amplitudes. In contrast to [42], we ambidextrously perform half-Mellin
transforms over the scales of the $\mu,\tilde{\mu}$ spinor instead of
performing them on the original spinors
$\operatorname{\lambda},\operatorname{\tilde{\lambda}}$. Just as was the case
in momentum space, the two choices are equivalent, since under a change of
variable given by an inversion $t\mapsto\frac{1}{\bar{u}}$ the half-Mellin
transformed twistor eigenstate becomes

$\displaystyle\begin{split}|\mu,\operatorname{\tilde{\lambda}};k,s_{k},l\rangle:=&\frac{1}{2}\int_{\operatorname{\mathbb{R}}_{*}}\frac{dt}{\absolutevalue{t}}\absolutevalue{t}^{2k}\text{sgn}(t)^{s_{k}}|t\mu,\operatorname{\tilde{\lambda}};l\rangle\\\
=&\frac{1}{2}\int_{\operatorname{\mathbb{R}}_{*}}\frac{d\bar{u}}{\absolutevalue{\bar{u}}}\absolutevalue{\bar{u}}^{-2k}\text{sgn}(\bar{u})^{s_{k}}|\bar{u}^{-1}\mu,\operatorname{\tilde{\lambda}};l\rangle\\\
=&\frac{1}{2}\int_{\operatorname{\mathbb{R}}_{*}}\frac{d\bar{u}}{\absolutevalue{\bar{u}}}\absolutevalue{\bar{u}}^{-2k}\text{sgn}(\bar{u})^{s_{k}}\absolutevalue{\bar{u}}^{-2(l-1)}\text{sgn}(\bar{u})^{-2(l-1)}|\mu,\bar{u}\operatorname{\tilde{\lambda}};l\rangle\\\
=&\frac{1}{2}\int_{\operatorname{\mathbb{R}}_{*}}\frac{d\bar{u}}{\absolutevalue{\bar{u}}}\absolutevalue{\bar{u}}^{2\bar{h}}\text{sgn}(\bar{u})^{s_{\bar{h}}}|\mu,\bar{u}\operatorname{\tilde{\lambda}};l\rangle\,,\\\
\end{split}$ (10.1)

which is exactly the half-Mellin transform over the $\tilde{\lambda}$ spinor,
where we have used that $k=1-h$ and $h-\bar{h}=l$. A similar equivalence is
true for the half-Mellin transformed dual twistor eigenstate.

###  10.1 Pure Yang-Mills

We start off by deriving the three-point and four-point celestial twistor
amplitudes of pure Yang-Mills Theory.

####  10.1.1 Three Points: Mellin Transform of a Sgn Function

Three-Point $MHV$

When we analytically continue to $(2,2)$ signature we can write down a three-
point MHV amplitude with real external kinematics

$\displaystyle A^{--+}_{3}=\frac{\langle 12\rangle^{3}}{\langle
23\rangle\langle
31\rangle}\,\delta^{4}(\lambda^{\alpha}_{1}\tilde{\lambda}^{\dot{\alpha}}_{1}+\lambda^{\alpha}_{2}\tilde{\lambda}^{\dot{\alpha}}_{2}+\lambda^{\alpha}_{3}\tilde{\lambda}^{\dot{\alpha}}_{3})\,.$
(10.2)

As in [24] we perform the twistor or half-Fourier transforms ambidextrously.
In our conventions this corresponds to

$\displaystyle\tilde{A}^{--+}_{3}:=\frac{1}{(2\pi)^{3}}\int
d^{2}\tilde{\lambda}_{1}\,d^{2}\tilde{\lambda}_{2}\,d^{2}\lambda_{3}\,e^{i[\tilde{\mu}_{1}\tilde{\lambda}_{1}]}\,e^{i[\tilde{\mu}_{2}\tilde{\lambda}_{2}]}\,e^{i\langle\lambda_{3}\mu_{3}\rangle}A^{--+}_{3}\,.$
(10.3)

To perform the half-Fourier transforms we must regulate the poles appearing
from $\langle 23\rangle\langle 31\rangle$ in (10.2). Following [24], we
regulate these and subsequent poles with a principle value prescription, which
is chosen exactly to preserve the little group properties of the amplitude.
With this choice the three-point MHV twistor amplitude is given by

$\displaystyle\tilde{A}^{--+}_{3}:=\frac{(\pi
i)^{3}}{(2\pi)^{3}}\text{sgn}(\langle\lambda_{1}\lambda_{2}\rangle)\,\text{sgn}(\langle\lambda_{1}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{1}])\,\text{sgn}(\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\,.$
(10.4)

In order to map to Mellin space we use ambidextrous half-Mellin transforms
over the scales of the spinors $\tilde{\mu}_{1},\tilde{\mu}_{2},\mu_{3}$.

Symmetrised Amplitudes

As explained in Section 7.3 we can separately perform a Mellin transform over
positive re-scalings of the spinors and then build odd/even states via linear
combination of sums and differences over the parameters
$\bar{\epsilon}_{1},\bar{\epsilon}_{2},\epsilon_{3}\in\\{1,-1\\}$ associated
with flipping the signs of the spinors
$\tilde{\mu}_{1},\tilde{\mu}_{2},\mu_{3}$. We now perform the sums first and
then integrate to introduce some of the key points. At four points there are
many such linear combinations, so we will perform the reverse method. That is,
we integrate over positive rescalings first to derive a compact expression in
(10.27), from which we can easily (anti-)symmetrise the amplitudes.

For the three-point amplitude there are 8 possible combinations of even/odd
states under the flipping of the signs of
$\tilde{\mu}_{1},\tilde{\mu}_{2},\mu_{3}$. Following our formalism, we denote
these states as
$\tilde{A}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}$ for
$s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\in\\{0,1\\}$. Note that, from the
form of the three-point MHV twistor amplitude (10.4), when we flip the sign of
all the spinors $\tilde{\mu}_{1},\tilde{\mu}_{2},\mu_{3}$ the amplitude
remains invariant

$\displaystyle\tilde{A}^{--+}_{3}(-\tilde{\mu}_{1},-\tilde{\mu}_{2},-\mu_{3})=\tilde{A}^{--+}_{3}(\tilde{\mu}_{1},\tilde{\mu}_{2},\mu_{3})\,.$
(10.5)

The symmetrised amplitude

$\displaystyle\tilde{A}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}=\frac{1}{2^{3}}\sum_{\bar{\epsilon}_{1},\bar{\epsilon}_{2},\epsilon_{3}}\bar{\epsilon}_{1}^{s_{\bar{k}_{1}}}\bar{\epsilon}_{2}^{s_{\bar{k}_{2}}}\epsilon_{3}^{s_{k_{3}}}\tilde{A}^{--+}_{3}(\bar{\epsilon}_{1}\tilde{\mu}_{1},\bar{\epsilon}_{2}\tilde{\mu}_{2},\epsilon_{3}\mu_{3})\,,$
(10.6)

contains pairs related by (10.5) which can sum together or subtract to give
zero depending on whether $(-1)^{s_{\bar{k}_{1}}+s_{\bar{k}_{2}}+s_{k_{3}}}$
equals plus or minus one. This immediately implies if
$s_{\bar{k}_{1}}+s_{\bar{k}_{2}}+s_{k_{3}}=1\mod 2$ then the amplitude
vanishes, which can be checked explicitly. We hence have the following four
non-vanishing amplitudes

$\displaystyle\begin{split}\tilde{A}^{--+}_{3}\\{0,0,0\\}\,,\,\,\,\tilde{A}^{--+}_{3}\\{1,1,0\\}\,,\,\,\,\tilde{A}^{--+}_{3}\\{1,0,1\\}\,,\,\,\,\tilde{A}^{--+}_{3}\\{0,1,1\\}\,,\end{split}$
(10.7)

where using (10.6) we have

$\displaystyle\begin{split}\tilde{A}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}=\frac{2(\pi
i)^{3}}{2^{3}(2\pi)^{3}}\text{sgn}(\langle\lambda_{1}\lambda_{2}\rangle)\,&\bigg{(}\text{sgn}(\langle\lambda_{1}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{1}])\,\text{sgn}(\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\\\
&+(-1)^{s_{\bar{k}_{1}}}\text{sgn}(\langle\lambda_{1}\mu_{3}\rangle-[\tilde{\lambda}_{3}\tilde{\mu}_{1}])\,\text{sgn}(\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\\\
&+(-1)^{s_{\bar{k}_{2}}}\text{sgn}(\langle\lambda_{1}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{1}])\,\text{sgn}(\langle\lambda_{2}\mu_{3}\rangle-[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\\\
&+(-1)^{s_{k_{3}}}\text{sgn}(\langle\lambda_{1}\mu_{3}\rangle-[\tilde{\lambda}_{3}\tilde{\mu}_{1}])\,\text{sgn}(\langle\lambda_{2}\mu_{3}\rangle-[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\bigg{)}\,,\end{split}$
(10.8)

From the amplitudes (10.7) given by (10.8) we can recover the original
amplitude by summing them. These amplitudes can be re-cast as

$\displaystyle\begin{split}&\tilde{A}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}=\frac{(\pi
i)^{3}}{(2\pi)^{3}}\text{sgn}(\langle\lambda_{1}\lambda_{2}\rangle\langle\lambda_{1}\mu_{3}\rangle\langle\lambda_{2}\mu_{3}\rangle)\,\tilde{\delta}\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\,\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times
G_{s_{\bar{k}_{2}}+s_{k_{3}}}(\theta_{31}^{-1})G_{s_{\bar{k}_{1}}+s_{k_{3}}}(\theta_{32}^{-1})\,,\end{split}$
(10.9)

where we have included a Kronecker delta

$\tilde{\delta}\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\coloneqq\delta_{s_{\bar{k}_{1}}+s_{\bar{k}_{2}}+s_{k_{3}},0}\,,$
(10.10)

(with argument evaluated mod 2) which enforces the aforementioned condition
for the amplitude to be non-vanishing. We have also defined the variables

$\displaystyle\theta_{ij}=\frac{\langle\lambda_{j}\mu_{i}\rangle}{[\tilde{\lambda}_{i}\tilde{\mu}_{j}]}\,,$
(10.11)

and the (anti-)symmetrised sgn functions

$\displaystyle\begin{split}G_{0}(x):=\frac{1}{2}\bigg{(}\text{sgn}(1+x)+\text{sgn}(1-x)\bigg{)}\,,\\\
G_{1}(x):=\frac{1}{2}\bigg{(}\text{sgn}(1+x)-\text{sgn}(1-x)\bigg{)}\,,\end{split}$
(10.12)

which can be written in terms of Heaviside step functions as follows

$\displaystyle\begin{split}G_{0}(x)=&\Theta(1-\absolutevalue{x}),\\\
G_{1}(x)=&\text{sgn}(x)\Theta(\absolutevalue{x}-1).\end{split}$ (10.13)

We will use the above definitions (10.13) many times over when performing our
integrals.

The (2,2) signature celestial twistor amplitude in Mellin space, denoted with
a calligraphic symbol $\mathcal{A}$, is then given by an integral over $t_{i}$
the positive scale of $\mu_{i}$ with weight $2k_{i}$

$\displaystyle\begin{split}\tilde{\mathcal{A}}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}:=&\int_{0}^{\infty}\frac{d\bar{t}_{1}}{\bar{t}_{1}}\,\frac{d\bar{t}_{2}}{\bar{t}_{2}}\,\frac{dt_{3}}{t_{3}}\,\bar{t}_{1}^{2\bar{k}_{1}}\,\bar{t}_{2}^{2\bar{k}_{2}}\,t_{3}^{2k_{3}}\tilde{A}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}(t_{1}\tilde{\mu}_{1},\bar{t}_{2}\tilde{\mu}_{2},t_{3}\mu_{3}).\end{split}$
(10.14)

As such we are lead to consider the following integrals

$\displaystyle\begin{split}\mathcal{I}(\pm,\pm):=&\int_{0}^{\infty}\frac{d\bar{t}_{1}}{\bar{t}_{1}}\,\frac{d\bar{t}_{2}}{\bar{t}_{2}}\,\frac{dt_{3}}{t_{3}}\,\bar{t}_{1}^{2\bar{k}_{1}}\,\bar{t}_{2}^{2\bar{k}_{2}}\,t_{3}^{2k_{3}}\\\
&\quad\times\frac{1}{4}\bigg{(}\text{sgn}(1+\frac{\bar{t}_{1}}{t_{3}}\theta_{31}^{-1})\pm\text{sgn}(1-\frac{\bar{t}_{1}}{t_{3}}\theta_{31}^{-1})\bigg{)}\bigg{(}\text{sgn}(1+\frac{\bar{t}_{2}}{t_{3}}\theta_{32}^{-1})\pm\text{sgn}(1-\frac{\bar{t}_{2}}{t_{3}}\theta_{32}^{-1})\bigg{)}.\end{split}$
(10.15)

The general character of $\mathcal{I}(\pm,\pm)$ (and corresponding integrals
beyond three points) is very simple - it is a Mellin integral (over a positive
quadrant) of an integrand which is a piece-wise constant function in various
regions bounded by the planes
$\frac{\bar{t}_{i}}{t_{3}}=\absolutevalue{\theta_{3i}}$.

Another important characteristic of the integral $\mathcal{I}$, and its higher
point cousins, is that it is a conformal integral and as such one of the
degrees of freedom serves simply to parameterise an overall scale over which
we integrate. In addition, the integrand itself is invariant under overall
dilatations since it is a pure gluon amplitude. We can fix this scaling
redundancy by integrating it out. We choose to write $\bar{t}_{1}=t_{3}x_{1}$
and $\bar{t}_{2}=t_{3}x_{2}$ and then integrate over the overall scale $t_{3}$

$\displaystyle\begin{split}\mathcal{I}(\pm,\pm)=&\int_{0}^{\infty}\frac{dt_{3}}{t_{3}}t_{3}^{2\bar{k}_{1}+2\bar{k}_{2}+2k_{3}}\int_{0}^{\infty}\frac{dx_{1}\,dx_{2}}{x_{1}\,x_{2}}\,x_{1}^{2\bar{k}_{1}}\,x_{2}^{2\bar{k}_{2}}\,\\\
&\quad\times\frac{1}{4}\bigg{(}\text{sgn}(1+x_{1}\theta_{31}^{-1})\pm\text{sgn}(1-x_{1}\theta_{31}^{-1})\bigg{)}\bigg{(}\text{sgn}(1+x_{2}\theta_{32}^{-1})\pm\text{sgn}(1-x_{2}\theta_{32}^{-1})\bigg{)}\\\
=&\;2\pi i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\,\int_{0}^{\infty}\frac{dx_{1}\,dx_{2}}{x_{1}\,x_{2}}\,x_{1}^{2\bar{k}_{1}}\,x_{2}^{2\bar{k}_{2}}\\\
&\quad\times\frac{1}{4}\bigg{(}\text{sgn}(1+x_{1}\theta_{31}^{-1})\pm\text{sgn}(1-x_{1}\theta_{31}^{-1})\bigg{)}\bigg{(}\text{sgn}(1+x_{2}\theta_{32}^{-1})\pm\text{sgn}(1-x_{2}\theta_{32}^{-1})\bigg{)},\end{split}$
(10.16)

where we have recovered a dilatation invariance delta function using equation
(7.19) and the fact that the weights $\bar{k}_{1}$, $\bar{k}_{2}$, $k_{3}$ are
all pure imaginary. This is a special feature of the ambidextrous twistor
gluon amplitude and relies on the fact we have analytically continued from
(1,3) signature, so all the helicities are plus or minus one.

Now that we have integrated out an overall scale the remaining integrals over
$x_{1},x_{2}$ are separable and each has manifestly acquired a scale given by
$\theta_{31},\theta_{32}$ respectively. This separable behaviour persists at
four points.

From the three-point MHV celestial amplitude we are led to consider the
following Mellin transforms of the (anti-)symmetrised sgn functions
$G_{0},G_{1}$ in (10.13)

$\displaystyle\begin{split}\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}G_{s}(x_{1}\theta_{31}^{-1})=&\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\,\frac{1}{2}\bigg{(}\text{sgn}(1+x_{1}\theta_{31}^{-1})+(-1)^{s}\text{sgn}(1-x_{1}\theta_{31}^{-1})\bigg{)}\\\
=&\;\text{sgn}(\theta_{31})^{s}\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\Theta((-1)^{s}(1-x_{1}\absolutevalue{\theta_{31}}^{-1})).\end{split}$
(10.17)

An important property of the function $G_{0}$ is that it is zero for large
values $\absolutevalue{x_{1}}>\theta_{31}$ and so its Mellin transform has a
semi-infinite strip of definition where the integral is defined for $2k_{1}$
with real part greater than zero. On the other hand, the function $G_{1}$ is
zero for small values $\absolutevalue{x_{1}}<\theta_{31}$ and so its semi-
infinite strip of definition is where $2k_{1}$ has real part less than zero.
Note that the original sgn function, since it is $\mathcal{O}(1)$ everywhere,
has a Mellin transform which only marginally converges when its weight lies on
the imaginary axis and diverges otherwise. As such we can only re-organise the
integral (10.17) by integrating each sgn function separately if we also give a
regularisation prescription. We demonstrate a natural regularisation in the
next section in equation (10.21), which also commutes with the even/odd
projection in (10.17).

Continuing with our derivation, we compute the Mellin transform of $G_{0}$ and
$G_{1}$ functions - since all the weights $k_{i}$ are pure imaginary we must
regularise the weights to ensure they are within the corresponding strip of
definition

$\displaystyle\begin{split}&\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}+\epsilon}\,\frac{1}{2}\bigg{(}\text{sgn}(1+x_{1}\theta_{31}^{-1})+\text{sgn}(1-x_{1}\theta_{31}^{-1})\bigg{)}=\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}+\epsilon}\,\Theta\bigg{(}1-x_{1}\absolutevalue{\theta_{31}}^{-1}\bigg{)}\\\
=&\;\Theta(\theta_{31})\frac{\theta_{31}^{2\bar{k}_{1}+\epsilon}}{2\bar{k}_{1}+\epsilon}+\Theta(-\theta_{31})\frac{(-\theta_{31})^{2\bar{k}_{1}+\epsilon}}{2\bar{k}_{1}+\epsilon}=\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}+\epsilon},\\\
&\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}-\epsilon}\,\frac{1}{2}\bigg{(}\text{sgn}(1+x_{1}\theta_{31}^{-1})-\text{sgn}(1-x_{1}\theta_{31}^{-1})\bigg{)}=\text{sgn}(\theta_{31})\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}-\epsilon}\,\Theta\bigg{(}x_{1}\absolutevalue{\theta_{31}}^{-1}-1\bigg{)}\\\
=&-\Theta(\theta_{31})\frac{\theta_{31}^{2\bar{k}_{1}-\epsilon}}{2\bar{k}_{1}-\epsilon}+\Theta(-\theta_{31})\frac{(-\theta_{31})^{2\bar{k}_{1}-\epsilon}}{2\bar{k}_{1}-\epsilon}=\text{sgn}(-\theta_{31})\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}-\epsilon}.\end{split}$
(10.18)

So the even case gives an ‘advanced’ prescription for the pole in weight space
and the odd case has the ‘retarded’ prescription. Also the even/odd symmetry
is trivially carried by an overall sgn function and the remaining integral
simply depends on a manifestly positive scale at
$\absolutevalue{\theta_{31}}$. Finally, we have removed the appearance of
$\epsilon$ in the exponents since the only role it should play is to
regularise the pole.

