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The first boundary Virasoro singular vectors at central charge minus two

Example

At $p = 2$ and $d = 1$ , the boundary Virasoro central charge is $c = - 2$ and the weights are $a_{n} = ((4 n + 1)^{2} - 1) / 8$ and $b_{n} = ((4 n - 1)^{2} - 1) / 8$ . Their chain begins $a_{0} = 0 < b_{1} = 1 < a_{1} = 3 < b_{2} = 6 < a_{2} = 10$ .

Let $v_{0}$ be the highest vector in $M (- 2, 0)$ and $v_{1}$ the highest vector in $M (- 2, 1)$ . Their first singular vectors are $s_{1} = L_{- 1} v_{0}, s_{2} = (L_{- 1}^{2} - 2 L_{- 2}) v_{1} .$ The Verma map sending $v_{1}$ to $s_{1}$ sends $s_{2}$ to $s_{3} = (L_{- 1}^{3} - 2 L_{- 2} L_{- 1}) v_{0},$ a nonzero singular vector at global grade three. Thus the first embeddings are $M (- 2, 3) ↪ M (- 2, 1) ↪ M (- 2, 0)$ .

The resulting simple-module characters, through relative grade six, are $ch L (- 2, 0) = 1 + q^{2} + q^{3} + 2 q^{4} + 2 q^{5} + 4 q^{6} + O (q^{7}),$ $ch L (- 2, 1) = q (1 + q + q^{2} + 2 q^{3} + 3 q^{4} + 4 q^{5} + 6 q^{6} + O (q^{7})) .$ These finite coefficients illustrate the exact all-grade character formulas $(1 - q) / \prod_{m \geq 1} (1 - q^{m})$ and $(q - q^{3}) / \prod_{m \geq 1} (1 - q^{m})$ , respectively.

Statement references

Virasoro Verma modules and contravariant forms

PBW basis and grades of Virasoro Verma modules

Single-chain maximal submodules at boundary Virasoro central charges

Facts & Assumptions

Given: The two algebraic Verma modules and the fixed values $p = 2, d = 1$ .

In this verification, a singular vector means a nonzero $L_{0}$ -eigenvector annihilated by every $L_{m}$ with $m > 0$ . Thus the calculations below establish both the eigenvector and positive-mode conditions; [F1] gives the highest-vector relations used.

[F1]

Virasoro algebra in the $L_{m}$ convention gives $[L_{m}, L_{n}] = (m - n) L_{m + n} + (m^{3} - m) δ_{m + n, 0} C / 12$ ; Virasoro Verma modules and contravariant forms fixes the induced module and highest-vector relations, with $C = - 2$ in this example and $L_{m} v_{h} = 0$ for $m > 0$ and $L_{0} v_{h} = h v_{h}$ .

[F2]

PBW basis and grades of Virasoro Verma modules gives linearly independent negative-mode PBW monomials and grade dimensions $P (N)$ .

[F3]

Single-chain maximal submodules at boundary Virasoro central charges identifies each first embedded Verma as the maximal proper submodule and proves the exact simple-character formulas.

Verification

1.1F1F2algebra

Substitution into the boundary formulas gives $c = - 2$ and $a_{0} = 0, b_{1} = 1, a_{1} = 3, b_{2} = 6, a_{2} = 10$ . From [F1], $L_{1} s_{1} = [L_{1}, L_{- 1}] v_{0} = 2 L_{0} v_{0} = 0,$ and for $m \geq 2$ , $L_{m} s_{1} = (m + 1) L_{m - 1} v_{0} = 0$ . The vector is nonzero by [F2], so $s_{1}$ is singular.

1.2F1F2algebra

For a highest vector $v_{h}$ , commutation gives $L_{1} L_{- 1}^{2} v_{h} = (4 h + 2) L_{- 1} v_{h}, L_{1} L_{- 2} v_{h} = 3 L_{- 1} v_{h},$ $L_{2} L_{- 1}^{2} v_{h} = 6 h v_{h}, L_{2} L_{- 2} v_{h} = (4 h + c / 2) v_{h} .$ At $h = 1, c = - 2$ the coefficient pairs are $(6, 3)$ for each of $L_{1}, L_{2}$ , so both modes annihilate $s_{2}$ . For $m > 2$ , applying $L_{m}$ to a relative-grade-two vector lands below the Verma top and is zero. The distinct PBW terms make $s_{2} \neq 0$ , so it is singular.

2.1F2F3step 1.1step 1.2algebra

A singular vector at weight $h^{'}$ defines a Verma homomorphism by sending the new highest vector to that vector. Under [F2], this map is right multiplication by its nonzero negative-mode polynomial. PBW filters $U (n_{-})$ with associated graded the polynomial algebra on the negative modes, which has no zero divisors; hence right multiplication is injective. Applying this to $s_{1}$ and $s_{2}$ gives the stated embeddings, and the composite sends $v_{2}$ to $s_{3}$ . The PBW coefficients $1, - 2$ show $s_{3} \neq 0$ . The proved chain lemma [F3] supplies maximality and the exact simple characters.

3.1F1F2step 2.1algebra

Directly verify the composite singular vector. Commutators give $L_{1} L_{- 1}^{3} v_{0} = 6 L_{- 1}^{2} v_{0}, L_{1} L_{- 2} L_{- 1} v_{0} = 3 L_{- 1}^{2} v_{0},$ $L_{2} L_{- 1}^{3} v_{0} = 6 L_{- 1} v_{0}, L_{2} L_{- 2} L_{- 1} v_{0} = 3 L_{- 1} v_{0} .$ Thus $L_{1} s_{3} = L_{2} s_{3} = 0$ . Also $L_{3} L_{- 1}^{3} v_{0} = 24 L_{0} v_{0} = 0$ and $L_{3} L_{- 2} L_{- 1} v_{0} = 5 L_{1} L_{- 1} v_{0} = 10 L_{0} v_{0} = 0$ . For $m > 3$ , positive modes land below the top. Therefore $s_{3}$ is singular.

4.1F2F3algebra∎

For $M (- 2, 0)$ , the character from [F3] is $(1 - q) R (q)$ ; for $M (- 2, 1)$ it is $q (1 - q^{2}) R (q)$ , where $R (q) = \prod_{m \geq 1} (1 - q^{m})^{- 1}$ . The coefficients of $R$ through degree six are the partition numbers $1, 1, 2, 3, 5, 7, 11$ . Subtracting the shifted coefficients gives $1, 0, 1, 1, 2, 2, 4$ and, inside the factor $q$ , $1, 1, 1, 2, 3, 4, 6$ . These are the displayed expansions; the coefficient arithmetic is finite and exact.

The first boundary Virasoro singular vectors at central charge minus two

Example

Statement references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources