aainstitutetext: National Institute of Science Education and Research (NISER),
Bhubaneswar 752050, Odisha, Indiabbinstitutetext: Homi Bhabha National
Institute, Anushakti Nagar, Mumbai, India-400085ccinstitutetext: Department of
Physical Sciences,
IISER Kolkata, Mohanpur, West Bengal 741246, Indiaddinstitutetext: Department
of Theoretical Physics,
Tata Institute of Fundamental Research, Homi Bhabha Road, Mumbai 400005,
Indiaeeinstitutetext: Centre for High Energy Physics, Indian Institute of
Science, C.V. Raman Avenue, Bangalore 560012, India

# Celestial OPE in Self Dual Gravity

Shamik Banerjee${}^{\,a,b}$, Harshal Kulkarni${}^{\,c,d}$, Partha
Paul${}^{\,e}$<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

In this paper we compute the celestial operator product expansion between two
outgoing positive helicity gravitons in the self dual gravity. It has been
shown that the self dual gravity is a $w_{1+\infty}$-invariant theory whose
scattering amplitudes are one loop exact with all positive helicity gravitons.
Celestial $w_{1+\infty}$ symmetry is generated by an infinite tower of
(conformally soft) gravitons which are holomorphic conserved currents. We find
that at any given order only a finite number of $w_{1+\infty}$ descendants
contribute to the OPE. This is somewhat surprising because the spectrum of
conformal dimensions in celestial CFT is not bounded from below. However, this
is consistent with our earlier analysis based on the representation theory of
$w_{1+\infty}$. The phenomenon of truncation suggests that in some (unknown)
formulation the spectrum of conformal dimensions in the dual two dimensional
theory can be bounded from below.

##  1 Introduction

Celestial holography is a conjectured duality between quantum gravity in 4D
asymptotically flat spacetime and a quantum field theory on the 2D celestial
sphere Strominger:2017zoo ; Pasterski:2016qvg ; Pasterski:2017kqt . Symmetries
play an important role in this conjectured duality. The Lorentz group in 4D
acts on the 2D celestial sphere as the global conformal group. So the dual
theory should be a conformal field theory. Motivated by this, a new basis was
introduced Pasterski:2016qvg ; Pasterski:2017kqt ; Banerjee:2018gce  in which
the $S$-matrix elements transform like 2D conformal correlators. Besides the
two dimensional global conformal symmetry, CCFT has various infinite
dimensional current algebra symmetries Sachs:1962zza ; Strominger:2013jfa ; He
; Strominger:2014pwa ; Barnich:2009se ; Kapec:2016jld ; Kapec:2014opa ;
He:2017fsb ; Donnay:2020guq ; Stieberger:2018onx ; Banerjee:2022wht ;
Banerjee:2020zlg ; Banerjee:2021cly ; Gupta:2021cwo ; Guevara:2021abz ;
Strominger:2021lvk ; Himwich:2021dau ; Ball:2021tmb ; Adamo:2021lrv ;
Costello:2022wso ; Stieberger:2022zyk ; Stieberger:2023fju ; Ball:2023qim ;
Melton:2022fsf ; Banerjee:2020kaa ; Costello:2022upu ; Costello:2023vyy ;
Garner:2023izn ; Donnay:2021wrk ; Donnay:2022hkf .

Operator product expansion (OPE) is a central tool to study various aspects of
any CFT. In the context of celestial CFT also, OPE played an important role in
identifying new symmetries Guevara:2021abz ; Strominger:2021lvk ;
Banerjee:2020zlg , null states Banerjee:2020zlg ; Banerjee:2021cly ;
Banerjee:2019aoy ; Banerjee:2019tam ; Pano:2023slc ; Banerjee:2023zip ;
Banerjee:2020vnt ; Banerjee:2023rni ; Adamo:2022wjo ; Ren:2023trv ; Hu:2021lrx
; Ebert:2020nqf ; Pasterski:2021fjn  etc. It has also found applications in
the Bootstrap program Atanasov:2021cje ; Ghosh:2022net . Based on the
universal singular structure of the tree-level OPE between two positive
helicity gravitons, it was shown in Guevara:2021abz  that CCFT has an infinite
tower of soft symmetries which close into $w_{1+\infty}$ algebra
Strominger:2021lvk . Loop corrections to the tree level celestial OPEs have
been studied in Bhardwaj:2022anh ; Krishna:2023ukw .

In a previous paper Banerjee:2023zip , we have studied the implications of the
$w_{1+\infty}$ symmetry at the level of OPEs by using representation theory.
By studying the subleading terms in the OPE between two positive helicity
outgoing gravitons, we have shown that there should exist an infinite number
of theories which are invariant under $w_{1+\infty}$ algebra.

In this paper we derive the OPE in one such theory, known as the quantum self
dual gravity Ooguri:1991fp ; Chalmers:1996rq ; Bern:1998xc ; Bern:1998sv ;
Krasnov:2021sf  which was shown to be $w_{1+\infty}$ invariant in
Penrose:1968pr ; Penrose:1976pr ; Boyer:1985pn . Here we do a collinear
expansion of the known graviton scattering amplitudes in the self dual gravity
theory and extract the celestial OPE from there. For simplicity, we analyse
the $5$-point all plus amplitude in self dual gravity and factorize it in the
collinear limit through a $4$ point amplitude. The results we obtain are
consistent with what we proposed in Banerjee:2023zip  based on the
representtion theory of $w_{1+\infty}$. The rest of the paper is organised as
follows.

In section 2 we introduce notations and conventions used in this paper.
Section 3 briefly describes the $w_{1+\infty}$ algebra and how the whole tower
of the $w_{1+\infty}$-currents can be generated using the two $sl_{2}(R)$ sub-
algebras. In section 4 we briefly discuss about the scattering amplitudes in
quantum self dual gravity. Section 5 discusses how to extract the OPE between
two positive helicity outgoing gravitons from the 5-point one-loop self dual
amplitude. We start by simplifying the 5-point amplitude in the momentum space
and then Mellin transform it to get the celestial amplitudes. We then discuss
how to factorize each term order by order in the OPE limit of the celestial
amplitudes. The null states of the self dual gravity appearing at various
orders of the OPE and the invariance of the OPE under $w_{1+\infty}$ algebra
are discussed in Appendices I and J.

For the sake of completeness of the paper, we give a brief review of the
celestial amplitude in Appendix A. In Appendix B, we discuss the
parametrization of the $4$ and $5$-point delta functions which are useful in
our context of the OPE expansion. Appendices C and D discuss how to simplify
the $4$ and $5$-point amplitudes in momentum space using momentum conserving
delta functions and various identities of the spinor-helicity brackets. These
simplifications are done keeping in mind the fact that we want to factorize
the $5$-point amplitude in terms of the $4$-point amplitude in the OPE
expansion. Appendix E deals with the Mellin transformation of the $5$-point
amplitude. In Appendix F we discuss the conditions on the graviton primary
operators under the $w_{1+\infty}$ algebra which follow from the universal
structure of the OPE. In Appendix G, we list the transformation properties of
all the MHV null states under the action of ${sl_{2}(R)}_{V}$ and
$\overline{sl_{2}(R)}$ algebras which are required to show the
$w_{1+\infty}$-invariance of the self dual OPE. Appendix H briefly reviews the
construction of a general $w_{1+\infty}$-algebra invariant OPE and how one can
obtain an infinite family of $w_{1+\infty}$-algebra invariant theories.

##  2 Notations and Conventions

In this paper, we will work in the $(2,2)$ signature space-time, which is also
known as Klein space. The null momentum $p^{\mu}$ of a massless particle,
satisfying the onshell condition $p^{2}=0$, is parametrized as,

$\begin{split}p^{\mu}&=\epsilon\,q^{\mu}\\\
q^{\mu}&=\omega\\{1+z\bar{z},z+\bar{z},z-\bar{z},1-z\bar{z}\\}\end{split}$ (1)

where $\epsilon=\pm 1$ for outgoing and incoming particles respectively,
$(z,\bar{z})$ are two independent real variables and $\omega$ is any positive
number interpreted as the energy of the particle. In Klein space the null
infinity takes the form of a Lorentzian torus (known as the celestial torus)
times a null line. The Lorentz group in $(2,2)$ signature is given by
$SO(2,2)\simeq\frac{SL(2,\mathbb{R})_{L}\times
SL(2,\mathbb{R})_{R}}{\mathbb{Z}_{2}}$ and acts as the group of conformal
transformations on the celestial torus:

$\begin{gathered}SL(2,\mathbb{R})_{L}:\qquad z\to\frac{az+b}{cz+d},\
\bar{z}\to\bar{z},\ ad-bc=1,\\\
SL(2,\mathbb{R})_{R}:\qquad\bar{z}\to\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}},\
z\to z,\ \bar{a}\bar{d}-\bar{b}\bar{c}=1.\end{gathered}$ (2)

In our conventions the spinor-helicity variables are given by,

$\left<ij\right>=2\epsilon_{i}\epsilon_{j}\sqrt{\omega_{i}\omega_{j}}z_{ij},\
[ij]=2\sqrt{\omega_{i}\omega_{j}}\bar{z}_{ij}$ (3)

where $z_{ij}=z_{i}-z_{j}$ and we also have $2p_{i}\cdot
p_{j}=-\left<ij\right>[ij]$.

##  3 Review of $w_{1+\infty}$ Algebra

We start by reviewing the $w_{1+\infty}$ algebra which follows from the
universal singular terms in the OPE between two positive helicity outgoing
gravitons. Let $G^{+}_{\Delta}(z,\bar{z})$ denote the positive helicity
graviton conformal primary operator of dimension $\Delta$ at the point
$(z,\bar{z})$ on the celestial torus. The universal singular terms in the OPE
are given by

$\begin{gathered}G^{+}_{\Delta_{1}}(z_{1},\bar{z}_{1})G^{+}_{\Delta_{2}}(z_{2},\bar{z}_{2})=-\frac{\bar{z}_{12}}{z_{12}}\sum_{n=0}^{\infty}B\left(\Delta_{1}-1+n,\Delta_{2}-1\right)\frac{{\bar{z}_{12}}^{n}}{n!}\bar{\partial}^{n}G^{+}_{\Delta_{1}+\Delta_{2}}(z_{2},\bar{z}_{2})\end{gathered}$
(4)

Let us define an infinite family of positive helicity conformally soft
Donnay:2018neh ; Pate:2019mfs ; Fan:2019emx ; Nandan:2019jas ; Adamo:2019ipt ;
Puhm:2019zbl ; Guevara:2019ypd  gravitons Guevara:2021abz  as,

$H^{k}(z,\bar{z})=\lim_{\Delta\rightarrow
k}(\Delta-k)G^{+}_{\Delta}(z,\bar{z}),\ k=1,0,-1,-2,...$ (5)

with weights $\left(\frac{k+2}{2},\frac{k-2}{2}\right)$. It follows from the
OPE (4) that we can introduce the following truncated mode expansion

$H^{k}(z,\bar{z})=\sum_{m=\frac{k-2}{2}}^{\frac{2-k}{2}}\frac{H^{k}_{m}(z)}{\bar{z}^{m+\frac{k-2}{2}}}$
(6)

and the modes $H^{k}_{m}(z)$ are the conserved holomorphic currents. The
currents $H^{k}_{m}(z)$ can be further mode expanded in the $z$-variable to
get,

$H^{k}_{m}(z)=\sum_{\alpha\in\mathbb{Z}-\frac{k+2}{2}}\frac{H^{k}_{\alpha,m}}{z^{\alpha+\frac{k+2}{2}}}$
(7)

and one can show Guevara:2021abz  that the modes $H^{k}_{\alpha,m}$ satisfy
the algebra111Here we are assuming that $\kappa=\sqrt{32\pi G_{N}}=2$.

$\begin{gathered}\left[H^{k}_{\alpha,m},H^{l}_{\beta,n}\right]\\\
=-\left[n(2-k)-m(2-l)\right]\frac{\left(\frac{2-k}{2}-m+\frac{2-l}{2}-n-1\right)!}{\left(\frac{2-k}{2}-m\right)!\left(\frac{2-l}{2}-n\right)!}\frac{\left(\frac{2-k}{2}+m+\frac{2-l}{2}+n-1\right)!}{\left(\frac{2-k}{2}+m\right)!\left(\frac{2-l}{2}+n\right)!}H^{k+l}_{\alpha+\beta,m+n}\end{gathered}$
(8)

This is called the Holographic Symmetry Algebra (HSA). Now if we make the
following redefinition (or discrete light transformation)Strominger:2021lvk

$w^{p}_{\alpha,m}=\frac{1}{2}\left(p-m-1\right)!\left(p+m-1\right)!H^{-2p+4}_{\alpha,m}$
(9)

then (8) turns into the $w_{1+\infty}$ algebra222This is the wedge subalgebra
of $w_{1+\infty}$.

$\left[w^{p}_{\alpha,m},w^{q}_{\beta,n}\right]=\left[m(q-1)-n(p-1)\right]w^{p+q-2}_{\alpha+\beta,m+n}$
(10)

For our purpose it is more convenient to work with the HSA (8) rather than the
$w_{1+\infty}$ algebra. However, we continue to refer to the HSA as the $w$
algebra.

Now, in Banerjee:2023zip , it was shown that the whole tower of the
$w$-currents can be generated using the two $sl_{2}(R)$ sub-algebras. One of
them is $sl_{2}(R)_{V}$333Here V stands for vertical. Please see Fig.1 for an
explanation.  generated by the operators
$\\{H^{1}_{-\frac{1}{2},-\frac{1}{2}},\ H^{0}_{0,0},\
H^{-1}_{\frac{1}{2},\frac{1}{2}}\\}$,

$\begin{gathered}\left[H^{0}_{0,0},H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right]=H^{1}_{-\frac{1}{2},-\frac{1}{2}}\\\
\left[H^{0}_{0,0},H^{-1}_{\frac{1}{2},\frac{1}{2}}\right]=-H^{-1}_{\frac{1}{2},\frac{1}{2}}\\\
\left[H^{1}_{-\frac{1}{2},-\frac{1}{2}},H^{-1}_{\frac{1}{2},\frac{1}{2}}\right]=-H^{0}_{0,0}\end{gathered}$
(11)

The other $sl_{2}(R)$ sub-algebra is generated by the global (Lorentz)
conformal transformations $\\{H^{0}_{0,1},H^{0}_{0,0},H^{0}_{0,-1}\\}$. We
call this $\overline{sl_{2}(R)}$ because this acts only on the $\bar{z}$
coordinate. Now the $w$ symmetry is generated by the infinite number of soft
currents $\\{H^{k}_{p}(z)\\}$ where $k=1,0,-1,-2,...$ is the dimension
$(\Delta)$ of the soft operator and $\frac{k-2}{2}\leq p\leq-\frac{k-2}{2}$.
For a fixed $k$, the soft currents
$\\{H^{k}_{-\frac{k-2}{2}}(z),...,H^{k}_{\frac{k-2}{2}}(z)\\}$ transform in a
spin $\frac{2-k}{2}$ representation of the $\overline{sl_{2}(R)}$.

Figure 1: The figure shows the soft currents. The rows and the columns are
indexed by the $\overline{sl_{2}(R)}$ weights and the dimension
$(\Delta=k=1,0,-1,-2,...)$ of the conformally soft graviton
$H^{k}(z,\bar{z})$, which generates the currents sitting in a row,
respectively. $\overline{sl_{2}(R)}$ acts horizontally along a row and
$sl_{2}(R)_{V}$ acts vertically along a column. In this way they generate the
whole symmetry algebra starting from the current $H^{1}_{\frac{1}{2}}(z)$ on
the top left corner.

Now let us consider the currents
$\\{H^{1}_{\frac{1}{2}},H^{0}_{1},...,H^{k}_{\frac{2-k}{2}},...\\}$ with the
lowest $\overline{sl_{2}(R)}$ weights. These currents transform in an
irreducible highest weight representation of the $sl_{2}(R)_{V}$. This can be
seen from the following commutation relations following from (8),

$\begin{gathered}\left[H^{-1}_{\frac{1}{2},\frac{1}{2}},H^{k}_{\alpha,\frac{2-k}{2}}\right]=-\frac{1}{2}(k-2)(k-3)H^{k-1}_{\alpha+\frac{1}{2},\frac{2-(k-1)}{2}}\\\
\left[H^{0}_{0,0},H^{k}_{\alpha,\frac{2-k}{2}}\right]=(k-2)H^{k}_{\alpha,\frac{2-k}{2}}\\\
\left[H^{1}_{-\frac{1}{2},-\frac{1}{2}},H^{k}_{\alpha,\frac{2-k}{2}}\right]=-H^{k+1}_{\alpha-\frac{1}{2},\frac{2-(k+1)}{2}}\end{gathered}$
(12)

Therefore, starting from the current $H^{1}_{\frac{1}{2}}(z)$ we can generate
any other $w$ current by the combined action of the $\overline{sl_{2}(R)}$ and
$sl_{2}(R)_{V}$ (Fig.1).

##  4 Scattering Amplitudes in Quantum Self Dual Gravity

In this section, following Bern:1998xc  we briefly review the all-plus
helicity scattering amplitudes in quantum self dual gravity. In $(2,2)$
signature, self duality translates into the following condition on the Riemann
tensor,

$R_{\mu\nu\rho\sigma}=\frac{1}{2}{\varepsilon_{\mu\nu}}^{\alpha\beta}R_{\alpha\beta\rho\sigma}$
(13)

where $\varepsilon^{\mu\nu\alpha\beta}$ is the completely antisymmetric tensor
with $\varepsilon^{0123}=+1$. In order to maintain the reality condition on
the fields, the self dual gravity is described in either $(2,2)$ or $(0,4)$
signature. In Lorentzian $(1,3)$ signature, the condition (13) acquires an
extra factor of $i$ and contain no real solutions. At the classical level, the
linearized self dual solutions consist of positive helicity plane waves.

In this paper, we are interested in the collinear behaviour of gravitons in
the self dual gravity. At the tree level, we have only one non-trivial
amplitude: the three point $\overline{MHV}$ amplitude where, only one external
graviton has negative helicity and other two have positive helicity. The
appearance of the negative helicity graviton in the three point
$\overline{MHV}$ amplitude can be explained from the fact that the action
contains a Lagrange multiplier which is physically interpreted as the negative
helicity graviton.

At the one loop level, the only non-zero amplitudes are the ones with all plus
helicity gravitons with the minimum number of gravitons being four. These
amplitudes are both UV and IR finite. The only divergences of these amplitudes
are collinear and soft divergences. Our interest in this theory stems from the
fact that this is a non-trivial quantum theory which is known to be $w$
invariant.

