As the title suggests, I'm interested in deformation theory of vertex algebras and their representations.

In the paper https://arxiv.org/abs/1806.08754, the authors construct, for a vertex algebra $V$ with a module $M$, a family of cohomology spaces $H^{i}_{VA}(V, M)$. Naturally, they in fact construct a complex computing the above and call it vertex algebra cohomology (of $V$ with coefficients in $M$). Further they prove that low degree cohomology groups can be interpreted in the usual way (a version of singular vectors, derivations, extensions etc). In the case of the adjoint representation I believe we obtain a complex controlling the deformation theory of the vertex algebra, as one would hope.

I find the calculus of vertex algebras somewhat daunting at times and so I find the construction hard to follow. I'd like to understand it in a simple case, hopefully not so simple as to be completely degenerate.

Let $V$ then he a holomorphic vertex algebra, so that for all $v\in V$ the field $v(z)$ is an element of $End(V)[[z]]$. It is not hard to show that such a $V$ is equivalent to the data of a commutative algebra with a derivation. Switching to this language I'll write $(A, \delta)$ for such an object. What is the vertex algebra cohomology of $(A, \delta)$ with cohomolgy in the adjoint representation?

If I'm not mistaken, square zero deformations of $(A, \delta) $ are Hochschild cohomology classes $\gamma\in HH^{2}(A)$ such that $Lie_{\delta} (\gamma) =0$. Perhaps this generalizes in the obvious way to other cohomology groups, whatever they are. (Note that the answer should have the structure of a dgla and $Ker(Lie_{\delta})$ indeed has such a structure, in fact that of a Gerstenhaber algebra I believe.)


1 Answer 1


The complex considered on that article is a linear algebraic version of a complex constructed by Tamarkin in his ICM address


As he points out, in the case of deformations of vertex algebras, instead of a dgla, the complex that controls said deformations is a differential graded Lie conformal algebra. The construction of that complex in the linear algebraic setting is very similar to what we wrote in the article you cite, however there are some technical difficulties that one has to deal with and they have to do with finiteness conditions on the vertex algebra V as a C[T] module. In the case of Tamarkin he considered finitely generated D-modules (which no interesting vertex algebra is). The construction of this complex in the linear algebraic setting will eventually come out.

Another comment related to your question (the only one I see is about commutative vertex algebras) is the following. Given a filtered vertex algebra V such that its associated graded A is a Poisson vertex algebra, in particular it is a commutative algebra with a derivation. One can study vertex algebras that deform this given Poisson vertex algebra structure A. This computation was carried by Tamarkin in the case when A is a symmetric algebra. A commutative vertex algebra can be viewed as a Poisson vertex algebra with the zero Poisson bracket (the OPE is regular, or zero). If you look at deformations preserving this Poisson structure you'd get the Hochschild cohomology as you hint. However the second cohomology has more structure corresponding to cocycles deforming the OPE.

  • $\begingroup$ Thank you, I was hoping you'd answer this! My understanding of your remark re. my precise question is that we shouldn't expect to be able to compute the vertex deformations of a holom vertex algebra $(A, \delta)$ purely in terms of the hochschild calculus of $A$, since this would only compute holomorphic deformations and we should expect to have non-holomorphic ones too? This seems very reasonable to me. $\endgroup$
    – EBz
    Jul 2, 2019 at 10:59
  • 1
    $\begingroup$ Exactly. However take a look at the last computation in Tamarkin's cause it deals with a particular case (the free case) in some detail $\endgroup$ Jul 2, 2019 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.