p_{-1}+p_{1}\dot{p}_{-1}),$
(161)

the left-hand side of (159) is

LHS(159) $\displaystyle=\mp
2\sqrt{\mu^{2}+p_{1}p_{-1}}m\frac{\mu}{p_{-1}^{m+1}}\dot{p}_{-1}f_{m}^{\pm}-(m+1)\frac{\mu}{p_{-1}^{m}}f_{m-1}^{\pm}\cdot(\dot{p}_{1}p_{-1}+p_{1}\dot{p}_{-1})$
$\displaystyle\quad-(m+1)P_{m-1}\dot{p}_{1}+(m-1)P_{m+1}\dot{p}_{-1}$
$\displaystyle=\mp
2\sqrt{\mu^{2}+p_{1}p_{-1}}m\frac{\mu}{p_{-1}^{m+1}}\dot{p}_{-1}f_{m}^{\pm}-(m+1)\frac{\mu}{p_{-1}^{m}}f_{m-1}^{\pm}\cdot(\dot{p}_{1}p_{-1}+p_{1}\dot{p}_{-1})$
$\displaystyle\quad+(m+1)\frac{\mu}{p_{-1}^{m-1}}f_{m-1}^{\pm}\dot{p}_{1}-(m-1)\frac{\mu}{p_{-1}^{m+1}}f_{m+1}^{\pm}\dot{p}_{-1}$
(162)

The two terms containing $\dot{p}_{1}$ cancel each other, while the terms
proportional to $\dot{p}_{-1}$ are

$\displaystyle-\dot{p}_{-1}\frac{\mu}{p_{-1}^{m+1}}\left((m-1)f_{m+1}^{\pm}\pm
2m\sqrt{\mu^{2}+p_{1}p_{-1}}f_{m}^{\pm}+(m+1)p_{1}p_{-1}f_{m-1}^{\pm}\right),$
(163)

which is zero due to (54). This proves (159) and thus (77). Equation (79) is
proved in a similar way.

We consider next the partially gauge fixed Lagrangian

${\cal L}_{0}=\dot{x}^{0}P_{0}+\dot{x}^{1}p_{1}+\dot{x}^{-1}p_{-1},$ (164)

obtained from the full Lagrangian by setting $x^{m}=0$ for $|m|\geq 2$, and
which, as explained in the text, is just the standard Lagrangian for a massive
Poincaré particle. This Lagrangian has the gauge symmetry transformation
induced by the remaining first-class constraint $\phi_{0}$, associated with
reparametrization invariance, and the Poincaré invariance generated by
$p_{0}$, $p_{1}$, $p_{-1}$, $J$, $K_{+}$ and $K_{-}$, but also the infinite
set of symmetries given by the residual gauge transformations (111), (112),
which we write in the condensed notation

$\displaystyle\delta_{\text{res}}^{m}x^{1}$
$\displaystyle=\epsilon^{m}(m+1)\frac{P_{m-1}}{2P_{0}}=A_{m}(p_{1},p_{-1}),$
(165) $\displaystyle\delta_{\text{res}}^{m}x^{-1}$
$\displaystyle=-\epsilon^{m}(m-1)\frac{P_{m+1}}{2P_{0}}=B_{m}(p_{1},p_{-1}),$
(166)

with all the other variables $x^{0}$, $p_{0}$, $p_{1}$ and $p_{-1}$ invariant.
One has

$\displaystyle\delta_{\text{res}}{\cal L}_{0}$
$\displaystyle=\delta_{\text{res}}(\dot{x}^{1}p_{1}+\dot{x}^{-1}p_{-1})=p_{1}\frac{\mbox{d}}{\mbox{d}\tau}A_{m}+p_{-1}\frac{\mbox{d}}{\mbox{d}\tau}B_{m}=C_{m}\dot{p}_{1}+D_{m}\dot{p}_{-1},$
(167)

where

$\displaystyle C_{m}$ $\displaystyle=p_{1}\frac{\partial A_{m}}{\partial
p_{1}}+p_{-1}\frac{\partial B_{m}}{\partial p_{1}},$ (168) $\displaystyle
D_{m}$ $\displaystyle=p_{1}\frac{\partial A_{m}}{\partial
p_{-1}}+p_{-1}\frac{\partial B_{m}}{\partial p_{-1}}.$ (169)

The quasi-invariance of the Lagrangian, that is, the existence of a function
$F_{0}$ such that $\delta_{\text{res}}{\cal
L}_{0}=\frac{\mbox{d}}{\mbox{d}\tau}F_{0}$, is equivalent to

$\frac{\partial C_{m}}{\partial p_{-1}}=\frac{\partial D_{m}}{\partial
p_{1}},$ (170)

which boils down to

$\frac{\partial A_{m}}{\partial p_{-1}}=\frac{\partial B_{m}}{\partial
p_{1}},$ (171)

which in turn can be proved using the expressions of $P_{n}$, $P_{0}$ in terms
of $p_{1}$ and $p_{-1}$ and the properties of the functions $f_{n}^{\pm}$.

