TITLE: $H^4$ of the Monster
QUESTION [43 upvotes]: 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?

REPLY [25 votes]: 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.