The three-point MHV celestial twistor amplitudes are then given by a compact
formula

$\boxed{\begin{split}\tilde{\mathcal{A}}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}=&\frac{\pi}{4}\text{sgn}(\langle\lambda_{1}\lambda_{2}\rangle\langle\lambda_{1}\mu_{3}\rangle\langle\lambda_{2}\mu_{3}\rangle)\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\tilde{\delta}\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\\\
&\quad\times\text{sgn}(-\theta_{31})^{s_{\bar{k}_{1}}}\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}+(-1)^{s_{\bar{k}_{1}}}\epsilon}\text{sgn}(-\theta_{32})^{s_{\bar{k}_{2}}}\frac{\absolutevalue{\theta_{32}}^{2\bar{k}_{2}}}{2\bar{k}_{2}+(-1)^{s_{\bar{k}_{2}}}\epsilon}.\end{split}}$
(10.19)

The formula (10.19) has similarities with light transformed conformal
correlators that have appeared in the literature in [42, 63], up to
differences due to the newly defined light transform (8.3) with absolute
values and sgn functions and the specific normalisation we have chosen. We
give a formula for the unsymmetrised celestial twistor amplitude in (10.27)
which is closer to the light transformed conformal correlators of [42].

It is also a simple task to check conformal covariance of (10.19) for each
leg. Note first that we are explicitly using only the ambidextrous variables,
in this case $\bar{k}$ for negative helicity legs and $k$ for the positive
helicity legs. In general, for conformal covariance we require that if we sum
the exponents wherever a positive helicity leg $i$ appears in the
$\theta_{ik}$ ratios we have overall weight $-2k_{i}$, while for a negative
helicity leg $j$ we have overall weight $2\bar{k}_{j}$. Then conformal
covariance is guaranteed using the defining relations $k:=1-h$ and
$\bar{k}:=1-\bar{h}$ and the relations $\bar{k}_{1,2}=-h_{1,2}$ and
$k_{3}=-\bar{h}_{3}$ specific to the ambidextrous variables and pure gluon
amplitudes. Finally, for some legs we must also use dilatation invariance, for
example for leg 3 we use the condition $\bar{k}_{1}+\bar{k}_{2}+k_{3}=0$.

In the derivation of (10.19) we first performed the sum over incoming and
outgoing parameters in order to define even and odd parity states and
afterwards we performed the integrals over positive energies. In general,
performing the integrals first and then the sum afterwards will give the same
result only if the integrals are finite. That is, the Mellin transform of a
sum of two functions is equal to the sum of their Mellin transforms only if
there is an overlap in their strip of definition. We saw already that the
symmetrised and anti-symmetrised sgn functions had a strip of definition in
the right and left half planes respectively. However, the sgn function itself
has a Mellin transform which only marginally converges when the weight lies on
the imaginary axis and diverges otherwise. We show in the below section that
we can define a natural regularisation prescription to define the Mellin
transform of a sgn function and that this regularisation commutes with the
(anti-)symmetrisation procedure.

Unsymmetrised Amplitude

We now look at the case where we first integrate (10.4) over positive re-
scalings and then perform the even/odd projections afterwards. This will of
course give the same result as in the previous section as long as we are
careful in how we regularise marginally convergent Mellin transforms. We hence
consider the integral

$\displaystyle\begin{split}\mathcal{I}_{3pt}=&2\pi
i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\int_{0}^{\infty}\frac{dx_{1}\,dx_{2}}{x_{1}\,x_{2}}\,x_{1}^{2\bar{k}_{1}}\,x_{2}^{2\bar{k}_{2}}\text{sgn}(1+x_{1}\theta_{31}^{-1})\text{sgn}(1+x_{2}\theta_{32}^{-1}),\end{split}$
(10.20)

which is the unsymmetrised version of equation (10.16).

Again the integrals over $x_{1}$ and $x_{2}$ separate and are of identical
form after the replacement $1\mapsto 2$. Focusing on the $x_{1}$ integral,
since the sgn function is always $\mathcal{O}(1)$ its strip of definition is
when $2\bar{k}_{1}$ is pure imaginary and we must regulate in a similar manner
to the integral (7.20) for the delta function. Instead of regulating with a
factor of $e^{-\epsilon\absolutevalue{y}}$, we use a regularisation
prescription which is anchored at the scale set by $\theta_{31}$. That is we
define
$\absolutevalue{\absolutevalue{y}}:=\text{sgn}(y-\ln(\absolutevalue{\theta_{31}}))y$
which equals $y$ when $y>\ln(\absolutevalue{\theta_{31}})$ and $-y$ when
$y<\ln(\absolutevalue{\theta_{31}})$, then define the regulated integral

$\displaystyle\begin{split}\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\,\text{sgn}(1+x_{1}\theta_{31}^{-1})=&\int_{-\infty}^{\infty}dy\,e^{2\bar{k}_{1}y-\epsilon\absolutevalue{\absolutevalue{y}}}\,\text{sgn}(1+e^{y}\theta_{31}^{-1})\\\
=&\;\Theta(-\theta_{31})\left(\int_{-\infty}^{\ln(-\theta_{31})}dy\,e^{2\bar{k}_{1}y+\epsilon
y}-\int_{\ln(-\theta_{31})}^{\infty}dy\,e^{2\bar{k}_{1}y-\epsilon y}\right)\\\
&\;\;+\;\Theta(\theta_{31})\left(\int_{-\infty}^{\ln\theta_{31}}dy\,e^{2\bar{k}_{1}y+\epsilon
y}+\int_{\ln\theta_{31}}^{\infty}dy\,e^{2\bar{k}_{1}y-\epsilon y}\right)\\\
=&\;\Theta(-\theta_{31})\bigg{(}\frac{(-\theta_{31})^{2\bar{k}_{1}+\epsilon}}{2\bar{k}_{1}+\epsilon}+\frac{(-\theta_{31})^{2\bar{k}_{1}-\epsilon}}{2\bar{k}_{1}-\epsilon}\bigg{)}+\Theta(\theta_{31})\bigg{(}\frac{\theta_{31}^{2\bar{k}_{1}+\epsilon}}{2\bar{k}_{1}+\epsilon}-\frac{\theta_{31}^{2\bar{k}_{1}-\epsilon}}{2\bar{k}_{1}-\epsilon}\bigg{)}\\\
=&\;\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}\bigg{(}\frac{1}{2\bar{k}_{1}+\epsilon}+\text{sgn}(-\theta_{31})\frac{1}{2\bar{k}_{1}-\epsilon}\bigg{)},\end{split}$
(10.21)

where again we remove the $\epsilon$ in the exponents since it plays no
regularising role there.

Note that in the case $\theta_{31}>0$ the sgn gives minus one and so the
Mellin transform (10.21) is proportional to a delta function as expected.
While the case $\theta_{31}<0$ the sgn function gives plus one and the Mellin
transform (10.21) is proportional to the principal value pole. These are
defined as

$\displaystyle\begin{split}\delta(2\bar{k}_{1})\coloneqq\frac{1}{2\pi
i}\bigg{(}\frac{1}{2\bar{k}_{1}+\epsilon}-\frac{1}{2\bar{k}_{1}-\epsilon}\bigg{)}\,,\\\
P.V.\,\frac{1}{2\bar{k}_{1}}\coloneqq\frac{1}{2}\bigg{(}\frac{1}{2\bar{k}_{1}+\epsilon}+\frac{1}{2\bar{k}_{1}-\epsilon}\bigg{)}\,.\end{split}$
(10.22)

So we may write

$\displaystyle\begin{split}\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\,\text{sgn}(1+x_{1}\theta_{31}^{-1})=\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}\bigg{(}\pi
i\,\Theta(\theta_{31})\delta(\bar{k}_{1})+\Theta(-\theta_{31})\frac{1}{\bar{k}_{1}}\bigg{)}.\end{split}$
(10.23)

where it is understood that principal values are always implicit for a
$1/\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}$ pole.

Thus the above result (10.21) is the Mellin transform of a sgn function
depending on a single parameter. This corresponds to the decomposition of the
sgn function into even and odd parts as

$\displaystyle\text{sgn}(1+\theta_{31}^{-1})=\Theta(1-\absolutevalue{\theta_{31}}^{-1})+\text{sgn}(\theta_{31})\Theta(\absolutevalue{\theta_{31}}^{-1}-1).$
(10.24)

While it is not in general true that the Mellin transform of a sum is the sum
of their Mellin transforms (the strips of definition must also be compatible
such that all the integrals are convergent) the regularisation we have used is
precisely such that this is true and is motivated as such. In other words,
given the Mellin transform of a sgn function (10.21) it is simple to check
that if we take even and odd combinations we recover the results (10.18).

To completely confirm our results we perform the inverse Mellin transform

$\displaystyle\begin{split}\int_{-i\infty}^{i\infty}\frac{d(2\bar{k}_{1})}{2\pi
i}\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}\bigg{(}\frac{1}{2\bar{k}_{1}+\epsilon}+\text{sgn}(-\theta_{31})\frac{1}{2\bar{k}_{1}-\epsilon}\bigg{)},\end{split}$
(10.25)

where we have chosen a contour located at $a=0$ corresponding to a strip of
definition on the imaginary axis. Now the $\epsilon$-prescription in each term
gives poles in the $2\bar{k}_{1}$ plane to either the left or right of the
imaginary axis - in addition, we must condition this integral on the magnitude
of $\theta_{31}$ to determine how to close the contour. The case
$\absolutevalue{\theta_{31}}>1$ requires us to close the contour as an anti-
clockwise semicircle to the left such that the real part of $2\bar{k}_{1}$ is
less than zero and so will only pick up a $2\pi i$ residue contribution from
the first term, while for $\absolutevalue{\theta_{31}}<1$ we must close the
contour clockwise to the right and we get a contribution from the second term
and an extra minus sign. Hence we have

$\displaystyle\begin{split}\int_{-i\infty}^{i\infty}\frac{d(2\bar{k}_{1})}{2\pi
i}\bigg{(}\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}+\epsilon}}{2\bar{k}_{1}+\epsilon}+\text{sgn}(-\theta_{31})\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}-\epsilon}}{2\bar{k}_{1}-\epsilon}\bigg{)}=&\;\Theta(\absolutevalue{\theta_{31}}-1)-\text{sgn}(-\theta_{31})\Theta(1-\absolutevalue{\theta_{31}})\\\
=&\;\text{sgn}(1+\theta_{31}^{-1}).\end{split}$ (10.26)

Thus we see that the $\epsilon$ prescription is crucial for the inverse Mellin
transform and corresponds to a particular contour choice for each term done
independently. With this specific contour choice in mind we may drop the
$\epsilon$ prescription in (10.21) which corresponds to dropping delta
function terms in (10.23) which have singular support. The unsymmetrised
amplitude then takes a very simple form

$\boxed{\begin{split}\tilde{\mathcal{A}}^{--+}_{3}=&\frac{\pi}{4}\text{sgn}(\langle\lambda_{1}\lambda_{2}\rangle\langle\lambda_{1}\mu_{3}\rangle\langle\lambda_{2}\mu_{3}\rangle)\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\Theta(-\theta_{31})\Theta(-\theta_{32})\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{\bar{k}_{1}}\frac{\absolutevalue{\theta_{32}}^{2\bar{k}_{2}}}{\bar{k}_{2}},\end{split}}$
(10.27)

from which we may (anti-)symmetrise to give the amplitudes (10.19) on the
celestial torus. The above properties of Mellin transforms of sgn functions
and their regularisation will be used again at four points.

Three-Point $\overline{MHV}$

Here we summarise the results of the similar derivation of the three-point
$\overline{MHV}$ amplitude. The spacetime amplitude

$\displaystyle
A^{++-}_{3}=\frac{[12]^{3}}{[23][31]}\,\delta^{4}(\lambda^{\alpha}_{1}\tilde{\lambda}^{\dot{\alpha}}_{1}+\lambda^{\alpha}_{2}\tilde{\lambda}^{\dot{\alpha}}_{2}+\lambda^{\alpha}_{3}\tilde{\lambda}^{\dot{\alpha}}_{3})$
(10.28)

is mapped to an ambidextrous twistor amplitude given by the following half-
Fourier transforms

$\displaystyle\tilde{A}^{++-}_{3}:=\frac{1}{(2\pi)^{3}}\int
d^{2}\lambda_{1}\,d^{2}\lambda_{2}\,d^{2}\tilde{\lambda}_{3}\,e^{i\langle\lambda_{1}\mu_{1}\rangle}\,e^{i\langle\lambda_{2}\mu_{2}\rangle}\,e^{i[\tilde{\mu}_{3}\tilde{\lambda}_{3}]}A^{++-}_{3},$
(10.29)

which gives the result

$\displaystyle\tilde{A}^{++-}_{3}=\frac{(\pi
i)^{3}}{(2\pi)^{3}}\text{sgn}([\tilde{\lambda}_{1}\tilde{\lambda}_{2}])\,\text{sgn}(\langle\mu_{1}\lambda_{3}\rangle+[\tilde{\mu}_{3}\tilde{\lambda}_{1}])\,\text{sgn}(\langle\mu_{2}\lambda_{3}\rangle+[\tilde{\mu}_{3}\tilde{\lambda}_{2}]).$
(10.30)

The (2,2) signature three-point $\overline{MHV}$ celestial twistor amplitudes
follow from an analogous derivation to that in the section above, and are
given by

$\boxed{\begin{split}\tilde{\mathcal{A}}^{++-}_{3}\\{s_{k_{1}},s_{k_{2}},s_{\bar{k}_{3}}\\}=&\frac{\pi}{4}\text{sgn}([\tilde{\lambda}_{1}\tilde{\lambda}_{2}][\tilde{\mu}_{3}\tilde{\lambda}_{1}][\tilde{\mu}_{3}\tilde{\lambda}_{2}])\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\tilde{\delta}\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\\\
&\quad\times\text{sgn}(-\theta_{13})^{s_{k_{1}}}\frac{\absolutevalue{\theta_{13}^{-1}}^{2k_{1}}}{2k_{1}+(-1)^{s_{k_{1}}}\epsilon}\text{sgn}(-\theta_{23})^{s_{k_{2}}}\frac{\absolutevalue{\theta_{23}^{-1}}^{2k_{2}}}{2k_{2}+(-1)^{s_{k_{2}}}\epsilon}\,,\end{split}}$
(10.31)

where now the sums present in the delta functions are

$\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}=2k_{1}+2k_{2}+2\bar{k}_{3},\quad\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}=s_{k_{1}}+s_{k_{2}}+s_{\bar{k}_{3}}\,.$
(10.32)

The unsymmetrised amplitude from which the amplitudes are built is, dropping
delta function terms as before,

$\boxed{\begin{split}\tilde{\mathcal{A}}^{++-}_{3}=\frac{\pi}{4}\text{sgn}([\tilde{\lambda}_{1}\tilde{\lambda}_{2}][\tilde{\mu}_{3}\tilde{\lambda}_{1}][\tilde{\mu}_{3}\tilde{\lambda}_{2}])\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\Theta(-\theta_{13})\Theta(-\theta_{23})\frac{\absolutevalue{\theta_{13}^{-1}}^{2k_{1}}}{k_{1}}\frac{\absolutevalue{\theta_{23}^{-1}}^{2k_{2}}}{k_{2}}\,.\end{split}}$
(10.33)

####  10.1.2 Four Points: Mellin Transform of a Product of Sgn Functions

Four-Point ‘Alternating’ Amplitude

We now move on to consider four points. We will discover that much of the same
structure appears and identical methods can be used. At four points the $MHV$
and $\overline{MHV}$ amplitudes are one and the same, however we can consider
different ways in which we assign helicity to each leg. We have $A_{4}^{+-+-}$
which we call the ‘alternating’ helicity amplitude and $A_{4}^{++--}$ which we
call the ‘separated’ helicity amplitude. Following [24] we use the same
ambidextrous twistor space for both of these amplitudes, in which case we find
that both take their simplest, although qualitatively different form. We will
first consider the alternating helicity amplitude and map it to celestial
twistor space directly since the integrals are more straightforward. Once the
dust has settled, we will be able to write the separated amplitude in terms of
the alternating one. The alternating four-point amplitude is given by

$\displaystyle A^{+-+-}_{4}=\frac{\langle 24\rangle^{4}}{\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle
41\rangle}\,\delta^{4}(\lambda^{\alpha}_{1}\tilde{\lambda}^{\dot{\alpha}}_{1}+\lambda^{\alpha}_{2}\tilde{\lambda}^{\dot{\alpha}}_{2}+\lambda^{\alpha}_{3}\tilde{\lambda}^{\dot{\alpha}}_{3}+\lambda^{\alpha}_{4}\tilde{\lambda}^{\dot{\alpha}}_{4})\,.$
(10.34)

Once again, we map to twistor space in an ambidextrous manner by performing
the following half-Fourier transforms

$\displaystyle\tilde{A}^{+-+-}_{4}:=\frac{1}{(2\pi)^{4}}\int
d^{2}\lambda_{1}\,d^{2}\tilde{\lambda}_{2}\,d^{2}\lambda_{3}\,d^{2}\tilde{\lambda}_{4}\,e^{i\langle\lambda_{1}\mu_{1}\rangle}\,e^{i[\tilde{\mu}_{2}\tilde{\lambda}_{2}]}\,e^{i\langle\lambda_{3}\mu_{3}\rangle}\,e^{i[\tilde{\mu}_{4}\tilde{\lambda}_{4}]}A^{+-+-}_{4}\,.$
(10.35)

After regulating poles using a principal value prescription as before, the
resulting amplitude takes the following form [24]

$\displaystyle\begin{split}\tilde{A}^{+-+-}_{4}=&\;\frac{(\pi
i)^{4}}{(2\pi)^{4}}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle+[\tilde{\lambda}_{1}\tilde{\mu}_{2}])\,\text{sgn}(\langle\lambda_{4}\mu_{1}\rangle+[\tilde{\lambda}_{1}\tilde{\mu}_{4}])\\\
&\qquad\quad\times\text{sgn}(\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\,\text{sgn}(\langle\lambda_{4}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{4}])\\\
=&\;\frac{1}{16}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle)\\\
&\qquad\quad\times\text{sgn}(1+\theta_{12}^{-1})\,\text{sgn}(1+\theta_{14}^{-1})\text{sgn}(1+\theta_{32}^{-1})\,\text{sgn}(1+\theta_{34}^{-1})\,.\end{split}$
(10.36)

Symmetrised Amplitudes

We are now tasked with mapping this amplitude to Mellin space via ambidextrous
Mellin transforms. At four points we sum over the incoming/outgoing parameters
and there are 16 combinations of parity under flipping of the sign of the
spinors $\mu_{1},\tilde{\mu}_{2},\mu_{3},\tilde{\mu}_{4}$. So we define