The one loop all-plus $n$-graviton stripped amplitude in self dual gravity is
given by Bern:1998xc ,

$\displaystyle
A_{n}(1^{+},2^{+},...,n^{+})=-\frac{i}{(4\pi)^{2}960}\left(-\frac{\kappa}{2}\right)^{n}\sum_{\begin{subarray}{c}1\leq
a<b\leq n\\\ M,N\end{subarray}}h(a,M,b)h(b,N,a)tr^{3}[aMbN]$ (14)

where $a$ and $b$ are the external legs and $M$ and $N$ are two sets such that
$M\bigcup N={1,...a-1,a+1,...b-1,b+1,...n}$ and $M\bigcap N=\phi$. The sum is
over all possible $(a,b)$ and $(M,N)$, where $(M,N)$ and $(N,M)$ are not
distinguished. The trace is defined as,

$tr[aMbN]=\left<a|K_{M}|b\right]\left<b|K_{N}|a\right]+\left[a|K_{M}|b\right>\left[b|K_{N}|a\right>$
(15)

where $K_{M}=\sum_{i\in M}k_{i}$. The “half-soft” function $h$ is given by,

$\begin{gathered}h(a,\\{1,2,\cdots,n\\},b)=\frac{[12]}{\left<12\right>}\frac{\left<a|K_{1,2}|3\right]\left<a|K_{1,3}|4\right]\cdots\left<a|K_{1,n-1}|n\right]}{\left<23\right>\left<34\right>\cdots\left<n-1,n\right>\left<a1\right>\left<a2\right>\cdots\left<an\right>\left<1b\right>\left<nb\right>}+\mathcal{P}(2,3,\cdots,n)\end{gathered}$
(16)

where $K_{1,m}=\sum_{i=1}^{m}k_{i}$ and $\mathcal{P}(2,3,\cdots,n)$ represents
all permutations keeping the first leg fixed. Throughout this paper we will
set $\kappa=2$.

##  5 Graviton-Graviton OPE from Self Dual Amplitudes

In this section, we take the $4$ and $5$ point all plus amplitudes and express
them in the conformal primary basis by (modified) Mellin transformation. Then
we take the (collinear) OPE limit $(z_{45}\to 0,\bar{z}_{45}\to 0)$ in the
5-point amplitude with the aim of factorizing it into some differential
operators acting on the 4-point amplitude at every order in the
$(z_{45},\bar{z}_{45})$ expansion. Let us now closely look at the 4-point
amplitude, first in momentum space and then in Mellin space.444For the sake of
convenience of the reader we have moved some of the intermediate steps in the
calculations to an Appendix. We have refereed to the Appendix in the main text
whenever necessary.

###  5.1 4-Point Momentum Space Amplitude

From (14), the 4-point amplitude is given by,

$\displaystyle A_{4}(1^{+},2^{+},3^{+},4^{+})$ $\displaystyle=$
$\displaystyle-\frac{i}{(4\pi)^{2}960}\ B_{4}$ (17)

where

$B_{4}=\sum_{\begin{subarray}{c}1\leq a<b\leq 4\\\
M,N\end{subarray}}h(a,M,b)h(b,N,a)tr^{3}[aMbN]$ (18)

Using the explicit expressions for the trace and the “half-soft” functions,
$B_{4}$ can be easily evaluated and then simplified to get (see Appendix C for
details),

$\displaystyle
B_{4}=-2^{4}\left[\frac{\left<13\right>\left<23\right>([13][23])^{3}}{\left<15\right>^{2}\left<25\right>^{2}}+\left(2\leftrightarrow
3\right)+\left(1\leftrightarrow 3\right)\right]$ (19)

where we have relabelled 4 as 5. In terms of $(\omega,z,\bar{z})$ variables,
the above equation becomes

$\displaystyle
B_{4}=-2^{8}\left[\epsilon_{1}\epsilon_{2}\frac{\omega_{1}\omega_{2}\omega_{3}^{4}}{\omega_{5}^{2}}\frac{z_{13}z_{23}(\bar{z}_{13}\bar{z}_{23})^{3}}{z_{15}^{2}z_{25}^{2}}+\left(2\leftrightarrow
3\right)+\left(1\leftrightarrow 3\right)\right]$

This is the form of the 4-point momentum space amplitude that we use in
evaluating the Mellin transform and other manipulations.

###  5.2 4-Point Mellin Amplitude

The modified Mellin transform of the $n$-point amplitude is given by,

$\mathcal{M}_{n}\big{(}\\{u_{i},z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}=\prod_{i=1}^{n}\int_{0}^{\infty}d\omega_{i}\
\omega_{i}^{\Delta_{i}-1}e^{-i\sum_{i=1}^{n}\epsilon_{i}\omega_{i}u_{i}}A_{n}\big{(}\\{\omega_{i},z_{i},\bar{z}_{i},\sigma_{i}\\}\big{)}$
(20)

where $u$ can be thought of as a time coordinate and $\epsilon_{i}=\pm 1$ for
an outgoing (incoming) particle. Note that
$A_{n}\big{(}\\{\omega_{i},z_{i},\bar{z}_{i},\sigma_{i}\\}\big{)}$ in (20) is
the full momentum space amplitude including the momentum conserving delta
function. Using (20) we now Mellin transform the $4$-point momentum space
amplitude (17). Using the parametrization of 4-point delta function given by
(LABEL:4pt_delta), we get the full 4-point Mellin amplitude as

$\begin{gathered}\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{5}})=\frac{i}{(4\pi)^{2}960}2^{6}\frac{\Gamma(\Delta^{\prime})}{(i\mathcal{D})^{\Delta^{\prime}}}\delta(x-\bar{x})\prod_{k=1}^{3}\left(\epsilon_{k}\sigma_{k,1}\right)^{\Delta_{k}-1}\\\
\times\left[\mathcal{N}_{4}+\mathcal{N}_{4}\left(1\leftrightarrow
3\right)+\mathcal{N}_{4}\left(2\leftrightarrow 3\right)\right]\end{gathered}$
(21)

where $\Delta^{\prime}=\Delta_{1}+\Delta_{2}+\Delta_{3}+\Delta_{5}$ and

$\begin{split}\mathcal{N}_{4}&=\sigma_{1,1}\,\sigma_{2,1}\,\sigma_{3,1}^{4}\frac{z_{13}z_{23}(\bar{z}_{13}\bar{z}_{23})^{3}}{z_{15}^{2}z_{25}^{2}}\\\
\mathcal{D}&=\sum_{k=1}^{3}\sigma_{k,1}u_{k5}\end{split}$ (22)

$\mathcal{N}_{4}(1\leftrightarrow 3)$ and $\mathcal{N}_{4}(2\leftrightarrow
3)$ corresponds to $\mathcal{N}_{4}$ with the points (1,3) and (2,3)
interchanged, respectively. The expressions for $\sigma_{i,j}$ are given in
Appendix B. Note that when we interchange the points $(1,2,3)$ in
$\mathcal{N}_{4}$, only the first subscript in $\sigma_{i,j}$ changes, second
one remains unchanged.

###  5.3 5-point Amplitude in Self Dual Gravity

The 5-point one loop all plus helicity stripped amplitude (without the
momentum conservation delta function) is given by

$\displaystyle
A_{5}(1^{+},2^{+},3^{+},4^{+},5^{+})=\frac{i}{(4\pi)^{2}960}B_{5}$ (23)

where

$\displaystyle B_{5}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}1\leq a<b\leq 5\\\
M,N\end{subarray}}h(a,M,b)h(b,N,a)tr^{3}[aMbN]$ (24)

The above expression consists of 30 distinct terms in total. The expression of
$B_{5}$ has been explicitly computed and simplified in the Appendix D. Its
simplified form gives:

$\displaystyle
B_{5}=-8\left[\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}+\frac{[24]\left<13\right>\left<35\right>([13][35])^{3}}{\left<24\right>\left<12\right>\left<14\right>\left<25\right>\left<45\right>}+\frac{[15]\left<23\right>\left<34\right>([23][34])^{3}}{\left<15\right>\left<21\right>\left<25\right>\left<14\right>\left<54\right>}\right.$
$\displaystyle\left.\frac{[14]\left<23\right>\left<35\right>([23][35])^{3}}{\left<14\right>\left<21\right>\left<24\right>\left<15\right>\left<45\right>}+\frac{[45]\left<13\right>\left<23\right>([13][23])^{3}}{\left<45\right>\left<14\right>\left<15\right>\left<42\right>\left<52\right>}+\frac{[34]\left<25\right>\left<15\right>([15][25])^{3}}{\left<34\right>\left<13\right>\left<14\right>\left<32\right>\left<42\right>}\right.$
$\displaystyle\left.\frac{[35]\left<14\right>\left<24\right>([14][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<32\right>\left<52\right>}+\frac{[12]\left<34\right>\left<35\right>([34][35])^{3}}{\left<12\right>\left<41\right>\left<42\right>\left<15\right>\left<25\right>}+\frac{[12]\left<35\right>\left<45\right>([35][45])^{3}}{\left<12\right>\left<31\right>\left<32\right>\left<14\right>\left<24\right>}\right.$
$\displaystyle\left.\frac{[12]\left<34\right>\left<45\right>([34][45])^{3}}{\left<12\right>\left<31\right>\left<32\right>\left<15\right>\left<25\right>}\right]+\left(1\leftrightarrow
3\right)+\left(2\leftrightarrow 3\right)$ (25)

To avoid complication, we will not write down the Mellin transformation for
the full 5-point amplitude. Rather, we will first expand the 5-point amplitude
around $z_{45}=0,\ \bar{z}_{45}=0$ in momentum space and then Mellin transform
the individual terms in that expansion.

###  5.4 Expansion of the 5-Point Amplitude around $z_{45}=\bar{z}_{45}=0$ in
Momentum Space

By parameterizing (25) in terms of $\\{\omega,z,\bar{z}\\}$ one may think that
there are holomorphic singularities in the limit $z_{4}\to z_{5}$ which goes
like $\frac{1}{z_{45}}$. But this is not true. Clubbing together all the
twelve singular-looking terms, and rewriting them gives contributions only at
leading $\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})$ and higher orders (see
Appendix D for details). By parameterizing $\omega_{4}=t\,\omega_{P},\
\omega_{5}=(1-t)\omega_{P}$ we arrange all the terms in (25) in the following
way:

$B_{5}=-2^{7}\frac{\omega_{P}}{t(1-t)}\left(\frac{\bar{z}_{45}}{z_{45}}T_{L}+T_{\mathcal{O}(1)}+\bar{z}_{45}T_{\bar{z}}\right)+\textnormal{Higher
Order Terms}$ (26)

where

$\begin{gathered}T_{L}=\left[\epsilon_{1}\frac{z_{12}z_{25}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{35}^{2}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}}+\epsilon_{2}\frac{z_{12}z_{15}\bar{z}_{12}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{35}^{2}}\frac{\omega_{1}^{4}\omega_{2}}{\omega_{3}^{2}}+\epsilon_{3}\frac{z_{13}z_{15}\bar{z}_{13}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{25}^{2}}\frac{\omega_{1}^{4}\omega_{3}}{\omega_{2}^{2}}\right]\\\
+\left[\epsilon_{1}\epsilon_{2}\frac{z_{13}z_{23}\bar{z}_{13}^{3}\bar{z}_{23}^{3}}{z_{15}^{2}z_{25}^{2}}\frac{\omega_{1}\omega_{2}\omega_{3}^{4}}{\omega_{P}^{3}}+\left(1\leftrightarrow
3\right)+\left(2\leftrightarrow 3\right)\right]\end{gathered}$ (27)

The expressions for $T_{\mathcal{O}(1)},\ T_{\bar{z}}$ and the detailed
calculation about how we arrived at these expressions are given in the
Appendix D. The point we want to emphasize here is that, (26) is the expansion
of the 5-point amplitude around $z_{45}=\bar{z}_{45}=0$ in the momentum space.
One should not confuse it with the OPE expansion on the celestial torus, which
will be done in the following subsections. The terms
$T_{L},T_{\mathcal{O}(1)},\ T_{\bar{z}}$ contain energy factors
$\\{\omega_{1},\omega_{2},\omega_{3}\\}$ which will contribute to the OPE
expansion after Mellin transformation. On top of that we have 5-point momentum
conserving delta functions as well as other factors in the Mellin integral,
all of which will contribute in the OPE limit of the 5-point Mellin amplitude.
(26) is just a neat way of organizing the 5-point momentum space amplitude,
which allows us to easily extract the OPE from the 5-point celestial
amplitude.

###  5.5 Mellin Transformation of the 5-Point Amplitude and Extracting the
Graviton-Graviton OPE

Let us start with the modified Mellin transformation of $B_{5}$ given by:

$\displaystyle\widetilde{B}_{5}=\int_{0}^{\infty}\prod_{i=1}^{5}d\omega_{i}\,\omega_{i}^{\Delta_{i}-1}e^{-i\sum_{i=1}^{5}\epsilon_{i}\omega_{i}u_{i}}B_{5}\delta^{(4)}\left(\sum_{i=1}^{5}\epsilon_{i}\omega_{i}q_{i}\right)$
(28)

In the above equation for $B_{5}$, we use the expansion (26). Then using the
5-point delta function parametrization given in the Appendix B.2, we can
extract each term in the OPE factorization in the Mellin space. We now discuss
the terms order by order in the OPE expansion in Mellin space.

####  5.5.1 Leading Order

For convenience let’s take $\epsilon_{4}=\epsilon_{5}=+1$. Then the leading
order term in (28) is given by

$\displaystyle\tilde{B}_{5}|_{\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})}$
$\displaystyle=-2^{6}\frac{\bar{z}_{45}}{z_{45}}B(\Delta_{4}-1,\Delta_{5}-1)\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\delta(x-\bar{x})$
$\displaystyle\times\prod_{k=1}^{3}(\epsilon_{k}\sigma_{k,1})^{\Delta_{k}-1}\left[\mathcal{N}_{4}+\mathcal{N}_{4}\left(2\leftrightarrow
3\right)+\mathcal{N}_{4}\left(1\leftrightarrow 3\right)\right]$ (29)

where $\Delta=\sum_{i=1}^{5}\Delta_{i}$. This has been derived in detail in
Appendix E. Finally, taking care of the pre-factors, we can write down the
Mellin transformation of the complete 5-point amplitude $A_{5}$ (23) at
$\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})$:

$\begin{gathered}\mathcal{M}_{5}|_{\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})}=-\frac{i}{(4\pi)^{2}960}2^{6}B(\Delta_{4}-1,\Delta_{5}-1)\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\delta(x-\bar{x})\frac{\bar{z}_{45}}{z_{45}}\prod_{k=1}^{3}(\epsilon_{k}\sigma_{k,1})^{\Delta_{k}-1}\\\
\times\left[\mathcal{N}_{4}+\mathcal{N}_{4}\left(2\leftrightarrow
3\right)+\mathcal{N}_{4}\left(1\leftrightarrow 3\right)\right]\end{gathered}$
(30)

This gives us the 5-point Mellin amplitude at leading order. In terms of the
4-point Mellin amplitude
$\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{4}+\Delta_{5}})$
given by (21), we can write (30) as follows:

$\begin{gathered}\mathcal{M}_{5}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},4^{+}_{\Delta_{4}},5^{+}_{\Delta_{5}})=-\frac{\bar{z}_{45}}{z_{45}}B(\Delta_{4}-1,\Delta_{5}-1)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{4}+\Delta_{5}})+\cdots\end{gathered}$
(31)

Thus, at the level of OPE we have

$\boxed{G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})=-\frac{\bar{z}_{45}}{z_{45}}B(\Delta_{4}-1,\Delta_{5}-1)G^{+}_{\Delta_{4}+\Delta_{5}}(z_{5},\bar{z}_{5})+\cdots}$
(32)

####  5.5.2 $\mathcal{O}(1)$ Terms

Now we turn our attention to the $\mathcal{O}(1)$ terms in the five point
Mellin amplitude. The complete expression for the 5-point Mellin amplitude at
$\mathcal{O}(1)$ is given by (119),

$\begin{gathered}\mathcal{M}_{5}\big{|}_{\mathcal{O}(1)}=-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\sum_{k=0}^{4}B(\Delta_{4}+k-1,\Delta_{5}-1)\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta\left(x-\bar{x}\right)\end{gathered}$
(33)

where $\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})$ are some
functions of its arguments, but their explicit expressions are not important
for OPE factorizations. Now we take the leading conformal soft limit
${\Delta_{4}\to 1}$ in the above equation to get:

$\begin{gathered}\lim_{\Delta_{4}\to
1}(\Delta_{4}-1){\mathcal{M}}_{5}\big{|}_{\mathcal{O}(1)}=-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma\left(\sum_{i=1,i\neq
4}^{5}\Delta_{i}+1\right)}{\left(i\mathcal{D}\right)^{\sum_{i=1,i\neq
4}^{5}\Delta_{i}+1}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{i\lambda_{i}}\mathcal{F}^{(1)}_{0}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta(x-\bar{x})\end{gathered}$
(34)

Only the $k=0$ term in the sum in (33) survives because, in the $\Delta_{4}\to
1$ limit, $B(\Delta_{4}+k-1,\Delta_{5}-1)$ is non-singular for all $k>0$.

On the other hand, from the leading soft graviton theorem we know that

$\begin{gathered}\lim_{\Delta_{4}\to
1}(\Delta_{4}-1){\mathcal{M}}_{5}\big{|}_{\mathcal{O}(1)}=\mathcal{H}^{1}_{-\frac{3}{2},\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{5}})\end{gathered}$
(35)

To make things transparent, we have used $\mathcal{H}$-notations when the soft
modes are acting on the Mellin amplitudes as differential operators and the
number $5$ in the argument of $\mathcal{H}$ denotes that it’s a descendant of
the $5$-th conformal graviton primary. The consistency of the two equations
(34) and (35) implies that:

$\begin{gathered}-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma\left(\sum_{k=1,k\neq
4}^{5}\Delta_{k}+1\right)}{\left(i\mathcal{D}\right)^{\sum_{k=1,k\neq
4}^{5}\Delta_{k}+1}}\prod_{k=1}^{3}(\epsilon_{k}\sigma_{k,1})^{i\lambda_{k}}\mathcal{F}^{(1)}_{0}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta(x-\bar{x})\\\
=\mathcal{H}^{1}_{-\frac{3}{2},\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{5}})\end{gathered}$
(36)

Now, we can replace $\Delta_{5}$ by $\Delta_{4}+\Delta_{5}-1$ in (36) and then
use it in (33) to get:

$\begin{gathered}\mathcal{M}_{5}\big{|}_{\mathcal{O}(1)}=B(\Delta_{4}-1,\Delta_{5}-1)\mathcal{H}^{1}_{-\frac{3}{2},\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{4}+\Delta_{5}-1})\\\
-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\sum_{k=1}^{4}B(\Delta_{4}+k-1,\Delta_{5}-1)\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta\left(x-\bar{x}\right)\end{gathered}$
(37)

Here we have replaced the $\mathcal{F}^{(1)}_{0}$ dependent term in (33) in
terms of a soft graviton mode acting on the 4-point amplitude. Let us now
repeat the same procedure for $\mathcal{F}^{(1)}_{1}$.

