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S. Carnahan
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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.

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.

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.

lengthened
Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

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, meaning you will pick up 12th roots of unity from the associator). This suggests that pulling If you pull back along a pointed map $B(\mathbb{Z}/24\mathbb{Z}) \to BM$ corresponding to an element in class 24J yields(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.

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 then yields a $1/12$ discrepancy in $L_0$-eigenvalues (meaning you will pick up 12th roots of unity from the associator). This suggests that pulling back along a pointed map $B(\mathbb{Z}/24\mathbb{Z}) \to BM$ corresponding to an element in class 24J yields a cocycle of order 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 at the end, though.

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.

Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

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 then yields a $1/12$ discrepancy in $L_0$-eigenvalues (meaning you will pick up 12th roots of unity from the associator). This suggests that pulling back along a pointed map $B(\mathbb{Z}/24\mathbb{Z}) \to BM$ corresponding to an element in class 24J yields a cocycle of order 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 at the end, though.