By Abe T.

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This shows that Aut (SF) ∼ = Sp(2d, C). We note that the automorphism θ is in the center of Sp(2d, C) and θ is the center of Sp(2d, C). Therefore, Sp(2d, C)/ θ faithfully acts on SF + . We shall prove that Aut (SF + ) ∼ = Sp(2d, C)/ θ . We see that the characters SM (τ ) for M = SF ± , SF(θ )± are mutually distinct. This implies that for any g ∈ Aut (SF + ) and irreducible SF + -module M, the SF + -module (Mg , Y g ( · , z)) with Mg = M and Y g ( · , z) = Y(g(·) , z) is isomorphic to itself because SMg (τ ) = SM (τ ).

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5) = φ2 (τ ) which follow from the well known modular transformation lows πi η(τ + 1) = e 12 η(τ ), η − 1 τ = (−iτ )1/2 η(τ ). By using the formula we have the following proposition. 6) 790 T. 5 The modular transformations of SSF ± (τ ) and SSF(θ )± (τ ) with respect to the transformations τ → τ + 1 and τ → − τ1 are given by SSF ± (τ + 1) = e SSF ± 1 − τ = 1 2d+1 dπi 6 SSF(θ )+ (τ ) − SSF(θ )− (τ ) ± SSF(θ )± (τ + 1) = ±e− SSF(θ )± − 1 τ = SSF ± (τ ), dπi 12 (−iτ )d (SSF + (τ ) − SSF − (τ )), 2 SSF(θ )± (τ ), 1 SSF(θ )+ (τ ) + SSF(θ )− (τ ) ± 2d−1 SSF + (τ ) + SSF − (τ ) .

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