More generally, when the energy $\omega$ is high compared to the uniform norm $||\partial_z^2 C||_\infty$, one has

$$a_s(\omega, \zeta, \bar{\zeta}) = \frac{1}{\left(\frac{s}{2}\right)! \sqrt{2}} \left( \frac{i\sigma \partial_\zeta^2 C(\zeta, \bar{\zeta})}{\omega} \right)^{\frac{s}{2}} a(\omega, \zeta, \bar{\zeta}) + \mathcal{O}\left( \left( \frac{||\partial_z^2 C||_\infty}{\omega} \right)^{\frac{s}{2}+2} \right). \quad (9.10)$$

Thus the state (9.3) effectively appears as a spin-0 state at high energy, but is filled with higher-spin states in the infrared. In particular, the leading correction are spin-2 states which superficially seem to have a pole in the energy $\omega$: this is suggestive that they might carry some memory. However, if the full spin-2 state $a_2(\omega, \zeta, \bar{\zeta})$ really had a pole, these states would not be normalisable by themselves which would imply that the state (9.3) would also be of infinite norm. The norm of a state of the form (6.67) indeed reads

$$\langle \Phi, \Phi \rangle := \int \omega \, d\omega \int \frac{i}{2} \, d\zeta \wedge d\bar{\zeta} \int \frac{d\alpha}{2\pi} |\Phi(\omega, \zeta, \bar{\zeta}, \alpha)|^2 = \sum_{s \in \mathbb{Z}} \int \omega \, d\omega \int \frac{i}{2} \, d\zeta \wedge d\bar{\zeta} |a_s(\omega, \zeta, \bar{\zeta})|^2. \quad (9.11)$$

Fortunately, this problem does not arise: the state (9.3) is a perfectly well-behaved state inside the UIR of $BMS_4$. This is because if $a(\omega, \zeta, \bar{\zeta})$ has finite norm then so does its supertranslated counterpart $a(\omega, \zeta, \bar{\zeta})e^{i\langle K, P \rangle}$. Therefore, since the states of a BMS particle form a Hilbert space, the sum of these state has finite norm as well.

We are now left with the question of the average supermomentum of the state:

$$\begin{aligned} \langle \mathcal{P}(z, \bar{z}) \rangle &= \int \omega d\omega d\zeta d\bar{\zeta} \int \frac{d\alpha}{2\pi} \mathcal{P}(\zeta, \bar{\zeta}) |\Phi(\omega, \zeta, \bar{\zeta}, \alpha)|^2 \\ &= \int \omega d\omega \int \frac{d\alpha}{2\pi} \left( \omega + \frac{\sigma}{\omega} (e^{i2\alpha} \partial_z^2 + e^{-i2\alpha} \partial_{\bar{z}}^2) \right) |\Phi(\omega, z, \bar{z}, \alpha)|^2. \end{aligned} \quad (9.12)$$

We will suppose that $a(\omega, \zeta, \bar{\zeta})$ is a hard state in the sense that it is only supported at energies $\omega > \Lambda$ above a cut off $\Lambda \gg ||\partial_z^2 C||_\infty$, so that the average momentum is only influenced by the leading-order behaviour of the modulus square of (9.6)

$$\begin{aligned} |\Phi(\omega, z, \bar{z}, \alpha)|^2 = |a_0(\omega, z, \bar{z})|^2 &+ \left[ e^{2i\alpha} (\bar{a}_0(\omega, z, \bar{z})a_2(\omega, z, \bar{z}) + a_0(\omega, z, \bar{z})\bar{a}_{-2}(\omega, z, \bar{z})) + \text{c.c.} \right] \\ &+ \mathcal{O}\left( \left( \frac{||\partial_z^2 C||_\infty}{\omega} \right)^2 \right). \end{aligned} \quad (9.13)$$

In order to ensure the finiteness of the scalar field energy, we have at least $\omega^2|a(\omega, z, \bar{z})|^2 = \mathcal{O}(\frac{1}{\omega})$. Therefore, the average supermomentum is the sum

$$\langle \mathcal{P}(z, \bar{z}) \rangle = \langle P(z, \bar{z}) \rangle + \partial_z^2 \partial_{\bar{z}}^2 \langle \mathcal{N}(z, \bar{z}) \rangle$$

of a hard contribution which, at high energy, coincides with the average supermomentum of the hard state $a_0(\omega, z, \bar{z})$,

$$\langle P(z, \bar{z}) \rangle := \int \omega^2 d\omega \int \frac{d\alpha}{2\pi} |\Phi(\omega, z, \bar{z}, \alpha)|^2 = \int_{\Lambda}^{\infty} \omega^2 d\omega |a_0(\omega, z, \bar{z})|^2 + \mathcal{O}\left( \left( \frac{||\partial_z^2 C||_\infty}{\Lambda} \right)^2 \right) \quad (9.14)$$
complemented by an effective memory effect

$$
\begin{align*}
\partial_z^2 \partial_{\bar{z}}^2 \langle \mathcal{N}(z, \bar{z}) \rangle &:= \int \omega d\omega \int \frac{d\alpha}{2\pi} \left( \frac{\sigma}{\omega} (e^{i2\alpha} \partial_z^2 + e^{-i2\alpha} \partial_{\bar{z}}^2) \right) |\Phi(\omega, z, \bar{z}, \alpha)|^2 \\
&= \partial_{\bar{z}}^2 \left( \sigma \int d\omega \int \frac{d\alpha}{2\pi} e^{-i2\alpha} |\Phi(\omega, z, \bar{z}, \alpha)|^2 \right) + \text{c.c.} \\
&= \partial_{\bar{z}}^2 \left( \sigma \int_{\Lambda}^{\infty} d\omega (\bar{a}_0(\omega, z, \bar{z}) a_2(\omega, z, \bar{z}) + a_0(\omega, z, \bar{z}) \bar{a}_{-2}(\omega, z, \bar{z})) \right) + \text{c.c.} + \mathcal{O}\left(\left(\frac{\|\partial_z^2 C\|_{\infty}}{\Lambda}\right)^4\right) \\
&= \partial_{\bar{z}}^2 \left( -\sigma \partial_z^2 C(z, \bar{z}) \int_{\Lambda}^{\infty} d\omega |a(\omega, z, \bar{z})|^2 \frac{\sin(\omega C(z, \bar{z}))}{\omega} \right) + \text{c.c.} + \mathcal{O}\left(\left(\frac{\|\partial_z^2 C\|_{\infty}}{\Lambda}\right)^4\right).
\end{align*}
$$

