14
$\begingroup$

Given any vertex algebra $V$, you can give a particular quotient $\DeclareMathOperator\Zhu{Zhu}\Zhu V=V/\cdots$ an algebra structure using (a small amount of) the vertex algebra structure. As far as I can tell, two of its main properties are:

  1. If $V$ is nice enough, $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\Mod{Mod}\Rep(V)=\Mod(\Zhu V)$ (edit: not quite, see comment).
  2. If $V$ is a chiral quantisation of a Poisson space $X$, i.e. $\DeclareMathOperator\gr{gr}\gr V=\mathcal{O}(J_\infty X)$ with respect to the canonical ("Li") filtration on $V$, then $\gr\Zhu V=\mathcal{O}(X)$.

My question is whether there is a cleaner way of thinking about the Zhu algebra. In particular,

  1. Can you define the Zhu algebra of a factorisation algebra?
  2. What is the physics name for Zhu algebra of a $2$d CFT, and how does it behave? Is there a $2$d TQFT analogue?
  3. Is there a proof that the Zhu algebra of the affine vertex algebra is $\Zhu V^k(\mathfrak{g})=U(\mathfrak{g})$ using the BD Grassmannian?

A clean enough answer should be able to explain the two main properties above, as well as why the explicit formula for $\Zhu V$ is what it is. Thus, I don't think "$\Zhu V$ is the solution to $\Rep(V)=\Mod(?)$, which exists because …" is a satisfying enough answer.

$\endgroup$
1
  • 1
    $\begingroup$ I have been told by referees that it is strictly false in general that "Mod(Zhu(V)) = Rep(V)". Rather, there is an adjunction between these categories which is a bijection on simples but changes the extensions. At least, that's what I've been told --- I have not tried to understand the details myself. $\endgroup$ Commented Jun 1, 2023 at 12:17

1 Answer 1

9
$\begingroup$

This is a question I've asked around periodically and haven't heard a fully satisfying answer for, but I can report what I understood. Let me say off the bat that the nicest statement I'm aware of is in the paper Chiral Homology of elliptic curves and the Zhu algebra of van Ekeren and Heluani, so the reader might do better to just go there.

Personally I think an answer of the form "$\DeclareMathOperator\Zhu{Zhu}\Zhu(V)$ is the solution to $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\Mod{Mod}\Rep(V)=\Mod(?)$" would be fully satisfying, i.e., if $\Zhu$ were just given as a monadic construction on the category of modules of a vertex algebra. Sadly however this is not true — I'm not aware of any categorical statements of the form $\Rep(V)=\Mod(\Zhu(V))$, even for nice $V$, only a statement about irreducible graded modules, so this is not a good starting point for a characterization of $\Zhu(V)$.

In particular it's important for the story that $\Zhu(-)$ crucially uses the grading ($L_0$) which I'd consider a secondary structure on top of the "core" structure of a vertex algebra (=factorization algebra on the affine line). So it's unreasonable to ask for a version of the Zhu algebra for general factorization algebras (as I bugged people about at some point). The picture I'd like to imagine (as suggested by the paper of van Ekeren and Heluani I linked to above) is roughly that $\Zhu(V)$ is the observables on a "very long cylinder", as opposed to the observables on the punctured disc (the enveloping algebra $U(V)$) — i.e. you need to stretch out using rescaling and in the limit you get the Zhu algebra.

I think this is probably what the [close to original] description of the Zhu algebra by I. Frenkel and Y. Zhu means — rather than think of $\Zhu(V)$ as a quotient of $V$ (which I find confusing) they realize it as the algebra of "zero-modes" — a quotient of the degree zero part (from the $L_0$ grading) of the associative algebra of modes of vertex operators (called either $U(V)$ or $U(U(V))$ depending on your conventions) by the two-sided ideal built out of all products of degree $n$ and degree $-n$ operators.

Anyway as to your first two questions (I don't have a useful comment on the third):

  • No as mentioned Zhu algebras are not about factorization algebras in general but about $\mathbb C^*$-equivariant factorization algebras on the line.

  • A physics name is "zero modes", but it's a special form of dimensional reduction — we're passing from chiral observables in a 2d CFT to observables in a 1d QFT, AKA an associative algebra, by reducing radially on a cylinder AND working equivariantly for rotation.

  • In oriented 2d TFT, you could take the observables, as an $\operatorname{SO}(2)$-fixed $E_2$-algebra $A$, and perform an analogous dimensional reduction to topological quantum mechanics: you take its Hochschild homology, the associative algebra $U(A)=\int_{S^1}A$ (compactification on a cylinder) and then pass to $\operatorname{SO}(2)$-invariants to get an associative algebra (linear over $\mathbb C[\epsilon]=H^*({\operatorname B}{\operatorname{SO}}(2))$).

$\endgroup$
2
  • 2
    $\begingroup$ I would correct that Zhu(.) uses the grading L_0. This is just an unfortunate historic statement that has to do with the exponential change of coordinates. Any vertex algebra, conformal or not, has an associative quotient Z(V). When the vertex algebra is conformal this quotient is canonically isomorphic to Zhu's algebra via an exponential change of coordinates. You can take a look at arxiv.org/abs/1804.00020 for an short construction and explanation of this. More broadly, I think Zhu(V) is intrinsically related to the boundary of the moduli space M_{1,1} $\endgroup$ Commented May 10, 2023 at 13:35
  • $\begingroup$ @ReimundoHeluani Could you spell out your last sentence a bit more, even if just conjectural? It seems really interesting! $\endgroup$
    – Pulcinella
    Commented May 22, 2023 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .