Harish-Chandra character. We shall also use the heuristic notations

$\displaystyle\Theta_{{\cal F}_{\Delta,s}}(g)\equiv\operatorname{Tr}_{{\cal
F}_{\Delta,s}}g$ (5.1)

which only exist in the distribution sense. In this section, we will mainly
focus on $g=q^{D}$, namely $\Theta_{{\cal
F}_{\Delta,s}}(q)\equiv\Theta_{\Delta,s}(q^{D})$. The more general case, with
$q^{D}$ being decorated by angular momenta, i.e. $\Theta_{{\cal
F}_{\Delta,s}}(q,\bm{u})\equiv\Theta_{\Delta,s}(q^{D}e^{u_{1}M_{12}}e^{u_{2}M_{34}}\cdots)$,
will be discussed in appendix G.

###  5.2 Compute $\Theta_{{\cal F}_{\Delta,s}}(q)$

In this subsection, we show how to compute the character $\Theta_{{\cal
F}_{\Delta,s}}(q)$. Let’s start from the $s=0$ case. By definition, one would
find an orthonormal basis of ${\cal F}_{\Delta}$ and compute the matrix
element of $D$ with respect to this basis (which is actually done in the
appendix E with the basis being spherical harmonics). Then exponentiate the
infinite dimensional matrix $D$ and compute the trace. However, we can make
life much easier by directly using the ket basis $|x\rangle$. The action of
$q^{D}$ on $|x\rangle$ is

$\displaystyle q^{D}|x\rangle=q^{\bar{\Delta}}\left|qx\right\rangle=\int
d^{d}y\,q^{\bar{\Delta}}\delta^{d}(qx-y)|y\rangle$ (5.2)

from which we can directly read off the matrix elements of $q^{D}$ with
respect to the basis $|x\rangle$. The integral of the diagonal entries of
$q^{D}$ is localized on the fixed points of $q^{D}$, i.e. $0$ and $\infty$ or
equivalently the southern and northern poles when conformally mapped to
$S^{d}$. But the $x^{i}$ coordinate system only covers the $x=0$ point and
hence we would not be able to correctly capture the contribution from $\infty$
if we naively computed $\int\,d^{d}x\,q^{\Delta}\delta^{d}(qx-x)$. We can
bypass this problem by conjugating $q^{D}$ with special conformal
transformations, i.e. $q^{D}\to e^{-b\cdot K}q^{D}e^{b\cdot K}$. This
conjugation maps the two fixed points $0$ and $\infty$ to $0$ and
$\frac{b^{i}}{b^{2}}$ respectively. More explicitly, the action of $e^{-b\cdot
K}q^{D}e^{b\cdot K}$ on $|x\rangle$ is given by

$\displaystyle e^{-b\cdot K}q^{D}e^{b\cdot
K}|x\rangle=\Omega(q,x,b)^{\bar{\Delta}}|\tilde{x}(q,x,b)\rangle,\,\,\,\,\,\tilde{x}^{i}=q\frac{x^{i}+\left(q-1\right)x^{2}b^{i}}{1+2(q-1)\,x\cdot
b+\left(q-1\right)^{2}b^{2}x^{2}}$ (5.3)

where

$\displaystyle\Omega(q,x,b)=\frac{q}{1+2(q-1)\,x\cdot
b+\left(q-1\right)^{2}b^{2}x^{2}}$ (5.4)

Therefore the character $\Theta_{{\cal F}_{\Delta}}(q)$ should be

$\displaystyle\Theta_{{\cal F}_{\Delta}}(q)=\int
d^{d}x\,\Omega(q,x,b)^{\bar{\Delta}}\delta^{d}(\tilde{x}-x)$ (5.5)

This integral is localized at $x^{i}=0$ and $x^{i}=\frac{b^{i}}{b^{2}}$. At
these points, the scaling factor $\Omega(q,x,b)$ becomes

$\displaystyle\Omega(q,0,b)=q,\,\,\,\,\,\Omega(q,b^{i}/b^{2},b)=q^{-1}$ (5.6)

and the Jacobian associated to the map $x^{i}\to\tilde{x}^{i}-x^{i}$ becomes
$|q-1|^{d}$ and $|1-q^{-1}|^{d}$ respectively. Altogether, the integral in eq.
(5.5) yields

$\displaystyle\Theta_{{\cal
F}_{\Delta}}(q)=\frac{q^{\bar{\Delta}}}{|q-1|^{d}}+\frac{q^{-\bar{\Delta}}}{|1-q^{-1}|^{d}}=\frac{q^{\Delta}+q^{\bar{\Delta}}}{|1-q|^{d}}$
(5.7)

where the $b$ dependence drops out explicitly as expected from the conjugation
invariance.

######  Remark 5.4.

The character $\Theta_{{\cal F}_{\Delta}}(q)$ given by eq. (5.7) is symmetric
under $q\to q^{-1}$. This property, which holds for all ${\cal
F}_{\Delta,\bm{s}}$, is a result of the conjugation invariance of $\Theta$
since $D=L_{0,d+1}$ is mapped to $-D$ by the conjugation of $e^{i\pi
L_{d+1,i}}$. In view of this, we assume $\boxed{0<q<1}$ in the remaining part
of this section, without loss of generality.

The generalization of our computation to the spinning case is almost
straightforward. Let’s write the ket basis of ${\cal F}_{\Delta,s}$ as
$|x\rangle_{\alpha}$ where the index $\alpha$ carries the spin-$s$
representation of ${\text{SO}}(d)$. The action of $e^{-b\cdot
K}e^{tD}e^{b\cdot K}$ on this basis can be schematically expressed as

$\displaystyle e^{-b\cdot K}q^{D}e^{b\cdot K}|x\rangle_{\alpha}={\cal
O}_{\alpha\beta}(q,x,b)\Omega(q,x,b)^{\bar{\Delta}}|\tilde{x}(q,x,b)\rangle_{\beta}$
(5.8)

where ${\cal O}_{\alpha\beta}(q,x,b)$ is an ${\text{SO}}(d)$ rotation matrix
in the spin-$s$ representation. In general, ${\cal O}_{\alpha\beta}(q,x,b)$
takes a very sophisticated form but while evaluated at the two fixed points,
i.e. $x^{i}=0$ and $x^{i}=\frac{b^{i}}{b^{2}}$, it becomes an identity matrix.
Altogether, we have

$\displaystyle\Theta_{{\cal F}_{\Delta,s}}(q)$
$\displaystyle=\sum_{\alpha}\int d^{d}x\,{\cal
O}_{\alpha\alpha}(q,x,b)\Omega(q,x,b)^{\bar{\Delta}}\delta^{d}(\tilde{x}-x)$
$\displaystyle=\Theta_{{\cal
F}_{\Delta}}(q)\times\left(\sum_{\alpha}\,1\right)=D^{d}_{s}\,\frac{q^{\Delta}+q^{\bar{\Delta}}}{(1-q)^{d}}$
(5.9)

where $D^{d}_{s}$ is the dimension of the spin-$s$ representation of
${\text{SO}}(d)$. Thus
$D^{d}_{s}\,\frac{q^{\Delta}+q^{\bar{\Delta}}}{(1-q)^{d}}$ is the Harish-
Chandra character for both principal and complementary series.

###  5.3 Harish-Chandra characters of exceptional series

Now let’s proceed to compute the Harish-Chandra characters of exceptional
series ${\cal V}_{s,0}$ and ${\cal U}_{s,t}$. For ${\cal V}_{s,0}$, recall
that ${\cal V}_{s,t}\equiv{\rm Im\,}(z\cdot\partial_{x})^{s-t}\cong{\cal
F}_{1-s,t}/\ker(z\cdot\partial_{x})^{s-t}$, which yields

$\displaystyle\Theta_{{\cal V}_{s,0}}(q)=\Theta_{{\cal
F}_{1-s}}(q)-\Theta_{\ker(z\cdot\partial_{x})^{s}}(q)$ (5.10)

Since $\ker(z\cdot\partial_{x})^{s}$ carries the ${\mathbb{Y}}_{s-1}$ of
${\text{SO}}(1,d+1)$, $\Theta_{\ker(z\cdot\partial_{x})^{s}}(q)$ is nothing
but the usual ${\text{SO}}(d+2)$ character corresponding to the highest weight
representation ${\mathbb{Y}}_{s-1}$, denoted by
$\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1}}(q)$. Thus we obtain

$\displaystyle\Theta_{{\cal
V}_{s,0}}(q)=\frac{q^{1-s}+q^{d+s-1}}{(1-q)^{d}}-\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1}}(q)$
(5.11)

Similarly for ${\cal U}_{s,t}$, using the isomorphism ${\cal
U}_{s,t}\cong{\cal F}_{1-t,s}/{\cal V}_{s,t}$, we obtain

$\displaystyle\Theta_{{\cal
U}_{s,t}}(q)=D^{d}_{s}\frac{q^{1-t}+q^{d+t-1}}{(1-q)^{d}}-D^{d}_{t}\frac{q^{1-s}+q^{d+s-1}}{(1-q)^{d}}+\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1,t}}(q)$
(5.12)

In general, one can write out the characters like
$\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1,t}}(q)$ explicitly by using Weyl
character formula [53]. However, in order to compare with the paper [28], we
derive a slightly different expression (c.f. eq. (H.3)) for
$\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1,t}}(q)$ in appendix H. Plugging
this new expression into eq. (5.11) and eq. (5.12) yields

$\displaystyle\Theta_{{\cal
V}_{s,0}}(q)=\left[\frac{q^{1-s}+q^{d+s-1}}{(1-q)^{d}}\right]_{+},\,\,\,\,\,\Theta_{{\cal
U}_{s,t}}(q)=\left[D^{d}_{s}\frac{q^{1-t}+q^{d+t-1}}{(1-q)^{d}}-D^{d}_{t}\frac{q^{1-s}+q^{d+s-1}}{(1-q)^{d}}\right]_{+}$
(5.13)

where the “flipping” operator $[\,\,\,]_{+}$ acts on an arbitrary Laurent
series $\sum_{k}c_{k}q^{k}$ as

$\displaystyle\left[\sum_{k}c_{k}q^{k}\right]_{+}=\sum_{k<0}(-c_{k})q^{-k}+\sum_{k>0}c_{k}q^{k}$
(5.14)

We have checked that Harish-Chandra characters $\Theta_{{\cal F}_{\Delta,s}}$,
$\Theta_{{\cal V}_{s,0}}$ and $\Theta_{{\cal U}_{s,t}}$ agree with the results
in literature [38, 39, 50]. Some lower $d$ examples are

$\displaystyle d=3:\,\,\Theta_{{\cal
U}^{\pm}_{s,t}}(q)=\frac{(2s+1)q^{t+2}-(2t+1)q^{s+2}}{(1-q)^{3}}$
$\displaystyle d=4:\,\,\Theta_{{\cal
U}_{s,t}}(q)=2(s-t)(s+t+2)\frac{q^{2}}{(1-q)^{4}}=\left(D^{d}_{s,t+1}+D^{d}_{s,-(t+1)}\right)\frac{q^{2}}{(1-q)^{4}}$
(5.15)

where $D^{d}_{s,t+1}=D^{d}_{s,-(t+1)}=(s-t)(s+t+2)$. In contrast to the $d=3$
case, the small $q$ expansion of $\Theta_{{\cal U}_{s,t}}(q)$ for
${\text{SO}}(1,5)$ starts with a quadratic term, independent of the spin or
depth. Actually, by using eq. (5.13), one can check that this property
persists in any higher dimensions