##  Appendix C Invariance of the massless limit constraints under
superrotations

Using $\\{p_{n},K_{+}^{m}\\}=(n-m)p_{n+m}$ one has that the variation of
$\varphi_{n}$ under a superrotation induced by $K_{+}^{m}$ is

$\displaystyle\\{\varphi_{n},K_{+}^{m}\\}$ $\displaystyle=\\{p_{n}\pm(\mp
1)^{n}p_{-1}^{-n}(\sqrt{p_{1}p_{-1}})^{n+1},K_{+}^{m}\\}$
$\displaystyle=(n-m)p_{n+m}\pm(\mp)^{n}(-np_{-1}^{-n-1}(-1-m)p_{-1+m})(\sqrt{p_{1}p_{-1}})^{n+1}$
$\displaystyle\pm(\mp)^{n}p_{-1}^{-n}(n+1)(\sqrt{p_{1}p_{-1}})^{n}\frac{1}{2\sqrt{p_{1}p_{-1}}}\left(p_{1}(-1-m)p_{-1+m}+p_{-1}(1-m)p_{1+m}\right).$

Since we only have to deal with the case $m\geq 2$, one has that $n+m$, $-1+m$
and $1+m$ are all positive. Using $\varphi_{n+m}$, $\varphi_{-1+m}$ and
$\varphi_{1+m}$ one can express $p_{n+m}$, $p_{-1+m}$ and $p_{1+m}$ in terms
of $p_{1}$ and $p_{-1}$, and one obtains, after re-arranging terms and
extracting the common dependency of all the terms in $p_{1}$ and $p_{-1}$,

$\displaystyle\\{\varphi_{n},K_{+}^{m}\\}$
$\displaystyle\simeq(\mp)^{n+m+1}p_{-1}^{-n-m}(\sqrt{p_{1}p_{-1}})^{n+m+1}$
$\displaystyle\cdot\left(n-m-n(1+m)+(1+m)\frac{n+1}{2}-(1-m)\frac{n+1}{2}\right)=0.$

Similarly, from $\\{p_{n},K_{-}^{m}\\}=-(m+n)p_{n-m}$,

$\displaystyle\\{\varphi_{n},K_{-}^{m}\\}$ $\displaystyle=\\{p_{n}\pm(\mp
1)^{n}p_{-1}^{-n}(\sqrt{p_{1}p_{-1}})^{n+1},K_{-}^{m}\\}$
$\displaystyle=-(n+m)p_{n-m}\pm(\mp)^{n}(-np_{-1}^{-n-1}(-(m-1)p_{-1-m}))(\sqrt{p_{1}p_{-1}})^{n+1}$
$\displaystyle\pm(\mp)^{n}p_{-1}^{-n}(n+1)(\sqrt{p_{1}p_{-1}})^{n}\frac{1}{2\sqrt{p_{1}p_{-1}}}\left(p_{1}(-(m-1)p_{-1-m})+p_{-1}(-(m+1)p_{1-m})\right).$

Again, since we must only consider $m\geq 2$, both $p_{-1-m}$ and $p_{1-m}$
are BMS momenta with negative indexes, and can be expressed in terms of
$p_{1}$ and $p_{-1}$ using $\bar{\varphi}_{1+m}$ and $\bar{\varphi}_{m-1}$,
respectively. For $n-m\geq 0$, one can use $\varphi_{n-m}$ for $p_{n-m}$,
while for $n-m<0$ $p_{n-m}$ has negative index and can be expressed in terms
of $p_{1}$ and $p_{-1}$ using $\bar{\varphi}_{m-n}$. It turns out that in both
cases the term obtained from $p_{n-m}$ is the same, and one has

$\displaystyle\\{\varphi_{n},K_{-}^{m}\\}$
$\displaystyle\simeq(\mp)^{n+m+1}p_{1}^{-m}p_{-1}^{-n}(\sqrt{p_{1}p_{-1}})^{n+m+1}$
$\displaystyle\cdot\left(-(n+m)-n(m-1)+(m-1)\frac{n+1}{2}+(m+1)\frac{n+1}{2}\right)=0.$

It should be noticed that it is this case, the variation of a positive index
constraint under a negative index superrotation, the one that breaks the
invariance of the theory under superrotations in the massive case.

Due to the real character of the Poisson bracket, one will also have

$\displaystyle\\{\bar{\varphi}_{n},K_{-}^{m}\\}^{*}$
$\displaystyle=\\{\varphi_{n},K_{+}^{m}\\}\simeq 0,$
$\displaystyle\\{\bar{\varphi}_{n},K_{+}^{m}\\}^{*}$
$\displaystyle=\\{\varphi_{n},K_{-}^{m}\\}\simeq 0,$

and thus all the constraints are weakly invariant under all the
superrotations.