By taking the subleading conformal soft limit ${\Delta_{4}\to 0}$ in (37), we
get

$\begin{gathered}\lim_{\Delta_{4}\to
0}\Delta_{4}\mathcal{M}_{5}\big{|}_{\mathcal{O}(1)}=-(\Delta_{5}-2)\,\mathcal{H}^{1}_{-\frac{3}{2},\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{5}-1})\\\
-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma(\sum_{k=1,k\neq
4}^{5}\Delta_{k})}{\left(i\mathcal{D}\right)^{\sum_{k=1,k\neq
4}^{5}\Delta_{k}}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\mathcal{F}^{(1)}_{1}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta\left(x-\bar{x}\right)\end{gathered}$
(38)

Now, from subleading soft graviton theorem we know that:

$\begin{gathered}\lim_{\Delta_{4}\to
0}\Delta_{4}\mathcal{M}_{5}\big{|}_{\mathcal{O}(1)}=-\mathcal{H}^{0}_{-1,1}(5)\mathcal{H}^{1}_{-\frac{1}{2},-\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{5}-1})\end{gathered}$
(39)

Again, consistency of the two equations (38) and (39) gives us the function
$\mathcal{F}^{(1)}_{1}$ in terms of the leading and subleading soft modes.
Substituting this back in (37) results in

$\begin{gathered}\mathcal{M}_{5}\big{|}_{\mathcal{O}(1)}=\frac{\Gamma(\Delta_{4}+1)}{\Gamma(\Delta_{4})}B(\Delta_{4}-1,\Delta_{5}-1)\mathcal{H}^{1}_{-\frac{3}{2},\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{4}+\Delta_{5}-1})\\\
+B(\Delta_{4},\Delta_{5}-1)\mathcal{H}^{0}_{-1,1}(5)\left(-\mathcal{H}^{1}_{-\frac{1}{2},-\frac{1}{2}}(5)\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{4}+\Delta_{5}-1})\right)\\\
-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\sum_{k=2}^{4}B(\Delta_{4}+k-1,\Delta_{5}-1)\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta\left(x-\bar{x}\right)\end{gathered}$
(40)

We continue this process till all the $\mathcal{F}^{(1)}_{k}$’s have been
replaced by descendant correlation functions of the soft modes. From the above
equation (40), it is clear that to replace all the $\mathcal{F}^{(1)}_{k}$’s
by the descendant correlation functions of the soft modes, we have to go till
sub4leading order in the soft limits of $\Delta_{4}$. We only write the final
result here which is given by,

$\begin{gathered}\mathcal{M}_{5}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},4^{+}_{\Delta_{4}},5^{+}_{\Delta_{5}})\big{|}_{\mathcal{O}(1)}=\sum_{k=0}^{4}\frac{1}{(4-k)!}\frac{\Gamma(\Delta_{4}+4)}{\Gamma(\Delta_{4}+k)}B(\Delta_{4}+k-1,\Delta_{5}-1)\\\
\times\mathcal{H}^{1-k}_{\frac{k-3}{2},\frac{k+1}{2}}(5)\left(\mathcal{H}^{1}_{-\frac{1}{2},-\frac{1}{2}}(5)\right)^{k}\mathcal{M}_{4}(1^{+}_{\Delta_{1}},2^{+}_{\Delta_{2}},3^{+}_{\Delta_{3}},5^{+}_{\Delta_{4}+\Delta_{5}-1})\end{gathered}$
(41)

Now that we have factorized the $\mathcal{O}(1)$ terms in the 5-point Mellin
amplitude completely in terms of soft modes acting on the 4-point amplitude,
we can easily extract the $\mathcal{O}(1)$ graviton graviton OPE from the
above equation. It is given by,

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\mathcal{O}(1)}=\sum_{k=0}^{4}\frac{1}{(4-k)!}\frac{\Gamma(\Delta_{4}+4)}{\Gamma(\Delta_{4}+k)}B(\Delta_{4}+k-1,\Delta_{5}-1)\\\
\times
H^{1-k}_{\frac{k-3}{2},\frac{k+1}{2}}\left(H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)^{k}G^{+}_{\Delta_{4}+\Delta_{5}-1}(z_{5},\bar{z}_{5})\end{gathered}$
(42)

We can rewrite (42) using the null states of MHV-sector. From (136), it is
clear that all the soft modes $H^{1-k}_{\frac{k-3}{2},\frac{k+1}{2}}$ with
$k=1,\cdots,4$ can be replaced by the MHV null states $\\{\Phi_{k},\
k=1,\cdots,4\\}$. Thus, (41) in terms of the $\mathcal{O}(1)$ MHV null states
(136), becomes:

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\mathcal{O}(1)}=B(\Delta_{4}-1,\Delta_{5}-1)H^{1}_{-\frac{3}{2},\frac{1}{2}}G^{+}_{\Delta_{4}+\Delta_{5}-1}(z_{5},\bar{z}_{5})\\\
+\sum_{k=1}^{4}\frac{1}{(4-k)!}\frac{\Gamma(\Delta_{4}+4)}{\Gamma(\Delta_{4}+k)}B(\Delta_{4}+k-1,\Delta_{5}-1)\Phi_{k}(\Delta_{4}+\Delta_{5})\\\
\end{gathered}$ (43)

Thus we see that, the $\mathcal{O}(1)$ terms in the self dual OPE between two
positive helicity outgoing gravitons can completely be written in terms of the
$\mathcal{O}(1)$ MHV OPE and the $\mathcal{O}(1)$ null states of the MHV
sector.

Now, as discussed in section H, we can define a new basis for MHV null states
instead of $\Phi_{k}$’s. This new basis is given by (138). For our
convenience, let us write (138) here again,

$\begin{gathered}\Omega_{k}(\Delta)=\sum_{n=1}^{k}\frac{1}{(k-n)!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta+n-2)}\Phi_{n}(\Delta)\end{gathered}$
(44)

This basis has nice transformation properties under the $w$-algebra
Banerjee:2023zip , reviewed in section G. Represented in terms of this new
$\Omega$-basis, the graviton-graviton OPE (43) takes a very simple form,

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\mathcal{O}(1)}=B(\Delta_{4}-1,\Delta_{5}-1)H^{1}_{-\frac{3}{2},\frac{1}{2}}G^{+}_{\Delta_{4}+\Delta_{5}-1}(z_{5},\bar{z}_{5})\\\
+\sum_{k=1}^{4}B(\Delta_{4}+k-1,\Delta_{5}-1)\Omega_{k}(\Delta_{4}+\Delta_{5})\\\
\end{gathered}$ (45)

Thus we see that, the $\mathcal{O}(1)$ OPE between two positive helicity
outgoing gravitons in quantum self dual gravity truncates at $n=4$ of the
general $w$-invariant OPE (140). We now move on to $\mathcal{O}(\bar{z})$ OPE
in the next subsection.

####  5.5.3 $\mathcal{O}(\bar{z}_{45})$ Term

The soft modes that appear at order $\bar{z}_{45}$ from the $w$-algebra are
given by

$H^{k}_{-\frac{k+2}{2},-\frac{k}{2}},\ k=1,0,-1,\cdots$ (46)

Now, like the $\mathcal{O}(1)$ OPE we can factorize the
$\mathcal{O}(\bar{z}_{45})$ terms from the 5-point amplitude using the soft
limits and $w$-modes. The crucial difference from $\mathcal{O}(1)$ is that,
now we have to go one order higher in the soft limits than $\mathcal{O}(1)$.
We start by writing the $\mathcal{O}(\bar{z}_{45})$ term of the 5-point Mellin
amplitude given by (see (123)),

$\begin{gathered}\mathcal{M}_{5}\big{|}_{\mathcal{O}(\bar{z}_{45})}=-\frac{i}{(4\pi)^{2}960}2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\sum_{k=1}^{5}B(\Delta_{4}+k-1,\Delta_{5}-1)\mathcal{F}^{(\bar{z})}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\end{gathered}$
(47)

One can easily see from (47) that, to factorize the 5-point Mellin amplitude
completely, i.e, to replace all the functions
$\mathcal{F}^{(\bar{z})}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})$ by the
descendant correlation functions of soft modes, we have to continue taking the
soft limits in $\Delta_{4}$ till we reach $\Delta_{4}\to-4$. Thus the highest
soft modes that can appear in the OPE at $\mathcal{O}(\bar{z}_{45})$ are given
by $H^{-4}_{1,2}$. We have discussed how to factorize the amplitude at
$\mathcal{O}(1)$ in terms of the descendant correlators of the soft modes in
the previous section in detail. One has to repeat the same procedure for
$\mathcal{O}(\bar{z}_{45})$ as well. Without going into much detail we
directly write the $\mathcal{O}(\bar{z}_{45})$ OPE which is given by,

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\mathcal{O}(\bar{z}_{45})}=G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\text{MHV
at}\ \mathcal{O}(\bar{z}_{45})}\\\
+\sum_{k=1}^{4}\frac{1}{(n-k)!}\frac{\Gamma(\Delta_{4}+n+1)}{\Gamma(\Delta_{4}+k+1)}B(\Delta_{4}+k,\Delta_{5}-1)\Psi_{k}(\Delta_{4}+\Delta_{5}+1)\end{gathered}$
(48)

where

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\text{MHV
at}\
\mathcal{O}(\bar{z}_{45})}=B(\Delta_{4}-1,\Delta_{5}-1)\left[\frac{\Delta_{4}-1}{\Delta_{4}+\Delta_{5}-2}H^{0}_{-1,0}\left(-H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)\right.\\\
\left.+\Delta_{4}\,H^{1}_{-\frac{3}{2},-\frac{1}{2}}\right]G^{+}_{\Delta_{4}+\Delta_{5}-1}\end{gathered}$
(49)

and $\Psi_{k}(\Delta_{4}+\Delta_{5}+1)$ is given by

$\displaystyle\Psi_{k}(\Delta_{4}+\Delta_{5}+1)$ $\displaystyle=$
$\displaystyle\left[H^{-k}_{\frac{k-2}{2},\frac{k}{2}}\left(-H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)^{k+1}-\frac{(-1)^{k}}{k!}\frac{\Gamma(\Delta_{4}+\Delta_{5}+k-1)}{\Gamma(\Delta_{4}+\Delta_{5}-1)}H^{0}_{-1,0}\left(-H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)\right.$
(50)
$\displaystyle\left.-(-1)^{k}\frac{k}{(k+1)!}\frac{\Gamma(\Delta_{4}+\Delta_{5}+k-1)}{\Gamma(\Delta_{4}+\Delta_{5}-2)}H^{1}_{-\frac{3}{2},-\frac{1}{2}}\right]G^{+}_{\Delta_{4}+\Delta_{5}-1}$

In terms of the new basis defined in (139), the above OPE can again be written
in a very nice and simple form given by,

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\mathcal{O}(\bar{z}_{45})}=G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\text{MHV
at}\ \mathcal{O}(\bar{z}_{45})}\\\
+\bar{z}_{45}\sum_{k=1}^{4}B(\Delta_{4}+k,\Delta_{5}-1)\
\Pi_{k}(\Delta_{4}+\Delta_{5}+1)\end{gathered}$ (51)

Thus we see that the $\mathcal{O}(\bar{z}_{45})$ terms in the OPE again
truncate at $n=4$ of the general $w$-invariant OPE (140).

##  6 Discussion

Operator product expansion plays a very important role in any quantum field
theory and therefore it is important to understand the structure of OPE in the
celestial CFTs. In its current formulation, celestial CFTs differ from more
conventional CFTs in many ways. The primary difference is that the spectrum of
the operator dimensions in celestial CFTs is not bounded from below. Taken at
face value, this implies that the number of descendants that can appear at any
given order of the celestial OPE can be infinite. However, this is not a very
desirable feature and warrants further study.

In this paper, we have undertaken the task of computing the celestial OPE of
two positive helicity outgoing gravitons in the quantum self-dual gravity. It
is known that the self dual gravity enjoys $w$ invariance. Therefore, one
should be able to express the OPE in terms of $w$ descendants of the graviton
primary. This is what we have found. However, the most surprising fact which
comes out of our study is that at any given order the OPE contains only a
finite number of $w$ descendants. Therefore, the self dual gravity behaves
like any other CFT with a spectrum of operator dimensions bounded from below.

This raises some interesting questions. For example, we know that the
Holographic Symmetry Algebra (HSA) contains an infinite tower of holomorphic
currents $H^{k}_{n}(z)$ with $k$ going from $1$ to $-\infty$. Our calculation
shows that in the self dual theory at $\mathcal{O}(1)$ and at
$\mathcal{O}(\bar{z})$ the list of $w$ descendants truncate at $k=-3$ and
$k=-4$, respectively. However, this is somewhat unnatural given the fact that
the currents $H^{-3}_{n}(z)$ and $H^{-4}_{n}(z)$ do not play any distinguished
role in the algebra. Therefore, it is natural to wonder if there are other $w$
invariant theories where the truncation occurs at other values of $k$. This is
consistent with our earlier analysis Banerjee:2023zip  based on the $w$
algebra representation where we found that one can write down an infinite
number of consistent $w$ invariant OPEs where truncation happens at different
values of $k$. Therefore, truncation is not a reflection of $w$ symmetry. We
leave the construction of these theories as an interesting problem for the
future.

Before we end, we would like to point out that truncation means that the self
dual theory in many ways behave like theories with operator dimensions bounded
from below. So it is very likely that the self dual theory and the (tree-
level) MHV sector of GR can be reformulated in terms of celestial primary
operators with dimensions strictly bounded from below. Interesting proposals
along this line has been put forward in Cotler:2023qwh ; Freidel:2022skz  It
will be fascinating if they can be applied to the present problem.

##  7 Acknowledgements

SB would like to thank the participants of the Kickoff Workshop for the Simons
Collaboration on Celestial Holography for helpful comments. The work of SB is
partially supported by the Swarnajayanti Fellowship (File No-
SB/SJF/2021-22/14) of the Department of Science and Technology and SERB,
India. The work of HK is partially supported by the KVPY fellowship of the
Department of Science and Technology, Government of India. The work of PP is
supported by an IOE endowed Postdoctoral position at IISc, Bengaluru, India.

##  Appendix A Brief Review of Celestial or Mellin Amplitudes For Massless
Particles

The Celestial or Mellin amplitude for massless particles in four dimensions is
defined as the Mellin transformation of the $S$-matrix element, given by
Pasterski:2016qvg ; Pasterski:2017kqt

$\mathcal{M}_{n}\big{(}\\{z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}=\prod_{i=1}^{n}\int_{0}^{\infty}d\omega_{i}\
\omega_{i}^{\Delta_{i}-1}A_{n}\big{(}\\{\omega_{i},z_{i},\bar{z}_{i},\sigma_{i}\\}\big{)}$
(52)

where $\sigma_{i}$ denotes the helicity of the $i$-th particle and the on-
shell momenta are parametrized by (1). The scaling dimensions
$(h_{i},\bar{h}_{i})$ are defined as,

$h_{i}=\frac{\Delta_{i}+\sigma_{i}}{2},\quad\bar{h}_{i}=\frac{\Delta_{i}-\sigma_{i}}{2}$
(53)

Under the Lorentz transformation (2), the Mellin amplitude $\mathcal{M}_{n}$
transforms as,

$\mathcal{M}_{n}\big{(}\\{z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}=\prod_{i=1}^{n}\frac{1}{(cz_{i}+d)^{2h_{i}}}\frac{1}{(\bar{c}\bar{z}_{i}+\bar{d})^{2\bar{h}_{i}}}\mathcal{M}_{n}\bigg{(}\frac{az_{i}+b}{cz_{i}+d}\
,\frac{\bar{a}\bar{z}_{i}+\bar{b}}{\bar{c}\bar{z}_{i}+\bar{d}}\
,h_{i},\bar{h}_{i}\bigg{)}$ (54)

This is the familiar transformation law for the correlation function of
primary operators of weight $(h_{i},\bar{h}_{i})$ in a $2$-D CFT under the
global conformal group.

In Einstein gravity, the Mellin amplitude as defined in (52) usually diverges.
This divergence can be regulated by defining a modified Mellin amplitude as
Banerjee:2018gce ; Banerjee:2019prz ,

$\mathcal{M}_{n}\big{(}\\{u_{i},z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}=\prod_{i=1}^{n}\int_{0}^{\infty}d\omega_{i}\
\omega_{i}^{\Delta_{i}-1}e^{-i\sum_{i=1}^{n}\epsilon_{i}\omega_{i}u_{i}}A_{n}\big{(}\\{\omega_{i},z_{i},\bar{z}_{i},\sigma_{i}\\}\big{)}$
(55)

where $u$ can be thought of as a time coordinate and $\epsilon_{i}=\pm 1$ for
an outgoing (incoming) particle. Under (Lorentz) conformal tranansformation
the modified Mellin amplitude $\mathcal{M}_{n}$ transforms as,

$\mathcal{M}_{n}\big{(}\\{u_{i},z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}=\prod_{i=1}^{n}\frac{1}{(cz_{i}+d)^{2h_{i}}}\frac{1}{(\bar{c}\bar{z}_{i}+\bar{d})^{2\bar{h}_{i}}}\mathcal{M}_{n}\bigg{(}\frac{u_{i}}{|cz_{i}+d|^{2}}\
,\frac{az_{i}+b}{cz_{i}+d}\
,\frac{\bar{a}\bar{z}_{i}+\bar{b}}{\bar{c}\bar{z}_{i}+\bar{d}}\
,h_{i},\bar{h}_{i}\bigg{)}$ (56)

Under global space-time translation, $u\rightarrow
u+A+Bz+\bar{B}\bar{z}+Cz\bar{z}$, the modified Mellin amplitude is invariant,
i.e,

$\mathcal{M}_{n}\big{(}\\{u_{i}+A+Bz_{i}+\bar{B}\bar{z}_{i}+Cz_{i}\bar{z}_{i},z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}=\mathcal{M}_{n}\big{(}\\{u_{i},z_{i},\bar{z}_{i},h_{i},\bar{h}_{i}\\}\big{)}$
(57)

Now in order to make manifest the conformal nature of the dual theory living
on the celestial sphere it is useful to write the (modified) Mellin amplitude
as a correlation function of conformal primary operators. So let us define a
generic conformal primary operator as,

$\phi^{\epsilon}_{h,\bar{h}}(z,\bar{z})=\int_{0}^{\infty}d\omega\
\omega^{\Delta-1}a(\epsilon\omega,z,\bar{z},\sigma)$ (58)

where $\epsilon=\pm 1$ for an annihilation (creation) operator of a massless
particle of helicity $\sigma$. Under (Lorentz) conformal transformation the
conformal primary transforms like a primary operator of scaling dimension
$(h,\bar{h})$

$\phi^{\prime\epsilon}_{h,\bar{h}}(z,\bar{z})=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\mathcal{\phi}^{\epsilon}_{h,\bar{h}}\bigg{(}\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (59)

Similarly in the presence of the time coordinate $u$ we have,

$\phi^{\epsilon}_{h,\bar{h}}(u,z,\bar{z})=\int_{0}^{\infty}d\omega\
\omega^{\Delta-1}e^{-i\epsilon\omega u}a(\epsilon\omega,z,\bar{z},\sigma)$
(60)

Under (Lorentz) conformal transformations

$\phi^{\prime\epsilon}_{h,\bar{h}}(u,z,\bar{z})=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\mathcal{\phi}^{\epsilon}_{h,\bar{h}}\bigg{(}\frac{u}{|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (61)

In terms of (58), the Mellin amplitude can be written as the correlation
function of conformal primary operators

$\mathcal{M}_{n}=\bigg{\langle}{\prod_{i=1}^{n}\phi^{\epsilon_{i}}_{h_{i},\bar{h}_{i}}(z_{i},\bar{z}_{i})}\bigg{\rangle}$
(62)

Similarly using (60), the modified Mellin amplitude can be written as,

$\mathcal{M}_{n}=\bigg{\langle}{\prod_{i=1}^{n}\phi^{\epsilon_{i}}_{h_{i},\bar{h}_{i}}(u_{i},z_{i},\bar{z}_{i})}\bigg{\rangle}$
(63)

###  A.1 Comments on notation in the paper

Note that the conformal primaries carry an extra index $\epsilon$ which
distinguishes between an incoming and an outgoing particle. In this paper, for
notational simplicity, we omit this additional index unless this plays an
important role. So in most places we simply write the (modified) Mellin
amplitude as,

$\mathcal{M}_{n}=\bigg{\langle}{\prod_{i=1}^{n}\phi_{h_{i},\bar{h}_{i}}(z_{i},\bar{z}_{i})}\bigg{\rangle}$
(64)

or

$\mathcal{M}_{n}=\bigg{\langle}{\prod_{i=1}^{n}\phi_{h_{i},\bar{h}_{i}}(u_{i},z_{i},\bar{z}_{i})}\bigg{\rangle}$
(65)

Similarly in many places in the paper we denote a graviton primary of weight
$\Delta=h+\bar{h}$ by $G^{\sigma}_{\Delta}$ where $\sigma=\pm 2$ is the
helicity (= $h-\bar{h}$). Since we are considering pure gravity, we can
further simplify the notation to $G^{\pm}_{\Delta}$ by omitting the $2$.