$\displaystyle d\geq 5:\,\,\Theta_{{\cal U}_{s,t}}(q)=D^{d}_{s,t+1}q^{2}+{\cal
O}(q^{3})$ (5.16)

In [30], we give a physical explanation for this difference between $d=3$ and
$d\geq 4$, by counting quasinormal modes of spin-$s$ gauge fields in the
static path of $\text{dS}_{d+1}$ 111111The static patch metric is given by
$ds^{2}=-(1-r^{2})dt^{2}+\frac{dr^{2}}{1-r^{2}}+r^{2}d\Omega^{2}_{d-1}$, where
$0\leq r\leq 1$. The dS horizon is at $r=1$. A quasinormal mode in the static
patch is defined as an on-shell mode with the in-falling boundary condition
$e^{-i\omega(t-\rho)}$ near horizon, where $\rho=\tanh^{-1}(r)$.. More
generally, we show in [30] that Harish-Chandra characters encode precisely de
Sitter quasinormal modes as follows. Let $\varphi$ be a unitary field in
$\text{dS}_{d+1}$ and let $R$ be the corresponding ${\text{SO}}(1,d+1)$ UIR.
Let $\Sigma=\\{\omega_{QN}\\}$ be the set of quasinormal frequencies of
$\varphi$ and $d_{\omega_{QN}}$ be the degeneracy of quasinormal modes with
quasinormal frequency $\omega_{QN}$. Then the Harish-Chandra character
$\Theta_{R}(q)$ can be recovered as

$\displaystyle\Theta_{R}(q)=\sum_{\omega_{QN}\in\Sigma}\,d_{\omega_{QN}}\,q^{i\omega_{QN}},\,\,\,\,\,0<q<1$
(5.17)

where the condition $0<q<1$ is necessary for the infinite sum over
$\omega_{QN}$ to be convergent because ${\rm Im\,}(\omega_{QN})>0$. These
quasinormal modes are eigenmodes of the hermitian operator $-iD$ but with
complex eigenvalues $\omega_{QN}$. They do not belong to the single-particle
Hilbert space of $\varphi$. Instead, they appear as resonances [28].

######  Remark 5.5.

For the discrete series of ${\text{SO}}(1,2)$, we use the quotient space
realization ${\cal D}^{+}_{N}\bigoplus{\cal D}^{-}_{N}={\cal F}_{1-N}/P_{N}$
where $P_{N}$ carries the spin-($N-1$) representation of ${\text{SO}}(1,2)$

$\displaystyle\Theta_{{\cal D}^{+}_{N}\bigoplus{\cal
D}^{-}_{N}}(q)=\Theta_{{\cal
F}_{1-N}}(q)-\sum_{k=1-N}^{N-1}q^{k}=\frac{2\,q^{N}}{1-q}$ (5.18)

The character for each summand ${\cal D}^{\pm}_{N}$ is $\frac{q^{N}}{1-q}$
[53].

######  Remark 5.6.

In the $d=2$ case, according to footnote 9, we should include
$\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s-1,-t}}(q)$ on the R.H.S of eq.
(5.12) when $t\geq 1$:

$\displaystyle t\geq 1:\,\,\,\,\,\Theta_{{\cal
U}_{s,t}}=\frac{2q}{(1-q^{2})}\left(q^{t}+q^{-t}-q^{s}-q^{-s}\right)+\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s-1,t}}(q)+\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s-1,-t}}(q)=0$
(5.19) $\displaystyle t=0:\,\,\,\,\,\Theta_{{\cal
U}_{s,0}}=\frac{4\,q-q^{1+s}-q^{1-s}}{(1-q)^{2}}+\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s-1}}=\frac{2\,q}{(1-q)^{2}}$
(5.20)

where we have used
$\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s-1,t}}(q)=\frac{q}{(1-q^{2})}\left(q^{s}+q^{-s}-q^{t}-q^{-t}\right)$
and
$\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s-1}}(q)=\frac{q}{(1-q)^{2}}\left(q^{s}+q^{-s}-2\right)$.
The vanishing of $\Theta_{{\cal U}_{s,t}}$ for $t\geq 1$ is expected due to
remark 4.4.

##  6 ${\text{SO}}(1,d+1)$ vs. ${\text{SO}}(2,d)$

In this section, we will comment on distinctions between UIRs of
${\text{SO}}(2,d)$ and ${\text{SO}}(1,d+1)$. ${\text{SO}}(2,d)$ is the
isometry group of $(d+1)$ dimensional AdS and the conformal group of
$\mathbb{R}^{1,d-1}$. It shares the same Lie algebra as ${\text{SO}}(1,d+1)$
but with different reality conditions (see, for example [76])

$\displaystyle
D^{\dagger}=D,\,\,\,\,\,P_{i}^{\dagger}=K_{i},\,\,\,\,\,M_{ij}^{\dagger}=-M_{ij}$
(6.1)

$D$ generates time translation in the global coordinates of AdS, playing the
role of a Hamiltonian. So the physically relevant representations of
${\text{SO}}(2,d)$ are the lowest-weight ones, where the Hamiltonian $D$ is
bounded from below. These representations are fixed by a primary state and the
corresponding representation spaces are spanned by primary states together
with their descendants. In contrast, primary states in the dS sense (e.g.
$|\bar{\Delta},0\rangle_{i_{1}\cdots i_{s}}$ in e.q. (4)) are not normalizable
and do not belong to the representation spaces ${\cal F}_{\Delta,s}$. The
${\text{SO}}(1,d+1)$ UIRs given by the list in section 4.7 are not lowest-
weight representations. We also want to mention that ${\text{SO}}(2,d)$ does
have UIRs that are not lowest-weight representations, for example principal
series representations, which describe continuous-spin fields [77, 78].

Since $D$ is bounded from below, the most natural way to define an
${\mathfrak{so}}(2,d)$ character is [79, 39]121212One can also turn on the
${\text{SO}}(d)$ generators, i.e.
$\chi_{R}(q,\bm{x})=\operatorname{Tr}_{R}q^{D}x_{1}^{L_{12}}x_{2}^{L_{34}}\cdots$

$\displaystyle\chi_{R}(q)\equiv\operatorname{Tr}_{R}q^{D},\,\,\,\,\,0<|q|<1$
(6.2)

where $R$ is a lowest-weight representation of ${\text{SO}}(2,d)$. Compared to
the Harish-Chandra character of ${\text{SO}}(1,d+1)$, the trace in (6.2) makes
perfect sense given $0<|q|<1$ but $\chi_{R}(q)$ fails to be a group character
because $q^{D}$ is not an element of ${\text{SO}}(2,d)$ for $0<|q|<1$. Some
examples of the ${\mathfrak{so}}(2,d)$ characters are [39]:

  * •

The scalar representation of scaling dimension
$\Delta>\max\\{0,\frac{d-2}{2}\\}$, which describes a scalar field in
$\text{AdS}_{d+1}$ of mass $m^{2}=\Delta(\Delta-d)$:

$\displaystyle\chi_{\Delta}(q)=\frac{q^{\Delta}}{(1-q)^{d}}$ (6.3)

It takes a similar form as the Harish-Chandra character $\Theta_{{\cal
F}_{\Delta}}(q)$ of ${\text{SO}}(1,d+1)$, except the absence of the
$q^{\bar{\Delta}}$ term. The difference arises from the fact that in AdS we
should choose either standard quantization or alternate quantization.

  * •

The spin-$s$ representation of scaling dimension $\Delta=d+s-2$, which
describes a massless spin-$s$ gauge field in $\text{AdS}_{d+1}$ or a spin-$s$
conversed current in boundary $\text{CFT}_{d}$:

$\displaystyle\chi_{s}(q)=\frac{D^{d}_{s}q^{d+s-2}-D^{d}_{s-1}q^{d+s-1}}{(1-q)^{d}}$
(6.4)

It is very different from its ${\text{SO}}(1,d+1)$ counterpart

$\displaystyle\Theta_{{\cal
F}_{s,s-1}}(q)=\chi_{s}(q)+\frac{D_{s}^{d}q^{2-s}-D^{d}_{s-1}q^{1-s}}{(1-q)^{d}}+\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1,s-1}}(q)$
(6.5)

In [30], we attribute the discrepancy between $\Theta_{{\cal F}_{s,s-1}}(q)$
and $\chi_{s}(q)$ to gauge-invariant quasinormal modes sourced by a boundary
spin-$s$ gauge field.

## Acknowledgments

I am grateful to Frederik Denef, Austin Joyce and Bob Penna for discussions
and explanations. I especially thank Dionysios Anninos, Frederik Denef and
Albert Law for a thorough reading and comments on the draft. Z.S. is supported
by NSF grant PHY-1914860 and the Gravity Initiative at Princeton University.

##  Appendix A UIRs of ${\text{SL}}(2,\mathbb{R})$

In this appendix, we review the UIRs of ${\text{SL}}(2,\mathbb{R})$, following
the constructions in section 3.1. The group ${\text{SL}}(2,\mathbb{R})$
consists of real $2\times 2$ matrices with determinant 1. Its Lie algebra
$\mathfrak{sl}(2,\mathbb{R})$ is isomorphic to ${\mathfrak{so}}(1,2)$. The
latter is spanned by three generators $\\{P,K,D\\}$, satisfying the
commutation relation (3.1). We will also use them as a basis of
$\mathfrak{sl}(2,\mathbb{R})$. In the defining representation of
${\text{SL}}(2,\mathbb{R})$, the generators $\\{P,K,D\\}$ can be realized as
the following $2\times 2$ matrices

$\displaystyle P=\begin{pmatrix}0&-1\\\
0&0\end{pmatrix},\,\,\,\,\,K=\begin{pmatrix}0&0\\\
1&0\end{pmatrix},\,\,\,\,\,D=\frac{1}{2}\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}$ (A.1)

With this choice, $L_{0}=-\frac{i}{2}(P+K)=-\frac{\sigma_{2}}{2}$, where
$\sigma_{2}$ is the second Pauli matrix.

For ${\text{SO}}(1,2)$, $e^{2\pi iL_{0}}$ is the identity element and hence
the spectrum of $L_{0}$ must be integers in any ${\text{SO}}(1,2)$
representation. In contrast, $e^{2\pi iL_{0}}=-\small\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$ in ${\text{SL}}(2,\mathbb{R})$. As an involutory central
element, $e^{2\pi iL_{0}}$ is either 1 (even “parity”) or $-1$ (odd “parity”)
in an irreducible representation of ${\text{SL}}(2,\mathbb{R})$. The UIRs of
${\text{SL}}(2,\mathbb{R})$ with an even parity are equivalent to UIRs of
${\text{SO}}(1,2)$. Since the latter have been analyzed extensively in section
3, we will focus on parity odd UIRs in this appendix. In these
representations, the eigenvalues of $L_{0}$ are half-integers. Denote the
eigenstates of $L_{0}$ by $|r\rangle$, i.e. $L_{0}|r\rangle=r|r\rangle$, where
$r\in\mathbb{Z}+\frac{1}{2}$. The action of $L_{\pm}$ on these states are
given by (3.5) and (3.6), with $n$ replaced by $r$:

$\displaystyle L_{\pm}|r\rangle=(r\pm\Delta)|r\pm 1\rangle$ (A.2)

The reality condition $L^{\dagger}_{\pm}=L_{\mp}$ implies

$\displaystyle\frac{\langle r+1|r+1\rangle}{\langle
r|r\rangle}=\frac{r+\bar{\Delta}}{r+\Delta^{*}}$ (A.3)

For $\lambda_{r}\equiv\frac{r+\bar{\Delta}}{r+\Delta^{*}}$ to be real,
$\Delta$ lies on either the real line or $\frac{1}{2}+i\mathbb{R}$, i.e.