$\displaystyle\tilde{A}^{+-+-}_{4}\\{s_{k_{1}},s_{\bar{k}_{2}},s_{k_{3}},s_{\bar{k}_{4}}\\}=\frac{1}{2^{4}}\sum_{\epsilon_{1},\bar{\epsilon}_{2},\epsilon_{3},\bar{\epsilon}_{4}}\epsilon_{1}^{s_{k_{1}}}\bar{\epsilon}_{2}^{s_{\bar{k}_{2}}}\epsilon_{3}^{s_{k_{3}}}\bar{\epsilon}_{4}^{s_{\bar{k}_{4}}}\tilde{A}^{+-+-}_{4}(\epsilon_{1}\mu_{1},\bar{\epsilon}_{2}\tilde{\mu}_{2},\epsilon_{3}\mu_{3},\bar{\epsilon}_{4}\tilde{\mu}_{4})\,,$
(10.37)

where the $\\{s_{k_{1}},s_{\bar{k}_{2}},s_{k_{3}},s_{\bar{k}_{4}}\\}$ label
the parity for each spinor. As was the case at three points, if we flip the
sign of all the spinors
$\mu_{1},\tilde{\mu}_{2},\mu_{3},\tilde{\mu}_{4}\mapsto-\mu_{1},-\tilde{\mu}_{2},-\mu_{3},-\tilde{\mu}_{4}$
then the amplitude (10.36) is invariant thus for the amplitude (10.37) to be
non-zero we must have the even condition
$s_{k_{1}}+s_{\bar{k}_{2}}+s_{k_{3}}+s_{\bar{k}_{4}}=0\mod 2$. There are then
eight non-vanishing amplitudes, which by direct calculation are given by

$\displaystyle\begin{split}&\tilde{A}^{+-+-}_{4}\\{0,0,0,0\\}=\frac{1}{16}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle)\\\
&\qquad\qquad\qquad\quad\times\bigg{(}G_{0}(\theta_{12})G_{0}(\theta_{14})G_{0}(\theta_{32})G_{0}(\theta_{34})+G_{1}(\theta_{12})G_{1}(\theta_{14})G_{1}(\theta_{32})G_{1}(\theta_{34})\bigg{)},\\\
&\tilde{A}^{+-+-}_{4}\\{1,1,0,0\\}=\frac{1}{16}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle)\\\
&\qquad\qquad\qquad\quad\times\bigg{(}G_{1}(\theta_{12})G_{0}(\theta_{14})G_{0}(\theta_{32})G_{0}(\theta_{34})+G_{0}(\theta_{12})G_{1}(\theta_{14})G_{1}(\theta_{32})G_{1}(\theta_{34})\bigg{)},\\\
&\qquad\qquad\vdots\\\
&\tilde{A}^{+-+-}_{4}\\{1,1,1,1\\}=\frac{1}{16}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle)\\\
&\qquad\qquad\qquad\quad\times\bigg{(}G_{1}(\theta_{12})G_{0}(\theta_{14})G_{0}(\theta_{32})G_{1}(\theta_{34})+G_{0}(\theta_{12})G_{1}(\theta_{14})G_{1}(\theta_{32})G_{0}(\theta_{34})\bigg{)},\end{split}$
(10.38)

where we use the expressions (10.13) for the $G_{0}$ and $G_{1}$ functions.
The general formula for the symmetrised amplitude is

$\displaystyle\begin{split}\tilde{A}^{+-+-}_{4}\\{s_{k_{1}},s_{\bar{k}_{2}},s_{k_{3}},s_{\bar{k}_{4}}\\}=&\frac{1}{16}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle)\delta\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\\\
&\quad\quad\times\sum_{r_{ij}}G_{r_{12}}(\theta_{12})G_{r_{14}}(\theta_{14})G_{r_{32}}(\theta_{32})G_{r_{34}}(\theta_{34}),\end{split}$
(10.39)

where the sum in (10.39) gives the two unique terms with $r_{ij}\in{0,1}$ such
that

$\displaystyle\begin{split}&r_{12}+r_{14}=s_{k_{1}}\mod 2\,,\\\
&r_{12}+r_{32}=s_{\bar{k}_{2}}\mod 2\,,\\\ &r_{32}+r_{34}=s_{k_{3}}\mod
2\,,\\\ &r_{34}+r_{14}=s_{\bar{k}_{4}}\mod 2\,.\end{split}$ (10.40)

For example, if we want an amplitude that is odd in $\mu_{1}$
($s_{k_{1}}{=}1$) then we require a product of an odd number of the anti-
symmetric $G_{1}$ functions evaluated at the variables $\theta_{1i}$, which
contain $\mu_{1}$.

One can derive the various symmetrised four-point celestial twistor gluon
amplitudes by Mellin transforming (10.39), exactly as was done at three
points. We are led to consider the integrals

$\displaystyle\begin{split}\mathcal{I}(r_{12},r_{14},r_{32},r_{34})&:=\int_{0}^{\infty}\frac{dt_{1}\,d\bar{t}_{2}\,dt_{3}\,d\bar{t}_{4}}{t_{1}\bar{t}_{2}t_{3}\bar{t}_{4}}\,t_{1}^{2k_{1}}\,\bar{t}_{2}^{2\bar{k}_{2}}\,t_{3}^{2k_{3}}\,\bar{t}_{4}^{2\bar{k}_{4}}\\\
&\qquad\qquad\times
G_{r_{12}}\bigg{(}\frac{\bar{t}_{2}}{t_{1}}\theta^{-1}_{12}\bigg{)}G_{r_{14}}\bigg{(}\frac{\bar{t}_{4}}{t_{1}}\theta^{-1}_{14}\bigg{)}G_{r_{32}}\bigg{(}\frac{\bar{t}_{2}}{t_{3}}\theta^{-1}_{32}\bigg{)}G_{r_{34}}\bigg{(}\frac{\bar{t}_{4}}{t_{3}}\theta^{-1}_{34}\bigg{)}.\end{split}$
(10.41)

These integrals have a similar character to the integrals we met in the three-
point case. They are conformal Mellin integrals over a positive quadrant in
four dimensions, whose integrands are piece-wise constant functions in regions
bounded by hyperplanes through the origin, for example
$t_{1}=\bar{t}_{2}\absolutevalue{\theta^{-1}_{12}}$.

Rather than continue with the derivation of the symmetrised amplitudes, we
will instead derive the unsymmetrised amplitude by Mellin transforming
(10.36). With this, we can find the symmetrised celestial twistor amplitudes
through simple (anti-)symmetrisation.

Unsymmetrised Amplitude

We now calculate the unsymmetrised four-point celestial twistor amplitude in
its entirety, extending the three-point results (10.27) and (10.33). At four
points we are led to consider Mellin transforms of a product of sgn functions.
Once again, we must regularise the integrals such that the Mellin inverse
exists and the regularisation commutes with taking even and odd linear
combinations.

Performing ambidextrous Mellin transforms on the un-symmetrised amplitude
(10.36) leads to the integral

$\displaystyle\begin{split}\mathcal{I}_{4pt}&:=\int_{0}^{\infty}\frac{dt_{1}\,d\bar{t}_{2}\,dt_{3}\,d\bar{t}_{4}}{t_{1}\bar{t}_{2}t_{3}\bar{t}_{4}}\,t_{1}^{2k_{1}}\,\bar{t}_{2}^{2\bar{k}_{2}}\,t_{3}^{2k_{3}}\,\bar{t}_{4}^{2\bar{k}_{4}}\\\
&\qquad\qquad\times\text{sgn}\bigg{(}1+\frac{\bar{t}_{2}}{t_{1}}\theta^{-1}_{12}\bigg{)}\text{sgn}\bigg{(}1+\frac{\bar{t}_{2}}{t_{3}}\theta^{-1}_{32}\bigg{)}\text{sgn}\bigg{(}1+\frac{\bar{t}_{4}}{t_{1}}\theta^{-1}_{14}\bigg{)}\text{sgn}\bigg{(}1+\frac{\bar{t}_{4}}{t_{3}}\theta^{-1}_{34}\bigg{)}.\end{split}$
(10.42)

As usual, we can integrate out an overall scale which we choose to be $t_{1}$.
The resulting integral has a corresponding dilatation invariance delta
function and now manifestly depends on the scales
$\absolutevalue{\theta_{ij}}$:

$\displaystyle\begin{split}\mathcal{I}_{4pt}&=2\pi
i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\int_{0}^{\infty}\frac{dx\,dy\,dz}{xyz}\,x^{2\bar{k}_{2}}y^{2k_{3}}z^{2\bar{k}_{4}}\,\text{sgn}\bigg{(}1+x\,\theta^{-1}_{12}\bigg{)}\\\
&\qquad\qquad\times\text{sgn}\bigg{(}1+\frac{x}{y}\theta^{-1}_{32}\bigg{)}\text{sgn}\bigg{(}1+z\,\theta^{-1}_{14}\bigg{)}\text{sgn}\bigg{(}1+\frac{z}{y}\theta^{-1}_{34}\bigg{)},\\\
&=2\pi i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\int_{0}^{\infty}\frac{dy}{y}\,y^{2k_{3}}\,\operatorname{\mathcal{J}}_{2}(y)\operatorname{\mathcal{J}}_{4}(y),\end{split}$
(10.43)

where we have the Mellin integrals

$\displaystyle\begin{split}\operatorname{\mathcal{J}}_{2}(y):=\int_{0}^{\infty}\frac{dx}{x}\,x^{2\bar{k}_{2}}\,\text{sgn}\bigg{(}1+x\,\theta^{-1}_{12}\bigg{)}\text{sgn}\bigg{(}1+\frac{x}{y}\theta^{-1}_{32}\bigg{)}\\\
=\int_{-\infty}^{\infty}ds\,e^{2\bar{k}_{2}s-\epsilon\absolutevalue{\absolutevalue{s}}}\,\text{sgn}\bigg{(}1+e^{s}\,\theta^{-1}_{12}\bigg{)}\text{sgn}\bigg{(}1+\frac{e^{s}}{y}\theta^{-1}_{32}\bigg{)},\end{split}$
(10.44)

and similarly for $\operatorname{\mathcal{J}}_{4}(y)$. As in the three-point
case two of the integrals separate, and here the third integration over $y$
entwines them. In equation (10.44) above we also define a specific
regularisation which takes into account the two parameters of the integral

$\displaystyle\begin{split}\absolutevalue{\absolutevalue{s}}:=\frac{1}{2}\bigg{(}\text{sgn}\bigg{(}s-\ln(\absolutevalue{\theta_{12}})\bigg{)}+\text{sgn}\bigg{(}s-\ln(\absolutevalue{\theta_{32}})\bigg{)}\bigg{)}\,s,\end{split}$
(10.45)

such that when $s$ is greater than both $\ln(\absolutevalue{\theta_{12}})$ and
$\ln(\absolutevalue{\theta_{32}})$ we have
$\absolutevalue{\absolutevalue{s}}=s$, when $s$ lies between them we have
$\absolutevalue{\absolutevalue{s}}=0$, and when $s$ is less than both we have
$\absolutevalue{\absolutevalue{s}}=-s$.

To calculate (10.44) we must condition on the relative magnitude of
$\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}$. Choosing $y=0$ for
convenience since it can be reinstated at the end, we compute
$\operatorname{\mathcal{J}}_{2}(0)$ to be

$\displaystyle\begin{split}\operatorname{\mathcal{J}}_{2}(0)&:=\Theta\bigg{(}1-\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}\bigg{)}\bigg{[}\frac{(\absolutevalue{\theta_{32}})^{2\bar{k}_{2}}}{2\bar{k}_{2}+\epsilon}+\text{sgn}(\theta_{32})\bigg{(}\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}}{2\bar{k}_{2}}-\frac{(\absolutevalue{\theta_{32}})^{2\bar{k}_{2}}}{2\bar{k}_{2}}\bigg{)}\\\
&\qquad\qquad\qquad\qquad-\text{sgn}(\theta_{12}\theta_{32})\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}}{2\bar{k}_{2}-\epsilon}\bigg{]}\\\
&+\Theta\bigg{(}\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}-1\bigg{)}\bigg{[}\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}}{2\bar{k}_{2}+\epsilon}+\text{sgn}(\theta_{12})\bigg{(}\frac{(\absolutevalue{\theta_{32}})^{2\bar{k}_{2}}}{2\bar{k}_{2}}-\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}}{2\bar{k}_{2}}\bigg{)}\\\
&\qquad\qquad\qquad\qquad-\text{sgn}(\theta_{12}\theta_{32})\frac{(\absolutevalue{\theta_{32}})^{2\bar{k}_{2}}}{2\bar{k}_{2}-\epsilon}\bigg{]},\end{split}$
(10.46)

where we have also dropped the regulariser $\epsilon$ when it appears in the
exponents as usual.

Just as for the Mellin transform of a single sgn function, when
$\theta_{12},\theta_{32}>0$ we recover a contribution proportional to a delta
function, while if $\theta_{12},\theta_{32}<0$ we recover the principal value
pole while the other mixed cases give linear combinations of the two.

To confirm our results we now perform the inverse Mellin transform on (10.46).
In the result (10.46) we have poles with various $\epsilon$-prescriptions -
advanced, retarded and principal value - where poles written
$\frac{1}{2\bar{k}_{2}}$ are principal valued (10.22). These
$\epsilon$-prescriptions are crucial for performing the inverse Mellin
transform. When we integrate $2\bar{k}_{2}$ along the imaginary axis we will
close the contour to form a large semi-circle and pick up various possible
contributions from each of these regulated poles. The direction in which we
close the contour will vary term by term and is dictated by the condition that
the integral dies off on the semicircular arc; this depends on the the
magnitudes of $\absolutevalue{\theta_{12}}$ and $\absolutevalue{\theta_{32}}$.
For example, for terms involving $\absolutevalue{\theta_{12}}$ and for
$\absolutevalue{\theta_{12}}>1$ we close to the left (negative real part) and
we get a non-zero contribution if this term also has a $+\epsilon$
prescription. For terms involving $\absolutevalue{\theta_{32}}$ and for
$\absolutevalue{\theta_{32}}<1$ we close to the right (positive real part) and
we get a non-zero contribution if this term also has a $-\epsilon$
prescription. The principal value terms require some care. We pick up
contributions regardless of which way we close the contour. However, the
principal valued poles are multiplied by a difference of terms and the
residues from the two terms cancel if we close the contour in the same
direction for both terms and their residues sum if we close the contour in
opposite directions for each term.

In summary, if we perform an inverse Mellin transform on (10.46) we have six
cases for the magnitudes of
$\absolutevalue{\theta_{12}},\absolutevalue{\theta_{32}}$. Each case picks up
a contribution from exactly one of the terms in (10.46):

  1. 1.

$\absolutevalue{\theta_{12}}>\absolutevalue{\theta_{32}}>1$ gives
$\Theta\bigg{(}1-\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}\bigg{)}\,\Theta(\absolutevalue{\theta_{12}}-1)\,\Theta(\absolutevalue{\theta_{32}}-1)$

  2. 2.

$\absolutevalue{\theta_{12}}>1>\absolutevalue{\theta_{32}}$ gives
$\Theta\bigg{(}1-\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}\bigg{)}\,\Theta(\absolutevalue{\theta_{12}}-1)\,\Theta(1-\absolutevalue{\theta_{32}})\,\text{sgn}(\theta_{32})$

  3. 3.

$1>\absolutevalue{\theta_{12}}>\absolutevalue{\theta_{32}}$ gives
$\Theta\bigg{(}1-\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}\bigg{)}\,\Theta(1-\absolutevalue{\theta_{12}})\,\Theta(1-\absolutevalue{\theta_{32}})\,\text{sgn}(\theta_{12}\theta_{32})$

  4. 4.

$\absolutevalue{\theta_{32}}>\absolutevalue{\theta_{12}}>1$ gives
$\Theta\bigg{(}\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}-1\bigg{)}\,\Theta(\absolutevalue{\theta_{12}}-1)\,\Theta(\absolutevalue{\theta_{32}}-1)$

  5. 5.

$\absolutevalue{\theta_{32}}>1>\absolutevalue{\theta_{12}}$ gives
$\Theta\bigg{(}\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}-1\bigg{)}\,\Theta(1-\absolutevalue{\theta_{12}})\,\Theta(\absolutevalue{\theta_{32}}-1)\,\text{sgn}(\theta_{12})$

  6. 6.

$1>\absolutevalue{\theta_{32}}>\absolutevalue{\theta_{12}}$ gives
$\Theta\bigg{(}\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}-1\bigg{)}\,\Theta(1-\absolutevalue{\theta_{12}})\,\Theta(1-\absolutevalue{\theta_{32}})\,\text{sgn}(\theta_{12}\theta_{32})$

The sum of the above contributions gives the required result - that is we
recover the original function
$\text{sgn}(1+\theta_{12}^{-1})\,\text{sgn}(1+\theta_{32}^{-1})$.

Once again we can discard the $\epsilon$ prescriptions, and hence delta
function terms, with the proviso that there exists a series of contour
deformations of our inverse Mellin transform such that the Mellin inversion
theorem holds. Then we can find a simpler formula for the
$\operatorname{\mathcal{J}}_{2}(y)$ function, which is (reinstating $y$)

$\displaystyle\begin{split}\operatorname{\mathcal{J}}_{2}(y)&=2\Theta(-\theta_{12})\Theta(\theta_{32})\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}}{2\bar{k}_{2}}+2\,\Theta(\theta_{12})\Theta(-\theta_{32})\frac{(y\absolutevalue{\theta_{32}})^{2\bar{k}_{2}}}{2\bar{k}_{2}}\\\
&+2\,\Theta(-\theta_{12})\Theta(-\theta_{32})\,\text{sgn}\bigg{(}1-y\absolutevalue{\frac{\theta_{32}}{\theta_{12}}}\bigg{)}\bigg{[}\frac{(y\absolutevalue{\theta_{32}})^{2\bar{k}_{2}}}{2\bar{k}_{2}}-\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}}{2\bar{k}_{2}}\bigg{]}.\end{split}$
(10.47)

Using this result we can then find the full un-symmetrised integral
$\mathcal{I}_{4pt}$ from which the unsymmetrised amplitude is simply

$\tilde{\mathcal{A}}_{4}^{+-+-}=\frac{1}{16}\text{sgn}(\langle\mu_{1}\lambda_{2}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\mu_{3}\lambda_{4}\rangle\langle\lambda_{4}\mu_{1}\rangle)\mathcal{I}_{4pt}$
(10.48)

We compute $\mathcal{I}_{4pt}$ by plugging in the function (10.47) into
integral (10.43). The resulting integral over $y$ consists of a collection of
products of sgn functions and so is of the exact same form as the
$\mathcal{J}$ integrals we just calculated. Hence we just iterate the method.

