††institutetext: Department of Mathematics, King’s College London,
The Strand, London WC2R 2LS, U.K.

# Fermions with $SU(1,n)$ Spacetime Symmetry

Joseph Smith<EMAIL_ADDRESS>
###### Abstract

We construct theories of free fermions in $(2n-1)$-dimensions with $SU(1,n)$
spacetime symmetry from the null reduction of fermions on a $2n$-dimensional
$\Omega$-deformed Minkowski background for $n=2$ and $n=3$. These play a role
in the 5d $SU(1,3)$-invariant theories that are conjectured to offer a full
description of certain 6d superconformal field theories. We find the
$(2n-1)$-dimensional manifestation of the supersymmetry of a free
$2n$-dimensional boson-fermion system, which we use to fix the fermion two-
point functions. It is then shown that the full $2n$-dimensional two-point
function can be recovered through resummation. Limits of the theories are
considered, and it is observed that both Galilean and Carrollian field
theories appear in different regimes. We confirm that the correlation
functions obey the $SU(1,n)$ Ward identities and the representations of the
fermions under this group are discussed.

##  1 Introduction

An interesting theme in modern research has been the study of field theories
without Lorentz symmetry. While these have long been considered in the context
of condensed matter physics, there have been numerous recent advances in their
application to problems in high energy physics. These include non-relativistic
limits of string theory and supergravity Gomis:2000bd ; Gomis:2005pg ;
Bergshoeff:2023ogz ; Oling:2022fft ; Harmark:2019upf , the identification of
non-relativistic limits of $\mathcal{N}=4$ super Yang-Mills with quantum
mechanical models Harmark:2014mpa ; Baiguera:2020jgy ; Baiguera:2022pll , and
the role of Carrollian field theories in flat-space holography Donnay:2022aba
; Donnay:2022wvx ; Bagchi:2022emh ; Duval:2014lpa ; Duval:2014uva , as well as
many other topics.

Though non-Lorentzian field theories have an existence outside of Lorentzian
theories, the two are often connected. For instance, non-Lorentzian field
theories can be constructed from a Lorentzian one through null reduction. One
of the lightcone directions in $D$-dimensional Minkowski spacetime is
periodically identified and a Kaluza-Klein reduction on this circle is
performed in a procedure known as Discrete Lightcone Quantization (DLCQ). The
theory can then be truncated at a given level, leaving us with a theory in
$D-1$ dimensions with a non-Lorentzian spacetime symmetry. If the theory we
started with was conformally invariant, the reduced theory has Schrödinger
symmetry Son:2008ye .

The $D-1$ dimensional theory we obtain at a given level is well-defined, but
the question of whether the full $D$-dimensional theory can be recovered from
the Kaluza-Klein tower is subtle Yamawaki:1998cy  and we may wonder if we can
find a similar procedure that resolves these issues. In Lambert:2020zdc , a
diffeomorphism of Minkowski spacetime in $D=2n$ dimensions was introduced
that, after a conformal compactification, maps to an $\Omega$-deformed
Nekrasov:2002qd  spacetime in which one of the null directions takes values in
a finite interval. Reducing a theory on this direction leaves us with a
Kaluza-Klein tower of theories with $SU(1,n)$ spacetime symmetry
Lambert:2021nol . Since the deformation is induced using a combination of
coordinate and Weyl transformations, we can map quantities computed in the
deformed theory to the original Lorentzian theory (and vice-versa). This is
unlike the DLCQ approach, where the procedure to recover the original theory
(decompactification of the null circle) is not invertible. Keeping the full
tower of $SU(1,n)$-invariant $(2n-1)$-dimensional field theories should allow
us, in principle, to be able to compute quantities in the original theory that
are inaccessible in other approaches. Even outside of this connection to
Lorentzian field theories, $SU(1,n)$ theories are a novel class of QFTs that
are worthy of further study, particularly with respect to their quantum
properties.

A noteworthy application of $SU(1,n)$ theories has been to the study of 6d
superconformal field theories (SCFTs). As such theories are not thought to
admit conventional 6d Lagrangian descriptions, most of what is known about
them can only be found from string constructions (see Heckman:2018jxk  for a
review). In Lambert:2019jwi ; Lambert:2020jjm  5d Lagrangian theories with
$SU(1,3)$ spacetime symmetry were introduced and conjectured to provide a
complete description of certain 6d SCFTs. While an exciting prospect, there
are still many non-trivial properties these theories must possess. For
example, the Lagrangians are only invariant under half the supercharges111Note
that we are referring to non-conformal supercharges here. of the 6d theories.
The proposal requires that the remaining half appear non-perturbatively, with
the supercharges taking us between levels of the Kaluza-Klein tower given by
sectors of different instanton number. This is difficult to verify and it
would be useful to study a toy model in which we can explicitly show how all
the $2n$-dimensional supercharges act in the reduced theory. This can be
provided by the reduction of a $2n$-dimensional supersymmetric theory where we
have access to the full tower of fields at the Lagrangian level. The simplest
example we can consider is a free theory comprising a complex scalar and a
Weyl spinor. Previous work Lambert:2020zdc ; Lambert:2021nol  has focused on
the properties of scalar fields with $SU(1,n)$ spacetime symmetry and
relatively little is known about fermions in these backgrounds. The
construction and analysis of such theories is therefore of interest, and will
be the focus of this work.

The paper is organised as follows. In section 2 we review the coordinate
transformation that induces the $\Omega$-deformation and find the action of a
$2n$-dimensional free Dirac fermion on this background, which we then reduce
along the null direction to obtain a tower of $(2n-1)$-dimensional theories.
In section 3 we examine the manifestation of $2n$-dimensional supersymmetry in
the $(2n-1)$-dimensional theories; we then use this to determine the fermionic
two-point functions using the scalar two-point functions determined in
Lambert:2021nol . The tower of $(2n-1)$-dimensional correlation functions are
resummed, reproducing the two-point function of the $2n$-dimensional theory.
In section 4 we examine a limit of the $(2n-1)$-dimensional theory that
reduces it to the DLCQ of a free massless fermion. We find the two-point
functions of the $(2n-1)$-dimensional theory in this limit and show that the
original $2n$-dimensional two-point function can still be recovered. We also
comment on a related limit that gives a theory with Carroll symmetry. The
reduction of a 2d fermion is considered in section 5. In section 6 we show
that the $(2n-1)$-dimensional two-point functions satisfy the Ward identities
associated with $\mathfrak{su}(1,n)$ spacetime symmetry. Finally, in section 7
we discuss our results and prospects for future work. Various additional
material is collected in the appendices. In appendix A we outline our
conventions for fermions and gamma matrices. In appendix B we provide a
derivation of the generator of the local Lorentz transformation associated
with our coordinate transformation. In appendix C the coefficients of the two-
point functions for $SU(1,n)$-invariant scalar fields are computed and shown
to satisfy an infinite tower of Green’s function-like equations that descend
from the $2n$-dimensional Green’s function equation.

##  2 Reduction of the Fermionic Action

Following the philosophy of Lambert:2021nol , we will construct fermions with
$SU(1,n)$ spacetime symmetry from the reduction of free fermions in
$2n$-dimensional Minkowski spacetime along a null interval222The fermion
action for generic $n$ was presented in passing in Lambert:2021nol , but key
details about the transformation of the fields that will be necessary to
reconstruct the $2n$-dimensional theory were not emphasised.. This leads to an
infinite tower of decoupled theories in $(2n-1)$-dimensions. Depending on
one’s motivation we can either truncate to a finite set of fields that possess
$SU(1,n)$ spacetime symmetry or we can keep the whole Kaluza-Klein tower,
allowing us to reconstruct the original $2n$-dimensional theory. We will take
the latter approach, which will impose extra constraints on the moding of our
fields: such constraints can be dropped if one is only interested in
intrinsically defined $(2n-1)$-dimensional theories. Our goal will ultimately
be to show that the Kaluza-Klein tower of $(2n-1)$-dimensional fermions is
equivalent to the original $2n$-dimensional theory and that no information is
lost in this procedure.

###  2.1 Coordinate Transformation

Let us review the coordinate transformation introduced in Lambert:2021nol . We
start with Minkowski spacetime in $D=2n$ dimensions and split our coordinates
into lightcone directions $(\hat{x}^{+},\hat{x}^{-})$ and transverse
directions $\hat{x}^{i}$. Though we will keep our dimension general at this
stage, we will be interested in the cases $n=2$ and $n=3$ as these are the
only (even) dimensions in which we have both transverse coordinates and known
interacting CFTs; for this reason, when explicit expressions are required we
will only provide them for these cases.

We work with the coordinate transformation

$\displaystyle\hat{x}^{+}$ $\displaystyle=2R\tan\left(\frac{x^{+}}{2R}\right)\
,$ (1a) $\displaystyle\hat{x}^{-}$
$\displaystyle=x^{-}+\frac{x^{2}}{4R}\tan\left(\frac{x^{+}}{2R}\right)\ ,$
(1b) $\displaystyle\hat{x}^{i}$
$\displaystyle=x^{i}-R\Omega_{ij}x^{j}\tan\left(\frac{x^{+}}{2R}\right)\ ,$
(1c)

defined in terms of a lengthscale $R$ and an antisymmetric matrix
$\Omega_{ij}$ satisfying

$\Omega_{ij}\Omega_{ik}=\frac{1}{R^{2}}\delta_{jk}$ (2)

that acts on the transverse coordinates. We will use the notation

$x^{i}x^{i}\equiv x^{2}$ (3)

throughout when this causes no confusion. For convenience, we will choose
$\Omega_{ij}$ to be

$\Omega^{(4d)}_{ij}=\frac{1}{R}\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$ (4)

in 4d and

$\Omega^{(6d)}_{ij}=\frac{1}{R}\begin{pmatrix}0&1&0&0\\\ -1&0&0&0\\\
0&0&0&1\\\ 0&0&-1&0\end{pmatrix}$ (5)

in 6d333We note that with this choice $\Omega^{(6d)}$ is self-dual, in
contrast to the anti-self-dual choice made in previous work. when performing
explicit computations.

With this choice the coordinate $x^{+}$ takes values in the interval $(-\pi
R,\pi R)$. After a Weyl transformation of the metric we can compactify the
interval through the inclusion of its endpoints, extending the range of
$x^{+}$ to $[-\pi R,\pi R]$. The transformation of the metric takes the form

$\displaystyle\hat{g}_{\mu\nu}d\hat{x}^{\mu}d\hat{x}^{\nu}$
$\displaystyle=-2d\hat{x}^{+}d\hat{x}^{-}+d\hat{x}^{i}d\hat{x}^{i}$
$\displaystyle=\Delta^{2}\left(-2dx^{+}\left(dx^{-}+\frac{1}{2}\Omega_{ij}x^{j}dx^{i}\right)+dx^{i}dx^{i}\right)$
$\displaystyle=\Delta^{2}g_{\mu\nu}dx^{\mu}dx^{\nu}\ ,$ (6)

with

$\Delta=\sec\left(\frac{x^{+}}{2R}\right)\ .$ (7)

From our perspective the choice of the scale $R$ is arbitrary; however, we
note that in the holographic realisation of this geometry Lambert:2019jwi  $R$
is the $AdS_{2n+1}$ radius.

Our goal is to work with spinor fields on the $\Omega$-deformed geometry
defined above, so we need to define vielbeins $\hat{e}$ and $e$ for the
metrics $\hat{g}$ and $g$. If we apply a combination of the coordinate
transformation (1) and a Weyl transformation by $\Delta$ to $\hat{e}$ we will
in general not obtain $e$, but a vielbein related to it by a local Lorentz
transformation $\Lambda\in SO(1,2n-1)$ given by

$\tensor{\Lambda}{{}^{a}_{b}}=\Delta\frac{\partial
x^{\mu}}{\partial\hat{x}^{\nu}}e^{a}_{\mu}\hat{e}^{\nu}_{b}\ .$ (8)

Unlike in the bosonic case, $\Lambda$ will enter into the transformation of a
spinor field and will be crucial in relating correlation functions of the
theory defined on the $\Omega$-deformed background to those of the original
Minkowski theory.

A convenient choice for the vielbeins is

$\displaystyle\hat{e}^{0}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(d\hat{x}^{+}+d\hat{x}^{-}\right)\ ,$
(9a) $\displaystyle\hat{e}^{1}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(d\hat{x}^{+}-d\hat{x}^{-}\right)\ ,$
(9b) $\displaystyle\hat{e}^{i}$ $\displaystyle=d\hat{x}^{i}\ ,$ (9c)

for the metric $\hat{g}$, and

$\displaystyle e^{0}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(dx^{+}+dx^{-}+\frac{1}{2}\Omega_{ij}x^{j}dx^{i}\right)\
,$ (10a) $\displaystyle e^{1}$
$\displaystyle=\frac{1}{\sqrt{2}}\left(dx^{+}-dx^{-}-\frac{1}{2}\Omega_{ij}x^{j}dx^{i}\right)\
,$ (10b) $\displaystyle e^{i}$ $\displaystyle=dx^{i}\ ,$ (10c)

for the metric $g$. With these in hand, a brief calculation shows the
components of $\Lambda$ are

$\displaystyle\tensor{\Lambda}{{}^{0}_{0}}$
$\displaystyle=\frac{1}{2}\left(\cos\left(\frac{x^{+}}{2R}\right)+\sec\left(\frac{x^{+}}{2R}\right)\left[1+\frac{x^{2}}{8R^{2}}\right]\right)\
,$ (11a) $\displaystyle\tensor{\Lambda}{{}^{1}_{0}}$
$\displaystyle=\frac{1}{2}\left(\cos\left(\frac{x^{+}}{2R}\right)-\sec\left(\frac{x^{+}}{2R}\right)\left[1+\frac{x^{2}}{8R^{2}}\right]\right)\
,$ (11b) $\displaystyle\tensor{\Lambda}{{}^{0}_{1}}$
$\displaystyle=\frac{1}{2}\left(\cos\left(\frac{x^{+}}{2R}\right)-\sec\left(\frac{x^{+}}{2R}\right)\left[1-\frac{x^{2}}{8R^{2}}\right]\right)\
,$ (11c) $\displaystyle\tensor{\Lambda}{{}^{1}_{1}}$
$\displaystyle=\frac{1}{2}\left(\cos\left(\frac{x^{+}}{2R}\right)+\sec\left(\frac{x^{+}}{2R}\right)\left[1-\frac{x^{2}}{8R^{2}}\right]\right)\
,$ (11d) $\displaystyle\tensor{\Lambda}{{}^{i}_{0}}$
$\displaystyle=\tensor{\Lambda}{{}^{i}_{1}}=\frac{1}{2\sqrt{2}}\left(\cos\left(\frac{x^{+}}{2R}\right)\Omega_{ij}x^{j}-\sin\left(\frac{x^{+}}{2R}\right)\frac{x^{i}}{R}\right)\
,$ (11e) $\displaystyle\tensor{\Lambda}{{}^{0}_{j}}$
$\displaystyle=-\tensor{\Lambda}{{}^{1}_{j}}=\frac{1}{2\sqrt{2}}\sec\left(\frac{x^{+}}{2R}\right)\,\Omega_{jk}x^{k}\
,$ (11f) $\displaystyle\tensor{\Lambda}{{}^{i}_{j}}$
$\displaystyle=\cos\left(\frac{x^{+}}{2R}\right)\,\delta_{ij}+\sin{\left(\frac{x^{+}}{2R}\right)}\,R\Omega_{ij}\
.$ (11g)

We are interested in transforming spinor fields from Minkowski spacetime to
our new coordinate system. In doing so we will require a spinor representation
of the local Lorentz transformation $\Lambda$. The simplest spinor
representation to compute is the Dirac representation, where $\Lambda$ is
represented by

$S[\Lambda]=\operatorname{Exp}\left(\frac{1}{4}\lambda^{ab}\gamma_{ab}\right)\
.$ (12)

Calculating this requires knowledge of the Lie algebra element
$\lambda\in\mathfrak{so}(1,2n-1)$ that generates $\Lambda$, which in our
conventions is given by

$\Lambda=\operatorname{Exp}\left(\lambda\right)\ .$ (13)

While it is not immediately obvious what form this should take, a derivation
based on the accidental isomorphism between $Spin(1,3)$ and $SL(2,\mathbb{C})$
(given in appendix B) for the $n=2$ case or an inspired guess shows that the
correct element is

$\displaystyle\lambda^{01}$
$\displaystyle=\ln\cos\left(\frac{x^{+}}{2R}\right)\ ,$ (14a)
$\displaystyle\lambda^{ij}$ $\displaystyle=\frac{1}{2}\Omega_{ij}x^{+}\ ,$
(14b)
$\displaystyle\lambda^{0i}=-\lambda^{1i}=-\frac{1}{2\sqrt{2}R\sin\left(\frac{x^{+}}{2R}\right)}\Bigg{[}x^{i}\left(\cos\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)+\sin\left(\frac{x^{+}}{2R}\right)\frac{x^{+}}{2R}\right)$
$\displaystyle+R\Omega_{ij}x^{j}\left(\sin\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)-\cos\left(\frac{x^{+}}{2R}\right)\frac{x^{+}}{2R}\right)\Bigg{]}$
$\displaystyle\ .$ (14c)

It is then a straightforward task to compute the Dirac spinor representation
of $\Lambda$; using the gamma matrices outlined in appendix A we find

$\displaystyle S^{(4d)}$ $\displaystyle=\begin{pmatrix}S^{(4d)}_{+}&0\\\
0&S^{(4d)}_{-}\end{pmatrix}\ ,$ (15a) $\displaystyle S^{(4d)}_{+}$
$\displaystyle=\frac{e^{-\frac{ix^{+}}{4R}}}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}3+e^{\frac{ix^{+}}{R}}-\frac{u}{\sqrt{2}R}&-1+e^{\frac{ix^{+}}{R}}-\frac{u}{\sqrt{2}R}\\\
-1+e^{\frac{ix^{+}}{R}}+\frac{u}{\sqrt{2}R}&3+e^{\frac{ix^{+}}{R}}+\frac{u}{\sqrt{2}R}\end{pmatrix}\
,$ (15b) $\displaystyle S^{(4d)}_{-}$
$\displaystyle=\frac{e^{\frac{ix^{+}}{4R}}}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}3+e^{-\frac{ix^{+}}{R}}+\frac{\bar{u}}{\sqrt{2}R}&1-e^{-\frac{ix^{+}}{R}}-\frac{\bar{u}}{\sqrt{2}R}\\\
1-e^{-\frac{ix^{+}}{R}}+\frac{\bar{u}}{\sqrt{2}R}&3+e^{-\frac{ix^{+}}{R}}-\frac{\bar{u}}{\sqrt{2}R}\end{pmatrix}\
,$ (15c)

for the representation in 4d, and

$\displaystyle S^{(6d)}[\Lambda]$
$\displaystyle=\begin{pmatrix}S^{(6d)}_{+}&0\\\ 0&S^{(6d)}_{-}\end{pmatrix}\
,$ (16a) $\displaystyle S^{(6d)}_{+}$
$\displaystyle=\frac{1}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}3+e^{\frac{ix^{+}}{R}}&-1+e^{\frac{ix^{+}}{R}}&0&0\\\
-1+e^{\frac{ix^{+}}{R}}&3+e^{\frac{ix^{+}}{R}}&0&0\\\
0&0&3+e^{-\frac{ix^{+}}{R}}&1-e^{-\frac{ix^{+}}{R}}\\\
0&0&1-e^{-\frac{ix^{+}}{R}}&3+e^{-\frac{ix^{+}}{R}}\\\ \end{pmatrix}$
$\displaystyle\hskip
14.22636pt+\frac{1}{4\sqrt{2}R\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}-u&-u&v&-v\\\
u&u&-v&v\\\ \bar{v}&\bar{v}&\bar{u}&-\bar{u}\\\
\bar{v}&\bar{v}&\bar{u}&-\bar{u}\end{pmatrix}\ ,$ (16b) $\displaystyle
S_{-}^{(6d)}$
$\displaystyle=\frac{e^{-\frac{ix^{+}}{2R}}}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}3+e^{\frac{ix^{+}}{R}}-\frac{u}{\sqrt{2}R}&-1+e^{\frac{ix^{+}}{R}}-\frac{u}{\sqrt{2}R}&-\frac{\bar{v}}{\sqrt{2}R}&\frac{\bar{v}}{\sqrt{2}R}\\\
-1+e^{\frac{ix^{+}}{R}}+\frac{u}{\sqrt{2}R}&3+e^{\frac{ix^{+}}{R}}+\frac{u}{\sqrt{2}R}&\frac{\bar{v}}{\sqrt{2}R}&-\frac{\bar{v}}{\sqrt{2}R}\\\
0&0&0&0\\\ 0&0&0&0\end{pmatrix}$ $\displaystyle\hskip
14.22636pt+\frac{e^{\frac{ix^{+}}{2R}}}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}0&0&0&0\\\
0&0&0&0\\\
-\frac{v}{\sqrt{2}R}&-\frac{v}{\sqrt{2}R}&3+e^{-\frac{ix^{+}}{R}}+\frac{\bar{u}}{\sqrt{2}R}&1-e^{-\frac{ix^{+}}{R}}-\frac{\bar{u}}{\sqrt{2}R}\\\
-\frac{v}{\sqrt{2}R}&-\frac{v}{\sqrt{2}R}&1-e^{-\frac{ix^{+}}{R}}+\frac{\bar{u}}{\sqrt{2}R}&3+e^{-\frac{ix^{+}}{R}}-\frac{\bar{u}}{\sqrt{2}R}\\\
\end{pmatrix}\ ,$ (16c)

for the 6d case. These are written in terms of the complex coordinates

$\displaystyle u=x^{2}+ix^{3}\ ,$ (17a) $\displaystyle v=x^{5}+ix^{4}\ .$
(17b)

Our choice of gamma matrices makes the chiral decomposition of the Dirac
representation manifest, which will be convenient when working with Weyl
spinors.

The last quantity we require is the spin connection
$\tensor{\omega}{{}_{a}{}_{b}}$ of (10). A quick computation gives

$\displaystyle\omega_{0i}=\frac{1}{2\sqrt{2}}\Omega_{ij}dx^{j}\ ,$ (18a)
$\displaystyle\omega_{1i}=\frac{1}{2\sqrt{2}}\Omega_{ij}dx^{j}\ ,$ (18b)
$\displaystyle\omega_{ij}=-\frac{1}{2}\Omega_{ij}dx^{+}\ .$ (18c)

###  2.2 The Action for General Dimensions

With the geometric quantities calculated above in hand, we are ready to find
the action of a fermion on the $\Omega$-deformed background. The action for a
free massless Dirac spinor on a curved background with metric $\hat{g}$ is

$S=-\int
d^{D}\hat{x}\,\hat{e}\hat{\bar{\psi}}\gamma^{a}\hat{e}_{a}^{\mu}\hat{\nabla}_{\mu}\hat{\psi}\
.$ (19)

We are interested in performing a coordinate transformation of the form
considered above, in which we write the transformed metric as

$\hat{g}_{\mu\nu}(\hat{x})d\hat{x}^{\mu}d\hat{x}^{\nu}=\Delta^{2}(x)g_{\mu\nu}(x)dx^{\mu}dx^{\nu}\
.$ (20)

Specifying a vielbein $e$ for $g$, we can use the Lorentz transformation in
eq. (8) to define the spinor field transformation

$\hat{\psi}(\hat{x})=\Delta^{-\frac{D-1}{2}}S^{-1}[\Lambda]\psi(x)\ .$ (21)

It can be shown that this choice leaves the form of the action invariant
Shapiro:2016pfm , so we are left with

$S=-\int d^{D}x\,e\bar{\psi}\gamma^{a}e^{\mu}_{a}\nabla_{\mu}\psi\ .$ (22)

Let us now specialise to the transformation (1) in $D=2n$ dimensions. We can
substitute (10) and (18) into the action to obtain the $\Omega$-deformed
spinor action

$S=i\int
d^{2n}x\,\psi^{{\dagger}}\Bigg{[}\sqrt{2}P_{+}\left(\partial_{+}+\frac{1}{8}\Omega_{ij}\gamma_{ij}\right)\psi+\sqrt{2}P_{-}\partial_{-}\psi+\gamma_{0i}D_{i}\psi\Bigg{]}\
,$ (23)

where we work with the projectors
$P_{\pm}=\frac{1}{2}\left(\mathbbm{1}\pm\gamma_{01}\right)$ and have defined
the differential operator

$D_{i}=\partial_{i}-\frac{1}{2}\Omega_{ij}x^{j}\partial_{-}\ .$ (24)

In this form the complete symmetry group of the action after null reduction
will not be manifest: we can remedy this by introducing the projections of
$\psi$ with respect to $P_{\pm}$, which we denote by

$P_{+}\psi=\chi\;,\;P_{-}\psi=\lambda\ .$ (25)

The action is then

$S=i\int
d^{2n}x\,\Bigg{[}\sqrt{2}\lambda^{{\dagger}}\partial_{-}\lambda+\chi^{{\dagger}}\gamma_{0i}D_{i}\lambda+\lambda^{{\dagger}}\gamma_{0i}D_{i}\chi+\sqrt{2}\chi^{{\dagger}}\left(\partial_{+}+\frac{1}{8}\Omega_{ij}\gamma_{ij}\right)\chi\Bigg{]}\
.$ (26)

We would like to reduce our theory on the null interval. To do this, we take
the fields to have mode expansions of the form

$\displaystyle\lambda(x^{+},x^{-},x)=\sum_{k\in\mathcal{S}}\lambda_{k}(x^{-},x)e^{-\frac{ikx^{+}}{R}}\
,$ (27a)
$\displaystyle\chi(x^{+},x^{-},x)=\sum_{k\in\mathcal{S}}\chi_{k}(x^{-},x)e^{-\frac{ikx^{+}}{R}}\
.$ (27b)

The set of allowed mode numbers $\mathcal{S}$ is fixed by requiring that the
field obtained from a coordinate transformation of the Minkowski spinor,

$\hat{\psi}(\hat{x})=\psi^{\prime}(x)\ ,$ (28)

can be written as the Fourier series

$\psi^{\prime}(x^{+},x^{-},x)=\sum_{l\in\mathbb{Z}}\psi^{\prime}_{l}(x^{-},x)e^{-\frac{ilx^{+}}{R}}\
,$ (29)

mirroring the analogous prescription for a scalar field Lambert:2021nol . This
amounts to demanding consistency of the equation

$\Delta^{-\frac{d-1}{2}}S^{-1}[\Lambda]\sum_{k\in\mathcal{S}}\psi_{k}(x^{-},x)e^{-\frac{ikx^{+}}{R}}=\sum_{l\in\mathbb{Z}}\psi^{\prime}_{l}(x^{-},x)e^{-\frac{ilx^{+}}{R}}\
.$ (30)

As the Lorentz transformation is dependent on $x^{+}$ the details of this will
depend on the spacetime dimension. Substituting the mode expansion into (26)
allows us to perform the integral over $x^{+}$, giving

$S=2\pi iR\sum_{k\in\mathcal{S}}\int
d^{2n-1}x\Bigg{[}\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}+\chi_{k}^{{\dagger}}\gamma_{0i}D_{i}\lambda_{k}+\lambda_{k}^{{\dagger}}\gamma_{0i}D_{i}\chi_{k}+\sqrt{2}\chi^{{\dagger}}\left(\frac{1}{8}\Omega_{ij}\gamma_{ij}-\frac{ik}{R}\right)\chi_{k}\Bigg{]}\
.$ (31)

This action was first constructed in Lambert:2021nol . We note that if we were
only interested in using the reduction to construct a fermion with $SU(1,n)$
spacetime symmetry (i.e. if we were to truncate to the pair
$(\lambda_{k},\chi_{k})$) then we can relax the condition on $k$ and instead
take it to be any real number.

####  2.2.1 The Action for 5d

Let us now specialise (31) to specific dimensions, namely $D=6$ and $D=4$.
While most of (31) is independent of dimension, the term involving
$\Omega_{ij}\gamma_{ij}$ is not. Expanding out the expression for $D=6$ allows
us to rewrite this as

$\Omega_{ij}\gamma_{ij}=\frac{2}{R}\gamma_{23}\left(\mathbbm{1}+\gamma_{01}\gamma_{*}\right)\
,$ (32)

where our convention for $\gamma_{*}$ is given in appendix A. Let us now
impose the condition that our fields are Weyl spinors with negative chirality:
from this we see that

$\Omega_{ij}\gamma_{ij}\chi=0$ (33)

and the final term in the action vanishes, leaving the simplified 5d action

$S^{(5d)}=2\pi iR\sum_{k\in\mathcal{S}^{(5d)}}\int
d^{5}x\Bigg{[}\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}+\chi_{k}^{{\dagger}}\gamma_{0i}D_{i}\lambda_{k}+\lambda_{k}^{{\dagger}}\gamma_{0i}D_{i}\chi_{k}-\frac{\sqrt{2}ik}{R}\chi^{{\dagger}}_{k}\chi_{k}\Bigg{]}\
.$ (34)

Using the explicit form (16) of the spinor representation, the consistency
condition (30) for $\mathcal{S}^{(5d)}$ can be written as

$\sum_{l\in\mathbb{Z}}\psi^{\prime}_{l}e^{-\frac{ilx^{+}}{R}}=\left(e^{\frac{2ix^{+}}{R}}A+e^{\frac{ix^{+}}{R}}B+C+e^{-\frac{ix^{+}}{R}}D\right)\sum_{k\in\mathcal{S}^{(5d)}}\psi_{k}e^{-\frac{ikx^{+}}{R}}$
(35)

where $A,B,C,D$ are $x^{+}$-independent matrices whose form will not be
important. It immediately follows from this that the set of allowed modes is
just $\mathcal{S}^{(5d)}=\mathbb{Z}$.

####  2.2.2 The Action for 3d

We can do the exact same analysis for $D=4$. However, we now have

$\Omega_{ij}\gamma_{ij}=\frac{2i}{R}\gamma_{01}\gamma_{*}\ ,$ (36)

so the action of the matrix on the field $\chi$ is

$\Omega_{ij}\gamma_{ij}\chi=\pm\frac{2i}{R}\chi\ ,$ (37)

depending on the chirality of our fields. Unlike in the 5d case no
simplification occurs by choosing a particular sign; for definiteness we will
take our fields to have positive chirality from here onwards. The effect of
this is to shift $k$ to $k-\frac{1}{4}$ in the action. For this reason it will
be useful to redefine our field by an $x^{+}$-dependent phase, where we take

$\hat{\psi}(\hat{x})=\Delta^{-\frac{3}{2}}S_{+}^{-1}[\Lambda]e^{-\frac{ix^{+}}{4R}}\psi(x)$
(38)

in place of (21) to simplify the action. With this, the action of the 3d
theory is

$S^{(3d)}=2\pi iR\sum_{k\in\mathcal{S}^{(3d)}}\int
d^{3}x\Bigg{[}\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}+\chi_{k}^{{\dagger}}\gamma_{0i}D_{i}\lambda_{k}+\lambda_{k}^{{\dagger}}\gamma_{0i}D_{i}\chi_{k}-\frac{\sqrt{2}ik}{R}\chi_{k}^{{\dagger}}\chi_{k}\Bigg{]}\
.$ (39)

Let us determine the set $\mathcal{S}^{(3d)}$. Examining the consistency
condition (30) using the explicit form (15) for the spinor representation and
the modified transformation (38) leads to

$\sum_{l\in\mathbb{Z}}\psi_{l}^{\prime}e^{-\frac{ilx^{+}}{R}}=\left(e^{\frac{ix^{+}}{R}}A+B+e^{-\frac{ix^{+}}{R}}C\right)\sum_{k\in\mathcal{S}^{(3d)}}\psi_{k}e^{-\frac{ikx^{+}}{R}}$
(40)

in terms of $x^{+}$-independent matrices $A,B,C$, so as for the 5d theory we
have $\mathcal{S}^{(3d)}=\mathbb{Z}$.

###  2.3 A Simplification for Non-Zero $k$

The action of the level-$k$ mode takes the form

$S_{k}=2\pi iR\int
d^{2n-1}x\Bigg{[}\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}+\chi_{k}^{{\dagger}}\gamma_{0i}D_{i}\lambda_{k}+\lambda_{k}^{{\dagger}}\gamma_{0i}D_{i}\chi_{k}-\frac{\sqrt{2}ik}{R}\chi_{k}^{{\dagger}}\chi_{k}\Bigg{]}$
(41)

in both 3d and 5d. The field $\chi_{k}$ plays a remarkably different role
depending on whether $k=0$ or $k\neq 0$. In the former case, the final term in
the action vanishes. The field $\chi_{0}$ is then a spinorial Lagrange
multiplier imposing the constraint

$\gamma_{0i}D_{i}\lambda_{k}=0\ .$ (42)

However, when $k\neq 0$ $\chi_{k}$ has the algebraic equation of motion

$\chi_{k}=-\frac{iR}{\sqrt{2}k}\gamma_{0i}D_{i}\lambda_{k}\ .$ (43)

We can substitute this back into the action to obtain

$S_{k}=\frac{\sqrt{2}\pi R^{2}}{k}\int
d^{2n-1}x\Bigg{[}\frac{2i}{R}\lambda_{k}^{{\dagger}}\left(k-\Gamma\right)\partial_{-}\lambda_{k}+\lambda_{k}^{{\dagger}}D^{2}\lambda_{k}\Bigg{]}\
,$ (44)

where we’ve defined the matrix

$\Gamma=\frac{iR\Omega_{ij}\gamma_{ij}}{4}\ .$ (45)

As mentioned above, the properties of this matrix are dimension dependent.
Using (36) we see that in 3d we have

$\Gamma\lambda_{k}=\frac{1}{2}\lambda_{k}\ ,$ (46)

so the action is just

$S^{(3d)}_{k}=\frac{\sqrt{2}\pi R^{2}}{k}\int
d^{3}x\Bigg{[}\frac{2i}{R}\left(k-\frac{1}{2}\right)\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}+\lambda_{k}^{{\dagger}}D^{2}\lambda_{k}\Bigg{]}\
.$ (47)

In 5d it has the nice property that

$P_{-}\Gamma^{2}=P_{-}\ ,$ (48)

so we can define the pseudo-projector

$P_{\pm}^{\Gamma}=\frac{1}{2}\left(\mathbbm{1}\pm\Gamma\right)$ (49)

that behaves as a projector when acting on the subspace defined by projecting
with $P_{-}$. Defining the fields

$\lambda_{k}^{\pm}=P^{\Gamma}_{\pm}\lambda_{k}\ ,$ (50)

the action decomposes as

$S^{(5d)}_{k}=\frac{\sqrt{2}\pi R^{2}}{k}\int
d^{5}x\Bigg{[}\frac{2i(k-1)}{R}\lambda_{k}^{+{\dagger}}\partial_{-}\lambda^{+}_{k}+\frac{2i(k+1)}{R}\lambda_{k}^{-{\dagger}}\partial_{-}\lambda^{-}_{k}+\lambda_{k}^{+{\dagger}}D^{2}\lambda^{+}_{k}+\lambda_{k}^{-{\dagger}}D^{2}\lambda^{-}_{k}\Bigg{]}\
.$ (51)

By counting the degrees of freedom in our fields we observe that both
$\lambda_{k}$ in the 4d case and $\lambda_{k}^{\pm}$ in the 6d case are
complex Grassmann numbers. It is then no surprise that their actions are
almost identical to the level-$k$ action for the $SU(1,n)$-invariant free
complex scalar field

$S^{(\phi)}_{k}=2\pi R\int
d^{2n-1}x\Bigg{[}\frac{2ik}{R}\bar{\phi}_{k}\partial_{-}\phi_{k}+\bar{\phi}_{k}D^{2}\phi_{k}\Bigg{]}$
(52)

obtained in Lambert:2021nol . The novel part of this is the shift in $k$; we
will see later when we examine the Ward identities satisfied by the two-point
functions that this is due to the non-trivial transformations of the fields
under spatial rotations.

We can use this form of the action to predict the form of the
$\lambda_{k}\lambda_{k}^{{\dagger}}$ two-point function without doing any
detailed computations. Defining the $(2n-1)$-dimensional scalar two-point
function

$\displaystyle
G_{k}^{((2n-1)d)}\equiv\bra{\Omega}\phi_{k}(x_{1}^{-},x_{1})\bar{\phi}_{k}(x_{2}^{-},x_{2})\ket{\Omega}$
(53)

for the action (52), matching the normalisation of the fields in the above
actions suggests that the fermion two-point functions should be related to the
scalar two-point function by

$\bra{\Omega}\lambda_{k}(x_{1}^{-},x_{1})\lambda_{k}^{{\dagger}}(x_{2}^{-},x_{2})\ket{\Omega}=\frac{\sqrt{2}k}{R}P_{-}G_{k-\frac{1}{2}}^{(3d)}$
(54)

in 3d, and

$\bra{\Omega}\lambda_{k}(x_{1}^{-},x_{1})\lambda_{k}^{{\dagger}}(x_{2}^{-},x_{2})\ket{\Omega}=\frac{\sqrt{2}k}{R}P_{-}\left(P_{+}^{\Gamma}G_{k-1}^{(5d)}+P_{-}^{\Gamma}G_{k+1}^{(5d)}\right)$
(55)

in 5d. The projection operators have been added by hand to project onto the
subspaces in which each field lives. We will show in the next section that
this intuition is correct.

##  3 SUSY and Correlation Functions

###  3.1 Supersymmetry in the Reduced Theory

Since the fermionic theory we are considering is free, the quantities of
interest are the two-point functions of the fields. There are many ways we
could compute these, but it will prove convenient to fix these using the
supersymmetry present in a free theory with a single complex scalar and
fermion and the known two-point functions for an $SU(1,n)$ scalar field
Lambert:2021nol .

We start with the action for a free theory in Minkowski spacetime consisting
of complex scalar and Weyl spinor fields,

$S=-\int
d^{2n}\hat{x}\left(\eta^{\mu\nu}\hat{\partial}_{\mu}\bar{\hat{\phi}}\hat{\partial}_{\nu}\hat{\phi}+\bar{\hat{\psi}}\gamma^{\mu}\hat{\partial}_{\mu}\psi\right)\
.$ (56)

This is invariant under the transformation

$\displaystyle\delta\hat{\phi}=\bar{\hat{\xi}}\hat{\psi}\ ,$ (57a)
$\displaystyle\delta\hat{\psi}=-\hat{\partial}_{\mu}\hat{\phi}\gamma^{\mu}\hat{\xi}\
,$ (57b)

where $\hat{\xi}$ is a Weyl spinor. We can then use the coordinate
transformation (1) and perform the reduction along $x^{+}$, leaving us with an
infinite tower of scalar and spinor theories in $(2n-1)$-dimensions. Since we
still have a full description of the system (albeit in terms of an infinite
number of fields) after transforming and taking the null reduction, the
supersymmetry of the $2n$-dimensional theory will be present in some form in
the $(2n-1)$-dimensional theory. As the supercharge transforms as a
$2n$-dimensional spinor, its components will generically develop dependence on
$x^{+}$ and will therefore not commute with the generator of translations
along the compactified null direction. This means that we expect some fraction
of the supersymmetry transformations in the reduced theory will take us
between levels of the Kaluza-Klein tower.

####  3.1.1 3d Supersymmetry

Let us first look at the 3d theory. Combining the scalar field action (52) for
$n=2$ and $k^{(\phi)}\in\mathbb{Z}+\frac{1}{2}$ with the fermion action (39)
gives

$\displaystyle S=2\pi R\sum_{k\in\mathbb{Z}}\int d^{3}x\Bigg{[}$
$\displaystyle\frac{2i}{R}\left(k-\frac{1}{2}\right)\bar{\phi}_{k-\frac{1}{2}}\partial_{-}\phi_{k-\frac{1}{2}}+\bar{\phi}_{k-\frac{1}{2}}D^{2}\phi_{k-\frac{1}{2}}+i\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}$
(58)
$\displaystyle+i\chi_{k}^{{\dagger}}\gamma_{0i}D_{i}\lambda_{k}+i\lambda_{k}^{{\dagger}}\gamma_{0i}D_{i}\chi_{k}+\frac{\sqrt{2}k}{R}\chi_{k}^{{\dagger}}\chi_{k}\Bigg{]}\
.$

We introduce a 4d Weyl spinor parameter $\epsilon$, from which we can define
the $\gamma_{01}$ chiral components

$\displaystyle\xi$ $\displaystyle=P_{+}\epsilon\ ,$ (59a) $\displaystyle\eta$
$\displaystyle=P_{-}\epsilon\ .$ (59b)

A short computation shows that the transformations

$\displaystyle\delta\phi_{k-\frac{1}{2}}$
$\displaystyle=\eta^{{\dagger}}\lambda_{k}\ ,$ (60a)
$\displaystyle\delta\lambda_{k}$
$\displaystyle=-\frac{\sqrt{2}k}{R}\phi_{k-\frac{1}{2}}\eta\ ,$ (60b)
$\displaystyle\delta\chi_{k}$
$\displaystyle=i\gamma_{0i}D_{i}\phi_{k-\frac{1}{2}}\eta\ ,$ (60c)

and

$\displaystyle\delta\phi_{k+\frac{1}{2}}$
$\displaystyle=\xi^{{\dagger}}\chi_{k}\ ,$ (61a)
$\displaystyle\delta\lambda_{k}$
$\displaystyle=i\gamma_{0i}D_{i}\phi_{k+\frac{1}{2}}\xi\ ,$ (61b)
$\displaystyle\delta\chi_{k}$
$\displaystyle=-i\sqrt{2}\partial_{-}\phi_{k+\frac{1}{2}}\xi\ ,$ (61c)

both leave the action invariant. The commutators of these transformations on
the scalar modes are

$\displaystyle[\delta_{\eta_{1}},\delta_{\eta_{2}}]\phi_{k-\frac{1}{2}}$
$\displaystyle=\frac{\sqrt{2}k}{R}\left(\eta_{1}^{{\dagger}}\eta_{2}-\eta_{2}^{{\dagger}}\eta_{1}\right)\phi_{k-\frac{1}{2}}\
,$ (62a)
$\displaystyle[\delta_{\xi_{1}},\delta_{\xi_{2}}]\phi_{k-\frac{1}{2}}$
$\displaystyle=i\sqrt{2}\left(\xi_{1}^{{\dagger}}\xi_{2}-\xi_{2}^{{\dagger}}\xi_{1}\right)\partial_{-}\phi_{k-\frac{1}{2}}\
,$ (62b) $\displaystyle[\delta_{\eta},\delta_{\xi}]\phi_{k-\frac{1}{2}}$
$\displaystyle=i\xi^{{\dagger}}\gamma_{0i}\eta
D_{i}\phi_{k-\frac{3}{2}}-i\eta^{{\dagger}}\gamma_{0i}\xi
D_{i}\phi_{k+\frac{1}{2}}\ .$ (62c)

We observe that we have closure of the algebra for the commutators between
transformations of the same chirality, but that this breaks down for
commutators between transformations of mixed chirality. It naively appears
that these transformations are not symmetries of the theory. However, since we
have an infinite number of fields we can repackage them into a sum of the form

$\phi=\sum_{k\in\mathbb{Z}}e^{-\frac{ik\alpha}{R}}\phi_{k-\frac{1}{2}}$ (63)

in terms of an arbitrary parameter $\alpha$. We also rescale our spinor
parameters to

$\displaystyle\hat{\eta}$ $\displaystyle=e^{\frac{i\alpha(1+c)}{2R}}\eta\ ,$
(64a) $\displaystyle\hat{\xi}$ $\displaystyle=e^{-\frac{i\alpha(1-c)}{2R}}\xi\
,$ (64b)

where $c$ is a real constant. The action of the commutators on the field
$\phi$ using the rescaled spinor parameters is

$\displaystyle[\delta_{\hat{\eta}_{1}},\delta_{\hat{\eta}_{2}}]\phi$
$\displaystyle=i\sqrt{2}\left(\eta_{1}^{{\dagger}}\eta_{2}-\eta_{2}^{{\dagger}}\eta_{1}\right)\partial_{\alpha}\phi\
,$ (65a) $\displaystyle[\delta_{\hat{\xi}_{1}},\delta_{\hat{\xi}_{2}}]\phi$
$\displaystyle=i\sqrt{2}\left(\xi_{1}^{{\dagger}}\xi_{2}-\xi_{2}^{{\dagger}}\xi_{1}\right)\partial_{-}\phi\
,$ (65b) $\displaystyle[\delta_{\hat{\eta}},\delta_{\hat{\xi}}]\phi$
$\displaystyle=i\left(\xi^{{\dagger}}\gamma_{0i}\eta-\eta^{{\dagger}}\gamma_{0i}\xi\right)D_{i}\phi\
.$ (65c)

As expected, the algebra closes on the new field. We are able to identify
$\alpha$ with the coordinate of the emergent dimension. Note that for this
argument to work it is crucial that we retain the full Kaluza-Klein tower of
fields.

