# Symmetries of free massless particles and soft theorems

Shamik Banerjee  Institute of Physics,
Sachivalaya Marg, Bhubaneshwar, India-751005
and
Homi Bhabha National Institute, Anushakti Nagar, Mumbai, India-400085
<EMAIL_ADDRESS>
###### Abstract

In an earlier paper we have constructed a basis of massless single particle
quantum states which transform in the unitary principal series representation
of the four dimensional Lorentz group. The S-matrix written in this basis
gives rise to the Mellin transformed amplitude of Pasterski-Shao-Strominger
and its generalization. In this basis the particle can be thought of as living
on the null-infinity in the Minkowski space. In this paper we take some
preliminary steps to see how the connection between soft theorems and
symmetries work out in this picture. For simplicity we consider only the
leading soft photon and soft graviton theorems which are related to U(1) Kac-
Moody and supertranslations.

###### Contents

  1. I Introduction
  2. II conventions
  3. III Review
  4. IV Symmetries of free massless particles
    1. IV.1 (Super)translation
      1. IV.1.1 Global BMS
      2. IV.1.2 Local BMS
      3. IV.1.3 Space-time relaization
    2. IV.2 Global U(1) Kac-Moody
    3. IV.3 Discussion
  5. V S-Matrix and Soft Limits
    1. V.1 A Convenient Set Of Variables
    2. V.2 S-matrix
    3. V.3 Soft Limit
  6. VI Leading Soft Photon and Graviton
    1. VI.1 Soft Photon Limit
    2. VI.2 Soft Graviton Limit
  7. VII A slightly different treatment of the soft-operator
  8. VIII Discussion
  9. IX Acknowledgement

Soft Theorems Shamik Banerjee

##  I Introduction

In the pioneering works Strominger:2013lka ; Strominger:2013jfa  a remarkable
connection was found between the asymptotic symmetries of gauge and gravity
theories in flat space and the physics in the infrared. In particular, this
has established Strominger:2013lka ; Strominger:2013jfa ; He:2014cra ;
Strominger:2014pwa ; He  that Weinberg’s soft theorems Weinberg:1965nx  for
massless particles can be thought of as the Ward identity for infinite-
dimensional global symmetries acting on the null-infinity in Minkowski space-
time. Later this was extended to massive particles Kapec:2015ena ;
Campiglia:2015qka ; Campiglia:2015kxa  in which case the infinite dimensional
symmetries act on timelike infinity rather than null-infinity. In this paper
we focus only on the massless case.

In four dimensions the space of null-momentum directions is a two-sphere - the
so-called celestial sphere. The four dimensional Lorentz group
$SL(2,\mathbb{C})$ acts on the celestial sphere as the group of global
conformal transformations. Now in a related development Pasterski:2016qvg ;
Pasterski:2017kqt ; Pasterski:2017ylz  Pasterski, Shao and Strominger have
proposed an integral (Mellin) transform of massless (and massive) scattering
amplitude (S-matrix) with the property that the resulting amplitude transforms
under $SL(2,\mathbb{C})$ as the correlation function of products of
$SL(2,\mathbb{C})$ primaries living on the celestial sphere. From the space-
time point of view one gets the amplitude on the celestial sphere if, instead
of the plane waves, one uses the conformal primary wave-functions
Pasterski:2017kqt ; Cheung:2016iub ; deBoer:2003vf  to describe the external
particles. The study of Pasterski:2017kqt  also established that the conformal
primary wave-functions are closely related to the Unitary Principal Continuous
Series Representation of $SL(2,\mathbb{C})$ (Lorentz group) bargman ;
Gadde:2017sjg ; Hogervorst:2017sfd ; Simmons-Duffin:2017nub . The Mellin
transformed amplitude has been computed for several field theories at both
tree and loop level Pasterski:2017ylz ; Cardona:2017keg ; Lam:2017ofc ;
Banerjee:2017jeg ; Schreiber:2017jsr .

Now, one of the most attractive features of this approach Pasterski:2016qvg ;
Pasterski:2017kqt ; Pasterski:2017ylz  is that it suggests a way of going
directly from the momentum space to null-infinity. The connection between soft
theorems and asymptotic symmetries also suggests that one should be able to do
this. But this particular approach goes beyond the soft sector of the gauge or
gravity theory. In an earlier paper Banerjee:2018gce  we have studied this
question from the Hilbert space perspective. Following Pasterski:2016qvg ;
Pasterski:2017kqt ; Pasterski:2017ylz  we have constructed Banerjee:2018gce  a
basis of massless single particle quantum states which transform in the
unitary principal series representation of the Lorentz group. It turns out
that these states also transform nicely under global space-time translations.
From physical point of view a massless particle in this state can be thought
of as living on the null-infinity.

An interesting question to ask is that how one can see the connection between
soft-theorems and symmetries in this approach. In this paper we take
preliminary steps in this direction by rewriting Weinberg’s soft theorems in
this basis Pasterski:2016qvg ; Pasterski:2017kqt ; Pasterski:2017ylz ;
Banerjee:2018gce  and relating it to infinite dimensional global symmetries
Strominger:2013lka ; Strominger:2013jfa . Technically this is very similar to
Kapec:2017gsg  except that we are using a different basis.

##  II conventions

In this paper we consider only massless particles. It is well known that we
can associate a Hermitian matrix with a (null) four-vector $p$ in the
following way,

$p=\begin{pmatrix}p^{0}-p^{3}&p^{1}+ip^{2}\\\
p^{1}-ip^{2}&p^{0}+p^{3}\end{pmatrix}$ (1)

The determinant of the matrix gives the norm of the four vector, i.e, $\det
p=-p^{2}$. Now let $\Lambda$ be a $SL(2,\mathbb{C})$ matrix which acts on $p$
as,

$p\rightarrow p^{\prime}=\Lambda p\Lambda^{\dagger}=\begin{pmatrix}p^{\prime
0}-p^{\prime 3}&p^{\prime 1}+ip^{\prime 2}\\\ p^{\prime 1}-ip^{\prime
2}&p^{\prime 0}+p^{\prime 3}\end{pmatrix},\ \ \Lambda=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in SL(2,\mathbb{C})$ (2)

Under this action $p^{\prime}$ is again an Hermitian matrix and the
determinant does not change, i.e, $\det p=\det p^{\prime}$ and so the the norm
of the four-vectors $p$ and $p^{\prime}$ are the same. So Eq-2 represents a
Lorentz transformation. In this way we can associate to every
$SL(2,\mathbb{C})$ matrix $\Lambda$ a Lorentz transformation which we also
denote by $\Lambda$. To be more precise, the pair of $SL(2,\mathbb{C})$
matrices $\\{\pm\Lambda\\}$ get mapped to the same Lorentz transformation. In
terms of the components of the four vector $p$ we can write the Lorentz
transformation as,

$p^{\prime\mu}={\Lambda^{\mu}}_{\nu}(a,b,c,d)\ p^{\nu}$ (3)

where ${\Lambda^{\mu}}_{\nu}(a,b,c,d)$ is calculated from Eq-2.

Another important quantity is the complex number $z$ defined as,

$z=\frac{p^{1}+ip^{2}}{p^{0}+p^{3}},\ \ p^{2}=0$ (4)

Topologically a space-like cross-section of the null-cone is a two-sphere and
$z$ is the stereographic coordinate on that. So $z$ specifies the direction of
the null momentum vector. The space of null directions is also known as the
celestial sphere.

Under (Lorentz) $SL(2,\mathbb{C})$ transformation $z$ transforms as,

$z\rightarrow z^{\prime}=\frac{p^{\prime 1}+ip^{\prime 2}}{p^{\prime
0}+p^{\prime 3}}=\frac{az+b}{cz+d}$ (5)

So the Lorentz transformations act as global conformal transformations on
$S^{2}$.

In terms of $z$ a null vector $p$ can be parametrized as,

$p=E\bigg{(}1,\frac{z+\bar{z}}{1+z\bar{z}},\frac{-i(z-\bar{z})}{1+z\bar{z}},\frac{1-z\bar{z}}{1+z\bar{z}}\bigg{)},\
\ E=p^{0}$ (6)

An alternative parametrization is,

$p=\omega\bigg{(}1+z\bar{z},z+\bar{z},-i(z-\bar{z}),1-z\bar{z}\bigg{)}=\frac{E}{1+z\bar{z}}\bigg{(}1+z\bar{z},z+\bar{z},-i(z-\bar{z}),1-z\bar{z}\bigg{)}$
(7)

We will use both.