With the usual $\epsilon$ regularisation, which we then drop, we find the
unsymmetrised celestial twistor amplitude can be written compactly as

$\boxed{\begin{split}\tilde{\mathcal{A}}_{4}^{+-+-}=&\frac{\pi
i}{16}\text{sgn}(\langle\mu_{1}\lambda_{2}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\mu_{3}\lambda_{4}\rangle\langle\lambda_{4}\mu_{1}\rangle)\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\;|\theta_{12}|^{2\bar{k}_{2}}|\theta_{14}|^{2k_{3}+2\bar{k}_{4}}|\theta_{34}|^{-2k_{3}}\\\
&\times\bigg{[}\,\Theta(-\theta_{12})\Theta(-\theta_{32})\Theta(-\theta_{14})\Theta(-\theta_{34})\,\text{sgn}(1-|\theta|)\Big{(}V_{12}(\theta)+V_{34}(\theta)-V_{32}(\theta)-V_{14}(\theta)\Big{)}\\\
&-\Theta(-\theta_{12})\Theta(-\theta_{32})\Theta(\theta_{34})\Theta(-\theta_{14})\;V_{34}(\theta)-\Theta(\theta_{12})\Theta(-\theta_{32})\Theta(-\theta_{14})\Theta(-\theta_{34})\;V_{12}(\theta)\\\
&-\Theta(-\theta_{12})\Theta(\theta_{32})\Theta(-\theta_{34})\Theta(-\theta_{14})\;V_{32}(\theta)-\Theta(-\theta_{12})\Theta(-\theta_{32})\Theta(-\theta_{34})\Theta(\theta_{14})\;V_{14}(\theta)\bigg{]}\,,\end{split}}$
(10.49)

where $\theta$ is the ratio of conformally invariant cross-ratios

$\theta:=\frac{\theta_{12}\theta_{34}}{\theta_{14}\theta_{32}}=\frac{\frac{\langle\operatorname{\lambda}_{2}\mu_{1}\rangle\langle\operatorname{\lambda}_{4}\mu_{3}\rangle}{\langle\operatorname{\lambda}_{4}\mu_{1}\rangle\langle\operatorname{\lambda}_{2}\mu_{3}\rangle}}{\frac{[\operatorname{\tilde{\lambda}}_{1}\tilde{\mu}_{4}][\operatorname{\tilde{\lambda}}_{3}\tilde{\mu}_{2}]}{[\operatorname{\tilde{\lambda}}_{1}\tilde{\mu}_{2}][\operatorname{\tilde{\lambda}}_{3}\tilde{\mu}_{4}]}}=:\frac{r}{\bar{r}}\,,$
(10.50)

and the polynomial functions $V_{12}$ etc. are given by

$\displaystyle\begin{split}V_{12}(\theta)=&\frac{\absolutevalue{\theta}^{-2\bar{k}_{2}}}{2\bar{k}_{2}(2\bar{k}_{2}+2k_{3})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4})},\quad
V_{32}(\theta)=\frac{1}{2\bar{k}_{2}2k_{3}(2k_{3}+2\bar{k}_{4})},\\\
V_{14}(\theta)=&\frac{\absolutevalue{\theta}^{2k_{3}+2\bar{k}_{4}}}{2\bar{k}_{4}(2k_{3}+2\bar{k}_{4})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4})},\quad
V_{34}(\theta)=\frac{\absolutevalue{\theta}^{2k_{3}}}{2k_{3}2\bar{k}_{4}(2\bar{k}_{2}+2k_{3})}.\\\
\end{split}$ (10.51)

Equation (10.49) is the main result of this section. We have pulled out a
conformally covariant factor
$\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}\absolutevalue{\theta_{14}}^{2k_{3}+2\bar{k}_{4}}\absolutevalue{\theta_{34}}^{-2k_{3}}$
and the remaining function has support in various ‘channels’ dictated by the
signs of the variables $\theta_{ij}$. Since we have accounted for conformal
covariance, this function can only depend on the conformally invariant cross-
ratios $r,\bar{r}$. In fact, in each of these channels, we have simple
polynomial functions (with imaginary powers) depending only on the absolute
value of the ratio of conformally invariant cross-ratios. On the momentum
space celestial torus, momentum conservation at four points implied that the
conformal cross-ratios $z,\bar{z}$ are not independent but are actually equal
$z=\bar{z}$. Correspondingly, in twistor space, the cross-ratios $r,\bar{r}$
are now independent but the amplitude only depends on their ratio $\theta$.

Now we return to the task of finding the symmetrised celestial twistor
amplitudes. We have sixteen parity combinations for the integrals
$\mathcal{I}(r_{12},r_{32},r_{34},r_{14})$, two for each amplitude in (10.37).
From (10.49) we can find all of them via (anti-)symmetrisation. For example,
we can find the fully symmetric integral

$\begin{split}&\mathcal{I}(0,0,0,0)\\\ &=2\pi
i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}\absolutevalue{\theta_{14}}^{2k_{3}+2\bar{k}_{4}}\absolutevalue{\theta_{34}}^{-2k_{3}}\times\\\
&\\!\bigg{(}\Theta(\absolutevalue{\theta}-1)\bigg{[}-\frac{\absolutevalue{\theta}^{2k_{3}}}{(2k_{3}-\epsilon)(2\bar{k}_{4}+\epsilon)(2\bar{k}_{2}+2k_{3})}-\frac{\absolutevalue{\theta}^{-2\bar{k}_{2}}}{(2\bar{k}_{2}+\epsilon)(2\bar{k}_{2}+2k_{3})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4}+\epsilon)}\bigg{]}\\\
+&\Theta(1-\absolutevalue{\theta})\bigg{[}-\frac{1}{(2\bar{k}_{2}+\epsilon)(2k_{3}-\epsilon)(2k_{3}+2\bar{k}_{4})}-\frac{\absolutevalue{\theta}^{2k_{3}+2\bar{k}_{4}}}{(2\bar{k}_{4}+\epsilon)(2k_{3}+2\bar{k}_{4})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4}+\epsilon)}\bigg{]}\bigg{)}.\end{split}$
(10.52)

The integral $\mathcal{I}(1,1,1,1)$ takes a very similar form,

$\begin{split}&\mathcal{I}(1,1,1,1)\\\ &=2\pi
i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}\absolutevalue{\theta_{14}}^{2k_{3}+2\bar{k}_{4}}\absolutevalue{\theta_{34}}^{-2k_{3}}\,\text{sgn}(\theta_{12}\theta_{32}\theta_{34}\theta_{14})\times\\\
&\\!\bigg{(}\Theta(\absolutevalue{\theta}-1)\bigg{[}\frac{1}{(2\bar{k}_{2}-\epsilon)(2k_{3}+\epsilon)(2k_{3}+2\bar{k}_{4})}+\frac{\absolutevalue{\theta}^{2k_{3}+2\bar{k}_{4}}}{(2\bar{k}_{4}-\epsilon)(2k_{3}+2\bar{k}_{4})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4}-\epsilon)}\bigg{]}\\\
+&\Theta(1-\absolutevalue{\theta})\bigg{[}\frac{\absolutevalue{\theta}^{2k_{3}}}{(2k_{3}+\epsilon)(2\bar{k}_{4}-\epsilon)(2\bar{k}_{2}+2k_{3})}+\frac{\absolutevalue{\theta}^{-2\bar{k}_{2}}}{(2\bar{k}_{2}-\epsilon)(2\bar{k}_{2}+2k_{3})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4}-\epsilon)}\bigg{]}\bigg{)}.\end{split}$
(10.53)

In the above formulae, we have reinstated the $\epsilon$ regularisation
stemming from the result (10.46), which is required to prescribe the correct
contour for the inverse Mellin transform.

The two integrals $\mathcal{I}(0,0,0,0)$ and $\mathcal{I}(1,1,1,1)$ contribute
to the fully symmetrised celestial twistor amplitude

$\tilde{\mathcal{A}}^{+-+-}_{4}\\{0,0,0,0\\}=\frac{1}{16}\text{sgn}(\langle\mu_{1}\lambda_{2}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\mu_{3}\lambda_{4}\rangle\langle\lambda_{4}\mu_{1}\rangle)\tilde{\delta}\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}(\mathcal{I}(0,0,0,0)+\mathcal{I}(1,1,1,1)),$
(10.54)

which in turn is a light transformed correlator in the celestial CFT. A
similar fully symmetrised light transformed correlator was computed in [66] by
performing light transforms on a (2,2) celestial amplitude analytically
continued from (1,3) with summation over incoming/outgoing parameters. In that
paper the result was computed in terms of special functions, Fox H-functions
and Generalised I-functions, defined by Mellin-Barnes integrals; it would be
interesting to understand how these results are related to the discussion
here.

We find that the general structure of the formulas for $\mathcal{I}(0,0,0,0)$
and $\mathcal{I}(1,1,1,1)$ is repeated for all the symmetrised integrals. They
all contain the same basic functions $V_{12},V_{32}$ etc. inherited from the
unsymetrised amplitude (10.49). Unlike the unsymetrised amplitude (10.49),
which has support on a handful of channels dictated by the signs of the
$\theta_{ij}$, the symmetrised amplitudes have a clarified structure with only
two channels which are given by the magnitude of $\absolutevalue{\theta}$
being greater than or less than one.

The only differences between the various symmetrised amplitudes are: the
overall sgn functions which control the even and odd symmetry, which
polynomial contibutions are probed by the two ‘channels’
$\absolutevalue{\theta}>1$ or $\absolutevalue{\theta}<1$, and the $\epsilon$
regularisation of the poles in the weights. We note that, there are two terms
$\mathcal{I}(r_{12},r_{32},r_{34},r_{14})$ which have the same even/odd parity
and so contribute to the same amplitude - these two terms have opposite
arguments $r_{ij}=0\leftrightarrow r_{ij}=1$, opposite sgn function factors
present, opposite $\epsilon$ prescription for the poles and opposite
allocation of terms to each ‘channel’.

Finally we note that all of the symmetrised integrals
$\mathcal{I}(r_{12},r_{32},r_{34},r_{14})$ (10.41) can be encapsulated in the
general formula

$\boxed{\begin{split}&\mathcal{I}(r_{12},r_{32},r_{34},r_{14})=2\pi
i\;\text{sgn}\big{[}\langle\mu_{1}\lambda_{2}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\mu_{3}\lambda_{4}\rangle\langle\lambda_{4}\mu_{1}\rangle\big{]}\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\;|\theta_{12}|^{2\bar{k}_{2}}|\theta_{14}|^{2k_{3}+2\bar{k}_{4}}|\theta_{34}|^{-2k_{3}}\\\
&\times\prod_{(i,j):r_{ij}=1}\\!\\!\\!\\!\\!\\!(\text{sgn}(\theta_{ij}))\sum_{x,y}\bigg{[}\Big{(}\delta^{1}_{x}V_{x}(\theta)-\delta^{0}_{y}V_{y}(\theta)\Big{)}\Theta(|\theta|-1)+\Big{(}\delta^{1}_{y}V_{y}(\theta)-\delta^{0}_{x}V_{x}(\theta)\Big{)}\Theta(1-|\theta|)\bigg{]},\end{split}}$
(10.55)

where in the above the product is over the values of $(i,j)$ such that
$r_{ij}=1$, we sum over the variables
$x\in(r_{32},r_{14}),y\in(r_{12},r_{34})$, and we define
$V_{r_{ij}}(\theta):=V_{ij}(\theta)$. As a further example, for the case
$\mathcal{I}(1,0,1,0)$ we have only a $\Theta(1-|\theta|)$ term

$\begin{split}\mathcal{I}(1,0,1,0)&=2\pi
i\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\bigg{)}\,\text{sgn}(\theta_{12}\theta_{34})\,\Theta(1-\absolutevalue{\theta})\,\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}\absolutevalue{\theta_{14}}^{2k_{3}+2\bar{k}_{4}}\absolutevalue{\theta_{34}}^{-2k_{3}}\\\
&\times\bigg{(}\frac{\absolutevalue{\theta}^{2k_{3}}}{(2k_{3})(2\bar{k}_{4})(2\bar{k}_{2}+2k_{3})}+\frac{\absolutevalue{\theta}^{-2\bar{k}_{2}}}{(2\bar{k}_{2})(2\bar{k}_{2}+2k_{3})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4})}\\\
&-\frac{1}{(2\bar{k}_{2})(2k_{3})(2k_{3}+2\bar{k}_{4})}-\frac{\absolutevalue{\theta}^{2k_{3}+2\bar{k}_{4}}}{(2\bar{k}_{4})(2k_{3}+2\bar{k}_{4})(2\bar{k}_{2}+2k_{3}+2\bar{k}_{4})}\bigg{)}\,,\end{split}$
(10.56)

which can be checked directly by performing the Mellin integrals in (10.41).
Similarly for $\mathcal{I}(0,1,0,1)$ we have only a $\Theta(|\theta|-1)$ term.
The simplicity of the four-point results can be seen to be a consequence of
BCFW recursion relations in celestial twistor space which we explore in detail
in Section 11. The result for $\mathcal{I}$ decomposed into all the even/odd
parts $\mathcal{I}(r_{12},r_{32},r_{34},r_{14})$ is given in a companion
Mathematica file.

Four-Point ‘Separated’ Amplitude

Finally, the separated four-point amplitude $A^{YM}(++--)$ is given by

$\displaystyle A^{++--}_{4}=\frac{\langle 34\rangle^{4}}{\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle
41\rangle}\,\delta^{4}(\lambda^{\alpha}_{1}\tilde{\lambda}^{\dot{\alpha}}_{1}+\lambda^{\alpha}_{2}\tilde{\lambda}^{\dot{\alpha}}_{2}+\lambda^{\alpha}_{3}\tilde{\lambda}^{\dot{\alpha}}_{3}+\lambda^{\alpha}_{4}\tilde{\lambda}^{\dot{\alpha}}_{4}).$
(10.57)

We again perform the same twistor transforms as in (10.35) and so the twistor
amplitude is in the same ambidextrous basis and takes the following form [24]

$\displaystyle\begin{split}\tilde{A}^{++--}_{4}=&\frac{(\pi
i)^{4}}{(2\pi)^{4}}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle+[\tilde{\lambda}_{1}\tilde{\mu}_{2}])\,\text{sgn}(\langle\lambda_{4}\mu_{1}\rangle+[\tilde{\lambda}_{1}\tilde{\mu}_{4}])\\\
&\qquad\quad\times\delta^{(3)}(\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}])\,\text{sgn}(\langle\lambda_{4}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{4}])\\\
=&\frac{1}{16}\text{sgn}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle)\absolutevalue{\langle\lambda_{2}\mu_{3}\rangle}^{-4}\\\
&\qquad\quad\times\text{sgn}(1+\theta_{12}^{-1})\,\text{sgn}(1+\theta_{14}^{-1})\delta^{(3)}(1+\theta_{32}^{-1})\,\text{sgn}(1+\theta_{34}^{-1}).\end{split}$
(10.58)

The separated amplitude is hence
$\absolutevalue{\langle\lambda_{2}\mu_{3}\rangle}^{-4}$ multiplied by the
fourth derivative with respect to $\theta_{32}^{-1}$ of the alternating
amplitude above. Under the half-Mellin transforms this operator becomes
$\absolutevalue{\langle\lambda_{2}\mu_{3}\rangle}^{-4}(|\theta_{32}|^{2}\partial_{|\theta_{32}|})^{4}\,\exp(-2\partial_{\bar{k}_{2}})$,
which we then apply to (10.49). When acting on $\Theta$ functions containing
$\theta_{32}$, either directly or via the cross-ratio $\theta$, these
derivatives give (derivatives of) delta functions, which have only singular
support and so again we drop these terms for generic kinematics. Furthermore,
there are no powers of $\theta_{32}$ in the first line of (10.49). Thus we
only need to apply the derivatives to the powers of $\absolutevalue{\theta}$
inside the $V$ functions. This yields straightforward additional numerator
factors, since

$\displaystyle(|\theta_{32}|^{2}\partial_{|\theta_{32}|})^{4}\absolutevalue{\theta}^{n}=n(n-1)(n-2)(n-3)|\theta_{32}|^{4}|\theta|^{n}.$
(10.59)

Hence we can easily find the separated amplitude from the alternating one.

We leave the task of explicitly computing celestial twistor amplitudes beyond
four points to future work. However, we note that the method of producing
twistor amplitudes by directly half-Fourier transforming momentum space
amplitudes becomes more difficult as the number of particles and MHV degree
grows. In particular in the ambidextrous basis of [24] at five points and
beyond one encounters Grassmann integrals which do not disentangle into sgn
functions. We can Mellin transform these integrals and we recover a class of
integrals related to hyper-geometric functions and their generalisations
studied in [66, 63]. Nevertheless, the simplicity of ambidextrous twistor
amplitudes extends well beyond four points, see for example the six-point
‘alternating’ $(+,-,+,-,+,-)$ amplitude in [24]. This simplicity is largely
due to the existence of powerful BCFW formulae. The topic of BCFW recursion
relations is something we explore in Section 11 and is a promising road to
higher multiplicity celestial twistor amplitudes.

###  10.2 Pure Gravity

For the gravity case, we may follow similar arguments to those used above for
celestial twistor gluon amplitudes.

####  10.2.1 Three Points: Mellin Transform of a Mod Function

Three-Point $MHV$

For the three-point gravity amplitude we begin with the momentum space
expression

$\displaystyle M^{--+}_{3}=\frac{\langle 12\rangle^{6}}{\langle
23\rangle^{2}\langle
31\rangle^{2}}\,\delta^{4}(\lambda^{\alpha}_{1}\tilde{\lambda}^{\dot{\alpha}}_{1}+\lambda^{\alpha}_{2}\tilde{\lambda}^{\dot{\alpha}}_{2}+\lambda^{\alpha}_{3}\tilde{\lambda}^{\dot{\alpha}}_{3}).$
(10.60)

The ambidextrous twistor transform of this is

$\displaystyle\tilde{M}^{--+}_{3}:=\frac{1}{(2\pi)^{3}}\int
d^{2}\tilde{\lambda}_{1}\,d^{2}\tilde{\lambda}_{2}\,d^{2}\lambda_{3}\,e^{i[\tilde{\mu}_{1}\tilde{\lambda}_{1}]}\,e^{i[\tilde{\mu}_{2}\tilde{\lambda}_{2}]}\,e^{i\langle\lambda_{3}\mu_{3}\rangle}M^{--+}_{3},$
(10.61)

which leads to [24]

$\displaystyle\tilde{M}^{--+}_{3}=\frac{(\pi
i)^{3}}{(2\pi)^{3}}\Big{|}\langle\lambda_{1}\lambda_{2}\rangle\Big{|}\,\Big{|}\langle\lambda_{1}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{1}]\Big{|}\,\Big{|}\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}]\Big{|}.$
(10.62)

Analogous arguments to those given above for the YM case lead us to consider
only the amplitudes

$\displaystyle\begin{split}\tilde{M}^{--+}_{3}\\{0,0,0\\},\,\,\,\tilde{M}^{--+}_{3}\\{1,1,0\\},\,\,\,\tilde{M}^{--+}_{3}\\{1,0,1\\},\,\,\,\tilde{M}^{--+}_{3}\\{0,1,1\\},\end{split}$
(10.63)

where the (anti-)symmetrised amplitudes take the general form

$\displaystyle\begin{split}&\tilde{M}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}=\frac{(\pi
i)^{3}}{(2\pi)^{3}}\Big{|}\langle\lambda_{1}\lambda_{2}\rangle\langle\lambda_{1}\mu_{3}\rangle\langle\lambda_{2}\mu_{3}\rangle\Big{|}\,\delta\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\,H_{s_{\bar{k}_{2}}+s_{k_{3}}}(\theta_{31}^{-1})H_{s_{\bar{k}_{1}}+s_{k_{3}}}(\theta_{32}^{-1}),\end{split}$
(10.64)

where we recall the definition used above

$\displaystyle\theta_{ij}=\frac{\langle\lambda_{j}\mu_{i}\rangle}{[\tilde{\lambda}_{i}\tilde{\mu}_{j}]}$
(10.65)

and define the (anti-)symmetrised mod functions

$\displaystyle\begin{split}H_{0}(x):=\frac{1}{2}\bigg{(}\Big{|}1+x\Big{|}+\Big{|}1-x\Big{|}\bigg{)},\\\
H_{1}(x):=\frac{1}{2}\bigg{(}\Big{|}1+x\Big{|}-\Big{|}1-x\Big{|}\bigg{)}.\end{split}$
(10.66)

We then compute the following Mellin transforms

$\displaystyle\begin{split}&\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\,\frac{1}{2}\bigg{(}\Big{|}1+x_{1}\theta_{31}^{-1}\Big{|}+\Big{|}1-x_{1}\theta_{31}^{-1}\Big{|}\bigg{)}=\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}(2\bar{k}_{1}+1)}\,,\\\
&\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\,\frac{1}{2}\bigg{(}\Big{|}1+x_{1}\theta_{31}^{-1}\Big{|}-\Big{|}1-x_{1}\theta_{31}^{-1}\Big{|}\bigg{)}=\text{sgn}(-\theta_{31})\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}(2\bar{k}_{1}+1)}\,.\end{split}$
(10.67)

We are again, by analogy with the Yang-Mills case, adopting a regularisation
$\epsilon>0$ of the weights such that boundary terms at infinity do not
contribute. In the case of the Mellin transform of the antisymmetrised mod
function $H_{1}$, the function is $\mathcal{O}(x)$ for small $x$ and
$\mathcal{O}(1)$ for large values. Hence the strip of definition is for
$-1<\text{Re}(2\bar{k}_{1})<0$. In gravity, the helicity is now $-2$ for leg
one and the weight has real part $\text{Re}(2\bar{k}_{1})=-1$. Hence we can
regularise the weight with $2\bar{k}_{1}\rightarrow 2\bar{k}_{1}+\epsilon$ as
usual. In other words, to invert the Mellin transform we can choose a straight
line contour which goes to the right of the pole at minus one and to left of
the pole at zero.