Let us conclude this section by sketching a possible generalisation of the previous case to supermomenta $\mathcal{P} = P[\omega, \zeta, \bar{\zeta}] + \partial_z^2 \partial_{\bar{z}}^2 \mathcal{N}[\omega, \zeta, \bar{\zeta}, \alpha, \beta, \bar{\beta}]$ in a generic $SL(2, \mathbb{C})$-orbit, where the hard piece is $P[\omega, \zeta, \bar{\zeta}](z, \bar{z}) = \omega \delta^{(2)}(z - \zeta, \bar{z} - \bar{\zeta})$ as usual. For such a massless BMS particle, consider a particular state in the decomposition (6.57)

$$ \Phi(\omega, \zeta, \bar{\zeta}, \alpha, \beta, \bar{\beta}) = e^{i\lambda\alpha} \tilde{\Phi}_n(\omega, \zeta, \bar{\zeta}; \vec{\pi}) \quad (9.15) $$

which can be interpreted as the state of a massless Poincaré particle of helicity $\lambda$ and momentum $p_\mu = \omega q_\mu(\zeta, \bar{\zeta})$ in a given gravitational vacuum. The corresponding BMS state in another gravitational vacuum is obtained by acting with the corresponding shift $\mathcal{C}$. Concreteley, the state (9.15) picks a phase

$$ \langle \mathcal{P}, \mathcal{T} \rangle = \omega C(\zeta, \bar{\zeta}) + \langle \partial_z^2 \mathcal{N}[\omega, \zeta, \bar{\zeta}, \alpha, \beta, \bar{\beta}], \partial_z^2 C \rangle \quad (9.16) $$

due to (8.6). The first term on the right-hand side does not depend on the $ISO(2)$-ccordinates so it does not change the particle content but the second term does. In the Hilbert space topology, the space $\mathcal{O}(-3, 1)$ of spin-weighted densities is a Hilbert space and the Cauchy-Schwarz inequality implies that the second term can be bounded as follows

$$ \left| \langle \partial_z^2 \mathcal{N}[\omega, \zeta, \bar{\zeta}, \alpha, \beta, \bar{\beta}], \partial_z^2 C \rangle \right| \le \left\| \partial_z^2 \mathcal{N}[\omega, \zeta, \bar{\zeta}, \alpha, \beta, \bar{\beta}] \right\|_2 \left\| \partial_z^2 C \right\|_2, \quad (9.17) $$

for the $L^2$-norm $\|\cdot\|_2$ on the sphere.<fnref n="40" /> For a shift $[C] \in \mathcal{E}[1]/\mathbb{R}^{3,1}$ of gravitational vacua, the latter norm coincide with (8.4): $\langle\langle C, C\rangle\rangle = \|\partial_z^2 C\|_2$. By analogy with the previous case, one would expect the superposition of states in distinct vacua to be controlled by the distance between these gravitational vacua (in the sense of Corollary 8.3).

# 10 Conclusion and summary

Let us conclude by summarising our main results.

In Section 3, we introduced a Lorentz-invariant decomposition of supermomenta into hard and soft pieces (Proposition 3.1). This decomposition is based on the following important properties: i) a supermomentum is soft iff it projects onto a vanishing momentum (Proposition 2.14); ii) the hard representations of the BMS group coincide with usual Poincaré representations, in particular, they do not branch under restriction to a Poincaré subgroup (Corollary 5.5) since the little group of a hard supermomentum is identical to the little group of its associated momentum

<footnote n="40">⁴⁰The analogue of such a norm is a delicate issue in the nuclear topology, so it will not be investigated here.</footnote>
(Subsection 5.1). Some conceptual and technical subtleties related to the nonlinearity of the hard
vs soft decomposition (Propositions 3.4-3.6) and their relations with Weinberg soft factors, were
discussed in Subsection 3.2.

We emphasised in Section 5 that the non-linearity of the decomposition means that hard
states alone cannot fulfil conservation of supermomenta (Propositions 5.3-5.4); therefore, in order
to enforce this conservation law, other type of BMS particles need to be introduced. While hard
BMS particles coincide with Poincaré particles, realised in terms of their natural scattering data
(Subsection 5.4), in all other cases, a single BMS particle is in general a BMS multiplet of usual
Poincaré particles.

In Sections 6 and 7, we provided the explicit form of the wavefunctions of generic BMS
particles (Subsection 6.3 and 7.1). An explicit algorithm for the decomposition of these wave-
functions in suitable bases of functions (on the quotient of the Poincaré little group by the BMS
little group) was provided for the decomposition of the BMS particle in terms of Poincaré par-
ticles.<fnref n="41" /> In Section 6.5, we stressed that this interpretation of generic BMS particles in terms
of superposition of usual Poincaré particles crucially depend on the choice of gravity vacuum,
that is on the choice of a Poincaré subgroup $ISO_0(3,1) \subset BMS_4$ for the branching. For in-
stance, a state appearing as a simple scalar field in one gravity vacuum will be interpreted, in
another gravity vacuum, as a complicated quantum superposition of particles of all possible spins
(including, typically, continuous spins, in the massless case).

In Section 8, the wavefunctions of BMS particle introduced in [31] were discussed in terms
of the homogenous spaces appearing in the asymptotic quantisation program (the space of cuts
at null infinity, the space of gravitation vacua, etc). We stressed that there exists a Lorentz-
invariant notion of distance between gravitational vacua (Corollary 8.3). We also pointed out
that the Fock space of any countable collection of BMS particles is a separable Hilbert space
(Subsection 8.5).

In Section 9, we exhibited, on a particular example of state, a concrete high-energy limit
in which one can control the relative weight of the tower of higher-spin particles in the BMS
multiplet and which furthermore allowed for an average supermomentum including memory. It
remains to be seen if such types of limits may help to obtain a correspondence principle where
usual S-matrix results are recovered, in some suitable regime, from a BMS-invariant S-matrix.
At the end of this section, we briefly argued that the distance between gravitational vacua could
play a role in such a mechanism. We also constructed examples of normalisable *BMS* states
allowing for a pole in the energy, with the resulting $L^2$ norm being soften by the presence of
continuous-spin particles in the spectrum. The physical properties of such examples need to be
explored further.

Another important direction to explore in the future would be to perform a systematic
investigation of soft BMS particles (i.e. UIRs of BMS group with vanishing momentum) along
the same lines as what we did for generic BMS particles. Another work in progress is the
classification of the UIRs of the asymptotic symmetry group of quantum electrodynamics (QED)
with massless charged fields, i.e. the group $SO(3,1) \times (\mathbb{R}^{3,1} \times \mathcal{E}[0])$. This could provide an
alternative path for defining a Hilbert space of asymptotic multi-particle states suitable for an
infrared-finite scattering theory of massless QFTs, such as QED with massless charged fields or

<footnote n="41">⁴¹Propositions 6.6 and 6.8 were mentioned in [12] but the direct sum of usual massless representations was not mentioned explicitly. Furthermore, Propositions 6.10-6.11 are new and complete the analysis of all possible cases of branching in the massless case. Finally, Propositions 7.3-7.4 of McCarthy were recovered in an explicit way.</footnote>
Yang-Mills theory. This different road could perhaps circumvent some of the undesirable features mentioned e.g. in [83].

Acknowledgements

We are grateful to G. Barnich, R. Gicquaud, B. Oblak, E. Skvortsov, B. Valsesia for discusions.
We also would like to acknowledge decisive discussions with L. Donnay on the physical motiva-
tions underlying this work; in particular the present fresh perspective on BMS representations
would not have been pursued without her persistent insistence that BMS representations ought
to capture infrared physics.

This research was partially completed during the thematic programme ‘Carrollian Physics
and Holography’ at the Erwin Schrödinger Institute of Vienna,
# Appendices

## A Paneitz operator: image and kernel

In Appendix A.1, we provide a proof that Proposition 2.14 works for smooth densities. We prove that it extends to distributions in Appendix A.2. Finally, in Appendix A.3 we provide a proof of Proposition 2.14 in the Hilbert topology. We include the latter proof because this reasoning turns out to be instructive and closer to usual computations (in terms of Hermitian forms and Hilbert spaces).