$\displaystyle(i)\,\Delta=\frac{1}{2}+i\mu,\,\,\mu\in\mathbb{R},\,\,\,\,\,(ii)\,\Delta\in\mathbb{R}$
(A.4)

The positivity of $\lambda_{r}$ requires separate discussions for different
values of $\Delta$:

  * •

$\lambda_{r}$ is identically equal to 1 when
$\Delta=\frac{1}{2}+i\mu,\mu\not=0$. We can consistently choose $\langle
r|r\rangle=1$ for all $r\in\mathbb{Z}+\frac{1}{2}$ and hence the resulting
representation, denoted by ${\cal P}^{-}_{\Delta}$, is unitary. ${\cal
P}^{-}_{\Delta}$ is a parity odd principal series representation. The
irreducibility of ${\cal P}^{-}_{\Delta}$ is shown pictorially in fig. (A.1).
${\cal P}^{-}_{\Delta}$ and ${\cal P}^{-}_{\bar{\Delta}}$ are equivalent since
we can easily generalize the shadow transformation, c.f. (3.13), to the parity
odd case.

${\cdots}$${\left|-\frac{5}{2}\right\rangle}$${\left|-\frac{3}{2}\right\rangle}$${\left|-\frac{1}{2}\right\rangle}$${\left|\frac{1}{2}\right\rangle}$${\left|\frac{3}{2}\right\rangle}$${\left|\frac{5}{2}\right\rangle}$${\cdots}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$
Figure A.1: The action of $L_{\pm}$ in principal series ${\cal
P}^{-}_{\Delta}$.

  * •

When $\Delta\in\mathbb{R}$ but $\Delta\notin\mathbb{Z}+\frac{1}{2}$, $r$ takes
value in all half-integers and we need

$\displaystyle\lambda_{r}=\frac{(r+\frac{1}{2})^{2}-(\Delta-\frac{1}{2})^{2}}{(r+\Delta)^{2}}>0$
(A.5)

to hold for all $r\in\mathbb{Z}+\frac{1}{2}$. However,
$\lambda_{-\frac{1}{2}}<0$. So there does not exist unitary complementary
series with odd parity. The physical reason for the absence of these
representations is that the principal series representation ${\cal
P}^{-}_{\frac{1}{2}+i\mu}$ describes a Dirac fermion of mass $\mu$ in
$\text{dS}_{2}$. The Wick rotation $\mu\to i\mu$ yields a fermion with an
imaginary mass, spoiling the hermiticity of the Dirac action.

  * •

When $\Delta\in\mathbb{N}+\frac{1}{2}$, the vector space spanned by all
$|r\rangle$ is reducible, c.f. diagram (A.2), and contains two irreducible
$\mathfrak{sl}(2,\mathbb{R})$-invariant subspaces:

$\displaystyle\\{|r\rangle\\}_{r\geq\Delta}:\text{carries a lowest-weight
representation}\,\,{\cal D}^{-+}_{\Delta}$
$\displaystyle\\{|r\rangle\\}_{r\leq-\Delta}:\text{carries a highest-weight
representation}\,\,{\cal D}^{--}_{\Delta}$ (A.6)

where the first “-” in the superscripts of ${\cal D}^{-\pm}_{\Delta}$
indicates that these representations are parity odd.

${\cdots}$${\left|-\frac{7}{2}\right\rangle}$${\left|-\frac{5}{2}\right\rangle}$${\left|-\frac{3}{2}\right\rangle}$${\left|-\frac{1}{2}\right\rangle}$${\left|\frac{1}{2}\right\rangle}$${\left|\frac{3}{2}\right\rangle}$${\left|\frac{5}{2}\right\rangle}$${\left|\frac{7}{2}\right\rangle}$${\cdots}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$
Figure A.2: The action of $L_{\pm}$ for $\Delta=\frac{3}{2}$.
$\\{|\frac{3}{2}\rangle,|\frac{5}{2}\rangle,\cdots\\}$ furnishes an
irreducible lowest-weight representation and
$\\{|-\frac{3}{2}\rangle,|-\frac{5}{2}\rangle,\cdots\\}$ furnishes an
irreducible highest-weight representation.

It is clear that all $\lambda_{r}$ are positive when restricted to these two
subspaces. Thus ${\cal D}^{-\pm}_{\Delta}$ are also UIRs, belonging to
discrete series. In each ${\cal D}^{-\pm}_{\Delta}$, we choose the
normalization to be

$\displaystyle\langle r|r\rangle_{{\cal D}^{-\pm}_{\Delta}}=\frac{\Gamma(\pm
r+\bar{\Delta})}{\Gamma(\pm r+\Delta)}$ (A.7)

  * •

When $\Delta\in\mathbb{Z}_{<0}+\frac{1}{2}$, the vector space spanned by all
$|r\rangle$ contains only one irreducible
$\mathfrak{sl}(2,\mathbb{R})$-invariant subspace which furnishes a finite
dimensional nonunitary representation. Fig. (A.3) is an example for
$\Delta=-\frac{1}{2}$.

${\cdots}$${\left|-\frac{5}{2}\right\rangle}$${\left|-\frac{3}{2}\right\rangle}$${\left|-\frac{1}{2}\right\rangle}$${\left|\frac{1}{2}\right\rangle}$${\left|\frac{3}{2}\right\rangle}$${\left|\frac{5}{2}\right\rangle}$${\cdots}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$$\scriptstyle{L_{+}}$$\scriptstyle{L_{-}}$
Figure A.3: The action of $L_{\pm}$ when $\Delta=-\frac{1}{2}$. In this case,
$|\pm\frac{1}{2}\rangle$ span the spin-$\frac{1}{2}$ representation of
${\text{SL}}(2,\mathbb{R})$.

Altogether, the UIRs of ${\text{SL}}(2,\mathbb{R})$ (up to unitary
equivalence) can be summarized as follows:

  * •

The trivial representation.

  * •

For each $\Delta\in\frac{1}{2}+i\mathbb{R}_{>0}$, there exist two inequivalent
principal series representations ${\cal P}_{\Delta}$ and ${\cal
P}^{-}_{\Delta}$, with opposite parities. For $\Delta=\frac{1}{2}$, only the
parity even one, i.e. ${\cal P}_{\frac{1}{2}}$, exists.

  * •

For each $\frac{1}{2}<\Delta<1$, there exists one complementary series
representation ${\cal C}_{\Delta}$, with an even parity.

  * •

For each $\Delta\in\frac{1}{2}\mathbb{Z}_{>0}$, there exist two inequivalent
discrete series representations, which are complex conjugate of each other.
When $\Delta\in\mathbb{Z}_{>0}$, the two representations ${\cal
D}^{\pm}_{\Delta}$ are parity even, and when
$\Delta\in\mathbb{N}+\frac{1}{2}$, the two representations ${\cal
D}^{-\pm}_{\Delta}$ are parity odd.

In this list, the $\Delta=\frac{1}{2}$ point is very special because it
represents three inequivalent UIRs, one in the principal series with an even
parity, which describes a scalar field of mass $\frac{1}{2}$ in
$\text{dS}_{2}$, and two in the discrete series with an odd parity, which
describe the left-handed and right-handed Majorana-Weyl fermions in
$\text{dS}_{2}$.

##  Appendix B The explicit action of $\mathfrak{L}_{N}$ on
$\psi^{(N)}_{n}(x)$

In the section 3.2.4, we have defined an operator ${\mathfrak{L}}_{N}:{\cal
F}^{+}_{N}\to{\cal F}^{+}_{1-N}$ such that the composition
$\partial_{x}^{2N-1}\circ{\mathfrak{L}}_{N}$ gives the identity operator on
${\cal F}^{+}_{N}$. In this appendix, we aim to compute the action of
${\mathfrak{L}}_{N}$ on the basis $\psi^{(N)}_{n}(x)$ of ${\cal F}^{+}_{N}$,
c.f. eq.(3.42)

$\displaystyle\left({\mathfrak{L}}_{N}\psi^{(N)}_{n}\right)(x)=\frac{1}{\Gamma(2N-1)}\int_{\mathbb{R}+i\epsilon}\frac{dy}{2\pi
i}\,(x-y)^{2(N-1)}\log(x-y)\psi^{(N)}_{n}(y)$ (B.1)

(a) Original contour

(b) Deformed contour

Figure B.1: The contour deformation for the $y$-integral in eq. (B.1). The
black wavy line denotes the branch cut of $\log(x-y)$.

As shown in the fig. (B.1), we can deform the contour to go around the branch
cut and then $\log(x-y)$ is replaced by its discontinuity along the branch cut

$\displaystyle\left({\mathfrak{L}}_{N}\psi^{(N)}_{n}\right)(x)=-\frac{2^{N}}{\sqrt{2\pi}\,\Gamma(2N-1)}\int_{x}^{\infty}dy\,(y-x)^{2(N-1)}\frac{(1-iy)^{n-N}}{(1+iy)^{n+N}}$
(B.2)

Next, we rewrite $(1-iy)^{n-N}$ as a polynomial of $1+iy$ and evaluate the
$y$-integral for each monomial by using integration by part repeatedly

$\displaystyle\left({\mathfrak{L}}_{N}\psi^{(N)}_{n}\right)(x)$
$\displaystyle=\frac{(-)^{n-N+1}2^{N}}{\sqrt{2\pi}\Gamma(2N-1)}\sum_{k=0}^{n-N}(-2)^{k}\binom{n-N}{k}\int_{x}^{\infty}dy\,\frac{(y-x)^{2(N-1)}}{(1+iy)^{2N+k}}$
$\displaystyle=\frac{2^{N}i}{\sqrt{2\pi}}\frac{(-)^{n+1}}{1+ix}\sum_{k=0}^{n-N}\binom{n-N}{k}\frac{k!}{(2N+k-1)!}\left(\frac{-2}{1+ix}\right)^{k}$
(B.3)

To compute the remaining series of $k$, we can extend the sum to $k=1-2N$ and
subtract these extra terms by hand. The sum from $k=1-2N$ to $k=n-N$ is
nothing but a binomial expansion. Altogether, we obtain

$\displaystyle\left({\mathfrak{L}}_{N}\psi^{(N)}_{n}\right)(x)=i(-)^{N+1}\frac{\Gamma(n-N+1)}{\Gamma(N+n)}\psi^{(1-N)}_{n}(x)+\text{Pol}_{N,n}(x)$
(B.4)

where $\text{Pol}_{N,n}(x)$ is a polynomial in $x$ annihilated by
$\partial_{x}^{2N-1}$

$\displaystyle\text{Pol}_{N,n}(x)=\frac{2^{N-1}(-)^{n+1}i}{\sqrt{2\pi}}\frac{\Gamma(n-N+1)}{\Gamma(N+n)}\sum_{k=0}^{2(N-1)}\binom{n+N-1}{2(N-1)-k}\left(\frac{1+ix}{-2}\right)^{k}$
(B.5)