However, the case of the symmetrised $H_{0}$ function in gravity is more
subtle. The $H_{0}$ function is $\mathcal{O}(1)$ for small $x$ but is
$\mathcal{O}(x)$ for large $x$ and so the strip of definition does not exist
and the integral is strongly divergent regardless of the value of the weights.
This fact has been noted elsewhere in the literature e.g. [70, 71] \- in
gravity the Mellin transform does not in general converge and this had led to
the definition of a ‘regularised Mellin transform’ which importantly includes
an explicit factor of $e^{-\delta x}$ in its definition [70, 72]. We
implicitly use such a regularised Mellin transform and note that when we take
$\delta\rightarrow 0_{+}$ we recover the symmetric result of (10.67). One can
also consider taking an inverse Mellin transform of the result by integrating
over the weight $2\bar{k}_{1}$ to recover the original function $H_{0}$. In
this case, a suitable contour does exist, although it is no longer a vertical
straight line since there is no such strip of definition. Instead, a contour
which snakes to the left of the pole at minus one and then back around to the
right of the pole at zero allows us to pick up the correct residues and
recover $H_{0}$.

The three-point MHV celestial gravity twistor amplitude is then given by

$\boxed{\begin{split}\tilde{\mathcal{M}}^{--+}_{3}\\{s_{\bar{k}_{1}},s_{\bar{k}_{2}},s_{k_{3}}\\}=&\frac{\pi}{4}\Big{|}\langle\lambda_{1}\lambda_{2}\rangle\langle\lambda_{1}\mu_{3}\rangle\langle\lambda_{2}\mu_{3}\rangle\Big{|}\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\,+2\bigg{)}\delta\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\\\
&\quad\times\text{sgn}(-\theta_{31})^{s_{\bar{k}_{1}}}\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}(2\bar{k}_{1}+1)}\text{sgn}(-\theta_{32})^{s_{\bar{k}_{2}}}\frac{\absolutevalue{\theta_{32}}^{2\bar{k}_{2}}}{2\bar{k}_{2}(2\bar{k}_{2}+1)}\,.\end{split}}$
(10.68)

Formulae (10.67) together imply that the unsymmetrised integral is given by

$\displaystyle\int_{0}^{\infty}\frac{dx_{1}}{x_{1}}\,x_{1}^{2\bar{k}_{1}}\,\Big{|}1+x_{1}\theta_{31}^{-1}\Big{|}=2\,\Theta(-\theta_{31})\frac{|\theta_{31}|^{2\bar{k}_{1}}}{2\bar{k}_{1}(2\bar{k}_{1}+1)}\,,$
(10.69)

where as usual we drop terms proportional to a delta function in weight space.
Similar considerations mentioned in the above cases also apply here regarding
the regularisation of (10.69). Finally, we conclude that the unsymmetrised
amplitude is

$\boxed{\tilde{\mathcal{M}}^{--+}_{3}=\frac{\pi}{4}\Big{|}\langle\lambda_{1}\lambda_{2}\rangle\langle\lambda_{1}\mu_{3}\rangle\langle\lambda_{2}\mu_{3}\rangle\Big{|}\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\,+2\bigg{)}\,\Theta(-\theta_{31})\Theta(-\theta_{32})\frac{\absolutevalue{\theta_{31}}^{2\bar{k}_{1}}}{2\bar{k}_{1}(2\bar{k}_{1}+1)}\frac{\absolutevalue{\theta_{32}}^{2\bar{k}_{2}}}{2\bar{k}_{2}(2\bar{k}_{2}+1)}.}$
(10.70)

Three-Point $\overline{MHV}$

The $\tilde{\mathcal{M}}^{++-}_{3}$ amplitude is obtained by similar
manipulations, and is

$\boxed{\begin{split}\tilde{\mathcal{M}}^{++-}_{3}\\{s_{k_{1}},s_{k_{2}},s_{\bar{k}_{3}}\\}=&\frac{\pi}{4}\Big{|}[\tilde{\lambda}_{1}\tilde{\lambda}_{2}][\tilde{\mu}_{3}\tilde{\lambda}_{1}][\tilde{\mu}_{3}\tilde{\lambda}_{2}]\Big{|}\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\,+2\bigg{)}\delta\bigg{(}\sum_{i}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}\bigg{)}\\\
&\quad\times\text{sgn}(-\theta_{13})^{s_{k_{1}}}\frac{\absolutevalue{\theta_{13}^{-1}}^{2k_{1}}}{2k_{1}(2k_{1}+1)}\;\text{sgn}(-\theta_{23})^{s_{k_{2}}}\frac{\absolutevalue{\theta_{23}^{-1}}^{2\bar{k}_{2}}}{2\bar{k}_{2}(2\bar{k}_{2}+1)},\end{split}}$
(10.71)

while the unsymmetrised amplitude is

$\boxed{\tilde{\mathcal{M}}^{++-}_{3}=\frac{\pi}{4}\Big{|}[\tilde{\lambda}_{1}\tilde{\lambda}_{2}][\tilde{\mu}_{3}\tilde{\lambda}_{1}][\tilde{\mu}_{3}\tilde{\lambda}_{2}]\Big{|}\,\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\,+2\bigg{)}\,\Theta(-\theta_{13})\Theta(-\theta_{23})\frac{\absolutevalue{\theta_{13}^{-1}}^{2k_{1}}}{2k_{1}(2k_{1}+1)}\;\frac{\absolutevalue{\theta_{23}^{-1}}^{2\bar{k}_{2}}}{2\bar{k}_{2}(2\bar{k}_{2}+1)}.}$
(10.72)

####  10.2.2 Four Points: Mellin Transform of a Product of Mod Functions

Four-Point ‘Alternating’ Amplitude

Turning to four points, the ‘alternating’ four-point twistor amplitude takes
the following form [24]

$\displaystyle\begin{split}\tilde{M}^{+-+-}_{4}=&\frac{(\pi
i)^{4}}{(2\pi)^{4}}\Big{|}\langle\lambda_{2}\mu_{1}\rangle+[\tilde{\lambda}_{1}\tilde{\mu}_{2}]\Big{|}\,\Big{|}\langle\lambda_{4}\mu_{1}\rangle+[\tilde{\lambda}_{1}\tilde{\mu}_{4}]\Big{|}\Big{|}\langle\lambda_{2}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{2}]\Big{|}\,\Big{|}\langle\lambda_{4}\mu_{3}\rangle+[\tilde{\lambda}_{3}\tilde{\mu}_{4}]\Big{|}\\\
=&\frac{1}{16}\Big{|}\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle\Big{|}\Big{|}1+\theta_{12}^{-1}\Big{|}\,\Big{|}1+\theta_{14}^{-1}\Big{|}\Big{|}1+\theta_{32}^{-1}\Big{|}\,\Big{|}1+\theta_{34}^{-1}\Big{|}.\end{split}$
(10.73)

We now perform the Mellin transforms first, following the Yang-Mills case for
guidance, and only (anti-)symmetrise afterward. In evaluating the integrals
below, we will drop boundary terms at zero and infinity by using suitable
regularisations as per the discussion above. We will need the integral

$\displaystyle\mathcal{K}^{(a,b)}(y,k):=\int_{0}^{\infty}\frac{dx}{x}x^{2k}\Big{|}1+\frac{x}{a}\Big{|}\Big{|}1+\frac{x}{y\,b}\Big{|},$
(10.74)

which is the gravity analogue of the $\operatorname{\mathcal{J}}_{2}(y)$ and
$\operatorname{\mathcal{J}}_{4}(y)$ Yang-Mills integrals (10.44). Defining the
indefinite integral

$\displaystyle\mathcal{L}^{(a,b)}(x,y,k):=\int\frac{dx}{x}x^{2k}\Big{(}1+\frac{x}{a}\Big{)}\Big{(}1+\frac{x}{y\,b}\Big{)}$
$\displaystyle=\frac{x^{2k}}{2k}+\frac{x^{2k+1}}{2k+1}\bigg{(}\frac{1}{yb}+\frac{1}{a}\bigg{)}+\frac{x^{2k+2}}{2k+2}\frac{1}{yab}$
(10.75)

we find

$\displaystyle\begin{split}\mathcal{K}^{(a,b)}(y,k)=&2\,\Theta(b)\Theta(-a)\mathcal{L}^{(a,b)}(|a|,y,k)+2\,\Theta(-b)\Theta(a)\mathcal{L}^{(a,b)}(|b|y,y,k)\\\
&+\Theta(-b)\Theta(-a)\,\text{sgn}\bigg{(}1-y\absolutevalue{\frac{b}{a}}\bigg{)}\Big{(}2\,\mathcal{L}^{(a,b)}(|b|y,y,k)-2\,\mathcal{L}^{(a,b)}(|a|,y,k)\Big{)}.\end{split}$
(10.76)

We note the similarity with the analogous Yang-Mills result (10.47) the only
difference being the form of the indefinite integral (10.75).

The gravity amplitude is then given by

$\displaystyle\begin{split}\tilde{\mathcal{M}}^{+-+-}_{4}=&2\pi
i\;\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\,+4\bigg{)}\;\big{|}\langle\lambda_{2}\mu_{1}\rangle\big{|}\big{|}\langle\lambda_{4}\mu_{1}\rangle\big{|}\big{|}\langle\lambda_{2}\mu_{3}\rangle\big{|}\big{|}\langle\lambda_{4}\mu_{3}\rangle\big{|}\;\\\
&\qquad\times\int_{0}^{\infty}dy\;y^{2k_{3}-1}\mathcal{K}^{(\theta_{12},\theta_{32})}(y,\bar{k}_{2})\mathcal{K}^{(\theta_{14},\theta_{34})}(y,\bar{k}_{4}),\end{split}$
(10.77)

where just like at three points we have a shifted dilatation invariance delta
function. Dilatation invariance for both Yang-Mills and gravity is linked to
the mass dimension of a $n$-point amplitude which must be $-n$ in both cases.
However, for gravity not just the momentum but also the coupling,
$\kappa\coloneqq\sqrt{32\pi G_{N}}$, carries mass dimension minus one. Hence,
for an $n$ point amplitude with $n-2$ powers of $\kappa$, the total power of
spinors appearing is shifted by $2(n-2)$, and so the dilatation invariance
delta function for gravity is shifted by $2(n-2)$, matching our results at
three and four points.

Before stating the integrated result of (10.77), we first define the general
form of the weight dependent coefficients that appear in the gravity case

$\displaystyle
C_{k,l,m}=-\frac{1}{2k(2k+1)}\;\frac{1}{2l(2l+1)}\;\frac{1}{2m(2m+1)}\,.$
(10.78)

We also define the following functions of the conformal cross-ratios

$\displaystyle\begin{split}W^{\pm}_{12}(\theta)=&\;C_{\bar{k}_{2},\bar{k}_{2}+k_{3}-\frac{1}{2},\bar{k}_{2}+k_{3}+\bar{k}_{4}}\;|\theta|^{-2\bar{k}_{2}}\pm
C_{\bar{k}_{2}+\frac{1}{2},\bar{k}_{2}+k_{3},\bar{k}_{2}+k_{3}+\bar{k}_{4}+\frac{1}{2}}\;\absolutevalue{\theta}^{-2\bar{k}_{2}-1},\\\
W^{\pm}_{32}(\theta)=&\;\pm
C_{\bar{k}_{2}+\frac{1}{2},k_{3}-1,k_{3}+\bar{k}_{4}-\frac{1}{2}}\;|\theta|\;+C_{\bar{k}_{2},k_{3}-\frac{1}{2},k_{3}+\bar{k}_{4}}\,,\\\
W^{\pm}_{14}(\theta)=&\;\pm
C_{\bar{k}_{4}+\frac{1}{2},k_{3}+\bar{k}_{4},\bar{k}_{2}+k_{3}+\bar{k}_{4}+\frac{1}{2}}\;|\theta|^{2k_{3}+2\bar{k}_{4}+1}+C_{\bar{k}_{4},k_{3}+\bar{k}_{4}-\frac{1}{2},\bar{k}_{2}+k_{3}+\bar{k}_{4}}\;|\theta|^{2k_{3}+2\bar{k}_{4}}\,,\\\
W^{\pm}_{34}(\theta)=&\;C_{\bar{k}_{4},k_{3}-\frac{1}{2},\bar{k}_{2}+k_{3}}\;|\theta|^{2k_{3}}\pm
C_{\bar{k}_{4}+1/2,k_{3}-1,\bar{k}_{2}+k_{3}-\frac{1}{2}}\;|\theta|^{2k_{3}-1}\,.\end{split}$
(10.79)

Finally, computing (10.77) the four-point ‘alternating’ unsymmetrised
celestial twistor gravity amplitude is then

$\boxed{\begin{split}\tilde{\mathcal{M}}^{+-+-}_{4}&=\;\pi
i\;\delta\bigg{(}\sum_{i}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}\,+4\bigg{)}\;\big{|}(\langle\lambda_{2}\mu_{1}\rangle\langle\lambda_{4}\mu_{1}\rangle\langle\lambda_{2}\mu_{3}\rangle\langle\lambda_{4}\mu_{3}\rangle\big{|}\;|\theta_{12}|^{2\bar{k}_{2}}|\theta_{14}|^{2k_{3}+2\bar{k}_{4}}|\theta_{34}|^{-2k_{3}}\\\
&\times\bigg{[}\,\Theta(-\theta_{12})\Theta(-\theta_{32})\Theta(-\theta_{14})\Theta(-\theta_{34})\,\text{sgn}(1-|\theta|)\Big{(}W^{-}_{12}(\theta)+W^{-}_{34}(\theta)-W^{-}_{32}(\theta)-W^{-}_{14}(\theta)\Big{)}\\\
&\,\,\,\,-\Theta(-\theta_{12})\Theta(-\theta_{32})\Theta(\theta_{34})\Theta(-\theta_{14})\;W^{+}_{34}(\theta)-\Theta(\theta_{12})\Theta(-\theta_{32})\Theta(-\theta_{14})\Theta(-\theta_{34})\;W^{+}_{12}(\theta)\\\
&\,\,\,\,-\Theta(-\theta_{12})\Theta(\theta_{32})\Theta(-\theta_{34})\Theta(-\theta_{14})\;W^{+}_{32}(\theta)-\Theta(-\theta_{12})\Theta(-\theta_{32})\Theta(-\theta_{34})\Theta(\theta_{14})\;W^{+}_{14}(\theta)\bigg{]}.\end{split}}$
(10.80)

From this, we can derive the symmetrised amplitudes by considering
(anti-)symmetrisations which satisfy $\sum
s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}=0\mod 2$; the others
are zero. Note the remarkably similar structure to that of the Yang-Mills
amplitude (10.49). All the symmetrised amplitudes in gravity may be captured
by an analogous formula to (10.55) in the Yang-Mills case. Finally, the
‘separated’ amplitude with helicities $(++--)$ is then obtained from this by
derivatives with respect to $|\theta_{32}|$, as discussed in the Yang-Mills
case above.

##  11 BCFW Recursion

The BCFW-type recursion relations of [73] have been developed in twistor space
in [24], and here we will apply them in the celestial context. In this
section, we will generate celestial twistor recursion relations and show how
these can be used in practice to re-derive our results at four points.

For an $n$-point twistor amplitude, where the deformed legs $i$ and $j$ have
helicities $(+,-)$, the recursion relation of [24] is

$\displaystyle\begin{split}\tilde{A}_{n}=&\sum_{L,R}\text{sgn}(\langle\mu_{i}\lambda_{j}\rangle+[\tilde{\mu}_{j}\tilde{\lambda}_{i}])\int
d^{2}\lambda\;d^{2}\operatorname{\tilde{\lambda}}\;d^{2}\mu\;d^{2}\tilde{\mu}\;e^{i[\tilde{\lambda}\tilde{\lambda}_{i}]}e^{i\langle\lambda\lambda_{j}\rangle}\;\delta^{{}^{\prime\prime\prime}}\\!\big{(}\langle\operatorname{\lambda}\mu\rangle+[\operatorname{\tilde{\lambda}}\tilde{\mu}]\big{)}\;\\\
&\qquad\qquad\;\Bigg{(}\tilde{A}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},+);(\lambda,\tilde{\mu},-)\Big{]}\tilde{A}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-);(\mu,\tilde{\lambda},+)\Big{]}\\\
&\qquad\qquad\qquad\qquad\qquad+\tilde{A}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},+);(\mu,\tilde{\lambda},+)\Big{]}\tilde{A}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-);(\lambda,\tilde{\mu},-)\Big{]}\Bigg{)},\end{split}$
(11.1)

where we are writing the twistors in terms of spinors as
$W=(\mu,\tilde{\lambda}),Z=(\lambda,\tilde{\mu})$ and any external leg
variables that are not shifted in the amplitudes are not written explicitly.
Note we work in an ambidextrous basis which follows the helicity of the
particles, that is we use $W_{i},Z_{j}$ for the $(+,-)$ deformed legs. We also
sum over the helicity of the glued leg which gives the two terms in (11.1)
with different twistor variables - $(\mu,\tilde{\lambda})$ for plus helicity
and $(\lambda,\tilde{\mu})$ for minus helicity.