Let us assume that $\mathcal{E}[w]$ is the vector space of smooth conformal densities of weight $w$ on the celestial sphere. The topological dual $\mathcal{E}[w]'$ is spanned by the corresponding distributions (i.e. continuous linear forms on $\mathcal{E}[w]$). In particular, $\mathcal{E}[-2-w] \subset \mathcal{E}[w]'$ since smooth conformal densities of weight $-2-w$ can be thought as distributions acting on smooth conformal densities of weight $w$. In particular,

$$ \mathcal{E}[1] \subset \mathcal{E}[-3]', \qquad \mathcal{E}[-3] \subset \mathcal{E}[1]'. \qquad (A.1) $$

where $\mathcal{E}[1]'$ is the vector space of distributions corresponding to supermomenta. Let us denote $Ann(\mathbb{R}^{3,1}) \subset \mathcal{E}[-3]$ the space of smooth conformal densities of weight $-3$ annihilating the translation subspace $\mathbb{R}^{3,1} \subset \mathcal{E}[-1]$ and $(Ann(\mathbb{R}^{3,1}))'$ the corresponding dual space. One has

$$ \mathcal{E}[1]/\mathbb{R}^{3,1} \subset (Ann(\mathbb{R}^{3,1}))', \qquad Ann(\mathbb{R}^{3,1}) \subset (\mathcal{E}[1]/\mathbb{R}^{3,1})'. \qquad (A.2) $$

### A.1 Proof of Proposition 2.14 in the smooth case

For smooth densities, Proposition 2.14 can be expressed as the following

**Proposition A.1.** *The Paneitz operator*

$$ \partial^2 \bar{\partial}^2 : \mathcal{E}[1]/\mathbb{R}^{3,1} \xrightarrow{\sim} Ann(\mathbb{R}^{3,1}) \qquad (A.3) $$

*is an isomorphism in the smooth category.*

**Proof:** This can be checked very concretely by performing some computations in terms of the representative $f \in C^\infty(S^2)$, for which one can use spherical harmonics $Y_\ell^m$ as a basis. From [41, Eq. (3.24)] one has

$$ \partial^2 \bar{\partial}^2 Y_\ell^m = \gamma^{-3} \ell(\ell^2 - 1)(\ell + 2) Y_\ell^m. \qquad (A.4) $$

Therefore the kernel of the Paneitz operator is spanned precisely by the representatives with only spherical harmonics $\ell = 0$ and $\ell = 1$ in their decomposition. In other words, the kernel coincides with translations $\mathbb{R}^{3,1} \subset \mathcal{E}[1]$. Furthermore, the image of the Paneitz operator coincides with representatives whose spherical harmonics $\ell = 0$ and $\ell = 1$ vanish. Put differently, its image coincides with $Ann(\mathbb{R}^{3,1}) \subset \mathcal{E}[-3]$. This proves Proposition 2.14 for smooth densities. $\square$
**Remark A.2.** Proposition A.1 can also be extracted from [45, Sections III.3.3 and III.5.2]. More precisely, there exist two short exact sequences of SL(2, C) representations (cf. [45, Fig.4 on p.155]):

<figure><img src="image_7.png" /></figure>

and

<figure><img src="image_8.png" /></figure>

It is well-known that two short exact sequences of this form can be spliced together and are equivalent to a single long exact sequence (see e.g. [100, Exercise 13 on p.180]). Applying this splicing procedure to (A.5)-(A.6) gives a long sequence which, when restricting to real conformal densities, yields (2.20).

**A.2 Proof of Proposition 2.14 in the case of distributions**

By duality, we deduce from Proposition A.1 the following

**Proposition A.3.** *The Paneitz operator extends to an isomorphism*

$$ \bar{\partial}^2 \bar{\partial}^2 : (\mathrm{Ann}(\mathbb{R}^{3,1}))' \xrightarrow{\sim} (\mathcal{E}[1]/\mathbb{R}^{3,1})' . \qquad (A.7) $$

*between spaces of distributions.*

**Proof:** Let us first prove surjectivity. If $\beta \in (\mathcal{E}[1]/\mathbb{R}^{3,1})'$ we can define a distribution $\alpha \in (\mathrm{Ann}(\mathbb{R}^{3,1}))'$ via

$$ \langle \alpha, \psi \rangle := \langle \beta, \chi \rangle, \quad \text{since} \quad \forall \psi \in \mathrm{Ann}(\mathbb{R}^{3,1}), \exists! \chi \in \mathcal{E}[1]/\mathbb{R}^{3,1} : \psi = \bar{\partial}^2 \bar{\partial}^2 \chi, \qquad (A.8) $$

where we used Proposition A.1. Therefore,

$$ \langle \beta, \chi \rangle = \langle \alpha, \bar{\partial}^2 \bar{\partial}^2 \chi \rangle = \langle \bar{\partial}^2 \bar{\partial}^2 \alpha, \chi \rangle, \quad \forall \chi \in \mathcal{E}[1]/\mathbb{R}^{3,1}. \qquad (A.9) $$

Hence, $\beta = \bar{\partial}^2 \bar{\partial}^2 \alpha$.

Let us now prove injectivity. If $\alpha \in (\mathrm{Ann}(\mathbb{R}^{3,1}))'$ satisfies $\bar{\partial}^2 \bar{\partial}^2 \alpha = 0$, then

$$ \langle \bar{\partial}^2 \bar{\partial}^2 \alpha, \chi \rangle = \langle \alpha, \bar{\partial}^2 \bar{\partial}^2 \chi \rangle = 0, \quad \forall \chi \in \mathcal{E}[1]/\mathbb{R}^{3,1}. \qquad (A.10) $$

Since any $\psi \in \mathrm{Ann}(\mathbb{R}^{3,1})$ can always be written as $\psi = \bar{\partial}^2 \bar{\partial}^2 \chi$ for some $\chi \in \mathcal{E}[1]/\mathbb{R}^{3,1}$, we deduce that $\alpha = 0$. $\square$
From Proposition A.3, we can deduce that soft supermomenta always belong to the image of the Paneitz operator.

**Corollary A.4.** Any element $\mathcal{P} \in \mathcal{E}[-1]'$ which annihilates the translation subspace $\mathbb{R}^{3,1} \subset \mathcal{E}[-1]$ takes the form $\mathcal{P} = \bar{\partial}^2 \bar{\partial}^2 \mathcal{N}$ where $\mathcal{N} \in (\mathrm{Ann}(\mathbb{R}^{3,1}))'$.

**Proof:** Since $\mathcal{P}$ annihilates translations by assumption, it defines a distribution on $\mathcal{E}[-1]/\mathbb{R}^{3,1}$ that we will denote with the same letter: $\mathcal{P} \in (\mathcal{E}[1]/\mathbb{R}^{3,1})'$. By Proposition A.3, this means that there exists $\mathcal{N} \in (\mathrm{Ann}(\mathbb{R}^{3,1}))'$ such that $\mathcal{P} = \bar{\partial}^2 \bar{\partial}^2 \mathcal{N}$. $\square$

### A.3 Proof of Proposition 2.14 in the Hilbert topology

Let $\mathcal{T} \in \mathcal{E}[1]$ be a supertranslation and let us define the positive semi-definite quadratic form $\langle\cdot, \cdot\rangle$ on $\mathcal{E}[1]$ given explicitly by the expression (8.4). In order for $\langle\mathcal{T}, \mathcal{T}\rangle$ to exist, it is enough to assume that the representatives of supertranslations belong to the Sobolev space $H^2(S^2)$, i.e. $\mathcal{T} \in L^2(S^2)$ as well as all its derivatives up to second order. In particular, this works for smooth densities on the celestial sphere (and clearly does not work for distributions since their norm can be infinite).