##  Appendix B Parametrization of the Delta Functions

In this Appendix, we parametrize the 4-point and 5-point delta functions which
will be convenient for our purpose of extracting the OPE.

###  B.1 4-Point Delta Function

In $(2,2)$ split signature, the parametrization of the null momentum ($p_{i}$)
for $i$-th massless particle in terms of
$\left(\omega_{i},z_{i},\bar{z}_{i}\right)$ is given by

$p_{i}=\omega_{i}\\{1+z_{i}\bar{z}_{i},z_{i}+\bar{z}_{i},(z_{i}-\bar{z}_{i}),1-z_{i}\bar{z}_{i}\\},\qquad
p_{i}^{2}=0$ (66)

This allows us to write down the 4-point momentum conserving delta function in
the following way which is more convenient for us

$\displaystyle\delta^{(4)}\left(\sum_{i=1,\neq 4}^{5}\epsilon_{i}p_{i}\right)$
$\displaystyle=$ $\displaystyle\frac{1}{4}\delta\left(\sum_{i=1,\neq
4}^{5}\epsilon_{i}\omega_{i}\right)\delta\left(\sum_{i=1}^{3}\epsilon_{i}\omega_{i}z_{i5}\right)\delta\left(\sum_{i=1}^{3}\epsilon_{i}\omega_{i}\bar{z}_{i5}\right)\delta\left(\sum_{i=1}^{3}\epsilon_{i}\omega_{i}z_{i5}\bar{z}_{i5}\right)$
$\displaystyle=$
$\displaystyle\epsilon_{1}\epsilon_{2}\epsilon_{3}\epsilon_{5}\frac{1}{4\omega_{5}}\delta(\omega_{1}-\omega_{1}^{*})\delta(\omega_{2}-\omega_{2}^{*})\delta(\omega_{3}-\omega_{3}^{*})\delta(x-\bar{x})$

where $\epsilon_{i}=\pm 1$ for outgoing (incoming) particle and

$\displaystyle\omega_{i}^{*}$ $\displaystyle=$
$\displaystyle\epsilon_{5}\,\omega_{5}\,\epsilon_{i}\,\sigma_{i,1}$ (68)
$\displaystyle\sigma_{1,1}$ $\displaystyle=$
$\displaystyle-\frac{z_{25}\bar{z}_{35}}{z_{12}\bar{z}_{13}}$ (69)
$\displaystyle\sigma_{2,1}$ $\displaystyle=$
$\displaystyle\frac{z_{15}\bar{z}_{35}}{z_{12}\bar{z}_{23}}$ (70)
$\displaystyle\sigma_{3,1}$ $\displaystyle=$
$\displaystyle-\frac{z_{25}\bar{z}_{15}}{z_{23}\bar{z}_{13}}$ (71)
$\displaystyle x$ $\displaystyle=$ $\displaystyle
z_{12}z_{35}\bar{z}_{13}\bar{z}_{25},\
\bar{x}=z_{13}z_{25}\bar{z}_{12}\bar{z}_{35}$ (72)

The $\sigma_{i,1}$’s defined above satisfy the following identities on the
support of $\delta(x-\bar{x})$

$\displaystyle\sigma_{1,1}+\sigma_{2,1}+\sigma_{3,1}+1$ $\displaystyle=$
$\displaystyle 0$ (73) $\displaystyle
z_{15}\sigma_{1,1}+z_{25}\sigma_{2,1}+z_{35}\sigma_{3,1}$ $\displaystyle=$
$\displaystyle 0$ (74)
$\displaystyle\bar{z}_{15}\sigma_{1,1}+\bar{z}_{25}\sigma_{2,1}+\bar{z}_{35}\sigma_{3,1}$
$\displaystyle=$ $\displaystyle 0$ (75)

This representation for the 4-point delta function and the properties of
$\sigma_{i,1}$’s will be useful in extracting the OPE. Note that in this delta
function representation, we have indexed the four particles by $1,2,3,5$
because to extract the OPE, we take the 4 $\to$ 5 OPE limit in the 5-point
Mellin amplitude and then factorize it in terms of the 4-point Mellin
amplitude now indexed by $1,2,3,5$. This is a notation that we followed
throughout the paper.

###  B.2 5-Point Delta Function

We now write down the representation for the delta function for 5 particles.
For concreteness, we take $\epsilon_{4}=\epsilon_{5}=+1$. Since we are
interested in the OPE limit $4\to 5$, it is convenient to use the following
parametrization

$\omega_{4}=t\omega_{P},\ \omega_{5}=(1-t)\omega_{P}$ (76)

in representing the 5-point delta function. For the case of $n=5$ particles in
four spacetime dimensions we have four constraint equations coming from the
four components of the energy momentum conserving equations. We can solve
these four constraint equations for three energy variables
$\\{\omega_{1},\omega_{2},\omega_{3}\\}$ in terms of $\omega_{4}$ and
$\omega_{5}$. Thus, the representation of the 5-point delta function which is
better suited for our purposes of performing the OPE decomposition of the
Mellin amplitude in the (4, 5) channel, is given by Banerjee:2020zlg
555Please note that in Banerjee:2020zlg  the OPE factorization has been done
starting from the 6-point Mellin amplitude whereas in this paper it is done
starting from the 5-point amplitude. Thus, in parametrizing the 5-point delta
function in this paper, we have used the same methodology which was used for
6-point delta function in Banerjee:2020zlg .,

$\displaystyle\delta^{(4)}\left(\sum_{i=1}^{5}\epsilon_{i}\omega_{i}q_{i}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{4\omega_{P}}\delta(\omega_{1}-\omega_{1}^{*})\delta(\omega_{2}-\omega_{2}^{*})\delta(\omega_{3}-\omega_{3}^{*})$
$\displaystyle\times\delta\left(x-\bar{x}-tz_{45}\left(\frac{x}{z_{35}}-\frac{\bar{x}}{z_{25}}\right)-t\bar{z}_{45}\left(\frac{x}{\bar{z}_{25}}-\frac{\bar{x}}{\bar{z}_{35}}\right)+tz_{45}\bar{z}_{45}\left(\frac{x}{z_{35}\bar{z}_{25}}-\frac{\bar{x}}{z_{25}\bar{z}_{35}}\right)\right)$

where for $i=\\{1,2,3\\}$ we have

$\displaystyle\omega_{i}^{*}$ $\displaystyle=$
$\displaystyle\omega_{P}\tilde{\omega}_{i}^{*}$
$\displaystyle\tilde{\omega}_{i}^{*}$ $\displaystyle=$
$\displaystyle\epsilon_{i}\left(\sigma_{i,1}+tz_{45}\sigma_{i,2}+t\bar{z}_{45}\sigma_{i,3}+tz_{45}\bar{z}_{45}\sigma_{i,4}\right)$
(78)

and the $\sigma_{i,1},\ x,\ \bar{x}$ are given by (69)-(72). We also have

$\sigma_{i,2}=\frac{\partial\sigma_{i,1}}{\partial z_{5}},\
\sigma_{i,3}=\frac{\partial\sigma_{i,1}}{\partial\bar{z}_{5}},\
\sigma_{i,4}=\frac{\partial\sigma_{i,1}}{\partial
z_{5}\partial\bar{z}_{5}},\qquad\forall\,i=1,2,3.$ (79)

##  Appendix C Simplification of the 4-point Amplitude

In this Appendix, we simplify the 4-point self dual one loop amplitude in
momentum space which is used in section 5. We start with the equation (18) for
the 4-point amplitude:

$\displaystyle B_{4}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}1\leq a<b\leq 4\\\
M,N\end{subarray}}h(a,M,b)h(b,N,a)\text{\text{tr}}^{3}[aMbN]$ $\displaystyle=$
$\displaystyle
h(1,3,2)h(2,4,1)\text{tr}^{3}[1324]+h(1,2,3)h(3,4,1)\text{tr}^{3}[1234]+h(1,2,4)h(4,3,1)\text{tr}^{3}[1243]$
$\displaystyle+$ $\displaystyle
h(2,1,3)h(3,4,2)\text{tr}^{3}[2134]+h(2,1,4)h(4,3,2)\text{tr}^{3}[2143]+h(3,1,4)h(4,2,3)\text{tr}^{3}[3142]$

The trace function is given by

$\text{tr}[aMbN]=\left<a|K_{M}|b\right]\left<b|K_{N}|a\right]+\left[a|K_{M}|b\right>\left[b|K_{N}|a\right>$
(80)

For $M=\\{i\\},\ N=\\{l\\}$ we have

$\displaystyle\text{tr}[aibl]$ $\displaystyle=$
$\displaystyle\left<a|k_{i}|b\right]\left<b|k_{l}|a\right]+\left[a|k_{i}|b\right>\left[b|k_{l}|a\right>$
(81) $\displaystyle=$
$\displaystyle\left<ai\right>[ib]\left<bl\right>[la]+\left<bi\right>[ia]\left<al\right>[lb]$

From the above equation we can see that $\text{tr}[aibl]=\text{tr}[ialb]$.
Using this property of the trace function and the expression for the half soft
function

$h(a,i,b)=\frac{1}{\left<ai\right>^{2}\left<ib\right>^{2}}$ (82)

(C) can be simplified as,

$\displaystyle B_{4}$ $\displaystyle=$ $\displaystyle
2\left(h(1,3,2)h(2,4,1)\text{tr}^{3}[1324]+h(1,2,3)h(3,4,1)\text{tr}^{3}[1234]+h(1,2,4)h(4,3,1)\text{tr}^{3}[1243]\right)$

Now, using the momentum conservation for four particles in the trace functions
(81) and the explicit expressions of the half soft functions (82),
(LABEL:B_42) finally gives,

$\displaystyle B_{4}$ $\displaystyle=$
$\displaystyle-2^{4}\left(\frac{\left<13\right>\left<23\right>([13][23])^{3}}{\left<14\right>^{2}\left<24\right>^{2}}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right)$

As mentioned earlier, since we will index the 4 particles as $1,2,3,5$,
relabelling $4$ as $5$ in the above expression gives the following form of the
4-point amplitude in momentum space,

$\displaystyle B_{4}$ $\displaystyle=$
$\displaystyle-2^{4}\left(\frac{\left<13\right>\left<23\right>([13][23])^{3}}{\left<15\right>^{2}\left<25\right>^{2}}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right)$

##  Appendix D Simplification of the 5-point Amplitude

Similar to what was done for the 4-point case, we will now simplify the
5-point self dual one loop amplitude in momentum space which is used in
section 5.3 by considering the equation (24):

$\displaystyle B_{5}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}1\leq a<b\leq 5\\\
M,N\end{subarray}}h(a,M,b)h(b,N,a)\text{tr}^{3}[aMbN]$ (86) $\displaystyle=$
$\displaystyle
h(1,M,2)h(2,N,1)\text{tr}^{3}[1M2N]+h(1,M,3)h(3,N,1)\text{tr}^{3}[1M3N]+h(1,M,4)h(4,N,1)\text{tr}^{3}[1M4N]$
$\displaystyle+$ $\displaystyle
h(1,M,5)h(5,N,1)\text{tr}^{3}[1M5N]+h(2,M,3)h(3,N,2)\text{tr}^{3}[2M3N]+h(2,M,4)h(4,N,2)\text{tr}^{3}[2M4N]$
$\displaystyle+$ $\displaystyle
h(2,M,5)h(5,N,2)\text{tr}^{3}[2M5N]+h(3,M,4)h(4,N,3)\text{tr}^{3}[3M4N]+h(3,M,5)h(5,N,3)\text{tr}^{3}[3M5N]$
$\displaystyle+$ $\displaystyle h(4,M,5)h(5,N,4)\text{tr}^{3}[4M5N]$

The two sets $M$ and $N$ are such that $M\bigcup
N={1,...a-1,a+1,...b-1,b+1,...n}$ and $M\bigcap N=\phi$ and the sum is over
all possible $a,b$ and sets $(M,N)$, where $(M,N)$ and $(N,M)$ are not
distinguished. For 5-point amplitudes, with $M=\\{i,j\\},\ N=\\{l\\}$, the
trace function given by,

$\text{tr}[aMbN]=\left<a|K_{M}|b\right]\left<b|K_{N}|a\right]+\left[a|K_{M}|b\right>\left[b|K_{N}|a\right>$
(87)

becomes

$\displaystyle\text{tr}[a\\{i+j\\}b\\{l\\}]$ $\displaystyle=$
$\displaystyle\left<a|k_{i}+k_{j}|b\right]\left<b|k_{l}|a\right]+\left[a|k_{i}+k_{j}|b\right>\left[b|k_{l}|a\right>$
(88) $\displaystyle=$ $\displaystyle\text{tr}[aibl]+\text{tr}[ajbl]$
$\displaystyle=$
$\displaystyle(\left<ai\right>[ib]+\left<aj\right>[jb])\left<bl\right>[la]+(\left<bi\right>[ia]+\left<bj\right>[ja])\left<al\right>[lb]$

Now, using momentum conservation in the spinor notation

$\displaystyle\left<ai(\neq\\{a,b,j,k\\})\right>[i(\neq\\{a,b,j,k\\})b]+\left<aj(\neq\\{a,b,i,k\\})\right>[j(\neq\\{a,b,j,k\\})b]$
$\displaystyle+\left<ak(\neq\\{a,b,i,j\\})\right>[k(\neq\\{a,b,i,j\\})b]=0$
(89)

one can show that,

$\displaystyle\text{tr}[a\\{i+j\\}b\\{l\\}]$ $\displaystyle=$
$\displaystyle-2\left<al\right>[al]\left<bl\right>[bl]$

where each label is different. Thus we see that
$\text{tr}[a\\{i+j\\}b\\{l\\}]$ is independent of $\\{i,j\\}$. The half soft
functions needed for the simplification of the 5-point amplitude are given by,

$\displaystyle h(a,\\{i,j\\},b)$ $\displaystyle=$
$\displaystyle\frac{[ij]}{\left<ij\right>\left<ai\right>\left<aj\right>\left<ib\right>\left<jb\right>}$
(90) $\displaystyle h(a,\\{i\\},b)$ $\displaystyle=$
$\displaystyle\frac{1}{\left<ai\right>^{2}\left<ib\right>^{2}}$ (91)

Thus, we see that $h(a,\\{i,j\\},b)=h(a,\\{j,i\\},b)$.

Now, using the explicit form of the trace and half soft functions in terms of
spinor helicity brackets, we can write (86) as,

$\displaystyle
B_{5}=-8\left[\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}+\frac{[24]\left<13\right>\left<35\right>([13][35])^{3}}{\left<24\right>\left<12\right>\left<14\right>\left<25\right>\left<45\right>}+\frac{[15]\left<23\right>\left<34\right>([23][34])^{3}}{\left<15\right>\left<21\right>\left<25\right>\left<14\right>\left<54\right>}\right.$
$\displaystyle+\left.\frac{[14]\left<23\right>\left<35\right>([23][35])^{3}}{\left<14\right>\left<21\right>\left<24\right>\left<15\right>\left<45\right>}+\frac{[45]\left<13\right>\left<23\right>([13][23])^{3}}{\left<45\right>\left<14\right>\left<15\right>\left<42\right>\left<52\right>}+\frac{[34]\left<25\right>\left<15\right>([15][25])^{3}}{\left<34\right>\left<13\right>\left<14\right>\left<32\right>\left<42\right>}\right.$
$\displaystyle+\left.\frac{[35]\left<14\right>\left<24\right>([14][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<32\right>\left<52\right>}+\frac{[12]\left<34\right>\left<35\right>([34][35])^{3}}{\left<12\right>\left<41\right>\left<42\right>\left<15\right>\left<25\right>}+\frac{[12]\left<35\right>\left<45\right>([35][45])^{3}}{\left<12\right>\left<31\right>\left<32\right>\left<14\right>\left<24\right>}\right.$
$\displaystyle\left.+\frac{[12]\left<34\right>\left<45\right>([34][45])^{3}}{\left<12\right>\left<31\right>\left<32\right>\left<15\right>\left<25\right>}\right]+\left(1\leftrightarrow
3\right)+\left(2\leftrightarrow 3\right)$ (92)

Before simplifying this, first note that the first 4 terms (and hence a total
of 12 terms) in the above expression have the apparent form that seems to go
like $\sim\frac{1}{\langle 45\rangle}$. However, it cannot be true that the
5-point amplitude has a leading behaviour of $\sim\frac{1}{\langle
45\rangle}$. We will show that these terms add up to contribute to the leading
order ($\mathcal{O}(\frac{[45]}{\langle 45\rangle})$), $\mathcal{O}(1)$ and
higher orders as expected. Hence, to simplify further, let us first write down
these 12 terms explicitly,

$\displaystyle-\frac{B_{5}^{\textnormal{S}}}{8}$ $\displaystyle=$
$\displaystyle\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}+\frac{[35]\left<12\right>\left<24\right>([12][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<34\right>\left<54\right>}+\frac{[24]\left<13\right>\left<35\right>([13][35])^{3}}{\left<24\right>\left<12\right>\left<14\right>\left<25\right>\left<45\right>}$
$\displaystyle+$
$\displaystyle\frac{[34]\left<12\right>\left<25\right>([12][25])^{3}}{\left<34\right>\left<13\right>\left<14\right>\left<35\right>\left<45\right>}+\frac{[15]\left<23\right>\left<34\right>([23][34])^{3}}{\left<15\right>\left<21\right>\left<25\right>\left<14\right>\left<54\right>}+\frac{[35]\left<12\right>\left<14\right>([12][14])^{3}}{\left<35\right>\left<23\right>\left<25\right>\left<34\right>\left<54\right>}$
$\displaystyle+$
$\displaystyle\frac{[14]\left<23\right>\left<35\right>([23][35])^{3}}{\left<14\right>\left<21\right>\left<24\right>\left<15\right>\left<45\right>}+\frac{[34]\left<12\right>\left<15\right>([12][15])^{3}}{\left<34\right>\left<23\right>\left<24\right>\left<35\right>\left<45\right>}+\frac{[15]\left<23\right>\left<24\right>([23][24])^{3}}{\left<15\right>\left<31\right>\left<35\right>\left<14\right>\left<54\right>}$
$\displaystyle+$
$\displaystyle\frac{[25]\left<13\right>\left<14\right>([13][14])^{3}}{\left<25\right>\left<32\right>\left<35\right>\left<24\right>\left<54\right>}+\frac{[14]\left<23\right>\left<25\right>([23][25])^{3}}{\left<14\right>\left<31\right>\left<34\right>\left<15\right>\left<45\right>}+\frac{[24]\left<13\right>\left<15\right>([13][15])^{3}}{\left<24\right>\left<32\right>\left<34\right>\left<25\right>\left<45\right>}$

