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The vacuum Verma module has a level-one null vector

Statement

For every c∈C, the vacuum Verma module M(c,0) has a nonzero singular vector L−1v0 at grade one. Its Shapovalov norm is zero, it lies in the radical, and det⁡G1(c,0)=0. This is an algebraic null-vector example; it asserts no positivity or Hilbert-space structure.

Facts & Assumptions

Given: The algebraic module M(c,0) and its normalized form.

[F1]

Virasoro algebra in the Lm convention: [Lm,Ln]=(m−n)Lm+n+(m3−m)δm+n,0C/12.

[F2]

Virasoro Verma modules and contravariant forms: Lnv0=0 for n>0, L0v0=0, and Cv0=cv0; the normalized form is contravariant and its radical is the unique maximal proper submodule.

[F3]

PBW basis and grades of Virasoro Verma modules: each grade has its partition basis and L−1v0 is the nonzero basis vector for the unique partition of grade one.

Proof

1.1F1F2F3algebra

Put w=L−1v0. It is nonzero by [F3] and has grade one because [L0,L−1]=L−1 by [F1]. For n=1, L1w=[L1,L−1]v0=2L0v0=0. For every n≥2, Lnw=[Ln,L−1]v0=(n+1)Ln−1v0=0, since n−1>0. Thus w is singular. Also L0w=w, so it is not a second grade-zero vector.

2.1F1F2F3step 1.1algebra∎

Contravariance gives ⟨w,w⟩=⟨v0,L1L−1v0⟩=0, because L1L−1v0=2L0v0=0. If z is homogeneous of grade n, then L0w=w, L0z=nz, and contravariance under L0 gives (1−n)⟨w,z⟩=0. Thus w is orthogonal to every grade n≠1. Grade one is spanned by w by [F3], and its self-pairing is zero, so w is orthogonal to the whole algebraic direct sum and lies in the radical. At general highest weight h, the grade-one Gram matrix is the one-entry matrix (2h), so at h=0 its determinant is zero for every central charge c. A zero norm here is a bilinear-form statement and uses no positivity assumption.

Depends on

Dependency tree · 2 levels

Sources