24

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.

answered Jul 27 '17 at 8:44

Theo Johnson-Freyd

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18

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:
John F. R. Duncan and Sander Mack-Crane, The Moonshine module for Conwayâ€™s group, arXiv:1409.3829.
and in John Duncan's paper:
John F. R. Duncan, Super-...

16

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 of the 71 that Schellekens wrote down.
Each of these 71 Lie algebras is realised as the weight 1 piece of some holomorphic c=24 vertex algebra, but it is ...

11

Such an object is described in Dixon, Ginsparg, Harvey, Beauty and the Beast: superconformal symmetry in a monster module Comm. Math. Phys. Volume 119, Number 2 (1988), 221-241. A reasonably explicit construction is given in Huang's paper A nonmeromorphic extension of the moonshine module vertex operator algebra.
In short: the Leech lattice vertex algebra $...

10

This is rather an addition to my comment, as Scott Morrision already give the same answer.
Usually, what's meant by the Verlinde formula is that the fusion coefficients $N_ {ij}^k$ can be determined by the S-matrix by the formula:
$$
N_{ij}^k = \sum_l \frac{S_{jl}S_{il} (S^{-1})_{lk}}{S_{0l}}.
$$
While this formula looks mysterious, it basically says that ...

10

It's claimed on Page 54 of Bakalov and Kirillov's Lectures on Tensor Categories and Modular Functors that the formula in Scott's post above is first proven in Moore and Seiberg's Classical and Quantum Conformal Field Theory). I've only just grabbed that paper, but equation (A.7) gives a generalization of the formula above to "n-point function characters at ...

9

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 assumption that the sectors of a rational CFT form a modular tensor category (although I. Frenkel hadn't invented the name yet). However, chiral CFTs are much ...

7

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 Vertex Algebras and Relative Vertex Operators". The object has the name "abelian intertwining algebra", and standard examples come from applying the lattice ...

7

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 algebra is rational and $C_2$-cofinite, and this is conjecturally strong enough to obtain strong factorization properties for conformal blocks in higher genus. ...

7

This answer is related to my answer here: Duality between orbifold and quasi-Hopf algebra (twisted quantum doubles) and the comment by Scott.
Every finite group $G$ can be embedded in some symmetric group $S_n$ and then you can take for example
the $n$-fold product of the $E_8$ lattice VOA, for which the representation category is trivial. Then I guess the ...

7

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. For instance, on $\mathbb{A}^1$, only translation-invariant chiral algebras give rise to vertex algebras. You can see this looking just at commutative chiral ...

7

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 module) studied in the work of Bakalov, de Sole, Heluani and Kac. Is this true? If not what is the precise relation?
This is not exactly so. If you take $X=\mathbb{...

6

You might find "An Introduction to Conformal Field Theory" by M. Gaberdiel (arXiv:hep-th/9910156v2) useful. He has a brief discussion of how in some cases chiral algebras can be assembled into Conformal Field Theories and has some further references. The language of this review is somewhere in-between the CFT language of physicists and the more formal VOA ...

6

The main sources are Awata et al or Frenkel-Reshetikhin. In http://arxiv.org/pdf/q-alg/9507034v5.pdf section 4, you can see the q,t case. You can also look at http://arxiv.org/pdf/q-alg/9505025v1.pdf where the introduction gives more references to how this relates to the undeformed case. This gives the classical Virasoro as functions on opers on the ...

6

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 Doubles, and Permutation Orbifolds". In particular, if $G$ is a finite solvable group, then $D(G)$ (and more generally, any twist $D^\omega(G)$) is the ...

6

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 $d>n$ implies the same condition for $n < d$.

6

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 irreducible objects in a vertex tensor category of representations of a VOA (although if one does not know that the vertex tensor category structure exists, ...

5

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 assumption that $V$ is rational. At this point, we only know the modular tensor property when $V$ is rational, $C_2$-cofinite, of CFT type, and self-dual as a $V$-...

5

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 (similar to Zhu's proof of the linear independence of characters in Modular invariance of characters of vertex operator algebras) although I think this is not ...

4

Verlinde's formula for a modular tensor category says that that tensor product multiplicities $N_{abc}$ for simple objects are given by
$$ \frac{1}{\mathcal{D}^2} \sum_d \frac{S_{ad} S_{bd} S_{cd}}{S_{1d}} $$
where $\mathcal{D}^2$ is the sum of the squares of the dimensions of the simple objects, $S_{xy}$ is the $(x,y)$ entry of the S-matrix (normalized so ...

4

This was meant to be a comment to David's answer but it grew too much. At first I thought with David's intuition that your definition seemed as a sheaf of vertex algebras on X. But then I realized that I couldn't see the restriction maps. On the other hand, over each point $x \in X$ --restricting to constant maps to X -- you obtain an algebra over that ...

4

With David Ben Zvi's permission, I post the following answer that I got from him by email:
``this feels wrong to me, but not sure. A vertex algebroid
is just a truncation of the notion of sheaf of vertex algebras,
just as a Lie algebra is gotten out of the ≤1 filtered
piece of the enveloping algebra.. so unless your definition
is equivalent to a sheaf of ...

4

Edited: I have missed your "positive". The signature of the lattice modulo 8 depends on the form only (some people call this Brown invariant and van der Blij theorem; Nikulin below calls this just the signature of the form).
Otherwise (given the right signature), I would suggest that the map is surjective, and the easiest proof would be the known ...

4

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 coefficient of
$$(\mathbf{1}, Y(v, x)w) = (-x^{-2})^d (Y(v, x^{-1})\mathbf{1}, w) = (-x^{-2})^d (e^{x^{-1} L(-1)} v, w) = 0$$
since the weight of $v$ is greater ...

4

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 examples involve factorization spaces like the Beilinson-Drinfeld Grassmannian, and applications to the Geometric Langlands program. Another example is the concept of ...

4

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 $V_L$ into an honest VOA. I think what you say is still true using vertex algebras of indefinite lattices, but one has to be very careful with definitions to make ...

4

Regarding your first question, physicists are interested in classifying modular invariant partition functions for two-dimensional rational conformal field theories. Simple currents are a useful tool for constructing such partition functions. An early physics paper on the subject that develops this point of view is https://www.sciencedirect.com/science/...

4

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 operator algebra $V$ will be regular, provided that the categorical dimension $\mathrm{dim}_{\mathcal{C}}\,V\neq 0$, where $\mathcal{C}$ is the modular tensor ...

3

This may be of interest:
Monstrous BPS-Algebras and the Superstring Origin of Moonshine
by Natalie M. Paquette, Daniel Persson, Roberto Volpato
http://arxiv.org/abs/1601.05412

3

The paper Beauty and the Beast (open access) shows that the Moonshine module contains a copy of the super-Virasoro algebra, and so in some sense is already supersymmetric. I don't know how to interpret it in terms of a susy string, however.

answered May 20 '16 at 22:40

Theo Johnson-Freyd

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