As a quick consistency check, combining eq. (3.2.4) and (B.4), we obtain the
expected property

$\displaystyle\partial_{x}^{2N-1}\left({\mathfrak{L}}_{N}\psi^{(N)}_{n}\right)(x)=\psi^{(N)}_{n}(x)$
(B.6)

##  Appendix C Light scalars in $\text{dS}_{2}$

In this appendix, we quantize a light scalar field $\phi$ of scaling dimension
$\Delta=\frac{1}{2}+\nu,\nu\in(-\frac{1}{2},\frac{1}{2})$ in the conformal
global coordinates of $\text{dS}_{2}$, along the lines of subsection 3.3.
Expand $\phi$ as follows

$\displaystyle\phi=\sum_{n}\phi_{n}a_{n}+\phi_{n}^{*}a_{n}^{\dagger}$ (C.1)

where the mode functions are chosen to be

$\displaystyle\phi_{n}=\frac{\Gamma(n+\Delta)}{\sqrt{2}}\,e^{-in\,t}\,\mathbf{F}\left(\Delta,\bar{\Delta},n+1,\frac{1}{1+e^{2it}}\right)\frac{e^{-int}}{\sqrt{2\pi}}$
(C.2)

These modes are not unit normalized with respect to the Klein-Gordon inner
product (3.57)

$\displaystyle(\phi_{n},\phi_{m})_{\text{KG}}=\frac{\Gamma(n+\Delta)}{\Gamma(n+\bar{\Delta})}\delta_{nm}$
(C.3)

and hence the norm of single-particle states
$|n\rangle_{E}=a_{n}^{\dagger}|E\rangle$ becomes

${}_{E}\langle
n|m\rangle_{E}=\frac{\Gamma(n+\bar{\Delta})}{\Gamma(n+\Delta)}\delta_{nm}$
(C.4)

Since we have changed the $\Gamma$-function factor in the definition of
$\phi_{n}$ compared to the heavy scalar case, i.e.
$\Gamma(n+\bar{\Delta})\to\Gamma(n+\Delta)$, the action of ${\cal L}_{\pm}$ on
$\phi_{n}$ should change accordingly (but the action on $\phi^{*}_{n}$ remains
the same)

$\displaystyle{\cal L}_{\pm}\phi_{n}=(n\pm\bar{\Delta})\phi_{n\pm
1},\,\,\,\,\,{\cal L}_{\pm}\phi_{n}^{*}=-(n\mp\bar{\Delta})\phi^{*}_{n\mp 1}$
(C.5)

which yields

$\displaystyle[L_{\pm},a_{n}]=-(n\mp\Delta)a_{n\mp
1},\,\,\,\,\,[L_{\pm},a^{\dagger}_{n}]=(n\pm\Delta)a^{\dagger}_{n\pm 1}$ (C.6)

$L_{\pm}$ satisfying (C.6) are hermitian conjugate of each other. These
equations also complete the ${\mathfrak{so}}(1,2)$ action on the single-
particle Hilbert space ${\cal H}_{\Delta}=\text{span}\\{|n\rangle_{E}\\}_{n}$

$\displaystyle L_{\pm}|n\rangle_{E}=(n\pm\Delta)|n\pm
1\rangle_{E},\,\,\,\,\,L_{0}|n\rangle_{E}=|n\rangle_{E}$ (C.7)

With the inner product ${}_{E}\langle
m|n\rangle_{E}=\frac{\Gamma(n+\bar{\Delta})}{\Gamma(n+\bar{\Delta})}\delta_{nm}$
and the ${\mathfrak{so}}(1,2)$ action (C.7), we are able to identify the
single-particle Hilbert space ${\cal H}_{\Delta}$ as the complementary series
representation ${\cal C}_{\Delta}$ constructed in section 3.1.

Next we read off the leading behaviors of $\phi$ near the future boundary

$\displaystyle\phi\left(t=\frac{\pi}{2}-\delta,\varphi\right)\overset{\delta\to
0}{\approx}C_{\Delta}\,\delta^{\Delta}{\cal
O}_{\Delta}(\varphi)+C_{\bar{\Delta}}\,\delta^{\bar{\Delta}}{\cal
O}_{\bar{\Delta}}(\varphi)$ (C.8)

where
$C_{\Delta}=2^{\nu}\,\frac{\Gamma(-2\nu)}{\Gamma(\bar{\Delta})},C_{\bar{\Delta}}=2^{-\nu}\,\frac{\Gamma(2\nu)}{\Gamma(\Delta)}$
and

$\displaystyle{\cal
O}_{\Delta}(\varphi)=\sum_{n\in\mathbb{Z}}\frac{\Gamma(n+\Delta)}{\Gamma(n+\bar{\Delta})}\left((-i)^{n+\Delta}\frac{e^{-in\varphi}}{\sqrt{2\pi}}a_{n}+\text{h.c.}\right),\,\,\,\,\,{\cal
O}_{\bar{\Delta}}(\varphi)=\sum_{n\in\mathbb{Z}}\left((-i)^{n+\bar{\Delta}}\frac{e^{-in\varphi}}{\sqrt{2\pi}}a_{n}+\text{h.c.}\right)$
(C.9)

${\cal O}_{\Delta}$ and ${\cal O}_{\bar{\Delta}}$ defined this way are primary
operators of scaling dimension $\Delta$ and $\bar{\Delta}$ respectively,
satisfying eq. (3.3.3). Their two-point functions with respect to the
Euclidean vacuum are

$\displaystyle\langle E|{\cal O}_{\Delta}(\varphi_{1}){\cal
O}_{\Delta}(\varphi_{2})|E\rangle=\frac{N_{\bar{\Delta}}}{(1-\cos\varphi_{12})^{\Delta}},\,\,\,\,\,\langle
E|{\cal O}_{\bar{\Delta}}(\varphi_{1}){\cal
O}_{\bar{\Delta}}(\varphi_{2})|E\rangle=\frac{N_{\Delta}}{(1-\cos\varphi_{12})^{\bar{\Delta}}}$
$\displaystyle\langle E|{\cal O}_{\Delta}(\varphi_{1}){\cal
O}_{\bar{\Delta}}(\varphi_{2})|E\rangle=e^{-i\pi\nu}\delta(\varphi_{12}),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\langle
E|{\cal O}_{\bar{\Delta}}(\varphi_{1}){\cal
O}_{\Delta}(\varphi_{2})|E\rangle=e^{i\pi\nu}\delta(\varphi_{12})$ (C.10)

where the normalization constant $N_{\Delta}$ is defined in eq. (3.50). Define
the single-particle excitation of ${\cal O}_{\Delta}$ and ${\cal
O}_{\bar{\Delta}}$ as

$\displaystyle|\Delta,\varphi\rangle\equiv{\cal
O}_{\Delta}(\varphi)|E\rangle=\sum_{n}i^{n+\Delta}\frac{\Gamma(n+\Delta)}{\Gamma(n+\bar{\Delta})}\frac{e^{in\varphi}}{\sqrt{2\pi}}|n\rangle_{E}$
$\displaystyle|\bar{\Delta},\varphi\rangle\equiv{\cal
O}_{\bar{\Delta}}(\varphi)|E\rangle=\sum_{n}i^{n+\bar{\Delta}}\frac{e^{in\varphi}}{\sqrt{2\pi}}|n\rangle_{E}$
(C.11)

They can be used to decompose the identity operator on ${\cal H}_{\Delta}$

$\displaystyle\mathbb{1}_{{\cal
H}_{\Delta}}=e^{-i\pi\nu}\int\,d\varphi|\Delta,\varphi\rangle\langle\bar{\Delta},\varphi|=e^{i\pi\nu}\int\,d\varphi|\bar{\Delta},\varphi\rangle\langle\Delta,\varphi|$
(C.12)

which leads to the completeness of both $|\Delta,\varphi\rangle$ and
$|\bar{\Delta},\varphi\rangle$ in ${\cal H}_{\Delta}$

$\displaystyle\mathbb{1}_{{\cal H}_{\Delta}}$ $\displaystyle=\int
d\varphi_{1}d\varphi_{2}|\Delta,\varphi_{1}\rangle\langle\bar{\Delta},\varphi_{1}|\bar{\Delta},\varphi_{2}\rangle\langle\Delta,\varphi_{2}|$
$\displaystyle=N_{\Delta}\int
d\varphi_{1}d\varphi_{2}\frac{1}{(1-\cos\varphi_{12})^{\bar{\Delta}}}|\Delta,\varphi_{1}\rangle\langle\Delta,\varphi_{2}|$
$\displaystyle=N_{\bar{\Delta}}\int
d\varphi_{1}d\varphi_{2}\frac{1}{(1-\cos\varphi_{12})^{\Delta}}|\bar{\Delta},\varphi_{1}\rangle\langle\bar{\Delta},\varphi_{2}|$
(C.13)

Therefore $|\Delta,\varphi\rangle$ and $|\bar{\Delta},\varphi\rangle$ are
simply two different basis of the single-particle Hilbert space.

Switching to the stereographic coordinate of $S^{1}$, which is shown in fig.
(3.7), we define the $|x\rangle$ basis as

$\displaystyle|x\rangle=\left(\frac{2}{1+x^{2}}\right)^{\bar{\Delta}}|\bar{\Delta},\varphi(x)\rangle$
(C.14)

The ${\mathfrak{so}}(1,2)$ action on this basis is given by eq. (3.16). In
addition, the inner product of the $|x\rangle$-basis can be derived from (C)

$\displaystyle\langle
x|y\rangle=\frac{2^{\bar{\Delta}}}{(1+x^{2})^{\bar{\Delta}}}\frac{2^{\bar{\Delta}}}{(1+y^{2})^{\bar{\Delta}}}\frac{N_{\Delta}}{(1-\cos(\varphi(x)-\varphi(y)))^{\bar{\Delta}}}=\frac{2^{\bar{\Delta}}N_{\Delta}}{(x-y)^{2\bar{\Delta}}}$
(C.15)

which is consistent with eq. (3.2.2) for $c=N_{\Delta}$ and is also the same
as the shadow transformation $S_{\Delta}(x,y)$, c.f. (3.50).

##  Appendix D Induced representations

In this appendix, we review the method of induced representations and its
applications in the ${\text{SO}}(1,d+1)$ case. Let $H$ be a subgroup of $G$
and let $\rho$ be a representation of $H$ on some vector space $V$. The
induced representation $\text{ind}^{G}_{H}\rho$ of $G$ operates on the
following space of equivariant maps from $G$ to $V$

$\displaystyle\text{Map}_{H}(G,V)\equiv\left\\{\Psi:G\to
V|\Psi(gh)=\rho(h)^{-1}\Psi(g),\,\,g\in G,h\in H\right\\}$ (D.1)

by

$\displaystyle(g\circ\Psi)(g^{\prime})\equiv\Psi\left(g^{-1}g^{\prime}\right)$
(D.2)

In the case of $G={\text{SO}}(1,d+1)$, we take $H$ to be the minimal parabolic
subgroup ${\text{S}}\equiv{\text{N}}{\text{A}}{\text{M}}$ and $V$ to be the
space $\text{P}_{s}[z^{i}]$ of homogeneous polynomials in a null vector
$z^{i}$ of degree $s$, which is also equivalent to the space of symmetric and
traceless tensors of rank $s$. The representation $\rho=\rho_{\Delta,s}$ of S
is chosen to be

$\displaystyle\rho_{\Delta,s}\left(ne^{\lambda
D}m\right)f(z)=e^{-\lambda\Delta}f\left(\rho_{1}(m)^{-1}z\right)$ (D.3)

where $n\in{\text{N}}$, $m\in{\text{M}}$ and $\rho_{1}(m)$ is the fundamental
representation of ${\text{M}}={\text{SO}}(d)$, i.e.