We will focus on the case of Yang-Mills in the following. To obtain the
recursion relation for the Mellin-transformed twistor amplitudes, we perform
Mellin transforms on all the external variables, and inverse Mellin transforms
on all the variables inside the amplitudes on the right-hand side of the above
equation. Using the scaling properties of the amplitudes, the Mellin and
inverse Mellin transforms cancel for all the external variables on the right-
hand side except those labelled by $i$ and $j$. We also have the inverse
Mellin transform over $K,\bar{K}$ for the glued legs of each left and right
amplitude, we can choose a single pair of weights $K,\bar{K}$ since the
integral transform acts linearly on the BCFW expression. We thus obtain the
following BCFW relation

$\displaystyle\begin{split}\tilde{\mathcal{A}}_{n}=&\sum_{L,R}\int
d^{2}\lambda\;d^{2}\tilde{\lambda}\;d^{2}\mu\;d^{2}\tilde{\mu}\int\;\frac{dK\;d\bar{K}}{(2\pi
i)^{2}}\sum_{s_{K},s_{\bar{K}}\in\\{0,1\\}}\;e^{i[\tilde{\lambda}\tilde{\lambda}_{i}]}e^{i\langle\lambda\lambda_{j}\rangle}\;\delta^{{}^{\prime\prime\prime}}\Big{(}\langle\operatorname{\lambda}\mu\rangle+[\operatorname{\tilde{\lambda}}\tilde{\mu}]\Big{)}\;\\\
&\;\int\;\frac{d\kappa_{i}\;d\bar{\kappa}_{j}}{(2\pi
i)^{2}}\\!\\!\sum_{s_{\kappa_{i}},s_{\bar{\kappa}_{j}}\in\\{0,1\\}}\\!\\!\\!\\!\\!\\!I(i,j)\Bigg{(}\;\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},\kappa_{i});(\lambda,\tilde{\mu},\bar{K})\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},\bar{\kappa}_{j});(\mu,\tilde{\lambda},K)\Big{]}\\\
&\qquad\qquad\qquad\qquad+\;\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},\kappa_{i});(\mu,\tilde{\lambda},K)\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},\bar{\kappa}_{j});(\lambda,\tilde{\mu},\bar{K})\Big{]}\Bigg{)},\end{split}$
(11.2)

where for notational simplicity we have left implicit the discrete weights
$s_{K}$ etc. which follow suit with the continuous ones. We have also defined
the Mellin integrals $I(i,j)$ for legs $i$ and $j$ which we evaluate first
using the scale covariance of the celestial amplitudes

$\displaystyle\begin{split}\mathcal{I}(i,j)\coloneqq&\int_{\operatorname{\mathbb{R}}_{*}}\frac{dt_{i}}{\absolutevalue{t_{i}}}\;\absolutevalue{t_{i}}^{2(k_{i}-\kappa_{i})}\text{sgn}(t_{i})^{s_{k_{i}}-s_{\kappa_{i}}}\;\int_{\operatorname{\mathbb{R}}_{*}}\frac{d\bar{t}_{j}}{\absolutevalue{\bar{t}_{j}}}\;\absolutevalue{\bar{t}}_{j}^{2(\bar{k}_{j}-\bar{\kappa}_{j})}\text{sgn}(\bar{t}_{j})^{s_{\bar{k}_{j}}-s_{\bar{\kappa}_{j}}}\\\
&\qquad\qquad\qquad\qquad\qquad\text{sgn}\big{(}t_{i}\langle\mu_{i}\lambda_{j}\rangle+\bar{t}_{j}[\tilde{\lambda}_{i}\tilde{\mu}_{j}]\big{)}\\\
=&\;2\pi
i\,\delta(2(k_{i}+\bar{k}_{j}-\kappa_{i}-\bar{\kappa}_{j}))\,\tilde{\delta}(s_{k_{i}}+s_{\bar{k}_{j}}-s_{\kappa_{i}}-s_{\bar{\kappa}_{j}}+1)\,\text{sgn}(\langle\mu_{i}\operatorname{\lambda}_{j}\rangle)\absolutevalue{\theta_{ij}}^{2(\bar{k}_{j}-\bar{\kappa}_{j})}\\\
&\qquad\qquad\text{sgn}(-\theta_{ij})^{s_{\bar{k}_{j}}-s_{\bar{\kappa}_{j}}}\frac{1}{2(\bar{k}_{j}-\bar{\kappa}_{j})+(-1)^{s_{\bar{k}_{j}}-s_{\bar{\kappa}_{j}}}\epsilon}\\\
=&:2\pi
i\,\delta(2(k_{i}+\bar{k}_{j}-\kappa_{i}-\bar{\kappa}_{j}))\,\tilde{\delta}(s_{k_{i}}+s_{\bar{k}_{j}}-s_{\kappa_{i}}-s_{\bar{\kappa}_{j}}+1)\\\
&\qquad\qquad\qquad\,\text{sgn}(\langle\mu_{i}\operatorname{\lambda}_{j}\rangle)\,\mathcal{G}(\bar{k}_{j}-\bar{\kappa}_{j},s_{\bar{k}_{j}}-s_{\bar{\kappa}_{j}},\theta_{ij}).\end{split}$
(11.3)

which we have evaluated in terms of the function $\mathcal{G}(k,s,x)$ which is
just defined to be the Mellin transform over $\operatorname{\mathbb{R}}_{+}$
of $G_{s}(x)$ appearing in (10.17). Note again that, for readability, we are
using the notation $\tilde{\delta}$ for a Kronecker delta evaluated mod 2.

We continue performing all the Mellin transforms such that the BCFW relation
is purely in terms of projective integrals. We first write
$\delta^{{}^{\prime\prime\prime}}(x)\sim\int ds\;s^{3}\;e^{isx}$ and then by
rescaling $\mu,\tilde{\mu}$ by $1/s$ we perform the integral over $s$ which
generates the delta functions below in (11.4). In addition, we also break up
the BCFW measure as
$d^{2}\operatorname{\lambda}=du\absolutevalue{u}\langle\operatorname{\lambda}d\operatorname{\lambda}\rangle$,
$d^{2}\mu=dt\absolutevalue{t}\langle\mu d\mu\rangle$ etc., and perform the
integrals over the scales $u,\bar{u},t,\bar{t}$ which we can extract from the
left and right celestial subamplitudes. Now recall that we are using
ambidextrous variables for the glued legs and so we always have the relations
$\bar{h}=-k$ or $h=-\bar{k}$ which we used at three and four-point and are
specific to gluon amplitudes. Therefore we pull out the following factors in
the integrals over $u,\bar{u},t,\bar{t}$ 181818Performing these integrals
requires analytic continuation. That is we relax the conditions from the delta
function and compute the integrals in the region of $K,\bar{K}$ space where
they converge. We then analytically continue the final expression to the
region supported by the delta functions.

$\begin{split}\mathcal{I}(\bar{K},K)\coloneqq&\;2\pi
i\,\delta(2K+2\bar{K})\,\tilde{\delta}(s_{K}+s_{\bar{K}}+1)\int_{\operatorname{\mathbb{R}}_{*}}\\!du\,d\bar{u}\,dt\,d\bar{t}\,\absolutevalue{u}^{1+2\bar{K}}\text{sgn}(u)^{-s_{\bar{K}}}\absolutevalue{\bar{u}}^{1+2K}\text{sgn}(\bar{u})^{-s_{K}}\\\
&\qquad\times\absolutevalue{\bar{t}}^{1-2\bar{K}}\text{sgn}(\bar{t})^{-s_{\bar{K}}}\absolutevalue{t}^{1-2K}\text{sgn}(t)^{-s_{K}}\,e^{iu\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle}\,e^{i\bar{u}[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}]}\,e^{i\,tu\langle\operatorname{\lambda}\mu\rangle}\,e^{i\,\bar{t}\bar{u}[\operatorname{\tilde{\lambda}}\tilde{\mu}]}.\end{split}$
(11.4)

These integrals are of an identical form to (9.6), encountered when relating a
half-Fourier transform to a light transform. From (9.6) we define

$\mathcal{N}(a,b,x):=\int_{\operatorname{\mathbb{R}}_{*}}\,dt\,\absolutevalue{t}^{a-1}\text{sgn}(t)^{b}e^{itx}=2\pi\,i^{-b}\frac{\Gamma(a)}{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-a-b}{2})}\absolutevalue{x}^{-a}\text{sgn}(x)^{b},$
(11.5)

then we have

$\begin{split}\mathcal{I}(\bar{K},K)=&\;2\pi
i\,\delta(2K+2\bar{K})\,\tilde{\delta}(s_{K}+s_{\bar{K}}+1)\mathcal{N}(2-2K,-s_{K},\langle\operatorname{\lambda}\mu\rangle)\mathcal{N}(2-2\bar{K},-s_{\bar{K}},[\operatorname{\tilde{\lambda}}\tilde{\mu}])\\\
&\times\int_{\operatorname{\mathbb{R}}_{*}}\\!du\,d\bar{u}\,\absolutevalue{u}^{2K+2\bar{K}-1}\text{sgn}(u)^{-s_{K}-s_{\bar{K}}}\absolutevalue{\bar{u}}^{2\bar{K}+2K-1}\text{sgn}(\bar{u})^{-s_{K}-s_{\bar{K}}}\,e^{iu\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle}\,e^{i\bar{u}[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}]}.\end{split}$
(11.6)

Now we perform the final integrals and simplify on the constraint of the delta
functions,

$\begin{split}\mathcal{I}(\bar{K},K)=&\;(2\pi
i)^{3}\,\delta(2K+2\bar{K})\,\tilde{\delta}(s_{K}+s_{\bar{K}}+1)\,\mathcal{N}(2-2K,-s_{K},\langle\operatorname{\lambda}\mu\rangle)\\\
&\qquad\times\mathcal{N}(2-2\bar{K},-s_{\bar{K}},[\operatorname{\tilde{\lambda}}\tilde{\mu}])\,\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}])\\\
=&:(2\pi
i)^{3}\,\delta(2K+2\bar{K})\,\tilde{\delta}(s_{K}+s_{\bar{K}}+1)\,\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}])\,\mathcal{R}(2\bar{K},2K,s_{\bar{K}},s_{K}).\end{split}$
(11.7)

Thus we obtain a BCFW relation given by

$\displaystyle\begin{split}\tilde{\mathcal{A}}_{n}=&\;(2\pi
i)^{4}\text{sgn}(\langle\mu_{i}\operatorname{\lambda}_{j}\rangle)\sum_{L,R}\int\frac{dK\;d\bar{K}}{(2\pi
i)^{2}}\\!\\!\sum_{s_{K},s_{\bar{K}}\in\\{0,1\\}}\\!\\!\\!\\!\delta(2K+2\bar{K})\,\tilde{\delta}(s_{K}+s_{\bar{K}}+1)\\\
&\times\int\\!\frac{d\kappa_{i}\;d\bar{\kappa}_{j}}{(2\pi
i)^{2}}\\!\\!\\!\\!\sum_{s_{\kappa_{i}},s_{\bar{\kappa}_{j}}\in\\{0,1\\}}\\!\\!\\!\\!\\!\\!\delta(2(k_{i}\\!+\\!\bar{k}_{j}\\!-\\!\kappa_{i}\\!-\\!\bar{\kappa}_{j}))\,\tilde{\delta}(s_{k_{i}}\\!+\\!s_{\bar{k}_{j}}\\!-\\!s_{\kappa_{i}}\\!-\\!s_{\bar{\kappa}_{j}}\\!+1)\\\
&\times\mathcal{G}(\bar{k}_{j}-\bar{\kappa}_{j},s_{\bar{k}_{j}}-s_{\bar{\kappa}_{j}},\theta_{ij})\int\,\langle\operatorname{\lambda}d\operatorname{\lambda}\rangle\,[\operatorname{\tilde{\lambda}}d\operatorname{\tilde{\lambda}}]\,\langle\mu\,d\mu\rangle\,[\tilde{\mu}\,d\tilde{\mu}]\,\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}])\\\
&\times\Bigg{(}\mathcal{R}(2\bar{K},2K,s_{\bar{K}},s_{K})\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},\kappa_{i});(\lambda,\tilde{\mu},\bar{K})\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},\bar{\kappa}_{j});(\mu,\tilde{\lambda},K)\Big{]}\\\
&+\mathcal{R}(2K,2\bar{K},s_{K},s_{\bar{K}})\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},\kappa_{i});(\mu,\tilde{\lambda},K)\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},\bar{\kappa}_{j});(\lambda,\tilde{\mu},\bar{K})\Big{]}\Bigg{)}.\end{split}$
(11.8)

Having done all the scale integrals we now have a projective form of the BCFW
recursion relations in celestial twistor space. We can actually simplify the
above expression greatly since the integrals which glue the weights can be
performed against the delta functions. The left and right Yang-Mills
subamplitudes come with dilatation invariance delta functions and also
Kronecker deltas enforcing overall evenness. For example, for the first
product of amplitudes we have the delta functions

$\begin{split}\delta\bigg{(}2\bigg{(}\kappa_{i}+\sum_{m\in
L,m\not=i}\\!\\!\\!\mathrel{\mathop{\bar{k}}\limits^{\vbox to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{m}+\bar{K}\bigg{)}\bigg{)}\,\tilde{\delta}\bigg{(}s_{\kappa_{i}}+\sum_{m\in
L,m\not=i}\\!\\!s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{m}}+s_{\bar{K}}\bigg{)}\\\
\times\delta\bigg{(}2\bigg{(}\bar{\kappa}_{j}+\sum_{m\in
R,m\not=j}\\!\\!\\!\mathrel{\mathop{\bar{k}}\limits^{\vbox to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{m}+K\bigg{)}\bigg{)}\,\tilde{\delta}\bigg{(}s_{\bar{\kappa}_{j}}+\sum_{m\in
R,m\not=j}\\!\\!s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{m}}+s_{K}\bigg{)},\end{split}$
(11.9)

while for the second term we simply swap $K$ and $\bar{K}$. We now use the
delta functions in (11.9) to localise the sums and integrals over
$\kappa_{i},\bar{\kappa}_{j},s_{\kappa_{i}},s_{\bar{\kappa}_{j}}$. In addition
we have the four external delta functions in (11.8) and plugging in the
localised values of
$\kappa_{i},\bar{\kappa}_{j},s_{\kappa_{i}},s_{\bar{\kappa}_{j}}$ these give
the overall dilatation invariance delta functions of the amplitude
$\tilde{\mathcal{A}}_{n}$

$\begin{split}&\delta(2K+2\bar{K})\,\tilde{\delta}(s_{K}+s_{\bar{K}}+1)\delta\bigg{(}\sum_{a=1}^{n}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}+2\\!K+2\bar{K}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a=1}^{n}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}+\\!s_{K}\\!+\\!s_{\bar{K}}\\!+1\bigg{)}\\\
=&\;\delta\bigg{(}\sum_{a=1}^{n}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a=1}^{n}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\delta\bigg{(}2\\!K+2\bar{K}\bigg{)}\,\tilde{\delta}\bigg{(}\\!s_{K}\\!+\\!s_{\bar{K}}\\!+1\bigg{)}.\end{split}$
(11.10)

Hence, we have the following BCFW recursion relation

$\displaystyle\begin{split}\tilde{\mathcal{A}}_{n}=&(2\pi
i)^{2}\text{sgn}(\langle\mu_{i}\operatorname{\lambda}_{j}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\sum_{L,R}\int\frac{dK\;d\bar{K}}{(2\pi
i)^{2}}\\!\\!\\!\sum_{s_{K},s_{\bar{K}}\in\\{0,1\\}}\\\
&\times\delta\bigg{(}2\\!K+2\bar{K}\bigg{)}\,\tilde{\delta}\bigg{(}\\!s_{K}\\!+\\!s_{\bar{K}}\\!+1\bigg{)}\,\mathcal{G}(\bar{k}_{j}+k_{R}+\bar{K},s_{\bar{k}_{j}}+s_{k_{R}}+s_{\bar{K}},\theta_{ij})\\\
&\times\int\,\langle\operatorname{\lambda}d\operatorname{\lambda}\rangle\,[\operatorname{\tilde{\lambda}}d\operatorname{\tilde{\lambda}}]\,\langle\mu\,d\mu\rangle\,[\tilde{\mu}\,d\tilde{\mu}]\,\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}])\,\mathcal{R}(2K,2\bar{K},s_{K},s_{\bar{K}})\\\
&\times\Bigg{(}\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},-k_{L}-K);(\lambda,\tilde{\mu},K)\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-k_{R}-\bar{K});(\mu,\tilde{\lambda},\bar{K})\Big{]}\\\
&\quad+\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},-k_{L}-K);(\mu,\tilde{\lambda},K)\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-k_{R}-\bar{K});(\lambda,\tilde{\mu},\bar{K})\Big{]}\Bigg{)},\end{split}$
(11.11)

where in the above we have sub-amplitudes that are stripped of their
dilatation invariance delta functions and also we have redefined summation and
integration variables $(K,\bar{K},s_{K},s_{\bar{K}})$ for each term to pull
out common factors. We would also like to reiterate that all the weights in
the above are for the $\mu,\tilde{\mu}$ variables - the weights for the
$\operatorname{\lambda},\operatorname{\tilde{\lambda}}$ being found using the
fact the helicity is $\pm 1$ which implies the relations $h=-\bar{k}$ and
$\bar{h}=-k$. Although not written explictly above, the localised values of
the discrete weights follow that of the continuous ones.

We now perform the inverse Mellin transform over $K$ and $s_{K}$ using the
delta functions and arrive at the following BCFW recursion relation for
celestial twistor Yang-Mills amplitudes

$\boxed{\begin{split}\tilde{\mathcal{A}}_{n}=&\;2\pi
i\,\text{sgn}(\langle\mu_{i}\operatorname{\lambda}_{j}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\sum_{L,R}\int\frac{d\bar{K}}{2\pi
i}\\!\\!\sum_{s_{\bar{K}}\in\\{0,1\\}}\\\
&\times\int\,\langle\operatorname{\lambda}d\operatorname{\lambda}\rangle\,[\operatorname{\tilde{\lambda}}d\operatorname{\tilde{\lambda}}]\,\langle\mu\,d\mu\rangle\,[\tilde{\mu}\,d\tilde{\mu}]\,\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}])\\\
&\qquad\times\mathcal{R}\bigg{(}-2\bar{K},2\bar{K},-s_{K}-1,s_{K}\bigg{)}\,\mathcal{G}(\bar{k}_{j}+k_{R}+\bar{K},s_{\bar{k}_{j}}+s_{k_{R}}+s_{\bar{K}},\theta_{ij})\\\
&\times\Bigg{(}\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},-k_{L}+\bar{K});(\lambda,\tilde{\mu},-\bar{K},-\hat{s}_{\bar{K}})\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-k_{R}-\bar{K});(\mu,\tilde{\lambda},\bar{K},s_{\bar{K}})\Big{]}\\\
&+\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},-k_{L}+\bar{K});(\mu,\tilde{\lambda},-\bar{K},-\hat{s}_{\bar{K}})\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-k_{R}-\bar{K});(\lambda,\tilde{\mu},\bar{K},s_{\bar{K}})\Big{]}\Bigg{)},\end{split}}$
(11.12)

where we have again left the discrete weights implicit in the above expression
but they follow the same pattern as the continuous weights with one exception
- in the left sub-amplitudes the weight $s_{K}$ is localised to the shifted
value $-\hat{s}_{\bar{K}}:=-s_{\bar{K}}-1$.

We are left with a contour integral over $\bar{K}$ and a sum over the weights
$s_{\bar{K}}$, together with four projective integrals over
$\operatorname{\lambda},\operatorname{\tilde{\lambda}},\mu,\tilde{\mu}$.
Carrying these out will generate the full amplitude - we demonstrate this in
an example below by computing the four-point amplitude from a product of two
three-point amplitudes, after which we will make some general comments valid
at $n$-points.