Keeping terms only upto $\mathcal{O}(\bar{z}_{45})$, the first term above can
be rewritten as,

$\begin{gathered}\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}=-\frac{1}{\left<45\right>}\frac{[25]\left<12\right>^{3}\left<24\right>^{3}[12]^{3}[24]^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<13\right>^{2}\left<34\right>^{2}}+3\frac{[45]}{\left<45\right>}\frac{[25]\left<12\right>^{2}\left<15\right>\left<24\right>^{3}[12]^{3}[24]^{2}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<13\right>^{2}\left<34\right>^{2}}\\\
+3\frac{[25]\left<12\right>^{3}\left<24\right>^{2}[12]^{2}[15][24]^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<13\right>^{2}\left<34\right>^{2}}-3\left<45\right>\frac{[25]\left<12\right>^{2}[12][15]^{2}[24]^{3}}{\left<25\right>\left<15\right>\left<13\right>^{2}\left<34\right>^{2}}-9\left[45\right]\frac{[25]\left<12\right>\left<24\right>[12]^{2}[15][24]^{2}}{\left<25\right>\left<13\right>^{2}\left<34\right>^{2}}\end{gathered}$
(94)

Now we use a little trick to explicitly show that the terms in
(LABEL:sing_terms) add up to give ($\mathcal{O}(\frac{[45]}{\langle
45\rangle})$), $\mathcal{O}(1)$ and higher orders contributions. It involves
appropriately combining terms in the equation. To see this, note that the
first term in RHS of (94) and second term in RHS of (LABEL:sing_terms) can be
combined to get,

$\begin{gathered}-\frac{1}{\left<45\right>}\frac{[25]\left<12\right>^{3}\left<24\right>^{3}[12]^{3}[24]^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<13\right>^{2}\left<34\right>^{2}}+\frac{[35]\left<12\right>\left<24\right>([12][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<34\right>\left<54\right>}\\\
=-\frac{1}{\left<45\right>}\frac{\left<12\right>\left<24\right>([12][24])^{3}}{\left<13\right>^{2}\left<15\right>\left<25\right>\left<34\right>^{2}\left<35\right>}\left(\left<12\right>\left<24\right>\left<35\right>[25]+\left<13\right>\left<34\right>\left<25\right>[35]\right)\end{gathered}$
(95)

Note that although we are writing equalities everywhere, one should keep in
mind that there are higher order terms as well. However, here, and throughout
this paper, we will always write expressions keeping terms only upto
$\mathcal{O}(\bar{z}_{45})$. Now, using the Shouten Identity
$\left<24\right>\left<35\right>=\left<25\right>\left<34\right>+\left<23\right>\left<45\right>$
and momentum conservation equation, we can write the above equation as

$\begin{gathered}-\frac{1}{\left<45\right>}\frac{[25]\left<12\right>^{3}\left<24\right>^{3}[12]^{3}[24]^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<13\right>^{2}\left<34\right>^{2}}+\frac{[35]\left<12\right>\left<24\right>([12][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<34\right>\left<54\right>}\\\
=-\frac{1}{\left<45\right>}\frac{\left<12\right>\left<24\right>([12][24])^{3}}{\left<13\right>^{2}\left<15\right>\left<25\right>\left<34\right>^{2}\left<35\right>}\left(-\left<14\right>\left<25\right>\left<34\right>[45]+\left<12\right>\left<23\right>\left<45\right>[25]\right)\end{gathered}$
(96)

Hence the first two terms in (LABEL:sing_terms) give

$\begin{gathered}\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}+\frac{[35]\left<12\right>\left<24\right>([12][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<34\right>\left<54\right>}=3\frac{[45]}{\left<45\right>}\frac{[25]\left<12\right>\left<24\right>^{2}[12]^{3}[24]^{2}}{\left<25\right>\left<13\right>^{2}\left<34\right>^{2}}\\\
+3\frac{[25]\left<12\right>^{2}\left<24\right>[12]^{2}[15][24]^{3}}{\left<25\right>\left<15\right>\left<13\right>^{2}\left<34\right>^{2}}+\frac{[45]}{\left<45\right>}\frac{\left<12\right>\left<14\right>\left<24\right>[12]^{3}[24]^{3}}{\left<13\right>^{2}\left<15\right>\left<34\right>\left<35\right>}-\frac{\left<12\right>^{2}\left<23\right>\left<24\right>[25][12]^{3}[24]^{3}}{\left<13\right>^{2}\left<15\right>\left<25\right>\left<34\right>^{2}\left<35\right>}\\\
-3\left<45\right>\frac{[25]\left<12\right>^{2}[12][15]^{2}[24]^{3}}{\left<25\right>\left<15\right>\left<13\right>^{2}\left<34\right>^{2}}-9\left[45\right]\frac{[25]\left<12\right>\left<24\right>[12]^{2}[15][24]^{2}}{\left<25\right>\left<13\right>^{2}\left<34\right>^{2}}\end{gathered}$
(97)

Using momentum conservation again in the 4th term in the RHS of the above
equation, we finally get

$\begin{gathered}\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}+\frac{[35]\left<12\right>\left<24\right>([12][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<34\right>\left<54\right>}=3\frac{[45]}{\left<45\right>}\frac{[25]\left<12\right>\left<24\right>^{2}[12]^{3}[24]^{2}}{\left<25\right>\left<13\right>^{2}\left<34\right>^{2}}\\\
+\frac{[45]}{\left<45\right>}\frac{\left<12\right>\left<14\right>\left<24\right>[12]^{3}[24]^{3}}{\left<13\right>^{2}\left<15\right>\left<34\right>\left<35\right>}+2\frac{[25]\left<12\right>^{2}\left<24\right>[12]^{2}[15][24]^{3}}{\left<13\right>^{2}\left<25\right>\left<15\right>\left<34\right>^{2}}-\frac{[25]\left<12\right>^{2}\left<24\right>[12]^{2}[14][24]^{3}}{\left<13\right>^{2}\left<15\right>\left<25\right>\left<34\right>\left<35\right>}\\\
-3\left<45\right>\frac{[25]\left<12\right>^{2}[12][15]^{2}[24]^{3}}{\left<25\right>\left<15\right>\left<13\right>^{2}\left<34\right>^{2}}-9\left[45\right]\frac{[25]\left<12\right>\left<24\right>[12]^{2}[15][24]^{2}}{\left<25\right>\left<13\right>^{2}\left<34\right>^{2}}\end{gathered}$
(98)

Similarly, the 3rd and 4th terms in RHS of (LABEL:sing_terms) give

$\begin{gathered}\frac{[24]\left<13\right>\left<35\right>([13][35])^{3}}{\left<24\right>\left<12\right>\left<14\right>\left<25\right>\left<45\right>}+\frac{[34]\left<12\right>\left<25\right>([12][25])^{3}}{\left<34\right>\left<13\right>\left<14\right>\left<35\right>\left<45\right>}=3\frac{[45]}{\left<45\right>}\frac{[24]\left<12\right>\left<25\right>^{2}[12]^{3}[25]^{2}}{\left<24\right>\left<13\right>^{2}\left<35\right>^{2}}\\\
+\frac{[45]}{\left<45\right>}\frac{\left<12\right>\left<15\right>\left<25\right>[12]^{3}[25]^{3}}{\left<13\right>^{2}\left<14\right>\left<34\right>\left<35\right>}+2\frac{[24]\left<12\right>^{2}\left<25\right>[12]^{2}[14][25]^{3}}{\left<24\right>\left<14\right>\left<13\right>^{2}\left<35\right>^{2}}-\frac{[24]\left<12\right>^{2}\left<25\right>[12]^{2}[15][25]^{3}}{\left<13\right>^{2}\left<14\right>\left<24\right>\left<34\right>\left<35\right>}\\\
+3\left<45\right>\frac{[24]\left<12\right>^{2}[12][14]^{2}[25]^{3}}{\left<24\right>\left<14\right>\left<13\right>^{2}\left<35\right>^{2}}+9\left[45\right]\frac{[24]\left<12\right>\left<25\right>[12]^{2}[14][25]^{2}}{\left<24\right>\left<13\right>^{2}\left<35\right>^{2}}\end{gathered}$
(99)

As is clear from the above equations, we can combine the 12 terms of
(LABEL:sing_terms) in groups of two as shown above to see that the leading
order contribution coming from (LABEL:sing_terms) is indeed
$\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})$ instead of the apparent
$\mathcal{O}(\frac{1}{z_{45}})$.

Now, we rewrite the first 4 terms in (LABEL:sing_terms) in terms of
$\\{\omega_{i},z_{i},\bar{z}_{i}\\}$, and then expand around
$z_{45}=\bar{z}_{45}=0$. As mentioned earlier, we only keep terms upto
$\mathcal{O}(\bar{z}_{45})$ to get

$\begin{gathered}\frac{[25]\left<13\right>\left<34\right>([13][34])^{3}}{\left<25\right>\left<12\right>\left<15\right>\left<24\right>\left<54\right>}+\frac{[35]\left<12\right>\left<24\right>([12][24])^{3}}{\left<35\right>\left<13\right>\left<15\right>\left<34\right>\left<54\right>}+\frac{[24]\left<13\right>\left<35\right>([13][35])^{3}}{\left<24\right>\left<12\right>\left<14\right>\left<25\right>\left<45\right>}+\frac{[34]\left<12\right>\left<25\right>([12][25])^{3}}{\left<34\right>\left<13\right>\left<14\right>\left<35\right>\left<45\right>}\\\
=2^{4}\epsilon_{1}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}\omega_{4}\omega_{5}}\left(\omega_{4}+\omega_{5}\right)^{3}\frac{\bar{z}_{45}}{z_{45}}\frac{z_{12}z_{25}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{35}^{2}}\\\
-2^{4}\epsilon_{1}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}\omega_{4}\omega_{5}}\left[\left(\omega_{4}+\omega_{5}\right)^{3}-5\omega_{4}\omega_{5}\left(\omega_{4}+\omega_{5}\right)\right]\frac{z_{12}^{2}\bar{z}_{12}^{2}\bar{z}_{15}\bar{z}_{25}^{4}}{z_{13}^{2}z_{15}z_{35}^{2}}\\\
+2^{4}\epsilon_{1}\bar{z}_{45}\frac{z_{12}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{15}z_{35}^{3}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}\omega_{4}\omega_{5}}\left[-z_{15}z_{35}\,\omega_{4}\left(\omega_{4}^{2}+6\omega_{4}\omega_{5}-3\omega_{5}^{2}\right)+z_{25}z_{35}\left(-\omega_{4}^{3}+\omega_{5}^{3}\right)\right.\\\
\left.+z_{15}z_{25}\left(\omega_{4}^{3}+6\omega_{4}^{2}\omega_{5}+\omega_{5}^{3}\right)\right]+2^{4}\epsilon_{1}\bar{z}_{45}\frac{z_{12}^{2}\bar{z}_{12}^{2}\bar{z}_{25}^{3}}{z_{13}^{2}z_{15}z_{35}^{2}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}\omega_{4}\omega_{5}}\left[\bar{z}_{25}\,\omega_{4}\left(\omega_{4}^{2}-2\omega_{5}^{2}\right)\right.\\\
\left.+\bar{z}_{15}\left(3\omega_{4}^{3}-6\omega_{4}^{2}\omega_{5}-2\omega_{4}\omega_{5}^{2}+\omega_{5}^{3}\right)\right]+\cdots\end{gathered}$
(100)

The contribution from the other 8 terms in (LABEL:sing_terms) is simply
obtained by taking different permutations of 1,2 and 3 in the above
expression. Setting $\omega_{4}=t\omega_{P},\ \omega_{5}=(1-t)\omega_{P}$ and
collecting all the singular terms we finally get

$B_{5}^{\textnormal{S}}=-2^{7}\frac{\omega_{P}}{t(1-t)}\left(\frac{\bar{z}_{45}}{z_{45}}T^{\textnormal{S}}_{L}+T^{\textnormal{S}}_{\mathcal{O}(1)}+\bar{z}_{45}T^{\textnormal{S}}_{\bar{z}}\right)$
(101)

where

$\begin{gathered}T^{\textnormal{S}}_{L}=\left[\epsilon_{1}\frac{z_{12}z_{25}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{35}^{2}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}}+\epsilon_{2}\frac{z_{12}z_{15}\bar{z}_{12}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{35}^{2}}\frac{\omega_{1}^{4}\omega_{2}}{\omega_{3}^{2}}+\epsilon_{3}\frac{z_{13}z_{15}\bar{z}_{13}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{25}^{2}}\frac{\omega_{1}^{4}\omega_{3}}{\omega_{2}^{2}}\right]\end{gathered}$
(102)
$\begin{gathered}T^{\textnormal{S}}_{\mathcal{O}(1)}=-\left[\epsilon_{1}\frac{z_{12}^{2}\bar{z}_{12}^{2}\bar{z}_{15}\bar{z}_{25}^{4}}{z_{13}^{2}z_{15}z_{35}^{2}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}}+\epsilon_{2}\frac{z_{12}^{2}\bar{z}_{12}^{2}\bar{z}_{25}\bar{z}_{15}^{4}}{z_{23}^{2}z_{25}z_{35}^{2}}\frac{\omega_{1}^{4}\omega_{2}}{\omega_{3}^{2}}+\epsilon_{3}\frac{z_{13}^{2}\bar{z}_{13}^{2}\bar{z}_{35}\bar{z}_{15}^{4}}{z_{23}^{2}z_{35}z_{25}^{2}}\frac{\omega_{1}^{4}\omega_{3}}{\omega_{2}^{2}}\right][1-5t(1-t)]\end{gathered}$
(103)

and

$\begin{gathered}T^{\textnormal{S}}_{\bar{z}}=\epsilon_{1}\frac{z_{12}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{15}z_{35}^{3}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}}\left[-z_{15}z_{35}\,t\left(t^{2}+6t(1-t)-3(1-t)^{2}\right)+z_{25}z_{35}\left(-t^{3}+(1-t)^{3}\right)\right.\\\
\left.+z_{15}z_{25}\left(t^{3}+6t^{2}(1-t)+(1-t)^{3}\right)\right]+\epsilon_{1}\frac{z_{12}^{2}\bar{z}_{12}^{2}\bar{z}_{25}^{3}}{z_{13}^{2}z_{15}z_{35}^{2}}\frac{\omega_{1}\omega_{2}^{4}}{\omega_{3}^{2}}\left[\bar{z}_{25}\,t\left(t^{2}-2(1-t)^{2}\right)\right.\\\
\left.+\bar{z}_{15}\left(3t^{3}-6t^{2}(1-t)-2t(1-t)^{2}+(1-t)^{3}\right)\right]\\\
+\epsilon_{2}\frac{z_{12}\bar{z}_{12}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{25}z_{35}^{3}}\frac{\omega_{1}^{4}\omega_{2}}{\omega_{3}^{2}}\left[-z_{25}z_{35}\,t\left(t^{2}+6t(1-t)-3(1-t)^{2}\right)+z_{15}z_{35}\left(-t^{3}+(1-t)^{3}\right)\right.\\\
\left.+z_{25}z_{15}\left(t^{3}+6t^{2}(1-t)+(1-t)^{3}\right)\right]+\epsilon_{2}\frac{z_{12}^{2}\bar{z}_{12}^{2}\bar{z}_{15}^{3}}{z_{23}^{2}z_{25}z_{35}^{2}}\frac{\omega_{1}^{4}\omega_{2}}{\omega_{3}^{2}}\left[\bar{z}_{15}\,t\left(t^{2}-2(1-t)^{2}\right)\right.\\\
\left.+\bar{z}_{25}\left(3t^{3}-6t^{2}(1-t)-2t(1-t)^{2}+(1-t)^{3}\right)\right]\\\
+\epsilon_{3}\frac{z_{13}\bar{z}_{13}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{35}z_{25}^{3}}\frac{\omega_{1}^{4}\omega_{3}}{\omega_{2}^{2}}\left[-z_{35}z_{25}\,t\left(t^{2}+6t(1-t)-3(1-t)^{2}\right)+z_{15}z_{25}\left(-t^{3}+(1-t)^{3}\right)\right.\\\
\left.+z_{35}z_{15}\left(t^{3}+6t^{2}(1-t)+(1-t)^{3}\right)\right]+\epsilon_{3}\frac{z_{13}^{2}\bar{z}_{13}^{2}\bar{z}_{15}^{3}}{z_{23}^{2}z_{35}z_{25}^{2}}\frac{\omega_{1}^{4}\omega_{3}}{\omega_{2}^{2}}\left[\bar{z}_{15}\,t\left(t^{2}-2(1-t)^{2}\right)\right.\\\
\left.+\bar{z}_{35}\left(3t^{3}-6t^{2}(1-t)-2t(1-t)^{2}+(1-t)^{3}\right)\right]\end{gathered}$
(104)

Taking into account the other 18 terms (although note that at
$\mathcal{O}(\bar{z}_{45})$, only 12 of these contribute and the 9th and 10th
term in (92) and the ($1\leftrightarrow 3$) and ($2\leftrightarrow 3$)
permutation of those do not contribute at this order) in (92) we finally get
$B_{5}$ as

$B_{5}=-2^{7}\frac{\omega_{P}}{t(1-t)}\left(\frac{\bar{z}_{45}}{z_{45}}T_{L}+T_{\mathcal{O}(1)}+\bar{z}_{45}T_{\bar{z}}\right)+\cdots$
(105)

where we have neglected the higher order terms in the expansion of the RHS of
(92) around $z_{45}=\bar{z}_{45}=0$ in $\left(\omega,z,\bar{z}\right)$ space
and

$\begin{split}T_{L}&=T_{L}^{\textnormal{S}}+\left[\epsilon_{1}\epsilon_{2}\frac{\omega_{1}\omega_{2}\omega_{3}^{4}}{\omega_{P}^{3}}\frac{z_{13}z_{23}\left(\bar{z}_{13}\bar{z}_{23}\right)^{3}}{z_{15}^{2}z_{25}^{2}}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\\\
T_{\mathcal{O}(1)}&=T^{\textnormal{S}}_{\mathcal{O}(1)}+\left[\epsilon_{1}\epsilon_{2}\epsilon_{3}\frac{\omega_{1}\omega_{2}\omega_{P}^{2}}{\omega_{3}}\frac{\bar{z}_{35}(\bar{z}_{15}\bar{z}_{25})^{3}}{z_{35}z_{13}z_{23}}\\{(1-t)^{5}+t^{5}\\}+\epsilon_{1}\epsilon_{2}\frac{\omega_{P}\omega_{3}^{4}}{\omega_{1}\omega_{2}}\frac{\bar{z}_{12}z_{35}^{2}\bar{z}_{35}^{6}}{z_{12}z_{15}^{2}z_{25}^{2}}t^{2}(1-t)^{2}\right.\\\
&\left.\hskip 170.71652pt+(1\leftrightarrow 3)+(2\leftrightarrow 3)\right]\\\
T_{\bar{z}}&=T^{\textnormal{S}}_{\bar{z}}+\left[\epsilon_{1}\epsilon_{2}\frac{\omega_{1}\omega_{2}\omega_{3}^{4}}{\omega_{P}^{3}}\frac{z_{13}z_{23}\left(\bar{z}_{13}\bar{z}_{23}\right)^{3}}{z_{15}^{2}z_{25}^{2}}\left(\frac{1}{z_{15}}+\frac{1}{z_{25}}\right)+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\\\
&-\left[\epsilon_{1}\epsilon_{2}\epsilon_{3}\frac{\omega_{1}\omega_{2}\omega_{P}^{2}}{\omega_{3}}\frac{(\bar{z}_{15}\bar{z}_{25})^{3}}{z_{13}z_{23}z_{35}}(1-t)^{5}+3\epsilon_{1}\epsilon_{2}\epsilon_{3}\frac{\omega_{1}\omega_{2}\omega_{P}^{2}}{\omega_{3}}\frac{\bar{z}_{35}(\bar{z}_{15}\bar{z}_{25})^{2}(\bar{z}_{15}+\bar{z}_{25})}{z_{13}z_{23}z_{35}}t^{5}\right.\\\
&\left.\hskip 113.81102pt+\
\epsilon_{1}\epsilon_{2}\frac{\omega_{3}^{4}\omega_{P}}{\omega_{1}\omega_{2}}\frac{\bar{z}_{12}z_{35}^{2}\bar{z}_{35}^{5}}{z_{12}z_{15}^{2}z_{25}^{2}}t^{2}(1-t)^{2}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\end{split}$ (106)

We will now Mellin transform (105) and take the OPE limit $4\to 5$. We want to
emphasize here that after Mellin transformation, the higher order terms in the
OPE expansion of the Mellin amplitude may receive contribution from the lower
order terms. This is because of the fact that, after Mellin transformation the
Mellin amplitude will depend on $\tilde{\omega}^{*}_{i}$’s as well as the
delta function involving cross ratios coming from 5-point momentum conserving
delta function as parametrized in (LABEL:5pt_delta_fn). In the next subsection
we analyse this in detail and calculate the order by order terms in the OPE
expansion $4\to 5$ of the 5-point Mellin amplitude.