$\displaystyle\rho_{1}\left(e^{\frac{1}{2}\theta^{ij}M_{ij}}\right):z^{k}\to
z^{k}+\theta^{kj}z_{j}+{\cal O}(\theta^{2})$ (D.4)

The induced representation $\text{ind}^{G}_{{\text{S}}}\rho_{\Delta,s}$ then
acts on the space of polynomial-valued functions $\Psi(g,z)$ that satisfy

$\displaystyle\Psi(gs,z)=\rho_{\Delta,s}(s)^{-1}\Psi(g,z),\,\,\,\,\,s\in{\text{S}}$
(D.5)

Due to the Bruhat decomposition of ${\text{SO}}(1,d+1)$ [50] and the
equivariant condition (D.5), such a function $\Psi(g,z)$ is completely fixed
by its value on ${\widetilde{\text{N}}}$, which is geometrically a flat space
$\mathbb{R}^{d}$.131313The Bruhat decomposition says that
${\text{SO}}(1,d+1)$, up to a lower dimensional submanifold, can be written as
a product ${\widetilde{\text{N}}}{\text{S}}$ . This lower dimensional manifold
corresponds to the infinity point of $\mathbb{R}^{d}$ when we quotient out the
S dependence. Define $\psi(x,z)\equiv\Psi(e^{x\cdot P},z)$ and we will show
that the (infinitesimal version of) induced representation
$\text{ind}^{G}_{{\text{S}}}\rho_{\Delta,s}$ agrees with the CFT-type
construction, c.f. eq.(4).

  * •

Translations:

$\displaystyle\left(e^{a\cdot P}\circ\psi\right)(x,z)=\left(e^{a\cdot
P}\circ\Psi\right)(e^{x\cdot P},z)=\Psi(e^{(x-a)\cdot P},z)=\psi(x-a,z)$ (D.6)

Derivative with respect to $a^{i}$ yields
$\boxed{P_{i}\psi(x,z)=-\partial_{i}\psi(x,z)}$

  * •

Dilatations:

$\displaystyle\left(e^{\lambda D}\circ\psi\right)(x,z)=\Psi(e^{-\lambda
D}e^{x\cdot P},z)$ (D.7)

Using $e^{-\lambda D}P_{i}e^{\lambda D}=e^{-\lambda}P_{i}$, we obtain

$\displaystyle\left(e^{\lambda
D}\circ\psi\right)(x,z)=\rho_{\Delta,s}(e^{\lambda
D})\psi(e^{-\lambda}x,z)=e^{-\lambda\Delta}\psi(e^{-\lambda}x,z)$ (D.8)

which leads to $\boxed{D\psi(x,z)=-(x\cdot\partial_{x}+\Delta)\psi(x,z)}$.

  * •

Rotations $m=e^{\frac{1}{2}\theta^{ij}M_{ij}}$:

$\displaystyle m\circ\psi(x,z)=\rho_{\Delta,s}(m)\Psi(m^{-1}e^{x\cdot
P}m,z)=\Psi(e^{x_{m}\cdot P},\rho_{1}(m)^{-1}z)$ (D.9)

where $x_{m}$ is defined via $m^{-1}e^{x\cdot P}m=e^{x_{m}\cdot P}$.
Infinitesimally, we have

$\displaystyle x_{m}^{i}=x^{i}-\theta^{ij}x_{j}+{\cal
O}(\theta^{2}),\,\,\,\,\,\rho_{1}(m)^{-1}z^{i}=z^{i}-\theta^{ij}z_{j}+{\cal
O}(\theta^{2})$ (D.10)

and hence
$\boxed{M_{ij}\psi(x,z)=(x_{i}\partial_{j}-x_{j}\partial_{i}+z_{i}\partial_{z^{j}}-z_{j}\partial_{z^{i}})\psi(x,z)}$.

  * •

Special conformal transformations:

$\displaystyle\left(e^{b\cdot K}\circ\psi\right)(x,z)=\Psi(e^{-b\cdot
K}e^{x\cdot P},z)$ (D.11)

For our purpose, we can replace $e^{-b\cdot K}$ by $1-b\cdot K$ and then use

$\displaystyle-e^{-x\cdot P}\,b\cdot K\,e^{x\cdot P}=-b\cdot K-2x\cdot
b\,D+2b^{i}x^{j}M_{ij}+(x^{2}b^{i}-2x\cdot b\,x^{i})P_{i}$ (D.12)

which yields

$\displaystyle\left(e^{b\cdot K}\circ\psi\right)(x,z)$
$\displaystyle=\Psi\left(e^{((1-2x\cdot b)x+x^{2}b)\cdot P}e^{-b\cdot
K}e^{-2x\cdot bD}e^{2b_{i}x_{j}M_{ij}},z\right)+{\cal O}(b^{2})$
$\displaystyle=e^{-2\Delta x\cdot b}\psi((1-2x\cdot b)x+x^{2}b,2\,x\cdot
z\,b-2\,b\cdot z\,x)+{\cal O}(b^{2})$ (D.13)

Take derivative with respect to $b^{i}$ and we obtain

$\displaystyle\boxed{K_{i}\psi(x,z)=\left(x^{2}\partial_{i}-2x_{i}(x\cdot\partial_{x}+\Delta)+2(x\cdot
z\,\partial_{z^{i}}-z_{i}\,x\cdot\partial_{z})\right)\psi(x,z)}$ (D.14)

##  Appendix E Irreducibility of ${\cal F}_{\Delta}$

In this appendix we give an elementary proof about the irreducibility of
${\cal F}_{\Delta}$ for a generic $\Delta$. The argument heavily relies on the
result of subsection 4.1 where we have learned that the ${\text{SO}}(d+1)$
content of ${\cal F}_{\Delta}$ consists of all single-row representations
${\text{SO}}(d+1)$. Assuming the existence of a nontrivial
${\text{SO}}(1,d+1)$ subspace ${\cal W}_{\Delta}$ of ${\cal F}_{\Delta}$, it
admits an ${\text{SO}}(d+1)$ decomposition of the following form

$\displaystyle{\cal W}_{\Delta}=\bigoplus_{n\in\sigma}{\mathbb{Y}}_{n}$ (E.1)

where $\sigma$ is a nontrivial subset of $\mathbb{N}$ and ${\mathbb{Y}}_{n}$
is the spin-$n$ representation of ${\text{SO}}(d+1)$. We will show that this
assumption is invalid because given an arbitrary wavefunction $\psi(x)$ in
certain $Y_{n}$, by acting dilatation operator on $\psi(x)$ repeatedly, it can
get components in any ${\text{SO}}(d+1)$ content of ${\cal F}_{\Delta}$.

Pick any $\psi(x)\in{\cal F}_{\Delta}$. According to the analysis of
subsection 4.1, it is a function $\hat{\psi}$ on $S^{d}$ (in stereographic
coordinate) multiplied by a Weyl factor, i.e.
$\psi(x)=\left(\frac{2}{1+x^{2}}\right)^{\Delta}\hat{\psi}(x)$. Switching to
spherical coordinates $x^{i}=\omega^{i}\cot\frac{\theta}{2},\omega^{i}\in
S^{d-1}$, the action of dilatation operator on $\hat{\psi}(\theta,\omega)$
becomes

$\displaystyle
D\,\hat{\psi}(\theta,\omega)=\left(\sin\theta\partial_{\theta}+\Delta\cos\theta\right)\hat{\psi}(\theta,\omega)$
(E.2)

The function $\hat{\psi}$ can be expanded in terms of spherical harmonics on
$S^{d}$

$\displaystyle Y_{n\ell\bm{m}}(\theta,\omega)={\cal
N}_{n\ell}\,\sin^{\ell}\theta\,C^{\ell+\frac{d-1}{2}}_{n-\ell}(\cos\theta)Y_{\ell\bm{m}}(\omega),\,\,\,\,n\geq\ell$
(E.3)

where $Y_{\ell\bm{m}}(\omega)$ denote the normalized spherical harmonics on
$S^{d-1}$ and ${\cal N}_{n\ell}$ is a constant such that $Y_{n\ell\bm{m}}$ is
normalized

$\displaystyle{\cal
N}_{n\ell}=2^{\ell-1+\frac{d}{2}}\,\Gamma\left(\ell+\frac{d-1}{2}\right)\sqrt{\frac{(n-\ell)!(n+\frac{d-1}{2})}{\pi\,\Gamma(n+\ell+d-1)}}$
(E.4)

For a fixed $n$, all
$\left(\frac{2}{1+x^{2}}\right)^{\Delta}Y_{n\ell\bm{m}}(\theta,\omega)$ span
the ${\mathbb{Y}}_{n}$ part of ${\cal F}_{\Delta}$. Next, we need to figure
out the action of $D$ on all the spherical harmonics. This computation can be
done by using the following recurrence relations of Gegenbauer polynomials

$\displaystyle\partial_{x}C^{\lambda}_{n}(x)=2\lambda
C^{\lambda+1}_{n-1}(x),\,\,\,\,\,x\,C^{\lambda}_{n}(x)=\frac{n+1}{2(n+\lambda)}C^{\lambda}_{n+1}(x)+\frac{n-1+2\lambda}{2(n+\lambda)}C^{\lambda}_{n-1}(x)$
$\displaystyle(1-x)^{2}C^{\lambda+1}_{n-1}(x)=\frac{n+2\lambda}{2\lambda}xC^{\lambda}_{n}(x)-\frac{(n+1)}{2\lambda}C^{\lambda}_{n+1}(x)$
(E.5)

which yield

$\displaystyle\boxed{DY_{n\ell\bm{m}}=(\Delta+n)\,\sqrt{\frac{(n+1-\ell)(n+\ell+d-1)}{(2n+d-1)(2n+d+1)}}Y_{n+1,\ell\bm{m}}+(1-\bar{\Delta}-n)\,\sqrt{\frac{(n-\ell)(n+\ell+d-2)}{(2n+d-3)(2n+d-1)}}Y_{n-1,\ell\bm{m}}}$
(E.6)

The coefficients of $Y_{n\pm 1,\ell\bm{m}}$ are nonvanishing except
$\Delta\in\\{d,d+1,d+2,\cdots\\}\cup\\{0,-1,-2,\cdots\\}$. Due to the
reasoning given at the beginning of this appendix, the representation ${\cal
F}_{\Delta}$ is irreducible when $\Delta$ is away form these integers. For
$\Delta\in\\{d,d+1,d+2,\cdots\\}$, the subspace with ${\text{SO}}(d+1)$
content $\bigoplus_{n\geq 1-\bar{\Delta}}{\mathbb{Y}}_{n}$ is mapped to itself
by $D$ and this claim also holds for any $L_{0,i}$, because $L_{0,i}$ can be
related to $D=L_{0,d+1}$ by ${\text{SO}}(d+1)$ conjugations. Therefore, this
subspace is irreducible with respect to ${\text{SO}}(1,d+1)$. Using the same
argument, we can claim that when $\Delta\in\\{0,-1,-2,\cdots\\}$, the finite
dimensional subspace $\bigoplus_{0\leq n\leq-\Delta}{\mathbb{Y}}_{n}$ is
irreducible, carrying the spin $(-\Delta)$ representation of
${\text{SO}}(1,d+1)$. These finite dimensional representations are nonunitary
except the $\Delta=0$ one which is the trivial representation.