In fact we can carry out the integrals over $\mu,\tilde{\mu}$ above
immediately, since the pre-factor
$\mathcal{R}(-2\bar{K},2\bar{K},-s_{\bar{K}}-1,s_{\bar{K}})$ is given by

$\begin{split}\mathcal{R}&=4\pi^{2}i\frac{\Gamma(2+2\bar{K})\Gamma(2-2\bar{K})\absolutevalue{\langle\operatorname{\lambda}\mu\rangle}^{2\bar{K}-2}\text{sgn}(\langle\operatorname{\lambda}\mu\rangle)^{s_{\bar{K}}}\absolutevalue{[\operatorname{\tilde{\lambda}}\tilde{\mu}]}^{-2\bar{K}-2}\text{sgn}([\operatorname{\tilde{\lambda}}\tilde{\mu}])^{s_{\bar{K}}+1}}{\Gamma(\frac{3}{2}-\bar{K}-\frac{s_{\bar{K}}}{2})\Gamma(\bar{K}+\frac{s_{\bar{K}}}{2}-\frac{1}{2})\Gamma(2+\bar{K}+\frac{s_{\bar{K}}}{2})\Gamma(-\bar{K}-\frac{s_{\bar{K}}}{2}-1)},\end{split}$
(11.13)

and contains exactly the correct factors of
$\absolutevalue{\langle\operatorname{\lambda}\mu\rangle}$ etc. and the correct
normalisation such that the $\mu$ (respectively $\tilde{\mu}$) integral is a
light transform (8.3) (respectively dual light transform (8.5)) of the glued
legs. The light and dual light transforms are self inverse so we conclude that
after the $\mu,\tilde{\mu}$ integrals, the glued legs are expressed with
variables $\operatorname{\lambda},\operatorname{\tilde{\lambda}}$ and are only
half-Mellin transformed without any light transforms acting at all. We can
hence write down an alternative form of celestial BCFW recursion which takes
as input amplitudes with _all but one leg light transformed_ , and glues the
un-transformed legs resulting in an amplitude with all legs light transformed.
From (11.12) the alternative BCFW recursion relation is

$\boxed{\begin{split}\tilde{\mathcal{A}}_{n}=&\;2\pi
i\,\text{sgn}(\langle\mu_{i}\operatorname{\lambda}_{j}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\sum_{L,R}\int\frac{d\bar{K}}{2\pi
i}\\!\\!\sum_{s_{\bar{K}}\in\\{0,1\\}}\\\
&\times\mathcal{G}(\bar{k}_{j}+k_{R}+\bar{K},s_{\bar{k}_{j}}+s_{k_{R}}+s_{\bar{K}},\theta_{ij})\int\,\langle\operatorname{\lambda}d\operatorname{\lambda}\rangle\,[\operatorname{\tilde{\lambda}}d\operatorname{\tilde{\lambda}}]\,\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{j}\rangle[\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{i}])\\\
&\times\Bigg{(}\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},-k_{L}+\bar{K},+);(\lambda,\operatorname{\tilde{\lambda}},\bar{h}_{L},s_{\bar{h}_{L}},-)\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-k_{R}-\bar{K},-);(\operatorname{\lambda},\tilde{\lambda},h_{R},s_{h_{R}},+)\Big{]}\\\
+&\tilde{\mathcal{A}}_{L}\Big{[}(\mu_{i},\tilde{\lambda}_{i},-k_{L}+\bar{K},+);(\operatorname{\lambda},\tilde{\lambda},h_{L},s_{h_{L}},+)\Big{]}\tilde{\mathcal{A}}_{R}\Big{[}(\lambda_{j},\tilde{\mu}_{j},-k_{R}-\bar{K},-);(\lambda,\operatorname{\tilde{\lambda}},\bar{h}_{R},s_{\bar{h}_{R}},-)\Big{]}\Bigg{)},\end{split}}$
(11.14)

where the weights of the glued legs are

$\begin{split}\bar{h}_{L}=1+\bar{K}=h_{L}\quad\bar{h}_{R}=1-\bar{K}=h_{R}\,,\\\
s_{\bar{h}_{L}}=-s_{\bar{K}}-1=s_{h_{L}}\quad
s_{\bar{h}_{R}}=s_{\bar{K}}=s_{h_{R}}.\end{split}$ (11.15)

We conclude that the most economical BCFW recursion relation for celestial
twistor amplitudes happens to involve gluing sub-amplitudes with one leg in
the Mellin basis and the rest in the light transformed basis. This form of the
recursion may not necessarily be the most useful at high multiplicity since it
requires recursion on an additional set of amplitudes beyond those we are
immediately interested in. Nevertheless, the recursion relation (11.14) proves
very useful for demonstrating the four-point BCFW recursion relation as we
will see below, since the three-point sub-amplitudes can be easily computed
directly.

The BCFW recursion relations for gravity amplitudes take a very similar form.
The only difference to the BCFW expression comes from the changed dilatation
invariance delta function which, as we saw at three and four points, is
shifted by $2(n-2)$ due to the mass dimension carried by the powers of the
coupling $\kappa$ in an $n$ point gravity amplitude. Hence, in the gravity
case the delta functions of the sub-amplitudes (11.9) are shifted and so the
localised values of the continuous weights of legs $i$ and $j$ on the left and
right subamplitudes are shifted by $2(n_{L,R}-2)$ also, while the localised
values for the discrete weights are the same as for the Yang-Mills case.

Four-Point BCFW

To see how the recursion relation works in an explicit example, consider the
four-point amplitude $A(1^{+},2^{-},3^{+},4^{-})$ with a BCFW shift of legs
$(1^{+},4^{-})$. In this case the amplitude pairs that appear in the BCFW
recursion are

$A^{\overline{MHV}}_{L}(1^{+},2^{-},P^{+})A^{MHV}_{R}(P^{-},3^{+},4^{-})+A^{MHV}_{L}(1^{+},2^{-},P^{-})A^{\overline{MHV}}_{R}(P^{+},3^{+},4^{-}).$
(11.16)

As usual the second pair of amplitudes do not contribute since they vanish for
generic kinematics and so we consider just the first pair. The relevant three-
point $MHV$ and $\overline{MHV}$ celestial amplitudes with two legs light
transformed are derived in Appendix A and for the case at hand are given by

$\begin{split}&\tilde{\mathcal{A}}_{L}^{\overline{MHV}}\Big{[}(\mu_{1},\operatorname{\tilde{\lambda}}_{1},-\bar{k}_{2}+\bar{K},+),(\operatorname{\lambda}_{2},\tilde{\mu}_{2},\bar{k}_{2},-),(\operatorname{\lambda},\operatorname{\tilde{\lambda}},1+\bar{K},-s_{\bar{K}}-1,+)\Big{]}\\\
=&\;\delta(\langle\operatorname{\lambda}\operatorname{\lambda}_{2}\rangle)\,\text{sgn}([\operatorname{\tilde{\lambda}}_{1}\operatorname{\tilde{\lambda}}][\operatorname{\tilde{\lambda}}_{1}\tilde{\mu}_{2}])\,\text{sgn}(-\theta_{12})^{-s_{\bar{k}_{2}}+s_{\bar{K}}+1}\frac{\absolutevalue{\theta_{12}^{-1}}^{-2\bar{k}_{2}+2\bar{K}}}{-2\bar{k}_{2}+2\bar{K}+(-1)^{-s_{\bar{k}_{2}}+s_{\bar{K}}+1}\epsilon}\\\
&\qquad\qquad\times\mathcal{N}\big{(}2\bar{K},-s_{\bar{K}},[\operatorname{\tilde{\lambda}}\tilde{\mu}_{2}]\big{)},\end{split}$
(11.17)

and

$\begin{split}&\tilde{\mathcal{A}}_{R}^{MHV}\Big{[}(\mu_{3},\operatorname{\tilde{\lambda}}_{3},k_{3},+),(\operatorname{\lambda}_{4},\tilde{\mu}_{4},-k_{3}-\bar{K},-),(\operatorname{\lambda},\operatorname{\tilde{\lambda}},1-\bar{K},s_{\bar{K}},-)\Big{]}\\\
=&\;\delta([\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{3}])\,\text{sgn}(\langle\operatorname{\lambda}_{4}\operatorname{\lambda}\rangle\langle\mu_{3}\operatorname{\lambda}_{4}\rangle)\,\text{sgn}(-\theta_{34})^{-s_{k_{3}}-s_{\bar{K}}}\frac{\absolutevalue{\theta_{34}}^{-2k_{3}-2\bar{K}}}{-2k_{3}-2\bar{K}+(-1)^{-s_{k_{3}}-s_{\bar{K}}}\epsilon}\\\
&\qquad\qquad\times\mathcal{N}\big{(}-2\bar{K},s_{\bar{K}}+1,\langle\mu_{3}\operatorname{\lambda}\rangle\big{)}.\end{split}$
(11.18)

where we have stripped off the dilatation invariance delta functions since
they have already been used to localise the weights.

We now consider the product of these two amplitudes and perform the integrals
over $\operatorname{\lambda},\operatorname{\tilde{\lambda}}$. Using the delta
functions which come from three-point momentum conservation, we identify
collinear pairs $\operatorname{\lambda}=\operatorname{\lambda}_{2}$ and
$\operatorname{\tilde{\lambda}}=\operatorname{\tilde{\lambda}}_{3}$191919The
other BCFW pair gives vanishing contribution for four particle kinematics. We
can see this directly since it has delta functions which give collinear pairs
$\operatorname{\lambda}=\operatorname{\lambda}_{4}$ and
$\operatorname{\tilde{\lambda}}=\operatorname{\tilde{\lambda}}_{1}$ which
forces the vanishing of the
$\text{sgn}(\langle\operatorname{\lambda}\operatorname{\lambda}_{4}\rangle)\,\text{sgn}([\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{1}])$
factors in the BCFW relation.. The four-point amplitude is then given by

$\begin{split}\tilde{\mathcal{A}}_{4}^{+-+-}=&\;2\pi
i\,\text{sgn}(\langle\mu_{1}\operatorname{\lambda}_{4}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\int\frac{d\bar{K}}{2\pi
i}\\!\\!\sum_{s_{\bar{K}}\in\\{0,1\\}}\\\
&\times\mathcal{G}(k_{3}+\bar{k}_{4}+\bar{K},s_{k_{3}}+s_{\bar{k}_{4}}+s_{\bar{K}},\theta_{14})\,\text{sgn}(\langle\operatorname{\lambda}_{2}\operatorname{\lambda}_{4}\rangle)\,\text{sgn}([\operatorname{\tilde{\lambda}}_{3}\operatorname{\tilde{\lambda}}_{1}])\tilde{\mathcal{A}}_{L}^{\overline{MHV}}\,\tilde{\mathcal{A}}_{R}^{MHV}\\\
=&\;2\pi
i\,\text{sgn}([\operatorname{\tilde{\lambda}}_{1}\tilde{\mu}_{2}]\langle\mu_{3}\operatorname{\lambda}_{4}\rangle\langle\mu_{1}\operatorname{\lambda}_{4}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\int\frac{d\bar{K}}{2\pi
i}\\!\\!\sum_{s_{\bar{K}}\in\\{0,1\\}}\\\
&\times\mathcal{G}(k_{3}+\bar{k}_{4}+\bar{K},s_{k_{3}}+s_{\bar{k}_{4}}+s_{\bar{K}},\theta_{14})\,\text{sgn}(-\theta_{12})^{-s_{\bar{k}_{2}}+s_{\bar{K}}+1}\text{sgn}(-\theta_{34})^{-s_{k_{3}}-s_{\bar{K}}}\\\
&\quad\times\,\frac{\absolutevalue{\theta_{12}^{-1}}^{-2\bar{k}_{2}+2\bar{K}}}{-2\bar{k}_{2}+2\bar{K}+(-1)^{-s_{\bar{k}_{2}}+s_{\bar{K}}+1}\epsilon}\,\frac{\absolutevalue{\theta_{34}}^{-2k_{3}-2\bar{K}}}{-2k_{3}-2\bar{K}+(-1)^{-s_{k_{3}}-s_{\bar{K}}}\epsilon}\\\
&\quad\quad\times\,\mathcal{N}\big{(}2\bar{K},-s_{\bar{K}},[\operatorname{\tilde{\lambda}}_{3}\tilde{\mu}_{2}]\big{)}\,\mathcal{N}\big{(}-2\bar{K},s_{\bar{K}}+1,\langle\mu_{3}\operatorname{\lambda}_{2}\rangle\big{)}.\end{split}$
(11.19)

Now, using the Legendre Duplication Formula for gamma functions (B.16) we find
that

$\begin{split}&\mathcal{N}\big{(}2\bar{K},-s_{\bar{K}},[\operatorname{\tilde{\lambda}}_{3}\tilde{\mu}_{2}]\big{)}\,\mathcal{N}\big{(}-2\bar{K},s_{\bar{K}}+1,\langle\mu_{3}\operatorname{\lambda}_{2}\rangle\big{)}\\\
=&\;2\pi
i\frac{1}{2\bar{K}+(-1)^{s_{\bar{K}}}\epsilon}\absolutevalue{[\operatorname{\tilde{\lambda}}_{3}\tilde{\mu}_{2}]}^{-2\bar{K}}\text{sgn}([\operatorname{\tilde{\lambda}}_{3}\tilde{\mu}_{2}])^{-s_{\bar{K}}}\absolutevalue{\langle\mu_{3}\operatorname{\lambda}_{2}\rangle}^{2\bar{K}}\text{sgn}(\langle\mu_{3}\operatorname{\lambda}_{2}\rangle)^{s_{\bar{K}}+1}\\\
=&\;2\pi
i\frac{1}{2\bar{K}+(-1)^{s_{\bar{K}}}\epsilon}\text{sgn}(\langle\mu_{3}\operatorname{\lambda}_{2}\rangle)\,\text{sgn}(-\theta_{32})^{s_{\bar{K}}}\absolutevalue{\theta_{32}}^{2\bar{K}},\end{split}$
(11.20)

where we have also included a regularisation of the pole in $2\bar{K}$. As
usual, when we integrate over $\bar{K}$, this regularisation corresponds to a
contour along the imaginary axis which avoids the pole either to the left or
right. Using the above and the definition of the $\mathcal{G}$ function in
(11.3), the four-point amplitude is given by

$\boxed{\begin{split}\tilde{\mathcal{A}}_{4}^{+-+-}=&\;4\pi^{2}\,\text{sgn}(\langle\mu_{1}\operatorname{\lambda}_{2}\rangle\langle\mu_{3}\operatorname{\lambda}_{2}\rangle\langle\mu_{3}\operatorname{\lambda}_{4}\rangle\langle\mu_{1}\operatorname{\lambda}_{4}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{a}s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}\bigg{)}\int\frac{d\bar{K}}{2\pi
i}\\!\\!\sum_{s_{\bar{K}}\in\\{0,1\\}}\\\
&\times\text{sgn}(\theta_{12})^{s_{\bar{k}_{2}}+s_{\bar{K}}}\text{sgn}(\theta_{32})^{s_{\bar{K}}}\text{sgn}(\theta_{34})^{s_{k_{3}}+s_{\bar{K}}}\text{sgn}(\theta_{14})^{s_{k_{3}}+s_{\bar{k}_{4}}+s_{\bar{K}}}\\\
&\quad\times\,\frac{\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}\absolutevalue{\theta_{14}}^{2k_{3}+2\bar{k}_{4}}\absolutevalue{\theta_{34}}^{-2k_{3}}}{2k_{3}+2\bar{k}_{4}+2\bar{K}+(-1)^{s_{k_{3}}+s_{\bar{k}_{4}}+s_{\bar{K}}}\epsilon}\,\frac{\absolutevalue{\theta}^{-2\bar{K}}}{2\bar{K}+(-1)^{s_{\bar{K}}}\epsilon}\\\
&\quad\quad\times\frac{1}{2k_{3}+2\bar{K}+(-1)^{s_{k_{3}}+s_{\bar{K}}+1}\epsilon}\,\frac{1}{-2\bar{k}_{2}+2\bar{K}+(-1)^{s_{\bar{k}_{2}}+s_{\bar{K}}+1}\epsilon}.\end{split}}$
(11.21)

The sum over $s_{\bar{K}}$ gives two terms which exactly correspond to the
decomposition (10.54) which we previously found. The two terms are the unique
$\mathcal{I}(r_{12},r_{32},r_{34},r_{14})$ with the correct even/odd symmetry
and, as already noticed, these terms have the opposite $\epsilon$
prescriptions and sgn function factors.

The contour integral over $\bar{K}$ has contour along the pure imaginary axis
and the expression has poles at
$\bar{K}=(0,\bar{k}_{2},-k_{3}-\bar{k}_{4},-k_{3})$ with varying $\epsilon$
prescription away from the imaginary axis. It is straightforward to check that
the residues on these poles generate four terms involving the functions
$V_{32},V_{12},V_{14},V_{34}$ given in (10.51) which make up all four-point
symmetrised amplitudes. With regard to the choice of contours, the $\bar{K}$
dependence in the above is $|\theta|^{-\bar{2}K}$, where $\theta$ is the ratio
of cross-ratios. In order to avoid contributions at infinity the contour must
be chosen to close to the right with large positive $\bar{K}$ for
$|\theta|>1$, while for $|\theta|<1$ it must close to the left with large
negative $\bar{K}$. So for each case we pick up a subset of the poles, namely
those with $\epsilon$ prescription such that the poles lie in either the left
or right hand half plane. In this way the various functions
$V_{32},V_{12},V_{14},V_{34}$ are assigned to the channels $|\theta|>1$ or
$|\theta|<1$.

For example, the fully symmetrised amplitude has all the
$s_{\mathrel{\mathop{\bar{k}}\limits^{\vbox to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\\!{}_{a}}$ equal to zero and
so we have two terms

$\begin{split}\tilde{\mathcal{A}}_{4}^{+-+-}\\{0,0,0,0\\}&=4\pi^{2}\,\text{sgn}(\langle\mu_{1}\operatorname{\lambda}_{2}\rangle\langle\mu_{3}\operatorname{\lambda}_{2}\rangle\langle\mu_{3}\operatorname{\lambda}_{4}\rangle\langle\mu_{1}\operatorname{\lambda}_{4}\rangle)\delta\bigg{(}\sum_{a}2\mathrel{\mathop{\bar{k}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\textbf{(\quad)}$}\vss}}}\negmedspace\negmedspace{}_{a}\bigg{)}\\\
&\times\absolutevalue{\theta_{12}}^{2\bar{k}_{2}}\absolutevalue{\theta_{14}}^{2k_{3}+2\bar{k}_{4}}\absolutevalue{\theta_{34}}^{-2k_{3}}[I(0,0,0,0)+\text{sgn}(\theta_{12}\theta_{32}\theta_{34}\theta_{14})\,I(1,1,1,1)]\end{split}$
(11.22)

where, focusing on the contour integral $I(0,0,0,0)$ we have

$\begin{split}I(0,0,0,0)&:=\int\frac{d\bar{K}}{2\pi
i}\,\frac{\absolutevalue{\theta}^{-2\bar{K}}}{2k_{3}+2\bar{k}_{4}+2\bar{K}+\epsilon}\,\frac{1}{2\bar{K}+\epsilon}\frac{1}{2k_{3}+2\bar{K}-\epsilon}\,\frac{1}{-2\bar{k}_{2}+2\bar{K}-\epsilon}\end{split}$
(11.23)

which gives four terms, one for each residue, with two terms appearing in each
channel as dictated by the $\pm\epsilon$ prescription. This exactly matches
the four terms (10.52) appearing in the function $\mathcal{I}(0,0,0,0)$ found
in the four-point amplitude. The $I(1,1,1,1)$ contour integral has opposite
$\pm\epsilon$ prescription and takes a similar form which matches the terms in
$\mathcal{I}(1,1,1,1)$ in (10.53).