Under (Lorentz) $SL(2,\mathbb{C})$ transformation,

$E\rightarrow E^{\prime}=E\ \frac{|az+b|^{2}+|cz+d|^{2}}{1+z\bar{z}}$ (8)

In terms of $\omega$,

$\omega\rightarrow\omega^{\prime}=\omega|cz+d|^{2}$ (9)

Let us now say a few things about the unitary operators which implement
Lorentz transformation in Hilbert space. Let us consider a Lorentz
transformation $\Lambda$ which acts on space-time coordinates as,
$x^{\mu}\rightarrow x^{\prime\mu}={\Lambda^{\mu}}_{\nu}x^{\nu}$. Let
$U(\Lambda)$ denote the corresponding unitary operator acting on the Hilbert
space. For an infinitesimal transformation we write Weinberg:1995mt ,

$U(\Lambda)=1+\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}+........,\ \
{\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu},\ \
\omega_{\mu\nu}=\eta_{\mu\alpha}\ {\omega^{\alpha}}_{\nu}=-\omega_{\nu\mu},\ \
J^{\mu\nu\dagger}=J^{\mu\nu}$ (10)

Now the six generators $J^{\mu\nu}$ can be organised into rotation and boost
three-vectors given by $\vec{J}=(J_{1},J_{2},J_{3})$ and
$\vec{K}=(K_{1},K_{2},K_{3})$. $\vec{J}$ and $\vec{K}$ are both Hermitian. For
our purpose it will be more convenient to write them in terms of the
generators $(L_{0},L_{\pm 1},\bar{L}_{0},\bar{L}_{\pm 1})$ defined as
Banerjee:2018gce ,

$L_{0}=\frac{-J_{3}+iK_{3}}{2},\ \ \bar{L}_{0}=-{L_{0}}^{\dagger}$ (11)
$L_{1}=\frac{-J_{1}+iK_{1}}{2}+i\frac{-J_{2}+iK_{2}}{2},\ \
\bar{L}_{1}=-{L_{1}}^{\dagger}$ (12)
$L_{-1}=-\frac{-J_{1}+iK_{1}}{2}+i\frac{-J_{2}+iK_{2}}{2},\ \
\bar{L}_{-1}=-{L_{-1}}^{\dagger}$ (13)

The commutation relation between these generators can be written as,

$[L_{m},L_{n}]=(m-n)L_{m+n},\ \ [\bar{L}_{m},\bar{L}_{n}]=(m-n)\bar{L}_{m+n},\
\ m,n=0,\pm 1$ (14)

The hermiticity of $\vec{J}$ and $\vec{K}$ gives the following reality
condition,

$L_{n}^{\dagger}=-\bar{L}_{n}$ (15)

Now consider an infinitesimal Lorentz transformation $U(\Lambda)$ generated by
$L_{n}$ and $\bar{L}_{n}$. It can be written as

$U(\Lambda)=1+\alpha L_{n}+\bar{\alpha}\bar{L}_{n}+.......$ (16)

where $\alpha$ is a complex number and $\bar{\alpha}$ is the complex conjugate
of $\alpha$. Using the reality condition on $L_{n}$ and $\bar{L}_{n}$ it is
easy to see that $U(\Lambda)$ is unitary.

Now the $SL(2,\mathbb{C})$ transformations which correspond to $U(\Lambda)$
can be written in terms of $z$ as Banerjee:2018gce

$z\rightarrow z^{\prime}=z+\alpha z^{n},\ \ n=0,\pm 1$ (17)

So as usual $L_{1}$ generates translation, $L_{0}$ generates dilatation and
$L_{-1}$ generates special conformal transformations on the two-sphere
parametrized by $(z,\bar{z})$.

##  III Review

Let us now review the part of Pasterski:2016qvg ; Pasterski:2017kqt ;
Pasterski:2017ylz ; Banerjee:2018gce  which will be used in this paper.

Let us consider free massless particles in four dimensions. The Hilbert space
of these particles carries unitary representation of the four dimensional
Poincare group $ISO(3,1)$. For our purpose it will be useful to identify the
Lorentz (sub-)group with $SL(2,\mathbb{C})$.

Using this data one can construct an abstract three dimensional space-time
with coordinates denoted by $(u,z,\bar{z})$ on which the Poincare group acts
as,

$\Lambda(u,z,\bar{z})=\bigg{(}\frac{u\ (1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}}\
,\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (18)
$T(l)(u,z,\bar{z})=(u+f(z,\bar{z},l),z,\bar{z})$ (19)

where

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(20)

Here $\Lambda$ is a $SL(2,\mathbb{C})$ (Lorentz) transformation and $T(l)$ is
a global space-time translation by a four-vector $l^{\mu}$. There is no
Poincare invariant metric in this space but rather there is a conformal
structure.

This three dimensional space-time hosts free Heisenberg-picture creation and
annihilation fields defined as,

$A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\int_{0}^{\infty}dE\
E^{i\lambda}e^{iEu}\ a^{\dagger}(p,\sigma),\ \lambda\in\mathbb{R},\
2\sigma\in\mathbb{Z}$ (21)
$A_{\lambda,\sigma}(u,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1-i\lambda}\int_{0}^{\infty}dE\
E^{-i\lambda}e^{-iEu}\ a(p,\sigma),\ \lambda\in\mathbb{R},\
2\sigma\in\mathbb{Z}$ (22)

where $a^{\dagger}(p,\sigma)$ and $a(p,\sigma)$ are the momentum space
creation and annihilation operators for a massless particle with null momentum
$p$ and helicity $\sigma$.

The equations defining $A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})$ and
$A_{\lambda,\sigma}(u,z,\bar{z})$ on the $(u,z,\bar{z})$ space are analogous
to the equations

$\phi^{-}(x^{\mu})=\int\frac{d^{3}\vec{p}}{(2\pi)^{3}2|\vec{p}|}e^{-ip.x}a^{\dagger}(p)$
(23)
$\phi^{+}(x^{\mu})=\int\frac{d^{3}\vec{p}}{(2\pi)^{3}2|\vec{p}|}e^{ip.x}a(p)$
(24)

which define, for example, the creation and annihilation parts of a free
massless scalar (quantum) field on the Minkowski space-time. The only
difference is that now we have defined
$A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})$ and
$A_{\lambda,\sigma}(u,z,\bar{z})$ also for spinning massless particles.

Let us also define Lorentz tensors in this space as any function
$\phi_{h,\bar{h}}(u,z,\bar{z})$ which transforms under $SL(2,\mathbb{C})$ as,

$\phi^{\prime}_{h,\bar{h}}(u^{\prime},z^{\prime},\bar{z}^{\prime})\bigg{(}\frac{dz^{\prime}}{dz}\bigg{)}^{h}\bigg{(}\frac{d\bar{z}^{\prime}}{d\bar{z}}\bigg{)}^{\bar{h}}=\phi_{h,\bar{h}}(u,z,\bar{z})$
(25)

where $(h,\bar{h})$ are arbitrary complex numbers subject to the restriction
that $h-\bar{h}\in\mathbb{Z}/2$. $(u^{\prime},z^{\prime},\bar{z}^{\prime})$
are related to $(u,z,\bar{z})$ by Eq-(18).

Under $SL(2,\mathbb{C})$ (Lorentz) transformation creation and annihilation
fields transform homogeneously as tensors,

$U(\Lambda)\ A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})\ U(\Lambda)^{-1}\\\
=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\
A^{\dagger}_{\lambda,\sigma}\bigg{(}\frac{u\
(1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (26)
$U(\Lambda)\ A_{\lambda,\sigma}(u,z,\bar{z})\ U(\Lambda)^{-1}\\\
=\frac{1}{(cz+d)^{2\bar{h}^{*}}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2h^{*}}}\
A_{\lambda,\sigma}\bigg{(}\frac{u\
(1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (27)

where $(h,\bar{h})$ are little-group indices given by,

$h=\frac{1+i\lambda-\sigma}{2},\ \bar{h}=\frac{1+i\lambda+\sigma}{2}$ (28)

$*$ denotes complex conjugation.

Now under global space-time translation by a four-vector $l^{\mu}$ the fields
transform as,

$e^{-il.P}A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})e^{il.P}=A^{\dagger}_{\lambda,\sigma}(u+f(z,\bar{z},l),z,\bar{z})$
(29)
$e^{-il.P}A_{\lambda,\sigma}(u,z,\bar{z})e^{il.P}=A_{\lambda,\sigma}(u+f(z,\bar{z},l),z,\bar{z})$
(30)

where $P^{\mu}$ is the translation operator and $f(z,\bar{z},l)$ is given in
Eq-20 .

The (anti-)commutator of the creation and annihilation fields can be computed
and is given by,

$[A_{\lambda,\sigma}(u,z,\bar{z}),A^{\dagger}_{\lambda^{\prime},\sigma^{\prime}}(u^{\prime},z^{\prime},\bar{z}^{\prime})]_{\pm}=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}-u+i\epsilon)\big{)}^{i(\lambda^{\prime}-\lambda)}},\
\epsilon\rightarrow 0+$ (31)

Here we have used the relation,

$[a(p_{1},\sigma_{1}),a^{\dagger}(p_{2},\sigma_{2})]_{\pm}=(2\pi)^{3}2|\vec{p}_{1}|\delta^{3}({\vec{p}_{1}-\vec{p}_{2}})\delta_{\sigma_{1}\sigma_{2}}$
(32)

This commutator is Lorentz covariant and translation invariant.

This, at least in principle, complete the specification of the free field
theory on $(u,z,\bar{z})$ space because any other thing can be computed from
the basic commutator given in Eq-(31).

Now, as far as the action of the Poincare group is concerned, this abstract
three dimensional space-time can be realized as the null-infinity of the
Minkowski space with $(u,z,\bar{z})$ as Bondi coordinates. But this does not
play any major role in the following discussions.

##  IV Symmetries of free massless particles

The following discussion is strongly motivated by the remarkable connection
between soft-theorems and asymptotic symmetries. In particular, it seems that
if we want to use the approach of Pasterski:2016qvg ; Pasterski:2017kqt ;
Pasterski:2017ylz ; Banerjee:2018gce  to study the relation between soft-
theorems and symmetries, then we cannot think of these symmetries as
asymptotic. The simplest reason being that the starting point is not a theory
defined in Minkowski space-time.

The following discussion is a very preliminary attempt in this direction.
Loosely speaking we pick up the hard part of the asymptotic symmetry generator
(charge) and then turn off all interactions. In the non-interacting theory the
hard-charge by itself is a symmetry generator and so we can study it in its
own right in the free Hilbert space. This is also closer to the usual way of
doing perturbative quantum field theory where we start with a free theory with
certain symmetries and then ask what kind of interactions are allowed if the
symmetry remains intact in the interacting theory.

