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Introduction to Vertex Operator Algebras

AcceptedSubmitted by ChenqiMeng · via CLI · AI-assisted28 Oct 2026, 09:01 UTC

Accepted 28 Oct 2026, 09:02 UTC

math.QA — Quantum Algebra · Primary
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  1. Snapshot 261028.42575b803345 — Rename the vertex algebra chapter
  2. Snapshot 261028.d91af2ea7ee6 — Rename the vertex algebra chapter heading to Vertex algebras and define the operator commutator at …

Rename the vertex algebra chapter heading to Vertex algebras and define the operator commutator at first use. Simplify the weight-one Lie algebra proposition to closure, antisymmetry, and Jacobi, explicitly distinguishing the operator commutator from the state bracket. Remove the distracting discussion of derivations and faithfulness.

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d91af2ea7ee699ea1fc591849d2a815a5ae47a1ad1f1e7491e37f3e156a9cccd
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261028.4a5afd9e6ab7
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Vertex operator algebras

--- Base version+++ Proposed version@@ -232,38 +232,43 @@  ## The weight-one Lie algebra -::: proposition [Lie algebra and derivations] {@prop:voa-weight-one}+::: proposition [The weight-one Lie algebra] {@prop:voa-weight-one} If $V$ is of CFT type, then $V_1$ is a Lie algebra with bracket  $$-[a,b]=a_{(0)}b.-$$--Its zero modes act on $V$ by derivations:--$$-[a_{(0)},Y(b,z)]=Y(a_{(0)}b,z),-\qquad-[a_{(0)},b_{(0)}]=(a_{(0)}b)_{(0)}.-$$--The zero-mode action need not be faithful.-:::--::: proof-Weight compatibility gives $a_{(0)}b\in V_1$ for $a,b\in V_1$. Taking the $z^{-1}$ coefficient of skew-symmetry gives+[a,b]:=a_{(0)}b,\qquad a,b\in V_1.+$$+:::++::: proof+The bracket is bilinear, and weight compatibility gives $a_{(0)}b\in V_1$. Taking the $z^{-1}$ coefficient of skew-symmetry gives  $$ a_{(0)}b=-b_{(0)}a+T(b_{(1)}a). $$ -Here higher terms vanish by nonnegative weights, while $b_{(1)}a\in V_0=\mathbb C\mathbf1$, so the last term is zero. Taking $\operatorname{Res}_x\operatorname{Res}_z$ in @prop:jacobi-identity gives the field commutator, and then taking its zero mode gives the second identity. Acting on $d\in V_1$ yields--$$-[a,[b,d]]-[b,[a,d]]=[[a,b],d],-$$--which is the Jacobi identity. The field commutator also proves the derivation property on all of $V$.+Higher terms vanish by nonnegative weights. Since $b_{(1)}a\in V_0=\mathbb C\mathbf1$ and $T\mathbf1=0$, the bracket is antisymmetric.++Now we prove the Jacobi identity holds for the Lie algebra. Taking $\operatorname{Res}_x\operatorname{Res}_z\operatorname{Res}_w$ in @prop:jacobi-identity gives++$$+[a_{(0)},b_{(0)}]=(a_{(0)}b)_{(0)},+$$++where the bracket on the left is the commutator of endomorphisms. Applying this identity to $d\in V_1$ yields++$$+a_{(0)}(b_{(0)}d)-b_{(0)}(a_{(0)}d)+=(a_{(0)}b)_{(0)}d,+$$++or, in terms of the bracket on states,++$$+[a,[b,d]]-[b,[a,d]]=[[a,b],d].+$$++Together with antisymmetry, this is the Jacobi identity. :::  ## References
Accepted by ChenqiMeng on 28 Oct 2026, 09:02 UTC. The submitted snapshot is preserved.

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