# Holography on the Quantum Disk

###### Abstract

Motivated by recent study of DSSYK and the non-commutative nature of its bulk
dual, we review and analyze an example of a non-commutative spacetime known as
the quantum disk proposed by L. Vaksman. The quantum disk is defined as the
space whose isometries are generated by the quantum algebra
$U_{q}(\mathfrak{su}_{1,1})$. We review how this algebra is defined and its
associated group $SU_{q}(1,1)$ that it generates, highlighting its non-trivial
coproduct that sources bulk non-commutativity. We analyze the structure of
holography on the quantum disk and study the imprint of non-commutativity on
the putative boundary dual.

###### Contents

  1. 1 Introduction
  2. 2 A review of the quantum disk
    1. 2.1 $SU_{q}(1,1)$: the symmetry group of the quantum disk
    2. 2.2 $U_{q}(\mathfrak{su}_{1,1})$: generators of $SU_{q}(1,1)$
    3. 2.3 Coordinates on the quantum disk and their algebra
      1. 2.3.1 Disk coordinates
      2. 2.3.2 Differential $q$-calculus
  3. 3 Holography
    1. 3.1 Asymptotic analysis
    2. 3.2 Propagators
    3. 3.3 Boundary two point function
  4. 4 $q$-Conformal Quantum Mechanics
    1. 4.1 Correlation functions
      1. 4.1.1 Two point function
      2. 4.1.2 Operator Product Expansion
  5. 5 Discussion
    1. 5.1 $q$-Conformal quantum mechanics
    2. 5.2 $q$-Schwarzian action
    3. 5.3 UV lattice and Renormalization, UV/IR mixing, q affects both
    4. 5.4 q-Conformal Blocks
    5. 5.5 DSSYK at $q<1$ and large $\beta$
  6. A Introduction to Quantum Groups
    1. A.1 General Philosophy
    2. A.2 Hopf algebra
      1. A.2.1 Hopf algebra $\mathcal{O}(SL_{2}(\mathbb{R}))$
      2. A.2.2 Hopf algebra $\mathcal{O}(SL_{2}(\mathbb{R}))$
    3. A.3 Invariant elements
    4. A.4 Powers of the invariant distance
  7. B $R$ Matrix approach to $SL_{q}(2)$
  8. C Casimir and derivatives

##  1 Introduction

Everything in quantum gravity should be quantum, including spacetime. A notion
of a quantum spacetime is described by a non-commutative geometry i.e. one
whose coordinates do not commute with one another. Examples of such spacetimes
have appeared in several areas in theoretical physics including string
theory[1], matrix models [2] and quantum field theories on non-commutative
spaces [3, 4, 5, 6].

Recently, non-commutative geometry has showed up in the bulk picture of double
scaled SYK model [7, 8, 9, 10]. The double scaling limit is the limit where
the number of fermions $N$ and the order of the fermion interaction in the SYK
Hamiltonian $p$ are taken to infinity while keeping $\lambda\equiv 2p^{2}/N$
fixed [11]. The bulk picture that emerges is described in terms of chords
anchored to the boundary [7, 8, 12]. These chords provide an unusual,
primitive version of spacetime where distances in the bulk are measured in
units of $\lambda$, the effective Planck length in the bulk (or ratio between
the Planck and AdS scales). Smooth spacetime emerges in the small $\lambda$
limit in which the number of chords proliferate and smooth spacetime emerges.

At finite temperature of DSSYK, symmetry algebra of the spacetime is a
subalgebra of $U_{q}(\mathfrak{su}_{1,1})$ [10]. The “quantum” aspect of the
algebra is the presence of a non-trivial coproduct. The action of an
infinitesimal generator on elements of tensor product of
$U_{q}(\mathfrak{su}_{1,1})$ modules $f\otimes g$ does not satisfy the Leibniz
rule, but instead has a “tailed” action given by

$\displaystyle E[f\otimes g]=E[f]\otimes g+K[f]\otimes E[g],$ (1.1)

where $E,K\in U_{q}(\mathfrak{su}_{1,1})$. The discreteness of the chord
spacetime is intimately related to this coproduct structure.

Surprisingly, this non-trivial coproduct has a remnant in the
$\lambda\rightarrow 0$ limit at finite temperature, known as the large $p$
limit of SYK. As shown in [10], the two point function in this regime is not
conformal, yet it is related to a conformal correlator through a non-symmetric
coordinate transformation on the coordinates of the two points:

$\displaystyle\cos^{-2\Delta}\left[{\pi v\over
2}\left(1-{\theta_{+}-\theta_{-}\over\pi}\right)\right]=\sin^{-2\Delta}\left[{\varphi_{+}-\varphi_{-}\over
2}\right],$ (1.2)

where $\varphi_{\pm}=\pm{\pi\over 2}+v\left(\theta_{\pm}\mp{\pi\over
2}\right)$. The coordinates $\varphi_{\pm}$ are the coordinates on the “fake
disk” whose boundary length is $\beta/v$. The fake disk coordinates makes the
$SU(1,1)$ symmetry manifest. For instance, it is invariant under the action of
$E[\mathcal{O}]\equiv\cos\varphi\partial_{\varphi}\mathcal{O}-\Delta\sin\varphi\mathcal{O}$.
If we express this symmetry in terms of physical disk coordinates
$\theta_{\pm}$, the generator $E$ will have to act differently on the
$\mathcal{O}(\theta_{+})$ and $\mathcal{O}(\theta_{-})$ due of the asymmetry
in the transformation between $\varphi_{\pm}$ and $\theta_{\pm}$. However,
this asymmetry can be compensated by assigning a non-trivial coproduct to the
action of $E$ by introducing another operator $P$ whose actions in terms of
physical coordinates is

$\displaystyle
E[\mathcal{O}(\theta_{+})\mathcal{O}(\theta_{-})]=E[\mathcal{O}(\theta_{+})]\mathcal{O}(\theta_{-})+P[\mathcal{O}(\theta_{+})]P\cdot
E[\mathcal{O}(\theta_{-})],$ (1.3)

where

$\displaystyle E[\mathcal{O}(\theta)]\equiv v^{-1}\cos\left[{\pi\over
2}+v\left(\theta-{\pi\over
2}\right)\right]\partial_{\theta}\mathcal{O}(\theta)-\Delta\sin\left[{\pi\over
2}+v\left(\theta-{\pi\over 2}\right)\right]\mathcal{O}(\theta),$ (1.4)
$\displaystyle P[\mathcal{O}(\theta)]\equiv\mathcal{O}(\theta-{\pi(1-v)/v}).$
(1.5)

These observations motivate the study of standard examples involving a non-
trivial coproduct, hence this paper on the quantum disk. This paper will
analyze the structure of holography on a non-commutative version of the
hyperbolic disk, also known as the quantum disk that was studied extensively
by L. Vaksman [13, 14, 15, 16, 17, 18]. This is a non-commutative spacetime
whose symmetry group is $SU_{q}(1,1)$.

This topic is sufficiently foreign to our community that we felt that we
should dedicate a large fraction of this paper to review L. Vaksman’s seminal
works on the quantum disk. Hence, sections 2 is a pedagogical review of [13,
14, 15, 16, 17, 18] describing how to define the quantum disk and its symmetry
quantum group. In section 3, we describe the mechanics of holography,
extracting obundary correlation functions and induced quantum symmmetry
transformations of boundary primary operators. In section 4, we study symmetry
aspects of the putative boundary dual of the quantum disk. We find the three
point function using symmetry and match it to a bulk Witten diagram.

Our contribution will be in reframing his results from the perspective of
holography, especially in extracting boundary correlation functions from the
bulk. In section 4 we study symmetry aspects of the putative boundary dual of
the quantum disk. We find the three point function using symmetry and match it
to a bulk Witten diagram.

##  2 A review of the quantum disk

In this section we describe how to define the quantum disk [13, 14, 15, 16,
17, 18], a $q$-deformed version of the hyperbolic disk, as the space whose
symmetry group is the quantum group $SU_{q}(1,1)$ [19, 20, 21, 22]. The
coordinates on the quantum disk will be noncommutative, a property that
descends from the noncommutativity of the matrix elements of group elements of
$SU_{q}(1,1)$. We will not describe the intrinsic geometric structure of the
quantum disk, which is obscured by its noncommutative nature, but rather focus
on the properties of functions of its coordinates.

###  2.1 $SU_{q}(1,1)$: the symmetry group of the quantum disk

To understand the symmetry group of the quantum disk $SU_{q}(1,1)$, we start
with the quantum deformation of the $SL(2,\mathbb{C})$, called $SL_{q}(2)$.
Elements of $SL_{q}(2)$ are all $2\times 2$ matrices, whose entries satisfy
the following relations

$\displaystyle SL_{q}(2)=\left\\{\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\Bigg{|}\ \ \ \begin{matrix}ab=qba,\ \ \ bd=qdb,\ \ \
ad=da+(q-q^{-1})cb\\\ ac=qca,\ \ \ cd=qdc,\ \ \ bc=cb,\ \ \ ad-
qbc=1\end{matrix}\right\\},$ (2.1)

The combination $\mathrm{det}_{q}\equiv ad-qbc$ is called the quantum
determinant and belongs to the center of the $a,b,c,d$ algebra. Without the
condition $\mathrm{det}_{q}=1$, (2.1) defines $GL_{q}(2)$. The above relations
generate an ideal and are determined by a solution to an $R$-matrix equation
(B.2). We describe this in appendix B.

The equation (2.1) is the defining representaion of $SL_{q}(2)$ [23]. It is
sometimes called its “coordinate representation” since the matrix elements
define coordinates parameterizing group elements on the group manifold. Just
like normal groups, this group includes the identity element $e$ where
$a=d=1,b=c=0$, which satisfies the relations (2.1). Thus, one can define a map
$\epsilon$ that takes any group element $g\in SL_{q}(2)$ to the identity. It
is known as the co-unit $\epsilon(g)=e$.

We can define the inverse of the above elements as

$\displaystyle g=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\rightarrow
g^{-1}=\begin{pmatrix}d&-q^{-1}b\\\ -qc&a\end{pmatrix}.$ (2.2)

Formally, we have $g\cdot g^{-1}=g^{-1}\cdot g=e$. The map from a group
element to its inverse is known as the antipode $S:g\rightarrow g^{-1}$. On
the entries of matrix defining the group element we have

$\displaystyle S(a)=d,\ S(b)=-q^{-1}b,\ S(c)=-qc,\ S(d)=a.$ (2.3)

Group multiplication can also be defined for $SL_{q}(2)$. It is a map from the
matrix elements of a pair of group elements $g_{1}$ and $g_{2}$ to their
product $g_{1}g_{2}$ in a way that preserves the ideal (2.1). To define this,
first we should think of the matrix elements $a,b,c,d$ as functions on the
group manifold that assign a value (abstractly) to each group element, i.e.
the matrix elements of a group element $g$ are $a(g),b(g),c(g),d(g)$. Then,
the multiplication can be defined as a map $\hat{\times}$ that takes matrix
elements of two group elements to matrix elements of a product of these
elements

$\displaystyle\begin{pmatrix}a(g_{1})&b(g_{1})\\\
c(g_{1})&d(g_{1})\end{pmatrix}\hat{\times}\begin{pmatrix}a(g_{2})&b(g_{2})\\\
c(g_{2})&d(g_{2})\end{pmatrix}$ (2.4)
$\displaystyle=\begin{pmatrix}a(g_{1})\otimes a(g_{2})+b(g_{1})\otimes
c(g_{2})&a(g_{1})\otimes b(g_{2})+b(g_{1})\otimes d(g_{2})\\\ c(g_{1})\otimes
a(g_{2})+d(g_{1})\otimes c(g_{2})&c(g_{1})\otimes b(g_{2})+d(g_{1})\otimes
d(g_{2})\end{pmatrix}\equiv\begin{pmatrix}a(g_{1}g_{2})&b(g_{1}g_{2})\\\
c(g_{1}g_{2})&d(g_{1}g_{2})\end{pmatrix}.$

Thus we have mapped the matrix element functions $a,b,c,d$ to the functions
acting on two copies of the group manifold. This operation of mapping
functions on one copy of a space to the functions on two copies is called the
coproduct, and is denoted by $\Delta(\cdot)$. For $SL_{q}(2)$ we have

$\displaystyle\Delta(a)=a\otimes a+b\otimes c,$ (2.5)
$\displaystyle\Delta(b)=a\otimes b+b\otimes d,$ (2.6)
$\displaystyle\Delta(c)=c\otimes a+d\otimes c,$ (2.7)
$\displaystyle\Delta(d)=c\otimes b+d\otimes d.$ (2.8)

The coproduct is an algebra homomorphism.