##  Appendix D Casimirs of the Lorentz and Poincaré groups in BMS space

The action of the Lorentz generators $K_{\pm}$, $J$ on the $p_{n}$,
$n\in\mathbb{Z}$, provided by Poisson brackets,

$\displaystyle\delta_{J}p_{n}=\\{p_{n},J\\}=-np_{n},$ (172)
$\displaystyle\delta_{+}p_{n}=\\{p_{n},K_{+}\\}=-(1-n)p_{n+1},$ (173)
$\displaystyle\delta_{-}p_{n}=\\{p_{n},K_{-}\\}=-(1+n)p_{n-1},$ (174)

leads to the definition of infinite dimensional matrices acting on vectors
$(\ldots,p_{-2},p_{-1},p_{0},p_{1},p_{2},\ldots)$ which implement this action,
given by

$\displaystyle J_{nm}$ $\displaystyle=-n\delta_{nm},$ (175)
$\displaystyle(K_{+})_{nm}$ $\displaystyle=-(1-m)\delta_{n,m+1},$ (176)
$\displaystyle(K_{-})_{nm}$ $\displaystyle=-(1+m)\delta_{n,m-1}.$ (177)

Using the structure constants of the $SO(2,1)$ algebra in the $J,K_{+},K_{-}$
basis one can construct the Killing form of the Lie algebra, and from that the
quadratic Casimir, which is given by

$C^{L}_{2}=\frac{1}{2}J^{2}+\frac{1}{4}K_{+}K_{-}+\frac{1}{4}K_{-}K_{+}.$
(178)

Using the above matrices one immediately obtains

$\left(\frac{1}{2}J^{2}+\frac{1}{4}K_{+}K_{-}+\frac{1}{4}K_{-}K_{+}\right)_{nm}=1\cdot\delta_{nm}$
(179)

and hence this corresponds to an adjoint representation on the space of BMS
momenta.

Similarly, one can consider the action on the space of BMS coordinates
$x^{n}$, $n\in\mathbb{Z}$,

$\displaystyle\delta_{J}x^{n}=\\{x^{n},J\\}=nx^{n},$ (180)
$\displaystyle\delta_{+}x^{n}=\\{x^{n},K_{+}\\}=(2-n)x^{n-1},$ (181)
$\displaystyle\delta_{-}x^{n}=\\{x^{n},K_{-}\\}=(2+n)x^{n+1},$ (182)

which leads to the matrices

$\displaystyle\tilde{J}_{nm}$ $\displaystyle=n\delta_{nm},$ (183)
$\displaystyle(\tilde{K}_{+})_{nm}$ $\displaystyle=(2-m)\delta_{n,m-1},$ (184)
$\displaystyle(\tilde{K}_{-})_{nm}$ $\displaystyle=(2+m)\delta_{n,m+1}.$ (185)

Again

$\left(\frac{1}{2}\tilde{J}^{2}+\frac{1}{4}\tilde{K}_{+}\tilde{K}_{-}+\frac{1}{4}\tilde{K}_{-}\tilde{K}_{+}\right)_{nm}=1\cdot\delta_{nm},$
(186)

which shows that it also corresponds to an adjoint representation on the space
of coordinates.

With respect to the Poincaré group, the quadratic Casimir in $2+1$ in our
coordinates is

$C_{2}^{P}=p_{0}^{2}-p_{1}p_{-1},$ (187)

which, for our system and taking into account the constraint $\phi_{0}$, takes
value $-\mu^{2}$. One may wonder if it is possible to obtain a quadratic
Casimir involving the higher BMS momenta, of the form

$C_{2}=A_{mn}p_{m}p_{n},\quad A_{mn}=A_{nm}.$ (188)

Imposing the invariance under $J$ one gets

$\delta_{J}C_{2}=-A_{mn}p_{m}p_{n}(m+n)=0$ (189)

which implies that the only $A_{mn}$ that can be different from zero are those
corresponding to $m=-n$. Thus

$A_{mn}=A_{n}\delta_{m,-n},$ (190)

with $A_{n}=A_{-n}$ due to the symmetry of $A_{mn}$. Computing now the
variation under $K_{+}$ and using this form for $A_{mn}$ one gets

$\delta_{+}C_{2}=-p_{m+1}p_{-m}((1-m)A_{m}+(2+m)A_{m+1})=0.$ (191)

In order to equal to zero the coefficients of this sum over $m$ one must
notice that the terms corresponding to $m=n$ and $m=-1-n$ yield the same
product $p_{m+1}p_{-m}$. Taking this into account and using $A_{m}=A_{-m}$ one
gets the first order recurrence relation

$(1-m)A_{m}+(2+m)A_{m+1}=0,\quad m=0,1,2,\ldots.$ (192)

The invariance under $K_{-}$ does not add any new condition. For $m=0$, (192)
yields

$A_{0}+2A_{1}=0,$

from which $A_{1}=-\frac{1}{2}A_{0}$ and hence also
$A_{-1}=-\frac{1}{2}A_{0}$. For $m=1$, however, the relation is

$0\cdot A_{1}+3A_{2}=0,$

from which $A_{2}=0$ and thus $A_{-2}=0$. From this point, using the
recurrence for higher values of $m$ leads to $A_{m}=A_{-m}=0$ for
$m=2,3,\ldots$. The final result is then that the only quadratic Casimir of
the Poincaré group in BMS space is, up to a global constant, the standard one,
given by (187).

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