###  IV.1 (Super)translation

Let us consider free massless particles of helicity $\sigma$. We keep the
helicity unspecified because at this stage nothing really depends crucially on
helicity and the following discussion is purely algebraic. For simplicity let
us just consider bosons. The conclusions remain unchanged for fermions.

The Hamiltonian can be written as,

$H=\int d\mu(p)|\vec{p}|\ a^{\dagger}(p,\sigma)a(p,\sigma)$ (33)

where $d\mu(p)$ is the Lorentz invariant measure given by

$d\mu(p)=\frac{d^{3}\vec{p}}{(2\pi)^{3}2|\vec{p}|}$ (34)

and

$[a(p_{1},\sigma),a^{\dagger}(p_{2},\sigma)]=(2\pi)^{3}\
2|\vec{p}_{1}|\delta^{3}({\vec{p}_{1}-\vec{p}_{2}})$ (35)

Now parametrizing $p$ in terms of $(E,z,\bar{z})$ we can write,

$H=\int d\mu(p)\ E\ a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)$ (36)

where

$d\mu(p)=\frac{d^{3}\vec{p}}{(2\pi)^{3}2|\vec{p}|}=\frac{E^{2}dE}{(2\pi)^{3}2E}\
\frac{4d^{2}z}{(1+z\bar{z})^{2}},\ \ d^{2}z=d\real z\ d\imaginary z$ (37)

Let us now consider the following charges defined as,

$\boxed{T_{f}=\int d\mu(p)\ E\
f(z,\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)=T_{f}^{\dagger}}$
(38)

where $f(z,\bar{z})$ is an arbitrary real smooth function on the 2-sphere
parametrized by $(z,\bar{z})$. The operator $T_{f}$ has the following
properties :

1)

$[H,T_{f}]=0$ (39)

So the charges are conserved for any function $f(z,\bar{z})$.

2)

$[T_{f},T_{f^{\prime}}]=0$ (40)

for arbitrary $f$ and $f^{\prime}$.

3) The last important property is,

$[T_{f},a^{\dagger}(E,z,\bar{z},\sigma)]=Ef(z,\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)$
(41) $[T_{f},a(E,z,\bar{z},\sigma)]=-Ef(z,\bar{z})a(E,z,\bar{z},\sigma)$ (42)

Let us now consider the unitary operator $U_{f}=e^{-iH_{f}}$. This unitary
operator is a ”Supertranslation - operator” in the $(u,z,\bar{z})$ space.

In order to see this let us go back to the basic creation/annihilation fields
in the $(u,z,\bar{z})$ space given by,

$A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\int_{0}^{\infty}dE\
E^{i\lambda}e^{iEu}\ a^{\dagger}(p,\sigma),\ \lambda\in\mathbb{R},\
\sigma\in\mathbb{Z}$ (43)
$A_{\lambda,\sigma}(u,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1-i\lambda}\int_{0}^{\infty}dE\
E^{-i\lambda}e^{-iEu}\ a(p,\sigma),\ \lambda\in\mathbb{R},\
\sigma\in\mathbb{Z}$ (44)

Now it is easy to see using the commutator of $T_{f}$ with the momentum space
creation/annihilation operators that ,

$\begin{gathered}e^{iT_{f}}A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})e^{-iT_{f}}=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\int_{0}^{\infty}dE\
E^{i\lambda}e^{iE(u+f(z,\bar{z}))}\ a^{\dagger}(p,\sigma)\\\
=A^{\dagger}_{\lambda,\sigma}(u+f(z,\bar{z}),z,\bar{z})\end{gathered}$ (45)

For infinitesimal transformation,

$f(z,\bar{z})\frac{\partial A_{\lambda,\sigma}^{\dagger}}{\partial
u}=i[T_{f},A_{\lambda,\sigma}^{\dagger}]$ (46)

Similarly,

$\begin{gathered}e^{iT_{f}}A_{\lambda,\sigma}(u,z,\bar{z})e^{-iT_{f}}=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\int_{0}^{\infty}dE\
E^{i\lambda}e^{iE(u+f(z,\bar{z}))}\ a(p,\sigma)\\\
=A_{\lambda,\sigma}(u+f(z,\bar{z}),z,\bar{z})\end{gathered}$ (47)

and

$f(z,\bar{z})\frac{\partial A_{\lambda,\sigma}}{\partial
u}=i[T_{f},A_{\lambda,\sigma}]$ (48)

for infinitesimal $f$.

So we can see that the unitary transformation $U_{f}=e^{-iT_{f}}$ generates a
point transformation in the $(u,z,\bar{z})$ space given by,

$(u,z,\bar{z})\rightarrow(u+f(z,\bar{z}),z,\bar{z})$ (49)

where $f(z,\bar{z})$ is an arbitrary real smooth function. Now if we choose
$f(z,\bar{z})$ to be of the following form

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(50)

where $l^{\mu}$ is some four-vector, then we get back the global Minkowski
space translations. If we relax this condition then we get ”Supertranslations”
in the $(u,z,\bar{z})$ space. At this point we would like to emphasize that
this has nothing to do with asymptotic symmetry.

Let us now see explicitly what happens to the commutator of
$A_{\lambda,\sigma}$ and $A^{\dagger}_{\lambda,\sigma}$. The commutator is
given by,

$[A_{\lambda,\sigma}(u,z,\bar{z}),A^{\dagger}_{\lambda^{\prime},\sigma^{\prime}}(u^{\prime},z^{\prime},\bar{z}^{\prime})]=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}-u+i\epsilon)\big{)}^{i(\lambda^{\prime}-\lambda)}},\
\epsilon\rightarrow 0+$ (51)

Now using the expressions for
$e^{iT_{f}}A_{\lambda,\sigma}(u,z,\bar{z})e^{-iT_{f}}$ and
$e^{iT_{f}}A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})e^{-iT_{f}}$ we get,

$\begin{gathered}\left[e^{iT_{f}}A_{\lambda,\sigma}(u,z,\bar{z})e^{-iT_{f}},e^{iT_{f}}A^{\dagger}_{\lambda^{\prime},\sigma^{\prime}}(u^{\prime},z^{\prime},\bar{z}^{\prime})e^{-iT_{f}}\right]\\\
\\\
=\left[A_{\lambda,\sigma}(u+f(z,\bar{z}),z,\bar{z}),A^{\dagger}_{\lambda^{\prime},\sigma^{\prime}}(u+f(z^{\prime},\bar{z}^{\prime}),z^{\prime},\bar{z}^{\prime})\right]\\\
\\\
=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}+f(z^{\prime},\bar{z}^{\prime})-u-f(z,\bar{z})+i\epsilon)\big{)}^{i(\lambda^{\prime}-\lambda)}}\\\
\\\
=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}-u+i\epsilon)\big{)}^{i(\lambda^{\prime}-\lambda)}}\\\
\\\
=\left[A_{\lambda,\sigma}(u,z,\bar{z}),A^{\dagger}_{\lambda^{\prime},\sigma^{\prime}}(u^{\prime},z^{\prime},\bar{z}^{\prime})\right]\end{gathered}$
(52)

The $f$’s in the denominator cancel due to the presence of
$\delta^{2}(z-z^{\prime})$. Physically this is just the fact that a free
particle moves without changing its direction of motion and this makes the
free theory on $(u,z,\bar{z})$ space ultralocal in the spatial direction.

Let us now write down the commutation relation between the Lorentz generators
and the supertranslations and show that the algebra is isomorphic to the BMS
algebra without superrotations.

Let us start with the definition of the charge $T_{f}$ given by,

$T_{f}=\int d\mu(p)\ E\
f(z,\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)$ (53)

We will also need the following transformation properties,

$U(\Lambda)^{-1}a^{\dagger}(p,\sigma)U(\Lambda)=e^{i\sigma\theta(\Lambda,p)}a^{\dagger}(\Lambda^{-1}p,\sigma)$
(54)
$U(\Lambda)^{-1}a(p,\sigma)U(\Lambda)=e^{-i\sigma\theta(\Lambda,p)}a(\Lambda^{-1}p,\sigma)$
(55)

where the quantities appearing have their standard meaning Weinberg:1995mt .
Using this we can write,

$\begin{gathered}U(\Lambda)^{-1}T_{f}U(\Lambda)=\int d\mu(p)\ E\
f(z,\bar{z})U(\Lambda)^{-1}a^{\dagger}(E,z,\bar{z},\sigma)U(\Lambda)U(\Lambda)^{-1}a(E,z,\bar{z},\sigma)U(\Lambda)\\\
=\int d\mu(p)\ E\
f(z,\bar{z})a^{\dagger}(\Lambda^{-1}E,\Lambda^{-1}z,\Lambda^{-1}\bar{z},\sigma)a(\Lambda^{-1}E,\Lambda^{-1}z,\Lambda^{-1}\bar{z},\sigma)\\\
=\int d\mu(\Lambda^{-1}p)\ \Lambda\Lambda^{-1}E\
f(\Lambda\Lambda^{-1}z,\Lambda\Lambda^{-1}\bar{z})a^{\dagger}(\Lambda^{-1}E,\Lambda^{-1}z,\Lambda^{-1}\bar{z},\sigma)a(\Lambda^{-1}E,\Lambda^{-1}z,\Lambda^{-1}\bar{z},\sigma)\\\
=\int d\mu(p)\ \Lambda E\ f(\Lambda
z,\Lambda\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\\\ =\int
d\mu(p)\ \bigg{(}E\ \frac{|az+b|^{2}+|cz+d|^{2}}{1+z\bar{z}}\bigg{)}\
f(\Lambda
z,\Lambda\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\\\ =\int
d\mu(p)\ E\ \bigg{(}\frac{|az+b|^{2}+|cz+d|^{2}}{1+z\bar{z}}\ f(\Lambda
z,\Lambda\bar{z})\bigg{)}a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\\\
=\int d\mu(p)\ E\
f^{\prime}(z,\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\\\
=T_{f^{\prime}}\end{gathered}$ (56)

So the supertranslation charges transform as,

$\boxed{U(\Lambda)^{-1}T_{f}U(\Lambda)=T_{f^{\prime}},\ \
f^{\prime}(z,\bar{z})=\frac{|az+b|^{2}+|cz+d|^{2}}{1+z\bar{z}}\ f(\Lambda
z,\Lambda\bar{z})}$ (57)

Here $\Lambda z=\frac{az+b}{cz+d}$.