One is often interested in defining functions on the group manifold. This
space of functions is a vector space spanned by monomials
$a^{i}b^{j}c^{k}d^{l}$ with integer powers and is denoted as
$\mathcal{O}(SL_{q}(2))$. The notions of co-unit, antipode and coproduct
extend to this algebra in the obvious way. The co-unit implements
$\epsilon:\mathcal{O}(SL_{q}(2))\rightarrow\mathbb{C}$, while the coproduct
implements
$\Delta:\mathcal{O}(SL_{q}(2))\rightarrow\mathcal{O}(SL_{q}(2))\otimes\mathcal{O}(SL_{q}(2))$.
We can introduce “reversed” maps that are compatible with these maps. These
maps are defined as the unit that maps a constant to the identity element of
the algebra $\eta:\mathbb{C}\rightarrow\mathcal{O}(SL_{q}(2))$ (it also maps a
constant to the identity of the group $SL_{q}(2)$) and a multiplication called
the product that maps a tensor product to a single copy
$m:\mathcal{O}(SL_{q}(2))\otimes\mathcal{O}(SL_{q}(2))\rightarrow\mathcal{O}(SL_{q}(2))$
acting as as $m(a\otimes b)=ab$. These structures together make
$\mathcal{O}(SL_{q}(2))$ a Hopf algebra.

To get $SU_{q}(1,1)$ we need to impose reality conditions on $SL_{q}(2)$. For
$q=1$, both are Mobius transformations on the complex plane, but $SU(1,1)$
preserves the unit disk. Defining complex conjugation makes
$\mathcal{O}(SL_{q}(2))$ into a Hopf $*$-algebra with an involution $*$ that
satisfies $(ab)^{*}\rightarrow b^{*}a^{*}$, i.e. it’s an anti-involution. The
reality condition for $SU_{q}(1,1)$ is $g^{*}=\eta S(g)\eta$ where
$\eta=\mathop{\rm diag}\nolimits\\{1,-1\\}$ and so

$\displaystyle\begin{pmatrix}a^{*}&c^{*}\\\
b^{*}&d^{*}\end{pmatrix}=\begin{pmatrix}d&q^{-1}b\\\ qc&a\end{pmatrix},$ (2.9)

This defines a conjugation on all elements in $\mathcal{O}(SU_{q}(1,1))$.

###  2.2 $U_{q}(\mathfrak{su}_{1,1})$: generators of $SU_{q}(1,1)$

One can define a set of generators on the group manifold of $SU_{q}(1,1)$ that
formally implements an infinitesimal translation. These generators are usually
labelled $K,E,F$ and they satisfy an algebra known as
$U_{q}(\mathfrak{su}_{1,1})$ (more precisely, this is the vector space spanned
by polynomials of $K,E,F$). Their action on on elements of $SU_{q}(1,1)$ is
defined through the pairing

$\displaystyle q^{-1}\langle K,a\rangle=q\langle K,d\rangle=q^{-1/2}\langle
E,c\rangle=q^{1/2}\langle F,b\rangle=1\,,$ (2.10)

Hence, on a group element we have

$\displaystyle\langle K,g\rangle=\begin{pmatrix}q&0\\\
0&q^{-1}\end{pmatrix},\quad\langle E,g\rangle=\begin{pmatrix}0&0\\\
q^{1/2}&0\end{pmatrix},\quad\langle F,g\rangle=\begin{pmatrix}0&q^{-1/2}\\\
0&0\end{pmatrix}.$ (2.11)

The pairing extends to relate all elements of $U_{q}(\mathfrak{su}_{1,1})$ to
$\mathcal{O}(SU_{q}(1,1))$, and establishes a duality between the two
algebras. To do so, it must satisfy a set of constraints (see appendix (A)).

One constraint is it must respect the algebra relations (or ideal) in (2.1)
once the pairing is extended to all elements of $\mathcal{O}(SU_{q}(1,1))$.
For instance, we must have $\langle E,ac\rangle=q\langle E,ca\rangle$. There
are multiple ways of extending the pairing, one of which is to impose that

$\displaystyle\langle K,AB\rangle$ $\displaystyle=\langle K,A\rangle\langle
K,B\rangle,$ (2.12) $\displaystyle\langle E,AB\rangle$ $\displaystyle=\langle
E,A\rangle\langle 1,B\rangle+\langle K,A\rangle\langle E,B\rangle,$ (2.13)
$\displaystyle\langle F,AB\rangle$ $\displaystyle=\langle F,A\rangle\langle
K^{-1},B\rangle+\langle 1,A\rangle\langle F,B\rangle,$ (2.14)

for any $A,B\in\mathcal{O}(SL_{q}(2))$. The pairing with the identity
implements the co-unit, $\langle 1,B\rangle=\epsilon(B)$. These relations
provide a coproduct for $U_{q}(\mathfrak{su}_{1,1})$ given by

$\displaystyle\Delta(K)$ $\displaystyle=K\otimes K,$ (2.15)
$\displaystyle\Delta(E)$ $\displaystyle=E\otimes 1+K\otimes E,$ (2.16)
$\displaystyle\Delta(F)$ $\displaystyle=F\otimes K^{-1}+1\otimes F,$ (2.17)

provided that the pairing satisfies

$\displaystyle\langle X\otimes Y,A\otimes B\rangle=\langle X,A\rangle\langle
Y,B\rangle,$ (2.18)

where $X,Y\in U_{q}(\mathfrak{su}_{1,1})$. As a result, the pairing can be
shown to satisfy

$\displaystyle\langle\Delta(XY),A\otimes B\rangle=\langle X\otimes
Y,\Delta(AB)\rangle.$ (2.19)

The generators $K,E,F$ must satisfy a set of reality conditions compatible
with the reality condition of $\mathcal{O}(SU_{q}(1,1))$. This can be defined
provided the pairing satisfies

$\displaystyle\langle X^{*},A\rangle=\langle X,S(A)^{*}\rangle^{*}.$ (2.20)

Using the ${\bf coproduct}$ from $K,E,F$, the antipode relations (2.3) and the
relations (2.9), implies the following reality conditions

$\displaystyle K^{*}=K,\ E^{*}=-KF,\ F^{*}=-EK^{-1}.$ (2.21)

Further conditions on the pairing can be used to define a unit, co-unit,
antipode, a product for the algebra of $K,E,F$ that makes it a Hopf
$*$-algebra. We review these additional structures in appendix A.2. So far, we
have established a duality between two Hopf $*$-algebras
$\mathcal{O}(SU_{q}(1,1))$ and freely generated algebra of $K,E,F$.

Finally, the pairing (2.10) is not faithful and has a kernel generated by the
relations, or ideal, between $K,E,F$,

$\displaystyle KE=q^{2}EK,\ \ KF=q^{-2}FK,\ \ EF-FE={K-K^{-1}\over q-q^{-1}}.$
(2.22)

See appendix A.2.2 for a derivation of this kernel. Imposing these on the
algebra of $K,E,F$ defines the algebra $U_{q}(\mathfrak{su}_{1,1})$. From this
one can deduce a Casimir that commutes with the generators

$\displaystyle C_{q}=EF+{q^{-1}(K-1)+q(K^{-1}-1)\over(q^{-1}-q)^{2}}.$ (2.23)

The action of the Casimir on tensor product representations descends from the
coproduct of the generators, and is given by

$\displaystyle\Delta(C_{q})=C_{q}\otimes K^{-1}+K\otimes C_{q}+E\otimes
F+KF\otimes EK^{-1}-\frac{(q^{-1}+q)(K-1)\otimes(K^{-1}-1)}{(q-q^{-1})^{2}}.$
(2.24)

###  2.3 Coordinates on the quantum disk and their algebra

Having defined the symmetry group of the quantum disk and its associated
generators, we turn to defining the quantum disk itself. Recall that the
standard hyperbolic disk can be obtained from a quotient of the
$SL_{2}(\mathbb{R})$ group manifold, see [24]. Similarly, as discussed in
[16], the quantum disk can be defined as the quotient of $SU_{q}(1,1)$ by the
right action of $K$,

$\displaystyle SU_{q}(1,1)=\left\\{\begin{pmatrix}t_{11}&t_{12}\\\
t_{21}&t_{22}\end{pmatrix}\Bigg{|}\begin{matrix}t_{11}t_{12}=qt_{12}t_{11},\
t_{12}t_{22}=qt_{22}t_{12},\ t_{11}t_{21}=qt_{21}t_{11},\
t_{21}t_{22}=qt_{22}t_{21},\\\ t_{12}t_{21}=t_{21}t_{12},\
t_{11}t_{22}=t_{22}t_{11}+(q-q^{-1})t_{21}t_{12},\
t_{11}t_{22}-qt_{12}t_{21}=1,\ \ \end{matrix}\right\\}$
$\displaystyle\mathbb{D}_{q}=SU_{q}(1,1)/K.$ (2.25)

The action (or coaction) of $SU_{q}(1,1)$ on the disk correspond to left
matrix multiplication $t^{\prime}=g\times t$ that maps to the tensor product
as

$\displaystyle t_{11}\rightarrow t^{\prime}_{11}=a\otimes t_{11}+b\otimes
t_{21},$ (2.26) $\displaystyle t_{12}\rightarrow t^{\prime}_{12}=a\otimes
t_{12}+b\otimes t_{22},$ (2.27) $\displaystyle t_{21}\rightarrow
t^{\prime}_{21}=c\otimes t_{11}+d\otimes t_{21},$ (2.28) $\displaystyle
t_{22}\rightarrow t^{\prime}_{22}=c\otimes t_{12}+d\otimes t_{22}.$ (2.29)

The parameters $t_{ij}^{\prime}$ have the same algebra as $t_{ij}$. It will
useful later to note that
$\det\nolimits_{q}(t^{\prime})=\det\nolimits_{q}(g\otimes
t)=\det\nolimits_{q}g\otimes\det\nolimits_{q}t$ where
$\det\nolimits_{q}t=t_{11}t_{22}-qt_{12}t_{21}$.

It is useful to analyze the action of generators on $t_{ij}$. The action of
the generators on $t_{ij}$ is obtained using (2.11) implying the following
transformation rules

$\displaystyle K(t_{11})=qt_{11},\ K(t_{12})=qt_{12},\
K(t_{21})=q^{-1}t_{21},\ K(t_{22})=q^{-1}t_{22},$ (2.30) $\displaystyle
E(t_{11})=0,\ E(t_{12})=0,\ E(t_{21})=q^{1/2}t_{11},\
E(t_{22})=q^{1/2}t_{12},$ (2.31) $\displaystyle F(t_{11})=q^{-1/2}t_{21},\
F(t_{12})=q^{-1/2}t_{22},\ F(t_{21})=0,\ F(t_{22})=0.$ (2.32)

To obtain the action of $U_{q}(\mathfrak{sl}_{2})$ on arbitrary products and
powers of $t_{ij}$, we follow the same rules as those defined in the pairing
in the previous section. The coproduct then implements the action on products
of $t_{ij}$.

####  2.3.1 Disk coordinates

The quotient to get the quantum disk can be implemented defining the disk
coordinates $z=q^{-1}t^{-1}_{21}t_{11},z^{*}=t_{22}t_{12}^{-1}$, which are
indeed invariant under the rescaling of $t_{ij}$ by the action of $K$. For
these two coordinates to actually be conjugates we have to apply the
$SU_{q}(1,1)$ reality conditions $t_{11}^{*}=t_{22},t_{12}^{*}=qt_{21}$. This
also follows from the conjugation relations (2.9) and by conjugating either
side of the transformations $t_{ij}\rightarrow t_{ij}^{\prime}$. Under the
above transformation rules, we have

$\displaystyle z\rightarrow w=q^{-1}t^{\prime-1}_{21}t^{\prime}_{11}=$
$\displaystyle(c\otimes t_{11}+d\otimes t_{21})^{-1}(a\otimes t_{11}+b\otimes
t_{21})$ $\displaystyle=$ $\displaystyle q^{-1}(qcz+d)^{-1}(qaz+b),$ (2.33)

which is the usual Mobius transformation. The tensor product symbol is dropped
from the last equation for brevity. As a check, one can show that this mapping
preserves the boundary; suppose that $w,w^{*}$ are constructed from the matrix
$t^{\prime}$ just as $z,z^{*}$ are constructed from $t$. Boundary points in
$z$ correspond to $zz^{*}=1$ which satisfy $\det\nolimits_{q}t=0$ which
implies that $\det\nolimits_{q}t^{\prime}=0$ and hence $ww^{*}=1$.111Since the
disk, classical or quantum, is strictly an open set, it doesn’t include the
boundary and hence $\det\nolimits_{q}$ is allowed to vanish.

