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Witt algebra of Laurent polynomial vector fields

Definition

Let R=C[z,z−1] be the algebra of finite Laurent polynomials. Define Der⁡C(R) as the vector space of complex-linear maps D:R→R satisfying the Leibniz rule D(fg)=D(f)g+fD(g). Let ∂z be the linear map with ∂z(zq)=qzq−1 for all q∈Z. For each m∈Z, set dm=−zm+1∂z. The Witt algebra W is the complex span of the dm inside Der⁡C(R), with bracket the commutator of derivations. The modes are indexed by all integers; all linear combinations in W are finite, and no completion is intended.

Facts & Assumptions

Given: The Laurent-polynomial algebra R and the displayed derivations dm.

[F1]

Lie algebras over a field: a Lie algebra over C is a complex vector space with an alternating bilinear bracket satisfying Jacobi.

Proof

1.1givenalgebra

For a,b∈Z, ∂z(za+b)=(a+b)za+b−1=∂z(za)zb+za∂z(zb), so linearity and bilinearity prove the Leibniz rule on all Laurent polynomials. Multiplying a derivation by −zm+1 preserves the Leibniz rule because R is commutative; hence each dm belongs to Der⁡C(R). For every q∈Z, dm(zq)=−qzm+q, so dmdn(zq)=q(n+q)zm+n+q and dndm(zq)=q(m+q)zm+n+q. Therefore [dm,dn](zq)=q(n−m)zm+n+q=(m−n)dm+n(zq). Since Laurent monomials span R, this proves [dm,dn]=(m−n)dm+n and closure of W under the commutator.

1.2givenalgebra

If a finite relation ∑mamdm=0 holds, applying it to z gives −∑mamzm+1=0. Distinct Laurent monomials are linearly independent in R, so every am=0; thus the dm form a basis of W.

2.1F1step 1.1algebra∎

For endomorphisms A,B,C of any vector space, the commutator is alternating, and expanding [A,[B,C]]+[B,[C,A]]+[C,[A,B]] cancels each of the six triple compositions with its negative. Thus the commutator bracket satisfies [F1] on End⁡C(R); by the closure in step 1.1, its restriction makes W a Lie algebra.

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