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The two-variable elliptic genus is a topological invariant of almost-complex manifolds that takes values in power series. These power series turn out to describe weak Jacobi forms when the manifold i …

This is typically given by the commutant, or coset construction. You take the vector subspace of $\mathcal{R}_\text{vac}[\mathfrak{g}_k]$ spanned by vectors $v$ satisfying $Y(u,z)v \in \mathcal{R}_\t …

The terminology is explained earlier on that page and the previous page in the paper. On the same page, we see that they set $\mathfrak{g} = \mathfrak{su}(2) \times \mathfrak{su}(2)$, and let the suba …

This was answered by Marcel Bischoff in a comment. For any vertex operator algebra, there is an invariant called the "character" or "graded dimension", that captures precisely the isomorphism type of …

Let $\mathfrak{g}$ be a complex Lie algebra with a distinguished nonzero central element $x$, and let $V$ be an irreducible representation of $\mathfrak{g}$. The usual proof of Schur's lemma can be a …

I do not have a complete answer to your questions, but this is what I can say for now: Question 1: A classification is impossible (see the response to question 3). Question 2: Additional examples ar …

A lot has happened in the last four years, and we now have lots of positive results. The current state of knowledge is given in Evans-Gannon, "Reconstruction and Local Extensions for Twisted Group Do …

The answer is no: it is possible to have a vertex algebra map between vertex operator algebras of equal central charge that does not take preserve the distinguished conformal vectors. Given a vertex …

As others have mentioned, there are many CFTs, but we can narrow down our list by looking at conditions that select for interesting automorphism groups. Perhaps the easiest is to consider holomorphic …

Well, the claim is bogus, so you can't expect the proof to hold much water. On the other hand, it may be instructive to try filling in details to see why it fails. First of all, we can't define conf …

In general, you won't get a vertex tensor category, because you don't get well-defined unit behavior when you use conformal blocks on higher genus surfaces. Huang-Lepowsky assume the vertex operator …

If your VOA $V$ is not rational, then it is quite unlikely that its category of representations is a modular tensor category. That is, you can safely conclude that Theorem 3 contains an unstated assu …

For any integer lattice $L$, you can write a theta function $\theta_L$ as a generating function for lattice vectors of a given norm. That is, $$\theta_L(\tau) = \sum_{a \in L} q^{(a,a)/2}.$$ The quot …

It seems that the previous answers describe Verlinde's formula for a modular tensor category, or a slight weakening of that condition. Moore and Seiberg essentially proved the formula under the assum …

I'm going to try to connect the language of Jeff Harvey's answer to the language of your question. It seems to give an answer to your title question, but not to the question you asked in the main tex …