The action of the generators on $z,z^{*}$ can be deduced from their action on
$t_{ij}$ and using the coproduct, which gives

$\displaystyle K(z)=q^{2}z,\ E(z)=-q^{5/2}z^{2},\ F(z)=q^{-3/2},$ (2.34)
$\displaystyle K(z^{*})=q^{-2}z^{*},\ E(z^{*})=q^{1/2},\
F(z^{*})=-q^{1/2}z^{*2}.$ (2.35)

Again, the coproduct we can extend this action to powers of $z,z^{*}$. For
“(anti)holomorphic” functions we have

$\displaystyle K(g(z))=g(q^{2}z),\ E(g(z))=-q^{5/2}z^{2}{g(z)-g(q^{2}z)\over
z-q^{2}z},\ F(g(z))=q^{-3/2}{g(z)-g(q^{-2}z)\over z-q^{-2}z},$ (2.36)
$\displaystyle K(g(z^{*}))=g(q^{-2}z^{*}),\
E(g(z^{*}))=q^{1/2}{g(z^{*})-g(q^{-2}z^{*})\over z^{*}-q^{2}z^{*}},\
F(g(z))=-q^{1/2}z^{*2}{g(z^{*})-g(q^{2}z^{*})\over z^{*}-q^{2}z^{*}}.$ (2.37)

The quantum disk has the interesting property of being noncommutative, namely
that $z,z^{*}$ do not commute, which descends from the noncommutativity of the
matrix elements (2.25). Starting from
$t_{11}t_{22}=t_{22}t_{11}+(q-q^{-1})t_{21}t_{12}$, we can multiply from the
right(left) with $t_{21}^{-1}(t_{12}^{-1})$ respectively, and then use the
remaining relations in (2.25) to commute things around until we have an
expression involving only $z,z^{*}$. The resulting relation is

$\displaystyle z^{*}z=q^{2}zz^{*}+1-q^{2}.$ (2.38)

This relation can also be obtained using $R$-matrix methods. As discussed in
appendix B, the universal $R$-matrix for $U_{q}(\mathfrak{su}_{1,1})$ is given
by

$\displaystyle\breve{R}(\cdot)=\exp_{q^{2}}\left((q^{-1}-q)F\otimes
E\right)q^{-{H\otimes H\over 2}}(\cdot),$ (2.39)

which in particular gives

$\displaystyle z^{*}z=m\breve{R}(z\otimes z^{*})=q^{2}zz^{*}+1-q^{2},$ (2.40)

where the $m$ multiplies the elements of the tensor product together222We
define the algebra structure on the tensor products of two algebras $A_{1,2}$
as $(a_{1}\otimes b_{1})(a_{2}\otimes b_{2})=a_{1}a_{2}\otimes b_{1}b_{2}$..

The algebra of $z,z^{*}$ is denoted as $\mathbb{C}_{q}[z,z^{*}]$. An important
element of this algebra is an element $y=1-zz^{*}$ that satisfy the following
commutation relations

$\displaystyle zy=q^{-2}yz,\quad z^{*}y=q^{2}yz^{*}.$ (2.41)

It is useful to show connection of $\mathbb{C}_{q}[z,z^{*}]$ with the
$q$-deformed harmonic oscillator. Thus, we notice that

$\displaystyle\hat{a}=\frac{z}{\sqrt{1-q^{2}}},\quad\hat{a}^{\dagger}=\frac{z^{*}}{\sqrt{1-q^{2}}},\quad\hat{a}^{\dagger}\hat{a}-q^{2}\hat{a}\hat{a}^{\dagger}=1,$
(2.42)

thus any element of $f\in\mathbb{C}_{q}[z,z^{*}]$ could be thought as an
operator acting in the Hilbert space of $q$-deformed harmonic oscillator. That
Hilbert space is spanned by vectors $\ket{n},n\in\mathbb{N}_{\geq 0}$ and the
action is given by

$\displaystyle z\ket{n}=\ket{n+1},\quad
z^{*}\ket{n}=\left(1-q^{2n}\right)\ket{n-1},\quad y\ket{n}=q^{2n}\ket{n},$
(2.43)

one can show that this mapping is actually an isomorphism and any function
$\psi(y)$ of $y$ could be given just by giving the values of $\psi(q^{2n})$.
In some loose sense, the quantum disk is a set of concentric circles discrete
radii of constant $y$ that accumulate up to the unit circle.

####  2.3.2 Differential $q$-calculus

Various notions of differential calculus can be extended to the non-
commutative setting. One can define a notion of a differential operator $d$
that satisfies the usual Leibnez rule $dz^{2}=zdz+dz\,z$. The differential
form $dz$ satifies $d(dz)=0$, which follows from $d^{2}=0$. The action on
$z^{*}$ follows similarly.

The action of the generators of $U_{q}(\mathfrak{su}_{1,1})$ on the
differentials follows from the Leibnez rule along the commutations above to
give

$\displaystyle K(dz)=q^{2}dz,\ E(dz)=-q^{5/2}dzz(1+q^{-2}),\ F(dz)=0,$ (2.44)
$\displaystyle K(dz^{*})=q^{-2}dz^{*},\ E(dz^{*})=0,\
F(dz^{*})=-q^{1/2}dz^{*}z^{*}(1+q^{2}).$ (2.45)

The algebra between $dz,dz^{*},z,z^{*}$ follows from the $R$-matrix in (2.39).
We have

$\displaystyle zdz=q^{-2}dz\,z,\ z^{*}dz^{*}=q^{2}dz^{*}\,z^{*},$ (2.46)
$\displaystyle zdz^{*}=q^{-2}dz^{*}z,\ z^{*}dz=q^{2}dz\,z^{*}.$ (2.47)

The differentials allow us to define differential operators acting on any
function $h(z,z^{*})$ on the quantum disk through

$\displaystyle dh(z,z^{*})$ $\displaystyle=dz{1\over
q^{-2}z-z}\left({h(q^{-2}z,z^{*})-h(z,z^{*})}\right)+dz^{*}\left(h(q^{-2}z,z^{*})-h(q^{-2}z,q^{2}z^{*})\right){1\over
z^{*}-q^{2}z^{*}}$ $\displaystyle\equiv dz{\partial h\over\partial
z}+dz^{*}{\partial h\over\partial z^{*}}.$ (2.48)

This defines the derivative operators ${\partial\over\partial
z},{\partial\over\partial z^{*}}$. Holomorphic and anti-holomorphic
differentials can be defined as $\partial h=dz{\partial h\over\partial
z},\,\bar{\partial}h=dz^{*}{\partial h\over\partial z^{*}}$. This means we can
right $d=\partial+\bar{\partial}$. It’s not hard to check that these
derivatives satisfy the following algebra

$\displaystyle\ \ \ {\partial\over\partial
z}z^{*}=q^{2}z^{*}{\partial\over\partial z},\ {\partial\over\partial
z^{*}}z=q^{-2}z{\partial\over\partial z^{*}},$ (2.49)
$\displaystyle{\partial\over\partial z}z=1+q^{-2}z{\partial\over\partial z},\
{\partial\over\partial z^{*}}z^{*}=1+q^{2}z^{*}{\partial\over\partial z^{*}}.$
(2.50)

The Casimir can be expressed in terms of these derivatives. Using the form the
Casimir (2.23) and its coproduct(2.24), its given by

$\displaystyle C_{q}=q^{-1}\left(1-zz^{*}\right)^{2}\frac{\partial}{\partial
z}\frac{\partial}{\partial z^{*}}.$ (2.51)

We check in the appendix (C) that the two sides of this equation have the same
action on any monomial $z^{n}z^{*m}$.

Next we define how to perform integrals over the entire quantum disk. Such an
integral can be thought of as a map of functions on the quantum disk to the
complex numbers. Assuming we have a measure $d\nu$ that invariant under the
action of the generators $E,F,K$, then for any function we must impose

$\displaystyle\int_{\mathbb{D}_{q}}d\nu
E^{m}F^{n}h(z,z^{*})=\int_{\mathbb{D}_{q}}d\nu\left(K^{l}-1\right)h(z,z^{*})=0,$
(2.52)

for any $m,n,l>0$. Now suppose we want to compute the integral of an arbitrary
function $h(z,z^{*})$. Any such function can be expanded as

$\displaystyle
h(z,z^{*})=\sum_{k>0}h_{k}(y)z^{k}+h_{0}(y)+\sum_{k>0}h_{-k}(y)z^{*k}.$ (2.53)

Note that the terms with extra factors of $z,z^{*}$, i.e. the first and third
terms in the expansion are not invariant under the action of the $K$ and can
be expressed as $E^{a}\tilde{f}(y)$ and $F^{b}\hat{f}(y)$ for some functions
$\tilde{f}(y),\hat{f}(y)$, and hence their integral must vanish.333The lowbrow
version of this argument is to say that only the zero, or radial, mode
survives the integral on the disk. What remains is

$\displaystyle\int_{\mathbb{D}_{q}}d\nu h(z,z^{*})=\int_{\mathbb{D}_{q}}d\nu
h_{0}(y),$ (2.54)

i.e. the integral of a purely radial function. There are two ways of making
this formal expression more explicit. The first is to note that the spectrum
of $y$ is given by the discrete values $q^{2n}$ for all $n\geq 0$, and hence
this integral can be expressed as a sum over $h_{0}$ evaluated on these
points, namely

$\displaystyle\int_{\mathbb{D}_{q}}d\nu
h_{0}(y)=\sum_{n=0}^{\infty}a_{n}h_{0}(q^{2n}),$ (2.55)

for some coefficients $a_{n}$ which must be determined. This can be done be
considering the (vanishing) integral of $F$ acting on $zh_{0}(y)$. Since

$\displaystyle
F(zh_{0}(y))=q^{-3/2}h_{0}(y)-q^{1/2}(1-y){h_{0}(q^{-2}y)-h_{0}(y)\over
1-q^{2}}.$ (2.56)

The integral of the first and second terms on the right hand side must vanish,
and so we end up with the recursion relation and it’s corresponding solution.

$\displaystyle a_{n}={1\over
q^{-2}-1}\left((1-q^{2(n+1)})a_{n+1}-(1-q^{2n})a_{n}\right),\quad
a_{n}=q^{-2n}a_{0}.$ (2.57)

The value of $a_{0}$ is undetermined. However, if we set it to
$a_{0}=\pi(1-q^{2})$ then we reproduce the standard integral on the classical
hyperbolic disk in the limit $q\rightarrow 1$, which we will show below.

There’s a more intuitive but less precise way of arriving at this result.
First we need find a measure of this integral that’s invariant under the
action of $K,E,F$. One such measure is the following two-form

$\displaystyle\epsilon=-{2i}y^{-2}dz\wedge dz^{*}.$ (2.58)

This is a two form on the disk expressed in terms of the cotangent space
element $dz\wedge dz^{*}$. Note that (2.47) imply we have $dz\wedge
dz^{*}=-q^{-2}dz^{*}\wedge dz$.444The minus sign appears from commuting $d$
past $d$. This form motivates the replacement

$\displaystyle\int_{\mathbb{D}_{q}}d\nu=\alpha\int y^{-2}d_{q^{2}}y,$ (2.59)

where $\alpha$ is some constant and the measure $d_{q^{2}}y$ stands for
Jackson integral, that we are integrating along $y$ but in discrete steps
given by the difference of $y$ at two consecutive points, i.e.
$d_{q^{-2}}y=(q^{2n}-q^{2(n+1)})$. Then we have

$\displaystyle\int_{\mathbb{D}_{q}}d\nu
h_{0}(y)=k\sum_{n=0}^{\infty}h_{0}(q^{2n})q^{-4n}(q^{2n}-q^{2(n+1)})=k(1-q^{2})\sum_{n=0}^{\infty}h_{0}(q^{2n})q^{-2n}.$
(2.60)

This agrees with the (2.55) up to an overall constant.

An additional interesting representation of this integral uses the
representation (2.43), for that we notice that

$\displaystyle\int_{\mathbb{D}_{q}}d\nu h(z,z^{*})=(1-q^{2})\sum_{n}\langle
n|h(z,z^{*})|n\rangle q^{-2n}=\mathop{\rm
tr}\nolimits\left[h(z,z^{*})\frac{1}{1-z^{*}z}\right],$ (2.61)

such representation allows effectively compute the integrals on quantum disk.

As a check, we can compare this to the expression to classical hyperbolic disk
by taking $q\rightarrow 1$. Then we have $r^{2}=|z|^{2}=1-y=1-q^{2n}$, and
hence $q^{2n}-q^{2(n+1)}\approx 2rdr$ and we get

$\displaystyle\pi(1-q^{2})\sum_{n=0}^{\infty}h_{0}(q^{2n})q^{-2n}=\pi\int
h_{0}(1-r^{2}){2rdr\over(1-r^{2})^{2}},$ (2.62)

which is indeed the radial integral on the hyperbolic disk.