####  3.1.2 5d Supersymmetry

We can repeat the analysis performed above for the 5d theory. In this case,
the combined action is

$\displaystyle S=2\pi R\sum_{k\in\mathbb{Z}}\int d^{3}x\Bigg{[}$
$\displaystyle\frac{2ik}{R}\bar{\phi}_{k}\partial_{-}\phi_{k}+\bar{\phi}_{k}D^{2}\phi_{k}+i\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}$
(66)
$\displaystyle+i\chi_{k}^{{\dagger}}\gamma_{0i}D_{i}\lambda_{k}+i\lambda_{k}^{{\dagger}}\gamma_{0i}D_{i}\chi_{k}+\frac{\sqrt{2}k}{R}\chi_{k}^{{\dagger}}\chi_{k}\Bigg{]}\
.$

Our supersymmetry parameter $\epsilon$ is a 6d Weyl spinor, which we split
into the components $(\xi,\eta)$ defined by

$\displaystyle\xi=P_{+}\epsilon\ ,$ (67a) $\displaystyle\eta=P_{-}\epsilon\ .$
(67b)

It will be necessary to further split $\eta$ as

$\eta_{\pm}=P_{\pm}^{\Gamma}\eta\ .$ (68)

It can then be shown that the transformations

$\displaystyle\delta\phi_{k}$ $\displaystyle=\xi^{{\dagger}}\chi_{k}\ ,$ (69a)
$\displaystyle\delta\lambda_{k}$ $\displaystyle=i\gamma_{0i}D_{i}\phi_{k}\xi\
,$ (69b) $\displaystyle\delta\chi_{k}$
$\displaystyle=-i\sqrt{2}\partial_{-}\phi_{k}\xi\ ,$ (69c)

and

$\displaystyle\delta\phi_{k}$
$\displaystyle=\eta_{\pm}^{{\dagger}}\lambda_{k\pm 1}\ ,$ (70a)
$\displaystyle\delta\lambda_{k}$ $\displaystyle=-\frac{\sqrt{2}k}{R}\phi_{k\mp
1}\eta_{\pm}\ ,$ (70b) $\displaystyle\delta\chi_{k}$
$\displaystyle=i\gamma_{0i}D_{i}\phi_{k\mp 1}\eta_{\pm}\ ,$ (70c)

leave the action invariant. The commutators of these transformations on
$\phi_{k}$ are

$\displaystyle[\delta_{\xi_{1}},\delta_{\xi_{2}}]\phi_{k}$
$\displaystyle=i\sqrt{2}\left(\xi_{1}^{{\dagger}}\xi_{2}-\xi_{2}^{{\dagger}}\xi_{1}\right)\partial_{-}\phi_{k}\
,$ (71a) $\displaystyle[\delta_{\eta_{1\pm}},\delta_{\eta_{2\pm}}]\phi_{k}$
$\displaystyle=\frac{\sqrt{2}(k\pm
1)}{R}\left(\eta_{1\pm}^{{\dagger}}\eta_{2\pm}-\eta_{2\pm}^{{\dagger}}\eta_{1\pm}\right)\phi_{k}\
,$ (71b) $\displaystyle[\delta_{\eta_{1\pm}},\delta_{\eta_{2\mp}}]\phi_{k}$
$\displaystyle=0\ ,$ (71c)
$\displaystyle[\delta_{\xi},\delta_{\eta_{\pm}}]\phi_{k}$
$\displaystyle=i\left(\eta_{\pm}^{{\dagger}}\gamma_{0i}\xi D_{i}\phi_{k\pm
1}-\xi^{{\dagger}}\gamma_{0i}\eta_{\pm}D_{i}\phi_{k\mp 1}\right)\ .$ (71d)

As in the 3d case, we do not have closure of the algebra between
transformations of mixed $\gamma_{01}$ chirality. However, if we again work
with the sum

$\phi=\sum_{k\in\mathbb{Z}}e^{-\frac{ik\alpha}{R}}\phi_{k}$ (72)

and the rescaled spinor variables444We are again free to redefine all spinor
variables by an additional factor like in (64), though we will choose not to
do this.

$\displaystyle\hat{\xi}$ $\displaystyle=\xi\ ,$ (73a)
$\displaystyle\hat{\eta}_{\pm}$
$\displaystyle=e^{\pm\frac{i\alpha}{R}}\eta_{\pm}\ ,$ (73b)

the commutators become

$\displaystyle[\delta_{\hat{\xi}_{1}},\delta_{\hat{\xi}_{2}}]\phi$
$\displaystyle=\sqrt{2}i\left(\xi_{1}^{{\dagger}}\xi_{2}-\xi_{2}^{{\dagger}}\xi_{1}\right)\partial_{-}\phi\
,$ (74a)
$\displaystyle[\delta_{\hat{\eta}_{1\pm}},\delta_{\hat{\eta}_{2\pm}}]\phi$
$\displaystyle=\sqrt{2}i\left(\eta_{1\pm}^{{\dagger}}\eta_{2\pm}-\eta_{2\pm}^{{\dagger}}\eta_{1\pm}\right)\left(\partial_{\alpha}\mp\frac{i}{R}\right)\phi\
,$ (74b)
$\displaystyle[\delta_{\hat{\eta}_{1\pm}},\delta_{\hat{\eta}_{2\mp}}]\phi$
$\displaystyle=0\ ,$ (74c)
$\displaystyle[\delta_{\hat{\xi}},\delta_{\hat{\eta}_{\pm}}]\phi$
$\displaystyle=i\left(\eta_{\pm}^{{\dagger}}\gamma_{0i}\xi-\xi^{{\dagger}}\gamma_{0i}\eta_{\pm}\right)D_{i}\phi\
,$ (74d)

and we recover closure of the algebra.

###  3.2 Correlation Functions in the Reduced Theory

We would like to find the two-point functions of the fermionic theory. Up to a
factor, these are solutions to the classical equation of motion and are not
hard to find. However, we need to fix the overall normalisation. The simplest
way to do this is to exploit the fact that the theory we’re interested in can
be extended to the supersymmetric theory discussed above, so the normalisation
of the fermion two-point functions can be fixed in terms of the known
normalisation of scalar correlation functions found in Lambert:2021nol . A
derivation of this normalisation, with discussion of how it relates to the
Green’s function equations of the reduced theory, is given in appendix C.

####  3.2.1 3d Correlation Functions

We first discuss the correlation functions of the 3d theory. As shown in
appendix C.1, the non-zero correlation functions of the modes of a complex
scalar field are

$\bra{\Omega}\phi_{k}(x^{-},x)\bar{\phi}_{k}(0)\ket{\Omega}=-\frac{i}{8\pi^{2}R}\frac{1}{\sqrt{z\bar{z}}}\left(\frac{z}{\bar{z}}\right)^{k}$
(75)

for $k>0$, and

$\bra{\Omega}\bar{\phi}_{k}(x^{-},x)\phi_{k}(0)\ket{\Omega}=\frac{i}{8\pi^{2}R}\frac{1}{\sqrt{z\bar{z}}}\left(\frac{z}{\bar{z}}\right)^{-k}$
(76)

for $k<0$, where we have defined the complex variable $z$ as

$z=x^{-}+\frac{ix^{2}}{4R}\ .$ (77)

As our theory is invariant under the action of the supercharges $Q_{+}$ and
$Q_{-}$, the vacuum $\ket{\Omega}$ satisfies

$Q_{+}\ket{\Omega}=Q_{-}\ket{\Omega}=Q_{+}^{{\dagger}}\ket{\Omega}=Q_{-}^{{\dagger}}\ket{\Omega}=0\
.$ (78)

This implies the correlation function identity

$\bra{\Omega}\\{Q_{-},\phi_{k-\frac{1}{2}}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\\}\ket{\Omega}=0\
,$ (79)

which after expanding and simplifying gives us the fermion two-point function

$\displaystyle\bra{\Omega}\lambda_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=\frac{\sqrt{2}k}{R}P_{-}\bra{\Omega}\phi_{k-\frac{1}{2}}(x^{-},x)\bar{\phi}_{k-\frac{1}{2}}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{4\sqrt{2}\pi^{2}R^{2}}\frac{1}{\sqrt{z^{2}}}\left(\frac{z}{\bar{z}}\right)^{k}P_{-}$
(80)

for $k>0$ and zero for $k$ outside this range. Proceeding similarly for other
combinations of supercharges acting on fields gives the non-vanishing two-
point functions

$\displaystyle\bra{\Omega}\lambda_{k}(x^{-},x)\chi_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{8\pi^{2}R^{2}}\frac{\gamma_{0i}x^{i}}{z\sqrt{\bar{z}^{2}}}\left(\frac{z}{\bar{z}}\right)^{k}P_{+}\
,$ (81a)
$\displaystyle\bra{\Omega}\chi_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{8\pi^{2}R^{2}}\frac{\gamma_{0i}x^{i}}{\sqrt{z^{2}}\bar{z}}\left(\frac{z}{\bar{z}}\right)^{k}P_{-}\
,$ (81b)
$\displaystyle\bra{\Omega}\chi_{k}(x^{-},x)\chi_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=\frac{1}{4\sqrt{2}\pi^{2}R}\frac{1}{z\sqrt{\bar{z}^{2}}}\left(\frac{z}{\bar{z}}\right)^{k}\left(k-(k+1)\frac{z}{\bar{z}}\right)P_{+}\
,$ (81c)

for $k\geq 0$, and

$\displaystyle\bra{\Omega}\lambda_{k}^{{\dagger}}(x^{-},x)\lambda_{k}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{4\sqrt{2}\pi^{2}R^{2}}\frac{1}{\sqrt{\bar{z}^{2}}}\left(\frac{z}{\bar{z}}\right)^{-k}P_{-}\
,$ (82a)
$\displaystyle\bra{\Omega}\lambda_{k}^{{\dagger}}(x^{-},x)\chi_{k}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{8\pi^{2}R^{2}}\frac{\gamma_{0i}x^{i}}{\sqrt{z^{2}}\bar{z}}\left(\frac{z}{\bar{z}}\right)^{-k}P_{-}\
,$ (82b)
$\displaystyle\bra{\Omega}\chi_{k}^{{\dagger}}(x^{-},x)\lambda_{k}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{8\pi^{2}R^{2}}\frac{\gamma_{0i}x^{i}}{z\sqrt{\bar{z}^{2}}}\left(\frac{z}{\bar{z}}\right)^{-k}P_{+}\
,$ (82c)
$\displaystyle\bra{\Omega}\chi_{k}^{{\dagger}}(x^{-},x)\chi_{k}(0)\ket{\Omega}$
$\displaystyle=-\frac{1}{4\sqrt{2}\pi^{2}R}\frac{1}{\sqrt{z^{2}}\bar{z}}\left(\frac{z}{\bar{z}}\right)^{-k}\left(k-(k+1)\frac{\bar{z}}{z}\right)P_{+}\
,$ (82d)

for $k\leq-1$. We note that the correlation functions should always read in
matrix form, i.e.

$\bra{\Omega}(\lambda_{k}^{{\dagger}})^{\beta}\left(\chi_{k}\right)_{\alpha}\ket{\Omega}\equiv\tensor{\left(\bra{\Omega}\lambda_{k}^{{\dagger}}\chi_{k}\ket{\Omega}\right)}{{}_{\alpha}^{\beta}}\
.$ (83)

As we have found the two-point functions of the modes, we can now resum them
to obtain the two-point function of the original theory. We will be
considering time-ordered two-point functions, where the time-ordering is taken
with respect to the coordinate $x^{-}$. The two-point function of the resummed
field $\lambda$ is given by

$\displaystyle\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\sum_{k\in\mathbb{Z}}e^{-\frac{ikx^{+}}{R}}\bra{\Omega}T\\{\lambda_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\sum_{k=1}^{\infty}\bigg{(}e^{-\frac{ikx^{+}}{R}}\Theta(x^{-})\bra{\Omega}\lambda_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle\hskip
42.67912pt-e^{\frac{ikx^{+}}{R}}\Theta(-x^{-})\bra{\Omega}\lambda_{-k}^{{\dagger}}(-x^{-},-x)\lambda_{-k}(0)\ket{\Omega}\bigg{)}\
.$ (84)

We have used translational invariance of the vacuum to move all coordinate
dependence to the first field. Substituting in the mode correlation functions
and taking $\sqrt{z^{2}}=z$ gives

$\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}=-\frac{iP_{-}}{4\sqrt{2}\pi^{2}R^{2}}\frac{1}{z}\sum_{k=1}^{\infty}\left(\Theta(x^{-})k\left(\frac{ze^{-\frac{ix^{+}}{R}}}{\bar{z}}\right)^{k}-\Theta(-x^{-})k\left(\frac{ze^{-\frac{ix^{+}}{R}}}{\bar{z}}\right)^{-k}\right)\
.$ (85)

However, these sums do not converge: we need to specify an $i\epsilon$
prescription. As $x^{-}$ has a definite sign in each term, we can take

$x^{+}\to x^{+}-ix^{-}\epsilon\equiv x^{+}_{\epsilon}$ (86)

for some infinitesimal $\epsilon>0$. We will explain the relation of this to
the usual $i\epsilon$ prescription below. The sums are now convergent and
(suppressing the $\epsilon$ terms for now) we find the result

$\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}=-\frac{iP_{-}}{4\sqrt{2}\pi^{2}R^{2}}\frac{\bar{z}}{\left(\bar{z}e^{\frac{ix^{+}}{2R}}-ze^{-\frac{ix^{+}}{2R}}\right)^{2}}\
.$ (87)

The same manipulations for our other two-point functions lead to the resummed
functions

$\displaystyle\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=-\frac{iP_{-}}{8\pi^{2}R^{2}}\frac{\gamma_{0i}x^{i}}{\left(\bar{z}e^{\frac{ix^{+}}{2R}}-ze^{-\frac{ix^{+}}{2R}}\right)^{2}}\
,$ (88a)
$\displaystyle\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=-\frac{iP_{+}}{8\pi^{2}R^{2}}\frac{\gamma_{0i}x^{i}}{\left(\bar{z}e^{\frac{ix^{+}}{2R}}-ze^{-\frac{ix^{+}}{2R}}\right)^{2}}\
,$ (88b)
$\displaystyle\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=-\frac{iP_{+}}{2\sqrt{2}\pi^{2}R}\frac{e^{\frac{ix^{+}}{2R}}\sin\left(\frac{x^{+}}{2R}\right)}{\left(\bar{z}e^{\frac{ix^{+}}{2R}}-ze^{-\frac{ix^{+}}{2R}}\right)^{2}}\
.$ (88c)

This is equivalent to the correlation function

$\bra{\Omega}T\\{\psi(x^{+},x^{-},x)\psi^{{\dagger}}(0)\\}\ket{\Omega}=-\frac{i}{8\pi^{2}R^{2}}\frac{1}{\left(\bar{z}e^{\frac{ix^{+}}{2R}}-ze^{-\frac{ix^{+}}{2R}}\right)^{2}}\left(2\sqrt{2}Re^{\frac{ix^{+}}{2R}}\sin\left(\frac{x^{+}}{2R}\right)P_{+}+\sqrt{2}\bar{z}P_{-}+\gamma_{0i}x^{i}\right)$
(89)

for the full spinor field $\psi$. Recalling the transformation555Note that we
should replace $x^{+}$ with $x_{\epsilon}^{+}$ in the transformation. (38),
the two-point function of the original Minkowski theory can be found from

$\displaystyle\bra{\Omega}T\\{\hat{\psi}(\hat{x}^{+},\hat{x}^{-},\hat{x})\hat{\psi}^{{\dagger}}(0)\\}\ket{\Omega}=e^{-\frac{ix^{+}}{4R}}\Delta^{-\frac{3}{2}}(x)S_{+}^{-1}[\Lambda_{x}]\bra{\Omega}T\\{\psi(x^{+},x^{-},x)\psi^{{\dagger}}(0)\\}\ket{\Omega}\
,$ (90)

which after evaluating and using

$\frac{\cos\left(\frac{x^{+}}{2R}\right)}{ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}}=\frac{-2iR}{-2\hat{x}^{+}\hat{x}^{-}+\hat{x}^{i}\hat{x}^{i}}$
(91)

gives

$\displaystyle\bra{\Omega}T\\{\hat{\psi}(\hat{x}^{+},\hat{x}^{-},\hat{x})\hat{\psi}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=-\frac{i\cos^{2}\left(\frac{x^{+}}{2R}\right)}{8\pi^{2}R^{2}}\frac{\sqrt{2}\hat{x}^{+}P_{+}+\sqrt{2}\hat{x}^{-}P_{-}+\gamma_{0i}\hat{x}^{i}}{\left(\bar{z}e^{\frac{ix^{+}}{2R}}-ze^{-\frac{ix^{+}}{2R}}\right)^{2}}$
$\displaystyle=-\frac{i}{2\pi^{2}}\frac{\gamma_{\mu}\gamma_{0}\hat{x}^{\mu}}{\left(-2\hat{x}^{+}\hat{x}^{-}+\hat{x}^{i}\hat{x}^{i}\right)^{2}}\
,$ (92)

or equivalently

$\bra{\Omega}T\\{\hat{\psi}(\hat{x}^{+},\hat{x}^{-},\hat{x})\hat{\bar{\psi}}(0)\\}\ket{\Omega}=\frac{\gamma_{\mu}\hat{x}^{\mu}}{2\pi^{2}\left(-2\hat{x}^{+}\hat{x}^{-}+\hat{x}^{i}\hat{x}^{i}\right)^{2}}\
.$ (93)

We recognise this as the two-point correlation for a free fermion in 4d
Minkowski spacetime when written in lightcone coordinates, with the
canonically normalised action

$S=-\int d^{4}\hat{x}\,\hat{\bar{\psi}}\gamma^{\mu}\partial_{\mu}\hat{\psi}\
.$ (94)

We should justify the strange coordinate-dependent $i\epsilon$ prescription
used to ensure convergence of the sum over modes. We recover $\epsilon$ in our
final correlation function by taking $\hat{x}^{\mu}$ to
$\hat{x}^{\mu}_{\epsilon}$, where $\hat{x}^{\mu}_{\epsilon}$ is the coordinate
transformation (1) after replacing $x^{+}$ with $x^{+}_{\epsilon}$. If we
examine the denominator of the correlation function as $R\to\infty$, we see
that

$\displaystyle\lim_{R\to\infty}\left(\frac{1}{(-2\hat{x}^{+}_{\epsilon}\hat{x}^{-}_{\epsilon}+\hat{x}^{i}_{\epsilon}\hat{x}^{i}_{\epsilon})^{2}}\right)$
$\displaystyle=\frac{1}{\left(-2\left(x^{+}-ix^{-}\epsilon\right)x^{-}+x^{i}x^{i}\right)^{2}}$
$\displaystyle=\frac{1}{\left(-2x^{+}x^{-}+x^{i}x^{i}+i\epsilon^{\prime}\right)^{2}}\
,$ (95)

where we’ve defined

$\epsilon^{\prime}=2(x^{-})^{2}\epsilon\ .$ (96)

We recognise this as the standard $i\epsilon$ prescription for a time-ordered
two-point function, justifying our procedure666Note that we should really
redefine our final expression in terms of a constant infinitesimal
$\epsilon^{\prime}$ so that the prescription is also valid when $x^{-}=0$.. We
can also consider the resummed correlation function (89) in the transformed
theory. Replacing $x^{+}$ by $x^{+}_{\epsilon}$ and expanding the denominator
to first order in $\epsilon$ gives

$\bra{\Omega}T\\{\psi(x^{+},x^{-},x)\psi^{{\dagger}}(0)\\}\ket{\Omega}=\frac{i\left(2\sqrt{2}Re^{\frac{ix^{+}}{2R}}\sin\left(\frac{x^{+}}{2R}\right)P_{+}+\sqrt{2}\bar{z}P_{-}+\gamma_{0i}x^{i}\right)}{2\pi^{2}\left(x^{2}\cos\left(\frac{x^{+}}{2R}\right)-4Rx^{-}\sin\left(\frac{x^{+}}{2R}\right)+i\epsilon^{\prime}\right)^{2}}\
,$ (97)

so we see that this is the natural $i\epsilon$ prescription in the theory with
action (26). Further motivation can be found from the analogous computation in
2d, which is performed in section 5, where the same prescription is found by
ensuring integrals over momenta are well-defined.

