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An explicit logarithmic conformal Heisenberg module

Example

For the rank-one Heisenberg conformal VOA, take $F_{1} = C [x_{1}, x_{2}, \dots] \otimes E$ with $E = C e_{0} \oplus C e_{1}$ , $N e_{1} = e_{0}$ , $N e_{0} = 0$ , and $a_{0} = I + N$ . The positive modes are $a_{r} = r \partial_{x_{r}}$ and negative modes are $a_{- r} = x_{r}$ . Its fields are the normal-ordered divided derivatives in the A-page construction, not constant dual-number fields. It is an admissible, grading-restricted logarithmic module, and on its lowest generalized weight space

$a_{0} ∣_{E} = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}), L_{F} (0) ∣_{E} = (\begin{array}{cc} \frac{1}{2} & 1 \\ 0 & \frac{1}{2} \end{array}) .$

Facts & Assumptions

Given: The concrete zero-mode matrix and oscillator action above.

[F1]

The Heisenberg Jordan Fock family is grading-restricted; it is logarithmic exactly when λ is nonzero proves the full conformal-VOA module construction, module Jacobi, truncation, admissibility and generalized weight spaces, with $L_{F} (0) = D + I / 2 + N$ at $λ = 1$ .

[F2]

Logarithmic generalized module requires grading restriction and a nontrivial Jordan chain for $L_{F} (0)$ .

Verification

1.1givenF1F2algebra

In the ordered basis $e_{0}, e_{1}$ , $N$ has matrix $(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ and square zero. Thus $(I + N)^{2} / 2 = I / 2 + N$ , giving the two displayed matrices by [F1]. In particular $(L_{F} (0) - 1 / 2) e_{1} = e_{0} \neq 0$ and $(L_{F} (0) - 1 / 2) e_{0} = 0$ . These are an actual length-two conformal Jordan chain, not just a nonsemisimple action of an unrelated algebra element.

2.1givenF1F2step 1.1algebra∎

On a weighted-degree- $d$ monomial $p$ , $L_{F} (0) (p \otimes e_{0}) = (d + 1 / 2) p \otimes e_{0}$ and $L_{F} (0) (p \otimes e_{1}) = (d + 1 / 2) p \otimes e_{1} + p \otimes e_{0}$ . The generalized weights are exactly $d + 1 / 2$ , with finite spaces $P_{d} \otimes E$ and lower bound $1 / 2$ , by [F1]. Therefore the module meets every grading restriction and is logarithmic by [F2]. Formally on this pair, exponentiating the nilpotent part terminates and gives $z^{L_{F} (0)} (p \otimes e_{1}) = z^{d + 1 / 2} (p \otimes e_{1} + (\log z) p \otimes e_{0})$ ; this explains the terminology while the module fields themselves remain the ordinary Laurent fields already proved in [F1].

Depends on

Used by

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Sources

Tomoyuki Arakawa and Anne Moreau, Arc Spaces and Vertex Algebras (standard reference, not scraped)