Finally, we discuss two useful relations that could be used in the computation
of the integrals. The first is integration by parts that could be expressed as

$\displaystyle\int_{\mathbb{D}_{q}}d\nu
Eh(z,z^{*})f(z,z^{*})=-\int_{\mathbb{D}_{q}}d\nu Kh(z,z^{*})Ef(z,z^{*})\,,$
(2.63)

that uses the action of $E$ on the product of two functions via coproduct and
the invariance of the integral with respect to the action $E$. The second is
the notion of a delta-function that can be defined as the element
$\delta_{z,\xi}\in\mathbb{C}_{q}[z,z^{*}]\otimes\mathbb{C}_{q}[\xi,\xi^{*}]$
that satisfies

$\displaystyle f(z,z^{*})=\int d\nu_{\xi}\delta_{z,\xi}f(\xi,\xi^{*}).$ (2.64)

##  3 Holography

This section will focus on holographic aspects of the quantum disk, focusing
on deriving boundary anchored propagators. We will restrict our analysis to
scalar fields. These fields will be arbitrary functions of $z,z^{*}$, and
hence are elements of $\mathbb{C}_{q}[z,z^{*}]$. The formulas (2.34, 2.35,
2.36, 2.37) will give us the action of the generators $E,F,K$ on the fields,
and the Casimir (2.23, 2.51) will provide the wave equation.

###  3.1 Asymptotic analysis

Suppose we have a field $\phi(z,z^{*})$ on the the quantum disk. This field is
an element $\phi\in\mathbb{C}_{q}[z,z^{*}]$. Without loss of generality, we
can use the commutation relations of $z,z^{*}$ to express it as

$\displaystyle\phi(z,z^{*})=\sum_{k>0}\phi_{k}(y)z^{k}+\phi_{0}(y)+\sum_{k>0}\phi_{-k}(y)z^{*k},$
(3.1)

where $y=1-zz^{*}$. We demand that the dynamics respects the $SU_{q}(1,1)$
symmetry of the quantum disk, which implies that its action must be invariant
under infinitesimal transformations generated by $K,E,F$. The simplest
$U_{q}(\mathfrak{sl}_{2})$ invariant wave equation linear in the field is the
sum of the Casimir and mass term

$\displaystyle\left(-\Box_{q}+m^{2}\right)\phi=0,\quad\mathrm{with}\ \
\Box_{q}=C=-q^{-1}(1-zz^{*})^{2}{\partial\over\partial
z}{\partial\over\partial z^{*}}.$ (3.2)

where we used the expression of the Casimir in (2.51) in the $z,z^{*}$
coordinates system. This wave equation follows from the action

$\displaystyle
S=\int_{\mathbb{D}_{q}}\left(-\phi\Box_{q}\phi+m^{2}\phi^{2}\right)d\nu.$
(3.3)

It is instructive to see how to integrate this by parts to express the kinetic
term as a square. With the appropriate measure and definition of $\Box$, we
write this integral as

$\displaystyle-\int_{\mathbb{D}_{q}}\phi\Box_{q}\phi\,d\nu$
$\displaystyle=q^{2}\int_{\mathbb{D}_{q}}\phi(z,z^{*})y^{2}{\partial\over\partial
z}{\partial\over\partial z^{*}}\phi(z,z^{*})\,y^{-2}dz\wedge dz^{*}$
$\displaystyle=\int_{\mathbb{D}_{q}}\phi(z,z^{*})y^{2}{\partial\over\partial
z^{*}}{\partial\over\partial z}\phi(z,z^{*})\,y^{-2}dz\wedge dz^{*}$
$\displaystyle=\int_{\mathbb{D}_{q}}\phi(z,z^{*})dz\wedge
dz^{*}{\partial\over\partial z^{*}}{\partial\over\partial z}\phi(z,z^{*})$
$\displaystyle=\int_{\mathbb{D}_{q}}\left(dz^{*}{\partial\over\partial
z^{*}}\phi(z,z^{*})\right)\wedge\left(dz{\partial\over\partial
z}\phi(z,z^{*})\right),$ (3.4)

where in the first step we commute the derivatives which pick up a factor of
$q^{-2}$. In the second step we move the measure through to the middle at no
expense. In the third, we commute the two factors of the measure and
integrated by part using that $\int\bar{\partial}\left(f_{1}dzf_{2}\right)=0$.

Studying the asymptotic solutions reveals some interesting aspects of fields
on the quantum disk. Without any sources on the boundary, we should consider
solutions that fall off near the boundary. Without loss of generality, we can
consider

$\displaystyle\phi(z,z^{*})\approx\left(f(z)+f^{*}(z^{*})\right)\,y^{\Delta}+...$
(3.5)

Note that on the boundary we can make the coefficient function to be only a
function of $z$ since the boundary satisfies $zz^{*}=1$. Since $y=0$ at the
boundary, using (2.24) the leading form of the wave equation reduces to

$\displaystyle{(1-q^{2\Delta})(1-q^{2-2\Delta})\over(1-q^{2})^{2}}+m^{2}=0.$
(3.6)

There are several interesting aspects of this expression. First, the left term
is the eigenvalue of the Laplacian (since the solution is an eigenvector
asymptotically) and was also obtained in [16]. As discussed in [25], to obtain
a solution with the boundary condition (3.5) that is real and non-singular
everywhere in the bulk, we must pick the principal series representation which
sets $\Delta={1\over 2}+i\rho$. The eigenvalue becomes

$\displaystyle\lambda_{\rho}={1+q^{2}-2q\cos\left(2\rho\log
q\right)\over(1-q^{2})^{2}},$ (3.7)

and hence the Laplacian is bounded [16, 25]

$\displaystyle{1\over(1+q)^{2}}\leq\Box_{q}\leq{1\over(1-q)^{2}}.$ (3.8)

The boundedness of the Laplacian implies that theory is finite, at least
classically, and in particular has no UV divergences at coincident points. In
the $q\to 1$ we restore the Breitenlohner-Freedman bound.

The second interesting fact about (3.2) is that the bound on eigenvalues
places a bound on masses of (classical) fields on this space! For $q\approx
1$, we see that the mass is upper bounded by $\ln q$, the unit of discreteness
of the radial direction, which would be the Planck scale from the perspective
of DSSYK.

###  3.2 Propagators

We will think of the propagator as a function on a tensor product of two
copies of the quantum disk with coordinates $z,z^{*}$ and $w,w^{*}$. Hence,
the propagator will be the following element

$\displaystyle G(z,z^{*};w,w^{*})=\sum_{m,n,k,l}A_{mn,kl}z^{m}z^{*n}\otimes
w^{k}w^{*l}.$ (3.9)

The task would to find solutions that satisfy

$\displaystyle-q^{-1}(1-zz^{*})^{2}{\partial\over\partial
z}{\partial\over\partial z^{*}}\phi(z,z^{*})+m^{2}\phi(z,z^{*})=J(z,z^{*}),$
(3.10) $\displaystyle\ \ \ \
\phi(z,z^{*})=\int_{\mathbb{D}_{q}}d\nu_{w}G(z,z^{*};w,w^{*})J(w,w^{*}).$
(3.11)

Similar to the classical AdS case, the bulk-to-bulk propagator on the quantum
disk has to be a function of a $U_{q}(\mathfrak{su}_{1,1})$ invariant
distance. This is a function defined on the tensor product of two quantum
disks that is annihilated by the actions of $E^{n},F^{m},(K^{k}-1)$ for any
$n,m,k$ acting on both coordinates. The distance is given by

$\displaystyle
d(z,z^{*};w,w^{*})=(1-zz^{*})^{-1}{\times_{\mathrm{op}}}(1-q^{-2}zw^{*}){\times_{\mathrm{op}}}(1-z^{*}w){\times_{\mathrm{op}}}(1-ww^{*})^{-1},$
(3.12)

where the $\times_{\mathrm{op}}$ is the ‘opposite multiplication’ symbol
indicating that multiplication on the first factor (the quantum disk
parameterized by $z,z^{*}$) includes a swap:
$f(z,z^{*}){\times_{\mathrm{op}}}g(z,z^{*})=g(z,z^{*})f(z,z^{*})$, where the
expression on the right involves standard multiplication. This means the
expanded form of the distance is

$\displaystyle
d(z,z^{*};w,w^{*})=(1-q^{-2}zw^{*}-z^{*}w+q^{-2}z^{*}zw^{*}w)(1-zz^{*})^{-1}(1-ww^{*})^{-1}.$
(3.13)

This opposite multiplication is important because it takes
$U_{q}(\mathfrak{su}_{1,1})$ invariant functions and returning another
invariant function, which follows from the following theorem[25].

###### Theorem 1

If $I_{1}=\sum_{i}a_{i}\otimes b_{i},\ I_{2}=\sum_{j}c_{j}\otimes d_{j}$ are
$U_{q}(\mathfrak{su}_{1,1})$ invariant, then $I_{12}\equiv
I_{1}\times_{\mathrm{op}}I_{2}\equiv\sum_{i,j}c_{j}a_{i}\otimes b_{i}d_{j}$ is
also $U_{q}(\mathfrak{su}_{1,1})$ invariant.

This theorem is proven in the appendix (A.3). This theorem is important for us
since we want to take powers of the invariant distance. The invariant distance
raised to an arbitrary positive power (in the opposite multiplication sense)
is given by

$\displaystyle d^{m}$
$\displaystyle=(1-zz^{*})^{-m}{\times_{\mathrm{op}}}(q^{-2m}zw^{*};q^{2})_{m}{\times_{\mathrm{op}}}(q^{-2(m-1)}z^{*}w;q^{2})_{m}{\times_{\mathrm{op}}}(1-ww^{*})^{-m},$
(3.14)

where we used the $q$-Pochhammer symbol defined as

$\displaystyle(a;q)_{m}=\prod_{j=0}^{m-1}(1-aq^{j}).$ (3.15)

See appendix A.4 for a derivation. It will be convenient to write the
invariant distance by implementing of the opposite multiplcations as follows

$\displaystyle d^{m}$
$\displaystyle=(1-zz^{*})^{-m}(1-ww^{*})^{-m}\left[(q^{-2m}zw^{*};q^{2})_{m}{\times_{\mathrm{op}}}(q^{-2(m-1)}z^{*}w;q^{2})_{m}\right],$
(3.16)

where the opposite multiplication takes place only within the square brackets.
Inverse powers of the distance will be defined as

$\displaystyle(d^{m})^{-1}$
$\displaystyle=\left[(q^{-2m}zw^{*};q^{2})_{m}{\times_{\mathrm{op}}}(q^{-2(m-1)}z^{*}w;q^{2})_{m}\right]^{-1}(1-zz^{*})^{m}(1-ww^{*})^{m}.$
(3.17)

Note that this expression disagrees with one derived in [25, 16]; we found
their expression to not be invariant while ours is. Thus, the general
propagator is of the form

$\displaystyle G(z,w)=\sum^{\infty}_{m=0}g_{m}\left(d^{m}(z,w)\right)^{-1},$
(3.18)

for some coefficients $g_{m}$. Only negative powers are included since the
fields are required to vanish near the boundary (in the absence of sources).

###  3.3 Boundary two point function

Near the boundary we expect to have
$\phi\sim\mathcal{O}(z,z^{*})(1-zz^{*})^{\Delta}$ as $(1-zz^{*})$ is taken to
0. Hence, the boundary limit of the bulk correlation function is of the form

$\displaystyle\langle\phi(z,z^{*})\phi(w,w^{*})\rangle=\langle\mathcal{O}(z,z^{*})\mathcal{O}(w,w^{*})\rangle(1-zz^{*})^{\Delta}(1-ww^{*})^{\Delta}.$
(3.19)

Thus we can read off the boundary correlation function using (3.17). Note that
the bulk coordinates are noncommutative while their boundary limits are
standard commutative variables. Hence, the boundary limit of bulk functions is
a many to one map and cannot be reversed unambiguously. Furthermore, the
commutativity of the boundary coordinates makes the $\times_{\mathrm{op}}$
redundant. Hence, the boundary two point function is

$\displaystyle\langle\mathcal{O}(\varphi)\mathcal{O}(\theta)\rangle$
$\displaystyle=(q^{-2\Delta}e^{i(\varphi-\theta)};q^{2})^{-1}_{\Delta}(q^{-2(\Delta-1)}e^{-i(\varphi-\theta)};q^{2})^{-1}_{\Delta},$
$\displaystyle=(e^{i(\varphi-\theta)};q^{2})_{-\Delta}(q^{2}e^{-i(\varphi-\theta)};q^{2})_{-\Delta}.$
(3.20)

As a check, the $q\rightarrow 1$ limit indeed limits to the 1-dimensional
conformal two point function,

$\displaystyle\langle\mathcal{O}(\varphi)\mathcal{O}(\theta)\rangle\rightarrow\left({1\over\sin\left[{\theta-\varphi\over
2}\right]}\right)^{2\Delta}.$ (3.21)

As a final comment we can write down the bulk to boundary propagator. By
invariance and the boundary condition
$\phi\sim\mathcal{O}(z,z^{*})(1-zz^{*})^{\Delta}$, the bulk to bonundary
propagator must be

$\displaystyle
K_{\Delta}\equiv\langle\phi(z,z^{*})\mathcal{O}(w,w^{*})\rangle=\left[(q^{-2\Delta}zw^{*};q^{2})_{\Delta}{\times_{\mathrm{op}}}(q^{-2(\Delta-1)}z^{*}w;q^{2})_{\Delta}\right]^{-1}(1-zz^{*})^{\Delta},$
(3.22)

where $w=e^{i\theta}$. Amusingly, both the bulk to boundary and boundary to
boundary propagators were studied by Vaksman [25, 16] as structures with
interesting transformation properties and not motivated by the context at
hand.

##  4 $q$-Conformal Quantum Mechanics

In this section we turn away from the bulk and analyze the properties of the
putative boundary quantum mechanical system with $SU_{q}(1,1)$ symmetry. Note
that while the action of the $SU_{q}(1,1)$ on $e^{i\theta}$ at the boundary is
the same as that of $SU(1,1)$ (up to constant factors) they acting differently
on powers $\left(e^{i\theta}\right)^{n}$ where $SU_{q}(1,1)$ acts with a
nontrivial coproduct.

Boundary primary operators are defined through the extrapolate dictionary from
the expansion of bulk operators near the boundary,
$\phi\sim\mathcal{O}_{\Delta}(e^{i\theta})y^{\Delta}$. This expansion
determines the transformation of boundary operator under $SU_{q}(1,1)$ in the
following way. The procedure is to act on the bulk field with a symmetry
generator, e.g. $E(\phi)$ and then perform the same expansion

$\displaystyle E(\phi)=E[\mathcal{O}(z)y^{\Delta}]\equiv
E_{\partial}[\mathcal{O}(z)]y^{\Delta},$ (4.1)

and hence

$\displaystyle
E_{\partial}[\mathcal{O}(z)]=E[\mathcal{O}(z)y^{\Delta}]y^{-\Delta},$ (4.2)

where $z\rightarrow e^{i\theta}$. The transformations under
$U_{q}(\mathfrak{su}_{1,1})$ are

$\displaystyle K_{\partial}[\mathcal{O}(z)]$
$\displaystyle=\mathcal{O}(q^{2}z),$ (4.3) $\displaystyle
E_{\partial}[\mathcal{O}(z)]$
$\displaystyle=-q^{5/2}z^{2}{\mathcal{O}(z)-\mathcal{O}(q^{2}z)\over
z-q^{2}z}+q^{1/2}{q^{2\Delta}-1\over q^{-2}-1}z\mathcal{O}(q^{2}z),$ (4.4)
$\displaystyle F_{\partial}[\mathcal{O}(z)]$
$\displaystyle=q^{-3/2}{\mathcal{O}(q^{-2}z)-\mathcal{O}(z)\over
q^{-2}z-z}+q^{1/2}{1-q^{-2\Delta}\over 1-q^{2}}z^{-1}\mathcal{O}(z).$ (4.5)

These will provide Ward identities from which we can extract the correlation
functions. Note that the action of the generators on the boundary satisfies
the same coproduct as that in the bulk.

The analysis of the correlation functions simplifies considerably when working
in the Fourier basis where we expand local operators as

$\displaystyle\mathcal{O}(z)=\sum_{n=-\infty}^{\infty}\\!\\!\mathcal{O}_{n}z^{-n},\
\ \mathcal{O}_{n}=\oint{dz\over 2\pi i}\,\mathcal{O}(z)z^{n-1}.$ (4.6)

The transformation rules (4.3), (4.4),(4.5) imply the Fourier modes transform
as

$\displaystyle K_{\partial}[\mathcal{O}_{n}]$
$\displaystyle=q^{-2n}\mathcal{O}_{n},$ (4.7) $\displaystyle
E_{\partial}[\mathcal{O}_{n}]$ $\displaystyle=q^{1/2}{q^{2\Delta-2n-2}-1\over
q^{-2}-1}\mathcal{O}_{n+1}=q^{\frac{1}{2}}\left[1+n-\Delta\right]_{q^{-2}}\mathcal{O}_{n+1},$
(4.8) $\displaystyle F_{\partial}[\mathcal{O}_{n}]$
$\displaystyle=q^{1/2}{q^{2n-2}-q^{-2\Delta}\over
1-q^{2}}\mathcal{O}_{n-1}=-q^{\frac{1}{2}-2\Delta}\left[\Delta+n-1\right]_{q^{2}}\mathcal{O}_{n-1}.$
(4.9)

###  4.1 Correlation functions

Summetry under $SU_{q}(1,1)$ implies that correlation functions are invariant
under the action of any of its generators, namely that

$\displaystyle\langle
E^{a}F^{b}(K^{c}-1)\left[\mathcal{O}(x_{1})...\mathcal{O}(x_{m})\right]\rangle=0.$
(4.10)

The strategy we follow in computing the position space correlation functions
is by first expanding in the Fourier basis as

$\displaystyle\langle\mathcal{O}(x_{1})...\mathcal{O}(x_{m})\rangle=\sum_{n_{1},..,n_{m}}\langle\mathcal{O}_{n_{1}}...\mathcal{O}_{n_{m}}\rangle
x_{1}^{n_{1}}...x_{m}^{n_{m}},$ (4.11)

and then determine the coefficients
$\langle\mathcal{O}_{n_{1}}...\mathcal{O}_{n_{m}}\rangle$ by symmetry before
performing the sum. Such approach was considered before by LeClair and Bernard
[26] for $SL_{q}(2,\mathbb{R})$ case. We’ll consider a few cases below for
$SU_{q}(1,1)$ group, the group of isometries of quantum disk.

####  4.1.1 Two point function

Our task is to determine the Fourier mode overlap
$\langle\mathcal{O}_{m}\mathcal{O}_{n}\rangle$. The simplest constraint comes
the action of $K_{\partial}-1$ requiring that
$\langle\mathcal{O}_{m}\mathcal{O}_{n}\rangle=d_{m}\delta_{m,-n}$. We will set
$d_{0}=1$. Under $E$ we get the constraint $\langle
E_{\partial}[\mathcal{O}_{m}]\mathcal{O}_{n}\rangle+\langle
K_{\partial}[\mathcal{O}_{m}]E_{\partial}[\mathcal{O}_{n}]\rangle=0$, where we
used the coproduct action of $E_{\partial}$. Taking $m=-(n+1)$, we obtain the
recursion relation

$\displaystyle\langle\mathcal{O}_{-(n+1)}\mathcal{O}_{n+1}\rangle=\alpha_{n}\langle\mathcal{O}_{-n}\mathcal{O}_{n}\rangle.$
(4.12)

where

$\displaystyle\alpha_{n}\equiv{q^{-\Delta-n-1}-q^{\Delta+n-1}\over
q^{\Delta-n-1}-q^{-\Delta+n+1}}.$ (4.13)

This can be solved using the boundary condition
$\langle\mathcal{O}_{0}\mathcal{O}_{0}\rangle=1$, and to find the Fourier
coefficients

$\displaystyle\langle\mathcal{O}_{m}\mathcal{O}_{n}\rangle=\delta_{m,-n}q^{-2\Delta
n}\frac{(q^{2\Delta};q^{2})_{n}}{(q^{-2\Delta+2};q^{2})_{n}}$ (4.14)

We notice that $\braket{O_{n}O_{-n}}\sim q^{-2\Delta n},n\to\infty$ and
$\braket{O_{n}O_{-n}}\sim q^{-(2-2\Delta)n},n\to-\infty$. The sum can be
recasted in the following compact way

$\displaystyle\langle\mathcal{O}(x_{1})\mathcal{O}(x_{2})\rangle=\sum_{n}\frac{(q^{2\Delta};q^{2})_{n}}{(q^{-2\Delta+2};q^{2})_{n}}\left(q^{-2\Delta}\frac{x_{1}}{x_{2}}\right)^{n}$
(4.15)

And using Ramanujan ${}_{1}\psi_{1}$ identity [27] we get

$\displaystyle\langle\mathcal{O}(x_{1})\mathcal{O}(x_{2})\rangle=\frac{(q^{2};q^{2})_{\infty}(q^{-4\Delta+2};q^{2})_{\infty}}{(q^{-2\Delta+2};q^{2})_{\infty}^{2}}\frac{(x_{1}/x_{2};q^{2})_{\infty}}{(q^{-2\Delta}x_{1}/x_{2};q^{2})_{\infty}}\frac{(q^{2}x_{2}/x_{1};q^{2})_{\infty}}{(q^{-2\Delta+2}x_{2}/x_{1};q^{2})_{\infty}}$
(4.16)

Or more compactly

$\displaystyle\langle\mathcal{O}(x_{1})\mathcal{O}(x_{2})\rangle=\frac{(q^{2};q^{2})_{\infty}(q^{-4\Delta+2};q^{2})_{\infty}}{(q^{-2\Delta+2};q^{2})_{\infty}^{2}}(x_{1}/x_{2};q^{2})_{-\Delta}(q^{2}x_{2}/x_{1};q^{2})_{-\Delta}$
(4.17)

Up to the overall constant factor, this is identical to the result obtained
from the quantum disk in (3.20) after setting
$x_{1}=e^{i\varphi},x_{2}=e^{i\theta}$.

####  4.1.2 Operator Product Expansion

In this subsection we will discuss briefly the operator product expansion.
Thus, assume that we have two operators $\mathcal{O}_{1,2}(x)$ and we can
expand through other local operators $\mathcal{O}_{i}(x)$ using the rule

$\displaystyle\mathcal{O}_{1}(x)\mathcal{O}_{2}(y)=\sum_{i}\lambda^{i}_{12}P_{12i}(x,y,E_{\partial})\mathcal{O}_{i}(y),$
(4.18)

The functions $P_{12i}$ roughly speaking should give Clebsch-Gordan
coefficients and should be fixed by the symmetries [28, 29, 30]. The
coefficients $\lambda^{i}_{12}$ are the OPE coefficients and determined by the
theory. Here we will try to compute for representation (4.3),(4.4), (4.5) in
the mode expansion. Thus, we want to find the coefficients $p_{n}$ that would
give the following decomposition

$\displaystyle\sum_{n}p_{n}\mathcal{O}_{1,n}\mathcal{O}_{2,-n}=\mathcal{O}_{3,0}.$
(4.19)

Using asymptotics of the two-point function below (4.14), we should demand
that the coefficients should decay faster than $p_{n}\sim
q^{(\Delta_{1}+\Delta_{2})n},n\to\infty$ and $p_{n}\sim
q^{(2-\Delta_{1}-\Delta_{2})|n|},n\to-\infty$.

The coefficients $p_{n}$ could be found with the use of the following trick.
We note that the operator on the right hand side is an eigenvector of the
Casimir, $C_{q}$. Thus, acting with the Casimir on both sides of this equation
gives the following recursion relation for $p_{n}$

$\displaystyle
q^{2+2\Delta_{1}}\left(q^{2\Delta_{1}}-q^{2n}\right)\left(q^{2\Delta_{2}}-q^{2n}\right)p_{n-1}+q^{2\Delta_{2}}(q^{2(n+\Delta_{1})}-1)(q^{2(n+\Delta_{2})}-1)p_{n+1}+$
$\displaystyle\left(q^{2(1+n+\Delta_{1})}+q^{2(1+n+\Delta_{2})}-q^{2(1+\Delta_{1}+\Delta_{2})}-q^{2(\Delta_{1}+\Delta_{2})}-q^{2(2n+\Delta_{1}+\Delta_{2})}-q^{2(1+2n+\Delta_{1}+\Delta_{2})}+\right.$
$\displaystyle\left.q^{2(n+2\Delta_{1}+\Delta_{2})}+q^{2(n+2\Delta_{2}+\Delta_{1})}\right)p_{n}=(1-q^{2\Delta_{3}})(1-q^{2-2\Delta_{3}})p_{n}.$
(4.20)

This is a somewhat complicated recursion relation to solve, but we considered
the following natural guess for the three point function which we checked
indeed satisfies this relation as well as (4.10). The three point function is
given by

$\displaystyle\hskip
142.26378pt\braket{\mathcal{O}_{1}(x_{1})\mathcal{O}_{2}(x_{2})\mathcal{O}_{3}(x_{3})}=$
$\displaystyle=$
$\displaystyle(q^{2\Delta_{12,3}}x_{1}/x_{2},q^{2})_{-\Delta_{12,3}}(q^{2\Delta_{12,3}}x_{2}/x_{3},q^{2})_{-\Delta_{23,1}}(q^{2+2\Delta_{23,1}}x_{3}/x_{1},q^{2})_{-\Delta_{13,2}}\times$
$\displaystyle(q^{2-2\Delta_{12,3}}x_{2}/x_{1},q^{2})_{-\Delta_{12,3}}(q^{2-2\Delta_{12,3}}x_{3}/x_{2},q^{2})_{-\Delta_{23,1}}(q^{-2\Delta_{23,1}}x_{1}/x_{3},q^{2})_{-\Delta_{13,2}},$
(4.21)

where $\Delta_{ij,k}=\frac{1}{2}\left(\Delta_{1}+\Delta_{2}-\Delta_{3}\right)$
and $\Delta_{i,j}=\Delta_{i}-\Delta_{j}$. Expanding this three point function
would allow us to extract $P_{i,j,k}$.

