(Indeed from j∈Z (−X) j 2 + j∈Z X j 2 = 2 j∈Z X 4j 2 and the Jacobi identity 13.6(b) we obtain
Ψ(X) −2 Ψ(X 2 ) + Ψ(−X) −2 Ψ(X 2 ) = 2Ψ(−X 4 ) −2 Ψ(X 8 ).
We then replace X by wX.) Thus we have
c = 2Ψ q (X 4 )Ψ((wX) 2 )Ψ(−X 4 ) −2 Ψ(X 8 ) + 2Ψ q (X 4 )Ψ((wX) 2 ) 2 Ψ(−wX) −1 Ψ(wX) −1 .
The product of the last two factors above can be rewritten using 3.16(f) with X replaced by wX. We obtain
c = 2Ψ q (X 4 )Ψ((wX) 2 )Ψ(−X 4 ) −2 Ψ(X 8 ) + 2Ψ q (X 4 )Ψ((wX) 2 ) −1 Ψ(X 4 ).
Combining the results above we obtain
n∈2N α n X n = a + b ′ + b ′′ + c = 3Ψ q (X 4 )Ψ((wX) 2 )Ψ(−X 4 ) −2 Ψ(X 8 ) + 3Ψ q (X 4 )Ψ((wX) 2 ) −1 Ψ(X 4 ) + 2Ψ q (X 4 )Ψ(X 2 ) −1 Ψ(X 4 ). (c)
4. Counting **data** in the dual group
4.1. Let s ∈ Ξ be such that ∆ := ∆ s ∈ A N 0 . We show: (a) β ψ s = ψ −s ∈ Sβ ∆ . It is enough to show that if U ∈ U • then [[⊕ λ∈U V −s λ ]] = N/2 + [[⊕ λ∈U V s λ ]]. This follows from 2.1(a) since ⊕ λ∈U V −s λ = ⊕ λ∈−U V s λ = ⊕ λ∈U −1 V s λ is a subspace of V complementary to ⊕ λ∈U V s λ . 4.2. Let A (resp. A N ) be the set of unordered pairs (∆, ∆ ′ ) where ∆, ∆ ′ are elements of A (resp. A N ) such that ∆ ′ = β ∆. For k ∈ {0
, 1, 2} let A N k be the set of all (∆, ∆ ′ ) ∈ A N such that exactly k of the numbers n ∆ 1 , n ∆ −1 are nonzero. Let A N, = (resp. A N,= ) be the set of all (∆, ∆ ′ ) ∈ A N such that ∆ = ∆ ′ (resp. ∆ = ∆ ′ ).