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Zhu zero-mode identities on singular vectors of an arbitrary module

Statement

For a conformal vertex algebra V of CFT type and an arbitrary left V-module M define $Ω (M) = {v \in M : a_{(n)}^{M} v = 0 for every homogeneous a \in V and every integer n \geq w t (a)} .$ For homogeneous a put $o_{M} (a) = a_{(w t (a) - 1)}^{M}$ , and extend linearly to V. Every $o_{M} (a)$ preserves $Ω (M)$ , and, for the circle and star operations and O(V) defined in the Zhu quotient construction, $o_{M} (O (V)) Ω (M) = 0, o_{M} (a * b) v = o_{M} (a) o_{M} (b) v (v \in Ω (M)), o_{M} (1) = {i d}_{M} .$ No finite-dimensionality, admissible grading, semisimplicity or diagonalizable L(0) is assumed for M.

Statement references

Jacobi and residue calculus from the state-field locality axioms

Zhu algebra of a CFT-type vertex operator algebra

Module for an ordinary vertex algebra

Facts & Assumptions

Given: The module and the explicitly defined subspace $Ω (M)$ . All equalities below are applied to a fixed vector v in that subspace.

[F1]

Jacobi and residue calculus from the state-field locality axioms gives the commutator and iterate mode identities from module Jacobi, including generalized binomial coefficients for negative integer upper indices.

[F2]

Zhu algebra of a CFT-type vertex operator algebra gives $a \circ b = \sum_{j = 0}^{r} (\binom{r}{j}) a_{(j - 2)} b$ and $a * b = \sum_{j = 0}^{r} (\binom{r}{j}) a_{(j - 1)} b$ for $a \in V_{r}$ ; each nonzero summand has its actual conformal weight supplied by F1.

[F3]

Module for an ordinary vertex algebra supplies vacuum and lower truncation. No grading condition on M is used.

Proof

1.1givenF1F3algebra

Let $a \in V_{r}$ , $c \in V_{s}$ and $n \geq s$ . By F1, $c_{(n)}^{M} o_{M} (a) v = o_{M} (a) c_{(n)}^{M} v + \sum_{i \geq 0} (\binom{n}{i}) (c_{(i)} a)_{(n + r - 1 - i)}^{M} v$ . The first term is zero by the definition of $Ω (M)$ . A nonzero state $c_{(i)} a$ has weight $s + r - i - 1$ , and its mode index is at least that weight because $n \geq s$ ; thus every term in the sum is zero too. The commutator sum is finite by state truncation. This proves invariance of $Ω (M)$ under every zero mode.

1.2algebra

For an integer t with $0 \leq t < r$ and either $d = 1$ or $d = 2$ , the polynomial $P_{t} (j) = (\binom{j - d}{t})$ has degree t. Consequently $\sum_{j = 0}^{r} (- 1)^{j} (\binom{r}{j}) (\binom{j - d}{t}) = 0$ . To verify this cancellation directly, put $Δ P (j) = P (j + 1) - P (j)$ . Iterating the elementary binomial recursion gives $Δ^{r} P (0) = \sum_{j = 0}^{r} (- 1)^{r - j} (\binom{r}{j}) P (j)$ . A difference reduces polynomial degree by at least one, so it is zero after r differences when $t < r$ . The generalized binomial $(\binom{j - d}{t}) = (j - d) (j - d - 1) \dots (j - d - t + 1) / t!$ is a polynomial also at $j = 0, 1$ ; no positive-index restriction is imposed.

2.1F1F2step 1.2algebra

Take homogeneous $a \in V_{r}, b \in V_{s}$ . For the jth circle summand, set $m = j - 2$ and $n = r + s - j$ , the zero-mode index of $a_{(j - 2)} b$ . F1 gives first-product terms $(- 1)^{t} (\binom{j - 2}{t}) a_{(j - 2 - t)}^{M} b_{(r + s - j + t)}^{M} v$ , all zero because $r + s - j + t \geq s$ . Its second-product term has the fixed b-index $n + m - t = r + s - 2 - t$ and coefficient $- (- 1)^{t + j} (\binom{j - 2}{t})$ . Terms t≥r are zero since $a_{(t)}^{M} v = 0$ . Summing the remaining terms over $0 \leq j \leq r$ with coefficients $(\binom{r}{j})$ gives zero by step 1.2 with $d = 2$ . Thus $o_{M} (a \circ b) v = 0$ . All rearrangements are finite: j has finite range, and the surviving second terms have $0 \leq t < r$ . Bilinearity proves the O(V) assertion.

3.1F1F2F3step 1.1step 1.2step 2.1algebra∎

For the jth star summand instead set $m = j - 1$ and $n = r + s - j - 1$ . The first-product b-index is $r + s - j - 1 + t$ , which is at least s except when $j = r$ and $t = 0$ . This unique surviving term has coefficient one and is $a_{(r - 1)}^{M} b_{(s - 1)}^{M} v$ . The second-product b-index is again $r + s - 2 - t$ and its coefficient after summing j is $- (- 1)^{t} \sum_{j = 0}^{r} (- 1)^{j - 1} (\binom{r}{j}) (\binom{j - 1}{t})$ . It vanishes for $t < r$ by step 1.2 with $d = 1$ , while t≥r vanishes by the defining annihilation of v. Hence $o_{M} (a * b) v = o_{M} (a) o_{M} (b) v$ . If $r = 0$ , the $t < r$ sums are empty and the same first-product argument still has exactly the one surviving term; thus no exceptional weight-zero case is omitted. Finally $1_{(- 1)}^{M} = {i d}_{M}$ by vacuum. Together with step 1.1 these prove all the asserted identities.

Depends on

Dependency tree · 2 levels

Sources

Dong, Li and Mason, Vertex operator algebras and associative algebras, Theorem 3.2 and its n=0 coefficient calculation