3j-symbol for the modular double SLq(2, ℝ) revisited

Modular double of quantum group SLq(2, ℝ) with |q| = 1 has a series of selfadjoint irreducible representations πs parameterized by s ∊ ℝ+. Ponsot and Teschner in [Comm. Math. Phys. 224 (2001) 613] considered a decomposition of the tensor product πs1 ⊗ πs2 into irreducibles. In our paper we give more detailed derivation and some new proofs.


Introduction
Conformal Field Theory is one of the main sources of quantum groups. The very first example of deformed algebra G q of functions on SL(2) was given by the monodromy matrix for the quantized Lax operator of the Liouville model [1]. The variable τ , entering the deformation parameter q = e iπτ played the role of the coupling constant. The duality τ → −1/τ observed in [2,3] was formalized in [4] in the notion of modular double.
The irreducible representations of the modular double of SL q (2, R), introduced in [4], were investigated in [5,6]. In particular in a remarkable paper [5] the problem of the decomposition of the tensor product is solved.
There is an intriguing connection of the representations of the modular double of SL q (2, R) and primary fields of the Liouville model. Both the representation π s of the modular double and vertex operators V α (x) = exp(αφ(x)) are labeled by the same number α = 1/2 + is, s > 0. Apparently there should be a correspondence between the operator expansion of V α 1 (x 1 )V α 2 (x 2 ) and the decomposition of π s 1 ⊗ π s 2 into irreducibles. Some indications on such connection can be found in [7]. However the work in this direction is still to be done. Having this in mind we decided to rederive the results of the paper of Ponsot and Teschner [5] and supply more details of derivations and proofs. Our paper is a complete exposition of the talks of the second author (L.D.F) in the early summer of 2012. The first author (S.E.D) joined the company in the late summer and his contribution let to important improvement of the full exposition. and similarly forẼ,F andK.
We shall use the irreducible representations which are equivalent to the used in [5,6]. Let u and v realize the Weyl relations uv = q 2 vu and act in L 2 (R) by the explicit formulae Here ω and ω are half-periods which substitute periods 1 and τ τ = ω /ω, ωω = −1/4, ω = ω + ω We shall consider the case τ > 0, so that ω and ω are pure imaginary with positive imaginary part. Operators u and v are unbounded and can be defined on the dense domain D consisting of the entire functions f (x), rapidly vanishing at infinity along the lines Im x = const. For instance we can take f (x) in the form for positive α, arbitrary complex β and polynomial P (x). The operators u and v are nonnegative and essentially selfadjoint. The representation π s is given by formulae The representation for the second triple is given by similar formulae in terms ofũ,ṽ andZ andZ = exp (−iπs/ω ). Thusũ,ṽ andZ are obtained from u, v and Z by interchange ω ω . Operatorsũ,ṽ have the same domain and are nonnegative and essentially selfadjoint. Let us note that there is the second regime for τ which could be called real form for SL q (2). It is the case |τ | = 1 or ω = −ω. In this case the involution interchanges pairs u, v andũ,ṽ by u * =ũ, v * =ṽ. This regime has many interesting features. In particular it corresponds to the value of central charge of Liouville model in the interval between 1 and 25. However in this paper we shall consider only regime τ > 0.

Modular quantum dilogarithm
We shall use this term for the function where the contour goes above the singularity at t = 0. This function has long history, different names, normalizations and applications. The normalization used in this paper is used in [10,11].
The term modular quantum dilogarithm is used for the function The adjective "modular" is due to the symmetry of Φ(u) after exchange ω ω and term "dilogarithm" is due to the asymptotic for τ → 0 where Li 2 (u) = ∞ n=1 u n /n 2 containing the Euler dilogarithm. Finally "quantum" is a conventional term for q-deformation.
The main property is the functional equation and similar one for the shift withq. In terms of function γ(x) these equations look as follows We shall need also the reflection formula and formula for the complex conjugation Asymptotic behaviour is γ(x) → 1 for Re(x) → +∞ and reflection formula (3.2) can be used to get asymptotic for Re(x) → −∞.
The function γ(x) has a pole at the point x = −ω and zero at the point x = +ω . The first terms of the series expansions are [10,11,14] γ(−ω + z) = − 1 2πic Comparison of (3.2) and (3.4) gives There are main integral identities [10,11] The last equation indicates that A(s) is an operator of convolution with kernel of the form It is advisable to make Fourier transform. Let F be operator We have uF = F v and vF = F u −1 and after conjugation by F we get from (3.1) We move v −1 and v to the right and cancel. After that we obtain the equations which are equivalent and differs by the change u → q 2 u so that really we have only one equation

It is evident that solution of this equation iŝ
and finally The unitarity of the operator A(s) is evident due to (3.3).

The main equation for the 3j-symbol
In this section we shall solve the system of equations for the function S(x 1 , x 2 , x 3 |s 1 , s 2 , s 3 ) where the operators e 12 , f 12 and K 12 act on variables x 1 and x 2 by and e 3 , f 3 , K 3 acts on variable x 3 . These operators can be obtained from the transposition u = u, v = v −1 and are given by It is already clear that function S realizes the decomposition of the representation π s 1 ⊗ π s 2 into irreducibles. More will be said in the end of the section. Following the previous section we shall often use variables allows to exclude v 2 from the equation (4.1a) and v −1 2 from the equation (4.1b) to get the system of equations for v −1 1 S/S and v 3 S/S as follows: More explicitly these equations read and can be supplemented by which is a corollary of (4.2). Let us look for solution in the form and due to (3.1) the choice reduces the equations to The dual equations, corresponding to the interchange ω ω , have the same solution (4.3) and after that the Ansatz reduce the freedom to double periodic function S 0 , which has to be constant.
Thus the solution of the equations (4.1) is given by The appearance of ω is due to sign factors in (4.3), ω and the singularities here have to be understood as ω → ω − i0, which will be explained in the course of the proof of completeness. The expression for S, equivalent to (4.4), was given in [5] without derivation. Now we can interpret the result in more details. The solution exists for any triplet of real s 1 , s 2 , s 3 and is unique up to normalization constant. This means that the representation with "spin" s 3 enter the tensor product π s 1 ⊗ π s 2 once for any s 3 . This can be formalized by the relation π s 1 ⊗ π s 2 = +∞ 0 ds 3 ρ(s 3 )π s 3 (4.5) We can consider S (x 1 x 2 x 3 ) as the kernel of the integral operator S defined by and equations (4.1) in operator form are e 12 S = Se 3 , f 12 S = Sf 3 , K 12 S = SK 3 (4.6) The complex conjugate function S (x 1 x 2 x 3 ) has interpretation as the kernel of the projection operator The measure ρ(s) in (4.5) should be found from normalization condition for the kernel S(x 1 , x 2 , x 3 ), which will be obtained in the last section.