The comments in the proof of Proposition 2.13 already ensure the exactness on the left in (2.20): $\mathrm{Im}(q) = \mathrm{Ker}(\bar{\partial}^2 \bar{\partial}^2)$. So one should prove the exactness on the right: $\mathrm{Im}(\bar{\partial}^2 \bar{\partial}^2) = \mathrm{Ker}(\pi)$. The inner product (2.17) implies that $\mathrm{Im}(\bar{\partial}^2 \bar{\partial}^2) \subseteq \mathrm{Ker}(\pi)$ since any supermomentum of the form $\bar{\partial}^2 \bar{\partial}^2 \mathcal{T}_1$ annihilates all translations $\mathcal{T}_2$, that is to say $\langle \bar{\partial}^2 \bar{\partial}^2 \mathcal{T}_1, \mathcal{T}_2 \rangle = 0$ for any $\mathcal{T}_1 \in \mathcal{E}[1]$ and $\mathcal{T}_2 \in \mathrm{Im}(q) \simeq \mathbb{R}^{3,1}$. Since the scalar product (2.17) is non-degenerate on the quotient $\mathcal{E}[1]/\mathbb{R}^{3,1}$, we have that $\mathrm{Im}(\bar{\partial}^2 \bar{\partial}^2) \simeq \mathcal{E}[1]/\mathbb{R}^{3,1}$ is a Hilbert space [45, Section III.6.5] and, by Fréchet-Riesz theorem,

$$ \begin{gathered} \mathcal{E}[1]/\mathbb{R}^{3,1} \to (\mathcal{E}[1]/\mathbb{R}^{3,1})' = \mathrm{Ann}(\mathbb{R}^{3,1}) \\ \mathcal{T} \mapsto \langle\mathcal{T}, \cdot\rangle = \langle \bar{\partial}^2 \bar{\partial}^2 \mathcal{T}, \cdot \rangle \end{gathered} \quad (A.11) $$

is an isomorphism. This concludes the proof in the Hilbert topology.

## B Distributional identity for hard supermomenta

In Appendices B.1 and B.2 below, we provide proofs for Proposition 3.4 and Proposition 3.5, respectively.

### B.1 Proof of Proposition 3.4: identity for massless hard supermomenta

For completeness, let us explain the origin of the identity (3.8). It is based on the Cauchy-Pompeiu formula

$$ \mathcal{T}(\zeta, \bar{\zeta}) = \frac{1}{2\pi i} \oint_{\partial D} \frac{\mathcal{T}(z, \bar{z})}{z - \zeta} dz + \frac{1}{2\pi i} \int_{D} dz \wedge d\bar{z} \frac{\partial_{\bar{z}} \mathcal{T}(z, \bar{z})}{z - \zeta} \quad (B.1) $$

where $D \subset \mathbb{C}$ is an open domain (topologically a disk) inside the complex plane and $\zeta \in D$. Making the replacement $\mathcal{T} \mapsto (\bar{z} - \zeta) \partial_{\bar{z}} \mathcal{T}(z, \bar{z})$ in (B.1) one obtains

$$ \begin{align} 0 &= \frac{1}{2\pi i} \oint \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \partial_{\bar{z}} \mathcal{T}(z, \bar{z}) dz + \frac{1}{2\pi i} \int dz \wedge d\bar{z} \frac{\partial_{\bar{z}} [(\bar{z} - \bar{\zeta}) \partial_{\bar{z}} \mathcal{T}(z, \bar{z})]}{z - \zeta} \tag{B.2} \\ &= \frac{1}{2\pi i} \oint \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \partial_{\bar{z}} \mathcal{T}(z, \bar{z}) dz + \frac{1}{2\pi i} \int dz \wedge d\bar{z} \frac{\partial_{\bar{z}} \mathcal{T}(z, \bar{z})}{z - \zeta} + \frac{1}{2\pi i} \int dz \wedge d\bar{z} \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \nonumber \end{align} $$
where the domain $D$ and the contour $\partial D$ have been and will be left implicit from now on. Subtracting (B.2) to (B.1), one obtains

$$ \begin{aligned} & \int \frac{i}{2} dz \wedge d\bar{z} \left( \mathcal{T}(z, \bar{z}) \delta^{(2)}(z - \zeta, \bar{z} - \bar{\zeta}) - \frac{1}{\pi} \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \right) \\ & = \frac{1}{2\pi i} \oint \left( \frac{\mathcal{T}(z, \bar{z})}{z - \zeta} - \partial_{\bar{z}} \mathcal{T}(z, \bar{z}) \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \right) dz, \end{aligned} \quad (B.3) $$

which prove (3.8).

Since $\mathcal{T}(z, \bar{z}) = |\hat{z}|^{-2} \hat{\mathcal{T}}(\hat{z}, \bar{\hat{z}})$ is a smooth globally-defined weight-one density on the celestial sphere, we have

$$ \begin{aligned} & \mathcal{T}(z, \bar{z}) = |\hat{z}|^{-2} \hat{\mathcal{T}}(\hat{z}, \bar{\hat{z}}) \quad (B.4) \\ & = |\hat{z}|^{-2} \left( \hat{\mathcal{T}}(0, 0) + \hat{z} \partial_{\hat{z}} \hat{\mathcal{T}}(0, 0) + \bar{\hat{z}} \partial_{\bar{\hat{z}}} \hat{\mathcal{T}}(0, 0) + \hat{z}\bar{\hat{z}} \partial_{\hat{z}}\partial_{\bar{\hat{z}}} \hat{\mathcal{T}}(0, 0) + \frac{\bar{\hat{z}}^2}{2} \partial_{\bar{\hat{z}}}^2 \hat{\mathcal{T}}(0, 0) + \frac{\hat{z}^2}{2} \partial_{\hat{z}}^2 \hat{\mathcal{T}}(0, 0) + \mathcal{O}(|\hat{z}|^3) \right) \\ & = z\bar{z} \hat{\mathcal{T}}(0, 0) + \bar{z} \partial_{\bar{z}} \hat{\mathcal{T}}(0, 0) + z \partial_{\hat{z}} \hat{\mathcal{T}}(0, 0) + \partial_{\hat{z}}\partial_{\bar{z}} \hat{\mathcal{T}}(0, 0) + \frac{\bar{z}}{2\bar{z}} \partial_{\bar{z}}^2 \hat{\mathcal{T}}(0, 0) + \frac{\bar{z}}{2z} \partial_{\hat{z}}^2 \hat{\mathcal{T}}(0, 0) + \mathcal{O}(|z|^{-1}). \end{aligned} $$