Now since the global translation generators $P^{\mu}$ are already part of the
supertranslations the above commutation relation already contains
$U(\Lambda)^{-1}P^{\mu}U(\Lambda)={\Lambda^{\mu}}_{\nu}P^{\nu}$. It is
instructive to write the global translation generators in terms of $T_{f}$. In
terms of creation/annihilation operators we can write,

$P^{\mu}=\int d\mu(p)\ p^{\mu}\
a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma),$ (58)

Now,

$p=E\bigg{(}1,\frac{z+\bar{z}}{1+z\bar{z}},\frac{-i(z-\bar{z})}{1+z\bar{z}},\frac{1-z\bar{z}}{1+z\bar{z}}\bigg{)}$
(59)

So we get,

$P^{0}=H=T_{f},\ \ f=1$ (60) $P^{1}=T_{f},\ \ f=\frac{z+\bar{z}}{1+z\bar{z}}$
(61) $P^{2}=T_{f},\ \ f=\frac{-i(z-\bar{z})}{1+z\bar{z}}$ (62) $P^{3}=T_{f},\
\ f=\frac{1-z\bar{z}}{1+z\bar{z}}$ (63)

Let us now write down the following linear combinations,

$P^{0}+P^{3}=T_{00},\ \ f=\frac{2z^{0}\bar{z}^{0}}{1+z\bar{z}}$ (64)
$P^{0}-P^{3}=T_{11},\ \ f=\frac{2z^{1}\bar{z}^{1}}{1+z\bar{z}}$ (65)
$P^{1}+iP^{2}=T_{10},\ \ f=\frac{2z^{1}\bar{z}^{0}}{1+z\bar{z}}$ (66)
$P^{1}-iP^{2}=T_{01},\ \ f=\frac{2z^{0}\bar{z}^{1}}{1+z\bar{z}}$ (67)

####  IV.1.1 Global BMS

Now to write down the generators of supertranslation we have to expand the
function $f(z,\bar{z})$ in a basis. The natural basis to do this is the
spherical harmonics on the sphere Sachs:1962zza . So we can define,

$T_{lm}=T_{f},\ \ f\sim Y_{lm}(\theta,\phi)$ (68)

We can use the basic relation Eq-57 to calculate the commutator between the
Lorentz generators and the $T_{lm}$. But for simplicity we will not do so.

####  IV.1.2 Local BMS

It is simpler to compute the commutator between Lorentz generators and
$T_{pq}$ defined as Barnich:2009se ,

$T_{f}=T_{pq},\ \ f=\frac{2z^{p}\bar{z}^{q}}{1+z\bar{z}}$ (69)

where $(p,q)$ are integers. The corresponding generators will be denoted by
$T_{pq}$.

Let us also note that the reality condition on the generators is given by,

$T_{pq}^{\dagger}=T_{qp}$ (70)

So the Hermitian charge can be written as,

$T=\sum_{p,q}C_{pq}T_{pq},\ \ C_{pq}^{*}=C_{qp}$ (71)

Let us now write down the commutators between the Lorentz generators
$(L_{0},L_{\pm 1},\bar{L}_{0},\bar{L}_{\pm 1})$ and $T_{pq}$. In order to this
we have to use the relation derived earlier,

$U(\Lambda)^{-1}T_{f}U(\Lambda)=T_{f^{\prime}},\ \
f^{\prime}(z,\bar{z})=\frac{|az+b|^{2}+|cz+d|^{2}}{1+z\bar{z}}\ f(\Lambda
z,\Lambda\bar{z})$ (72)

Now it is much simpler to work with the function $F(z,\bar{z})$ defined as,

$F(z,\bar{z})=(1+z\bar{z})f(z,\bar{z})$ (73)

In terms of $F$, the charge can be written as,

$\begin{gathered}T_{F}=T_{f}=\int d\mu(p)\ E\
f(z,\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\\\ =\int
d\mu(p)\ \frac{E}{1+z\bar{z}}\ \
(1+z\bar{z})f(z,\bar{z})a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\\\
=\int d\mu(p)\ \frac{E}{1+z\bar{z}}\ F(z,\bar{z})\
a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)\end{gathered}$ (74)

Under (Lorentz) $SL(2,\mathbb{C})$ transformation $T_{F}(=T_{f})$ transforms
as,

$U(\Lambda)^{-1}T_{F}U(\Lambda)=T_{F^{\prime}},\ \
F^{\prime}(z,\bar{z})=|cz+d|^{2}\ F(\Lambda z,\Lambda\bar{z})$ (75)

Now consider the infinitesimal $SL(2,\mathbb{C})$ transformation generated by
$(L_{n},\bar{L}_{n})$,

$U(\Lambda)=1+\alpha L_{n}+\bar{\alpha}\bar{L}_{n},\ \
U(\Lambda)^{-1}=1-\alpha L_{n}-\bar{\alpha}\bar{L}_{n}$ (76)

where $\bar{\alpha}$ is the complex conjugate of $\alpha$. In terms of the $z$
coordinate this generates the transformation,

$z\rightarrow\Lambda z=z+\alpha z^{n}$ (77)

Now the generators $T_{pq}$ correspond to
$F_{pq}(z,\bar{z})=2z^{p}\bar{z}^{q}$, i.e,

$\boxed{T_{pq}=\int d\mu(p)\ \frac{E}{1+z\bar{z}}\ 2z^{p}\bar{z}^{q}\
a^{\dagger}(E,z,\bar{z},\sigma)a(E,z,\bar{z},\sigma)=T_{qp}^{\dagger}}$ (78)

Now we use the infinitesimal version of the relation given in Eq-75

$(1-\alpha L_{n}-\bar{\alpha}\bar{L}_{n})T_{F}(1+\alpha
L_{n}+\bar{\alpha}\bar{L}_{n})=T_{F^{\prime}}$ (79)
$F^{\prime}(z,\bar{z})=\bigg{(}1-\frac{n+1}{2}(\alpha
z^{n}+\bar{\alpha}\bar{z}^{n})\bigg{)}F\big{(}z+\alpha
z^{n},\bar{z}+\bar{\alpha}\bar{z}^{n}\big{)}$ (80)

and replace $F(z,\bar{z})$ with $F_{pq}(z,\bar{z})=2z^{p}\bar{z}^{q}$ and
$T_{F}$ with $T_{pq}$. Now after simple manipulations we get the following
commutators,

$\boxed{[L_{n},T_{pq}]=\bigg{(}\frac{n+1}{2}-p\bigg{)}T_{p+n,q},\ \
[\bar{L}_{n},T_{pq}]=\bigg{(}\frac{n+1}{2}-q\bigg{)}T_{p,q+n},\ \
p,q\in{\mathbb{Z}}}$ (81)

The rest of the commutators are,

$\boxed{[L_{m},L_{n}]=(m-n)L_{m+n},\ \
[\bar{L}_{m},\bar{L}_{n}]=(m-n)\bar{L}_{m+n},\ \
[T_{pq},T_{p^{\prime}q^{\prime}}]=0,\ \ [L_{m},\bar{L}_{n}]=0,\ \
L_{n}^{\dagger}=-\bar{L}_{n}}$ (82)

Here $m,n=0,\pm 1$.

This is a subalgebra of the extended BMS algebra, including superrotations,
found in Barnich:2009se .

####  IV.1.3 Space-time relaization

Now since we know how the charges act on creation and annihilation operators
in momentum space we can also also write down the action of the unitary
operator $e^{-iT_{f}}$ on quantum fields defined on Minkowski space. Let us
consider a massless scalar field in Minkowski space. The field corresponding
to this is,

$\phi(x)=\int d\mu(p)\big{(}e^{ip.x}a(p)+e^{-ip.x}a^{\dagger}(p)\big{)}$ (83)

So the transformed field can be written as,

$\phi_{f}(x)=e^{iT_{f}}\phi(x)e^{-iT_{f}}=\int
d\mu(p)\big{(}e^{ip.x}e^{iE_{p}f(z,\bar{z})}a(p)+e^{-ip.x}e^{-iE_{p}f(z,\bar{z})}a^{\dagger}(p)\big{)},\
\ E_{p}=|\vec{p}|$ (84)

Now if we choose $f(z,\bar{z})$ to be of the following form,

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(85)

then

$\phi_{f}(x^{\mu})=\phi(x^{\mu}+l^{\mu})$ (86)

So it is a space-time translation by four-vector $l^{\mu}$. But, there are
infinitely many functions $f(z,\bar{z})$ for which there is no such simple
geometric interpretation in terms of space-time. In fact if we think of $a$
and $a^{\dagger}$ as complex conjugate c-numbers then $\phi_{f}$ is a solution
of the massless Klein-Gordon equation if $\phi$ is. So it maps classical
solutions to classical solutions as is expected for a symmetry transformation.