####  3.2.2 5d Correlation Functions

The same procedure can be used to find the correlation functions of the 5d
theory. As shown in appendix C.2, the non-zero scalar two-point functions are

$\bra{\Omega}\phi_{k}(x^{-},x)\bar{\phi}_{k}(0)\ket{\Omega}=-\frac{k}{16\pi^{3}R^{2}}\frac{1}{z\bar{z}}\left(\frac{z}{\bar{z}}\right)^{k}$
(98)

for $k>0$, and

$\bra{\Omega}\bar{\phi}_{k}(x^{-},x)\phi_{k}(0)\ket{\Omega}=\frac{k}{16\pi^{3}R^{2}}\frac{1}{z\bar{z}}\left(\frac{z}{\bar{z}}\right)^{-k}$
(99)

for $k<0$. The only difference between this and the 3d case is that, in 5d,
the action of $Q_{-}$ decomposes into the two subspaces defined by $\Gamma$,
with the value of $k$ shifting in opposite directions depending on the
subspace. This requires that the correlation function identities involving
$Q_{-}$ contain two scalar fields in order to not project onto either
subspace. For example, to get the $\lambda_{k}\lambda_{k}^{{\dagger}}$ two-
point function we must consider the identity

$\bra{\Omega}\\{Q_{-},(\phi_{k+1}+\phi_{k-1})\lambda_{k}^{{\dagger}}\\}\ket{\Omega}=0\
;$ (100)

from this we find

$\bra{\Omega}\lambda_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\ket{\Omega}=\frac{\sqrt{2}k}{R}P_{-}\Big{(}P_{+}^{\Gamma}\bra{\Omega}\phi_{k-1}(x^{-},x)\bar{\phi}_{k-1}(0)\ket{\Omega}+P_{-}^{\Gamma}\bra{\Omega}\phi_{k+1}(x^{-},x)\bar{\phi}_{k+1}(0)\ket{\Omega}\Big{)}\
,$ (101)

which after substituting in the appropriate scalar two-point functions is

$\bra{\Omega}\lambda_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\ket{\Omega}=-\frac{k}{8\sqrt{2}\pi^{3}R^{3}}P_{-}\left(\frac{z}{\bar{z}}\right)^{k}\left(\frac{(k+1)P_{-}^{\Gamma}}{\bar{z}^{2}}+\frac{(k-1)P_{+}^{\Gamma}}{z^{2}}\right)$
(102)

for $k>0$ and zero otherwise.

Proceeding similarly, the other non-vanishing two-point functions of the modes
are

$\displaystyle\bra{\Omega}\lambda_{k}(x^{-},x)\chi_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=\frac{ik}{32\pi^{3}R^{3}}\gamma_{0i}P_{+}\left(\frac{1}{z\bar{z}}\right)^{2}\left(\frac{z}{\bar{z}}\right)^{k}\Bigg{[}ix^{i}\Big{(}z(1+k)-\bar{z}(1-k)\Big{)}$
$\displaystyle\hskip
110.96556pt+R\Omega_{ij}x^{j}\Big{(}z(1+k)+\bar{z}(1-k)\Big{)}\Bigg{]}\ ,$
(103a)
$\displaystyle\bra{\Omega}\chi_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=\frac{ik}{32\pi^{3}R^{3}}\gamma_{0i}P_{-}\left(\frac{1}{z\bar{z}}\right)^{2}\left(\frac{z}{\bar{z}}\right)^{k}\Bigg{[}ix^{i}\Big{(}z(1+k)-\bar{z}(1-k)\Big{)}$
$\displaystyle\hskip
110.96556pt-R\Omega_{ij}x^{j}\Big{(}z(1+k)+\bar{z}(1-k)\Big{)}\Bigg{]}\ ,$
(103b)
$\displaystyle\bra{\Omega}\chi_{k}(x^{-},x)\chi_{k}^{{\dagger}}(0)\ket{\Omega}$
$\displaystyle=\frac{ik}{8\sqrt{2}\pi^{3}R^{2}}P_{+}\left(\frac{1}{z\bar{z}}\right)^{2}\left(\frac{z}{\bar{z}}\right)^{k}\Big{(}(1+k)z+(1-k)\bar{z}\Big{)}\
,$ (103c)

for $k>0$, and

$\displaystyle\bra{\Omega}\lambda_{k}^{{\dagger}}(x^{-},x)\lambda_{k}(0)\ket{\Omega}$
$\displaystyle=-\frac{k}{8\sqrt{2}\pi^{3}R^{3}}P_{-}\left(\frac{z}{\bar{z}}\right)^{-k}\left(\frac{(k+1)P_{-}^{\Gamma}}{z^{2}}+\frac{(k-1)P_{+}^{\Gamma}}{\bar{z}^{2}}\right)\
,$ (104a)
$\displaystyle\bra{\Omega}\chi_{k}^{{\dagger}}(x^{-},x)\lambda_{k}(0)\ket{\Omega}$
$\displaystyle=\frac{ik}{32\pi^{3}R^{3}}\gamma_{0i}P_{+}\left(\frac{1}{z\bar{z}}\right)^{2}\left(\frac{z}{\bar{z}}\right)^{-k}\Bigg{[}ix^{i}\Big{(}\bar{z}(1+k)-z(1-k)\Big{)}$
$\displaystyle\hskip
110.96556pt+R\Omega_{ij}x^{j}\Big{(}\bar{z}(1+k)+z(1-k)\Big{)}\Bigg{]}\ ,$
(104b)
$\displaystyle\bra{\Omega}\lambda_{k}^{{\dagger}}(x^{-},x)\chi_{k}(0)\ket{\Omega}$
$\displaystyle=\frac{ik}{32\pi^{3}R^{3}}\gamma_{0i}P_{-}\left(\frac{1}{z\bar{z}}\right)^{2}\left(\frac{z}{\bar{z}}\right)^{-k}\Bigg{[}ix^{i}\Big{(}\bar{z}(1+k)-z(1-k)\Big{)}$
$\displaystyle\hskip
110.96556pt-R\Omega_{ij}x^{j}\Big{(}\bar{z}(1+k)+z(1-k)\Big{)}\Bigg{]}\ ,$
(104c)
$\displaystyle\bra{\Omega}\chi_{k}^{{\dagger}}(x^{-},x)\chi_{k}(0)\ket{\Omega}$
$\displaystyle=-\frac{ik}{8\sqrt{2}\pi^{3}R^{2}}P_{+}\left(\frac{1}{z\bar{z}}\right)^{2}\left(\frac{z}{\bar{z}}\right)^{-k}\Big{(}(1+k)\bar{z}+(1-k)z\Big{)}\
,$ (104d)

for $k<0$.

We can resum the mode two-point functions using the same $i\epsilon$
prescription outlined above to obtain the two-point functions of the 6d
fields. A short computation gives

$\displaystyle\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}=\frac{P_{-}}{4\sqrt{2}\pi^{3}R^{3}}\frac{\cos\left(\frac{x^{+}}{2R}\right)\left(x^{-}-\frac{ix^{2}}{4R}\Gamma\right)-\sin\left(\frac{x^{+}}{2R}\right)\left(ix^{-}\Gamma+\frac{x^{2}}{4R}\right)}{\left(ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}\right)^{3}}\
,$ (105a)
$\displaystyle\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}=\frac{\gamma_{0i}P_{+}}{8\pi^{3}R^{3}}\frac{x^{i}\cos\left(\frac{x^{+}}{2R}\right)+R\Omega_{ij}x^{j}\sin\left(\frac{x^{+}}{2R}\right)}{\left(ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}\right)^{3}}\
,$ (105b)
$\displaystyle\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}=\frac{\gamma_{0i}P_{-}}{8\pi^{3}R^{3}}\frac{x^{i}\cos\left(\frac{x^{+}}{2R}\right)-R\Omega_{ij}x^{j}\sin\left(\frac{x^{+}}{2R}\right)}{\left(ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}\right)^{3}}\
,$ (105c)
$\displaystyle\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}=\frac{P_{+}}{2\sqrt{2}\pi^{3}R^{2}}\frac{\sin\left(\frac{x^{+}}{2R}\right)}{{\left(ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}\right)^{3}}}\
.$ (105d)

We can combine these into the natural 6d field $\psi$, yielding the two-point
function

$\displaystyle\bra{\Omega}T\\{\psi(x^{+},x^{-},x)\psi^{{\dagger}}(0)\\}\ket{\Omega}=\frac{1}{8\pi^{3}R^{3}}$
$\displaystyle\frac{1}{\left(ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}\right)^{3}}\Bigg{[}\cos\left(\frac{x^{+}}{2R}\right)\bigg{(}\sqrt{2}P_{-}\left(x^{-}-\frac{ix^{2}}{4R}\Gamma\right)+\gamma_{0i}x^{i}\bigg{)}$
$\displaystyle+\sin\left(\frac{x^{+}}{2R}\right)\bigg{(}2\sqrt{2}RP_{+}-\sqrt{2}P_{-}\left(ix^{-}\Gamma+\frac{x^{2}}{4R}\right)-R\Omega_{ij}\gamma_{1i}x^{j}\bigg{)}\Bigg{]}\
.$ (106)

Using the transformation (21) we find that the original Minkowski field
$\hat{\psi}$ has the two-point function

$\displaystyle\bra{\Omega}T\\{\hat{\psi}(\hat{x}^{+},\hat{x}^{-},\hat{x})\hat{\psi}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\Delta^{-\frac{5}{2}}(x)S^{-1}[\Lambda_{x}]\bra{\Omega}T\\{\psi(x^{+},x^{-},x)\psi^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{1}{8\pi^{3}R^{3}}\frac{\cos^{3}\left(\frac{x^{+}}{2R}\right)}{\left(ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}\right)^{3}}\Bigg{[}2\sqrt{2}R\tan\left(\frac{x^{+}}{2R}\right)P_{+}$
$\displaystyle+\sqrt{2}\left(x^{-}+\frac{x^{2}}{4R}\tan\left(\frac{x^{+}}{2R}\right)\right)P_{-}+\gamma_{0i}\left(x^{i}-\tan\left(\frac{x^{+}}{2R}\right)R\Omega_{ij}x^{j}\right)\Bigg{]}$
(107)

which after tidying up gives

$\bra{\Omega}T\\{\hat{\psi}(\hat{x}^{+},\hat{x}^{-},\hat{x})\hat{\psi}^{{\dagger}}(0)\\}\ket{\Omega}=-\frac{i}{\pi^{3}}\frac{\gamma_{\mu}\gamma_{0}\hat{x}^{\mu}}{\left(-2\hat{x}^{+}\hat{x}^{-}+\hat{x}^{i}\hat{x}^{i}\right)^{3}}\
,$ (108)

or equivalently

$\bra{\Omega}T\\{\hat{\psi}(\hat{x}^{+},\hat{x}^{-},\hat{x})\bar{\hat{\psi}}(0)\\}\ket{\Omega}=\frac{1}{\pi^{3}}\frac{\gamma_{\mu}\hat{x}^{\mu}}{\left(-2\hat{x}^{+}\hat{x}^{-}+\hat{x}^{i}\hat{x}^{i}\right)^{3}}\
.$ (109)

As in the 4d case, this is the usual two-point function of a Minkowski spinor
field with canonically normalised action

$S=-\int d^{6}\hat{x}\,\bar{\hat{\psi}}\gamma^{\mu}\partial_{\mu}\hat{\psi}\
.$ (110)

##  4 Limits of $SU(1,n)$ Fermions

###  4.1 The DLCQ Limit of $SU(1,n)$ Fermions

One of the novel properties of the $SU(1,n)$-invariant theories we have
analysed is that they contain the DLCQ of the original Minkowski theory as a
limiting case. We can find this using the procedure described in
Lambert:2020zdc , which we briefly review. Recall that we fixed the moding of
theory by asking that the ’physical’ fields777I.e. those obtained from a
diffeomorphism of the original Minkowski fields. admitted a Fourier series
expansion on the interval $x^{+}\in[-\pi R,\pi R]$. If we choose some
$N\in\mathbb{N}$ we can instead ask that the physical fields are periodic with
period $2\pi R_{+}$, where we define $R_{+}$ as

$R_{+}=\frac{R}{N}\ .$ (111)

In the fermionic case of interest the moding of both the physical and Weyl-
transformed fields matched, so the transformed fields will also be periodic
under a $2\pi R_{+}$ shift in $x^{+}$. The set of allowed modes for our fields
is restricted to $\mathcal{S}=N\mathbb{Z}$; in other words, we can express
each mode number $k$ as

$k=k_{+}N$ (112)

for some $k_{+}\in\mathbb{Z}$. This means the Kaluza-Klein tower takes the
form

$\psi(x^{+},x^{-},x)=\sum_{k_{+}\in\mathbb{Z}}e^{-\frac{ik_{+}}{R_{+}}}\psi_{k_{+}}(x^{-},x)\
.$ (113)

As taking the limit $N\to\infty$ with $R_{+}$ fixed sends $R\to\infty$ the
coordinate transformation (1) is trivial and the $\Omega$-deformation of the
spacetime metric disappears. This leaves us with a field theory on flat
Minkowski spacetime where our fields are periodic along a finite null
interval: we recognise this as the DLCQ of the theory.

There are two ways to find the two-point functions of the DLCQ theory. We can
either start with the theory in the DLCQ limit and find its correlation
functions, or we can directly take the limit of the $SU(1,n)$ fermionic two-
point functions. Both methods will produce the same results888For a more
detailed discussion of this point for a scalar field theory see appendix C.3.,
so for convenience we will work with the latter approach. Let us consider the
action (41) for a single level of the reduced theory. Making the replacements
(111) and (112), we see that we need to rescale our fields to

$\displaystyle\lambda_{k}(x^{-},x)$
$\displaystyle=\frac{1}{\sqrt{N}}\lambda_{k_{+}}(x^{-},x)\ ,$ (114a)
$\displaystyle\chi_{k}(x^{-},x)$
$\displaystyle=\frac{1}{\sqrt{N}}\chi_{k_{+}}(x^{-},x)\ ,$ (114b)

for the $N\to\infty$ limit to be well-defined. With this rescaling the action
becomes

$S_{k}\to 2\pi R\int
d^{2n-1}x\left(\sqrt{2}\lambda_{k}^{{\dagger}}\partial_{-}\lambda_{k}+\chi^{{\dagger}}\gamma_{0i}\partial_{i}\lambda_{k}+\lambda_{k}^{{\dagger}}\gamma_{0i}\partial_{i}\chi_{k}-\frac{\sqrt{2}ik}{R}\chi_{k}^{{\dagger}}\chi_{k}\right)$
(115)

as we take $N\to\infty$, where we have dropped all pluses for convenience. We
note in passing that this is invariant under the Schrödinger group.

Applying this limit to the 3d and 5d two-point functions gives

$\displaystyle\bra{\Omega}T\\{\lambda_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\sqrt{2}\,\mathcal{T}^{(n)}_{k}P_{-}}{4\pi R}\ ,$ (116a)
$\displaystyle\bra{\Omega}T\\{\lambda_{k}(x^{-},x)\chi_{k}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\mathcal{T}^{(n)}_{k}\gamma_{0i}x^{i}P_{+}}{4\pi
R\,x^{-}}\ ,$ (116b)
$\displaystyle\bra{\Omega}T\\{\chi_{k}(x^{-},x)\lambda_{k}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\mathcal{T}^{(n)}_{k}\gamma_{0i}x^{i}P_{-}}{4\pi
R\,x^{-}}\ ,$ (116c)
$\displaystyle\bra{\Omega}T\\{\chi_{k}(x^{-},x)\chi_{k}^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\sqrt{2}P_{+}\left(\pi\mathcal{T}_{k}^{(n)}x^{2}-(n-1)\mathcal{T}_{k}^{(n-1)}\right)}{8\pi^{2}R\,(x^{-})^{2}}\
,$ (116d)

where we’ve defined

$\mathcal{T}_{k}^{(\alpha)}=\begin{cases}\Theta(x^{-})\left(\frac{-ik}{2\pi
R\,x^{-}}\right)^{\alpha-1}\exp\left(\frac{ikx^{2}}{2R\,x^{-}}\right)&k>0\\\
0&k=0\\\ -\Theta(-x^{-})\left(\frac{-ik}{2\pi
R\,x^{-}}\right)^{\alpha-1}\exp\left(\frac{ikx^{2}}{2R\,x^{-}}\right)&k<0\end{cases}\
.$ (117)

We can resum these to obtain the two-point function of a free massless fermion
on Minkowski spacetime with a periodic null direction. We have

$\displaystyle\sum_{k\in\mathbb{Z}}e^{-\frac{ikx^{+}}{R}}\mathcal{T}_{k}^{(\alpha)}$
$\displaystyle=\left(\frac{-i}{2\pi
R\,x^{-}}\right)^{\alpha-1}\sum_{k=1}^{\infty}\left(\Theta(x^{-})k^{\alpha-1}q^{k}+(-1)^{\alpha}\Theta(-x^{-})k^{\alpha-1}q^{-k}\right)$
$\displaystyle=\left(\frac{-i}{2\pi
R\,x^{-}}\right)^{\alpha-1}\Big{(}\Theta(x^{-})\mathrm{Li}_{1-\alpha}(q)+(-1)^{\alpha}\Theta(-x^{-})\mathrm{Li}_{1-\alpha}(q^{-1})\Big{)}$
(118)

in terms of the polylogarithm

$\mathrm{Li}_{s}(z)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{s}}\ ,$ (119)

where we have defined

$q=\exp\left(\frac{i}{2Rx^{-}}\left(-2x^{+}x^{-}+x^{2}\right)\right)\equiv\exp\left(\frac{ix_{\mu}x^{\mu}}{2R\,x^{-}}\right)$
(120)

and used the usual $i\epsilon$ prescription (86) to ensure convergence of the
sum. When $\alpha\geq 2$ we can use the identity

$\mathrm{Li}_{-s}(z)=(-1)^{s-1}\mathrm{Li}_{-s}(z^{-1})$ (121)

valid for $s\in\mathbb{N}$ to simplify this to

$\sum_{k\in\mathbb{Z}}e^{-\frac{ikx^{+}}{R}}\mathcal{T}_{k}^{(\alpha)}=\left(\frac{-i}{2\pi
R\,x^{-}}\right)^{\alpha-1}\mathrm{Li}_{1-\alpha}(q)\ .$ (122)

We will also require the $\alpha=1$ case; we can then use the explicit form

$\mathrm{Li}_{0}(q)=\frac{q}{1-q}$ (123)

to obtain

$\sum_{k\in\mathbb{Z}}e^{-\frac{ikx^{+}}{R}}\mathcal{T}_{k}^{(1)}=\frac{i}{2}\left(\cot\left(\frac{x_{\mu}x^{\mu}}{4R\,x^{-}}\right)+i\big{[}\Theta(x^{-})-\Theta(-x^{-})\big{]}\right)\
.$ (124)

With these results in hand, we find the first three resummed two-point
functions are

$\displaystyle\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\sqrt{2}P_{-}}{4\pi R}\left(\frac{-i}{2\pi
R\,x^{-}}\right)^{n-1}\mathrm{Li}_{1-n}(q)\ ,$ (125a)
$\displaystyle\bra{\Omega}T\\{\lambda(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\gamma_{0i}x^{i}P_{+}}{4\pi R\,x^{-}}\left(\frac{-i}{2\pi
R\,x^{-}}\right)^{n-1}\mathrm{Li}_{1-n}(q)\ ,$ (125b)
$\displaystyle\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\lambda^{{\dagger}}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\gamma_{0i}x^{i}P_{-}}{4\pi R\,x^{-}}\left(\frac{-i}{2\pi
R\,x^{-}}\right)^{n-1}\mathrm{Li}_{1-n}(q)\ ,$ (125c)

for both $n=2$ and $n=3$. The form of the $\chi\chi^{{\dagger}}$ correlation
function is dimension dependent; a brief calculation gives

$\displaystyle\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}=-$
$\displaystyle\frac{\sqrt{2}iP_{+}}{16\pi^{2}R(x^{-})^{2}}\bigg{(}\frac{x^{2}}{x^{-}R}\mathrm{Li}_{-1}(q)+\cot\left(\frac{x_{\mu}x^{\mu}}{4R\,x^{-}}\right)$
$\displaystyle+i\big{[}\Theta(x^{-})-\Theta(-x^{-})\big{]}\bigg{)}$ (126)

when $n=2$, and

$\bra{\Omega}T\\{\chi(x^{+},x^{-},x)\chi^{{\dagger}}(0)\\}\ket{\Omega}=-\frac{\sqrt{2}P_{+}}{32\pi^{3}R^{2}(x^{-})^{3}}\left(\frac{x^{2}}{x^{-}R}\mathrm{Li}_{-2}(q)-4i\,\mathrm{Li}_{-1}(q)\right)$
(127)

when $n=3$. If we decompactify the null direction by taking $R\to\infty$, the
large-$R$ expansions

$\displaystyle\mathrm{Li}_{-s}(q)$
$\displaystyle=s!\left(\frac{2iR\,x^{-}}{x_{\mu}x^{\mu}}\right)^{s+1}+O(R^{s})\
,$ (128a) $\displaystyle\cot\left(\frac{x_{\mu}x^{\mu}}{4R\,x^{-}}\right)$
$\displaystyle=\frac{4R\,x^{-}}{x_{\mu}x^{\mu}}+O(1)\ ,$ (128b)

for $s\in\mathbb{N}$ give

$\lim_{R\to\infty}\left(\bra{\Omega}T\\{\psi(x^{+},x^{-},x)\psi^{{\dagger}}(0)\\}\ket{\Omega}\right)=\frac{i(n-1)!}{2\pi^{n}R\left(x_{\mu}x^{\mu}\right)^{n}}\left(\sqrt{2}x^{-}P_{-}+\sqrt{2}x^{+}P_{+}+\gamma_{0i}x^{i}\right)\
.$ (129)

We see that we have recovered the normal Minkowski spacetime two-point
functions (3.2.1) when $n=2$ and (108) when $n=3$.

While both the $SU(1,n)$ and DLCQ methods can reproduce the Minkowski
spacetime two-point function of the free fermion through the resummation of a
Kaluza-Klein tower of lower dimensional two-point functions, there are
conceptual differences between the approaches. Most notably, while we had to
take the $R\to\infty$ limit to recover the two-point function from the DLCQ
correlation functions, no such limit was taken on the $SU(1,n)$ side; the
original theory was recovered through a diffeomorphism and Weyl
transformation, so the mapping between the two is invertible. We also see that
the $SU(1,n)$ mode two-point functions exhibit a spatial fall-off, unlike the
DLCQ mode two-point functions. It is interesting to ask whether this leads to
any improvements in computing quantities in the quantum theory: however, we
leave this question to future work.

###  4.2 Hints of a Carrollian Limit

We have seen that taking $R\to\infty$ with the ratio $\frac{k}{R}$ fixed leads
to an action with Schrödinger symmetry at each level in the Kaluza-Klein
tower. It is therefore natural to ask what happens in the opposite limit, in
which we define

$\displaystyle R$ $\displaystyle=\omega R_{+}\ ,$ (130a) $\displaystyle k$
$\displaystyle=\omega k_{+}\ ,$ (130b)

and take the limit $\omega\to 0$ with $R_{+}$ and $k_{+}$ fixed. Let us first
consider this for the complex scalar action (52). First rescaling the field to

$\phi_{k}=\frac{\tilde{\phi}_{k}}{\sqrt{2\pi R}}$ (131)

gives (dropping the pluses, tildes, and field subscript to simplify notation)

$\displaystyle S_{k}^{(\phi)}=\int d^{2n-1}x\bigg{(}$
$\displaystyle\frac{2ik}{R}\bar{\phi}\partial_{-}\phi-\partial_{i}\bar{\phi}\partial_{i}\phi+\frac{1}{2\omega}\Omega_{ij}x^{j}\Big{(}\partial_{i}\bar{\phi}\partial_{-}\phi$
$\displaystyle+\partial_{-}\bar{\phi}\partial_{i}\phi\Big{)}-\frac{x^{2}}{4\omega^{2}R^{2}}\partial_{-}\bar{\phi}\partial_{-}\phi\bigg{)}\
.$ (132)

We see that there are two sets of terms that diverge; we must deal with both
of these to recover a sensible theory in the $\omega\to 0$ limit. As the
$O(\omega^{-2})$ term is a square, the most general way to remove the leading
divergence is to impose

$x^{i}\partial_{-}\phi=0\ .$ (133)

If we additionally require that our field is smooth we can drop the factor of
$x^{i}$, and we shall assume this is true from here onwards. The constraint
also removes the divergences arising from the $O(\omega^{-1})$ terms and
imposing it is enough to render the theory finite as $\omega\to 0$. We
therefore find the action

$S_{k}^{(\phi)}=\int
d^{2n-1}x\left(\bar{H}\partial_{-}\phi+\partial_{-}\bar{\phi}H-\partial_{i}\bar{\phi}\partial_{i}\phi\right)$
(134)

after taking the limit, where $H$ is a Lagrange multiplier field implementing
the constraint. This is the action of a ’spacelike’999Also referred to as
’magnetic’ deBoer:2023fnj . scalar field with conformal Carroll spacetime
symmetry Baiguera:2022lsw .

Let us do the same with the fermion action (41), where we assume that $k$ is
non-zero. If we again rescale our fields as in (131), this becomes

$\displaystyle S_{k}=i\int d^{2n-1}x\bigg{(}$
$\displaystyle\sqrt{2}\lambda^{{\dagger}}\partial_{-}\lambda+\chi^{{\dagger}}\gamma_{0i}\partial_{i}\lambda+\lambda^{{\dagger}}\gamma_{0i}\partial_{i}\chi-\frac{\sqrt{2}ik}{R}\chi^{{\dagger}}\chi$
$\displaystyle-\frac{1}{2\omega}\Omega_{ij}x^{j}\left(\chi^{{\dagger}}\gamma_{0i}\partial_{-}\lambda+\lambda^{{\dagger}}\gamma_{0i}\partial_{-}\chi\right)\bigg{)}\
.$ (135)

In contrast to the scalar case all divergent terms are of the same order in
$\omega$. To obtain a finite action we must impose

$\Omega_{ij}x^{j}\left(\chi^{{\dagger}}\gamma_{0i}\partial_{-}\lambda+\lambda^{{\dagger}}\gamma_{0i}\partial_{-}\chi\right)=0\
.$ (136)

This is quite a complicated constraint, and we will restrict ourselves to
working with the subset of field configurations that satisfy the simpler
conditions

$\displaystyle\partial_{-}\lambda$ $\displaystyle=0\ ,$ (137a)
$\displaystyle\partial_{-}\chi$ $\displaystyle=0\ .$ (137b)

It is valid to ask if there are any interesting solutions missed by imposing
these stronger constraints, though we will not pursue this here. Introducing
fermionic Lagrange multiplier fields $\rho$ and $\eta$ for our constraints
gives the action

$S_{k}=i\int
d^{2n-1}x\bigg{(}\chi^{{\dagger}}\gamma_{0i}\partial_{i}\lambda+\lambda^{{\dagger}}\gamma_{0i}\partial_{i}\chi-\frac{\sqrt{2}ik}{R}\chi^{{\dagger}}\chi+\eta^{{\dagger}}\partial_{-}\lambda+\partial_{-}\lambda^{{\dagger}}\eta+\rho^{{\dagger}}\partial_{-}\chi+\partial_{-}\chi^{{\dagger}}\rho\bigg{)}$
(138)

in the $\omega\to 0$ limit, from which we see that the equation of motion for
$\chi$ is now

$\chi=\frac{iR}{\sqrt{2}k}\left(\partial_{-}\rho-\gamma_{0i}\partial_{i}\lambda\right)\
.$ (139)

As this is still algebraic in $\chi$ we can substitute it back into the action
to obtain

$S_{k}=\frac{R}{\sqrt{2}k}\int
d^{2n-1}x\left(\partial_{-}\rho^{{\dagger}}\partial_{-}\rho+\eta^{{\dagger}}\partial_{-}\lambda+\partial_{-}\lambda^{{\dagger}}\eta-\partial_{i}\lambda^{{\dagger}}\partial_{i}\lambda\right)\
.$ (140)

We recognise the terms involving $\lambda$ as those of a ’spacelike’
Carrollian field, while the action for $\rho$ is that of a ’timelike’101010Or
’electric’. Carrollian field.

The limit $R\to 0$ is somewhat less natural than the previously studied
$R\to\infty$ limit. For instance, the correlation functions of both the scalar
and fermion $SU(1,n)$ theories vanish once it is taken. It would be
interesting to see if the limit (possibly modified) can be used to make non-
trivial statements about Carrollian field theories using quantities computed
in $SU(1,n)$ theories, or if the relationship between the two is limited to
the construction of Carrollian actions from their $SU(1,n)$ counterparts.

##  5 Reduction of the 2d Fermion

So far we have only been interested in theories in $D=4$ and $D=6$. As there
are no known interacting CFTs when $D>6$ there is little point in analysing
higher-dimensional cases. This still leaves us with the exceptional case
$D=2$, which we examine now.

The coordinate transformation (1) behaves differently in 2d as it does in
other dimensions since there are no transverse directions, and hence no
$\Omega$-deformation. In this case, the transformation is a conformal
transformation of Minkowski spacetime and the resulting field theory remains a
2d Lorentzian CFT. The null reduction is then a geometrically-implemented DLCQ
and falls somewhere between the methods of sections 3.2 and 4.1. We would like
to determine to what extent the correlation functions of the 2d theory can be
recovered from the reduced 1d theory.

Let us be more precise. We start with the 2d Minkowski metric in lightcone
coordinates,

$ds^{2}=-2d\hat{x}^{+}d\hat{x}^{-}\ .$ (141)

The coordinate transformation is

$\displaystyle\hat{x}^{+}$ $\displaystyle=2R\tan\left(\frac{x^{+}}{2R}\right)\
,$ (142a) $\displaystyle\hat{x}^{-}$ $\displaystyle=x^{-}\ ,$ (142b)

so the metric becomes

$ds^{2}=\sec^{2}\left(\frac{x^{+}}{2R}\right)\left(-2dx^{+}dx^{-}\right)\ .$
(143)

As usual, the coordinate $x^{+}$ takes values in the interval $(-\pi R,\pi
R)$; after a Weyl transformation we can then include the endpoints and take
$x^{+}\in[-\pi R,\pi R]$.

Our field theory is taken to be a single Majorana fermion $\hat{\psi}$ with
the action

$S=-\int
d\hat{x}^{+}d\hat{x}^{-}\bar{\hat{\psi}}\gamma^{\mu}\partial_{\mu}\hat{\psi}\
.$ (144)

Using $\gamma_{*}=\gamma_{01}$ and defining the chiral components

$\displaystyle\frac{1}{2}\left(\mathbbm{1}+\gamma_{*}\right)\hat{\psi}$
$\displaystyle=\hat{\chi}\xi_{1}\ ,$ (145a)
$\displaystyle\frac{1}{2}\left(\mathbbm{1}-\gamma_{*}\right)\hat{\psi}$
$\displaystyle=\hat{\lambda}\xi_{2}\ ,$ (145b)

in terms of orthonormal spinors $\\{\xi_{i}\\}$111111Note that this the fields
$\hat{\lambda}$ and $\hat{\chi}$ are real Grassmann numbers and possess no
spinor indices., we can rewrite the action as

$S=i\int
d\hat{x}^{+}d\hat{x}^{-}\Bigg{[}\hat{\chi}\partial_{\hat{+}}\hat{\chi}+\hat{\lambda}\partial_{\hat{-}}\hat{\lambda}\Bigg{]}\
.$ (146)

It’s then clear that the action is invariant if the fields transform as

$\displaystyle\hat{\chi}(\hat{x})$ $\displaystyle=\chi(x)\ ,$ (147a)
$\displaystyle\hat{\lambda}(\hat{x})$
$\displaystyle=\cos\left(\frac{x^{+}}{2R}\right)\lambda(x)\ ,$ (147b)

under (142). As discussed in section 2.2, we should fix the decomposition of
$\chi$ and $\lambda$ into modes by requiring that $\hat{\chi}$ and
$\hat{\lambda}$ admit Fourier series expansions. This gives

$\displaystyle\chi(x)$
$\displaystyle=\sum_{l\in\mathbb{Z}}e^{-\frac{ilx^{+}}{R}}\chi_{l}\ ,$ (148a)
$\displaystyle\lambda(x)$
$\displaystyle=\sum_{k\in\mathbb{Z}+\frac{1}{2}}e^{-\frac{ikx^{+}}{R}}\lambda_{k}\
.$ (148b)

In terms of the modes, the action becomes

$S=2\pi iR\sum_{l=1}^{\infty}\int
dx^{-}\lambda_{-(l-\frac{1}{2})}\partial_{-}\lambda_{l-\frac{1}{2}}+2\pi\sum_{l=1}^{\infty}l\int
dx^{-}\chi_{-l}\chi_{l}\ .$ (149)

We see that, unlike in higher dimensions, one of the chiral components of our
field has become non-dynamical upon performing the mode decomposition. This is
expected: recall that we are working with the DLCQ of a 2d theory, which is
known to freeze the chiral sector along which we perform the reduction (as
pointed out in Balasubramanian:2009bg ). From here onwards we will restrict
our attention to $\lambda$.

If we Fourier transform on $x^{-}$ with the conventions

$\lambda_{k}(x^{-})=\int\frac{d\omega}{2\pi}e^{-i\omega
x^{-}}\tilde{\lambda}_{k}(\omega)\ ,$ (150)

the action simplifies to

$S_{\lambda}=2\pi
R\sum_{l=1}^{\infty}\int\frac{d\omega}{2\pi}\,\omega\tilde{\lambda}_{-(l-\frac{1}{2})}(-\omega)\tilde{\lambda}_{l-\frac{1}{2}}(\omega)\
.$ (151)

We can add the source terms

$S_{\xi}=\sum_{l=-\infty}^{\infty}\int
dx^{-}\,\xi_{-(l-\frac{1}{2})}\lambda_{l-\frac{1}{2}}$ (152)

to the action, and after making the field redefinition

$\tilde{\lambda}_{k}(\omega)=\tilde{\eta}_{k}(\omega)-\frac{\xi_{k}(\omega)}{\pi
R\omega}$ (153)

we find

$S_{\lambda}+S_{\xi}=\sum_{l=1}^{\infty}\int\frac{d\omega}{2\pi}\Bigg{[}2\pi
R\omega\,\tilde{\eta}_{-(l-\frac{1}{2})}(-\omega)\tilde{\eta}_{l-\frac{1}{2}}(\omega)-\frac{1}{2\pi
R\omega}\tilde{\xi}_{-(l-\frac{1}{2})}(-\omega)\tilde{\xi}_{l-\frac{1}{2}}(\omega)\Bigg{]}\
.$ (154)

The partition function of the theory is then

$\mathcal{Z}[\xi]=\mathcal{Z}[0]\exp\left(-\frac{i}{2\pi
R}\sum_{l=1}^{\infty}\int\frac{d\omega}{2\pi}\,\frac{1}{\omega}\tilde{\xi}_{-(l-\frac{1}{2})}(-\omega)\tilde{\xi}_{l-\frac{1}{2}}(\omega)\right)\
.$ (155)

We can take derivatives of this and Fourier transform back to obtain the time-
ordered two-point functions

$\displaystyle\bra{\Omega}T\\{\lambda_{k}(x^{-})\lambda_{-k}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{i}{R}\int\frac{d\omega_{1}d\omega_{2}}{(2\pi)^{2}}\,e^{-i\omega_{1}x^{-}}\frac{\delta(\omega_{1}+\omega_{2})}{\omega_{2}}$
$\displaystyle=-\frac{i}{2\pi R}\int\frac{d\omega}{2\pi}\,\frac{e^{-i\omega
x^{-}}}{\omega}$ (156)

for both positive and negative $k$. As it stands, this integral is ill-
defined; we can fix this using a $k$-dependent $i\epsilon$ prescription

$\int\frac{d\omega}{2\pi}\,\frac{e^{-i\omega
x^{-}}}{\omega}\to\int\frac{d\omega}{2\pi}\,\frac{e^{-i\omega
x^{-}}}{\omega+\frac{ik\epsilon}{R}}\ .$ (157)

This can then be easily evaluated using the residue theorem, from which we get

$\bra{\Omega}T\\{\lambda_{k}(x^{-})\lambda_{-k}(0)\\}\ket{\Omega}=-\frac{e^{-\frac{k\epsilon
x^{-}}{R}}\Theta(x^{-})}{2\pi R}$ (158)

for $k>0$, and

$\bra{\Omega}T\\{\lambda_{k}(x^{-})\lambda_{-k}(0)\\}\ket{\Omega}=\frac{e^{\frac{k\epsilon
x^{-}}{R}}\Theta(-x^{-})}{2\pi R}$ (159)

for $k<0$. These are equivalent to the conditions

$\lambda_{k}(x^{-})\ket{\Omega}=0\;\;\forall\ k>0$ (160)

and the correlation functions

$\bra{\Omega}\lambda_{k}(x^{-})\lambda_{-k}(0)\ket{\Omega}=-\frac{e^{-\frac{k\epsilon\absolutevalue{x^{-}}}{R}}}{2\pi
R}$ (161)

for positive $k$.

Let us use this to reconstruct the two-point function of $\hat{\lambda}$. We
have

$\displaystyle\bra{\Omega}T\\{\hat{\lambda}(\hat{x}^{+},\hat{x}^{-})\hat{\lambda}(0)\\}\ket{\Omega}=$
$\displaystyle\cos\left(\frac{x^{+}}{2R}\right)\sum_{l=0}^{\infty}\Bigg{[}e^{-\frac{ix^{+}}{R}\left(l+\frac{1}{2}\right)}\bra{\Omega}T\\{\lambda_{l+\frac{1}{2}}(x^{-})\lambda_{-(l+\frac{1}{2})}(0)\\}\ket{\Omega}$
$\displaystyle+e^{\frac{ix^{+}}{R}\left(l+\frac{1}{2}\right)}\bra{\Omega}T\\{\lambda_{-(l+\frac{1}{2})}(x^{-})\lambda_{l+\frac{1}{2}}(0)\\}\ket{\Omega}\Bigg{]}\
,$ (162)

so substituting in the mode two-point functions gives

$\displaystyle\bra{\Omega}T\\{\hat{\lambda}(\hat{x}^{+},\hat{x}^{-})\hat{\lambda}(0)\\}\ket{\Omega}$
$\displaystyle=\frac{\cos\left(\frac{x^{+}}{2R}\right)}{2\pi
R}\sum_{l=0}^{\infty}\Bigg{[}-\Theta(x^{-})\sqrt{q}q^{l}+\Theta(-x^{-})\frac{q^{-l}}{\sqrt{q}}\Bigg{]}$
$\displaystyle=-\frac{\cos\left(\frac{x^{+}}{2R}\right)}{2\pi
R}\frac{\sqrt{q}}{1-q}\ ,$ (163)

where we’ve defined

$q=e^{-\frac{i}{R}\left(x^{+}-ix^{-}\epsilon\right)}\ .$ (164)

We note in passing that the $\epsilon$ shift of the exponent ensures
convergence of the sum. After some rearrangement, we recover the result
(suppressing the $\epsilon$ contribution)

$\bra{\Omega}T\\{\hat{\lambda}(\hat{x}^{+},\hat{x}^{-})\hat{\lambda}(0)\\}\ket{\Omega}=\frac{i}{2\pi\hat{x}^{+}}$
(165)

for the two-point function of a left-moving free fermion arising from the
action (144). While we have managed to reconstruct the left-moving sector of
the theory, it is impossible to reconstruct the right-moving sector in this
way. We should contrast this with the higher-dimensional cases previously
discussed, where it was found that the full two-point function can be
reconstructed from the null reduction.

##  6 $\mathfrak{su}(1,n)$ Ward Identities

While we have discussed both the fermion action and its associated two-point
functions in some depth, we have not yet shown that the theories really are
invariant under the spacetime symmetry algebra $\mathfrak{su}(1,n)$ as we have
claimed multiple times. We will now remedy this by showing that the two-point
functions satisfy the corresponding Ward identities.

Let us briefly review the $\mathfrak{su}(1,n)$ spacetime symmetry algebra and
its associated Ward identities before showing that they are satisfied in our
theory. First, consider the 3d theory. The algebra $\mathfrak{su}(1,2)$ is
generated by the basis $\\{P_{-},P_{i},B,T,G_{i},K\\}$ and can be centrally
extended by the element $P_{+}$. Physically, $P_{-}$ and $P_{i}$ form a non-
Abelian algebra of translations, $B$ is the generator of spatial rotations,
$T$ generates Lifshitz scalings, and $G_{i}$ and $K$ generate non-Lorentzian
special conformal transformations. We identify $P_{+}$ as the generator of
translations in the compactified direction in the original theory. The 5d
theory is almost identical; the only addition necessary to obtain
$\mathfrak{su}(1,3)$ is an $\mathfrak{su}(2)$ algebra with generators
$\\{J^{\alpha}\\}$. These commute with $B$ and generate the remaining spatial
rotations. The Lie brackets between the generators are given in
Lambert:2021nol , though we will instead work with anti-Hermitian operators
for convenience.

To simplify our discussion we will mostly work with primary operators
$\\{\mathcal{O}_{k}\\}$ that obey the conditions

$\displaystyle[G_{i},\mathcal{O}_{k}(0)]=[K,\mathcal{O}_{k}(0)]=0\ .$ (166)

Borrowing from the nomenclature of Lorentzian CFTs, all operators that do not
satisfy these conditions are referred to as descendants. The non-derivative
part of the action of the algebra on a primary operator at the origin is then

$\displaystyle[\mathcal{O}(0),T]$ $\displaystyle=\Delta\mathcal{O}(0)\ ,$
(167a) $\displaystyle[\mathcal{O}(0),P_{+}]$ $\displaystyle=-ip\mathcal{O}(0)\
,$ (167b) $\displaystyle[\mathcal{O}(0),B]$
$\displaystyle=r_{\mathcal{O}}[B]\mathcal{O}(0)\ ,$ (167c)
$\displaystyle[\mathcal{O}(0),J^{\alpha}]$
$\displaystyle=r_{\mathcal{O}}[J^{\alpha}]\mathcal{O}(0)\ ,$ (167d)

where we have assumed $\\{r_{\mathcal{O}}[J^{\alpha}]\\}$ form a finite-
dimensional irreducible representation of $\mathfrak{su}(2)$, and
$r_{\mathcal{O}}[B]$ is a representation of $\mathfrak{u}(1)$. We will take
$\Delta,k\in\mathbb{C}$.

The Ward identities for the two-point function of primary operator
$\mathcal{O}_{1}$ and $\mathcal{O}_{2}$ are then Lambert:2020zdc

$\displaystyle 0=$
$\displaystyle\left(p_{1}+p_{2}\right)\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168a) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\left(\frac{\partial}{\partial
x^{-}_{a}}\right)\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168b) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\left(\frac{\partial}{\partial
x^{i}_{a}}+\frac{1}{2}\Omega_{ij}x_{a}^{j}\frac{\partial}{\partial
x_{a}^{-}}\right)\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168c) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\left(2x_{a}^{-}\frac{\partial}{\partial
x^{-}_{a}}+x_{a}^{i}\frac{\partial}{\partial
x_{a}^{i}}+\Delta_{a}\right)\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168d) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\left(-R\Omega_{ij}x_{a}^{i}\frac{\partial}{\partial
x_{a}^{j}}+r_{a}[B]\right)\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168e) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\left(\bar{\eta}^{\alpha}_{ij}x_{a}^{i}\frac{\partial}{\partial
x_{a}^{j}}+r_{a}[J^{\alpha}]\right)\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168f) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\Bigg{(}\left(\frac{1}{2}\Omega_{ij}x_{a}^{-}x_{a}^{j}-\frac{1}{8R^{2}}x_{a}^{2}x_{a}^{i}\right)\frac{\partial}{\partial
x_{a}^{-}}+x_{a}^{-}\frac{\partial}{\partial
x_{a}^{i}}+\frac{1}{4}\bigg{(}2\Omega_{ik}x_{a}^{k}x_{a}^{j}$
$\displaystyle+2\Omega_{jk}x_{a}^{k}x_{a}^{i}-\Omega_{ij}x_{a}^{2}\bigg{)}\frac{\partial}{\partial
x_{a}^{j}}+\frac{1}{2}\Delta_{a}\Omega_{ij}x_{a}^{j}-ip_{a}x_{a}^{i}$
$\displaystyle+\left(\frac{n+1}{n-1}\right)\frac{x_{a}^{i}r_{a}[B]}{2R}-\frac{1}{2}\Omega_{ik}\bar{\eta}^{\alpha}_{jk}x_{a}^{j}r_{a}[J^{\alpha}]\Bigg{)}\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168g) $\displaystyle 0=$
$\displaystyle\sum_{a=1}^{2}\Bigg{(}\left(2(x_{a}^{-})^{2}-\frac{x_{a}^{4}}{8R^{2}}\right)\frac{\partial}{\partial
x_{a}^{-}}+\left(\frac{1}{2}\Omega_{ij}x_{a}^{j}x_{a}^{2}+2x_{a}^{-}x_{a}^{i}\right)\frac{\partial}{\partial
x_{a}^{i}}+2\Delta_{a}x_{a}^{-}$ $\displaystyle-
ip_{a}x_{a}^{2}+\left(\frac{n+1}{n-1}\right)\frac{x_{a}^{2}r_{a}[B]}{2R}-\frac{1}{2}x_{a}^{i}x_{a}^{j}\Omega_{ik}\bar{\eta}^{\alpha}_{jk}r_{a}[J^{\alpha}]\Bigg{)}\bra{\Omega}\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\ket{\Omega}\
,$ (168h)

where any terms with either $r_{a}[J^{\alpha}]$ or $\bar{\eta}_{ij}^{\alpha}$
are ignored for the $n=2$ identities. Note that the condition that the
operators are primary has only been used in the final two identities.

It will be convenient to introduce the complex coordinate

$z_{12}=x^{-}_{1}-x_{2}^{-}+\frac{1}{2}\Omega_{ij}x_{1}^{i}x_{2}^{j}+\frac{i}{4R}(x_{1}-x_{2})^{2}\
,$ (169)

which is the translated version of (77). We then immediately have

$\displaystyle\sum_{a=1}^{2}\left(\frac{\partial}{\partial
x_{a}^{-}}\right)f(z_{12},\bar{z}_{12})$ $\displaystyle=0\ ,$ (170)
$\displaystyle\sum_{a=1}^{2}\left(\frac{\partial}{\partial
x^{i}_{a}}+\frac{1}{2}\Omega_{ij}x_{a}^{j}\frac{\partial}{\partial
x_{a}^{-}}\right)f(z_{12},\bar{z}_{12})$ $\displaystyle=0\ ,$ (171)
$\displaystyle\sum_{a=1}^{2}\left(-R\Omega_{ij}x_{a}^{i}\frac{\partial}{\partial
x_{a}^{j}}\right)f(z_{12},\bar{z}_{12})$ $\displaystyle=0\ ,$ (172)
$\displaystyle\sum_{a=1}^{2}\left(\bar{\eta}^{\alpha}_{ij}x_{a}^{i}\frac{\partial}{\partial
x_{a}^{j}}\right)f(z_{12},\bar{z}_{12})$ $\displaystyle=0\ ,$ (173)

for any function $f$. Using the notation $\mathcal{L}_{i}$ and $\mathcal{L}$
for the differential operators in (168g) and (168h), similar computations give

$\displaystyle\mathcal{L}_{i}f(z_{12},\bar{z}_{12})$
$\displaystyle=\frac{i}{2R}(x_{1}-x_{2})^{i}\left(z_{12}\partial-\bar{z}_{12}\bar{\partial}\right)f+\frac{1}{2}\Omega_{ij}(x_{1}+x_{2})^{j}\left(z_{12}\partial+\bar{z}_{12}\bar{\partial}\right)f\
,$ (174) $\displaystyle\mathcal{L}f(z_{12},\bar{z}_{12})$
$\displaystyle=2(x_{1}^{-}+x_{2}^{-})\left(z_{12}\partial+\bar{z}_{12}\bar{\partial}\right)f+\frac{i}{2R}\left(x_{1}^{2}-x_{2}^{2}\right)\left(z_{12}\partial-\bar{z}_{12}\bar{\partial}\right)f\
.$ (175)

###  6.1 $\mathfrak{su}(1,2)$ Invariance in 3d

Let us compute the constraints imposed by the Ward identities on the
correlation functions of the 3d theory and check that they possess
$\mathfrak{su}(1,2)$ spacetime symmetry. We will focus on $k>0$, but the
extension to $k<0$ is obvious; we shall comment on the $k=0$ case later. In
this regime, the equation of motion (43) for $\chi_{k}$ allows it to be
expressed in terms of derivatives of $\lambda_{k}$. In other words, $\chi_{k}$
is a descendant field. This means it is sufficient to focus on the two-point
function of the primary field $\lambda_{k}$. The field $\lambda_{k}$ only has
a single complex degree of freedom, so we can take $r_{\lambda_{k}}[B]$ to be
an irreducible representation of $\mathfrak{u}(1)$.

Assuming translational invariance of our theory (which is simple to check from
the action), the two-point function we determined previously can be written as

$G^{\lambda\lambda}_{k}(x_{1}^{-},x_{1};x_{2}^{-},x_{2})=-\frac{ik}{4\sqrt{2}\pi^{2}R^{2}}z_{12}^{k-1}\bar{z}_{12}^{-k}P_{-}\
.$ (176)

The non-trivial Ward identities are then

$\displaystyle 0=$ $\displaystyle\
p_{\lambda_{k}}+p_{\lambda^{{\dagger}}_{k}}\ ,$ (177a) $\displaystyle 2=$
$\displaystyle\ \Delta_{\lambda_{k}}+\Delta_{\lambda^{{\dagger}}_{k}}\ ,$
(177b) $\displaystyle 0=$ $\displaystyle\
r_{\lambda_{k}}[B]+r_{\lambda_{k}^{{\dagger}}}[B]\ ,$ (177c) $\displaystyle
0=$ $\displaystyle\
\frac{i}{R}\left(k-\frac{1}{2}-Rp_{\lambda_{k}}-\frac{3i}{2}r_{\lambda_{k}}[B]\right)\left(x_{1}-x_{2}\right)^{i}$
$\displaystyle\
-\frac{1}{2}(1-\Delta_{\lambda_{k}})\Omega_{ij}\left(x_{1}-x_{2}\right)^{j}\
,$ (177d) $\displaystyle 0=$ $\displaystyle\
\frac{i}{R}\left(k-\frac{1}{2}-Rp_{\lambda_{k}}-\frac{3i}{2}r_{\lambda_{k}}[B]\right)\left(x_{1}^{2}-x_{2}^{2}\right)$
$\displaystyle\ -2(1-\Delta_{\lambda_{k}})\left(x_{1}^{-}-x_{2}^{-}\right)\ ,$
(177e)

where we have used the first three identities to simplify the final two.

If we take the Hermitian conjugate of (167) we see that we must have

$\displaystyle\Delta_{\lambda_{k}^{{\dagger}}}$
$\displaystyle=\left(\Delta_{\lambda_{k}}\right)^{*}\ ,$ (178a) $\displaystyle
p_{\lambda_{k}^{{\dagger}}}$ $\displaystyle=-\left(p_{\lambda_{k}}\right)^{*}\
,$ (178b) $\displaystyle r_{\lambda_{k}^{{\dagger}}}[B]$
$\displaystyle=\left(r_{\lambda_{k}}[B]\right)^{*}\ .$ (178c)

The first Ward identity is then just the statement that $p_{\lambda_{k}}$ is
real, and the third reduces to

$r_{\lambda_{k}}[B]=i\alpha_{\lambda_{k}}$ (179)

for some real $\alpha_{\lambda_{k}}$. This means that our charges need to
satisfy

$\displaystyle\Delta_{\lambda_{k}}$ $\displaystyle=1\ ,$ (180a) $\displaystyle
k-\frac{1}{2}$
$\displaystyle=Rp_{\lambda_{k}}-\frac{3}{2}\alpha_{\lambda_{k}}\ .$ (180b)

While the requirements of $\mathfrak{su}(1,2)$ symmetry completely fix the
value of $\Delta$, this is not true for $p$ and $\alpha$. We can see this from
the algebra: since $P_{+}$ is central, we have the one-parameter family of
reparameterisations

$\displaystyle B^{\prime}$ $\displaystyle=B+cP_{+}\ ,$ (181a) $\displaystyle
P_{+}^{\prime}$ $\displaystyle=P_{+}+\frac{3c}{2R}P_{+}\ ,$ (181b)

that leave the algebra invariant. This corresponds to the shift

$\displaystyle\alpha_{\lambda_{k}}^{\prime}$
$\displaystyle=\alpha_{\lambda_{k}}+cp_{\lambda_{k}}\ ,$ (182a) $\displaystyle
p_{\lambda_{k}}^{\prime}$
$\displaystyle=p_{\lambda_{k}}+\frac{3c}{2R}p_{\lambda_{k}}\ ,$ (182b)

in the charges. It’s easy to see that these transformations leave (180b)
invariant. We are therefore allowed to choose $p_{\lambda_{k}}$ to be any
(real) convenient value. It is natural to take

$p_{\lambda_{k}}=\frac{k+\delta}{R}\ ,$ (183)

where $\delta$ is a real number that is independent of $k$; with this choice
the representation of $B$ is the same for all $\lambda_{k}$, giving

$\alpha_{\lambda_{k}}=\frac{2\delta+1}{3}\ .$ (184)

Since all Ward identities except (168h) and $\eqref{eq: Gi WI}$ are also valid
for non-primary operators, they should still hold for our other two-point
functions. Let us consider this for

$G^{\lambda\chi}_{k}(x_{1}^{-},x_{1};x_{2}^{-},x_{2})=-\frac{ik}{8\pi^{2}R^{2}}\gamma_{0i}(x_{1}-x_{2})^{i}z_{12}^{k-1}\bar{z}_{12}^{-(k+1)}P_{+}\
.$ (185)

The non-trivial identities are

$\displaystyle p_{\lambda_{k}}+p_{\chi_{k}^{{\dagger}}}$ $\displaystyle=0\ ,$
(186a) $\displaystyle\Delta_{\lambda_{k}}+\Delta_{\chi_{k}^{{\dagger}}}$
$\displaystyle=3\ ,$ (186b) $\displaystyle
r_{\lambda_{k}}[B]+r_{\chi_{k}^{{\dagger}}}[B]$ $\displaystyle=i\ .$ (186c)

Using the results for $\lambda_{k}$, we find

$\displaystyle p_{\chi_{k}}$ $\displaystyle=p_{\lambda_{k}}\ ,$ (187a)
$\displaystyle\Delta_{\chi_{k}}$ $\displaystyle=2\ ,$ (187b) $\displaystyle
r_{\chi_{k}}[B]$ $\displaystyle=\frac{2i(\delta-1)}{3}\ .$ (187c)

The computation for the other two-point functions proceeds similarly, and one
finds that the identities are satisfied with these charges.

The situation is slightly different when $k=0$. The equation of motion for
$\chi_{0}$ does not relate it to $\lambda_{0}$, so at this level both
$\lambda_{0}$ and $\chi_{0}$ are primary fields. We should therefore check
that all Ward identities are satisfied for

$G^{\chi\chi}_{0}(x_{1}^{-},x_{1};x_{2}^{-},x_{2})=-\frac{P_{+}}{4\sqrt{2}\pi^{2}R\,\bar{z}_{12}^{2}}\
.$ (188)

Taking $p_{\chi_{0}}$ and $\Delta_{\chi_{0}}$ to be real, the non-trivial
conditions are

$\displaystyle\Delta_{\chi_{0}}$ $\displaystyle=2\ ,$ (189a) $\displaystyle
Rp_{\chi_{0}}+\frac{3i}{2}r_{\chi_{0}}[B]$ $\displaystyle=1\ .$ (189b)

If we ask that $p_{\chi_{0}}$ is consistent with our parameterisation (183)
for $k=0$ the second condition is

$r_{\chi_{0}}[B]=\frac{2i(\delta-1)}{3}\ ,$ (190)

so the charges found for $k\neq 0$ are consistent with $\chi_{0}$’s status as
a primary field.

It is interesting to ask which value of $\delta$ naturally occurs when we
obtain the theory through the reduction of a 4d Lorentzian theory. As $B$
descends from the spatial rotations of the 4d theory, which are generated by
the matrices $\\{\gamma_{ij}\\}$, the only possible tensorial structure is

$\displaystyle r_{\psi_{k}}[B]$
$\displaystyle=-\frac{R\beta_{\psi_{k}}}{2}\Omega_{ij}\gamma_{ij}$
$\displaystyle=i\beta_{\psi_{k}}\gamma_{01}\ ,$ (191)

where anti-Hermiticity of $r_{\psi_{k}}[B]$ constrains $\beta_{\psi_{k}}$ to
be real. The charges of $\lambda_{k}$ and $\chi_{k}$ under the rotation must
then differ by a sign, which forces us to take

$\delta=\frac{1}{4}$ (192)

and hence

$\displaystyle p_{\lambda_{k}}$
$\displaystyle=\frac{1}{R}\left(k+\frac{1}{4}\right)\ ,$ (193)
$\displaystyle\alpha_{\lambda_{k}}$ $\displaystyle=\frac{1}{2}\ .$ (194)

###  6.2 $\mathfrak{su}(1,3)$ Invariance in 5d

We will now perform the same analysis for the 5d theory. As the correlation
functions vanish when $k=0$ we only need to focus on the $k>0$ modes, meaning
the only primary fields are the $\lambda_{k}$ with translated correlation
functions

$\displaystyle G^{\lambda\lambda}_{k}(x_{1}^{-},x_{1};x_{2}^{-},x_{2})$
$\displaystyle=-\frac{k}{8\sqrt{2}\pi^{3}R^{3}}P_{-}\left((k+1)z_{12}^{k}\bar{z}_{12}^{-(k+2)}P_{-}^{\Gamma}+(k-1)z_{12}^{k-2}\bar{z}_{12}^{-k}P_{+}^{\Gamma}\right)$
$\displaystyle\equiv
P_{-}\left(\mathcal{A}z_{12}^{k}\bar{z}_{12}^{-(k+2)}P_{-}^{\Gamma}-\mathcal{B}z_{12}^{k-2}\bar{z}_{12}^{-k}P_{+}^{\Gamma}\right)\
.$ (195)

Our intuition from the calculation in the 3d case leads us to the assumption
that $p_{\lambda_{k}}$ and $\Delta_{\lambda_{k}}$ are real. This will turn out
to be correct, and we will assume it from here onwards for convenience. A
crucial difference from the 3d case is that $\lambda_{k}$ has more than one
degree of freedom, so we can no longer assume $r_{\lambda_{k}}[B]$ is
irreducible.

The non-trivial constraints imposed by the $\mathfrak{su}(1,3)$ Ward
identities are

$\displaystyle 2=$ $\displaystyle\ \Delta_{\lambda_{k}}\ ,$ (196a)
$\displaystyle 0=$ $\displaystyle\
r_{\lambda_{k}}[B]P_{-}P_{-}^{\Gamma}+P_{-}P_{-}^{\Gamma}r^{{\dagger}}_{\lambda_{k}}[B]=r_{\lambda_{k}}[B]P_{-}P_{+}^{\Gamma}+P_{-}P_{+}^{\Gamma}r^{{\dagger}}_{\lambda_{k}}[B]\
,$ (196b) $\displaystyle 0=$ $\displaystyle\
r_{\lambda_{k}}[J^{\alpha}]P_{-}P_{-}^{\Gamma}+P_{-}P_{-}^{\Gamma}r^{{\dagger}}_{\lambda_{k}}[J^{\alpha}]=r_{\lambda_{k}}[J^{\alpha}]P_{-}P_{+}^{\Gamma}+P_{-}P_{+}^{\Gamma}r^{{\dagger}}_{\lambda_{k}}[J^{\alpha}]\
,$ (196c) $\displaystyle 0=$ $\displaystyle\
\frac{i\left(k-Rp_{\lambda_{k}}\right)(x_{1}-x_{2})^{i}}{R}G_{k}^{\lambda\lambda}+\left(\frac{\delta_{ij}}{R}r_{\lambda_{k}}[B]-\frac{1}{2}\Omega_{ik}\bar{\eta}^{\alpha}_{jk}r_{\lambda_{k}}[J^{\alpha}]\right)x_{1}^{j}G_{k}^{\lambda\lambda}$
$\displaystyle\
+x_{2}^{j}G_{k}^{\lambda\lambda}\left(\frac{\delta_{ij}}{R}r^{{\dagger}}_{\lambda_{k}}[B]-\frac{1}{2}\Omega_{ik}\bar{\eta}^{\alpha}_{jk}r^{{\dagger}}_{\lambda_{k}}[J^{\alpha}]\right)+\frac{i(x_{1}-x_{2})^{i}P_{-}}{R}\bigg{(}\mathcal{A}z_{12}^{k}\bar{z}_{12}^{-(k+2)}P_{-}^{\Gamma}$
$\displaystyle\
-\mathcal{B}z_{12}^{k-2}\bar{z}_{12}^{-k}P_{+}^{\Gamma}\bigg{)}\ ,$ (196d)
$\displaystyle 0=$ $\displaystyle\
\frac{i\left(k-Rp_{\lambda_{k}}\right)(x_{1}^{2}-x_{2}^{2})}{R}G_{k}^{\lambda\lambda}+\left(\frac{\delta_{ij}}{R}r_{\lambda_{k}}[B]-\frac{1}{2}\Omega_{ik}\bar{\eta}^{\alpha}_{jk}r_{\lambda_{k}}[J^{\alpha}]\right)x_{1}^{i}x_{1}^{j}G_{k}^{\lambda\lambda}$
$\displaystyle\
+x_{2}^{i}x_{2}^{j}G_{k}^{\lambda\lambda}\left(\frac{\delta_{ij}}{R}r^{{\dagger}}_{\lambda_{k}}[B]-\frac{1}{2}\Omega_{ik}\bar{\eta}^{\alpha}_{jk}r^{{\dagger}}_{\lambda_{k}}[J^{\alpha}]\right)+\frac{i(x_{1}^{2}-x_{2}^{2})P_{-}}{R}\bigg{(}\mathcal{A}z_{12}^{k}\bar{z}_{12}^{-(k+2)}P_{-}^{\Gamma}$
$\displaystyle\
-\mathcal{B}z_{12}^{k-2}\bar{z}_{12}^{-k}P_{+}^{\Gamma}\bigg{)}\ ,$ (196e)

where we have used the first equality to simplify the final two. The natural
tensor structures to put forward for our representations are

$\displaystyle r_{\lambda_{k}}[B]$ $\displaystyle=i\alpha_{k}\Gamma\ ,$ (197a)
$\displaystyle r_{\lambda_{k}}[J^{\alpha}]$
$\displaystyle=\beta_{k}\bar{\eta}^{\alpha}_{ij}\gamma_{ij}\ ,$ (197b)

up to some constants $\alpha_{k}$ and $\beta_{k}$. Since
$\bar{\eta}^{\alpha}_{ij}\gamma_{ij}$ and $\Gamma$ commute, as can be checked
using the properties of contractions between self-dual and anti-self-dual ’t
Hooft symbols tHooft:1976snw , the second and third Ward identities are
automatically satisfied if both $\alpha_{k}$ and $\beta_{k}$ are real.
However, we have the additional requirement that
$\\{r_{\lambda_{k}}[J^{\alpha}]\\}$ forms a representation of the
$\mathfrak{su}(2)$ algebra

$[J^{\alpha},J^{\beta}]=-2\epsilon^{\alpha\beta\gamma}J^{\gamma}\ .$ (198)

Since

$\big{[}r_{\lambda_{k}}[J^{\alpha}],r_{\lambda_{k}}[J^{\beta}]\big{]}=-8\beta^{2}\epsilon^{\alpha\beta\gamma}r_{\lambda_{k}}[J^{\gamma}]\
,$ (199)

we see that this is only true if

$r_{\lambda_{k}}[J^{\alpha}]=\frac{1}{4}\bar{\eta}^{\alpha}_{ij}\gamma_{ij}\
.$ (200)

Using the identities

$\displaystyle\bar{\eta}^{\alpha}_{ij}\bar{\eta}^{\alpha}_{kl}$
$\displaystyle=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}-\epsilon_{ijkl}\
,$ (201a) $\displaystyle\gamma_{ijkl}$
$\displaystyle=-\epsilon_{ijkl}\gamma_{01}\gamma_{*}\ ,$ (201b)

a short computation shows

$\Omega_{ik}\bar{\eta}^{\alpha}_{jk}r_{\lambda_{k}}[J^{\alpha}]P_{-}=\Omega_{ik}\bar{\eta}^{\alpha}_{jk}P_{-}r^{{\dagger}}_{\lambda_{k}}[J^{\alpha}]=0\
,$ (202)

so the terms involving $r_{\lambda_{k}}[J^{\alpha}]$ in the final two Ward
identities vanish. As both have a near-identical structure, let us focus on
the first: the remaining terms evaluate to

$\frac{i\left(k-p_{\lambda_{k}}\right)(x_{1}-x_{2})^{i}}{R}G_{k}^{\lambda\lambda}+\frac{i(x_{1}-x_{2})^{i}P_{-}}{R}\left(1-\alpha_{\lambda_{k}}\right)\bigg{(}\mathcal{A}z_{12}^{k}\bar{z}_{12}^{-(k+2)}P_{-}^{\Gamma}-\mathcal{B}z_{12}^{k-2}\bar{z}_{12}^{-k}P_{+}^{\Gamma}\bigg{)}=0\
,$ (203)

and we must therefore take

$\displaystyle p_{\lambda_{k}}$ $\displaystyle=\frac{k}{R}\ ,$ (204a)
$\displaystyle\alpha_{\lambda_{k}}$ $\displaystyle=1\ .$ (204b)

As with the 3d case, the $\mathfrak{u}(1)$ charges of our theory are not
uniquely defined by the Ward identities. We can add any multiple of the
identity to $r_{\lambda_{k}}[B]$ in the form

$r_{\lambda_{k}}[B]=i\left(\Gamma+c\mathbbm{1}\right)$ (205)

with $c\in\mathbb{R}$ without spoiling any of its required properties; the
only change is to shift the value of $p_{\lambda_{k}}$ to

$p_{\lambda_{k}}=\frac{k+c}{R}\ .$ (206)

The discussion about the rotation generator inherited from the higher
dimensional theory in the 3d theory also applies here; if we require that
$r_{\lambda_{k}}[B]$ is a linear combination of the spinorial rotation
generators $\\{\gamma_{ij}\\}$ then we must impose $c=0$, which we will do for
the remainder of this section.