Making use of this asymptotic behavior, the boundary term on the right-hand side of (B.3) can then be explicitly computed and gives a non zero contribution

$$ \frac{1}{2\pi i} \oint \left( \frac{\mathcal{T}(z, \bar{z})}{z - \zeta} - \partial_{\bar{z}} \mathcal{T}(z, \bar{z}) \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \right) dz = \bar{\zeta}\zeta \hat{\mathcal{T}}(0, 0) + \bar{\zeta}\partial_{\bar{z}}\hat{\mathcal{T}}(0, 0) + \zeta\partial_{\hat{z}}\hat{\mathcal{T}}(0, 0) + \partial_{\hat{z}}\partial_{\bar{z}}\hat{\mathcal{T}}(0, 0). $$

To see this, first notice that, in the limit that we are considering, of infinite radius for our circular contour, the correction term $\mathcal{O}(|z|^{-1})$ in the expression of $\mathcal{T}(z, \bar{z})$ decays too fast to contribute to the boundary terms. The other contributions can then be computed by writing $z = re^{i\theta}$ and performing the integral in $\theta$ at fixed radius $r$. Therefore

$$ \begin{aligned} & \int \frac{i}{2} dz \wedge d\bar{z} \left( \mathcal{T}(z, \bar{z}) \delta^{(2)}(z - \zeta, \bar{z} - \bar{\zeta}) - \frac{1}{\pi} \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \frac{\bar{z} - \bar{\zeta}}{z - \zeta} \right) \\ & = \bar{\zeta}\zeta \hat{\mathcal{T}}(0, 0) + \bar{\zeta}\partial_{\bar{z}}\hat{\mathcal{T}}(0, 0) + \zeta\partial_{\hat{z}}\hat{\mathcal{T}}(0, 0) + \partial_{\hat{z}}\partial_{\bar{z}}\hat{\mathcal{T}}(0, 0) \quad (B.5) \\ & = \int \frac{i}{2} dz \wedge d\bar{z} \mathcal{T}(z, \bar{z}) q^{\mu}(\zeta, \bar{\zeta}) \mathcal{D}_{\mu}(\delta^{(2)}(z - \infty)). \end{aligned} $$

## B.2 Proof of Proposition 3.5: identity for massive hard supermomenta

Let us start from

$$ \int \frac{i}{2} dz \wedge d\bar{z} \mathcal{T}(z, \bar{z}) \frac{m^4}{\pi (q(z, \bar{z}) \cdot p)^3} = \frac{1}{\pi} \int \frac{i}{2} dz \wedge d\bar{z} \mathcal{T}(z, \bar{z}) \partial_{\bar{z}}^2 \left( \frac{(\partial_z q(z, \bar{z}) \cdot p)^2}{2 q(z, \bar{z}) \cdot p} \right) $$

where we used the valid equality (3.13) between functions on the complex plane. Integrating by part, we obtain

$$ \begin{aligned} & \int \frac{i}{2} dz \wedge d\bar{z} \left( \mathcal{T}(z, \bar{z}) \frac{m^4}{\pi (q \cdot p)^3} - \frac{1}{\pi} \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \frac{(\partial_z q \cdot p)^2}{2 q \cdot p} \right) \\ & = \frac{1}{2\pi i} \oint \left( \mathcal{T}(z, \bar{z}) \partial_{\bar{z}} \left( \frac{(\partial_z q \cdot p)^2}{2 q \cdot p} \right) - \partial_{\bar{z}} \mathcal{T}(z, \bar{z}) \frac{(\partial_z q \cdot p)^2}{2 q \cdot p} \right) dz, \end{aligned} \quad (B.6) $$

where, as in Appendix B.1, we implicitly integrate on the left-hand side on an open disk $D \subset \mathbb{C}$ inside the complex plane while the contour integral on the right-hand side is on the circle $\partial D$.
Since $\mathcal{T}(z, \bar{z}) = |\hat{z}|^{-2}\hat{\mathcal{T}}(\hat{z}, \bar{\hat{z}})$ is a smooth globally-defined weight-one density on the celestial sphere, we have the asymptotic behavior (B.4). On the other hand, making use of

$$ q(z, \bar{z}) \cdot p = (p^3 - p^0) + (p^1 - ip^2)z + (p^1 + ip^2)\bar{z} - (p^0 + p^3)|z|^2, \quad (B.7) $$

$$ \partial_z q(z, \bar{z}) \cdot p = (p^1 - ip^2) - (p^0 + p^3) \bar{z}, \quad (B.8) $$

we obtain

$$ \frac{(\partial_z q(z, \bar{z}) \cdot p)^2}{q(z, \bar{z}) \cdot p} = \frac{\bar{z}}{z} \left( -(p^0 + p^3) + \bar{z}^{-1}(p^1 - ip^2) - z^{-1}(p^1 + ip^2) \right. \\ \left. + |z|^{-2}(p^0 - p^3) - z^{-2}\frac{(p^1 + ip^2)^2}{p^0 + p^3} + O(|z|^{-3}) \right). \quad (B.9) $$

Using the above expansion as well as the expansion (B.4) of the supertranslation $\mathcal{T}(z, \bar{z})$, the boundary term on the right-hand side of (B.6) can then be explicitly computed in the limit where the boundary becomes an infinitely large circle. First one should check that, in the limit of large radius that we are considering, the subleading terms $O(|z|^{-3})$ in these asymptotic expansions do not contribute. It then turns out that several contributions of the leading terms vanish due to the angular integration $\int d\theta e^{i\theta} = \frac{1}{|z|} \oint dz$ in (B.6). Nevertheless, some terms survive the final angular integration and yield a non-zero contribution

$$ \frac{1}{2\pi i} \oint \left( \mathcal{T}(z, \bar{z}) \partial_{\bar{z}} \left( \frac{(\partial_z q \cdot p)^2}{2q \cdot p} \right) - \partial_{\bar{z}} \mathcal{T}(z, \bar{z}) \frac{(\partial_z q \cdot p)^2}{2q \cdot p} \right) dz \quad (B.10) \\ = \frac{p^3 - p^0}{2} \hat{T}(0, 0) - \frac{p^1 - ip^2}{2} \partial_{\hat{z}} \hat{T}(0, 0) - \frac{p^1 + ip^2}{2} \partial_{\bar{z}} \hat{T}(0, 0) - \frac{p^0 + p^3}{2} \partial_{\hat{z}} \partial_{\bar{z}} \hat{T}(0, 0). $$

Therefore we conclude that

$$ \int \frac{i}{2} dz \wedge d\bar{z} \left( \mathcal{T}(z, \bar{z}) \frac{-m^4}{\pi(q \cdot p)^3} + \frac{1}{\pi} \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \frac{(\partial_z q \cdot p)^2}{2q \cdot p} \right) = \int \frac{i}{2} dz \wedge d\bar{z} \, \mathcal{T}(z, \bar{z}) \, p^\mu \mathcal{D}_\mu (\delta^{(2)}(z - \infty)). $$

## B.3 Proof of Proposition 3.6: identity for massive and massless momenta

Since $\partial_z^2 q_\mu(z, \bar{z}) = 0$ for all $z \in \mathbb{C}$, we have

$$ \partial_z^2 \left( (q \cdot p) \ln |q \cdot p| \right) = \frac{(\partial_z q \cdot p)^2}{q \cdot p} \quad (B.11) $$