Now suppose we demand that an interacting theory of massless scalar fields
should also have this global symmetry. Arbitrary interaction will generically
break this symmetry, but Strominger:2013jfa ; He  have shown that if we couple
the massless particle to dynamical gravity then this symmetry is preserved.
This symmetry is now realized as the group of large diffeomorphisms and
becomes the conventional BMS supertranslations. So now, as a part of a much
bigger (gauge) symmetry group, the infinite dimensional global symmetry is
realized locally geometrically in space-time. What this means is that we can
write down actions, as integral over space-time of a local Lagrangian density,
which are invariant under diffeomorphism and the infinite dimensional global
symmetry is a subset of such diffeomorphisms. This should be contrasted with
the situation in $(u,z,\bar{z})$ space where supertranslations are realized
locally from the very beginning.

###  IV.2 Global U(1) Kac-Moody

Now consider a charged massless scalar particle. The usual expression for the
conserved charge of the particle can be written as,

$Q=e\int d\mu(p)\big{(}a^{\dagger}(p)a(p)-b^{\dagger}(p)b(p)\big{)}$ (87)

where the notation is standard. The commutators are,

$[Q,a^{\dagger}(p)]=ea^{\dagger}(p),\ \ [Q,a(p)]=-ea(p)$ (88)
$[Q,b^{\dagger}(p)]=-eb^{\dagger}(p),\ \ [Q,b(p)]=eb(p)$ (89)

Now just like supertranslation we can consider the following charges given by,

$Q_{f}=e\int
d\mu(p)f(z,\bar{z})\big{(}a^{\dagger}(p)a(p)-b^{\dagger}(p)b(p)\big{)}=Q_{f}^{\dagger}$
(90)

$Q_{f}$ is obviously conserved and $[Q_{f},Q_{f^{\prime}}]=0$ for all $f$. The
commutator of the charge with the cretaion-annihilation operators is given by,

$[Q_{f},a^{\dagger}(p)]=ef(z,\bar{z})a^{\dagger}(p),\ \
[Q_{f},a(p)]=-ef(z,\bar{z})a(p)$ (91)
$[Q_{f},b^{\dagger}(p)]=-ef(z,\bar{z})a^{\dagger}(p),\ \
[Q_{f},b(p)]=ef(z,\bar{z})a(p)$ (92)

On the creation and annihilation fields on the $(u,z,\bar{z})$ space the
unitary operator $e^{-iQ_{f}}$ acts like,

$e^{-iQ_{f}}A_{\lambda}^{\dagger}(u,z,\bar{z})e^{iQ_{f}}=e^{-ief(z,\bar{z})}A_{\lambda}^{\dagger}(u,z,\bar{z}),\
\
e^{-iQ_{f}}A_{\lambda}(u,z,\bar{z})e^{iQ_{f}}=e^{ief(z,\bar{z})}A_{\lambda}(u,z,\bar{z})$
(93)
$e^{-iQ_{f}}B_{\lambda}^{\dagger}(u,z,\bar{z})e^{iQ_{f}}=e^{ief(z,\bar{z})}B_{\lambda}^{\dagger}(u,z,\bar{z}),\
\
e^{-iQ_{f}}B_{\lambda}(u,z,\bar{z})e^{iQ_{f}}=e^{-ief(z,\bar{z})}B_{\lambda}(u,z,\bar{z})$
(94)

So in the $(u,z,\bar{z})$ charges act by local phase multiplication. So we get
an infinite dimensional global $U(1)$ symmetry. There is one copy of $U(1)$ at
every point of the sphere.

Let us now find out the Lorentz transformation properties of the charges. By
the same method, that we have applied to supertranslation, we get,

$U(\Lambda)^{-1}Q_{f}U(\Lambda)=Q_{f^{\prime}},\ \
f^{\prime}(z,\bar{z})=f(\Lambda z,\Lambda\bar{z})$ (95)

So the total charge given by $Q_{f=1}$ is a scalar. Now we define the
”moments”,

$Q_{pq}=e\int
d\mu(p)z^{p}\bar{z}^{q}\big{(}a^{\dagger}(p)a(p)-b^{\dagger}(p)b(p)\big{)}=Q_{qp}^{\dagger},\
\ (p,q)\in\mathbb{Z}$ (96)

Now as we have already discussed in the case of supertranslation, to get
globally defined charges we should expand $f(z,\bar{z})$ in speherical
harmonics on the $S^{2}$. But it is easier to work with the above charges.
This is also in the spirit of Strominger:2013lka .

The commutator of $Q_{pq}$ with the Lorentz generators $(L_{0},L_{\pm
1},\bar{L}_{0},\bar{L}_{\pm 1})$, as following from Eq-95 is given by,

$\boxed{[L_{n},Q_{pq}]=-p\ Q_{p+n,q}\ ,\ \ [\bar{L}_{n},Q_{pq}]=-q\ Q_{p,q+n}\
,\ \ [Q_{pq},Q_{p^{\prime}q^{\prime}}]=0}$ (97)

This is a level zero $U(1)$ Kac-Moody algebra except that we have confined
ourselves to the global (Lorentz) conformal algebra. To connect it to the more
familiar ”holomorphic” and ”antiholomorphic” charges, we can define
$Q_{p}=Q_{p,q=0}$ and $\bar{Q}_{q}=Q_{p=0,q}$ which satisfy,

$[L_{n},Q_{p}]=-pQ_{p+n},\ \ [\bar{L}_{n},Q_{p}]=0,\ \ [Q_{p},Q_{q}]=0$ (98)
$[L_{n},\bar{Q}_{q}]=0\ ,\ \ [\bar{L}_{n},\bar{Q}_{q}]=-qQ_{q+n},\ \
[\bar{Q}_{p},\bar{Q}_{q}]=0$ (99) $[Q,\bar{Q}]=0$ (100)

One can repeat the same exercise with $f(z,\bar{z})$ expanded in spherical
harmonics. It will be interesting to do that.

The most important question of course is how constraining the symmetries are
once we go to the interacting theory. For example the infinite dimensional
global $U(1)$ symmetry of a free charged scalar particle is preserved if we
couple it to photons He:2014cra ; Strominger:2017zoo . Then it is realized as
large gauge transformations at null-infinity. So it is sufficient to have
photons in the interacting theory. But is it also necessary ?

###  IV.3 Discussion

If we confine ourselves to the Hilbert space only then the statement that a
free massless particle has infinite symmetries has, strictly speaking, no
content. The Hilbert space of any free particle already has an infinite number
of conserved quantities given by the number operators $a^{\dagger}(p)a(p)$. We
can take any arbitrary linear combination of the number operators and generate
another conserved quantity. A priori there is no reason to prefer one linear
combination over the other. But we know that some of them are distinguished by
the fact that they are related to space-time symmetries and are also
symmetries of interacting theories. For example, Global space-time
translations belong to this class but there are many more. So how do we decide
which symmetries are useful ?

The connection between soft-theorems and symmetries suggest that in the free
Hilbert space of massless particles we consider hermitian operators of the
form,

$Q_{K}=\sum_{\sigma,\sigma^{\prime}}\int\int
d\mu(p)d\mu(p^{\prime})a(p,\sigma)^{\dagger}K(\sigma
p,\sigma^{\prime}p^{\prime})a(p^{\prime},\sigma^{\prime})$ (101)

where $K$ is a hermitian kernel which we assume to be local in momentum space,
i.e, it is a sum of derivatives of a finite number of momentum space delta
function. Now let us impose the following reasonable conditions :

1) The hermitian operators $Q_{K}$ corresponding to different $K$’s should
form a closed algebra under commutation. The algebra may be finite or infinite
dimensional. In the two examples we have studied they are infinite
dimensional.

2) It seems that as far as the free theory is concerned we can think in terms
of (quantum) fields either on Minkowski space or on the $(u,z,\bar{z})$ space.
The action of $Q_{K}$ on the fields should be local in at least one
description. Supertranslations are examples of this. The action of $T_{f}$ for
a generic $f$ is local only in the $(u,z,\bar{z})$ space whereas global
translations are local in both descriptions.

$Q_{K}$’s satisfying at least the above two conditions may reasonably be said
to form a symmetry algebra of a free massless particle. Now, heuristically
speaking, free particles are maximally symmetric objects. So classifying all
possible kernels $K$ subject to the constraint that the charges $Q_{K}$
satisfy the above two conditions can possibly give us the global symmetries
that a putative interacting theory of massless particles can have. This is a
mathematically well-posed problem and it will be interesting to know the
answer.