Going back to the bulk picture, one could wonder if the three point function
can be obtained from a bulk computation (it must by symmetry, but good to
check explicitly). For that we just need to compute the following
integral555The product of invariant elements should be taken according to the
rule (A.3) but since $x_{1,2,3}$ are commutative elements we can just multiply
directly.

$\displaystyle
G(x_{1},x_{2},x_{3})=\int_{\mathbb{D}_{q}}d\nu_{z}K_{\Delta_{1}}(z,x_{1})K_{\Delta_{2}}(z,x_{2})K_{\Delta_{3}}(z,x_{3}).$
(4.22)

$G(x_{1},x_{2},x_{3})$ is $U_{q}(\mathfrak{su}_{1,1})$ invariant since
$K_{\Delta_{1,2,3}}$ and the integral itself are $U_{q}(\mathfrak{su}_{1,1})$
invariant. To see that more explicitly, we notice that the integrand is an
element of $\mathbb{C}_{q}[z,z^{*}]\otimes\mathbb{C}[x_{1},x_{2},x_{3}]$ and
since it is $U_{q}(\mathfrak{su}_{1,1})$ invariant we have

$\displaystyle
0=\int_{\mathbb{D}_{q}}d\nu_{z}\Delta(E)K_{\Delta_{1}}(z,x_{1})K_{\Delta_{2}}(z,x_{2})K_{\Delta_{3}}(z,x_{3})$
$\displaystyle=\int_{\mathbb{D}_{q}}d\nu_{z}\left(E_{z}+K_{z}E_{x}\right)K_{\Delta_{1}}(z,x_{1})K_{\Delta_{2}}(z,x_{2})K_{\Delta_{3}}(z,x_{3})=E_{x}G(x_{1},x_{2},x_{3})=0.$
(4.23)

Checking that the Witten diagram works for general dimension is complicated,
but we managed to check it numerically for negative dimensions,
$\Delta_{1,2,3}=-1,-2$.

##  5 Discussion

We conclude with some outlook, remarks, and some open problems for future
work.

###  5.1 $q$-Conformal quantum mechanics

It would be interesting to revisit the problem of conformal quantum mechanics
but where the symmetry group is $SL_{q}(2)$ instead of $SL(2,\mathbb{R})$. In
fact, it is simple to generalize the conformal quantum mechanical system [31]
to a $q$-conformal system by replacing the time derivatives with a finite
difference derivative,

$\displaystyle{\cal L}=\left[\partial_{t}Q(t)\right]^{2}+{g\over
Q^{2}(t)}\rightarrow\left[D_{q}Q(t)\right]^{2}+{g\over Q^{2}(t)},$ (5.1)

where

$\displaystyle D_{q}Q(t)={Q(t)-Q(q^{2}t)\over t-q^{2}t}.$ (5.2)

The operator $Q$ has dimension $-1/2$, independent of $q$. It is unclear if
this is a feature or a bug, but the theory is non-local in time and perhaps
makes more sense in Euclidean time. Another unclear aspect is whether the time
integral should be a regular continuous one or a “Jackson integral” evaluated
a discrete time steps of at powers of $q^{2}$. Both choices are consistent
with $SL_{q}(2)$ symmetry.

###  5.2 $q$-Schwarzian action

One could wonder if the quantum disk is a solution to a $q$-deformed analogue
of JT gravity that localizes on $q$-AdS2 but with a boundary
reparameterization mode. The Lagrangian of this mode would be what one would
define as a $q$-deformed Schwarzian, i.e. a differential (finite difference)
operator that’s invariant under infinitesimal $SL_{2}$ transformation but with
a non-trivial coproduct. See [32, 33] for proposals on the $q$-deformed
Schwarzian.

The task is to find a (discrete) differential operators that’s invariant under
the action of $U_{q}(\mathfrak{su}_{1,1})$, assuming we define their action on
$t(u)$ to satisfy

$\displaystyle K\Big{(}f[t(u)]\Big{)}$ $\displaystyle=f[q^{2}t(u)],$ (5.3)
$\displaystyle E\Big{(}f[t(u)]\Big{)}$
$\displaystyle=-q^{1/2}t^{2}(u){f[t(u)]-f[q^{2}t(u)]\over t(u)-q^{2}t(u)},$
(5.4) $\displaystyle F\Big{(}f[t(u)]\Big{)}$
$\displaystyle=q^{1/2}{f[q^{-2}t(u)]-f[t(u)]\over q^{-2}t(u)-t(u)}.$ (5.5)

A related problem is to understand how to perform an arbitrary
reparameterization of non-commutative coordinates. These transformations would
be generated by a $q$-deformed Virasoro. Studying the central extension of
this algebra would be one way of understanding the $q$-Schwarzian. We leave
this for future work.

###  5.3 UV lattice and Renormalization, UV/IR mixing, q affects both

The interplay of between the discreteness of space and the renormalization
group is an old and well studied problem, see for instance [34]. The upshot is
that renormalization group doesn’t permit Lorentz invariance to emerge in the
IR if it is broken in the UV due to the existence of relevant operators that
can be generated under the RG flow. A natural question is whether Lorentz
invariance is preserved in the IR if the UV is invariant under a $q-$deformed
version. The analogue for the quantum disk is $SU_{q}(1,1)$ in the UV and
$SU(1,1)$ in the IR.

One potential hurdle is the problem of UV/IR mixing associated to non-
commutative spaces, discussed for instance here [1]. In fact, the quantum disk
has a hint of this in the spectrum of the Laplacian,

$\displaystyle(1+q)^{-2}\leq\Box_{q}\leq(1-q)^{-2},$ (5.6)

where both the IR and UV bounds are controlled by the same paramter $q$. One
would need to compute loop diagrams to truly settle this.

###  5.4 q-Conformal Blocks

Using the ideas similar to the usual conformal field theory we can use the
operator product expansion and assosiativity to constrain the possible
$SU_{q}(1,1)$ invariant field theories. For that, we must to study the four-
point functions

$\displaystyle\braket{O_{1}(z_{1})O_{2}(z_{2})O_{3}(z_{3})O_{4}(z_{4})}.$
(5.7)

As in the case of the usual CFTs we can not constrain the possible form of
this expression, but we can expand it in terms of conformal blocks that
satisfy the following equation

$\displaystyle
C_{q}G_{\Delta}(z_{1},z_{2},z_{3},z_{4})=C_{\Delta}G_{\Delta}(z_{1},z_{2},z_{3},z_{4}),$
(5.8)

it would be interesting to study the possible solutions of this equation.

###  5.5 DSSYK at $q<1$ and large $\beta$

One of the motivations of this work was to study a toy model that resembles
the putative bulk dual of DSSYK since it too has a noncommutative nature. A
natural question is whether the quantum disk emerges from some regime of
DSSYK.

There’s some evidence that this might be the case. It was noted in [10] that
invariance under $U_{q}(\mathfrak{su}_{1,1})$ emerges in the low temperature
limit of DSSYK at $q<1$. This symmetry fixes the form of the two point
function. The exact expression of the two point function was found in [7], so
it simply a matter of taking its low temperature limit. We leave this for
future work.

## Acknowledgements

We thank Micha Berkooz, Mikhail Isachenkov, Henry Lin, Simon Lin, Ohad
Mamroud, Alexei Milekhin and Yifan Wang for insightful discussion and
correspondence. F.K.P. is currently a Simons Junior Fellow at New York
University and supported by a grant 855325FP from the Simons Foundation.

##  Appendix A Introduction to Quantum Groups

###  A.1 General Philosophy

This appendix contains what hopefully will be an accessible and pedagogical
review of quantum groups. For that, we will start with the general philosophy
or idea that lead to the creation of quantum groups. Usually when we study
groups or other classical manifolds we have an idea that it is some sort of
geometric object that could be thought as some surface in higher dimensional
flat space. Instead, it is more convenient here to use the language of
algebraic geometry which analyzes functions on the manifold whose structures
are reflected in the algebra of functions. Thus, if we have a topological
space $\mathcal{M}$ we will denote $\mathcal{O}(\mathcal{M})$ as a set of all
functions $f\in\mathcal{O}(\mathcal{M}),f:\mathcal{M}\to\mathbb{C}$ on the
given topological manifold. This set automatically has the structure of
commutative algebra. We have the following operations and special element
$\mathbf{1}$

$\displaystyle\forall
f_{1,2}\in\mathcal{O}(\mathcal{M}),\quad(f_{1}+f_{2})(x)=f_{1}(x)+f_{2}(x),\quad(f_{1}\cdot
f_{2})(x)=f_{1}(x)f_{2}(x)\quad\forall x\in\mathcal{M}\,,$
$\displaystyle\mathbf{1}(x)=1,\quad\forall
x\in\mathcal{M},\quad\mathbf{1}\cdot f=f$ (A.1)

it is easy to check that $\mathbf{1}$ is indeed a unit in the algebra
$\mathcal{O}(\mathcal{M})$. By studying the algebraic properties of this
commutative algebra we can understand the geometric properties of the
topological manifold $\mathcal{M}$. For instance, the ideals of
$\mathcal{O}(\mathcal{M})$ will correspond to submanifolds and so on.

###  A.2 Hopf algebra

Now assume that $\mathcal{M}$ is not only just a topological manifold but also
a group. Then the algebra of functions $\mathcal{O}(\mathcal{M})$ gets
additional structures that encodes the group structure of this topological
group. Thus, we have now a special element $e\in\mathcal{M}$, the identity,
and the two additional operations, inverse and product. Evaluating a function
on the manifold on the element $e$ we can think as a map from
$\mathcal{O}(\mathcal{M})$ to complex functions. Such operations we will
denote as the co-unit

$\displaystyle\eta:\mathcal{O}(\mathcal{M})\to\mathbb{C},\quad\eta(f)=f(e).$
(A.2)

The existence of a group product means we can map any function of a single
variable to a function of two variables

$\displaystyle\Delta:\mathcal{O}(\mathcal{M})\to\mathcal{O}(\mathcal{M})\otimes\mathcal{O}(\mathcal{M}),\quad\Delta(f)(g_{1},g_{2})=f(g_{1}g_{2}),\quad\forall
g_{1,2}\in\mathcal{M}$ (A.3)

where we assumed we can make a replacement
$\mathcal{O}(\mathcal{M}\times\mathcal{M})=\mathcal{O}(\mathcal{M})\otimes\mathcal{O}(\mathcal{M})$.
The map $\Delta$ is the coproduct. Associativity of a group product suggests a
coassociativy of the coproduct $(\Delta\otimes{\rm id})\Delta=({\rm
id}\otimes\Delta)\Delta$. The group axioms imply that the following relation
must hold

$\displaystyle({\rm id}\otimes\eta)\Delta=(\eta\otimes{\rm id})\Delta={\rm
id}\,.$ (A.4)

If a set has a co-unit $\eta$ and coproduct then it is a coalgebra. If an
algebra has the structure of a coalgebra and these structures are compatible
in the sense

$\displaystyle\Delta(ab)=\Delta(a)\Delta(b),\quad\eta(ab)=\eta(a)\eta(b)\,,$
(A.5)

then the set is a bialgebra.

The inverse allows to define an additional structure that maps the algebra of
functions back to itself

$\displaystyle S:\mathcal{O}(\mathcal{M})\to\mathcal{O}(\mathcal{M}),\quad
S(f)(g)=f(g^{-1}),\quad\forall g\in\mathcal{M}\,,$ (A.6)

where the operation $S$ is the antipode. The group axioms imply that the
following relation must hold

$\displaystyle m(S\otimes{\rm id})\Delta=m(S\otimes{\rm id})\Delta=\eta\,.$
(A.7)

Finally, if we have a bialgebra with the antipode $S$ satisfying the above
relation then the bialgebra is a Hopf algebra. We see that the algebra of
functions on any topological group satisfies these relations. This results in
a commutative Hopf algebras.666However, it is not cocommutative because
$f(g_{1}g_{2})\neq f(g_{2}g_{1})$ for general $f$. This defines a group. On
the other hand, a quantum group is defined by a non-commutative Hopf algebra.

