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The Monster group $M$ acts on the moonshine vertex algebra $V^\natural$.

Because $V^\natural$ is a holomorphic vertex algebra (i.e., it has a unique irreducible module), there is a corresponding cohomology class $c\in H^3(M;S^1)=H^4(M;\mathbb Z)$ associated to this action.

Roughly speaking, the construction of that class goes as follows:

  • For every $g\in M$, pick an irreducible twisted module $V_g$ (there is only one up to isomorphism).
  • For every pair $g,h\in M$, pick an isomorphism $V_g\boxtimes V_h \to V_{gh}$,
  • where $\boxtimes$ denotes the fusion of twisted reps.
  • Given three elements $g,h,k\in M$, the cocycle $c(g,h,k)\in S^1$ is the discrepancy between $$ (V_g\boxtimes V_h)\boxtimes V_k \to V_{gh}\boxtimes V_k \to V_{ghk}\qquad\text{and}\qquad V_g\boxtimes (V_h\boxtimes V_k) \to V_g\boxtimes V_{hk} \to V_{ghk} $$

I think that not much known about $H^4(M,\mathbb Z)$. But is anything maybe known about that cohomology class? Is it non-zero? Assuming it is non-zero, would that have any implications?

More importantly: what is the meaning of that class?

Ben McKay
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André Henriques
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    Clearly, it's some sort of associator for the 2-gerbe sitting over the delooping of $M$.... ;-) but I reckon you'd have guessed this yourself. Breen's 1994 Asterisque volume may be useful here. – David Roberts Jul 01 '11 at 00:25
  • BTW, very nice question :) – David Roberts Jul 01 '11 at 00:25
  • Has there been recent progress on this question? – Theo Johnson-Freyd Apr 12 '17 at 00:05
  • @TheoJohnson-Freyd As of today, I don't think anything publicly available has changed. Naturally, it is a topic of active interest. – S. Carnahan Apr 12 '17 at 08:11
  • @S.Carnahan great. I ask because I believe I have made some progress on the question, but don't want to reproduce work that has already been done that I missed – Theo Johnson-Freyd Apr 13 '17 at 16:01
  • @TheoJohnson-Freyd Well, there is semi-public information: Miyamoto announced at a conference last year that he has a proof that the Monster-fixed-point subVOA of $V^\natural$ is $C_2$-cofinite and rational. This more or less implies the conditional arguments in my answer are valid. – S. Carnahan Apr 16 '17 at 01:59
  • @S.Carnahan I might be misremembering --- I can no longer find the video --- but I thought I saw an old MSRI lecture of yours in which you announced a similar result? – Theo Johnson-Freyd Apr 18 '17 at 16:28
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    @TheoJohnson-Freyd Perhaps you were watching the 57th minute of my talk here: http://www.birs.ca/events/2016/5-day-workshops/16w5009/videos – S. Carnahan Apr 19 '17 at 08:22
  • Ah, that's why I couldn't find it on msri's website. – Theo Johnson-Freyd Apr 19 '17 at 17:08

2 Answers2

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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 level 288, and the corresponding irreducible twisted modules have a character whose expansion is in powers of $q^{1/288}$. Fusion in a cyclic group generated by a 24J element then yields a $1/12$ discrepancy in $L_0$-eigenvalues, meaning you will pick up 12th roots of unity from the associator. If you pull back along a pointed map $B(\mathbb{Z}/24\mathbb{Z}) \to BM$ corresponding to an element in class 24J (i.e., if you forget about twisted modules outside this cyclic group) you get a cocycle of order 12. This is the largest order you can get by this method - everything else divides 12. I don't know how the cocycles corresponding to different cyclic groups fit together.

I don't know if you've seen Mason's paper, Orbifold conformal field theory and cohomology of the monster, but it is about related stuff. I don't understand how he got his meta-theorem with the number 48 at the end, though.

As far as implications or meaning of the cocycle, all I can say is that the automorphism 2-group of the category of twisted modules of $V^\natural$ has the monster as its truncation, and its 2-group structure is nontrivial. I've heard some speculation about twisting monster-equivariant elliptic cohomology, but I don't understand it. If you believe in AdS/CFT, this might say something about pure quantum gravity in 3 dimensions, but I have no idea what that would be.

Update Nov 2, 2011: I was at a conference in September, where G. Mason pointed out to me that $H^4(M,\mathbb{Z})$ probably contains an element of order 8, and therefore also $\mathbb{Z}/24\mathbb{Z}$. I believe the argument was the following: there is an order 8 element $g$ whose centralizer in the monster acts projectively on the unique irreducible $g$-twisted module of the monster vertex algebra $V^\natural$, such that one needs to pass to a cyclic degree 8 central extension to get an honest action. Rather than just looking at $L_0$-eigenvalues, one needs to examine character tables to eliminate smaller central extensions here. Naturally, like the claims I described before, the validity of this argument depends on some standard conjectures about the structure of twisted modules.

It seems that the relevant group-theoretic computation may have been known to S. Norton for quite some time. In his 2001 paper From moonshine to the monster that reconstructed information about the monster from a revised form of the generalized moonshine conjecture, he explicitly included a 24th root-of-unity trace ambiguity. I had thought perhaps he just liked the number 24 more than 12, but now I am leaning toward the possibility that he had a good reason.

S. Carnahan
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  • "I've heard some speculation about twisting monster-equivariant elliptic cohomology, but I don't understand it": from whom? – André Henriques Jul 02 '11 at 21:47
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    from Jacob and Nora, but their stories seem to differ. – S. Carnahan Jul 03 '11 at 14:33
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    @Scott, you probably know this. What is the maximal subgroup of the monster such that the orbifold of the moonshine still contains the $L(1/2,0)^{\oplus 48}$ frame? – Marcel Bischoff Mar 09 '16 at 20:28
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    @MarcelBischoff In theory the answer can be extracted from http://arxiv.org/abs/q-alg/9707008 but it is probably easier if you just ask Robert Griess (at Michigan). I think "what is the pointwise stabilizer of $L(1/2,0)^{\otimes 48}$ in the monster VOA?" is a slightly clearer version of your question. – S. Carnahan Mar 11 '16 at 05:10
  • Thanks! The reason I ask is because for this subgroup I in principle know how to find the cohomology class, I did not try yet, though. – Marcel Bischoff Mar 11 '16 at 05:16
  • What means "the conjugacy class 24J (made from certain elements of order 24) has a character of level 288, and the corresponding irreducible twisted modules have a character whose expansion is in powers of q..." ? Map from Conjugacy class to irreps ? as i want here :https://mathoverflow.net/questions/271067/monstrous-langlands-mckay-or-what-is-bijection-between-conjugacy-classes-and-irr ? – Alexander Chervov Jul 27 '17 at 15:59
  • @AlexanderChervov It does not mean what you seem to want. Level $N$ refers to the McKay-Thompson series, which is a modular function invariant under some $\Gamma_0(N)$. – S. Carnahan Jul 29 '17 at 16:49
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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.

Theo Johnson-Freyd
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