26
votes

### $H^4$ of the Monster

In arXiv:1707.08388, I calculate that the cohomology class you described has order 24 and that it is not a characteristic class in the ordinary sense.

18
votes

### What's the supersymmetric analogue of the Monster group?

There is a super analog constructed just as you describe with the Conway group $Co_0$ replacing the Monster and commuting with the superconformal algebra. The construction is described in detail in:
...

18
votes

Accepted

### Are umbral moonshine and umbral calculus connected?

We used the word "umbral" because umbral moonshine involves mock modular forms whose lack of modularity is characterized by another function known as the shadow of the mock modular form. ...

16
votes

Accepted

### 71, the Monster, and c = 24 CFTs

Schellekens' enumeration is exhaustive in the following sense: the degree 1 subspace of the meromorphic CFT/vertex algebra is naturally a Lie algebra, and it is known that this Lie algebra must be one ...

12
votes

### Why "holomorphic" vertex algebra?

Those in a hurry can skip to the bottom. Those with some leisure time can read the details.
To explain the answer, it is helpful to remember what vertex algebras are. Suppose you have a quantum field ...

12
votes

Accepted

### Why are VOA characters modular forms (geometrically)?

I'm sure someone here can handwave the intuition behind these statements much better than me. This handwaving version goes like this, one is interested in computing vacuum 1-point functions on a torus,...

9
votes

### What are advantages of chiral algebras over vertex algebras?

Some comments:
It is not necessarily true that chiral algebras are essentially conformal vertex algebras, as chiral algebras are allowed to vary over the curve in a way that vertex algebras are not. ...

9
votes

### Chiral homology for the Virasoro algebra and/or affine Lie algebra

There isn't much written up on how to compute the higher chiral homology of vertex algebras. Besides the original work of Beilinson and Drinfeld that covers in particular the universal cases of ...

8
votes

Accepted

### Factorization and vertex algebra cohomology

If $X$ is taken to a be a formal disc $D$, then we obtain a vertex algebra. I believe in this case the above construction produces the vertex algebra cohomology (with coefficients in the adjoint ...

8
votes

Accepted

### What are braided vertex algebras?

For the case of vector spaces graded by an abelian group (with braiding determined by an abelian 3-cocycle following Joyal-Street), this was done by Dong and Lepowsky in their 1993 book "Generalized ...

8
votes

### What is the Zhu algebra of a vertex algebra "really"?

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 ...

7
votes

Accepted

### Do we have a braided tensor category for vertex algebra modules by using conformal blocks on an arbitary compact Riemann Surface?

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 ...

7
votes

### $\text{Rep}(D(G))$ as representation category of a vertex operator algebra

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 ...

7
votes

### Defining extended TQFTs *with point, line, surface, â€¦ operators*

Yes there's a very natural way to incorporate defects of arbitrary dimension into the formalism of extended topological field theory, and this is a vital part of the structure of TQFT. For example in ...

6
votes

Accepted

### Linear independence of genus-one correlation functions

A proof of this property can be found in Proposition 2.2 of Huang's paper Vertex operator algebras and the Verlinde conjecture. This proof actually uses Zhu's algebras to simplify the discussions (...

6
votes

Accepted

### Simple current extensions in VOA theory and CFTs

I'm not a physicist, so I can't say anything authoritative on your first question, but I can say something about questions 2. and 3.
You are correct that in VOA theory, simple currents are invertible ...

6
votes

### Annihilation operators in a vertex algebra

Suppose there exists $v,w \in V(d)$ such that $v_{(d)}w \neq 0$. And now consider $(Tv)_{(d)}w = -2d \,v_{(d)}w \neq 0$. Notice also that by skew-symmetry your condition being true for $d>n$ ...

6
votes

### q-series identity related to Jackson-Slater, proof required

The following proof is due to George E. Andrews (who is now a coauthor on the mentioned paper).
Consider the following families of polynomials indexed by $n$
$$
S_n = \sum_{k \geq 0} q^{2k^2} \binom{...

6
votes

### Proofs of the Frobenius characteristic map

$\def\one{\mathbb{1}}$ Here is a direct attack. First, we give a concise definition of the Frobenius character map, then we check that it is a map of rings.
Let $(\alpha_1, \alpha_2, \dotsc)$ be any ...

6
votes

### Are umbral moonshine and umbral calculus connected?

Despite the authors not intending to imply a link to umbral calculus per se, there is indeed a connection of the umbral calculus / Sheffer polynomial operator calculus to monstrous moonshine as I ...

5
votes

Accepted

### When two vertex (operator) algebras can be patched-up to a full CFT on a genus 0 surface?

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 ...

5
votes

### What are advantages of chiral algebras over vertex algebras?

The main advantage of chiral algebras over vertex algebras is that they admit "very functorial" definitions, and this helps more general concepts and constructions appear naturally. The usual ...

5
votes

Accepted

### Deformations of Vertex Algebras

The complex considered on that article is a linear algebraic version of a complex constructed by Tamarkin in his ICM address
https://arxiv.org/abs/math/0304211
As he points out, in the case of ...

5
votes

Accepted

### coset of affine Lie algebra

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}_\...

5
votes

### What is the Zhu algebra of a lattice vertex algebra?

I'm expanding Reimundo Heluani's link, which gives the answer when $L$ is an even positive definite lattice. Write $\mathfrak{h}=L\otimes_{\mathbf{Z}}\mathbf{C}$. Every $\alpha\in L$ gives two ...

5
votes

### Proofs of the Frobenius characteristic map

$\newcommand\Schur{\mathrm{Schur}}\newcommand\op{^\text{op}}\newcommand\FinVect{\mathrm{FinVect}}\DeclareMathOperator\Rep{Rep}\newcommand\Vect{\mathrm{Vect}}\DeclareMathOperator\GL{GL}\...

4
votes

Accepted

### Modular tensor category associated to an even integral lattice and the lattice automorphism

Edit: I've thought about this question again, and I think the answer is more positive than what I said in an earlier version.
I will assume $L$ is positive-definite, since we need that to make $V_L$ ...

4
votes

Accepted

### Examples of simple vertex operator algebras (VOAs)

I expect there will never be a classification of simple VOAs, unless perhaps one is only sorting according to very rough criteria. This is because there are too many of them - even the rational case ...

4
votes

Accepted

### Are extensions of regular vertex operator algebras also regular?

Update: I think it will be useful to have a more coherently written answer, since I have learned more since my original response. Under the assumptions of the question, the answer is yes, the vertex ...

4
votes

Accepted

### Annihilation operators in a vertex algebra

The answer seems to be yes for quasi-primary $v$ if $V$ has a suitable invariant bilinear form. Then one can identify $v_{(n)} w$ with its pairing with the vacuum, and obtains it as the appropriate ...

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