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Aν implies that σ (A j ) < knqν /2π b for each j = 1, . . , ν , since the trace of any positive definite matrix is positive. 64) implies that each of the terms of the sum for g(A) has a factor of the form f1 (A1 ) = f (A1 ) with σ (A1 ) < knqν /2π b, which is zero. 29, G = 0, and so F = 0. Now we can prove that the subspace N of cusp forms of M is finite-dimensional. Since entries of positive semidefinite matrices A = (aαβ ) satisfy the inequalities aαα ± 2aαβ + aβ β ≥ 0, it follows that the number of positive semidefinite even matrices A of order n with σ (A) ≤ 2N does not exceed the bound (N + 1)n (2N + 1)n(n−1)/2 .

Hence the function H(Z) attains its maximum µ at some point Z0 = X0 + iY0 of Dn . Since H is Γ -invariant, we conclude that µ is the maximum of H on H, that is, H(Z) ≤ H(Z0 ) = µ for all Z ∈ H. Let us set Zt = Z0 + tE, where t = u + iv is a complex parameter, and consider the function h(t) = F(Zt )e−π iλ σ (Zt ) = ∑ f (A)eπ i(σ (AZ0 )+t σ (A))−λ σ (Z0 +tE)) ∑ f (A)eπ i(σ (AZ0 −λ Z0 )) eπ it(σ (A)−λ n) = h (w), A∈E, A>0 = A∈E, A>0 where w = eπ it and λ satisfies λ n = 1 + [kn/2π bn ], with [α ] denoting the greatest integer not exceeding α .

If M, M ∈ S, then M, M ∈ S and so MM ∈ S. From the obvious fact that X and Y are abelian groups under the addition of matrices, and from easily verified inclusions YY ⊂ Y, XX ⊂ X, XY ⊂ X, YX ⊂ X we conclude that MM ∈ R. Thus, MM ∈ S ∩ R = S. 1) implies that M −1 ∈ S if M ∈ S. Hence S is a subgroup of S. Now we shall show that H = M iE M∈S . 3 Transformations of Lobachevsky Half-Spaces x y y −x Let Z = 49 + iv ∈ H. 31) √1 v 0 and MZ iE = Z. 30) is contained in the right-hand side. 32) E. Thus, it is sufficient to check that σ (X ) = 0.

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