Beyond four points we will encounter similar steps and the BCFW recursion
relations will allow us to iteratively build amplitudes using previous results
as the inputs. Given the amplitudes with up to $n$ particles, we can build the
$n+1$-point amplitude by a sum of terms involving the gluing of two sub-
amplitudes, with integrals over the homogeneous coordinates of the glued legs
on the celestial torus. Given the similarity of these integrals to the
integrals in the twistor BCFW recursion relations of [24], we might expect a
diagrammatic formalism à la Hodges [74] to exist. Once the gluing is complete
the contour integral would simply build terms in a combinatorial fashion by
evaluating residues on the poles of the BCFW expression. We leave the
exploration of such higher multiplicity questions to future work.

##  12 Conclusions

In this paper we have computed ambidextrous light transformed correlators by
deriving celestial twistor amplitudes. This involved the formulation and
understanding of the Fourier, twistor, shadow, light and Mellin transforms in
their appropriate signature spacetimes. We proved relations between these
transforms, summarised by the commuting diagrams in Figures 1, 3, 4, before
their application to derive three and four-point YM and gravity amplitudes,
and a general BCFW recursion relation in sections 10 and 11. In concluding we
would like to highlight some areas for further research which are prompted by
our results.

We found a concise expression for the four-point YM amplitude in (10.55) and
it would be interesting to relate this to the four-point results of [66] for
four light-transformed fields. For gravity, the four-point alternating
amplitude, given in (10.80), shows a striking similarity in structure to its
YM analogue - clarifying the physics behind this may lead to insights
applicable to more general cases. The application of the double copy for
celestial twistor amplitudes (c.f. [75, 42]) should also be explored in this
context.

Extending these results to more general amplitudes is clearly a next step.
Higher point results could naturally be derived from the twistor celestial
recursion relation, given in (11.12) or (11.14). The structure of contour
integrals and cross-ratios that was apparent in our BCFW derivation of the
four-point amplitude would be expected to persist at higher points and it
would be interesting to pursue this via direct calculations. In particular the
$n$-point MHV amplitudes may be amenable to this approach. It would also be of
interest to understand if the integrands in these recursion relations have
some geometric interpretation, perhaps analogous to the diagrammatic rules and
identities described in [24] for the twistor amplitudes.

Developing the OPEs of light transformed correlators via celestial twistor
amplitudes should provide insights into the construction of higher point
amplitudes as well as elucidate the structure of celestial conformal field
theory, along with factorisation, conformal blocks, etc. These OPEs should
follow from the consequences of collinear limits for the amplitudes presented
here. The extension of this work to supersymmetry can be done using the
results in [12, 13] \- since the anticommuting superspace coordinate scales
under the little group, the chiral Mellin transform is modified for each of
the superspace component fields and hence amplitudes. The twistor space super-
amplitudes described in [24] neatly package together their component sub-
amplitudes and one might expect similar simplifications for their celestial
analogues. There would then exist super-shadow and super-light transforms that
are related to super-Fourier and super-twistor transforms. Thus the celestial
twistor superamplitudes could be found by Mellin transforming twistor space
superamplitudes.

The relative simplicity and interesting structure of the twistor celestial
amplitudes derived in this paper would seem to provide encouraging signs for
developing this approach in these directions.

Acknowledgements: We would like to thank Andreas Brandhuber and Gabriele
Travaglini for initial collaboration on this project and many interesting
discussions throughout. GRB and JG would like to thank the Kavli Institute for
Theoretical Physics at the University of California, Santa Barbara, where
their research was supported in part by the National Science Foundation under
Grant No. PHY- 1748958. This work was supported by the Science and Technology
Facilities Council (STFC) Consolidated Grants ST/P000754/1 “String theory,
gauge theory & duality” and ST/T000686/1 “Amplitudes, strings and duality”,
and by the European Union’s Horizon 2020 research and innovation programme
under the Marie Skłodowska-Curie grant agreement No. 764850 “SAGEX”. The work
of GRB and JG is supported by STFC quota studentships.

##  Appendix A All But One Leg Celestial Twistor Amplitudes

In this appendix, we present three-point amplitudes with all but one leg half-
Fourier transformed and also their celestial versions found via ambidextrous
half-Mellin transforms. These amplitudes feature in a form of the BCFW
recursion relations (11.14) for celestial twistor amplitudes and we have used
them above to derive the four-point ambidextrous celestial twistor amplitude
via BCFW recursion.

Ambidextrously half-Fourier transforming legs one and two of an
$\overline{MHV}$ three-point amplitude we have

$\begin{split}A^{\overline{MHV}}\big{[}(\mu_{1},\operatorname{\tilde{\lambda}}_{1},+),(\operatorname{\lambda}_{2},\tilde{\mu}_{2},-),(\operatorname{\lambda},\operatorname{\tilde{\lambda}},+)\big{]}=&\;\delta(\langle\operatorname{\lambda}\operatorname{\lambda}_{2}\rangle)\,\text{sgn}([\operatorname{\tilde{\lambda}}_{1}\operatorname{\tilde{\lambda}}])\,\text{sgn}(\langle\mu_{1}\operatorname{\lambda}_{2}\rangle+[\tilde{\mu}_{2}\operatorname{\tilde{\lambda}}_{1}])\\\
\qquad&\times\frac{\langle\operatorname{\lambda}_{2}\xi\rangle}{\langle\xi\operatorname{\lambda}\rangle}\,\exp(i\frac{\langle\operatorname{\lambda}\xi\rangle}{\langle\xi\operatorname{\lambda}_{2}\rangle}[\tilde{\mu}_{2}\operatorname{\tilde{\lambda}}])\end{split}$
(A.1)

which can be derived by an identical method to that laid out in [24]. We have
written the result in terms of a reference spinor $\xi$ on which the amplitude
does not depend. This property is a result of the presence of the delta
function
$\delta(\langle\operatorname{\lambda}\operatorname{\lambda}_{2}\rangle)$ which
is the remnant of three-point momentum conservation. We now half-Mellin
transform over the scales of the spinors
$\mu_{1},\tilde{\mu}_{2},\operatorname{\lambda}$ with weights
$k_{1},\bar{k}_{2},h$ and their corresponding discrete weights. Since we have
already performed many such Mellin transforms of exponentials and sgn
functions we can refer to previous calculations, for example (10.18) and
(9.6), and we find

$\begin{split}&A^{\overline{MHV}}\big{[}(\mu_{1},\operatorname{\tilde{\lambda}}_{1},k_{1},+),(\operatorname{\lambda}_{2},\tilde{\mu}_{2},\bar{k}_{2},-),(\operatorname{\lambda},\operatorname{\tilde{\lambda}},h,+)\big{]}\\\
\qquad=&\;\delta(\langle\operatorname{\lambda}\operatorname{\lambda}_{2}\rangle)\,\text{sgn}([\operatorname{\tilde{\lambda}}_{1}\operatorname{\tilde{\lambda}}][\tilde{\mu}_{2}\operatorname{\tilde{\lambda}}_{1}])\,\frac{\langle\operatorname{\lambda}_{2}\xi\rangle}{\langle\xi\operatorname{\lambda}\rangle}\text{sgn}(-\theta_{12})^{s_{k_{1}}}\frac{\absolutevalue{\theta_{12}^{-1}}^{2k_{1}}}{2k_{1}+(-1)^{s_{k_{1}}}\epsilon}\\\
&\qquad\times\delta\bigg{(}\sum_{i=1}^{2}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}+2-2h\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{i=1}^{2}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}+s_{h}\bigg{)}\mathcal{N}\Big{(}2h-2,s_{h}+1,\frac{\langle\operatorname{\lambda}\xi\rangle}{\langle\xi\operatorname{\lambda}_{2}\rangle}[\tilde{\mu}_{2}\operatorname{\tilde{\lambda}}]\Big{)},\end{split}$
(A.2)

where the function $\mathcal{N}(a,b,x)$ is defined in the main text (11.5).

Since in a celestial amplitude the spinors act as homogeneous coordinates, the
delta function
$\delta(\langle\operatorname{\lambda}\operatorname{\lambda}_{2}\rangle)$
actually implies $\operatorname{\lambda},\operatorname{\lambda}_{2}$ are
equivalent as homogeneous coordinates and so we may write

$\begin{split}&A^{\overline{MHV}}\big{[}(\mu_{1},\operatorname{\tilde{\lambda}}_{1},k_{1},+),(\operatorname{\lambda}_{2},\tilde{\mu}_{2},\bar{k}_{2},-),(\operatorname{\lambda},\operatorname{\tilde{\lambda}},h,+)\big{]}\\\
=\;&\delta(\langle\operatorname{\lambda}\operatorname{\lambda}_{2}\rangle)\,\text{sgn}([\operatorname{\tilde{\lambda}}_{1}\operatorname{\tilde{\lambda}}][\operatorname{\tilde{\lambda}}_{1}\tilde{\mu}_{2}])\,\text{sgn}(-\theta_{12})^{s_{k_{1}}}\frac{\absolutevalue{\theta_{12}^{-1}}^{2k_{1}}}{2k_{1}+(-1)^{s_{k_{1}}}\epsilon}\\\
&\qquad\times\delta\bigg{(}\sum_{i=1}^{2}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}+2-2h\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{i=1}^{2}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}+s_{h}\bigg{)}\mathcal{N}\Big{(}2h-2,s_{h}+1,[\operatorname{\tilde{\lambda}}\tilde{\mu}_{2}]\Big{)},\end{split}$
(A.3)

and the apparent dependence on the reference spinor drops out. The $MHV$
amplitude is completely analogous. We half-Fourier transform legs 3 and 4 and
then ambidextrously half-Mellin transform and find

$\begin{split}&A^{MHV}\big{[}(\mu_{3},\operatorname{\tilde{\lambda}}_{3},k_{3},+),(\operatorname{\lambda}_{4},\tilde{\mu}_{4},\bar{k}_{4},-),(\operatorname{\lambda},\operatorname{\tilde{\lambda}},\bar{h},-)\big{]}\\\
=\;&\delta([\operatorname{\tilde{\lambda}}\operatorname{\tilde{\lambda}}_{3}])\,\text{sgn}(\langle\operatorname{\lambda}_{4}\operatorname{\lambda}\rangle\langle\mu_{3}\operatorname{\lambda}_{4}\rangle)\,\text{sgn}(-\theta_{34})^{s_{\bar{k}_{4}}}\frac{\absolutevalue{\theta_{34}}^{2\bar{k}_{4}}}{2\bar{k}_{4}+(-1)^{s_{\bar{k}_{4}}}\epsilon}\\\
&\qquad\times\delta\bigg{(}\sum_{i=1}^{2}2\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.4pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}+2-2\bar{h}\bigg{)}\,\tilde{\delta}\bigg{(}\sum_{i=1}^{2}s_{\mathrel{\mathop{\bar{k}_{i}}\limits^{\vbox
to0.0pt{\kern
0.7pt\hbox{$\scriptstyle\\!\\!\textbf{(\quad)}$}\vss}}}}+s_{\bar{h}}\bigg{)}\mathcal{N}\Big{(}2\bar{h}-2,s_{\bar{h}}+1,\langle\mu_{3}\operatorname{\lambda}\rangle\Big{)}.\end{split}$
(A.4)

##  Appendix B From Shadow to Fourier

In this Appendix we complete the proof that a chiral Mellin transform of a
Fourier transform gives the shadow transform. This requires us to compute the
integral $\mathcal{I}$ in (5.11) given by

$\mathcal{I}:=\int_{\mathbb{C}^{*}}\frac{d\bar{\sigma}}{\bar{\sigma}}\wedge\frac{d\sigma}{\sigma}\sigma^{2-2h}\bar{\sigma}^{2-2\bar{h}}e^{i(\sigma+\bar{\sigma})}\,.$
(B.1)

Computing $\mathcal{I}$ completes the proof of the commuting diagram Figure
(1). However, as an additional check, we also prove the diagram in the reverse
direction by inverse chiral Mellin transforming the shadowed conformal primary
to recover a Fourier transformed momentum eigenstate.

Computing $\mathcal{I}$

To compute $\mathcal{I}$ we first change variables $\sigma=Re^{i\phi}$

$\begin{split}\mathcal{I}=&\;2i\int_{0}^{\infty}dR\,R^{3-2\Delta}\int_{0}^{2\pi}d\phi\,\left(e^{i\phi}\right)^{-2J}e^{2iR\cos{\phi}}\\\
=&\;2i\int_{0}^{\infty}dR\,R^{3-2\Delta}\mathcal{I}_{\phi}\,.\end{split}$
(B.2)

We can compute the angular integral as a contour integral using $z=e^{i\phi}$,
giving

$\mathcal{I}_{\phi}=-i\oint\frac{dz}{z}\,z^{-2J}e^{iR(z+z^{-1})}=-i\sum_{n=0}^{\infty}\frac{(iR)^{n}}{n!}\oint\frac{dz}{z}\,z^{-2J}\left(z+z^{-1}\right)^{n}\,.$
(B.3)

This is also a modified Bessel function of the first kind. Now we use the
binomial theorem and find that the contour integal has a residue contribution
of $2\pi i$ at $z=0$ only when $n+2J$ is even and $n-|2J|\geq 0$ and for the
single term in the binomial sum with $k=J+\frac{n}{2}\geq 0$. The integral is
then given by the sum

$\mathcal{I}_{\phi}=2\pi\sum_{\begin{subarray}{c}n\geq\lvert 2J\lvert\\\
n+2J\,\text{even}\end{subarray}}^{\infty}\\!\\!\frac{(iR)^{n}}{(\frac{n}{2}+J)!(\frac{n}{2}-J)!}=2\pi
i^{2\absolutevalue{J}}\sum_{l=0}^{\infty}(-1)^{l}\frac{R^{2l+2\absolutevalue{J}}}{l!(l+2\absolutevalue{J})!}\,,$
(B.4)

which we have rewritten with $l:=\frac{n}{2}-\absolutevalue{J}$. This sum
corresponds to a generalised hypergeometric function, since the ratio of any
two consecutive terms is a rational function in $l$. Labelling the terms in
the above sum by $a_{l}$ we have

$\frac{a_{l+1}}{a_{l}}=\frac{-R^{2}}{(l+1)(l+2\absolutevalue{J}+1)},\quad
a_{0}=2\pi i^{2\absolutevalue{J}}R^{2\absolutevalue{J}},$ (B.5)

hence the angular integral is given by

$\mathcal{I}_{\phi}=2\pi
i^{2\absolutevalue{J}}R^{2\absolutevalue{J}}\,{}_{0}F_{1}(\\_;2\absolutevalue{J}+1;-R^{2})\,.$
(B.6)

The sum (B.4) converges for all finite values of $R^{2}$ by the ratio test.

To evaluate the $R$ integral we will find it easier to work with the explicit
sum form of $\mathcal{I}_{\phi}$

$\begin{split}\mathcal{I}=4\pi
i\,i^{2\absolutevalue{J}}\int_{0}^{\infty}dRR^{3-2\Delta}\sum_{l=0}^{\infty}(-1)^{l}\frac{R^{2l+2\absolutevalue{J}}}{l!(l+2\absolutevalue{J})!}\,.\end{split}$
(B.7)

We now regulate the $R$ integral using $e^{-\delta R}$ and then take
$\delta\rightarrow 0^{+}$202020Ramanujan’s master theorem also gives the same
answer [76].. We also commute the infinite sum with the integral since both
now converge

$\begin{split}\mathcal{I}=&\;4\pi
i\,i^{2\absolutevalue{J}}\sum_{l=0}^{\infty}\frac{(-1)^{l}}{l!(l+2\absolutevalue{J})!}\int_{0}^{\infty}R^{3-2\Delta+2l+2\absolutevalue{J}}e^{-\delta
R}\\\ =&\;4\pi
i\,i^{2\absolutevalue{J}}\sum_{l=0}^{\infty}\frac{(-1)^{l}}{l!(l+2\absolutevalue{J})!}\frac{\Gamma(4-2\Delta+2l+2\absolutevalue{J})}{\delta^{4-2\Delta+2l+2\absolutevalue{J}}}\,.\end{split}$
(B.8)

Once again this sum corresponds to a hypergeometric function: labelling the
terms in the sum by $b_{l}$ we have

$\frac{b_{l+1}}{b_{l}}=\frac{(5-2\Delta+2l+2\absolutevalue{J})(4-2\Delta+2l+2\absolutevalue{J})}{(l+1)(l+2\absolutevalue{J}+1)}\frac{-1}{\delta^{2}},\quad
b_{0}=\frac{4\pi
i\,i^{2\absolutevalue{J}}\Gamma(4-2\Delta+2\absolutevalue{J})}{(2\absolutevalue{J})!\delta^{4-2\Delta+2\absolutevalue{J}}}\,.$
(B.9)

Thus the integral $\mathcal{I}$ is given by a type ${}_{2}F_{1}$
hypergeometric function. The sum converges for
$\absolutevalue{\frac{-4}{\delta^{2}}}<1/2$ and is on the principal branch

$\mathcal{I}=\frac{4\pi
i\,i^{2\absolutevalue{J}}\Gamma(4-2\Delta+2\absolutevalue{J})}{(2\absolutevalue{J})!\delta^{4-2\Delta+2\absolutevalue{J}}}\;{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{\frac{5}{2}-\Delta+\absolutevalue{J},\,\,2-\Delta+\absolutevalue{J}}{2\absolutevalue{J}+1};\frac{-4}{\delta^{2}}\right).$
(B.10)

In order to tame this expression we must use a series of transformations and
identities. The first of these is a Pfaff transformation,

${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};z\right)=(1-z)^{-b}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{b,c-a}{c};\frac{z}{z-1}\right)\,,$
(B.11)

and applying this to (B.10) gives

$\begin{split}\mathcal{I}=\frac{4\pi
i\,i^{2\absolutevalue{J}}\Gamma(4-2\Delta+2\absolutevalue{J})}{(2\absolutevalue{J})!(\delta+2)^{4-2\Delta+2\absolutevalue{J}}}{}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{2-\Delta+\absolutevalue{J},\,\,-\frac{3}{2}+\Delta+\absolutevalue{J}}{2\absolutevalue{J}+1};\frac{1}{1+\frac{\delta^{2}}{4}}\right)\,.\end{split}$
(B.12)

At this point we are safe to take the limit $\delta\rightarrow 0$ and we
obtain

$\mathcal{I}=4\pi
i\,\frac{i^{\absolutevalue{J}}}{(2\absolutevalue{J})!}2^{2\Delta-2\absolutevalue{J}-4}\Gamma(4-2\Delta+2\absolutevalue{J}){}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{2-\Delta+\absolutevalue{J},\,\,-\frac{3}{2}+\Delta+\absolutevalue{J}}{2\absolutevalue{J}+1};1\right)\,.$
(B.13)

We can now evaluate the ${}_{2}F_{1}$ at the point $1$ in terms of objects we
are very familiar with: gamma functions

${}_{2}F_{1}\left(\genfrac{}{}{0.0pt}{}{a,b}{c};1\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\,,$
(B.14)

which holds if $\real(c-a-b)>0$. This condition is satisfied for us since