Let us use the Ward identities valid for non-primary two-point functions to
fix the representations of $\chi_{k}$. Using the correlation function

$\displaystyle G^{\lambda\chi}_{k}(x_{1}^{-},x_{1};x_{2}^{-},x_{2})$
$\displaystyle=\frac{ik}{32\pi^{3}R^{3}}\gamma_{0i}P_{+}z_{12}^{k-2}\bar{z}_{12}^{-(k+2)}\Big{(}z_{12}(1+k)\left(i\delta_{ij}+R\Omega_{ij}\right)$
$\displaystyle\hskip
28.45274pt-\bar{z}_{12}(1-k)\left(i\delta_{ij}-R\Omega_{ij}\right)\Big{)}(x_{1}-x_{2})^{j}$
$\displaystyle\equiv\gamma_{0i}P_{+}\Big{(}\mathcal{A}\left(i\delta_{ij}+R\Omega_{ij}\right)-\mathcal{B}\left(i\delta_{ij}-R\Omega_{ij}\right)\Big{)}(x_{1}-x_{2})^{j}\
,$ (207)

we get the constraints

$\displaystyle 5=$ $\displaystyle\ \Delta_{\lambda_{k}}+\Delta_{\chi_{k}}\ ,$
(208a) $\displaystyle 0=$ $\displaystyle\
\gamma_{0i}P_{+}R\Omega_{ij}\Big{(}\mathcal{A}\left(i\delta_{jk}+R\Omega_{jk}\right)-\mathcal{B}\left(i\delta_{jk}-R\Omega_{jk}\right)\Big{)}(x_{1}-x_{2})^{k}$
$\displaystyle\
+r_{\lambda_{k}}[B]G^{\lambda\chi}_{k}+G^{\lambda\chi}_{k}r_{\chi_{k}}^{{\dagger}}[B]\
,$ (208b) $\displaystyle 0=$ $\displaystyle\
-\gamma_{0i}P_{+}\bar{\eta}^{\alpha}_{ij}\Big{(}\mathcal{A}\left(i\delta_{jk}+R\Omega_{jk}\right)-\mathcal{B}\left(i\delta_{jk}-R\Omega_{jk}\right)\Big{)}(x_{1}-x_{2})^{k}$
(208c) $\displaystyle\
+r_{\lambda_{k}}[J^{\alpha}]G^{\lambda\chi}_{k}+G^{\lambda\chi}_{k}r_{\chi_{k}}^{{\dagger}}[J^{\alpha}]\
.$ (208d)

It’s then easy to check that these are satisfied if

$\displaystyle\Delta_{\chi_{k}}$ $\displaystyle=3\ ,$ (209a) $\displaystyle
r_{\chi_{k}}[B]$ $\displaystyle=i\Gamma\ ,$ (209b) $\displaystyle
r_{\chi_{k}}[J^{\alpha}]$
$\displaystyle=\frac{1}{4}\bar{\eta}^{\alpha}_{ij}\gamma_{ij}\ .$ (209c)

Since the rotation generators are the same as those for $\lambda_{k}$ they
automatically form representation of $\mathfrak{u}(1)\oplus\mathfrak{su}(2)$.
It is straightforward to check that the valid Ward identities for the other
two-point functions are consistent with these assignments.

##  7 Conclusion

In this paper we have analysed fermions in $2n-1$ dimensions with $SU(1,n)$
spacetime symmetry. These were constructed through the null reduction of the
$2n$-dimensional free fermion CFT on an $\Omega$-deformed Minkowski
background. The two-point functions of the theory for $n=2$ and $n=3$ were
found using the $(2n-1)$-dimensional remnants of the supersymmetry of a free
complex scalar and free fermion in $2n$ dimensions, and the known two-point
functions for a scalar field with $SU(1,n)$ symmetry. The full non-compact
$2n$-dimensional two-point function was then reconstructed from the Kaluza-
Klein tower of $(2n-1)$-dimensional fields. The analogous transformation for a
2d fermion was also considered, and it was found we could only reconstruct the
left-moving fermion’s correlation function. We examined the $R\to\infty$ DLCQ
limit of the $(2n-1)$-dimensional two-point functions, showing that the
$2n$-dimensional two-point function could be recovered through resummation,
and commented on the $R\to 0$ limit, which gave a theory with Carroll
symmetry. Finally, it was shown that that the Ward identities for $SU(1,n)$
spacetime symmetry are satisfied at each level in the Kaluza-Klein tower.

Though we have only discussed free field theories, it is interesting to
compare the way in which $2n$-dimensional supersymmetry arises in the reduced
theories with the supersymmetric structure of the 5d $SU(1,3)$-invariant
interacting field theories found in Lambert:2019jwi ; Lambert:2020jjm . For
example, if we truncate the action of the 3d theory to the terms involving the
fields $(\lambda_{k},\chi_{k},\phi_{k+\frac{1}{2}})$ then our theory is
invariant under the $\xi$-type supercharges, while the $\eta$-type
supercharges of the full theory disappear. Similarly, if we only include the
fields $(\lambda_{k},\chi_{k},\phi_{k-\frac{1}{2}})$ this conclusion is
reversed. The same story is true in 5d, where retaining only fields with the
same value of $k$ preserves the $\xi$-type supercharges and mixing levels
preserves the $\eta$-type supercharges. As discussed in the introduction, this
is exactly what occurs in the interacting theories; the Lagrangians only
include a finite number of fields and are invariant under half the
supercharges of the 6d theories they are conjectured to describe. The fact
that the full higher-dimensional supercharge is recovered in our free theory
lends weight to the proposal that the same happens when the interacting
theories are treated non-perturbatively, as must be the case for the theories
to provide complete descriptions of 6d SCFTs. We leave it to future work to
show that the full superconformal symmetries of the free theory match the
structure of the 5d interacting theories, i.e. that the conformal supercharges
are present at each level in the reduced theory.

So far, the aforementioned 5d theories are the only known interacting field
theories with $SU(1,n)$ spacetime symmetry for $n\geq 2$. It would be
interesting to find other examples; the methods outlined in this work and in
Lambert:2021nol  allow any 4d Lagrangian CFT to be reduced on the
$\Omega$-deformed background, yielding a 3d theory with an infinite number of
fields and $SU(1,2)$ spacetime symmetry at each level. The task would then be
to try and find a non-trivial truncation of the theory to a finite number of
fields. The obvious candidate 4d theory is $\mathcal{N}=4$ SYM, which could
help elucidate the field-theoretic formulation of Spin Matrix Theories with
$PSU(1,2|3)$ symmetry Baiguera:2022pll . A larger class of theories that could
also be studied are Lagrangian 4d $\mathcal{N}=2$ SCFTs, which were classified
in Bhardwaj:2013qia . Finding the structure of the reduced theories associated
with these could shed light on applications of $SU(1,2)$ theories to non-
Lagrangian 4d $\mathcal{N}=2$ SCFTs, such as the class $\mathcal{S}$ theories
Gaiotto:2009we ; Gaiotto:2009hg .

###### Acknowledgements.

We thank Neil Lambert for the suggestion of this topic and many helpful
discussions. J.S. is supported by the STFC studentship ST/W507556/1.

##  Appendix A Fermion Conventions

We will follow the spinor conventions of Freedman:2012zz  throughout. In
particular, we define the Dirac conjugate of a spinor field by

$\bar{\psi}=i\psi^{{\dagger}}\gamma^{0}\ ,$ (210)

the covariant derivative of a spinor field by

$\nabla_{\mu}\psi=\left(\partial_{\mu}+\frac{1}{4}\omega_{\mu}^{ab}\gamma_{ab}\right)\
,$ (211)

and the chiral matrix $\gamma_{*}$ by

$\gamma_{*}=(-i)^{n-1}\gamma_{0}\gamma_{1}...\gamma_{n-1}\ .$ (212)

When we need an explicit basis for the gamma matrices, we shall use

$\displaystyle\gamma^{(4d)}_{0}$
$\displaystyle=\begin{pmatrix}0&\mathbbm{1}_{2}\\\
-\mathbbm{1}_{2}&0\end{pmatrix}\ ,$ (213a) $\displaystyle\gamma^{(4d)}_{i}$
$\displaystyle=\begin{pmatrix}0&\sigma_{i}\\\ \sigma_{i}&0\end{pmatrix}\ ,$
(213b)

in 4d, where $\\{\sigma_{i}\\}$ are the Pauli matrices, and

$\displaystyle\gamma^{(6d)}_{0}$
$\displaystyle=\begin{pmatrix}0&0&0&\mathbbm{1}_{2}\\\
0&0&\mathbbm{1}_{2}&0\\\ 0&-\mathbbm{1}_{2}&0&0\\\
-\mathbbm{1}_{2}&0&0&0\end{pmatrix}\ ,$ (214a)
$\displaystyle\gamma^{(6d)}_{1}$
$\displaystyle=\begin{pmatrix}0&0&0&\sigma_{1}\\\ 0&0&-\sigma_{1}&0\\\
0&-\sigma_{1}&0&0\\\ \sigma_{1}&0&0&0\end{pmatrix}\ ,$ (214b)
$\displaystyle\gamma^{(6d)}_{2}$
$\displaystyle=\begin{pmatrix}0&0&0&\sigma_{2}\\\ 0&0&-\sigma_{2}&0\\\
0&-\sigma_{2}&0&0\\\ \sigma_{2}&0&0&0\end{pmatrix}\ ,$ (214c)
$\displaystyle\gamma^{(6d)}_{3}$
$\displaystyle=\begin{pmatrix}0&0&0&\sigma_{3}\\\ 0&0&-\sigma_{3}&0\\\
0&-\sigma_{3}&0&0\\\ \sigma_{3}&0&0&0\end{pmatrix}\ ,$ (214d)
$\displaystyle\gamma_{4}^{(6d)}$
$\displaystyle=\begin{pmatrix}0&0&\mathbbm{1}_{2}&0\\\
0&0&0&\mathbbm{1}_{2}\\\ \mathbbm{1}_{2}&0&0&0\\\
0&\mathbbm{1}_{2}&0&0\end{pmatrix}\ ,$ (214e) $\displaystyle\gamma_{5}^{(6d)}$
$\displaystyle=\begin{pmatrix}0&0&-i\mathbbm{1}_{2}&0\\\
0&0&0&i\mathbbm{1}_{2}\\\ i\mathbbm{1}_{2}&0&0&0\\\
0&-i\mathbbm{1}_{2}&0&0\end{pmatrix}\ ,$ (214f)

in 6d; it is simple to check that these both satisfy

$\\{\gamma_{a},\gamma_{b}\\}=2\eta_{ab}\ .$ (215)

##  Appendix B Derivation of the Lorentz Transformation Generator

Let us derive the $\mathfrak{so}(1,3)$ element corresponding to the
transformation (11). We will do this using the accidental isomorphism

$SO(1,3)\cong\frac{SL(2,\mathbb{C})}{\mathbb{Z}_{2}}\ .$ (216)

of the 4d Lorentz group. Explicitly, given $\Lambda\in SO(1,3)$ we have two
corresponding $SL(2,\mathbb{C})$ group elements

$S[\Lambda]=\pm\frac{\tensor{\Lambda}{{}^{a}_{b}}\sigma_{a}\bar{\sigma}^{b}}{\sqrt{\det\left(\tensor{\Lambda}{{}^{a}_{b}}\sigma_{a}\bar{\sigma}^{b}\right)}}\
,$ (217)

with

$\sigma_{a}=(\mathbbm{1},\sigma_{i})\;,\;\bar{\sigma}_{a}=(-\mathbbm{1},\sigma_{i})\
.$ (218)

Since we require that $\Lambda=\mathbbm{1}$ maps to the identity on both
sides, we must pick the positive sign. Using the components of $\Lambda$ given
in (11), we find

$S[\Lambda]=\frac{e^{-\frac{ix^{+}}{4R}}}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\begin{pmatrix}3+e^{\frac{ix^{+}}{R}}-\frac{u}{\sqrt{2}R}&-1+e^{\frac{ix^{+}}{R}}-\frac{u}{\sqrt{2}R}\\\
-1+e^{\frac{ix^{+}}{R}}+\frac{u}{\sqrt{2}R}&3+e^{\frac{ix^{+}}{R}}+\frac{u}{\sqrt{2}R}\end{pmatrix}\
.$ (219)

The matrix $S[\Lambda]$ is the (positive-chirality) Weyl-spinor representation
of $\Lambda$, so if $\Lambda$ is generated by the Lie algebra element
$\lambda$, i.e.

$\Lambda=\operatorname{Exp}\left(\lambda\right)\ ,$ (220)

then we have

$S[\Lambda]=\operatorname{Exp}\left(\frac{1}{4}\lambda^{ab}\sigma_{ab}\right)\
,$ (221)

using the notation

$\sigma_{ab}=\frac{1}{2}\left(\sigma_{a}\bar{\sigma}_{b}-\sigma_{b}\bar{\sigma}_{a}\right)\
.$ (222)

It will be useful to rewrite the exponent in terms of three complex variables
$\\{\lambda_{i}\\}$ by defining

$\displaystyle\lambda^{ab}\sigma_{ab}$
$\displaystyle=2(\lambda^{01}+i\lambda^{23})\sigma_{1}+2(\lambda^{02}+i\lambda^{31})\sigma_{2}+2(\lambda^{03}+i\lambda^{12})\sigma_{3}$
$\displaystyle\equiv 2\sum_{i}\lambda_{i}\sigma_{i}\ .$ (223)

It is then easy to take the exponential of this, and we obtain

$S[\Lambda]=\begin{pmatrix}\cosh\left(\frac{v}{2}\right)+\frac{\lambda_{3}\sinh\left(\frac{v}{2}\right)}{v}&\frac{(\lambda_{1}-i\lambda_{2})\sinh\left(\frac{v}{2}\right)}{v}\\\
\frac{(\lambda_{1}+i\lambda_{2})\sinh\left(\frac{v}{2}\right)}{v}&\cosh\left(\frac{v}{2}\right)-\frac{\lambda_{3}\sinh\left(\frac{v}{2}\right)}{v}\end{pmatrix}$
(224)

with

$v=\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}}\ .$ (225)

Comparing this with (219) gives the three relations

$\displaystyle\frac{\lambda_{3}\sinh\left(\frac{v}{2}\right)}{v}=-\frac{u\,e^{-\frac{ix^{+}}{4R}}}{4R\sqrt{2\cos\left(\frac{x^{+}}{2R}\right)}}\
,$ (226a)
$\displaystyle\frac{i\lambda_{2}\sinh\left(\frac{v}{2}\right)}{v}=\frac{u\,e^{-\frac{ix^{+}}{4R}}}{4R\sqrt{2\cos\left(\frac{x^{+}}{2R}\right)}}\
,$ (226b)
$\displaystyle\cosh\left(\frac{v}{2}\right)=\frac{e^{-\frac{ix^{+}}{4R}}\left(3+e^{\frac{ix^{+}}{R}}\right)}{4\sqrt{\cos\left(\frac{x^{+}}{2R}\right)}}\
.$ (226c)

As the first two imply

$\lambda_{3}=-i\lambda_{2}\ ,$ (227)

we have $v=\lambda_{1}$. Its value is then fixed by considering the final
relation; a quick calculation shows that

$\lambda_{1}=\ln\cos\left(\frac{x^{+}}{2R}\right)+\frac{ix^{+}}{2R}\ .$ (228)

Substituting this back into the other two relations and reorganising gives the
Lie algebra element’s components

$\displaystyle\lambda^{01}$
$\displaystyle=\ln\cos\left(\frac{x^{+}}{2R}\right)\ ,$ (229a)
$\displaystyle\lambda^{23}$ $\displaystyle=\frac{x^{+}}{2R}\ ,$ (229b)
$\displaystyle\lambda^{03}=\lambda^{31}=\frac{1}{2\sqrt{2}R\sin\left(\frac{x^{+}}{2R}\right)}\Bigg{[}$
$\displaystyle
x^{2}\left(\sin\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)-\frac{x^{+}}{2R}\cos\left(\frac{x^{+}}{2R}\right)\right)$
$\displaystyle-x^{3}\left(\cos\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)+\frac{x^{+}}{2R}\sin\left(\frac{x^{+}}{2R}\right)\right)\Bigg{]}\
,$ (229c)
$\displaystyle\lambda^{12}=-\lambda^{02}=\frac{1}{2\sqrt{2}R\sin\left(\frac{x^{+}}{2R}\right)}\Bigg{[}$
$\displaystyle
x^{3}\left(\sin\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)-\frac{x^{+}}{2R}\cos\left(\frac{x^{+}}{2R}\right)\right)$
$\displaystyle+x^{2}\left(\cos\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)+\frac{x^{+}}{2R}\sin\left(\frac{x^{+}}{2R}\right)\right)\Bigg{]}\
.$ (229d)

We can rewrite this covariantly as

$\displaystyle\lambda^{01}$
$\displaystyle=\ln\cos\left(\frac{x^{+}}{2R}\right)\ ,$ (230a)
$\displaystyle\lambda^{0i}=-\lambda^{1i}=-\frac{1}{2\sqrt{2}R\sin\left(\frac{x^{+}}{2R}\right)}\Bigg{[}$
$\displaystyle
x^{i}\left(\cos\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)+\frac{x^{+}}{2R}\sin\left(\frac{x^{+}}{2R}\right)\right)$
$\displaystyle+R\Omega_{ij}x^{j}\left(\sin\left(\frac{x^{+}}{2R}\right)\ln\cos\left(\frac{x^{+}}{2R}\right)-\frac{x^{+}}{2R}\cos\left(\frac{x^{+}}{2R}\right)\right)\Bigg{]}\
,$ (230b) $\displaystyle\lambda^{ij}$
$\displaystyle=\frac{\Omega_{ij}x^{+}}{2}\ .$ (230c)

A straightforward calculation shows that if we use extend this to 6d and use
it as the generator for a 6d Lorentz group element we recover (11), so this is
indeed the correct generator to consider.