Thus, making use of Proposition 3.5,

$$ \int \frac{i}{2} dz \wedge d\bar{z} \, \mathcal{T}(z, \bar{z}) P(z, \bar{z}) \\ = -\frac{1}{\pi} \int \frac{i}{2} dz \wedge d\bar{z} \, \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \frac{(\partial_z q(z, \bar{z}) \cdot p)^2}{2q(z, \bar{z}) \cdot p} + (p^\mu \mathcal{D}_\mu \mathcal{T})(z = \infty) \\ = -\frac{1}{2\pi} \int \frac{i}{2} dz \wedge d\bar{z} \, \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \, \partial_z^2 \left( (q(z, \bar{z}) \cdot p) \ln |q(z, \bar{z}) \cdot p| \right) + (p^\mu \mathcal{D}_\mu \mathcal{T})(z = \infty) $$

integrating by parts gives

$$ \int \frac{i}{2} dz \wedge d\bar{z} \left( \mathcal{T}(z, \bar{z}) P(z, \bar{z}) + \frac{1}{2\pi} \partial_z^2 \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) q(z, \bar{z}) \cdot p \ln |q(z, \bar{z}) \cdot p| \right) - (p^\mu \mathcal{D}_\mu \mathcal{T})(z = \infty) \quad (B.12) \\ = -\frac{1}{2i} \oint \left( \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \partial_z \left( (q(z, \bar{z}) \cdot p) \ln |q(z, \bar{z}) \cdot p| \right) - \partial_z \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) (q(z, \bar{z}) \cdot p) \ln |q(z, \bar{z}) \cdot p| \right) d\bar{z}. $$
Now, on the one hand (B.7) implies that

$$ (q(z, \bar{z}) \cdot p) \ln |q(z, \bar{z}) \cdot p| = -(p^0 + p^3) |z|^2 \ln (|p^0 + p^3| |z|^2) + \mathcal{O}(|z| \ln(|z|)) \quad (B.13) $$

and, on the other hand, $\bar{\partial}^2 \mathcal{T}$ is a smooth spin-weighted density of weight ($w = -1, s = 2$) which behaves asymptotically as

$$ \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) = \frac{z}{\bar{z}^3} \partial_{\bar{z}}^2 \hat{\mathcal{T}}(0, 0) + \mathcal{O}(|z|^{-3}). \quad (B.14) $$

Here again, in the limit of large radius that we are considering, the subleading terms in these asymptotic expansions do not contribute to the boundary term (B.12). Finally, a careful calculation shows that the contribution of the leading order terms vanishes due to the angular integration.

## B.4 Proof of Corollary 3.7

Let {$p_i^\mu$} be a collection of momenta satisfying momentum conservation, $\sum_i p_i^\mu = 0$. Let us consider the combination $S = -\frac{1}{2\pi} \sum_i (q \cdot p_i) \ln |q \cdot p_i|$. From (B.7), we can write

$$ q(z, \bar{z}) \cdot p_i = -(p_i^0 + p_i^3) |z|^2 (1 + F_i(z, \bar{z})), \quad (B.15) $$

where $F_i(z, \bar{z}) = \mathcal{O}(|z|^{-1})$ are functions which are smooth around $z = 0$. Therefore,

$$ S(z, \bar{z}) = -\frac{1}{2\pi} \sum_i q(z, \bar{z}) \cdot p_i \ln \left[ (|p_i^0 + p_i^3| |z|^2 (1 + F_i(z, \bar{z}))) \right] \quad (B.16) $$

$$ = -\frac{1}{2\pi} \sum_i q(z, \bar{z}) \cdot p_i \ln (|p_i^0 + p_i^3|) - \frac{1}{2\pi} \sum_i q(z, \bar{z}) \cdot p_i \ln (1 + F_i(z, \bar{z})) \quad (B.17) $$

where we used momentum conservation to drop the extra term

$$ \sum_i q(z, \bar{z}) \cdot p_i \ln (|z|^2) = \ln (|z|^2) q(z, \bar{z}) \cdot (\sum_i p_i) = 0. \quad (B.18) $$

One sees that both terms in (B.17) define smooth densities of weight one around $z = \infty$ and, therefore, so does $S$.

## C Haar mesure on $SL(2, \mathbb{C})$

We here parametrises elements $g \in SL(2, \mathbb{C})$ as in (6.25).

The Killing metric on $SL(2, \mathbb{C})$ is given by

$$ g_{\text{Killing}} = \text{Re} \left[ \text{Tr} \left[ (g^{-1}dg) \odot (g^{-1}dg) \right] \right] $$

where $\odot$ is the symmetric tensor product (which we most often leave implicit in the expression of the metric, as is common use in physics literature). We have

$$ \text{Tr}[(g^{-1}dg) \odot (g^{-1}dg)] = \text{Tr}[(N^{-1}dN)^2] + \text{Tr}[L^{-1}dL]^2 + 2 \text{Tr}[(N^{-1}dN) \odot (dL L^{-1})] $$
and direct computation gives:

$$N^{-1}dN = \begin{pmatrix} -\frac{1}{2}\omega^{-1}d\omega & \omega d\zeta \\ 0 & \frac{1}{2}\omega^{-1}d\omega \end{pmatrix} \qquad (C.1)$$

$$L^{-1}dL = \begin{pmatrix} \frac{i}{2}d\alpha & 0 \\ d\beta - ie^{i\frac{\alpha}{2}}\beta d\alpha & -\frac{i}{2}d\alpha \end{pmatrix} \qquad (C.2)$$

$$dL L^{-1} = \begin{pmatrix} \frac{i}{2}d\alpha & 0 \\ e^{-i\alpha}d\beta & -\frac{i}{2}d\alpha \end{pmatrix} \qquad (C.3)$$

and thus

$$\mathrm{Tr}[(L^{-1}dL)^2] = -\frac{1}{2}d\alpha^2 \qquad (C.4)$$

$$\mathrm{Tr}[(N^{-1}dN)^2] = \frac{1}{2}\omega^{-2}d\omega^2 \qquad (C.5)$$

$$\mathrm{Tr}[(N^{-1}dN) \odot (dLL^{-1})] = -\frac{i}{2}\omega^{-1}d\omega d\alpha + \omega e^{-i\alpha}d\zeta d\beta \qquad (C.6)$$

and therefore

$$g_{\mathrm{Killing}} = -\frac{1}{2}d\alpha^2 + \frac{1}{2}\omega^{-2}d\omega^2 + \omega(e^{-i\alpha}d\zeta d\beta + e^{i\alpha}d\bar{\zeta} d\bar{\beta})$$

Therefore

$$\det(g_{\mathrm{Killing}}) = -2 \times \left(\frac{1}{2}\omega^{-2}\right) \times \omega^4 = \omega^2 \qquad (C.7)$$

and thus the volume element of the Killing metric reads

$$\mu_{\mathrm{Haar}} = \sqrt{\det(g_{\mathrm{Killing}})} \, d^6 X = \frac{1}{2} \omega \, d\omega \, d\zeta_1 \, d\zeta_2 \, d\alpha \, d\beta_1 \, d\beta_2, \qquad (C.8)$$

where $\zeta = \zeta_1 + i\zeta_2$ and $\beta = \beta_1 + i\beta_2$.