##  V S-Matrix and Soft Limits

###  V.1 A Convenient Set Of Variables

In practice it is much simpler to write the creation and annihilation fields
in terms of the new variables defined as,

$U=u(1+z\bar{z}),\ \omega=\frac{E}{1+z\bar{z}}$ (102)

Under $SL(2,\mathbb{C})$ (Lorentz) transformation

$\Lambda(U,z,\bar{z})=\bigg{(}\frac{U}{|cz+d|^{2}},\ \frac{az+b}{cz+d},\
\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (103)

and under space-time translation by a 4-vector $l^{\mu}$,

$T(l)(u,z,\bar{z})=(u+g(z,\bar{z},l),z,\bar{z})$ (104)

where

$g(z,\bar{z},l)=(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}=(1+z\bar{z})f(z,\bar{z},l)$
(105)

In terms of these new variables we can write,

$A^{\dagger}_{\lambda,\sigma}(U,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}d\omega\
\omega^{i\lambda}e^{i\omega U}\ a^{\dagger}(p,\sigma)$ (106)

and

$A_{\lambda,\sigma}(U,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}d\omega\
\omega^{-i\lambda}e^{-i\omega U}\ a(p,\sigma)$ (107)

The null momentum $p$ is now parametrized as,

$p=\omega(1+z\bar{z},z+\bar{z},-i(z-\bar{z}),1-z\bar{z})=\frac{E}{1+z\bar{z}}(1+z\bar{z},z+\bar{z},-i(z-\bar{z}),1-z\bar{z})$
(108)

###  V.2 S-matrix

Let us now consider a $M\rightarrow N$ massless particle scattering. The
quantum state of a particle is specified by its null four-momentum $p$,
helicity $\sigma$ and an extra label $n$ which carries information about other
internal quantum numbers like charge. The corresponding creation and
annihilation operators are denoted by $a^{\dagger}(p,\sigma,n)$ and
$a(p,\sigma,n)$. Since we are considering a scattering process, there are also
the labels $in$ and $out$ depending on whether the particle is incoming or
outgoing.

Let us now introduce the $in$ and $out$ creation and annihilation free-fields
in the $(U,z,\bar{z})$ space, defined as

$A^{\dagger
in/out}_{n,\lambda,\sigma}(U,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}d\omega\
\omega^{i\lambda}e^{i\omega U}\ a^{\dagger}_{in/out}(p,\sigma,n)$ (109)
$A^{in/out}_{n,\lambda,\sigma}(U,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}d\omega\
\omega^{-i\lambda}e^{-i\omega U}\ a_{in/out}(p,\sigma,n)$ (110)

Using these we can write,

$\begin{gathered}\tilde{A}\big{(}\\{n,\lambda,\sigma,U,z,\bar{z}\\}\big{)}=\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
S_{N\leftarrow M}\big{(}\\{p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (111)

where $\ket{\Omega}$ is the Poincare vacuum and the S-matrix is defined as
usual by,

$S_{N\leftarrow M}\big{(}\\{p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}=\bra{\Omega}\prod_{i=M+1}^{M+N}a_{out}(p_{i},\sigma_{i},n_{i})\prod_{j=1}^{M}a^{\dagger}_{in}(p_{j},\sigma_{j},n_{j})\ket{\Omega}$
(112)

We will refer to L.H.S of Eq-111 as the $\tilde{A}$-amplitude. This is
essentially the integral transform of the S-matrix and its generalization
which appears in Pasterski:2016qvg ; Pasterski:2017kqt ; Pasterski:2017ylz ;
Banerjee:2018gce . As we will see this form of the L.H.S is more convenient
for discussing the soft limit.

It will be interesting to construct the interpolating fields on this space but
this is beyond the scope of the present paper.

###  V.3 Soft Limit

Let us now consider a $M\rightarrow N+1$ scattering process. We have added an
extra particle in the final state with momentum $p$ and helicity $\sigma$
which will be taken to be soft at the end. The corresponding
$\tilde{A}$-amplitude can be written as,

$\begin{gathered}\bra{\Omega}A^{out}_{\lambda,\sigma}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}d\omega\omega^{-i\lambda}e^{-i\omega
U}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
S_{N+1\leftarrow M}\big{(}\\{p,\sigma\ ;\ p_{i},\sigma_{i},n_{i}\\},out\
\big{|}\ \\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (113)

In the definition of the $\tilde{A}$-amplitude all the external legs of the
S-matrix are integrated over. So in order to take the soft limit we have to
make the leg corresponding to the would-be soft particle free. This can be
done by inverting the transformation at the soft leg. The resulting formula
can be written as,

$\begin{gathered}\sqrt{2\pi^{2}}\int_{-\infty}^{\infty}d\lambda\
\omega^{i\lambda-1}e^{i\omega
U}\bra{\Omega}A^{out}_{\lambda,\sigma}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
S_{N+1\leftarrow M}\big{(}\\{p,\sigma\ ;\ p_{i},\sigma_{i},n_{i}\\},out\
\big{|}\ \\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (114)

We can see that the outgoing leg of the S-matrix with momentum $p$ and
helicity $\sigma$ is now free. Now one can take the soft limit by multiplying
both sides with appropriate factors of $\omega$ and taking $\omega$ to zero.
Let us denote this factor by $f(\omega)$.

$\begin{gathered}\lim_{\omega\rightarrow
0+}f(\omega)\sqrt{2\pi^{2}}\int_{-\infty}^{\infty}d\lambda\
\omega^{i\lambda-1}e^{i\omega
U}\bra{\Omega}A^{out}_{\lambda,\sigma}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
\lim_{\omega\rightarrow 0+}f(\omega)S_{N+1\leftarrow M}\big{(}\\{p,\sigma\ ;\
p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (115)

We will this formula to study the soft-limit.

##  VI Leading Soft Photon and Graviton

For leading soft photon and graviton we have to take $f(\omega)=\omega$. With
this the Eq-115 becomes,

$\begin{gathered}\lim_{\omega\rightarrow
0+}\sqrt{2\pi^{2}}\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}\bra{\Omega}A^{out}_{\lambda,\sigma}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
\lim_{\omega\rightarrow 0+}\omega S_{N+1\leftarrow M}\big{(}\\{p,\sigma\ ;\
p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (116)

Now consider the operator

$J_{\sigma,\omega}(z,\bar{z})=\sqrt{2\pi^{2}}\
\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}A^{out}_{\lambda,\sigma}(U,z,\bar{z})=\omega a^{out}(p,\sigma)$ (117)

We have omitted the $U$ dependence in $J_{\sigma,\omega}$ because it is equal
to the $U$-independent annihilation operator $\omega a^{out}(p,\sigma)$. We
can also work by keeping the $U$ dependence and then the soft theorems will
imply that the $U$ dependence is trivial. In fact this is more appropriate
from our point of view. We discuss this in detail in a later section.

Now using the $SL(2,\mathbb{C})$ transformation property of
$A_{\lambda,\sigma}$ we get,

$U(\Lambda)J_{\sigma,\omega}(z,\bar{z})U(\Lambda)^{-1}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}J_{\sigma,\omega|cz+d|^{2}}\bigg{(}\frac{az+b}{cz+d},\
\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (118)

where

$h=\frac{1+\sigma}{2},\ \bar{h}=\frac{1-\sigma}{2}$ (119)

Now if we take the soft limit $\omega\rightarrow 0+$ we can define the soft-
operator Kapec:2017gsg  $J_{\sigma}(z,\bar{z})=J_{\sigma,\omega=0}(z,\bar{z})$
whose Lorentz transformation property is

$U(\Lambda)J_{\sigma}(z,\bar{z})U(\Lambda)^{-1}=\frac{1}{(cz+d)^{1+\sigma}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{1-\sigma}}J_{\sigma}\bigg{(}\frac{az+b}{cz+d},\
\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (120)

We will now write Weinberg’s soft theorems in terms of the soft operator
$J_{\sigma}$ for $\sigma=+1$ and $\sigma=+2$.

###  VI.1 Soft Photon Limit

For photons $\sigma=\pm 1$ and we denote the corresponding soft photon
operators by $J_{\pm}(z,\bar{z})$ whose $SL(2,\mathbb{C})$ transformation
property is given by,

$U(\Lambda)J_{+}(z,\bar{z})U(\Lambda)^{-1}=\frac{1}{(cz+d)^{2}}\
J_{+}\bigg{(}\frac{az+b}{cz+d},\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)},\
\sigma=+1$ (121)

and

$U(\Lambda)J_{-}(z,\bar{z})U(\Lambda)^{-1}=\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2}}\
J_{-}\bigg{(}\frac{az+b}{cz+d},\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)},\
\sigma=-1$ (122)

So $J_{\pm}$ transform like $(1,0)$ and $(0,1)$ (quasi-) primary operators of
a 2-D CFT.

Let us now write down Weinberg’s soft-photon theorem in terms of $J_{\pm}$.

Let us consider a positive helicity ($\sigma=+1$) outgoing soft photon. In
this case Weinberg’s soft-photon theorem reduces to He:2014cra ;
Strominger:2017zoo ,

$\begin{gathered}\lim_{\omega\rightarrow 0+}\omega S_{N+1\leftarrow
M}\big{(}\\{p,\sigma=+1\ ;\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\\\
=\frac{1}{\sqrt{2}}\bigg{[}\sum_{i=M+1}^{M+N}\frac{Q_{i}}{z-z_{i}}-\sum_{j=1}^{M}\frac{Q_{j}}{z-z_{j}}\bigg{]}S_{N\leftarrow
M}\big{(}\\{\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (123)

where $Q_{p}$ is the charge of the $p$-th particle. Substituting this in
Eq-143 we get the following relation,

$\begin{gathered}\lim_{\omega\rightarrow 0+}\sqrt{2\pi^{2}}\
\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}\bra{\Omega}A^{out}_{\lambda,\sigma=+1}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{\sqrt{2}}\bigg{[}\sum_{i=M+1}^{M+N}\frac{Q_{i}}{z-z_{i}}-\sum_{j=1}^{M}\frac{Q_{j}}{z-z_{j}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}$
(124)

In terms of the soft-photon operator $J_{+}(z,\bar{z})$ Weinberg’s soft-photon
theorem takes the following form,

$\boxed{\begin{gathered}\bra{\Omega}J_{+}(z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{\sqrt{2}}\bigg{[}\sum_{i=M+1}^{M+N}\frac{Q_{i}}{z-z_{i}}-\sum_{j=1}^{M}\frac{Q_{j}}{z-z_{j}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}}$
(125)

The leading soft photon theorem appears in this form in Strominger:2013lka
written in a different basis of states. This has the form of a $U(1)$ Kac-
Moody ward identity in a 2-D CFT Strominger:2013lka ; He:2014cra ;
Cheung:2016iub  except that now the fields depend on an extra coordinate given
by $U$. But the $U$ direction does not play any role in this case.

