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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A commutative differential algebra gives a vertex algebra

Example

Let V=C[u] with its usual polynomial addition and multiplication, vacuum 1=1, and derivation T=d/du. Define Y(a,z)b=(ezTa)b=(∑r≥0zrTrar!)b. Then this state–field map makes V a vertex algebra. The series is finite for each a, so it is also a useful example where formal lower truncation is immediate.

Facts & Assumptions

Given: V=C[u], T=d/du, and the state-field convention of Vertex algebra in the state–field convention.

[A1]

T is the coefficientwise formal derivative on polynomials and obeys T(ab)=(Ta)b+a(Tb); the multiplication on V is commutative with unit 1.

[F1]

Vertex algebra in the state–field convention requires lower truncation, Y(1,z)=id⁡, T1=0, Y(a,z)1=ezTa, translation covariance, and locality.

Proof

1.1A1F1given

If a=0, then ezTa=0; otherwise let d=deg⁡a, so Tra=0 for r>d and ezTa is a polynomial in z. Hence Y(a,z)b has finite support in nonnegative powers for every b, and its coefficients define a field. Direct substitution gives Y(1,z)b=b and Y(a,z)1=ezTa. Also T1=0.

2.1A1F1step 1.1algebra

For any b∈V, the derivation law in [A1] gives [T,Y(a,z)]b=T((ezTa)b)−(ezTa)Tb=(TezTa)b=∂zY(a,z)b, so translation covariance holds.

3.1A1F1algebra∎

Each Y(a,z) acts by multiplication by ezTa. Multiplication operators commute because V is commutative, so [Y(a,z),Y(c,w)]=0 for every a,c∈V. Thus locality holds with exponent N=0. All axioms of [F1] are verified.

Depends on

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Dependency tree · two levels

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Sources