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Lower truncation alone does not imply locality

Statement refuted

Every pair of lower-truncated fields on a complex vector space is mutually local.

Facts & Assumptions

Given: The field and locality conventions of Formal distributions and fields on a vector space.

[F1]

Formal distributions and fields on a vector space: A(z)=∑nA(n)z−n−1 is a field if for every vector v, A(n)v=0 for all sufficiently large n; locality requires (z−w)N[A(z),B(w)]=0 for some N≥0.

Proof

1.1F1givenconstruct

Let V=C2 with basis v1,v2. Define constant fields A(z)=E12 and B(w)=E21, where E12v2=v1, E12v1=0, E21v1=v2, and E21v2=0. Their only nonzero modes are A(−1)=E12 and B(−1)=E21, so both are fields by [F1].

2.1F1step 1.1algebra∎

Their commutator is [A(z),B(w)]=E11−E22≠0. For every integer N≥0, (z−w)N(E11−E22) has nonzero coefficient zN(E11−E22), including N=0. Thus no locality exponent exists, although both fields are lower-truncated; the refuted implication fails.

Depends on

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