H^4 of the Monster - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:38:02Z http://mathoverflow.net/feeds/question/69222 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69222/h4-of-the-monster H^4 of the Monster André Henriques 2011-06-30T22:35:16Z 2011-11-02T10:22:40Z <p>The Monster group $M$ acts on the moonshine vertex algebra $V^\natural$.<br></p> <p>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.</p> <blockquote> <p>Roughly speaking, the construction of that class goes as follows:<br> &nbsp;&nbsp;&nbsp;&bull; For every $g\in M$, pick an irreducible twisted module $V_g$ (there is only one up to isomorphism).<br> &nbsp;&nbsp;&nbsp;&nbsp;&bull; For every pair $g,h\in M$, pick an isomorphism $V_g\boxtimes V_h \to V_{gh}$,<br> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where $\boxtimes$ denotes the fusion of twisted reps.<br> &nbsp;&nbsp;&nbsp;&nbsp;&bull; Given three elements $g,h,k\in M$, the cocycle $c(g,h,k)\in S^1$ is the discrepancy between $$(V_g\boxtimes V_h)\boxtimes V_k \to V_{gh}\boxtimes V_k \to V_{ghk}\qquad\text{and}\qquad V_g\boxtimes (V_h\boxtimes V_k) \to V_g\boxtimes V_{hk} \to V_{ghk}$$</p> </blockquote> <p>I think that not much known about $H^4(M,\mathbb Z)$...<br> But is anything maybe known about that cohomology class? Is it non-zero?<br> Assuming it is non-zero, would that have any implications?...</p> <p>More importantly: what is the <i>meaning</i> of that class?</p> http://mathoverflow.net/questions/69222/h4-of-the-monster/69233#69233 Answer by S. Carnahan for H^4 of the Monster S. Carnahan 2011-07-01T05:36:58Z 2011-11-02T10:22:40Z <p>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.</p> <p>I don't know if you've seen Mason's paper, <a href="http://www.newton.ac.uk/programmes/NST/Mason.pdf" rel="nofollow">Orbifold conformal field theory and cohomology of the monster</a>, but it is about related stuff. I don't understand how he got his meta-theorem with the number 48 at the end, though.</p> <p>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.</p> <p><strong>Update Nov 2, 2011:</strong> 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. </p> <p>It seems that the relevant group-theoretic computation may have been known to S. Norton for quite some time. In his 2001 paper <em><a href="http://books.google.co.jp/books?id=GYmU_9EKQCEC&amp;lpg=PA163&amp;pg=PA163#v=onepage" rel="nofollow">From moonshine to the monster</a></em> 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.</p>