##  Appendix C Normalisation of Scalar Two-Point Functions

###  C.1 3d Scalar Fields

Let us consider the action for a mode of a free complex scalar field,

$S_{k}=2\pi R\int
d^{3}x\Bigg{[}\frac{2ik}{R}\bar{\phi}_{k}\partial_{-}\phi_{k}+\bar{\phi}_{k}D^{2}\phi_{k}\Bigg{]}\
.$ (231)

We will ultimately require $k\in\mathbb{Z}+\frac{1}{2}$ but shall only
constrain it to be real for now. The equation of motion of the mode is

$\mathcal{L}_{k}\phi_{k}\equiv\frac{2ik}{R}\partial_{-}\phi_{k}+D^{2}\phi_{k}=0\
,$ (232)

which has the obvious spherically-symmetric solution

$\phi_{k}=c_{k}\frac{1}{\sqrt{z\bar{z}}}\left(\frac{z}{\bar{z}}\right)^{k}$
(233)

away from the origin. The time-ordered two-point function will satisfy the
equation of motion away from coincident insertion points, so we will also have

$G_{k}\equiv\bra{\Omega}T\\{\phi_{k}(x^{-},x)\bar{\phi}_{k}(0)\\}\ket{\Omega}=c_{k}\frac{1}{\sqrt{z\bar{z}}}\left(\frac{z}{\bar{z}}\right)^{k}$
(234)

up to a time-ordering factor that we will include below. We would like to fix
the constant $c_{k}$. Standard QFT lore tells us that the time-ordered two-
point function is the Green’s function of $\mathcal{L}$, and we should have

$\mathcal{L}_{k}G_{k}=\frac{i}{2\pi R}\delta(x^{-})\delta^{(2)}(x)\ .$ (235)

Let us calculate the left hand side of the Green’s function equation. This is
most easily done by introducing the coordinates

$z=re^{i\theta}$ (236)

with $r\in[0,\infty)$ and $\theta\in(0,\pi)$. With these, integration of
spherically-symmetric functions over spacetime takes the form

$\int_{\mathbb{R}^{3}}d^{3}x=4\pi
R\int_{0}^{\infty}dr\int_{0}^{\pi}d\theta\,r\ ,$ (237)

and the two-point function is

$G_{k}=\frac{c_{k}F_{k}(\theta)e^{2ik\theta}}{r}\ .$ (238)

We have introduced the time-ordering factor $F_{k}(\theta)$ defined by

$F_{k}(\theta)=\begin{cases}1&\theta\in(\alpha,\beta)\\\
0&\text{otherwise}\end{cases}\ ,$ (239)

which restricts us to the interval with endpoints $\alpha$ and $\beta$. We
will keep these arbitrary at this point.

The operator $\mathcal{L}_{k}$ is given in $(r,\theta)$ coordinates by

$\displaystyle\mathcal{L}_{k}=$
$\displaystyle-\frac{i}{R}\Bigg{[}\frac{1}{2}\left(e^{i\theta}-e^{-i\theta}\right)\left(r\partial_{r}^{2}+\partial_{r}+\frac{1}{r}\partial_{\theta}^{2}\right)-e^{i\theta}\left(k-\frac{1}{2}\right)\left(\partial_{r}+\frac{i}{r}\partial_{\theta}\right)$
$\displaystyle-e^{-i\theta}\left(k+\frac{1}{2}\right)\left(\partial_{r}-\frac{i}{r}\partial_{\theta}\right)\Bigg{]}\
.$ (240)

As $G_{k}$ is not differentiable at $r=0$ and $\theta=\alpha,\beta$, we need
to introduce the notion of the distributional derivative121212See
Friedlander_Joshi_2003  for an introduction to working with differentiation of
distributions. to interpret (235). Let $G$ and $f$ be some smooth functions on
$\mathbb{R}^{3}$, and define the action of $G$ on $f$ by

$G[f]=\int_{\mathbb{R}^{3}}d^{3}x\,Gf\ .$ (241)

As $G$ is smooth, the derivative $\mathcal{L}_{k}G$ is well-defined; we can
then act with this on $f$ and integrate by parts to obtain

$\displaystyle\mathcal{L}_{k}G[f]=-2\pi
i\int_{0}^{\infty}dr\int_{0}^{\pi}d\theta\,G\bigg{[}$
$\displaystyle(e^{i\theta}-e^{-i\theta})\left(r^{2}\partial_{r}^{2}f+3r\partial_{r}f+\partial_{\theta}^{2}f\right)+2i(e^{i\theta}+e^{-i\theta})\partial_{\theta}f$
$\displaystyle+2i\left(e^{i\theta}\left(k-\frac{1}{2}\right)-e^{-i\theta}\left(k+\frac{1}{2}\right)\right)\partial_{\theta}f$
$\displaystyle+2\left(e^{i\theta}\left(k-\frac{1}{2}\right)+e^{-i\theta}\left(k+\frac{1}{2}\right)\right)r\partial_{r}f\bigg{]}\
.$ (242)

We see that this expression makes sense even when $G$ is non-differentiable;
we then define the regularised distributional derivative of $G_{k}$ through
the action on a test function $f$, with

$(\mathcal{L}_{k}G_{k}[f])_{\epsilon}=\int_{\epsilon}^{\infty}dr\int_{0}^{\pi}G_{k}\mathcal{L}_{k}^{{\dagger}}f\
,$ (243)

where $\mathcal{L}_{k}^{{\dagger}}$ is defined by the previous expression.

We can now subsitute the expression (238) for $G_{k}$ in this expression. As
$G_{k}$ was chosen to satisfy $\mathcal{L}_{k}G_{k}=0$ away from the singular
points, the integral will only pick up boundary contributions. A brief
computation gives

$\displaystyle(\mathcal{L}_{k}G_{k}[f])_{\epsilon}=-2\pi
ic_{k}\int_{\epsilon}^{\infty}dr\int_{\alpha}^{\beta}d\theta\Bigg{[}$
$\displaystyle\frac{\partial}{\partial
r}\bigg{(}\left(e^{i(2k+1)\theta}-e^{i(2k-1)\theta}\right)(r\partial_{r}f+2f)$
$\displaystyle+2\left(e^{i(2k+1)\theta}\left(k-\frac{1}{2}\right)+e^{i(2k-1)\theta}\left(k+\frac{1}{2}\right)\right)f\bigg{)}$
$\displaystyle+\frac{\partial}{\partial\theta}\bigg{(}\left(e^{i(2k+1)\theta}-e^{i(2k-1)\theta}\right)\frac{1}{r}\partial_{\theta}f\bigg{)}\Bigg{]}$
(244)

which evaluates to

$\displaystyle(\mathcal{L}_{k}G_{k}[f])_{\epsilon}=$ $\displaystyle 2\pi
c_{k}\delta_{0}[f]\bigg{[}e^{i(2k+1)\theta}+e^{i(2k-1)\theta}\bigg{]}_{\theta=\alpha}^{\theta=\beta}$
$\displaystyle+4\pi
c_{k}\int_{\epsilon}^{\infty}\frac{dr}{r}\bigg{[}e^{2ik\theta}\sin\theta\partial_{\theta}f\bigg{]}_{\theta=\alpha}^{\theta=\beta}+O(\epsilon)\
.$ (245)

We see that for generic values of $\alpha$ and $\beta$ we get a non-delta
function term that survives as we take $\epsilon\to 0$. The most obvious
solution to this is to take $\alpha=0$ and $\beta=\pi$: this mirrors the time-
ordered two-point function of a Lorentzian CFT, where no $\Theta$-functions
are present. With this choice, we can take $\epsilon\to 0$ to find

$\mathcal{L}_{k}G_{k}=-4\pi c_{k}\left(1+e^{2\pi
ik}\right)\delta(x^{-})\delta^{(2)}(x)\ .$ (246)

For generic $k$ we have no issues; comparing with (235) then gives

$c_{k}=-\frac{i}{8\pi^{2}R\left(1+e^{2\pi ik}\right)}$ (247)

as the coefficient of the two-point function. However, the right hand side of
(246) vanishes when $k\in\mathbb{Z}+\frac{1}{2}$, which are precisely the
values we are interested in!

To find the resolution of this, we need to think about the 4d origin of the
theory; since $\ket{\Omega}$ is the vacuum of a free Lorentzian scalar, it
must be annihilated by all lowering operators. In particular, this applies to
the modes arising from the reduction on $x^{+}$, as these are eigenstates of
$p_{+}$. As discussed in Lambert:2021nol , there is a slight subtlety here:
the split into raising and lowering operators on $\ket{\Omega}$ is naturally
defined for the modes of $\hat{\phi}$, which for the 3d theory are related to
the modes of $\phi$ by

$\hat{\phi}_{k+\frac{1}{2}}=\frac{1}{2}\left(\phi_{k}+\phi_{k+1}\right)\ .$
(248)

Requiring our theory makes sense as the reduction of a well-defined 4d theory
imposes

$\hat{\phi}_{l}\ket{\Omega}=\bar{\hat{\phi}}_{-l}\ket{\Omega}=0\ \forall\,l>0\
,$ (249)

which is equivalent to the conditions

$\phi_{k}\ket{\Omega}=\bar{\phi}_{-k}\ket{\Omega}=0\ \forall\,k>-\frac{1}{2}\
.$ (250)

Taking time-ordering with respect to $x^{-}$, we then have

$\bra{\Omega}T\\{\phi_{k}(x^{-},x)\bar{\phi}_{k}(0)\\}\ket{\Omega}=\begin{cases}\Theta(x^{-})\bra{\Omega}\phi_{k}(x^{-},x)\bar{\phi}_{k}(0)\ket{\Omega}&k>0\\\
\Theta(-x^{-})\bra{\Omega}\bar{\phi}_{k}(0)\phi_{k}(x^{-},x)\ket{\Omega}&k<0\end{cases}\
.$ (251)

The two-point functions of the modes are related to the two-point function of
the 4d field by

$\bra{\Omega}T\\{\hat{\phi}(\hat{x})\hat{\bar{\phi}}(0)\\}\ket{\Omega}=\cos\left(\frac{x^{+}}{2R}\right)\sum_{k\in\mathbb{Z}+\frac{1}{2}}e^{-\frac{ikx^{+}}{R}}G_{k}(x^{-},x)\
.$ (252)

The 4d field satisfies the Green’s function equation

$\hat{\square}\bra{\Omega}T\\{\hat{\phi}(\hat{x})\hat{\bar{\phi}}(0)\\}\ket{\Omega}=i\delta^{(4)}(\hat{x})\
,$ (253)

where the d’Alembertian $\hat{\square}=\hat{\partial}^{2}$ is given by

$\hat{\square}=\cos^{2}\left(\frac{x^{+}}{2R}\right)\left(-2\left(\partial_{+}+\frac{\tan\left(\frac{x^{+}}{2R}\right)}{2R}\right)\partial_{-}+D^{2}\right)$
(254)

after transforming to un-hatted coordinates. The factor of
$\cos^{2}\left(\frac{x^{+}}{2R}\right)$ is evaluated on a delta function, and
is therefore unity. Substituting the expansion (252) into the equation (253)
and using the periodic delta function

$\delta(x^{+})=\frac{1}{2\pi R}\sum_{l\in\mathbb{Z}}e^{-\frac{ilx^{+}}{R}}$
(255)

to reduce on $x^{+}$, we find the infinite tower of relations

$\mathcal{L}_{l-\frac{1}{2}}G_{l-\frac{1}{2}}+\mathcal{L}_{l+\frac{1}{2}}G_{l+\frac{1}{2}}=\frac{i}{\pi
R}\delta(x^{-})\delta^{(2)}(x)$ (256)

for $l\in\mathbb{Z}$.

We see that the set of equations 256 impose weaker constraints than the set
235. However, they share one important property. If we examine the equation
for $l=0$ away from $x^{-}=0$, we have

$\mathcal{L}_{-\frac{1}{2}}G_{-\frac{1}{2}}+\mathcal{L}_{\frac{1}{2}}G_{\frac{1}{2}}=0\
.$ (257)

However, $G_{\frac{1}{2}}$ and $G_{-\frac{1}{2}}$ are only non-trivial for
non-overlapping ranges of $x^{-}$. This means that if we restrict our
attention to $x^{-}>0$ we find the equation

$\mathcal{L}_{\frac{1}{2}}G_{\frac{1}{2}}=0\ ,$ (258)

whereas if we focus on $x^{-}<0$ we find the analogous statement for
$G_{-\frac{1}{2}}$. This propagates through the tower of equations, and we see
that all the two-point functions satisfy the classical equations of motion
away from singularities.

We will now use the modified set of equation to fix $c_{k}$. It is necessary
to consider the cases $l>0$, $l=0$, and $l<0$ separately. Let us start with
$l>0$, which encompasses the two-point functions with $k>0$. In this regime,
both correlators will be of the form

$G_{k}=\begin{cases}\frac{c_{k}e^{2ik\theta}}{r}&\theta\in\left(0,\frac{\pi}{2}\right)\\\
0&\text{otherwise}\end{cases}\ .$ (259)

Using (C.1), we have

$(\mathcal{L}_{k}G_{k}[f])_{\epsilon}=-4c_{k}\pi\left(\delta_{0}[f]-\int_{\epsilon}^{\infty}\frac{dr}{r}\,e^{ik\pi}\left(\partial_{\theta}f\right)_{\theta=\frac{\pi}{2}}\right)+O(\epsilon)\
.$ (260)

Since this means the left hand side of (256) becomes (after taking
$\epsilon\to 0$)

$\mathcal{L}_{l-\frac{1}{2}}G_{l-\frac{1}{2}}[f]+\mathcal{L}_{l+\frac{1}{2}}G_{l+\frac{1}{2}}[f]=-4\pi(c_{l-\frac{1}{2}}+c_{l+\frac{1}{2}})\delta_{0}[f]-4\pi
i(c_{l-\frac{1}{2}}-c_{l+\frac{1}{2}})\int_{0}^{\infty}\frac{dr}{r}e^{il\pi}\left(\partial_{\theta}f\right)_{\theta=\frac{\pi}{2}}\
,$ (261)

we see that we must take

$c_{k}=-\frac{i}{8\pi^{2}R}$ (262)

for $k>0$. The calculation for $l<0$ is almost identical; as we have

$G_{k}=\begin{cases}\frac{c_{k}e^{2ik\theta}}{r}&\theta\in\left(\frac{\pi}{2},\pi\right)\\\
0&\text{otherwise}\end{cases}$ (263)

for $k<0$, (C.1) evaluates to

$(\mathcal{L}_{k}G_{k}[f])_{\epsilon}=4\pi
c_{k}\left(\delta_{0}[f]-\int_{\epsilon}^{\infty}\frac{dr}{r}\,e^{ik\pi}\left(\partial_{\theta}f\right)_{\theta=\frac{\pi}{2}}\right)+O(\epsilon)\
.$ (264)

After taking $\epsilon\to 0$ we see that (256) gives

$\mathcal{L}_{l-\frac{1}{2}}G_{l-\frac{1}{2}}[f]+\mathcal{L}_{l+\frac{1}{2}}G_{l+\frac{1}{2}}[f]=4\pi(c_{l-\frac{1}{2}}+c_{l+\frac{1}{2}})\delta_{0}[f]+4\pi
i(c_{l-\frac{1}{2}}-c_{l+\frac{1}{2}})\int_{0}^{\infty}\frac{dr}{r}e^{il\pi}\left(\partial_{\theta}f\right)_{\theta=\frac{\pi}{2}}\
,$ (265)

so the conditions are satisfied if

$c_{k}=\frac{i}{8\pi^{2}R}$ (266)

for $k<0$. Finally, we must check the case $l=0$, which links the two previous
results. Using our prior calculations we have

$\mathcal{L}_{-\frac{1}{2}}G_{-\frac{1}{2}}[f]+\mathcal{L}_{\frac{1}{2}}G_{\frac{1}{2}}[f]=4\pi\left(c_{-\frac{1}{2}}-c_{\frac{1}{2}}\right)\delta_{0}[f]+4\pi
i\left(c_{-\frac{1}{2}}+c_{\frac{1}{2}}\right)\int_{0}^{\infty}\frac{dr}{r}\,\left(\partial_{\theta}f\right)_{\theta=\frac{\pi}{2}}\
.$ (267)

Substituting (262) and (266) into this gives

$\mathcal{L}_{-\frac{1}{2}}G_{-\frac{1}{2}}[f]+\mathcal{L}_{\frac{1}{2}}G_{\frac{1}{2}}[f]=\frac{i}{\pi
R}\delta_{0}[f]\ ,$ (268)

which is consistent with (256).

As a test of these results, we can resum (252) and check that we get the
correct result. Using the $i\epsilon$ prescription described in section 3.2 we
find

$\displaystyle\bra{\Omega}T\\{\hat{\phi}(\hat{x})\hat{\bar{\phi}}(0)\\}\ket{\Omega}$
$\displaystyle=-\frac{i\cos\left(\frac{x^{+}}{2R}\right)}{8\pi^{2}R}\sum_{k=0}^{\infty}\left(\Theta(x^{-})\left(\frac{ze^{-\frac{ix^{+}}{R}}}{\bar{z}}\right)^{k+\frac{1}{2}}-\Theta(-x^{-})\left(\frac{ze^{-\frac{ix^{+}}{R}}}{\bar{z}}\right)^{-\left(k+\frac{1}{2}\right)}\right)$
$\displaystyle=\frac{i\cos\left(\frac{x^{+}}{2R}\right)}{8\pi^{2}R}\frac{1}{ze^{-\frac{ix^{+}}{2R}}-\bar{z}e^{\frac{ix^{+}}{2R}}}$
$\displaystyle=\frac{1}{4\pi^{2}\left(-2\hat{x}^{+}\hat{x}^{-}+\hat{x}^{2}\right)}\
,$ (269)

which is the two-point function of a canonically normalised 4d massless scalar
field, as hoped.

###  C.2 5d Scalar Fields

We can perform the exact same analysis for the reduction of a 6d scalar field
to a 5d theory. We will use the same spherically-symmetric coordinates as in
the 3d case. The two-point function obtained from the classical equations of
motion is

$G_{k}=\frac{c_{k}F_{k}(\theta)e^{2ik\theta}}{r^{2}}\ ,$ (270)

where we will be interested in taking $k\in\mathbb{Z}$. Integrals over
spherically-symmetric functions take the form

$\int_{\mathbb{R}^{5}}d^{5}x=\left(4\pi
R\right)^{2}\int_{0}^{\infty}\int_{0}^{\pi}drd\theta\,r^{2}\sin\theta\ .$
(271)

In these coordinates, the classical equation of motion is

$\displaystyle\mathcal{L}_{k}G_{k}=\frac{-2i}{R}\Bigg{[}$
$\displaystyle\frac{1}{4}\left(e^{i\theta}-e^{-i\theta}\right)\left(r\partial_{r}^{2}G+\partial_{r}G+\frac{1}{r}\partial_{\theta}^{2}G\right)-\frac{1}{2}e^{i\theta}(k-1)\left(\partial_{r}G+\frac{i}{r}\partial_{\theta}G\right)$
$\displaystyle-\frac{1}{2}e^{-i\theta}(k+1)\left(\partial_{r}G-\frac{i}{r}\partial_{\theta}G\right)\Bigg{]}\
.$ (272)

The same procedure as above allows us to define the regularised distributional
derivative of $G_{k}$ as

$\displaystyle(\mathcal{L}_{k}G_{k}[f])_{\epsilon}=-4\pi^{2}R\int_{\epsilon}^{\infty}dr\int_{0}^{\pi}d\theta\,G_{k}\Bigg{[}$
$\displaystyle\left(e^{i\theta}-e^{-i\theta}\right)^{2}\left(r^{3}\partial_{r}^{2}f+5r^{2}\partial_{r}f+4rf+r\partial_{\theta}^{2}f\right)$
$\displaystyle+2(r^{2}\partial_{r}f+2rf)\left((k-1)(e^{2i\theta}-1)+(k+1)(1-e^{-2i\theta})\right)$
$\displaystyle+4i\left(e^{2i\theta}-e^{-2i\theta}\right)r\partial_{\theta}f-4\left(e^{2i\theta}+e^{-2i\theta}\right)rf$
$\displaystyle+2ir\partial_{\theta}f\Big{(}(k-1)(e^{2i\theta}-1)-(k+1)(1-e^{-2i\theta})\Big{)}$
$\displaystyle-4rf\left((k-1)e^{2i\theta}-(k+1)e^{-2i\theta}\right)\Bigg{]}\
.$ (273)

Subsituting in our expression for $G_{k}$ gives

$\displaystyle\left(\mathcal{L}_{k}G_{k}[f]\right)_{\epsilon}=-4\pi^{2}R\,c_{k}\int_{\epsilon}^{\infty}dr\int_{\alpha}^{\beta}d\theta\Bigg{[}$
$\displaystyle\frac{\partial}{\partial
r}\bigg{(}e^{2i(k+1)\theta}\left(r\partial_{r}f+2(k+1)f\right)+e^{2i(k-1)\theta}\left(r\partial_{r}f-2(k-1)f\right)$
$\displaystyle-2e^{2ik\theta}\left(r\partial_{r}f+2f\right)\bigg{)}+\frac{\partial}{\partial\theta}\bigg{(}\frac{1}{r}\partial_{\theta}f\bigg{(}e^{2i(k+1)\theta}$
$\displaystyle+e^{2i(k-1)\theta}-2e^{2ik\theta}\bigg{)}\bigg{)}\Bigg{]}\ ,$
(274)

which localises to boundary contributions as expected. When none of the
exponents vanish (i.e. $k\neq-1,0,1$) this is

$\left(\mathcal{L}_{k}G_{k}[f]\right)_{\epsilon}=8\pi^{2}R\,c_{k}\Bigg{[}e^{2ik\theta}\bigg{(}\delta_{0}[f]\left(\sin
2\theta+\frac{i}{k}\right)-\left(\cos
2\theta-1\right)\int_{\epsilon}^{\infty}\frac{dr}{r}\,\partial_{\theta}f\bigg{)}\Bigg{]}_{\theta=\alpha}^{\theta=\beta}+O(\epsilon)\
,$ (275)

with the exceptional cases

$\displaystyle(\mathcal{L}_{1}G_{1})_{\epsilon}$
$\displaystyle=4\pi^{2}R\,c_{1}\Bigg{[}e^{2i\theta}\bigg{(}i\delta_{0}[f]\left(2-e^{2i\theta}\right)-2\left(\cos
2\theta-1\right)\int_{\epsilon}^{\infty}\frac{dr}{r}\,\partial_{\theta}f\bigg{)}\Bigg{]}_{\theta=\alpha}^{\theta=\beta}+O(\epsilon)\