##  Appendix F Normalizability of the inner product on ${\cal U}_{s,t}$

In the subsection 4.5, we introduced an inner product for the exceptional
series representation ${\cal U}_{s,t}={\cal F}_{1-t,s}/{\rm
Im\,}(z\cdot\partial_{x})^{s-t}$:

$\displaystyle(\psi,\varphi)_{{\cal
U}{s,t}}=\int\,d^{d}x_{1}d^{d}x_{2}\psi(x_{1})^{*}_{i_{1}\cdots
i_{s}}S^{+}_{1-t,s}(x_{12})_{i_{1}\cdots i_{2},j_{1}\cdots
j_{2}}\varphi(x_{2})_{j_{1}\cdots j_{s}}$ (F.1)

where $\psi_{i_{1}\cdots i_{s}},\varphi_{j_{1}\cdots j_{s}}\in{\cal
F}_{1-t,s}$ and $S^{+}_{1-t,s}$ defines the shadow transformation ${\cal
S}^{+}_{1-t,s}$ from ${\cal F}_{1-t,s}$ to ${\cal F}_{d+t-1,s}$, whose Fourier
transformation is given by eq. (4.2) in the index-free formalism:

$\displaystyle
S^{+}_{1-t,s}(p;z,w)=p^{d+2t-2}\sum_{\ell=t+1}^{s}\frac{(\ell-t)_{s-\ell}}{(d+t+\ell-2)_{s-\ell}}\Pi^{s\ell}(\hat{p};z,w)$
(F.2)

However, we did not justify the normalizability of this inner product. In this
appendix, we will prove that the inner product (F.1) is normalizable, by a
detailed study of the asymptotic behaviors of wavefunctions in ${\cal
F}_{1-t,s}$.

By definition, the large $x$ behavior of $\psi(x,z)\in{\cal F}_{1-t,s}$ is

$\displaystyle\psi(x,z)\stackrel{{\scriptstyle
x\to\infty}}{{\approx}}\sum_{k=0}^{\infty}x^{2(t-1)}C_{k}\left(\tilde{x}^{i},R(x)^{i}_{\,\,j}z^{j}\right)$
(F.3)

where $C_{k}(u,v)$ is a homogeneous polynomial of degree $k$ in the first
argument and degree $s$ in the second. With some simple power counting, one
can learn that the first $t$ polynomials, i.e. $\\{C_{0},C_{1},\cdots
C_{t-1}\\}$, might lead to divergence at small $p$ in (F.1). We will show this
does not happen because all $x^{2(t-1)}C_{k}$ with $0\leq k\leq t-1$ lie in
the image of $(z\cdot\partial_{x})^{s-t}$, which is annihilated by ${\cal
S}^{+}_{1-t,s}$. In particular, it suffices to show the following proposition:

######  Proposition F.1.

Let $P_{k}(x^{i})$ be an arbitrary degree-$k$ homogeneous polynomial in
$x^{i}$ and define $\hat{P}_{k}(x,v)=x^{2(t-1-k)}P_{k}(x^{i})v^{i_{1}}\cdots
v^{i_{s}}$, where $t\in\\{0,1,\cdots s-1\\}$ and
$v^{i}=R(x)^{i}_{\,\,j}z^{j}=\frac{2x\cdot z}{x^{2}}x^{i}-z^{i}$. When
$k=0,1,\cdots t-1$, $\hat{P}_{k}(x,v)$ can be expressed as
$\hat{P}_{k}(x,v)=(z\cdot\partial_{x})^{s-t}Q_{k}(x,v)$ for some polynomial
$Q_{k}(x,v)$.

In order to prove the proposition F.1, we need some lemmas:

######  Lemma F.2.

$\displaystyle(z\cdot\partial_{x})^{n}\frac{x^{i_{1}}\cdots
x^{i_{n}}}{x^{2}}=\frac{(-)^{n}\,n!}{x^{2}}v^{i_{1}}\cdots v^{i_{n}}$ (F.4)

Proof: Since the L.H.S is a totally symmetric tensor, it suffices to compute
$(z\cdot\partial_{x})^{n}\frac{(x\cdot\xi)^{n}}{x^{2}}$, where $\xi_{i}$ is an
auxiliary vector:

$\displaystyle(z\cdot\partial_{x})^{n}\frac{(x\cdot\xi)^{n}}{x^{2}}$
$\displaystyle=\sum_{\ell=0}^{n}\binom{n}{\ell}(z\cdot\partial_{x})^{n-\ell}(x\cdot\xi)^{n}\,(z\cdot\partial_{x})^{\ell}\frac{1}{x^{2}}$
$\displaystyle=\frac{n!}{x^{2}}\sum_{\ell=0}^{n}\binom{n}{\ell}(z\cdot\xi)^{n-\ell}\left(\frac{-2x\cdot
z\,x\cdot\xi}{x^{2}}\right)^{\ell}=\frac{(-)^{n}\,n!}{x^{2}}(v\cdot\xi)^{n}$
(F.5)

Stripping off $\xi_{i}$, we recover eq. (F.4).

######  Lemma F.3.

For a degree-$k$ polynomial $P_{k}(x)$, we have

$\displaystyle
P_{k}(x)(z\cdot\partial_{x})^{n+k}g(x)=(z\cdot\partial_{x})^{n}\left\\{\sum_{\ell=0}^{k}(-)^{\ell}\binom{n+\ell-1}{\ell}(z\cdot\partial_{x})^{\ell}P_{k}(x)\,(z\cdot\partial_{x})^{k-\ell}g(x)\right\\}$
(F.6)

Proof: Define

$\displaystyle
A(\bm{a})=\sum_{\ell=0}^{k}a_{\ell}(z\cdot\partial_{x})^{\ell}P_{k}(x)\,(z\cdot\partial_{x})^{k-\ell}g(x)$
(F.7)

We want to find $\bm{a}=(a_{0},a_{1},\cdots,a_{k})$ such that $A(\bm{a})$ is
mapped to $P_{k}(x)(z\cdot\partial_{x})^{n+k}g(x)$ under
$(z\cdot\partial_{x})^{n}$. Let’s first act $z\cdot\partial_{x}$ on
$A(\bm{a})$ and expand the result as follows

$\displaystyle
z\cdot\partial_{x}A(\bm{a})=\sum_{\ell=0}^{k}a^{(1)}_{\ell}(z\cdot\partial_{x})^{\ell}P_{k}(x)\,(z\cdot\partial_{x})^{k-\ell+1}g(x)$
(F.8)

where $a^{(1)}_{\ell}=a_{\ell}+a_{\ell-1}$ for $1\leq\ell\leq k$ and
$a_{0}^{(1)}=a_{0}$. The relation between $\bm{a}$ and $\bm{a}^{(1)}$ can be
captured by a $(k+1)\times(k+1)$ matrix $\Lambda$, i.e.
$\bm{a}^{(1)}=(\mathbb{1}+\Lambda)\,\bm{a}$, where

$\displaystyle\Lambda=\begin{pmatrix}0&0&0&\cdots&0&0\\\ 1&0&0&\cdots&0&0\\\
0&1&0&\cdots&0&0\\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\
0&0&0&\cdots&1&0\end{pmatrix}$ (F.9)

Then finding the coefficients $(a_{0},a_{1},\cdots a_{k})$ becomes a pure
linear algebra problem

$\displaystyle(1+\Lambda)^{n}\begin{pmatrix}a_{0}\\\ a_{1}\\\ \cdots\\\
a_{k}\end{pmatrix}=\begin{pmatrix}1\\\ 0\\\ \cdots\\\ 0\end{pmatrix}$ (F.10)

Since $(1+\Lambda)$ is invertible, the desired $\bm{a}$ exists and is given by
the first column of $(1+\Lambda)^{-n}$. The matrix $(1+\Lambda)^{-n}$ can be
computed by a formal Taylor expansion and the Taylor series truncates because
$\Lambda^{k+1}=0$:

$\displaystyle(1+\Lambda)^{-n}=\sum_{\ell=0}^{k}(-)^{\ell}\binom{n+\ell-1}{\ell}\Lambda^{\ell}$
(F.11)

Combining (F.11) and $(\Lambda^{\ell})_{i,1}=\delta_{i,\ell+1}$, we obtain

$\displaystyle a_{\ell}=(-)^{\ell}\binom{n+\ell-1}{\ell}$ (F.12)

and this finishes our proof of (F.6).

Now we start proving the proposition F.1. Using lemma F.2, we rewrite
$\hat{P}_{k}(x,v)$ as

$\displaystyle\hat{P}_{k}(x,v)=\frac{(-)^{s-t+k}}{(s-t+k)!}P_{k}(x)(x^{2}v^{i_{1}})\cdots(x^{2}v^{i_{t-k}})(z\cdot\partial_{x})^{s-t+k}\frac{x^{i_{t+1-k}}\cdots
x^{i_{s}}}{x^{2}}$ (F.13)

where $x^{2}v^{i_{j}}$ can be moved to the right of
$(z\cdot\partial_{x})^{s-t+k}$ since $z\cdot\partial_{x}(x^{2}v^{i})=0$

$\displaystyle\hat{P}_{k}(x,v)=P_{k}(x)(z\cdot\partial_{x})^{s-t+k}g(x,v),\,\,\,\,\,g(x,v)=\frac{(-)^{s-t+k}}{(s-t+k)!}x^{2(t-k-1)}v^{i_{1}}v^{i_{t-k}}x^{i_{t+1-k}}\cdots
x^{i_{s}}$ (F.14)

Then proposition F.1 immediately follows from lemma F.3.

##  Appendix G Computation of the decorated Harish-Chandra characters

In section 5.2, we have computed the Harish-Chandra characters $\Theta_{{\cal
F}_{\Delta,s}}(g)=\operatorname{Tr}_{{\cal F}_{\Delta,s}}g$ for $g$ sitting on
the one-dimensional subgroup generated by the boost $D$. In this appendix, we
will consider more general $g$ by also turning on angular momenta in the trace
$\operatorname{Tr}_{{\cal F}_{\Delta,s}}$, i.e.

$\displaystyle\Theta_{{\cal F}_{\Delta,s}}(q,\bm{u})=\operatorname{Tr}_{{\cal
F}_{\Delta,s}}\left(q^{D}\prod_{i=1}^{r}e^{u_{i}J_{i}}\right)$ (G.1)

where $r=\left\lfloor\frac{d}{2}\right\rfloor$, and $J_{i}=M_{2i-1,i}$ are
generators of the Cartan subalgebra of ${\text{SO}}(d)$.

