content_image/298425.jpg content_image/298423.jpg content_image/298424.jpg
Lemma A2. Let Z = diag ( z$_{1}$, ..., z$_{r}$ ) , where z$_{1}$, ..., z$_{r}$ are complex variables, and let b$_{1}$, ..., b$_{p}$ be complex constants. Then 

The series on the right-hand side of this equality converges uniformly over Ω$_{ρ}$ = {Z : max$_{j}$$_{≤}$$_{r}$ | z$_{j}$ | ≤ ρ − $^{1}$} , for any ρ > max$_{s}$$_{≤}$$_{p}$ | b$_{s}$ | . r p 

Proof: Macdonald (1995, Chapter VI, § 10) shows that ∏ j =1 ∏ s =1 (1 − b$_{s}$z$_{j}$ ) − 1 /α = ∑ ∞ k =0 ∑ κ ⊢ k P ( α ) κ ( B ) Q ( α ) κ ( Z ) , where Q ( α ) κ = c ( κ,α ) c ′ ( κ,α $_{)}$P ( α ) κ . This result together with (35) imply (36). The uniform convergence in (36) follows from the fact that func- tion ∏ r j =1 ∏ p s $_{=1}$(1 − b$_{s}$z$_{j}$ ) − 1 /α is analytic in an open region that includes Ω$_{ρ}$. □ We are now ready to prove Lemma 1. Consider the right-hand side of (1), which 

we will denote as RHS. We will assume that max$_{s}$$_{≤}$$_{p}$ | b$_{s}$ | < 1 and that the contour K is the unit circle in the complex plane. That these assumptions are without loss of generality follows from the fact that the value of RHS does not change under the transformation Z → ϕ Z , B → ϕB, and A → ϕ − $^{1}$A, where ϕ is any positive num- ber, and under a deformation of K into the unit circle (because such a deformation leaves the contour in the region of the analyticity of the integrand). With these as- sumptions, and noting that the component ∏ r $_{j>i}$( z$_{j}$ − z$_{i}$ ) 2 /α of ω ( α $^{)}$( A, B, Z ) equals ( − 1) r ( r − 1) / (2 α ) ∏ r j =1 z ( r − 1) /α j ∏ r j ̸ = i ( 1 − z$_{i}$z − 1 j ) 1 /α , we can rewrite RHS for α = 2 /β, where β is any positive integer, in the form of the torus scalar product 

<img src='content_image/298425.jpg'>

<img src='content_image/298423.jpg'>

where 

<img src='content_image/298424.jpg'>

