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Additional info for A Z 2-orbifold model of the symplectic fermionic vertex operator superalgebra

Example text

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 ). In particular, for any g ∈ Aut (SF + ), there exists a unique SF + module isomorphism fg : SF − → (SF − )g up to nonzero scalar multiple. Actually, Symplectic fermionic vertex operator superalgebra 791 if fg is another SF + -module isomorphism from SF − to (SF − )g then fg−1 ◦ fg is in Hom SF + (SF − , SF − ) ∼ = C.

W for any 1 ≤ i = j < d. We shall show that U is an A(SF + )-submodule of W. Firstly, we have [hk,k ]U ⊂ U for any 1 ≤ k ≤ d. U = 0 for any xi,j = ei,j , hi,j , f i,j . U ⊂ U for any xk,l = ek,l , hk,l , f k,l . U = 0. U = 0. [f i,j ] = [f i,k ] ∗ ([hi,i ] − [hj,j ]). U ⊂ U. W is closed under the actions of [ek,l ], [hk,l ], [f k,l ] for any 1 ≤ k, l ≤ d and that it is an A(SF + )-submodule of W. W is zero. Consequently we have [f i,j ] = 0 on W. One can also prove that [ei,j ] = [hi,j ] = 0 on W for 1 ≤ i = j ≤ d.

Since dim h = 2d, SF[2d + 1] = 0. Thus we have a sequence of SF + -submodules 0 = SF[2d + 1] ⊂ SF[2d] ⊂ SF[2d − 1] ⊂ · · · ⊂ SF[0] = SF. By definition, for any ψ ∈ h, ψ(0) SF[r] ⊂ SF[r + 1]. Thus ψ(0) acts trivially on the quotient SF[r]/SF[r + 1]. Therefore, SF[r]/SF[r + 1] ∼ = SF ⊗ (r (h)/r+1 (h)) as left SF-modules. Since dim(r (h)/r+1 (h)) = 2d r , SF[r]/SF[r + 1] is a direct sum copies of SF as an SF-module. of 2d r We note that SF is naturally an SF + -module. We determine the space (SF) of singular vectors in SF as an SF + -module.

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A Z 2-orbifold model of the symplectic fermionic vertex operator superalgebra by Abe T.

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