##  Appendix E Mellin Transformation of the 5-Point Amplitude

For the discussion of this Appendix, the prefactor $\frac{i}{(4\pi)^{2}960}$
in (23) is not important. Thus we only Mellin transform $B_{5}$ and keep terms
only upto $\mathcal{O}(\bar{z}_{45})$. Substituting (105) in (28) we get:

$\displaystyle\widetilde{B}_{5}=-2^{7}\int_{0}^{\infty}\prod_{i=1}^{5}d\omega_{i}\,\omega_{i}^{\Delta_{i}-1}e^{-i\sum_{i=1}^{4}\epsilon_{i}\omega_{i}u_{i5}}\frac{\omega_{P}}{t(1-t)}\left(\frac{\bar{z}_{45}}{z_{45}}T_{L}(\omega_{1},\omega_{2},\omega_{3},\omega_{P})+T_{\mathcal{O}(1)}(\omega_{1},\omega_{2},\omega_{3},\omega_{P})\right.$
$\displaystyle\left.+\bar{z}_{45}T_{\bar{z}}(\omega_{1},\omega_{2},\omega_{3},\omega_{P})\right)\delta^{(4)}\left(\sum_{i=1}^{5}\epsilon_{i}\omega_{i}q_{i}\right)\
\ \ \ $ (107)

where $T_{L},\ T_{\mathcal{O}(1)}$ and $\ T_{\bar{z}}$ are given by (106) and
we have kept their $\\{\omega\\}$ dependence explicit for our convenience.
Also we have used momentum conservation in the exponential. Now using the
parametrization (LABEL:5pt_delta_fn), we can perform the
$(\omega_{1},\omega_{2},\omega_{3})$ integrals to obtain:

$\begin{gathered}\widetilde{B}_{5}=-2^{5}\int_{0}^{1}dt\,t^{\Delta_{4}-2}(1-t)^{\Delta_{5}-2}\int_{0}^{\infty}d\omega_{P}\,\omega_{P}^{\Delta_{4}+\Delta_{5}-1}\prod_{i=1}^{3}(\omega^{*}_{i})^{\Delta_{i}-1}e^{-i\sum_{i=1}^{3}\epsilon_{i}\omega^{*}_{i}u_{i5}-i\omega_{P}t\,u_{45}}\\\
\times\left(\frac{\bar{z}_{45}}{z_{45}}T_{L}(\omega^{*}_{1},\omega^{*}_{2},\omega^{*}_{3},\omega_{P})+T_{\mathcal{O}(1)}(\omega^{*}_{1},\omega^{*}_{2},\omega^{*}_{3},\omega_{P})+\bar{z}_{45}T_{\bar{z}}(\omega^{*}_{1},\omega^{*}_{2},\omega^{*}_{3},\omega_{P})\right)\\\
\times\delta\left(x-\bar{x}-tz_{45}\left(\frac{x}{z_{35}}-\frac{\bar{x}}{z_{25}}\right)-t\bar{z}_{45}\left(\frac{x}{\bar{z}_{25}}-\frac{\bar{x}}{\bar{z}_{35}}\right)+tz_{45}\bar{z}_{45}\left(\frac{x}{z_{35}\bar{z}_{25}}-\frac{\bar{x}}{z_{25}\bar{z}_{35}}\right)\right)\end{gathered}$
(108)

Now from (78) and the explicit expressions of $T_{L},\ T_{\mathcal{O}(1)}$ and
$T_{\bar{z}}$ given by (106) one can see that

$\begin{split}T_{L}(\omega^{*}_{1},\omega^{*}_{2},\omega^{*}_{3},\omega_{P})&=\omega_{P}^{3}\,\mathcal{T}_{L}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})\\\
T_{\mathcal{O}(1)}(\omega^{*}_{1},\omega^{*}_{2},\omega^{*}_{3},\omega_{P})&=\omega_{P}^{3}\,\mathcal{T}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})\\\
T_{\bar{z}}(\omega^{*}_{1},\omega^{*}_{2},\omega^{*}_{3},\omega_{P})&=\omega_{P}^{3}\,\mathcal{T}_{\bar{z}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})\end{split}$
(109)

where

$\begin{split}\mathcal{T}_{L}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})&=\mathcal{T}_{L}^{\textnormal{S}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})+\left[\epsilon_{1}\epsilon_{2}\,\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}(\tilde{\omega}^{*}_{3})^{4}\frac{z_{13}z_{23}\left(\bar{z}_{13}\bar{z}_{23}\right)^{3}}{z_{15}^{2}z_{25}^{2}}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\\\
\mathcal{T}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})&=\mathcal{T}^{\textnormal{S}}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})+\left[\epsilon_{1}\epsilon_{2}\epsilon_{3}\frac{\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}}{\tilde{\omega}^{*}_{3}}\frac{\bar{z}_{35}(\bar{z}_{15}\bar{z}_{25})^{3}}{z_{35}z_{13}z_{23}}\\{(1-t)^{5}+t^{5}\\}\right.\\\
&\left.\hskip
85.35826pt+\epsilon_{1}\epsilon_{2}\frac{(\tilde{\omega}^{*}_{3})^{4}}{\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}}\frac{\bar{z}_{12}z_{35}^{2}\bar{z}_{35}^{6}}{z_{12}z_{15}^{2}z_{25}^{2}}t^{2}(1-t)^{2}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\\\
\mathcal{T}_{\bar{z}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})&=\mathcal{T}^{\textnormal{S}}_{\bar{z}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})+\left[\epsilon_{1}\epsilon_{2}\,\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}(\tilde{\omega}^{*}_{3})^{4}\frac{z_{13}z_{23}\left(\bar{z}_{13}\bar{z}_{23}\right)^{3}}{z_{15}^{2}z_{25}^{2}}\left(\frac{1}{z_{15}}+\frac{1}{z_{25}}\right)+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\\\
&-\left[\epsilon_{1}\epsilon_{2}\epsilon_{3}\frac{\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}}{\tilde{\omega}^{*}_{3}}\frac{(\bar{z}_{15}\bar{z}_{25})^{3}}{z_{13}z_{23}z_{35}}(1-t)^{5}+3\epsilon_{1}\epsilon_{2}\epsilon_{3}\frac{\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}}{\tilde{\omega}^{*}_{3}}\frac{\bar{z}_{35}(\bar{z}_{15}\bar{z}_{25})^{2}(\bar{z}_{15}+\bar{z}_{25})}{z_{13}z_{23}z_{35}}t^{5}\right.\\\
&\left.\hskip 113.81102pt+\
\epsilon_{1}\epsilon_{2}\frac{(\tilde{\omega}^{*}_{3})^{4}}{\tilde{\omega}^{*}_{1}\tilde{\omega}^{*}_{2}}\frac{\bar{z}_{12}z_{35}^{2}\bar{z}_{35}^{5}}{z_{12}z_{15}^{2}z_{25}^{2}}t^{2}(1-t)^{2}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\end{split}$ (110)

and
$\mathcal{T}_{L}^{\textnormal{S}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3}),\mathcal{T}^{\textnormal{S}}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})$
and
$\mathcal{T}^{\textnormal{S}}_{\bar{z}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})$
are given by (102), (103) and (104) respectively with
$\\{\omega_{1},\omega_{2},\omega_{3}\\}$ replaced by
$\\{\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3}\\}$.
Now we can perform the $\omega_{P}$ integral in (108) and obtain:

$\begin{gathered}\widetilde{B}_{5}=-2^{5}\Gamma(\Delta)\int_{0}^{1}dt\,t^{\Delta_{4}-2}(1-t)^{\Delta_{5}-2}\frac{\prod_{i=1}^{3}(\tilde{\omega}^{*}_{i})^{\Delta_{i}-1}}{\left[i\left(\sum_{i=1}^{3}\epsilon_{i}\tilde{\omega}^{*}_{i}u_{i5}+t\,u_{45}\right)\right]^{\Delta}}\\\
\times\left(\frac{\bar{z}_{45}}{z_{45}}\mathcal{T}_{L}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})+\mathcal{T}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})+\bar{z}_{45}\mathcal{T}_{\bar{z}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})\right)\\\
\times\delta\left(x-\bar{x}-tz_{45}\left(\frac{x}{z_{35}}-\frac{\bar{x}}{z_{25}}\right)-t\bar{z}_{45}\left(\frac{x}{\bar{z}_{25}}-\frac{\bar{x}}{\bar{z}_{35}}\right)+tz_{45}\bar{z}_{45}\left(\frac{x}{z_{35}\bar{z}_{25}}-\frac{\bar{x}}{z_{25}\bar{z}_{35}}\right)\right)\end{gathered}$
(111)

where $\Delta=\sum_{i=1}^{5}\Delta_{i}$. We now expand the above equation
around $z_{45}=\bar{z}_{45}=u_{45}=0$.

###  E.1 Evaluating the Leading Order Contribution

It is clear from (111) that the leading order term goes as
$\sim\frac{\bar{z}_{45}}{z_{45}}$ and the contribution to the leading order
can come only from the term containing
$\mathcal{T}_{L}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})$.
At leading order we have $\tilde{\omega}^{*}_{i}=\epsilon_{i}\sigma_{i,1}$.
Thus, the leading order term of $\widetilde{B}_{5}$ is given by:

$\begin{gathered}\widetilde{B}_{5}|_{\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})}=-2^{5}\frac{\bar{z}_{45}}{z_{45}}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\int_{0}^{1}dt\,t^{\Delta_{4}-2}(1-t)^{\Delta_{5}-2}\mathcal{T}_{L}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})\delta\left(x-\bar{x}\right)\end{gathered}$
(112)

where $\mathcal{D}=\left(\sum_{i=1}^{3}\sigma_{i,1}u_{i5}\right)$. From (102)
and the first equation of (106) we have

$\begin{gathered}\mathcal{T}_{L}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})=\left[\frac{z_{12}z_{25}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{35}^{2}}\frac{\sigma_{1,1}\sigma_{2,1}^{4}}{\sigma_{3,1}^{2}}+\frac{z_{12}z_{15}\bar{z}_{12}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{35}^{2}}\frac{\sigma_{1,1}^{4}\sigma_{2,1}}{\sigma_{3,1}^{2}}+\frac{z_{13}z_{15}\bar{z}_{13}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{25}^{2}}\frac{\sigma_{1,1}^{4}\sigma_{3,1}}{\sigma_{2,1}^{2}}\right]\\\
+\left[\sigma_{1,1}\sigma_{2,1}\sigma_{3,1}^{4}\frac{z_{13}z_{23}\left(\bar{z}_{13}\bar{z}_{23}\right)^{3}}{z_{15}^{2}z_{25}^{2}}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]\end{gathered}$ (113)

Now, using (69)-(72), one can show that

$\begin{split}\frac{z_{12}z_{25}\bar{z}_{12}^{3}\bar{z}_{25}^{3}}{z_{13}^{2}z_{35}^{2}}\frac{\sigma_{1,1}\sigma_{2,1}^{4}}{\sigma_{3,1}^{2}}&=\frac{z_{12}z_{13}(\bar{z}_{12}\bar{z}_{13})^{3}}{z_{25}^{2}z_{35}^{2}}\sigma_{1,1}^{4}\sigma_{2,1}\sigma_{3,1}\\\
\frac{z_{12}z_{15}\bar{z}_{12}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{35}^{2}}\frac{\sigma_{1,1}^{4}\sigma_{2,1}}{\sigma_{3,1}^{2}}&=\frac{z_{12}z_{23}(\bar{z}_{12}\bar{z}_{23})^{3}}{z_{15}^{2}z_{35}^{2}}\sigma_{1,1}\sigma_{2,1}^{4}\sigma_{3,1}\\\
\frac{z_{13}z_{15}\bar{z}_{13}^{3}\bar{z}_{15}^{3}}{z_{23}^{2}z_{25}^{2}}\frac{\sigma_{1,1}^{4}\sigma_{3,1}}{\sigma_{2,1}^{2}}&=\frac{z_{13}z_{23}(\bar{z}_{13}\bar{z}_{23})^{3}}{z_{15}^{2}z_{25}^{2}}\sigma_{1,1}\sigma_{2,1}\sigma_{3,1}^{4}\end{split}$
(114)

Using the above relations, we can simplify (113) to get

$\displaystyle\mathcal{T}_{L}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})$
$\displaystyle=2\left[\sigma_{1,1}\sigma_{2,1}\sigma_{3,1}^{4}\frac{z_{13}z_{23}\left(\bar{z}_{13}\bar{z}_{23}\right)^{3}}{z_{15}^{2}z_{25}^{2}}+(1\leftrightarrow
3)+(2\leftrightarrow 3)\right]$
$\displaystyle=2\left[\mathcal{N}_{4}+\mathcal{N}_{4}(1\leftrightarrow
3)+\mathcal{N}_{4}(2\leftrightarrow 3)\right]$ (115)

where the second equality follows from (22). Since this is independent of $t$,
we can easily carry out the $t$-integral in (112) to get

$\displaystyle\tilde{B}_{5}|_{\mathcal{O}(\frac{\bar{z}_{45}}{z_{45}})}$
$\displaystyle=-2^{6}\frac{\bar{z}_{45}}{z_{45}}B(\Delta_{4}-1,\Delta_{5}-1)\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\delta(x-\bar{x})$
$\displaystyle\times\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\left[\mathcal{N}_{4}+\mathcal{N}_{4}\left(1\leftrightarrow
3\right)+\mathcal{N}_{4}\left(2\leftrightarrow 3\right)\right]$ (116)

This precisely gives the equation (29).

###  E.2 Evaluating the $\mathcal{O}(1)$ Contribution

From (111), we can see that the $\mathcal{O}(1)$ contribution to the 5-point
amplitude essentially comes only from the term containing
$\mathcal{T}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})$
when $\tilde{\omega}^{*}_{i}$’s take their leading order value given by
$\epsilon_{i}\sigma_{i,1}$. Let us write the Mellin integral at order 1:

$\begin{gathered}\widetilde{B}_{5}\big{|}_{\mathcal{O}(1)}=-2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\int_{0}^{1}dt\,t^{\Delta_{4}-2}(1-t)^{\Delta_{5}-2}\mathcal{T}_{\mathcal{O}(1)}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})\delta\left(x-\bar{x}\right)\end{gathered}$
(117)

We will not attempt to take the explicit expressions of
$\mathcal{T}_{\mathcal{O}(1)}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})$
and Mellin integrate it. Rather we will take a different approach which is
more helpful for our purpose of the OPE factorization. Firstly, from the
second equation of (106) we observe that
$\mathcal{T}_{\mathcal{O}(1)}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})$
is a polynomial in $t$ with the highest power being $4$. We use this fact and
write
$\mathcal{T}_{\mathcal{O}(1)}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})$
as

$\mathcal{T}_{\mathcal{O}(1)}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})=\sum_{k=0}^{4}t^{k}\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})$
(118)

The explicit expressions for the functions
$\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})$ can be read out
from the second equation of (106). However, they are not relevant for our
discussions and hence we will not write them explicitly. Using (118), we can
easily evaluate the integral (117) to get,

$\begin{gathered}\widetilde{B}_{5}\big{|}_{\mathcal{O}(1)}=-2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\sum_{k=0}^{4}B(\Delta_{4}+k-1,\Delta_{5}-1)\mathcal{F}^{(1)}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\delta\left(x-\bar{x}\right)\end{gathered}$
(119)

This is the expression we have used in section (5.5.2).