## D Action of supertranslations in generic massless BMS representations

This appendix provides the proof of Proposition 6.5.

By definition, supertranslations act via a phase $\langle P, T \rangle$ where the supermomenta takes the form (6.31). This produces the action (6.51) where the phase $\varphi$ is given by the integral of the supertranslation $T$ multiplied by the soft part in (6.31). Integrating by part the Paneitz operator, one obtains

$$\begin{align*}
\varphi &= \int dz^2 \, \partial_z^2 \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \left| \frac{\partial z'}{\partial z} \right|^{-1} \mathcal{N}(z', \bar{z}'), \\
&= \int dz'^2 \, \partial_z^2 \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \left| \frac{\partial z}{\partial z'} \right|^{3} \mathcal{N}(z', \bar{z}'), \\
&= \omega^{-3} \int dz'^2 \, \partial_z^2 \partial_{\bar{z}}^2 \mathcal{T}(z, \bar{z}) \left| 1 + \beta z' \right|^{-6} \mathcal{N}(z', \bar{z}'),
\end{align*}$$
where (6.29) was used in the last equality. Moreover, if one does not integrate by part in the phase $\varphi$, one can also write

$$
\begin{align*}
\varphi &= \int dz^2 \mathcal{T}(z, \bar{z}) \left|\frac{\partial z'}{\partial z}\right|^3 \partial_{z'}^2 \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}'), & z' &= \frac{\omega(z - \zeta)}{e^{i\alpha} - \beta\omega(z - \zeta)} \\
&= \int dz'^2 \mathcal{T}(z, \bar{z}) \left|\frac{\partial z}{\partial z'}\right|^{-1} \partial_{z'}^2 \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}'), & z &= \frac{(e^{i\alpha} + \omega\zeta\beta) z' + \omega\zeta}{\omega(1 + \beta z')} \\
&= \omega \int dz'^2 \mathcal{T}(z, \bar{z}) \left|1 + \beta z'\right|^2 \partial_{z'}^2 \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}'),
\end{align*}
$$

where again (6.29) was used in the last step. Integrating by part gives

$$
\begin{align*}
\varphi &= \omega \int dz'^2 \partial_{z'}^2 (\mathcal{T}(z, \bar{z}) |1 + \beta z'|^2) \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}') \\
&= \omega \int dz'^2 \left( |1 + \beta z'|^2 \partial_{z'}^2 \mathcal{T}(z, \bar{z}) + 2\beta(1 + \bar{\beta}\bar{z}') \partial_{z'} \mathcal{T}(z, \bar{z}) \right) \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}') \\
&= \omega \int dz'^2 |1 + \beta z'|^2 \left( \frac{\partial^2 z}{\partial z'^2} \partial_z \mathcal{T}(z, \bar{z}) + \left(\frac{\partial z}{\partial z'}\right)^2 \partial_z^2 \mathcal{T}(z, \bar{z}) + \frac{\partial z}{\partial z'} \frac{2\beta}{1 + \beta z'} \partial_z \mathcal{T}(z, \bar{z}) \right) \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}') \\
&= \omega \int dz'^2 |1 + \beta z'|^2 \left(\frac{\partial z}{\partial z'}\right)^2 \partial_z^2 \mathcal{T}(z, \bar{z}) \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}') \\
&= \frac{e^{2i\alpha}}{\omega} \int dz'^2 \left(\frac{1 + \bar{\beta}\bar{z}'}{1 + \beta z'}\right)^2 \partial_z^2 \mathcal{T}(z, \bar{z}) \partial_{\bar{z}'}^2 \mathcal{N}(z', \bar{z}').
\end{align*}
$$

E Continuous-spin representations of the Poincaré group

We here briefly recap the “continuous-spin” UIRs of the Poincaré group (for more details, see e.g. the reviews [81, 82] and refs therein). Before that, it is useful to describe the UIRs of the massless little group, i.e. the UIRs of the Euclidean group (see e.g. [101, Chapter 9] for a basic introduction).

E.1 Unitary irreducible representations of the Euclidean group

The Euclidean group $ISO(2) = SO(2) \times \mathbb{R}^2$ is a semidirect product of the rotation group $SO(2)$ acting on the Abelian normal subgroup $\mathbb{R}^2$. Therefore, the corresponding UIRs are again obtained from Wigner's method of induced representations.

The first step of this procedure is to pick an eigenvalue of the translation generators of the
subgroup $\mathbb{R}^2 \subset ISO(2)$: this is nothing but a vector $\vec{\pi} \in \mathbb{R}^2$. The second step amounts to pick a
UIR of the stabiliser of this vector. Accordingly, there are two cases to distinguish:

1. $\vec{\pi} = \vec{0}$ : The degenerate case $\vec{\pi} = \vec{0}$ corresponds to representations which are unfaithful UIRs of $ISO(2)$ but which are faithful UIRs of $SO(2)$ instead. Therefore, the UIRs are labeled by a single number $\lambda \in \mathbb{R}$ and they pick a phase $e^{i\lambda\theta}$ under a rotation of angle $\theta$. This label may be called the “helicity”. This label $\lambda$ is clearly an integer for single-valued representations and a half-integer for double-valued representations of $ISO(2)$, i.e. $\lambda \in \frac{1}{2}\mathbb{Z}$ for UIRs of $ISO(2)$.

2. $\vec{\pi} \neq \vec{0}$ : The generic case $\vec{\pi} \neq \vec{0}$ corresponds to faithful UIRs of $ISO(2)$. Since the stabiliser in $SO(2)$ of a plane vector is the identity, the second step in the method of induced representations is absent. Therefore, these UIRs of $ISO(2)$ are only labeled by the norm
$\mu := |\vec{\pi}| \in \mathbb{R}^+$ (since the orbits of a vector in the plane under rotations are labeled by the radius of the vector).

Several remarks are in order because they will be instrumental for the identification of continuous-spin modes inside wavefunctions of massless BMS particles in Section 6.4.

**Remark E.1.** In practice, the UIRs in the second case are carried by square-integrable functions on the circle of radius $\mu$ in the plane of vectors $\vec{\pi}$. However, as always there are infinitely many equivalent realisations of a given UIR. For instance, the UIR can be represented by wavefunctions $\Psi(\pi_1, \pi_2)$ on the plane with support on the circle $(\pi_1)^2 + (\pi_2)^2 = \mu^2$. Equivalently, in dual space they are square-integrable solutions $\Psi(\beta_1, \beta_2)$ of the Helmholtz equation

$$ (\partial_{\beta_1}^2 + \partial_{\beta_2}^2 + \mu^2)\Psi = 0. \qquad (E.1) $$