###  VI.2 Soft Graviton Limit

Let us now consider a positive helicity ($\sigma=+2$) outgoing soft-graviton.
In this case Weinberg’s soft graviton theorem can be cast in the following
form He ,

$\begin{gathered}\lim_{\omega\rightarrow 0+}\omega S_{N+1\leftarrow
M}\big{(}\\{p,\sigma=+2\ ;\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\\\
=\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{\omega_{j}(\bar{z}-\bar{z}_{j})}{z-z_{j}}-\sum_{i=M+1}^{M+N}\frac{\omega_{i}(\bar{z}-\bar{z}_{i})}{z-z_{i}}\bigg{]}S_{N\leftarrow
M}\big{(}\\{\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (126)

where $\kappa=\sqrt{32\pi G_{N}}$. We have also parametrized an arbitrary null
momentum $p_{\alpha}$ as,

$p_{\alpha}=\omega_{\alpha}\big{(}1+z_{\alpha}\bar{z}_{\alpha},z_{\alpha}+\bar{z}_{\alpha},-i(z_{\alpha}-\bar{z}_{\alpha}),1-z_{\alpha}\bar{z}_{\alpha}\big{)}$
(127)

Now substituting this in Eq-143 we get

$\begin{gathered}\lim_{\omega\rightarrow
0+}\sqrt{2\pi^{2}}\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}\bra{\Omega}A^{out}_{\lambda,\sigma=+2}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{\omega_{j}(\bar{z}-\bar{z}_{j})}{z-z_{j}}-\sum_{i=M+1}^{M+N}\frac{\omega_{i}(\bar{z}-\bar{z}_{i})}{z-z_{i}}\bigg{]}S_{N\leftarrow
M}\big{(}\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (128)

Now in the same way we define the positive helicity soft-graviton operator
$O_{+}(U,z,\bar{z})$ by

$O_{+}(z,\bar{z})=\lim_{\omega\rightarrow 0}\omega
a^{out}(p,\sigma=+2)=\lim_{\omega\rightarrow 0+}\sqrt{2\pi^{2}}\
\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}A^{out}_{\lambda,\sigma=+2}(U,z,\bar{z})$ (129)

Using this operator we can write Eq-128 as,

$\begin{gathered}\bra{\Omega}O_{+}(z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
\bigg{\\{}\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{\omega_{j}(\bar{z}-\bar{z}_{j})}{z-z_{j}}-\sum_{i=M+1}^{M+N}\frac{\omega_{i}(\bar{z}-\bar{z}_{i})}{z-z_{i}}\bigg{]}S_{N\leftarrow
M}\big{(}\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\bigg{\\}}\end{gathered}$ (130)

Now applying $\bar{\partial}_{z}$ to both sides we get,

$\begin{gathered}\bra{\Omega}\bar{\partial}_{z}O_{+}(z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
\bigg{\\{}\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{\omega_{j}}{z-z_{j}}-\sum_{i=M+1}^{M+N}\frac{\omega_{i}}{z-z_{i}}\bigg{]}S_{N\leftarrow
M}\big{(}\\{p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\bigg{\\}}\\\ \\\
=\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{-i\partial/\partial
U_{j}}{z-z_{j}}+\sum_{i=M+1}^{M+N}\frac{-i\partial/\partial
U_{i}}{z-z_{i}}\bigg{]}\\\
\bigg{[}\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
S_{N\leftarrow M}\big{(}\ p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\bigg{]}\\\ \\\
=\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{-i\partial/\partial
U_{j}}{z-z_{j}}+\sum_{i=M+1}^{M+N}\frac{-i\partial/\partial
U_{i}}{z-z_{i}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}$
(131)

Therefore Weinberg’s soft-graviton theorem takes the following form,

$\begin{gathered}\bra{\Omega}\bar{\partial}O_{+}(z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{\kappa}{2}\bigg{[}\sum_{j=1}^{M}\frac{-i\partial/\partial
U_{j}}{z-z_{j}}+\sum_{i=M+1}^{M+N}\frac{-i\partial/\partial
U_{i}}{z-z_{i}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}$
(132)

Now using the relation

$\bar{\partial}_{z}\frac{1}{z-w}=2\pi\delta^{2}(z-w)$ (133)

it is easy to check that the insertion of the operator $T(g)$, given by

$T(g)=\frac{i}{\pi\kappa}\int
d^{2}zg(z,\bar{z})\bar{\partial}^{2}O_{+}(z,\bar{z})$ (134)

leads to

$\boxed{\begin{gathered}\bra{\Omega}T(g)\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\bigg{[}\sum_{j=1}^{M}g(z_{j},\bar{z}_{j})\frac{\partial}{\partial
U_{j}}+\sum_{i=M+1}^{M+N}g(z_{i},\bar{z}_{i})\frac{\partial}{\partial
U_{i}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}}$
(135)

This relation means that the insertion of the operator $T(g)$ generates a
point transformation in the $(U,z,\bar{z})$ space, given by,

$\big{(}U,z,\bar{z}\big{)}\rightarrow\big{(}U+g(z,\bar{z}),z,\bar{z}\big{)}$
(136)

where $g(z,\bar{z})$ is an arbitrary function on the plane. If we think of the
$(U,z,\bar{z})$ space as the null-infinity in Minkowski space then these
transformations can be identified with BMS supertranslations
Strominger:2013jfa ; Bondi:1962px ; Barnich:2009se .

Now as Eq-135 shows the $\tilde{A}$-amplitudes are not generically invariant
under this transformation Strominger:2013jfa . But there are four
transformations, corresponding to four global space-time translations, under
which $\tilde{A}$ remains unchanged. To see this let us first note that in
terms of the variable, $U=u(1+z\bar{z})$, global space-time translation by a
4-vector $l^{\mu}$ is given by,

$U\rightarrow
U+(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z},\
z\rightarrow z,\ \bar{z}\rightarrow\bar{z}$ (137)

Now let $g(z,\bar{z},l)$ denote the function $g(z,\bar{z})$ corresponding to
the global space-time translation by 4-vector $l^{\mu}$. Then,

$g(z,\bar{z},l)=(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}$
(138)

It is also easy to see that,

${\bar{\partial}}^{2}g(z,\bar{z},l)={\partial}^{2}g(z,\bar{z},l)=0$ (139)

Using $g(z,\bar{z},l)$ we can write the identity given in Eq-135 as,

$\begin{gathered}\frac{i}{\pi\kappa}\int d^{2}z\
g(z,\bar{z},l)\bar{\partial}^{2}\bra{\Omega}O_{+}(z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\bigg{[}\sum_{j=1}^{M}g(z_{j},\bar{z}_{j},l)\frac{\partial}{\partial
U_{j}}+\sum_{i=M+1}^{M+N}g(z_{i},\bar{z}_{i},l)\frac{\partial}{\partial
U_{i}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}$
(140)

Now assuming that we are allowed to integrate by parts we get,

$\begin{gathered}\frac{i}{\pi\kappa}\int
d^{2}z\bar{\partial}^{2}g(z,\bar{z},l)\bra{\Omega}O_{+}(z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\bigg{[}\sum_{j=1}^{M}g(z_{j},\bar{z}_{j},l)\frac{\partial}{\partial
U_{j}}+\sum_{i=M+1}^{M+N}g(z_{i},\bar{z}_{i},l)\frac{\partial}{\partial
U_{i}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\end{gathered}$
(141)

Now using $\bar{\partial}^{2}g(z,\bar{z},l)=0$ we get,

$\boxed{\bigg{[}\sum_{j=1}^{M}g(z_{j},\bar{z}_{j},l)\frac{\partial}{\partial
U_{j}}+\sum_{i=M+1}^{M+N}g(z_{i},\bar{z}_{i},l)\frac{\partial}{\partial
U_{i}}\bigg{]}\bra{\Omega}\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}=0}$
(142)

This is essentially the statement of the invariance of the $\tilde{A}$
amplitude (or the S-matrix in the Fock basis) under global space-time
translations.

##  VII A slightly different treatment of the soft-operator

Let us again consider the equation,

$\begin{gathered}\lim_{\omega\rightarrow
0+}\sqrt{2\pi^{2}}\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}\bra{\Omega}A^{out}_{\lambda,\sigma}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}\\\
=\frac{1}{(\sqrt{8\pi^{4}})^{M+N}}\bigg{(}\prod_{i=M+1}^{M+N}\int_{0}^{\infty}d\omega_{i}\omega_{i}^{-i\lambda_{i}}e^{-i\omega_{i}U_{i}}\bigg{)}\bigg{(}\prod_{j=1}^{M}\int_{0}^{\infty}d\omega_{j}\omega_{j}^{i\lambda_{j}}e^{i\omega_{j}U_{j}}\bigg{)}\\\
\lim_{\omega\rightarrow 0+}\omega S_{N+1\leftarrow M}\big{(}\\{p,\sigma\ ;\
p_{i},\sigma_{i},n_{i}\\},out\ \big{|}\
\\{p_{j},\sigma_{j},n_{j}\\},in\big{)}\end{gathered}$ (143)

and study the following operator.