###  E.3 Evaluating the Order $\mathcal{O}(\bar{z}_{45})$ Contribution

We apply the same strategy as the previous section here. However, we have to
be careful now as there will be contributions at $\mathcal{O}(\bar{z}_{45})$
from the lower order terms. Like $\mathcal{O}(1)$ terms, here also we are only
concerned about the $t$-dependence. Before proceeding further let us first
write down the expansion of different components in (111) around
$z_{45}=\bar{z}_{45}=u_{45}=0$. Keeping terms only upto
$\mathcal{O}(\bar{z}_{45})$ we have

$\begin{split}\tilde{\omega}_{i}^{*}&=\epsilon_{i}\left(\sigma_{i,1}+tz_{45}\sigma_{i,2}+t\bar{z}_{45}\sigma_{i,3}\right)\\\
\mathcal{T}_{L}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})&=\mathcal{T}_{L}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})+z_{45}\mathcal{T}_{L}^{(z)}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})+\bar{z}_{45}\mathcal{T}_{L}^{(\bar{z})}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\\\
\mathcal{T}_{\mathcal{O}(1)}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})&=\mathcal{T}_{\mathcal{O}(1)}(\epsilon_{1}\sigma_{1,1},\epsilon\sigma_{2,1},\epsilon_{3}\sigma_{3,1})+z_{45}\mathcal{T}_{\mathcal{O}(1)}^{(z)}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})+\bar{z}_{45}\mathcal{T}_{\mathcal{O}(1)}^{(\bar{z})}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\\\
\mathcal{T}_{\bar{z}}(\tilde{\omega}^{*}_{1},\tilde{\omega}^{*}_{2},\tilde{\omega}^{*}_{3})&=\mathcal{T}_{\bar{z}}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})+z_{45}\mathcal{T}_{\bar{z}}^{(z)}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})+\bar{z}_{45}\mathcal{T}_{\bar{z}}^{(\bar{z})}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\end{split}$
(120)

The explicit expressions for different $\mathcal{T}$’s are not required for
our discussions. For notational convenience, we will not write the arguments
of different $\mathcal{T}$’s and replace
$\mathcal{T}_{L,\mathcal{O}(1),\bar{z}}(\epsilon_{1}\sigma_{1,1},\epsilon_{2}\sigma_{2,1},\epsilon_{3}\sigma_{3,1})$
by $\mathcal{T}^{(0)}_{L,\mathcal{O}(1),\bar{z}}$. Let us first write down all
possible contributions to $\widetilde{B}_{5}$ at $\mathcal{O}(\bar{z}_{45})$.
From (111) we have,

$\begin{gathered}\widetilde{B}_{5}\big{|}_{\mathcal{O}(\bar{z}_{45})}=-2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\int_{0}^{1}dt\,t^{\Delta_{4}-2}(1-t)^{\Delta_{5}-2}\left[\left(\mathcal{T}^{(z)}_{L}+\mathcal{T}^{(\bar{z})}_{\mathcal{O}(1)}+\mathcal{T}^{(0)}_{\bar{z}}\right)\delta\left(x-\bar{x}\right)\right.\\\
\left.-t\left\\{\left(\frac{x}{z_{35}}-\frac{\bar{x}}{z_{25}}\right)\mathcal{T}^{(0)}_{L}+\mathcal{T}^{(0)}_{\mathcal{O}(1)}\left(\frac{x}{\bar{z}_{25}}-\frac{\bar{x}}{\bar{z}_{35}}\right)\right\\}\delta^{\prime}\left(x-\bar{x}\right)\right.\\\
\left.+t\left\\{(\Delta_{1}-1)\frac{\sigma_{1,2}}{\sigma_{1,1}}+(\Delta_{2}-1)\frac{\sigma_{2,2}}{\sigma_{2,1}}+(\Delta_{3}-1)\frac{\sigma_{3,2}}{\sigma_{3,1}}-\Delta\frac{\sum_{i=1}^{3}\sigma_{i,2}u_{i5}}{\mathcal{D}}\right\\}\mathcal{T}^{(0)}_{L}\delta\left(x-\bar{x}\right)\right.\\\
\left.+t\left\\{(\Delta_{1}-1)\frac{\sigma_{1,3}}{\sigma_{1,1}}+(\Delta_{2}-1)\frac{\sigma_{2,3}}{\sigma_{2,1}}+(\Delta_{3}-1)\frac{\sigma_{3,3}}{\sigma_{3,1}}-\Delta\frac{\sum_{i=1}^{3}\sigma_{i,3}u_{i5}}{\mathcal{D}}\right\\}\mathcal{T}^{(0)}_{\mathcal{O}(1)}\delta\left(x-\bar{x}\right)\right]\end{gathered}$
(121)

Now, by expanding the $\mathcal{T}$’s in (106) around $z_{45}=\bar{z}_{45}=0$
and keeping terms only upto $\mathcal{O}(\bar{z}_{45})$, one can check that
all the terms at different orders in the expansion are polynomial of $t$. The
highest degree of polynomial is $5$ and appears in
$\mathcal{T}^{(0)}_{\bar{z}}$ only. All the other $\mathcal{T}$’s have less
power of $t$. Thus we conclude that the terms in the parenthesis $[\cdots]$ in
(121) can be written as a polynomial of $t$ in the following way

$\begin{gathered}\left[\left(\mathcal{T}^{(z)}_{L}+\mathcal{T}^{\bar{z}}_{\mathcal{O}(1)}+\mathcal{T}^{(0)}_{\bar{z}}\right)\delta\left(x-\bar{x}\right)-t\left\\{\left(\frac{x}{z_{35}}-\frac{\bar{x}}{z_{25}}\right)\mathcal{T}^{(0)}_{L}+\mathcal{T}^{(0)}_{\mathcal{O}(1)}\left(\frac{x}{\bar{z}_{25}}-\frac{\bar{x}}{\bar{z}_{35}}\right)\right\\}\delta^{\prime}\left(x-\bar{x}\right)\right.\\\
\left.+t\left\\{(\Delta_{1}-1)\frac{\sigma_{1,2}}{\sigma_{1,1}}+(\Delta_{2}-1)\frac{\sigma_{2,2}}{\sigma_{2,1}}+(\Delta_{3}-1)\frac{\sigma_{3,2}}{\sigma_{3,1}}-\Delta\frac{\sum_{i=1}^{3}\sigma_{i,2}u_{i5}}{\mathcal{D}}\right\\}\mathcal{T}^{(0)}_{L}\delta\left(x-\bar{x}\right)\right.\\\
\left.+t\left\\{(\Delta_{1}-1)\frac{\sigma_{1,3}}{\sigma_{1,1}}+(\Delta_{2}-1)\frac{\sigma_{2,3}}{\sigma_{2,1}}+(\Delta_{3}-1)\frac{\sigma_{3,3}}{\sigma_{3,1}}-\Delta\frac{\sum_{i=1}^{3}\sigma_{i,3}u_{i5}}{\mathcal{D}}\right\\}\mathcal{T}^{(0)}_{\mathcal{O}(1)}\delta\left(x-\bar{x}\right)\right]\\\
=\sum_{k=1}^{5}t^{k}\mathcal{F}^{(\bar{z})}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\end{gathered}$
(122)

where once again the explicit expressions of
$\mathcal{F}^{(\bar{z})}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})$ are not
relevant for our discussions. Substituting (122) in (121) and performing the
$t$-integral, we finally get:

$\begin{gathered}\widetilde{B}_{5}\big{|}_{\mathcal{O}(\bar{z}_{45})}=-2^{5}\frac{\Gamma(\Delta)}{\left(i\mathcal{D}\right)^{\Delta}}\prod_{i=1}^{3}(\epsilon_{i}\sigma_{i,1})^{\Delta_{i}-1}\sum_{k=1}^{5}B(\Delta_{4}+k-1,\Delta_{5}-1)\mathcal{F}^{(\bar{z})}_{k}(\\{\epsilon_{i},z_{i},\bar{z}_{i}\\})\end{gathered}$
(123)

This is the form for the $\mathcal{O}(\bar{z}_{45})$ 5-point amplitude which
we use in the main text of this paper.

##  Appendix F $w$-Algebra Primaries

Let’s start with the universal term in the OPE between two positive helicity
hard gravitons given by,

$G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta}(0,0)=-\frac{1}{z}\sum_{n=0}^{\infty}B(\Delta_{1}-1+n,\Delta-1)\frac{\bar{z}^{n+1}}{n!}\bar{\partial}^{n}G^{+}_{\Delta+\Delta_{1}}(0,0)$
(124)

We now take the conformal soft limit, first by setting
$\Delta_{1}=k+\varepsilon$ and then taking $\varepsilon\to 0$ to get,

$\begin{gathered}\lim_{\varepsilon\to 0}\varepsilon
G^{+}_{k+\varepsilon}(z,\bar{z})G^{+}_{\Delta}(0,0)=-\frac{1}{z}\sum_{n=0}^{\infty}\left[\lim_{\varepsilon\to
0}\varepsilon
B(k-1+n+\varepsilon,\Delta-1)\right]\frac{\bar{z}^{n+1}}{n!}\bar{\partial}^{n}G^{+}_{\Delta+k}(0,0)\\\
\Rightarrow
H^{k}(z,\bar{z})G^{+}_{\Delta}(0,0)=-\frac{1}{z}\sum_{n=0}^{\infty}\left[\lim_{\varepsilon\to
0}\varepsilon
B(k-1+n+\varepsilon,\Delta-1)\right]\frac{\bar{z}^{n+1}}{n!}\bar{\partial}^{n}G^{+}_{\Delta+k}(0,0)\\\
\end{gathered}$ (125)

Next, we mode expand the soft graviton operator $H^{k}(z,\bar{z})$ on the LHS
of the above equation according to (6) and get,

$\begin{gathered}\sum_{m=\frac{k-2}{2}}^{\frac{2-k}{2}}\frac{H^{k}_{m}(z)}{\bar{z}^{m+\frac{k-2}{2}}}G^{+}_{\Delta}(0,0)=-\frac{1}{z}\sum_{n=0}^{\infty}\left[\lim_{\varepsilon\to
0}\varepsilon
B(k-1+n+\varepsilon,\Delta-1)\right]\frac{\bar{z}^{n+1}}{n!}\bar{\partial}^{n}G^{+}_{\Delta+k}(0,0)\end{gathered}$
(126)

By comparing the terms at order $\bar{z}^{n+1}$ on both the sides of the above
equation for $0\leq n\leq 1-k$, we get,

$\begin{gathered}H^{k}_{\frac{2-k}{2}-n-1}(z)G^{+}_{\Delta}(0,0)=-\frac{1}{z}\left[\lim_{\varepsilon\to
0}\varepsilon
B(k-1+n+\varepsilon,\Delta-1)\right]\frac{1}{n!}\bar{\partial}^{n}G^{+}_{\Delta+k}(0,0)\end{gathered}$
(127)

Now we use the holomorphic mode expansion (7) of the currents
$H^{k}_{\frac{2-k}{2}-n-1}(z)$ in the above equation and obtain,

$\begin{gathered}\sum_{\alpha}z^{-\alpha-\frac{k+2}{2}}H^{k}_{\alpha,\frac{2-k}{2}-n-1}G^{+}_{\Delta}(0,0)=-\frac{1}{z}\left[\lim_{\varepsilon\to
0}\varepsilon
B(k-1+n+\varepsilon,\Delta-1)\right]\frac{1}{n!}\bar{\partial}^{n}G^{+}_{\Delta+k}(0,0)\end{gathered}$
(128)

We can see from the above equation, that there is only a simple pole at $z=0$
on the RHS. Thus, the holomorphic singularity structure of the above equation
(128) tells us that the following conditions should hold,

$\begin{gathered}H^{k}_{-\frac{k+2}{2}+m,\frac{2-k}{2}-n-1}G^{+}_{\Delta}(0,0)=-\left[\lim_{\varepsilon\to
0}\varepsilon
B(k-1+n+\varepsilon,\Delta-1)\right]\frac{1}{n!}\bar{\partial}^{n}G^{+}_{\Delta+k}(0,0)\end{gathered}$
(129)

for $m=1$ and

$\begin{gathered}H^{k}_{-\frac{k+2}{2}+m,\frac{2-k}{2}-n-1}G^{+}_{\Delta}(0,0)=0\end{gathered}$
(130)

for $m>1$ and $0\leq n\leq 1-k$ with $k=1,0,-1,\cdots$.
Moreover, from (126), one can see that there is no term on the RHS that goes
like $\bar{z}^{0}$. Thus on the LHS, the coefficients of the $\bar{z}^{0}$
term should also vanish which gives the following condition,

$\begin{gathered}H^{k}_{\frac{2-k}{2}}(z)G^{+}_{\Delta}(0,0)=0\end{gathered}$
(131)

This equation implies

$H^{k}_{-\frac{k+2}{2}+m,\frac{2-k}{2}}G^{+}_{\Delta}(0,0)=0,\ m\geq 1$ (132)

##  Appendix G Transformation of the MHV Null States under ${sl_{2}(R)}_{V}$
and $\overline{sl_{2}(R)}$ Algebras

In this section of the Appendix, we list the transformation properties of all
the MHV null states appearing at different orders of the OPE between two
positive helicity outgoing gravitons under ${sl_{2}(R)}_{V}$ and
$\overline{sl_{2}(R)}$ algebras. Let us first write down their explicit
expressions in terms of the descendants of the $w$-algebra. We first write
down the actions of the $H^{-1}_{\frac{1}{2},\frac{1}{2}}$ on the null states
$\Phi_{k}(\Delta)$ given by (136) and $\Psi_{k}(\Delta)$ given by (137). They
are given by

$\begin{split}H^{-1}_{\frac{1}{2},\frac{1}{2}}\Phi_{k}(\Delta)&=-\frac{1}{2}(k+1)(k+2)\Phi_{k+1}(\Delta-1)-\frac{1}{2}(\Delta+k-3)(\Delta+k-4)\Phi_{k}(\Delta-1)\\\
&+\frac{(-1)^{k}}{k!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta-2)}\Phi_{1}(\Delta-1)\\\
H^{-1}_{\frac{1}{2},\frac{1}{2}}\Psi_{k}(\Delta)&=-\frac{1}{2}(k+2)(k-1)\Psi_{k+1}(\Delta-1)-\frac{1}{2}(\Delta+k-3)(\Delta+k-4)\Psi_{k}(\Delta-1)\\\
&-\frac{(-1)^{k}}{k!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta-2)}\Psi_{1}(\Delta-1)\\\
H^{-1}_{\frac{1}{2},\frac{1}{2}}\Omega_{k}(\Delta)&=\frac{1}{2}(k+1)(k+2)\Omega_{k}(\Delta-1)-\frac{1}{2}(\Delta-4)(\Delta-5)\Omega_{k}(\Delta-1)\\\
&-\frac{1}{2}(k+1)(k+2)\Omega_{k+1}(\Delta-1)\\\
H^{-1}_{\frac{1}{2},\frac{1}{2}}\Pi_{k}(\Delta)&=\frac{1}{2}k(k+1)\Pi_{k}(\Delta-1)-\frac{1}{2}(\Delta-4)(\Delta-5)\Pi_{k}(\Delta-1)\\\
&-\frac{1}{2}(k-1)(k+2)\Pi_{k+1}(\Delta-1)\end{split}$ (133)

The actions of $H^{1}_{-\frac{1}{2},-\frac{1}{2}}$ on the MHV null states are
given by,

$\begin{split}H^{1}_{-\frac{1}{2},-\frac{1}{2}}\Phi_{k}(\Delta)&=-\Phi_{k}(\Delta+1)-\Phi_{k-1}(\Delta+1)\\\
H^{1}_{-\frac{1}{2},-\frac{1}{2}}\Psi_{k}(\Delta)&=-\Psi_{k}(\Delta+1)-\Psi_{k-1}(\Delta+1)\\\
H^{1}_{-\frac{1}{2},-\frac{1}{2}}\Omega_{k}(\Delta)&=-\Omega_{k}(\Delta+1)\\\
H^{1}_{-\frac{1}{2},-\frac{1}{2}}\Pi_{k}(\Delta)&=-\Pi_{k+1}(\Delta+1)\end{split}$
(134)

The actions of $H^{0}_{0,1}$ on the MHV null states are given by,

$\begin{split}H^{0}_{0,1}\Phi_{k}(\Delta)&=0\\\
H^{0}_{0,1}\Psi_{k}(\Delta)&=(k+2)\Phi_{k+1}(\Delta-1)-2\frac{(-1)^{k}}{k!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta-2)}\Phi_{1}(\Delta-1)\\\
H^{0}_{0,1}\Omega_{k}(\Delta)&=0\\\
H^{0}_{0,1}\Pi_{k}(\Delta)&=-(\Delta+k-3)\Omega_{k}(\Delta-1)+(k+2)\Omega_{k+1}(\Delta-1)\end{split}$
(135)

In deriving the above transformation properties, we have used the algebra (8)
and the action of different operators on the primaries given in Appendix F.

##  Appendix H Review of General Structure of $w$-Invariant OPE

It was shown in Banerjee:2023zip , that the OPE between two positive helicity
outgoing graviton primaries of any $w$-invariant theory can always be written
in terms of the MHV OPE’s and its null states. The MHV null states that can
appear at $\mathcal{O}({z^{0}\bar{z}^{0}})$ and $\mathcal{O}({z^{0}\bar{z}})$
are given by Banerjee:2020zlg ; Banerjee:2021cly

$\Phi_{k}(\Delta)=\left[H^{1-k}_{\frac{k-3}{2},\frac{k+1}{2}}\left(-H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)^{k}-\frac{(-1)^{k}}{k!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta-2)}H^{1}_{-\frac{3}{2},\frac{1}{2}}\right]G^{+}_{\Delta-1}$
(136)

and

$\begin{gathered}\Psi_{k}(\Delta)=\bigg{[}H^{-k}_{\frac{k-2}{2},\frac{k}{2}}\left(-H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)^{k+1}-\frac{(-1)^{k}}{k!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta-2)}H^{0}_{-1,0}\left(-H^{1}_{-\frac{1}{2},-\frac{1}{2}}\right)\\\
-\frac{(-1)^{k}k}{(k+1)!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta-3)}H^{1}_{-\frac{3}{2},-\frac{1}{2}}\bigg{]}G^{+}_{\Delta-2}\end{gathered}$
(137)

respectively, where $k=1,2,3,\cdots,\infty$. However, it is more convenient to
work with the new basis defined by

$\begin{gathered}\Omega_{k}(\Delta)=\sum_{n=1}^{k}\frac{1}{(k-n)!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta+n-2)}\Phi_{n}(\Delta)\end{gathered}$
(138)

for the $\mathcal{O}({z^{0}\bar{z}^{0}})$ null states and similarly for the
$\mathcal{O}({z^{0}\bar{z}})$ null states the new basis is defined by,

$\Pi_{k}(\Delta)=\sum_{n=1}^{k}\frac{1}{(k-n)!}\frac{\Gamma(\Delta+k-2)}{\Gamma(\Delta+n-2)}\Psi_{n}(\Delta)$
(139)

There is another set of null states, which are of the Knizhnik-Zamolodchikov
type and decoupling of these null states give rise to differential equations
for the scattering amplitudes Banerjee:2020zlg ; Banerjee:2020vnt ;
Banerjee:2023rni ; Hu:2021lrx ; Fan:2022vbz ; Fan:2022kpp ; Hu:2022bpa . We
will discuss about these null states in the context of self dual gravity in
section I.2. Then, using these new basis (138) and (139) the OPE between two
positive helicity outgoing graviton primaries with dimensions $\Delta_{1}$ and
$\Delta_{2}$ of any $w$-invariant theory can always be written as,

$\begin{gathered}G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)=-\frac{\bar{z}}{z}B\left(\Delta_{1}-1,\Delta_{2}-1\right)G^{+}_{\Delta_{1}+\Delta_{2}}(0,0)\\\
+G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\text{MHV at}\
\mathcal{O}(z^{0}\bar{z}^{0})}+\sum_{p=1}^{n}B(\Delta_{1}-1+p,\Delta_{2}-1)\
\Omega_{p}(\Delta_{1}+\Delta_{2})\\\
+G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\text{MHV at}\
\mathcal{O}(z^{0}\bar{z}^{1})}+\bar{z}\sum_{p=1}^{n}B(\Delta_{1}+p,\Delta_{2}-1)\
\Pi_{p}(\Delta_{1}+\Delta_{2}+1)+\cdots\end{gathered}$ (140)

where $G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\text{MHV
at}\ \mathcal{O}(z^{0}\bar{z}^{0})}$ and
$G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\text{MHV at}\
\mathcal{O}(z^{0}\bar{z}^{1})}$ are the MHV OPEs at
$\mathcal{O}({z^{0}\bar{z}^{0}})$ and $\mathcal{O}({z^{0}\bar{z}})$
respectively. It has been shown in Pate:2019lpp  that the leading term in
$\bar{z}$ is uniquely determined by the $sl_{2}(R)_{V}$ invariance. Once the
leading term is known, the subleading terms in $\bar{z}$ of
$\mathcal{O}\left(\frac{\bar{z}^{q}}{z}\right)$, $q\geq 2$ are determined by
the $\overline{sl_{2}(R)}$ invariance.

