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29 votes
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$H^4$ of the Monster

There is some evidence from characters that $H^4(M,\mathbb{Z})$ contains $\mathbb{Z}/12\mathbb{Z}$. In particular, the conjugacy class 24J (made from certain elements of order 24) has a character of …
S. Carnahan's user avatar
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13 votes
Accepted

Verlinde's formula

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 …
S. Carnahan's user avatar
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11 votes
Accepted

Character of parity-twisted supersymmetric VOA module -- question inspired by the Stolz-Teic...

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 …
S. Carnahan's user avatar
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11 votes
1 answer
774 views

Is there a canonical map from the cohomology of orbifold chiral de Rham on an orbifold to th...

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 …
S. Carnahan's user avatar
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10 votes
Accepted

Codes, lattices, vertex operator algebras

I think the analogy you describe cannot be made precise with our current technology. For example, the word "functor" doesn't seem to have made an appearance yet in this context. If you have a code, …
S. Carnahan's user avatar
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8 votes

Conformal blocks in genus zero

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 …
S. Carnahan's user avatar
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7 votes
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Do we have a braided tensor category for vertex algebra modules by using conformal blocks on...

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 …
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7 votes
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Verlinde Formula and Theta Function Identities

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 …
S. Carnahan's user avatar
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7 votes
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Analogues of the Monster for central charges different from 24

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 …
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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 Do …
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7 votes
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The use of Schur's lemma for Lie algebras in physics (CFT)

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 …
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5 votes
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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 assu …
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5 votes

The Chern-Simons/Wess-Zumino-Witten correspondence

You are unlikely to find a proof of these claims, because Chern-Simons theory, as a quantum field theory in 3 dimensions, has not been precisely formulated mathematically. You can find some partial r …
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5 votes
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What happens to Virasoro at c=25?

I have an incomplete understanding of this, but I will try to say what I know. For each $c \in \mathbb{C}$, we define $Verma_c$ to be the category whose objects are Verma modules $V_{h,c}$ of central …
S. Carnahan's user avatar
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5 votes

Even lattices and binary codes

I don't have any references (e.g., Conway-Sloane) in front of me, so this will be rather basic. For all non-negative integers $n$, the set of even lattices $L$ satisfying $(2\mathbb{Z})^n \subseteq L …
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