Editing Vertex operator algebra (section)

==Commutative vertex algebras==
A vertex algebra <math>V</math> is commutative if all vertex operators <math>Y(u,z)</math> commute with each other.  This is equivalent to the property that all products <math>Y(u,z)v</math> lie in <math>V[[z]]</math>, or that <math>Y(u, z) \in \operatorname{End}[[z]]</math>. Thus, an alternative definition for a commutative vertex algebra is one in which all vertex operators <math>Y(u,z)</math> are regular at <math>z = 0</math>.{{sfn|Frenkel|Ben-Zvi|2001}}

Given a commutative vertex algebra, the constant terms of multiplication endow the vector space with a commutative and associative ring structure, the vacuum vector <math>1</math> is a unit and <math>T</math> is a derivation. Hence the commutative vertex algebra equips <math>V</math> with the structure of a commutative unital algebra with derivation.  Conversely, any commutative ring <math>V</math> with derivation <math>T</math> has a canonical vertex algebra structure, where we set <math>Y(u,z)v=u_{-1}vz^0=uv</math>, so that <math>Y</math> restricts to a map <math>Y:V \rightarrow \operatorname{End}(V)</math> which is the multiplication map <math>u \mapsto u \cdot</math> with <math>\cdot</math> the algebra product.  If the derivation <math>T</math> vanishes, we may set <math>\omega=0</math> to obtain a vertex operator algebra concentrated in degree zero.

Any finite-dimensional vertex algebra is commutative.
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! Proof
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This follows from the translation axiom. From <math>[T, Y(u,z)] = \partial_z Y(u,z)</math> and expanding the vertex operator as a power series one obtains
<math display = block> \operatorname{ad}T u_n = [T, u_n] = -nu_{n-1}.</math>
Then
<math display = block> \operatorname{ad}T^m u_n = (-1)^m n (n-1) \cdots (n - m + 1) u_{n-m}.</math>
From here, we fix <math>n</math> to always be non-negative. For <math>m > n</math>, we have <math>\operatorname{ad}T^m u_n = 0</math>.
Now since <math>V</math> is finite dimensional, so is <math>\operatorname{End}(V)</math>, and all the <math>u_n</math> are elements of <math>\operatorname{End}(V)</math>. So a finite number of the <math>u_n</math> span the vector subspace of <math>\operatorname{End}(V)</math> spanned by all the <math>u_n</math>. Therefore there's an <math>M</math> such that <math>\operatorname{ad}T^M u_n = 0</math> for all <math>u_n</math>. But also,
<math display = block>\operatorname{ad}T^M u_{M + n} = (-1)^m (M + n + 1)\cdots(n+1)u_n</math>
and the left hand side is zero, while the coefficient in front of <math>u_n</math> is non-zero. So <math>u_n = 0</math>. So <math>Y(u,z)</math> is regular. <math>\square</math>
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Thus even the smallest examples of noncommutative vertex algebras require significant introduction.