**Remark E.2.** Furthermore, one may equivalently consider realisations of the same UIRs of ISO(2) in terms of vector (or tensor) fields on the plane. Therefore, although these fields may have non-vanishing helicities, they would nevertheless carry an equivalent UIR, by assumption. This point will be important later on. For instance, consider for a given $s \in \mathbb{N}$ the rank-$s$ symmetric tensor fields with indices taking two values, $\Psi_{i_1 \dots i_s} = \partial_{\beta_{i_1}} \dots \partial_{\beta_{i_s}} \Psi$, defined as derivatives of the square-integrable solutions $\Psi(\beta_1, \beta_2)$ of the Helmholtz equation (E.1).$^{42}$ The correspondence between the tensor $\Psi_{i_1 \dots i_s}$ and the scalar $\Psi$ is one-to-one because the latter is assumed square-integrable (thus the polynomial solutions of the homogeneous equation $\partial_{\beta_{i_1}} \dots \partial_{\beta_{i_s}} \Psi = 0$ are not admissible). Introducing the complex variable $\beta = \beta_1 + i\beta_2$, one can equivalently consider the complex field $\tilde{\Psi}_{2s} := \partial_{\beta}^s \Psi$ defined as the $s$-th complex derivative of the square-integrable solutions $\Psi(\beta, \tilde{\beta})$ of the Helmholtz equation $(4\partial_{\beta}\partial_{\tilde{\beta}} + \mu^2)\Psi = 0$. Note that this field $\tilde{\Psi}_{2s}$ transforms as $\tilde{\Psi}_{2s} \rightarrow e^{is\theta}\tilde{\Psi}_{2s}$ under the rotation $\beta \rightarrow e^{i\theta}\beta$. Therefore the field $\Psi_{2s}$ has helicity $s$ but nevertheless carries a representation of ISO(2) equivalent to the representation carried by $\Psi$.

**Remark E.3.** Under translations $\tilde{\beta}' = \tilde{\beta} + \tilde{B}$, the Fourier transform $\Psi(\vec{\pi})$ of the function $\Psi(\tilde{\beta})$ merely picks a phase

$$ \Psi'(\vec{\pi}) = e^{i\vec{\pi} \cdot \vec{B}} \Psi(\vec{\pi}), \qquad (E.2) $$

while under rotations $\tilde{\beta}' = R_{\theta}(\tilde{\beta})$, it transforms as $\Psi'(\vec{\pi}) = \Psi(\vec{\pi}')$ where $\vec{\pi}' = R_{\theta}(\vec{\pi})$. For an irreducible representation, the function $\Psi(\vec{\pi})$ on the plane actually has support on the circle $S_{\mu}^1$ of radius $\mu$ in the plane, i.e. the circle $\vec{\pi}^2 = \mu^2$. It belongs to the Hilbert space $L^2(S_{\mu}^1)$ of square-integrable functions on the circle $S_{\mu}^1 \subset \mathbb{R}^2$. One can alternatively consider wavefunctions $\Psi(\vec{\pi})$ belonging to the Hilbert space $L^2(\mathbb{R}^2)$ of square-integrable functions on the plane and introduce polar coordinates $(\mu, \alpha)$ on the latter. In particular, a wavefunction $\Psi(\vec{\pi})$ with support on the circle $S_{\mu}^1$ is equivalent to a function $\Psi(\alpha; \vec{\pi}_0)$ of an angle only, while $\vec{\pi}_0$ is a choice of reference vector on the circle. The angle $\alpha$ is defined by $\vec{\pi} = R_{\alpha}(\vec{\pi}_0)$. Accordingly, rotations act by mere shifts $\alpha' = \alpha + \theta$ of the angle while translations (E.2) act as

$$ \Psi'(\alpha; \vec{\pi}_0) = e^{iR_{\alpha}(\vec{\pi}_0) \cdot \vec{B}} \Psi(\alpha; \vec{\pi}_0) \qquad (E.3) $$

This is again an equivalent realisation of the same UIR of ISO(2).

<footnote>$^{42}$This corresponds to the usual realisation of UIRs of ISO(2) in terms of symmetric tensor fields $\Phi_{i_1 \dots i_s}(\beta_1, \beta_2)$ which are divergenceless $\partial_{\beta_i} \Phi^{i}_{j_1 \dots j_{s-1}} = 0$ and solutions of the Helmholtz equation $(\partial_{\beta_1}^2 + \partial_{\beta_2}^2 + \mu^2)\Phi_{i_1 \dots i_s} = 0$. Hodge dualising all indices implies that $\Psi_{i_1 \dots i_s} := \epsilon_{i_1 j_1} \dots \epsilon_{i_{s_j} j_s} \Phi^{j_1 \dots j_s}$ obeys to the closure condition $\partial_{[i} \Psi_{j]k_1 \dots k_{s-1}} = 0$. By Poincaré lemma, this is equivalent to the exactness condition $\Psi_{i_1 \dots i_s} = \partial_{\beta_{i_1}} \dots \partial_{\beta_{i_s}} \Psi$.</footnote>
**Remark E.4.** The Hilbert space $L^2(S_\mu^1)$ of square-integrable functions on the circle is of course separable<sup>43</sup> (cf. the orthonormal basis of Fourier modes), although this is not manifest in terms of the plane-wave basis of solutions $e^{i\pi\cdot\beta}$ of the Helmholtz equation (E.1) (see e.g. [101, Sections 9.2-9.3] for the explicit relations between these two bases). The space $L^2(\mathbb{R}^2)$ of square-integrable functions on the plane is a separable Hilbert space which carries a unitary but reducible representation of $ISO(2)$. This representation obviously decomposes into a direct integral of UIRs of $ISO(2)$ as follows

$$L^2(\mathbb{R}^2) = \int_0^\infty d\mu L^2(S_\mu^1) \qquad (E.4)$$

Let us stress that $L^2(\mathbb{R}^2)$ is a separable Hilbert space (cf. the harmonic oscillator basis), although this may look in tension with the above decomposition since there is a continuous sum of UIRs on the right-hand-side. Nevertheless, there is no contradiction because the decomposition (E.4) is only natural in the plane-wave basis $e^{i\pi\cdot\beta}$ whose elements are not normalisable (so they do no provide an orthonormal basis which would be in contradiction with separability, cf. Footnote 43).

## E.2 Massless representations of the Poincaré group

The Poincaré little group of a null momentum $p_\mu$ is isomorphic to the Euclidean group, $\ell_p \simeq ISO(2)$. The step 2 in Theorem 4.1 implies that the massless UIRs of $ISO_0(3, 1)$ are induced from UIRs of the Euclidean subgroup $ISO(2) \subset SO_0(3, 1)$. Accordingly, one may introduce a vector $\pi$ describing the spinning degrees of freedom. As explained above, there are two cases to distinguish:

1. $\pi = 0$: These UIRs of the little group $ISO(2)$ are unfaithful. In fact, they are genuine UIRs of $SO(2)$ labeled by a single integer or half-integer $\lambda \in \frac{1}{2}\mathbb{Z}$ according to whether the representation is single or double valued. They induce UIRs of the Poincaré group $ISO_0(3, 1)$ which are the usual massless representations, where $\lambda$ is the helicity.

2. $\pi \neq 0$: These UIRs of $ISO(2)$ are labeled by the norm $\mu = |\pi| \in \mathbb{R}^+$. The corresponding UIRs of the Poincaré group $ISO_0(3, 1)$ are the continuous-spin representations. Their name originates from the fact that they are labeled by a continuous parameter: $\mu$. The only extra information is whether the UIR of $ISO_0(3, 1)$ is single or double valued.

<footnote n="43"><sup>43</sup> A Hilbert space is separable if and only if it has one countable orthonormal basis (see e.g. [57, Theorem 5.11] and [102, Theorem 3.52]). Actually, there is more: a Hilbert space is separable if and only if every orthonormal basis is countable (see e.g. [103, Problems 7 and 17]).</footnote>
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