It was shown in Banerjee:2023zip , that both the MHV null states
$\Omega_{k}(\Delta)$ and $\Pi_{k}(\Delta)$ form representations of
$sl_{2}(R)_{V}$. However, these representations are reducible because for any
integer $n\geq 0$, the subspaces spanned by
$\\{\Omega_{n+1}(\Delta),\Omega_{n+2}(\Delta),\cdots\\}$ and
$\\{\Pi_{n+1}(\Delta),\Pi_{n+2}(\Delta),\cdots\\}$ form a representation of
$sl_{2}(R)_{V}$. Hence we can get smaller representations spanned by the
states $\\{\Omega_{1}(\Delta),\Omega_{2}(\Delta),\cdots,\Omega_{n}(\Delta)\\}$
and $\\{\Pi_{1}(\Delta),\Pi_{2}(\Delta),\cdots,\Pi_{n}(\Delta)\\}$ if we set

$\begin{gathered}\Omega_{k+1}(\Delta)=0,\qquad k\geq n\geq 0,\\\
\Pi_{k+1}(\Delta)=0,\qquad k\geq n\geq 0.\end{gathered}$ (141)

Using the algebra (8), one can also check that the null states
$\Omega_{k}(\Delta)$ and $\Pi_{k}(\Delta)$ are primaries under
$\overline{sl_{2}(R)}$. Thus the conditions (141) are invariant under
$\overline{sl_{2}(R)}$, hence under whole $w$-algebra.

We have showed in section 3 that, the whole tower of $w$-currents can be
generated using two sub-algebras given by $\overline{sl_{2}(R)}$ and
$sl_{2}(R)_{V}$. Moreover, the conditions (141) are also invariant under
$\overline{sl_{2}(R)}$ and $sl_{2}(R)_{V}$, and hence under the full
$w$-algebra. Now, using these facts and the algebra (8), it is not hard to
show the OPE (140) is invariant under $w$-algebra. The important point we want
to emphasize about the OPE (140) is that the integer $n$ can take any
arbitrary value without breaking the $w$-invariance. Hence, there exists a
discrete infinite family of $w$-invariant OPEs. From (140) it is already clear
that $n=0$ gives the $MHV$-sector. In this paper, we have shown that $n=4$
gives the OPE of the quantum self-dual gravity theory which is known to be
$w$-invariant.

Now, the last thing we want to discuss in this section is that, the null
states $\\{\Omega_{1}(\Delta),\\\
\Omega_{2}(\Delta),\cdots,\Omega_{n}(\Delta)\\}$ are not completely
independent. For a given $n$, there is another set of
$\lceil{\frac{n}{2}}\rceil$666$\lceil{\frac{n}{2}}\rceil=\text{Smallest
integer}$ $\geq\frac{n}{2}$. nontrivial777There are of course the $n$ states
$\\{\Omega_{1}(\Delta),...,\Omega_{n}(\Delta)\\}$ which transform in a
representation of $sl_{2}(R)_{V}$ but, we cannot set them to zero because that
will lead us again to the MHV sector. states
$\\{\chi^{1}_{n}(\Delta),...,\chi_{n}^{\lceil{n/2}\rceil}(\Delta)\\}$ defined
as

$\begin{gathered}\chi^{1}_{n}(\Delta)=\sum_{p=1}^{n}\Omega_{p}(\Delta)\\\
\chi^{i}_{n}(\Delta)=\sum_{p=i}^{n}\prod_{q=i}^{2i-2}(p-q)\Omega_{p}(\Delta),\
i=2,3,...,\lceil{\frac{n}{2}}\rceil\end{gathered}$ (142)

which transform in a representation of the $sl_{2}(R)_{V}$ as a consequence of
(141). We can also set these states to zero

$\chi^{i}_{n}(\Delta)=0$ (143)

without violating the $sl_{2}(R)_{V}$ or $\overline{sl_{2}(R)}$ symmetry.

##  Appendix I Null States in Self Dual Gravity

In this Appendix, we will derive the null states of the self dual gravity
appearing at different orders of the OPE. We will first start with the OPE
between two positive helicity outgoing gravitons in the self dual gravity
derived in section 5.5. It is given by,

$\begin{gathered}G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})=-\frac{\bar{z}_{45}}{z_{45}}B\left(\Delta_{4}-1,\Delta_{5}-1\right)G^{+}_{\Delta_{4}+\Delta_{5}}(z_{5},\bar{z}_{5})\\\
+B\left(\Delta_{4}-1,\Delta_{5}-1\right)\,H^{1}_{-\frac{3}{2},\frac{1}{2}}G^{+}_{\Delta_{4}+\Delta_{5}-1}(z_{5},\bar{z}_{5})+\sum_{k=1}^{4}B(\Delta_{4}+k-1,\Delta_{5}-1)\Omega_{k}(\Delta_{4}+\Delta_{5})\\\
+\bar{z}_{45}\left[B(\Delta_{4}-1,\Delta_{5}-1)\,G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\text{MHV
at}\ \mathcal{O}(\bar{z}_{45})}\right.\\\
\left.+\sum_{k=1}^{4}B(\Delta_{4}+k,\Delta_{5}-1)\
\Pi_{k}(\Delta_{4}+\Delta_{5}+1)\right]+\cdots\end{gathered}$ (144)

where
$G^{+}_{\Delta_{4}}(z_{4},\bar{z}_{4})G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})\big{|}_{\text{MHV
at}\ \mathcal{O}(\bar{z}_{45})}$ is given by (49). We now derive the null
states appearing at $\mathcal{O}(1)$ and $\mathcal{O}(\bar{z}_{45})$.

###  I.1 Null States at $\mathcal{O}(1)$

We can see from (144) that at $\mathcal{O}(1)$ the OPE truncates at $k=4$. Now
we take the conformal soft limit $\Delta_{4}\to-4$ in (144). In this limit,
the soft descendant that appear at $\mathcal{O}(1)$ on the LHS of (144) is
given by $H^{-4}_{1,3}G^{+}_{\Delta_{5}}(z_{5},\bar{z}_{5})$. After taking the
same conformal soft limits on the RHS and comparing the results we get

$\Omega_{5}(\Delta)=\sum_{j=1}^{5}\frac{1}{(5-j)!}\frac{\Gamma(\Delta+3)}{\Gamma(\Delta+j-2)}\Phi_{j}(\Delta)=0$
(145)

where $\Phi_{j}(\Delta)$ are given by (136). Thus, we see that
$\Omega_{5}(\Delta)$ is a null state of the self dual gravity. Now we will
show the consistency of (145) under $w$-algebra. Under ${sl_{2}(R)}_{V}$,
$\Omega_{5}(\Delta)$ transforms as (133),(134),

$\begin{split}H^{1}_{-\frac{1}{2},-\frac{1}{2}}\Omega_{5}(\Delta)&=-\Omega_{5}(\Delta+1)\\\
H^{-1}_{\frac{1}{2},\frac{1}{2}}\Omega_{5}(\Delta)&=21\,\Omega_{5}(\Delta-1)-\frac{1}{2}(\Delta-4)(\Delta-5)\Omega_{5}(\Delta-1)-21\,\Omega_{6}(\Delta-1)\end{split}$
(146)

and $H^{0}_{0,0}=2\bar{L}_{0}$ is diagonal on these states. However,
$\Omega_{6}(\Delta-1)$ is also a null state of the theory and thus (145) is
invariant under ${sl_{2}(R)}_{V}$. One can also check that

$H^{0}_{0,1}\Omega_{5}(\Delta)=0$ (147)

Thus we see that (145) is also invariant under $\overline{sl_{2}(R)}$. Hence
we conclude that (145) is invariant under $w$-algebra.

There is another set of null states (142) at $\mathcal{O}(1)$ which can be
found using the commutativity property of the OPE together with the conformal
soft limits. In case of self dual gravity, they are explicitly given by,

$\begin{gathered}\chi^{1}_{4}(\Delta)=\sum_{p=1}^{4}\Omega_{p}(\Delta)\\\
\chi^{2}_{4}(\Delta)=\sum_{p=3}^{4}(p-2)\Omega_{p}(\Delta),\end{gathered}$
(148)

These null states also transform under the representation of ${sl_{2}(R)}_{V}$
and $\overline{sl_{2}(R)}$ algebra and as a consequence one can set them to 0
without violating the $w$-symmetry. The null states (148) play an important
role in showing the invariance of the Knizhnik-Zamolodchikov type null state
under $w$-algebra which will be discussed in the next subsection.

###  I.2 Null States at $\mathcal{O}(\bar{z}_{45})$: Knizhnik-Zamolodchikov
Type Null State

KZ type null state occur at $\mathcal{O}(z^{0}\bar{z}^{1})$ of the OPE. The
easiest way to derive it is to use the commutativity property of the OPE and
conformal soft limits together. So we start with the commutativity property of
the OPE given by,

$G^{+}_{\Delta_{1}}(z_{1},\bar{z}_{1})G^{+}_{\Delta_{2}}(z_{2},\bar{z}_{2})=G^{+}_{\Delta_{2}}(z_{2},\bar{z}_{2})G^{+}_{\Delta_{1}}(z_{1},\bar{z}_{1})$
(149)

Now we use the OPE (144) in (149), and take the leading conformal soft limits
$\Delta_{1}\to 1$. Then by comparing the terms at $\mathcal{O}(\bar{z}_{45})$
we get the following KZ type equation,

$\Xi_{4}(\Delta)=\xi(\Delta)+\sum_{k=1}^{4}\Pi_{k}(\Delta+1)=0$ (150)

where $\xi(\Delta)$ is the KZ type null state in the MHV sector given by
Banerjee:2020zlg

$\xi(\Delta)=\boxed{L_{-1}}G^{+}_{\Delta}+H^{0}_{0,-1}H^{1}_{-\frac{3}{2},\frac{1}{2}}G^{+}_{\Delta-1}+H^{0}_{-1,0}G^{+}_{\Delta}+(\Delta-1)\,H^{1}_{-\frac{3}{2},-\frac{1}{2}}G^{+}_{\Delta-1}$
(151)

We have used that $\chi^{1}_{4}(\Delta)$ is a null state in this theory to
arrive at the form (150). One can check that (150) is consistent under the
actions of $\overline{sl_{2}(R)}$ and $sl_{2}(R)_{V}$ generators. For example,

$H^{0}_{0,1}\Xi_{4}(\Delta)=6\,\Omega_{5}(\Delta)-(\Delta-3)\chi^{1}_{4}(\Delta)$
(152)

We have already shown that $\Omega_{5}(\Delta)$ and $\chi^{1}_{4}(\Delta)$ are
both null states in this theory, so we get,

$H^{0}_{0,1}\Xi_{4}(\Delta)=0$ (153)

Therefore $\Xi_{4}(\Delta)$ is an $\overline{sl_{2}(R)}$ primary.

Similarly, we have

$\begin{gathered}H^{-1}_{\frac{1}{2},\frac{1}{2}}\Xi_{4}(\Delta)=-\frac{1}{2}(\Delta-2)(\Delta-3)\Xi_{4}(\Delta-1)\\\
-9\,\Pi_{5}(\Delta)-H^{-1}_{\frac{1}{2},-\frac{1}{2}}\chi^{1}_{4}(\Delta)-H^{0}_{0,-1}\left((\Delta-1)\chi^{1}_{4}(\Delta-1)+\chi^{2}_{4}(\Delta-1)\right)\\\
\end{gathered}$ (154)

However, since $\Pi_{5}(\Delta),\chi^{1}_{4}(\Delta)$ and
$\chi^{2}_{4}(\Delta)$ are null states in the theory, we get

$H^{-1}_{\frac{1}{2},\frac{1}{2}}\Xi_{4}(\Delta)=-\frac{1}{2}(\Delta-2)(\Delta-3)\Xi_{4}(\Delta-1)$
(155)

Therefore, $\Xi_{4}(\Delta)$ transforms under a representation of the
$sl_{2}(R)_{V}$ and we can consistently set it to zero without violating the
$sl_{2}(R)_{V}$ symmetry. Hence, we conclude that (150) is indeed $w$
invariant. Decoupling of null states gives rise to differential equations
which the graviton scattering amplitudes in this theory have to satisfy.

##  Appendix J Invariance of the Self-Dual OPE Under $w$-Algebra

In Banerjee:2023zip , it was shown that the OPE (140) is invariant under
$w$-algebra for any arbitrary truncation in $n$, which has been reviewed in
Appendix H. We have shown in section 5.5 that self dual OPE truncates at $n=4$
of the general OPE (140). Thus, we can say that the invariance of the self
dual OPE under $w$-algebra is guaranteed. However, for the sake of
completeness of this paper and for the better readability, we will repeat the
same analysis here with focusing on the self dual OPE. As discussed in section
3, the whole $w$-algebra can be derived by the combined action of
${sl_{2}(R)}_{V}$ and $\overline{sl_{2}(R)}$. Thus it is enough to show the
invariance of the OPE under these two sub-algebras.

###  J.1 $w$-Invariance at $\mathcal{O}(1)$

Let us start with the $\mathcal{O}(1)$ OPE. We write it here again for the
readers convenience,

$\begin{gathered}G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\mathcal{O}(1)}=B(\Delta_{1}-1,\Delta_{2}-1)H^{1}_{-\frac{3}{2},\frac{1}{2}}G^{+}_{\Delta_{1}+\Delta_{2}-1}(0,0)\\\
+\sum_{k=1}^{4}B(\Delta_{1}+k-1,\Delta_{2}-1)\Omega_{k}(\Delta_{1}+\Delta_{2})\\\
\end{gathered}$ (156)

We now show that it is invariant under the two subalgebras ${sl_{2}(R)}_{V}$
and $\overline{sl_{2}(R)}$.

####  J.1.1  ${sl_{2}(R)}_{V}$ Invariance

To show the invariance of the OPE, we need the action of the $sl_{2}(R)_{V}$
on the MHV null states $\Omega_{k}(\Delta)$ that can appear at
$\mathcal{O}(1)$. These actions were computed in Banerjee:2023zip  and
reviewed in Appendix G. We also need the commutator algebra (8) along with the
action of these generators on the graviton primaries given by (see Appendix
F),

$\begin{split}H^{1}_{-\frac{1}{2},-\frac{1}{2}}G^{+}_{\Delta}(z,\bar{z})&=-G^{+}_{\Delta+1}(z,\bar{z})\\\
H^{-1}_{\frac{1}{2},\frac{1}{2}}G^{+}_{\Delta}(z,\bar{z})&=-\frac{1}{2}\left[(\Delta-2)(\Delta-3)+4(\Delta-2)\bar{z}\partial_{\bar{z}}+3\bar{z}^{2}\partial_{\bar{z}}^{2}\right]G^{+}_{\Delta-1}(z,\bar{z})\end{split}$
(157)

Using Appendix G, (8) and (157), it is not difficult to show that the
$\mathcal{O}(1)$ OPE (156) is invariant under
$H^{1}_{-\frac{1}{2},-\frac{1}{2}}$ whereas the action of
$H^{-1}_{\frac{1}{2},\frac{1}{2}}$ on both the sides of the OPE
$\eqref{fm_OPEO2}$ gives

$H^{-1}_{\frac{1}{2},\frac{1}{2}}\left(\textnormal{R.H.S}\ -\
\textnormal{L.H.S}\right)\ \textnormal{of}\
\eqref{fm_OPEO2}=-12B(\Delta_{1}+3,\Delta_{2}-1)\,\Omega_{5}(\Delta_{1}+\Delta_{2}-1)$
(158)

However, we have already shown in Appendix I.1, that $\Omega_{5}(\Delta)$ is a
null state of the self dual gravity appearing at $\mathcal{O}(1)$ of the OPE
and as a consequence we can set it to 0. Hence, we conclude that the
$\mathcal{O}(1)$ self dual OPE (156) is invariant under the ${sl_{2}(R)}_{V}$
algebra.

####  J.1.2  $\overline{sl_{2}(R)}$ Invariance

It was shown in Banerjee:2020zlg , that the OPE in the MHV-sector is invariant
under the action of $H^{0}_{0,1}$ 888$H^{0}_{0,1}\sim\bar{L}_{1}$. Also from
(135), we can see that the null states $\Omega_{k}(\Delta)$ are annihilated by
$H^{0}_{0,1}$. Therefore, we can say that $\mathcal{O}(1)$ self dual OPE (156)
is invariant under the $\overline{sl_{2}(R)}$ algebra.

###  J.2 $w$-Invariance at $\mathcal{O}(\bar{z})$

We now move on to showing the $w$-invariance of the self dual OPE at
$\mathcal{O}(\bar{z})$. Let us first write down the $\mathcal{O}(\bar{z})$ OPE
(51) again,

$\begin{gathered}G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\mathcal{O}(\bar{z})}=B(\Delta_{1}-1,\Delta_{2}-1)\,G^{+}_{\Delta_{1}}(z,\bar{z})G^{+}_{\Delta_{2}}(0,0)\big{|}_{\text{MHV
at}\ \mathcal{O}(\bar{z}_{45})}\\\
+\sum_{k=1}^{4}B(\Delta_{1}+k,\Delta_{2}-1)\
\Pi_{k}(\Delta_{1}+\Delta_{2}+1)\end{gathered}$ (159)

From the previous subsection, it is clear that the $w$-invariance of the OPE
at $\mathcal{O}(\bar{z})$ is guaranteed to follow if we can show that it is
invariant under the two subalgebras ${sl_{2}(R)}_{V}$ and
$\overline{sl_{2}(R)}$. Among the generators of these two subalgebras, we only
show the invariance of the OPE (159) under the actions of
$H^{-1}_{\frac{1}{2},\frac{1}{2}}$ and $H^{0}_{0,1}$. This is mainly because
the invariance of the OPE (159) under the rest of the generators are fairly
easy to show. By applying $H^{-1}_{\frac{1}{2},\frac{1}{2}}$ on both sides of
the OPE (159) we get,

$H^{-1}_{\frac{1}{2},\frac{1}{2}}\left(\textnormal{R.H.S}\ -\
\textnormal{L.H.S}\right)\ \textnormal{of}\
\eqref{ozi}=-9B(\Delta_{1}+4,\Delta_{2}-1)\,\Pi_{5}(\Delta_{1}+\Delta_{2})$
(160)

and for $H^{0}_{0,1}$ we have

$H^{(0)}_{0,1}\left(\textnormal{R.H.S}\ -\ \textnormal{L.H.S}\right)\
\textnormal{of}\
\eqref{ozi}=6B(\Delta_{4}+4,\Delta_{5}-1)\,\Omega_{5}(\Delta_{1}+\Delta_{2})$
(161)

However, from Appendix I, we know that both $\Pi_{5}(\Delta)$ and
$\Omega_{5}(\Delta)$ are the null states of the self dual gravity appearing at
$\mathcal{O}(\bar{z})$ and $\mathcal{O}(1)$ respectively. Thus, we conclude
that the $\mathcal{O}(\bar{z})$ OPE in self dual gravity is also invariant
under ${sl_{2}(R)}_{V}$ and $\overline{sl_{2}(R)}$, and hence under the whole
$w$-algebra.

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