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Extra resources for A Z 2-orbifold model of the symplectic fermionic vertex operator superalgebra
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Therefore, if v ∈ (SF), then for each i, vi is annihilated by the action of ψ(0) for any ψ ∈ h. Thus vi ∈ 2d (h) ⊂ SF[2d] and (SF) ∩ SF (1) ⊂ SF[2d]. 1) and hence (SF) = SF (0) ⊕ SF[2d](1) . 1 The socle soc(SF) of the SF + -module SF is SF[2d]. Proof Let M be an irreducible SF + -submodule of SF. Then (M) ⊂ (SF). 1). If M is generated from a vector in SF[2d](1) then it is an irreducible submodule of SF[2d] isomorphic to SF − . Suppose that M is generated from a vector u in SF (0) . By the classification of irreducible SF + -modules, M is isomorphic to SF + .
1θ → w− 1 ). Consequently, we see that W is isomorphic to either (SF + ) or (SF(θ )+ ). 15 For the vertex operator algebra SF + with d ≥ 2, any irreducible A(SF + )module is isomorphic to one in the list { (SF ± ), (SF(θ )± )}. Symplectic fermionic vertex operator superalgebra 787 5 Further structures of the vertex operator algebra SF + In this section we prove the irrationality of SF + by constructing reducible indecomposable SF + -modules, and determine the automorphism group of SF + . We also calculate the irreducible characters and their modular transformations.
Consider the SF-submodule SF[r] of SF generated from the subspace r (h) for any r ∈ Z≥0 . It is clear that SF[r] ∼ = SF ⊗ r (h) as vector spaces. 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.