Let’s start from the $s=0$ case. Following the derivation in section 5.2, we
first conjugate $q^{D}\prod_{i=1}^{r}e^{u_{i}J_{i}}$ by a special conformal
transformation

$\displaystyle\Theta_{{\cal F}_{\Delta}}(q,\bm{u})=\operatorname{Tr}_{{\cal
F}_{\Delta,s}}\left(e^{-b\cdot K}g_{0}e^{b\cdot
K}\right),\,\,\,\,\,g_{0}\equiv q^{D}\prod_{i=1}^{r}e^{u_{i}J_{i}}$ (G.2)

where $b^{i}\in\mathbb{R}^{d}$ is an arbitrary auxiliary vector. Then we need
to compute the action of $e^{-b\cdot K}g_{0}e^{b\cdot K}$ on $|x\rangle$. With
some straightforward but tedious computations, we obtain

$\displaystyle e^{-b\cdot K}g_{0}e^{b\cdot
K}|x\rangle=\Omega(q,\bm{u},x,b)^{\bar{\Delta}}|\tilde{x}(q,\bm{u},x,b)\rangle$
(G.3)

where

$\displaystyle\Omega(q,\bm{u},x,b)=\frac{q}{1-2\,x\cdot
b+b^{2}x^{2}+2\,q(b\cdot x_{\bm{u}}-b\cdot b_{\bm{u}}x^{2})+b^{2}q^{2}x^{2}}$
$\displaystyle\tilde{x}^{i}(q,\bm{u},x,b)=\Omega(q,\bm{u},x,b)(x^{i}_{\bm{u}}+x^{2}(q\,b^{i}-b^{i}_{\bm{u}}))$
(G.4)

Here $x_{\bm{u}}^{i}$ ($b^{i}_{\bm{u}}$) is related to $x^{i}$ ($b^{i}$) by a
rotation associated to $\prod_{i=1}^{r}e^{u_{i}J_{i}}$. More explicitly,
$x_{\bm{u}}^{i}=R(\bm{u})^{i}_{\,\,j}x^{j}$ where $R(\bm{u})$ is a block
diagonal matrix

$\displaystyle R(\bm{u})=\begin{cases}\text{diag}\left\\{\begin{pmatrix}\cos
u_{1}&\sin u_{1}\\\ -\sin u_{1}&\cos u_{1}\end{pmatrix},\begin{pmatrix}\cos
u_{2}&\sin u_{2}\\\ -\sin u_{2}&\cos
u_{2}\end{pmatrix},\cdots,\begin{pmatrix}\cos u_{r}&\sin u_{r}\\\ -\sin
u_{r}&\cos u_{r}\end{pmatrix}\right\\},&d=2r\\\
\text{diag}\left\\{\begin{pmatrix}\cos u_{1}&\sin u_{1}\\\ -\sin u_{1}&\cos
u_{1}\end{pmatrix},\begin{pmatrix}\cos u_{2}&\sin u_{2}\\\ -\sin u_{2}&\cos
u_{2}\end{pmatrix},\cdots,\begin{pmatrix}\cos u_{r}&\sin u_{r}\\\ -\sin
u_{r}&\cos u_{r}\end{pmatrix},1\right\\},&d=2r+1\end{cases}$ (G.5)

Given the action (G.3), the character $\Theta_{{\cal F}_{\Delta}}(q,\bm{u})$
can be expressed as

$\displaystyle\Theta_{{\cal F}_{\Delta}}(q,\bm{u})=\int
d^{d}x\,\Omega(q,\bm{u},x,b)^{\bar{\Delta}}\delta^{d}(\tilde{x}(q,\bm{u},x,b)-x)$
(G.6)

This integral is localized to the two fixed points of the map
$x\to\tilde{x}(q,\bm{u},x,b)$, namely $x^{i}=0$ and
$x^{i}=\frac{b^{i}}{b^{2}}$. At these points, the scaling factor
$\Omega(q,\bm{u},x,b)$ becomes

$\displaystyle\Omega(q,\bm{u},0,b)=q,\,\,\,\,\,\Omega(q,\bm{u},b^{i}/b^{2},b)=q^{-1}$
(G.7)

The remaining task is to figure out the Jacobian determinant associated to the
map $x^{i}\to\tilde{x}^{i}-x^{i}$ at these fixed points. At $x^{i}=0$, the
Jacobian matrix (denoted by $J_{1}$) is simply
$qR(\bm{u})^{i}_{\,\,j}-\delta^{i}_{j}$ and hence the corresponding Jacobian
is

$\displaystyle|\det J_{1}|=\prod_{i=1}^{r}(1-2q\cos
u_{i}+q^{2})\times\begin{cases}1,&d=2r\\\ |1-q|,&d=2r+1\end{cases}$ (G.8)

The Jacobian matrix at $x^{i}=\frac{b^{i}}{b^{2}}$ (denoted by $J_{2}$) takes
a more complicated form

$\displaystyle J_{2}=\frac{1}{q}(1-2B)R(\bm{u})(1-2B)-\mathbb{1}_{d}$ (G.9)

where $\mathbb{1}_{d}$ is the $d\times d$ identity matrix and
$B^{i}_{\,\,j}=\frac{1}{b^{2}}b^{i}b_{j}$. Although $J_{2}$ depends on $b^{i}$
explicitly, its determinant does not. Using $B^{2}=B$, one can immediately see
that $1-2B$ is an involutory matrix, which implies

$\displaystyle|\det
J_{2}|=|\det(q^{-1}R(\bm{u})-\mathbb{1}_{d})|=\prod_{i=1}^{r}(1-2q^{-1}\cos
u_{i}+q^{-2})\times\begin{cases}1,&d=2r\\\ |1-q^{-1}|,&d=2r+1\end{cases}$
(G.10)

Altogether, the decorated character $\Theta_{{\cal F}_{\Delta}}(q,\bm{u})$ is

$\displaystyle\Theta_{{\cal F}_{\Delta}}(q,\bm{u})$
$\displaystyle=\frac{q^{\bar{\Delta}}}{|\det
J_{1}|}+\frac{q^{-\bar{\Delta}}}{|\det
J_{2}|}=\frac{q^{\Delta}+q^{\bar{\Delta}}}{P_{d}(q,\bm{u})}$ (G.11)

where

$\displaystyle
P_{d}(q,\bm{u})=\prod_{i=1}^{r}(1-e^{iu_{i}}q)(1-e^{-iu_{i}}q)\times\begin{cases}1,&d=2r\\\
|1-q|,&d=2r+1\end{cases}$ (G.12)

Compared to the undecorated character $\Theta_{{\cal F}_{\Delta}}(q)$, c.f.
eq. (5.7), the only difference is that the denominator $|1-q|^{d}$ gets
replaced by $P_{d}(q,\bm{u})$. This result agrees with [39].

The derivation above can be straightforwardly generalized to spinning
representations ${\cal F}_{\Delta,s}$. We simply present the result here

$\displaystyle\Theta_{{\cal
F}_{\Delta,s}}(q,\bm{u})=\Theta^{{\text{SO}}(d)}_{{\mathbb{Y}}_{s}}(\bm{u})\frac{q^{\Delta}+q^{\bar{\Delta}}}{P_{d}(q,\bm{u})}$
(G.13)

where $\Theta^{{\text{SO}}(d)}_{{\mathbb{Y}}_{s}}(\bm{u})$ is the
${\text{SO}}(d)$ character corresponding to the spin-$s$ representation,
defined as the trace of $\prod_{i=1}^{r}e^{u_{i}J_{i}}$ over the spin-$s$
representation space (which, for example, is carried by
$|0\rangle_{i_{1}\cdots i_{s}}$ in our settings). Some lower $d$ examples of
$\Theta^{{\text{SO}}(d)}_{{\mathbb{Y}}_{s}}(\bm{u})$ are

$\displaystyle\Theta^{{\text{SO}}(3)}_{{\mathbb{Y}}_{s}}(u_{1})=\sum_{n=-s}^{s}e^{inu_{1}}=\frac{\sin((s+\frac{1}{2})u_{1})}{\sin(\frac{1}{2}u_{1})}$
(G.14)
$\displaystyle\Theta^{{\text{SO}}(4)}_{{\mathbb{Y}}_{s}}(u_{1},u_{2})=\frac{\cos((s+1)u_{1})-\cos((s+1)u_{2})}{\cos(u_{1})-\cos(u_{2})}$
(G.15)

##  Appendix H A new expression for certain $\text{SO}(d+2)$ characters

In this appendix, we show an interesting property of the ${\text{SO}}(d+2)$
characters $\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)$ corresponding
to two-row representations ${\mathbb{Y}}_{n,s}$. Let’s first give the
statement.

######  Claim 1.

The character $\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)$ has the same
polar part (i.e. all terms of negative powers in $x$) and the same constant
term as the following function

$\displaystyle\boxed{P^{d}_{n,s}(x)\equiv\frac{D^{d}_{s}\,x^{-n}-D^{d}_{n+1}\,x^{1-s}}{(1-x)^{d}}}$
(H.1)

Since $\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)$ is symmetric under
$x\leftrightarrow x^{-1}$, it is then completely encoded in the function
$P^{d}_{n,s}(x)$. In particular, when $s=0$, i.e. spin-$n$ representation, the
character $\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n}}(x)$ is encoded in

$\displaystyle\boxed{Q^{d}_{n}(x)\equiv\frac{x^{-n}}{(1-x)^{d}}}$ (H.2)

More explicitly, let $f(x)$ be a function with well-defined Laurent expansion
around $x=0$. Denote the polar part of $f(x)$ by $[f(x)]_{-}$ and denote the
polar part together with the constant term of $f(x)$ by $[f(x)]_{0}$. With
this notations, the claim implies that

$\displaystyle\boxed{\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)=\left[P^{d}_{n,s}(x)\right]_{0}+\left(\left.\left[P^{d}_{n,s}(x)\right]_{-}\right|_{x\to
x^{-1}}\right)}$ (H.3)

The claim can be proved rigorously by induction, using branching rules. Here
we want to present a more intriguing but less rigorous argument, that shows
the similarity and difference between the ${\mathbb{Y}}_{n,s}$ representation
of ${\text{SO}}(d+2)$ and the infinite dimensional lowest-weight
representations of ${\text{SO}}(1,d+1)$. In view of this, we need to rewrite
the ${\mathbb{Y}}_{n,s}$ representation in a CFT-style language as follows.
Denote the (antihermitian) generators of ${\mathfrak{so}}(d+2)$ by
$L_{AB},0\leq A,B\leq d+1$ which satisfy commutation relations

$\displaystyle[L_{AB},L_{CD}]=\delta_{BC}L_{AD}+\text{permutations}$ (H.4)

In the differential operator realization,
$L_{AB}=X_{A}\partial_{B}-X_{B}\partial_{A}$ where $X_{A}$ are coordinates in
$\mathbb{R}^{d+2}$. Mimicking the conformal algebra, we define the following
basis

$\displaystyle
M_{ij}=L_{ij},\,\,\,\,\,H=iL_{0,d+1},\,\,\,\,\,P_{i}=L_{d+1,i}+iL_{0i},\,\,\,\,\,K_{i}=-L_{d+1,i}+iL_{0i},$
(H.5)

where $1\leq i\leq d$. The new basis leads to some interesting commutators

$\displaystyle[M_{ij},H]=0,\,\,\,\,\,[H,P_{i}]=P_{i},\,\,\,\,[H,K_{i}]=-K_{i}\,\,\,\,\,[K_{i},P_{j}]=2M_{ij}-2\delta_{ij}H$
(H.6)