$J_{\sigma,\omega}(U,z,\bar{z})=\sqrt{2\pi^{2}}\
\int_{-\infty}^{\infty}d\lambda\ \omega^{i\lambda}e^{i\omega
U}A^{out}_{\lambda,\sigma}(U,z,\bar{z})$ (144)

This is the same as $\omega a^{out}(p,\sigma)$ but we do not want to use this
fact in this section. This is the reason why we have kept the $U$ dependence
in $J_{\sigma,\omega}$. The only thing that we want to use is the Lorentz
transformation property of the objects involved in Eq-143. We will show, using
the Lorentz transformation property of $A_{\lambda,\sigma}$, that in the soft
limit ($\omega\rightarrow 0+$) the soft operator defined as
$J_{\sigma}(U,z,\bar{z})=J_{\sigma,\omega=0}(U,z,\bar{z})$ transforms as a
tensor in the $(U,z,\bar{z})$ space. Then we invoke the soft theorem to show
that the tensor $J_{\sigma}(U,z,\bar{z})$ does not depend on $U$, i.e,
$\partial_{U}J_{\sigma}(U,z,\bar{z})=0$. This is an interesting phenomenon
because no such thing can happen in a Lorentz invariant theory defined on
Minkowski space. The condition that a non-trivial tensor function on Minkowski
space be independent of time is not Lorentz invariant. But the equation
$\partial_{U}J_{\sigma}(U,z,\bar{z})=0$ is in fact Lorentz invariant in
$(U,z,\bar{z})$ space. So the following demonstration can be thought of as an
overall consistency check of the procedure.

Using the transformation property of $A_{\lambda,\sigma}$ we get,

$U(\Lambda)J_{\sigma,\omega}(U,z,\bar{z})U(\Lambda)^{-1}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}J_{\sigma,\omega|cz+d|^{2}}\bigg{(}\frac{U}{|cz+d|^{2}},\frac{az+b}{cz+d},\
\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (145)

where $U(\Lambda)$ is a $SL(2,\mathbb{C})$ transformation and

$h=\frac{1+\sigma}{2},\ \bar{h}=\frac{1-\sigma}{2}$ (146)

Now let us take the soft limit $\omega\rightarrow 0+$ and define the soft
operator $J_{\sigma}(U,z,\bar{z})$ to be $J_{\sigma,\omega=0}(U,z,\bar{z})$.
Under $SL(2,\mathbb{C})$ transformation,

$U(\Lambda)J_{\sigma}(U,z,\bar{z})U(\Lambda)^{-1}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}J_{\sigma}\bigg{(}\frac{U}{|cz+d|^{2}},\frac{az+b}{cz+d},\
\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (147)

We can also study the transformation of $J_{\sigma,\omega}(U,z,\bar{z})$ under
global translation by a four vector $l^{\mu}$. The transformation property is
given by,

$e^{-il.P}J_{\sigma,\omega}(U,z,\bar{z})e^{il.P}=e^{-i\omega
g(z,\bar{z},l)}J_{\sigma,\omega}(U+g(z,\bar{z},l),z,\bar{z})$ (148)

where

$g(z,\bar{z},l)=(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}$
(149)

So in the soft limit $\omega\rightarrow 0+$ we get,

$e^{-il.P}J_{\sigma}(U,z,\bar{z})e^{il.P}=J_{\sigma}(U+g(z,\bar{z},l),z,\bar{z})$
(150)

So we can see that the soft-operator $J_{\sigma}(U,z,\bar{z})$ transforms like
an $ISL(2,\mathbb{C})$ primary as defined in Banerjee:2018gce .

Now soft theorem tells us that the correlator

$\bra{\Omega}J_{\sigma}(U,z,\bar{z})\prod_{i=M+1}^{M+N}A^{out}_{n_{i},\lambda_{i},\sigma_{i}}(U_{i},z_{i},\bar{z}_{i})\prod_{j=1}^{M}A^{\dagger
in}_{n_{j},\lambda_{j},\sigma_{j}}(U_{j},z_{j},\bar{z}_{j})\ket{\Omega}$ (151)

does not depend on $U$. So we can write
$\partial_{U}J_{\sigma}(U,z,\bar{z})=0$ as an operator equation because its
matrix elements between arbitrary in and out states is zero. Let us now show
that the condition $\partial_{U}J_{\sigma}(U,z,\bar{z})=0$ is in fact Lorentz
invariant.

We can prove this more generally for any tensor
$\phi_{h,\bar{h}}(U,z,\bar{z})$ which satisfies
$\partial_{U}\phi_{h,\bar{h}}(U,z,\bar{z})=0$.

Under $SL(2,\mathbb{C})$ transformation,

$\phi^{\prime}_{h,\bar{h}}(U^{\prime},z^{\prime},\bar{z}^{\prime})\bigg{(}\frac{dz^{\prime}}{dz}\bigg{)}^{h}\bigg{(}\frac{d\bar{z}^{\prime}}{d\bar{z}}\bigg{)}^{\bar{h}}=\phi_{h,\bar{h}}(U,z,\bar{z})$
(152)

where

$(U^{\prime},z^{\prime},\bar{z}^{\prime})=\bigg{(}\frac{U}{|cz+d|^{2}},\frac{az+b}{cz+d},\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$
(153)

Therefore,

$\phi^{\prime}_{h,\bar{h}}(U^{\prime},z^{\prime},\bar{z}^{\prime})=(cz+d)^{2h}(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}\phi_{h,\bar{h}}(U,z,\bar{z})$
(154)

and

$\frac{\partial}{\partial U^{\prime}}=\frac{\partial U}{\partial
U^{\prime}}\frac{\partial}{\partial U}+\frac{\partial z}{\partial
U^{\prime}}\frac{\partial}{\partial z}+\frac{\partial\bar{z}}{\partial
U^{\prime}}\frac{\partial}{\partial\bar{z}}=|cz+d|^{2}\frac{\partial}{\partial
U}$ (155)

Using this we get,

$\partial_{U^{\prime}}\phi^{\prime}_{h,\bar{h}}(U^{\prime},z^{\prime},\bar{z}^{\prime})=|cz+d|^{2}(cz+d)^{2h}(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}\
\partial_{U}\phi_{h,\bar{h}}(U,z,\bar{z})$ (156)

So $\partial_{U}\phi_{h,\bar{h}}$ transforms homogeneously under Lorentz
transformation. So in the $(U,z,\bar{z})$ space, if it is zero in one frame
then it is zero in any other frame. So
$J_{\sigma}(U,z,\bar{z})=J_{\sigma}(z,\bar{z})$.

This shows that in the $(U,z,\bar{z})$ space we have well-defined Lorentz
tensors of the form $\phi_{h,\bar{h}}(z,\bar{z})$.

##  VIII Discussion

In this paper we have tried to connect the lading soft theorems to symmetries
in the approach suggested by Pasterski:2016qvg ; Pasterski:2017kqt ;
Pasterski:2017ylz ; Banerjee:2018gce . This does This seems to raise more
questions than it answers. Let us list some of the pressing questions :

1) We have not dealt with the subleading soft theorems. There are fascinating
new results Cachazo:2014fwa ; Schwab:2014xua ; Bern:2014oka ; Broedel:2014fsa
; Sen:2017xjn ; Sen:2017nim ; Laddha:2017ygw ; Chakrabarti:2017ltl ;
Chakrabarti:2017zmh ; Laddha:2017vfh ; Laddha:2018rle ; Campoleoni:2017mbt  in
this subject. In particular it will be interesting to understand subleading
soft graviton theorem and its relation to superrotation Barnich:2009se ;
Banks:2003vp ; deBoer:2003vf ; Cheung:2016iub ; Kapec:2016jld ; Kapec:2014opa
; Kapec:2017gsg ; Pasterski:2015tva ; Campiglia:2014yka  from this point of
view.

2) The results of Pasterski:2016qvg ; Pasterski:2017kqt ; Pasterski:2017ylz
are also applicable to the massive particles. It will be interesting to extend
the results of Banerjee:2018gce  to the massive case SB .

3) We have worked with creation and annihilation free fields on the
$(u,z,\bar{z})$ space. In quantum field theory on Minkowski space we usually
work with quantum fields built out of both creation and annihilation
operators. This is required by causality and locality in space-time.

Can one define a reasonable notion of causality and locality in the
$(u,z,\bar{z})$ space ? This should be possible because the Poincare group
acts on this space. Using this, can one define local quantum (free) fields on
this space ? To what extent the interacting theory make sense ? If it makes
sense then its (global) symmetry structure should be closely related to the
(global) asymptotic symmetries of quantum gravity in flat space.

4) It will be interesting to better understand the unitary representation of
the four dimensional BMS group including superrotations Barnich:2014kra ;
Campoleoni:2016vsh ; Bagchi:2016geg .

5) Is there any possible holographic interpretation Pasterski:2016qvg ;
Cheung:2016iub ; deBoer:2003vf ; Kapec:2017gsg ; Bagchi:2016bcd  ?

##  IX Acknowledgement

I would like to thank especially Ashoke Sen for numerous enlightening
discussions on soft theorems and related subjects. I would also like to thank
Alok Laddha and Prahar Mitra for very helpful discussion on soft theorems and
related matters. Part of this work was presented in the ”First Spring Meeting
on Strings” held in NISER Bhubaneswar, India. I would like to thank the
organizers for holding this exciting meeting. I would also like to thank all
the participants of the meeting especially Jyotirmoy Bhattacharya, Sayantani
Bhattacharya, Bobby Ezhuthachan, Arnab Kundu, Jnan Maharana, Sudhakar Panda,
Ashoke Sen and Yogesh Srivastava for their valuable feedback.

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