####  A.2.1 Hopf algebra $\mathcal{O}(SL_{2}(\mathbb{R}))$

Before proceeding to the quantum case, let us consider first the most simple
classical group $SL_{2}(\mathbb{R})$ and see how the above definitions work. A
group $SL_{2}(\mathbb{R})$ could be defined as follows

$\displaystyle SL_{2}(\mathbb{R})=\left\\{\begin{pmatrix}t_{11}&t_{12}\\\
t_{21}&t_{22}\end{pmatrix}\in\mathbb{R}^{4}:t_{11}t_{22}-t_{12}t_{21}=1\right\\}\,,$
(A.8)

the elements of the matrix $t_{ij}$ could be thought as coordinates in the
sense they are maps from group elements $g$ to their matrix elements
$t_{ij}(g)$ in the above two dimensional representation, and they furnish a
basis for functions on the group. Hence, the algebra of functions is the
algebra of all possible polynomials in commutative variables $t_{ij}$ subject
to the relation $t_{11}t_{22}-t_{12}t_{21}=1$. This set is denoted by

$\displaystyle\mathcal{O}(SL_{2}(\mathbb{R}))=\mathbb{C}[t_{11},t_{12},t_{21},t_{22}]/\left(t_{11}t_{22}-t_{12}t_{21}-1\right)\,.$
(A.9)

The structures $\eta$, $S$ and $\Delta$ are determined by the group structure.
Roughly speaking, given the coordinates of a group element, these operations
compute the coordinates of the unit, product and inverse of this group element

$\displaystyle\eta(t_{ij})=\delta_{ij},\quad\Delta(t_{ij})=t_{ik}\otimes
t_{kj},\quad S(t_{ij})=\epsilon_{ik}\epsilon_{jl}t_{kl}\,.$ (A.10)

Let us expand briefly on the coproduct here. If we have two elements $g,h$
with coordinates $t_{ij}(g),t_{lm}(h)$ than the coproduct gives the
coordinates of their product

$\displaystyle\Delta(t_{ij})(g,h)=t_{ik}(g)t_{kj}(h)=g_{ik}h_{kj}=t_{ij}(gh)\,,$
(A.11)

and one can check that these operations satisfy the relations (A.1)-(A.7)
giving us a Hopf algebra $\mathcal{O}(SL_{2}(\mathbb{R}))$. Usually, along
with the group it is useful to consider a universal eneveloping algebra
$U(\mathfrak{sl}_{2})$ that could be thought as an algebra of the vector
fields generated by the left action of the group. Thus by considering the
infintesimal transformation we find three vector fields $E,F,H$ that are
subject to the following relations

$\displaystyle[H,E]=2E,\quad[H,F]=-2F,\quad[E,F]=H\,.$ (A.12)

$U(\mathfrak{sl}_{2})$ is a free generated algebra in variables $E,F,H$
subject to the above relations.

Since we are adopting an algebraic approach it would be useful to understand
how to derive this algebra starting from Hopf algebra
$\mathcal{O}(SL_{2}(\mathbb{R}))$. For that we notice first that
$U(\mathfrak{sl}_{2})$ is a Hopf algebra where the coalgebra structure is
defined as

$\displaystyle\eta(J_{i})=0,\quad\Delta(J_{i})=1\otimes J_{i}+J_{i}\otimes
1,\quad S(J_{i})=-J_{i},\quad J_{i}=E,F,H\,,$ (A.13)

then we notice that since $E,F,H$ are vector fields they have a natural action
on elements of the algebra $\mathcal{O}(SL_{2}(\mathbb{R}))$ which when
combined with co-unit defines a pairing between these two Hopf algebras

$\displaystyle\sum_{ijk}\langle c_{ijk}E^{i}F^{j}H^{k},f\rangle=\sum
c_{ijk}\eta(E^{i}F^{j}H^{k}f),\quad\sum c_{ijk}E^{i}F^{j}H^{k}\in
U(\mathfrak{sl}_{2}),f\in\mathcal{O}(SL_{2}(\mathbb{R})$ (A.14)

where we apply consequently vector fields $E,F,H$ on the given function
$f\in\mathcal{O}(SL_{2}(\mathbb{R}))$. This pairing is the key to
understanding of the relation between $U(\mathfrak{sl}_{2})$ and
$\mathcal{O}(SL_{2}(\mathbb{R}))$. First of all, we notice that this pairing
allows to restore the action of $U(\mathfrak{sl}_{2})$ on
$\mathcal{O}(SL_{2}(\mathbb{R}))$. Indeed $J(f)=(\langle
J,\cdot\rangle\otimes{\rm id})\Delta(f)$. Second of all, this pairing respects
the structure of the Hopf algebras namely

$\displaystyle\langle JK,f\rangle=(\langle J,\cdot\rangle\otimes\langle
F,\cdot\rangle)\Delta(f),\quad\langle
J,fg\rangle=(\langle\cdot,f\rangle\otimes\langle\cdot,g\rangle)\Delta(J)$
$\displaystyle\langle 1,f\rangle=\eta(f),\quad\langle J,1\rangle=\eta(J),\quad
f,g\in\mathcal{O}(SL_{2}(\mathbb{R})),J,K\in U(\mathfrak{sl}_{2})\,.$ (A.15)

Now, if we have two Hopf algebras $U,V$ that have a pairing $\langle,\rangle$
we would say these two Hopf algebras are dual to each other. We then define
that $U_{q}(\mathfrak{sl}_{2})$ is a dual algebra of
$\mathcal{O}(SL_{2}(\mathbb{R}))$. This same approach we will adopt for
quantum algebras.

####  A.2.2 Hopf algebra $\mathcal{O}(SL_{2}(\mathbb{R}))$

Having understood how from classical groups we can get the Hopf algebra, we
can forget about our base manifold and start immediately with some algebra.
Thus we consider the following algebra

$\displaystyle\mathcal{O}_{q}(SL_{2}(\mathbb{R}))=\mathbb{C}[t_{11},t_{12},t_{21},t_{22}]/I,$
$\displaystyle
I=\left(t_{11}t_{12}-qt_{12}t_{11},t_{11}t_{21}-qt_{21}t_{11},t_{12}t_{22}-qt_{22}t_{12},t_{21}t_{22}-qt_{22}t_{21},t_{12}t_{21}=t_{21}t_{12},\right.$
$\displaystyle\left.t_{11}t_{22}-t_{22}t_{11}=(q-q^{-1})t_{12}t_{21},t_{11}t_{22}-qt_{12}t_{21}-1\right),$
(A.16)

and polynomials generated by non-commutative variables $t_{ij}$. The structure
of the coalgebra is defined as

$\displaystyle\eta(t_{ij})=\delta_{ij},\quad\Delta(t_{ij})=t_{ik}\otimes
t_{kj},\quad S(t_{ij})=\epsilon_{il}\epsilon_{jm}q^{-\epsilon_{ij}}t_{lm}\,,$
(A.17)

from this we see that the quantum group as a group has a structure that
resembles the usual $SL_{2}(\mathbb{R})$ group with usual group product and
identity. So as a group a quantum group is no different from any other group
but the only difference is that coordinates are not commutative.

Now let us try to use the above procedure to construct the
$U_{q}(\mathfrak{sl}_{2})$. For that we will define the following elements
$E,F,K$ through the pairing, such that their pairing with basis elements is
defined as

$\displaystyle q^{-1}\langle K,t_{11}\rangle=q\langle
K,t_{22}\rangle=q^{1/2}\langle E,t_{12}\rangle=q^{-1/2}\langle
F,t_{21}\rangle=1\,,$ (A.18)

and the action on the product of two elements satisfy the following relation

$\displaystyle\langle E,ab\rangle=\langle E,a\rangle\eta(b)+\langle
K,a\rangle\langle E,b\rangle,$ $\displaystyle\langle F,ab\rangle=\langle
F,a\rangle\langle K^{-1},b\rangle+\eta(a)\langle F,b\rangle,$
$\displaystyle\langle K,ab\rangle=\langle K,a\rangle\langle K,b\rangle\,.$
(A.19)

one can check that such a definition is compatible with the ideal $I$, and
therefore defines a unique pairing on all elements of
$\mathcal{O}_{q}(SL_{2}(\mathbb{R}))$.

This pairing can be used to establish the following ideals of
$U_{q}(\mathfrak{sl}_{2})$

$\displaystyle I_{E}=KE-q^{2}EK,\quad I_{F}=K^{-1}F-q^{2}FK^{-1}\,,$ (A.20)

namely that $\langle I_{E,F},t_{ij}\rangle=0$ as we show below. Let us
concentrate on $I_{E}$. Now if we have a product of two elements we can
rewrite the action of $I_{E}$ using the coproduct

$\displaystyle\Delta(a)=a^{\prime}_{i}\otimes
a^{\prime\prime}_{i},\quad\Delta(b)=b^{\prime}_{j}\otimes
b^{\prime\prime}_{j},$
$\displaystyle\Delta(ab)=a^{\prime}_{i}b^{\prime}_{j}\otimes
a^{\prime\prime}_{i}b^{\prime\prime}_{j}\,.$ (A.21)

where $a,b$ (including the indexed and primed) $\in\mathcal{O}(SL_{q}(2))$.
These give

$\displaystyle\langle KE,ab\rangle=\langle
K,a^{\prime}_{i}b^{\prime}_{j}\rangle\langle
E,a^{\prime\prime}_{i}b^{\prime\prime}_{j}\rangle=\langle
K,a^{\prime}_{i}\rangle\langle K,b^{\prime}_{j}\rangle\left(\langle
E,a^{\prime\prime}_{i}\rangle\eta(b^{\prime\prime}_{j})+\langle
K,a^{\prime\prime}_{i}\rangle\langle E,b^{\prime\prime}_{j}\rangle\right)=$
$\displaystyle\langle KE,a\rangle\langle K,b\rangle+\langle K,a\rangle\langle
KE,b\rangle$ (A.22)

and

$\displaystyle\langle EK,ab\rangle=\langle
E,a^{\prime}_{i}b^{\prime}_{j}\rangle\langle
K,a^{\prime\prime}_{i}b^{\prime\prime}_{j}\rangle=\left(\langle
E,a^{\prime}_{i}\rangle\eta(b^{\prime}_{j})+\langle
K,a^{\prime}_{i}\rangle\langle E,b^{\prime}_{j}\rangle\right)\langle
K,a^{\prime\prime}_{i}\rangle\langle K,b^{\prime\prime}_{j}\rangle\ =$
$\displaystyle\langle EK,a\rangle\langle K,b\rangle+\langle K,a\rangle\langle
EK,b\rangle$ (A.23)

eventually yielding

$\displaystyle\langle I_{E},ab\rangle=\langle I_{E},a\rangle\langle
K,b\rangle+\langle I_{E},b\rangle\langle K,a\rangle,$ (A.24)

and therefore it is easy to see that $I_{E}\equiv 0$ (similarly for
$I_{F}\equiv 0$). Analogously, another ideal is

$\displaystyle P=EF-FE-\frac{K-K^{-1}}{q-q^{-1}}\,.$ (A.25)

Let us try to compute the pairing of this element with any element $t_{ij}$.
It is easy to check that

$\displaystyle\langle P,t_{ij}\rangle=0\,,$ (A.26)

and similarly all products $ab$. The most simple action is that of $K$

$\displaystyle\left\langle\frac{K-K^{-1}}{q-q^{-1}},ab\right\rangle=\frac{\langle
K,a\rangle\langle K,b\rangle-\langle K^{-1},a\rangle\langle
K^{-1},b\rangle}{q-q^{-1}}=$
$\displaystyle=\left\langle\frac{K-K^{-1}}{q-q^{-1}},a\right\rangle\left\langle
K^{-1},b\right\rangle+\left\langle
K,a\right\rangle\left\langle\frac{K-K^{-1}}{q-q^{-1}},b\right\rangle$ (A.27)

Then the action of $EF$ on those two elements is

$\displaystyle\langle EF,ab\rangle=\langle
E,a^{\prime}_{i}b^{\prime}_{j}\rangle\langle
F,a^{\prime\prime}_{i}b^{\prime\prime}_{j}\rangle=\left(\langle
E,a^{\prime}_{i}\rangle\eta(b^{\prime}_{j})+\langle
K,a^{\prime}_{i}\rangle\langle E,b^{\prime}_{j}\rangle\right)\left(\langle
F,a^{\prime\prime}_{i}\rangle\langle
K^{-1},b^{\prime\prime}_{j}\rangle+\eta(a^{\prime\prime}_{i})\langle
F,b^{\prime\prime}_{j}\rangle\right)=$ $\displaystyle\langle
E,a^{\prime}_{i}\rangle\langle F,a^{\prime\prime}_{i}\rangle\langle
K^{-1},b\rangle+\langle E,a\rangle\langle F,b\rangle+\langle
K,a^{\prime}_{i}\rangle\langle E,b^{\prime}_{j}\rangle\langle
F,a^{\prime\prime}_{i}\rangle\langle
K^{-1},b^{\prime\prime}_{j}\rangle+\langle K,a\rangle\langle
E,b^{\prime}_{j}\rangle\langle F,b^{\prime\prime}_{j}\rangle=$
$\displaystyle\langle EF,a\rangle\langle K^{-1},b\rangle+\langle
E,a\rangle\langle F,b\rangle+\langle KF,a\rangle\langle
EK^{-1},b\rangle+\langle K,a\rangle\langle EF,b\rangle\,,$