In particular, $P_{i}$ raises the eigenvalue of $H$ by 1 while $K_{i}$ lowers
the eigenvalue of $H$ by 1. Define complex (lightcone) coordinate
$z=X_{d+1}+iX_{0},\bar{z}=X_{d+1}-iX_{0}$ and then the differential operator
realization of $H,P_{i},K_{i}$ can be expressed as

$\displaystyle
H=z\partial_{z}-\bar{z}\partial_{\bar{z}},\,\,\,\,\,P_{i}=z\partial_{i}-2\,X_{i}\partial_{\bar{z}},\,\,\,\,\,K_{i}=-\bar{z}\partial_{i}+2\,X_{i}\partial_{z}$
(H.7)

First, let’s consider spin-$n$ representation of ${\text{SO}}(d+2)$ generated
by the lowest weight state $|lw\rangle$ that satisfies

$\displaystyle H|lw\rangle=-n|lw\rangle,\,\,\,\,\,K_{i}|lw\rangle=0,\,\,\,\,\
M_{ij}|lw\rangle=0$ (H.8)

A generic state in the Verma module generated by $|lw\rangle$ is a linear
combination of the descendants $P_{i_{1}}\cdots P_{i_{k}}|lw\rangle$. However,
some of the ${\mathfrak{so}}(d+2)$ descendants of $|lw\rangle$ are linearly
dependent. To see this more explicitly, let’s switch to the wavefunction
picture, where the lowest weight state $|lw\rangle$ corresponds to
$\psi_{lw}(z,\bar{z})=\bar{z}^{n}$ . One crucial observation is that all of
the states $\psi_{i_{1}\cdots i_{k}}\equiv P_{i_{1}}\cdots P_{i_{k}}\psi_{lw}$
with $k\leq n$ are linearly independent. To show this, it suffices to notice
that the top component of $\psi_{i_{1}\cdots i_{k}}$ in $X_{i}$ is
$(-2)^{k}X_{i_{1}}\cdots X_{i_{k}}\bar{z}^{n-k}$ for $k\leq n$ which arises
from the $-2X_{i}\partial_{\bar{z}}$ part of each $P_{i}$. It is clear that
$X_{i_{1}}\cdots X_{i_{k}}\bar{z}^{n-k}$ are nonvanishing and linearly
independent, and hence all $\psi_{i_{1}\cdots i_{k}}$ with $k\leq n$ are
linearly independent. Then it is almost trivial to construct the polar part
and the constant term of the character
$\Theta_{{\mathbb{Y}}_{n}}^{{\text{SO}}(d+2)}(x)\equiv\operatorname{Tr}_{{\mathbb{Y}}_{n}}x^{H}$

$\displaystyle\left[\Theta_{{\mathbb{Y}}n}^{{\text{SO}}(d+2)}(x)\right]_{0}=\sum_{k=0}^{n}\binom{d+k-1}{d-1}x^{k-n}=\left[\sum_{k=0}^{\infty}\binom{d+k-1}{d-1}x^{k-n}\right]_{0}=\left[Q^{d}_{n}(x)\right]_{0}$
(H.9)

This is exactly the single-row case of the claim of this appendix. Then the
full character can be recovered by using the $x\leftrightarrow x^{-1}$
symmetry. For example, due to this symmetry the coefficient of $x$ in
$\Theta_{{\mathbb{Y}}_{n}}^{{\text{SO}}(d+2)}(x)$ should be
$\binom{d+n-2}{d-1}$ rather than the native counting $\binom{d+n}{d-1}$. Even
more explicitly, we can find what are these $\binom{d+n-2}{d-1}$ states. The
$H=0$ eigenspace ${\cal H}^{(n)}_{0}$ is spanned by all $X_{i_{1}}\cdots
X_{i_{n}}$. Acting $P_{j}$ on them yields basis of the $H=1$ eigenspace ${\cal
H}^{(n)}_{1}$

$\displaystyle P_{j}X_{i_{1}}\cdots X_{i_{n}}=z\delta_{j(i_{1}}X_{i_{2}}\cdots
X_{i_{n})}$ (H.10)

Therefore ${\cal H}^{(n)}_{1}$ is spanned by $z\,X_{i_{1}}\cdots X_{i_{n-1}}$
and has dimension $\binom{d+n-2}{d-1}$.

Next, we consider more complicated representations like ${\mathbb{Y}}_{n,s}$,
for which we have a lowest weight state $|lw\rangle_{{i_{1}}\cdots i_{s}}$
that carries a spin-$s$ representation of ${\text{SO}}(d)$. As before,
$|lw\rangle_{{i_{1}}\cdots i_{s}}$ has quantum number $-n$ under $H$ and is
annihilated by $K_{i}$. To represent the lowest weight state as a
wavefunction, we need to introduce another copy of $\mathbb{R}^{d+2}$ with
coordinate $Y^{A}$ such that $L_{AB}$ is realized as

$\displaystyle
L_{AB}=X_{A}\partial_{X_{B}}-X_{B}\partial_{X_{A}}+Y_{A}\partial_{Y_{B}}-Y_{B}\partial_{Y_{A}}$
(H.11)

Define complex (lightcone) coordinate for the $Y$-space
$w=Y_{d+1}+iY_{0},\bar{w}=Y_{d+1}-iY_{0}$ and then $H,P_{i},K_{i}$ can be
expressed as

$\displaystyle H=z\partial_{z}-\bar{z}\partial_{\bar{z}}$
$\displaystyle+w\partial_{w}-\bar{w}\partial_{\bar{w}},\,\,\,\,\,P_{i}=z\partial_{X_{i}}-2\,X_{i}\partial_{\bar{z}}+w\partial_{Y_{i}}-2\,Y_{i}\partial_{\bar{w}}$
$\displaystyle
K_{i}=-\bar{z}\partial_{X_{i}}+2\,X_{i}\partial_{z}-\bar{w}\partial_{Y_{i}}+2\,Y_{i}\partial_{w}$
(H.12)

It is easy to check that the following wavefunction actually corresponds to
the lowest weight state

$\displaystyle\psi^{lw}_{i_{1}\cdots
i_{s}}=\bar{z}^{n-s}(\bar{z}Y-\bar{w}X)_{i_{1}}\cdots(\bar{z}Y-\bar{w}X)_{i_{s}}-\text{trace}$
(H.13)

Introduce a null $d$-vector $u_{i}$ and then $\psi^{lw}_{i_{1}\cdots i_{s}}$
can be more efficiently written as

$\displaystyle\psi_{ns}^{lw}(X_{A},Y_{B};u_{i})=\bar{z}^{n-s}(\bar{z}\,Y\cdot
u-\bar{w}\,X\cdot u)^{s}$ (H.14)

The claim
$\left[\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)\right]_{0}=\left[\frac{D^{d}_{s}x^{-n}-D^{d}_{n+1}x^{1-s}}{(1-x)^{d}}\right]_{0}$,
in particular the $x^{1-s}$ part on the R.H.S, signals that there should be a
constraint on the descendants $\psi^{lw}_{i_{1}\cdots i_{s}}$ at level
$(n+1-s)$. To show this, let’s focus on the component of
$\psi^{lw}_{i_{1}\cdots i_{s}}$ with the highest degree in $X_{i}$, which is
roughly $\bar{z}^{n-s}\bar{w}^{s}X_{i_{1}}\cdots X_{i_{s}}$ up to pure trace
component. The top $X_{i}$ components in the descendants of
$\psi^{lw}_{i_{1}\cdots i_{s}}$ come from $-2X_{i}\partial_{\bar{z}}\in P_{i}$
with $\partial_{\bar{z}}$ acting on $\bar{z}^{n-s}$, that is,

$\displaystyle P_{i_{1}}\cdots P_{i_{k}}\psi^{lw}_{j_{1}\cdots
j_{s}}\ni(-2)^{k}\left(\partial^{k}_{\bar{z}}\bar{z}^{n-s}\right)\,\bar{w}^{s}X_{i_{1}}\cdots
X_{i_{k}}X_{j_{1}}\cdots X_{j_{s}}$ (H.15)

Therefore for level $0\leq k\leq n-s$, all the naive descendants
$P_{i_{1}}\cdots P_{i_{k}}\psi^{lw}_{j_{1}\cdots j_{s}}$ are nonvanishing and
linearly independent. However at level $k=n+1-s$, one can easily check that
$(u\cdot P)^{k}\psi_{ns}^{lw}$ vanishes identically, which means the following
spin-$(n+1)$ wavefunction is actually zero

$\displaystyle P_{(i_{1}}\cdots P_{i_{n+1-s}}\psi^{lw}_{j_{1}\cdots
j_{s})}-\text{trace}=0$ (H.16)

To be more explicit, let’s take the two-form representation
${\mathbb{Y}}_{1,1}$ as an example. In this case, the lowest-weight
wavefunction is

$\displaystyle\psi^{lw}_{i}(X,Y)=\bar{z}\,Y_{i}-\bar{w}\,X_{i}$ (H.17)

and the states at the next level

$\displaystyle
P_{j}\psi^{lw}_{i}(X,Y)=2(X_{i}\,Y_{j}-X_{j}\,Y_{i})+(w\,\bar{z}-z\,\bar{w})\delta_{ij}$
(H.18)

where we only have antisymmetric and pure trace components and hence
$P_{(j}\psi^{lw}_{i)}-\frac{\delta_{ij}}{d}P_{k}\psi^{lw}_{k}=0$. Without the
constraint (H.16), the polar part of
$\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)$ would simply be encoded in
$D^{d}_{s}\frac{x^{-n}}{(1-x)^{d}}$, i.e. the result of $s=0$ multiplied by a
spin degeneracy. With this constraint, we should subtract the contribution of
itself together with its descendants, which is
$D^{d}_{n+1}\frac{x^{1-s}}{(1-x)^{d}}$ from a simple counting. Altogether, we
get

$\displaystyle\left[\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)\right]_{0}=\left[\frac{D^{d}_{s}x^{-n}-D^{d}_{n+1}x^{1-s}}{(1-x)^{d}}\right]_{0}=\left[P^{d}_{n,s}(x)\right]_{0}$
(H.19)

The claim 1 has some important corollaries. We first rewrite $P^{d}_{n,s}$ as

$\displaystyle P^{d}_{n,s}(x)=D^{d}_{s}Q^{d}_{n}(x)-D^{d}_{n+1}Q^{d}_{s-1}(x)$
(H.20)

Since $P^{d}_{n,s}(x)$ encodes the character of ${\mathbb{Y}}_{n,s}$ and
$Q^{d}_{n}(x)$ encodes the character of ${\mathbb{Y}}_{n}$, the elementary
equation (H.20) yields a highly nontrivial relation of characters

$\displaystyle\boxed{\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n,s}}(x)=D^{d}_{s}\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{n}}(x)-D^{d}_{n+1}\Theta^{{\text{SO}}(d+2)}_{{\mathbb{Y}}_{s-1}}(x)}$
(H.21)

It allows us to construct the Weyl characters of two-row representations by
only using the Weyl characters of single-row representations. Taking the limit
$x\to 1$ on both sides of (H.21), we obtain a nontrivial relation involving
the dimensions of different representations

$\displaystyle\boxed{D^{d+2}_{n,s}=D^{d}_{s}D^{d+2}_{n}-D^{d}_{n+1}D^{d+2}_{s-1}}$
(H.22)

This formula plays an important role in the derivation of the character
integral representation of one-loop partition functions of spinning fields in
$\text{dS}_{d+1}$ [28].

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