Undressing of the Casimir
To get the normalization for S it is useful to interpret it as an eigenfunction of the Casimir operator It is clear from (4.6), that as a function of x 1 and x 2 , S satisfies the equation where s 3 and x 3 play the role of parameters labeling eigenvalue and multiplicity. Explicitly C 12 can be written as We shall introduce a series of adjoint transformations of C 12 to reduce it to more simple form. These transformations we shall call "undressing". The first step is to use operator R 1 to cancel factors in front of v 1 and v 2 The solution is a multiplication operator by function The operator K 12 is invariant under transformation by R 1 , i.e R −1 1 K 12 R 1 = K 12 and C 12 transforms into It is clear that R 2 is similar to R 1 after interchange u 2 and v 2 , which is given by the Fourier transformation which respect variable x 2 , so that R 2 = F −1 2R 2 F 2 whereR 2 is a multiplication by Φ(Z 2 u 2 ). The operator K 12 is invariant under transformation R 2 and operator C 12 acquires the form Now we transform V 1 and V 2 to v 1 and v 2 by multiplication operator which leaves K 12 invariant and transform C 12 intõ Altogether the operator A = R 1 F −1 2R 2 F 2 R 3 givesC 12 from C 12 and leaves K 12 invariant, In more explicit form A −1 acts on the function of two variables f (x 1 , x 2 ) as follows: OperatorC 12 is much more simple then C 12 and the problem of simultaneous diagonalization ofC 12 and K 12 allows separation of variables. Consider equations for the corresponding eigenfunctions where we parameterize the eigenvalues by p and s 3 . The first equation allows to exclude v −1 1 from the second to get The general solution of the first equation is given by where x 21 = x 2 − x 1 and after substitution which eliminates the factor in front of v 2 , we get Introducing the new operators The operator in the LHS is well known in CFT. It appears as a trace of monodromy of Lax operator in the Liouville model [1]. In quantum Teichmüller thery it got the name of the length operator for geodesics. R. Kashaev has shown [12,13] that this operator has continues spectrum in the interval [2, ∞] with eigenvalues parameterized in the form Z + Z −1 with Z = exp − iπs ω , s ≥ 0 and eigenfunctions are given by The latter are even functions of s so that these can be considered for any s ∈ R. Kashaev proved [12] the orthogonality and completeness for φ(x, s) in the form . Incidentally, the same reflection coefficient appears in the discussion of the zero modes in the Liouville model in [15].
It is evident that integral operator P with the kernel defines a projection and U maps L 2 (R) into subspace P L 2 (R). However the natural completeness and now we can use them to formulate the orthogonality and completeness for the kernel S(x 1 , x 2 , x 3 ).

Undressing of the eigenfunctions
First of all we have to find out the connection between Ψ p (x 1 , x 2 ) and undressed eigenfunction A −1 S(x 1 , x 2 , x 3 ). The explicit expression for the undressed eigenfunction reads where undressing operator A −1 is given by (5.1). The t-integral is reduced to (3.5c) and can be calculated in explicit form so that we obtain Next step is the calculation of Fourier transformation with respect variable x 3 using (3.5b): γ(x 12 +s 2 +s 3 +ω ) This expression coincides with (5.5), up to overall normalization Z(s 1 , s 2 |s 3 , p), The special choice of the initial normalization constant S 0 given by where φ is real phase, leads to simplification Z(s 1 , s 2 |s 3 , p) = γ(p − s 3 ) so that we obtain properly normalized eigenfunctions Now we shall prove the orthogonality and completeness of the functions S(x 1 , x 2 , x 3 ) in the momentum representation It will be sufficient to prove the orthogonality and completeness for the undressed eigenfunctions S p (x 1 , x 2 ) because the dressed eigenfunction S(x 1 , x 2 , p) = AS p (x 1 , x 2 ) is obtained from the S p (x 1 , x 2 ) after action of the dressing unitary operator A. Due to unitarity this dressing operator effectively cancels out from considered relations. Let us begin from orthogonality. There the dressing operator A cancels out on the first step Note that the appearance of the second term containing δ (s + s ) and kernel of the intertwining operator (3.7) is the direct consequence of the equivalence of representations π s and π −s . The completeness for the undressed eigenfunction can be proven as follows Due to unitarity of the dressing operator the same relation holds for the dressed eigenfunctions,

Appendix A. Orthogonality and completeness of Kashaev eigenfunctions
We shall need the generalization of identity (3.5b) in the form where the integral converges under conditions The inverse formula is .4) and the contour goes below the singularity at s = 0.
Appendix A.1. Orthogonality We take eigenfunctions in the form and the singularities here have to be understood as ω → ω − i0 We have to calculate This function is even in λ and µ and it is sufficient to calculate it in one quadrant, say λ > 0 and µ < 0. First we transform the ratio of two γ-functions using (A.4), and calculate the x-integral using (A.2), and after these two steps arrive to the following expression for I(λ, µ): The ratio of γ-functions is reduced to the simple exponent γ(s + λ − µ + ω )γ(µ − s − λ − ω ) γ(s + ω )γ(−s − ω ) = e iπ(s+λ−µ+ω ) 2 −iπ(s+ω ) 2 due to reflection relation (3.2) so that we obtain for λ > 0 and µ < 0 The full answer is restored by the symmetry