Analogously for the opposite ordering

$\displaystyle\langle FE,ab\rangle=\langle
F,a^{\prime}_{i}b^{\prime}_{j}\rangle\langle
E,a^{\prime\prime}_{i}b^{\prime\prime}_{j}\rangle=\left(\langle
F,a^{\prime}_{i}\rangle\langle
K^{-1},b^{\prime}_{j}\rangle+\eta(a^{\prime}_{i})\langle
F,b^{\prime}_{j}\rangle\right)\left(\langle
E,a^{\prime\prime}_{i}\rangle\eta(b^{\prime\prime}_{j})+\langle
K,a^{\prime\prime}_{i}\rangle\langle E,b^{\prime\prime}_{j}\rangle\right)=$
$\displaystyle\langle F,a^{\prime}_{i}\rangle\langle
E,a^{\prime\prime}_{i}\rangle\langle K^{-1},b\rangle+\langle
F,a^{\prime}_{i}\rangle\langle K^{-1},b^{\prime}_{j}\rangle\langle
K,a^{\prime\prime}_{i}\rangle\langle E,b^{\prime\prime}_{j}\rangle+\langle
F,b\rangle\langle E,a\rangle+\langle F,b^{\prime}_{j}\rangle\langle
K,a\rangle\langle E,b^{\prime\prime}_{j}\rangle=$ $\displaystyle\langle
FE,a\rangle\langle K^{-1},b\rangle+\langle FK,a\rangle\langle
K^{-1}E,b\rangle+\langle F,b\rangle\langle E,a\rangle+\langle
FE,b\rangle\langle K,a\rangle\,,$

where we have used the properties of the coproduct $\Delta$ and co-unit
$\eta$. Combing the two above relations we get

$\displaystyle\langle P,ab\rangle=\langle P,a\rangle\langle
K^{-1},b\rangle+\langle K,a\rangle\langle P,b\rangle,$ (A.28)

therefore $P\equiv 0$. Together with $I_{E,F}$, these provide the definition
of quantum universal enveloping algebra $U_{q}(\mathfrak{sl}_{2})$.

###  A.3 Invariant elements

An important notion is that of invariant elements. To be more precise assume
we have two modules $\mathcal{M},\mathcal{N}$ of $U_{q}(\mathfrak{su}_{1,1})$.
An element $d\in\mathcal{M}\otimes\mathcal{N}$ is invariant if

$\displaystyle E(d)=F(d)=K(d)-d=0,\quad d=\sum_{i}m_{i}\otimes n_{i},$

where for example

$\displaystyle E(d)=\sum_{i}E(m_{i})\otimes n_{i}+K(m_{i})\otimes
E(n_{i})=0\,.$ (A.29)

If we assume that $\mathcal{M},\mathcal{N}$ are algebras and we have two
invariant elements $d_{1,2}$ then, as discussed in [13], we can construct a
new invariant element $d_{3}$ in the following way

$\displaystyle d_{1,2}=\sum_{i}m^{1,2}_{i}\otimes n^{1,2}_{i},\quad
d_{3}=\sum_{i,j}m^{1}_{i}m^{2}_{j}\otimes n^{2}_{j}n^{1}_{i}\,,$ (A.30)

the straightforward computation shows that this element is indeed invariant

$\displaystyle E(d_{3})=\sum_{i,j}\left[E(m^{1}_{i})m^{2}_{j}\otimes
n^{2}_{j}n^{1}_{i}+K(m^{1}_{i})E(m^{2}_{j})\otimes
n^{2}_{j}n^{1}_{i}+K(m^{1}_{i})K(m^{2}_{j})\otimes
E(n^{2}_{j})n^{1}_{i}\right.$
$\displaystyle\left.+K(m^{1}_{i})K(m^{2}_{j})\otimes
K(n^{2}_{j})E(n^{1}_{i})\right]=$ $\displaystyle\sum_{i}(K(m_{i}^{1})\otimes
1)E(d_{2})(1\otimes n_{i}^{1})+\sum_{i,j}(1\otimes
n^{2}_{j})E(d_{1})\left(m^{2}_{j}\otimes 1\right)=0$ (A.31)

###  A.4 Powers of the invariant distance

The steps of this proof follow that of [16] but differs in detail. Consider
the tensor product of two quantum disks paramterized by $t_{ij}$ and
$\tau_{ij}$. The following linear combinations of $\tau_{ij}\otimes t_{kl}$
are invariant under the action of $U_{q}(\mathfrak{su}_{1,1})$

$\displaystyle A$ $\displaystyle=t_{22}\tau_{11}-q^{-1}t_{12}\tau_{21},$
(A.32) $\displaystyle B$ $\displaystyle=t_{11}\tau_{22}-qt_{21}\tau_{12},$
(A.33)

where we have dropped the tensor product symbol for brevity. The disk
coordinates are related to these variable in the following way

$\displaystyle z$ $\displaystyle=q^{-1}t_{21}^{-1}t_{11},\
z^{*}=t_{22}t^{-1}_{12}$ (A.34) $\displaystyle w$
$\displaystyle=q^{-1}\tau_{21}^{-1}\tau_{11},\ w^{*}=\tau_{22}\tau^{-1}_{12}.$
(A.35)

The two invariants can be expressed in terms of the disk coordinates as
follows while implementing the opposite product on the first tensor factor

$\displaystyle A$
$\displaystyle=(-q^{-1}t_{12}\tau_{21})\times_{\mathrm{op}}(1-q^{2}z^{*}w),$
(A.36) $\displaystyle B$
$\displaystyle=(1-zw^{*})\times_{\mathrm{op}}(-qt_{21}\tau_{12}).$ (A.37)

Using the following relations that follow from the algebra (2.25)

$\displaystyle
t_{12}\tau_{21}\times_{\mathrm{op}}(z^{*}w)=q^{-2}(z^{*}w)\times_{\mathrm{op}}\tau_{21}t_{12},\
(zw^{*})\times_{\mathrm{op}}t_{21}\tau_{12}=q^{-2}t_{21}\tau_{12}\times_{\mathrm{op}}(zw^{*}),$
(A.38)

the powers of the invariants are

$\displaystyle A^{m}$
$\displaystyle=(-q^{-1}t_{12}\tau_{21})^{m}\times_{\mathrm{op}}(q^{2}z^{*}w;q^{2})_{m},$
(A.39) $\displaystyle B^{m}$
$\displaystyle=(zw^{*};q^{2})_{m}\times_{\mathrm{op}}(-qt_{21}\tau_{12})^{m}.$
(A.40)

Taking their opposite product we get

$\displaystyle B^{m}A^{m}$
$\displaystyle=(zw^{*};q^{2})_{m}\times_{\mathrm{op}}(-qt_{21}\tau_{12})^{m}\times_{\mathrm{op}}(-q^{-1}t_{12}\tau_{21})^{m}\times_{\mathrm{op}}(q^{2}z^{*}w;q^{2})_{m}$
(A.41)
$\displaystyle=q^{-2m}(zw^{*};q^{2})_{m}\times_{\mathrm{op}}(1-zz^{*})^{-m}(1-ww^{*})^{-m}\times_{\mathrm{op}}(q^{2}z^{*}w;q^{2})_{m}$
(A.42)

where we used $t_{12}t_{21}=-q^{-1}(1-zz^{*})^{-}1$ and similarly for $w$.
Commuting the middle factors across gives

$\displaystyle
B^{m}A^{m}=q^{-2m}(1-zz^{*})^{-m}\times_{\mathrm{op}}(q^{-2m}zw^{*};q^{2})_{m}\times_{\mathrm{op}}(q^{-2(m-1)}z^{*}w;q^{2})_{m}\times_{\mathrm{op}}(1-ww^{*})^{-m}$
(A.43)

The extra factors of $q$ inside the $q$-Pochhammers comes from commuting the
$(1-zz^{*})$ and $(1-ww^{*})$ through. For $m=1$ this is just the invariant
distance $d$ in (3.13). This constructs the appropriate invariant power of
$d$.

##  Appendix B $R$ Matrix approach to $SL_{q}(2)$

This non-commutative structure (2.1) defines a specific quantum deformation of
the $SL(2)$. These relations are self consistent, and any polynomial of the
matrix elements can be written as a sum over monomials with a given ordering
of the elements $a,b,c,d$ without changing the degree of the polynomial. These
two properties are guaranteed by the Yang-Baxter equation

$\displaystyle g^{i}_{\ k}\,g^{j}_{\ l}\,R^{kl}_{\ \ mn}=R^{ji}_{\ \
rs}\,g^{s}_{\ n}\,g^{r}_{\ m},$ (B.1)

where $g\in SL_{q}(2)$ and $R$ is a $4\times 4$ matrix of complex numbers
satisfying the Yang-Baxter (YB) equation

$\displaystyle R_{12}R_{23}R_{12}=R_{23}R_{12}R_{23},$ (B.2)

where $R_{12}=R^{i_{1}i_{2}}_{\ \ j_{1}j_{2}}\delta^{i_{3}}_{\ j_{3}}$ and the
others defined similarily. The flipped indices on the right hand side of (B.1)
is intentional, giving a relation between the two different orders of all the
matrix elements. 777This equation is sometimes written as $g^{i}_{\
k}\,g^{j}_{\ l}\,\hat{R}^{kl}_{\ \ mn}=\hat{R}^{ji}_{\ \ rs}\,g^{r}_{\
m}\,g^{s}_{\ n}$ where $\hat{R}=R\circ\tau$ and $\tau$ is a swap. For the case
at hand we have

$\displaystyle\begin{pmatrix}aa&ab&ba&bb\\\ ac&ad&bc&bd\\\ ca&cb&da&db\\\
cc&cd&dc&dd\end{pmatrix}\begin{pmatrix}q&0&0&0\\\ 0&1&q-q^{-1}&0\\\ 0&0&1&0\\\
0&0&0&q\end{pmatrix}=\begin{pmatrix}q&0&0&0\\\ 0&1&q-q^{-1}&0\\\ 0&0&1&0\\\
0&0&0&q\end{pmatrix}\begin{pmatrix}aa&ba&ab&bb\\\ ca&da&cb&db\\\
ac&bc&ad&bd\\\ cc&dc&cd&dd\end{pmatrix}$ (B.3)

The definition of commutation relations (B.1) allows the definition of the
space where the elements of the quantum groups act. Thus we introduce the
coordinate functions $x_{i}$ that satisfy the relation

$\displaystyle R^{kl}_{mn}x_{k}x_{l}=x_{m}x_{n},$ (B.4)

then one can check that if we introduce new objects $y_{k}=g^{i}_{k}x_{i}$
they also satisfy the same relation.

On general grounds, the solution of the YB equation is quite complicated, but
Drinfeld managed to find a universal solution when $R$ acts on a tensor
product of the represenations of quantum group [35]. Thus, if we have
$V_{1},V_{2}$ with some action of $U_{q}(\mathfrak{su}_{1,1})$ the $R$ matrix
is given as

$\displaystyle R=\exp_{q^{2}}\left((q^{-1}-q)E\otimes
F\right)q^{-\frac{H\otimes H}{2}}$ (B.5)

##  Appendix C Casimir and derivatives

Here we show that

$\displaystyle C_{q}=q^{-1}\left(1-zz^{*}\right)^{2}\frac{\partial}{\partial
z}\frac{\partial}{\partial z^{*}}$ (C.1)

Let us consider the action of Casimir operator on the following monomial

$\displaystyle C_{q}\left(z^{n}z^{*m}\right)=$ $\displaystyle
q^{3}[n]_{q^{2}}[m]_{q^{2}}z^{n+1}z^{*m+1}+q[n]_{q^{2}}[m]_{q^{2}}z^{n-1}z^{*m-1}-q(1+q^{2})[n]_{q^{2}}[m]_{q^{2}}z^{n}z^{*m}=$
$\displaystyle
q[n]_{q^{2}}[m]_{q^{2}}z^{n-1}\left(1-(1+q^{2})zz^{*}+q^{2}z^{2}z^{*2}\right)z^{*m-1}$
(C.2)

Let us compare it with the following operator

$\displaystyle q^{-1}\left(1-zz^{*}\right)^{2}\frac{\partial}{\partial
z}\frac{\partial}{\partial
z^{*}}(z^{n}z^{*m})=q^{-3}[m]_{q^{2}}[n]_{q^{2}}z^{n-1}y^{2}z^{*m-1}=$
$\displaystyle
q[m]_{q^{2}}[n]_{q^{2}}z^{n-1}\left(1-(1+q^{2})zz^{*}+q^{2}z^{2}z^{*2}\right)z^{*m-1}=C_{q}\left(z^{n}z^{*m}\right)$
(C.3)

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