### The Conway 2-Groups

#### Posted by John Baez

I recently bumped into this nice paper:

• Theo Johnson-Freyd and David Treumann, $\mathrm{H}^4(\mathrm{Co}_0,\mathbb{Z}) = \mathbb{Z}/24$.

which proves just what it says: the 4th integral cohomology of the Conway group $\mathrm{Co}_0$, in the sense of group cohomology, is $\mathbb{Z}/24$. I want to point out a few immediate consequences.

The Leech lattice $\Lambda \subset \mathbb{R}^{24}$ is, up rotations and reflections, the only lattice in 24 dimensions that is:

**unimodular**: the volume of $\mathbb{R}^{24}/\Lambda$ is 1,**even**: $v \cdot v$ is even for every vector $v$ in the lattice,

and:

- contains no
**roots**: that is, vectors with $v \cdot v = 2$.

It comes in two chiral forms: that is, no transformation of determinant $-1$ maps this lattice to itself.

The group of rotations that preserves the Leech lattice is called the Conway group $\mathrm{Co}_0$. It has 8,315,553,613,086,720,000 elements. It’s not simple, because the transformation $v \mapsto -v$ lies in the Conway group and commutes with everything else. If you mod out the Conway group by its center, which is just $\mathbb{Z}/2$, you get a finite simple group called $\mathrm{Co}_1$. But never mind, that’s just a distraction for now: I’m more interested in $\mathrm{Co}_0$.

In fact you don’t need to understand anything I just said, except $\mathrm{Co}_0$ is a group. As soon as Johnson-Freyd and Treumann tell us that

$\mathrm{H}^4(\mathrm{Co}_0,\mathbb{Z}) = \mathbb{Z}/24,$

we know that there are 24 inequivalent ways to extend this group $\mathrm{Co}_0$ to a 3-group. Let us call these the **Conway 3-groups**. (Beware: they may not all be inequivalent as 3-groups. They are inequivalent as *extensions*.)

We can think of a 3-group as a one-object weak 3-groupoid. In any of the Conway 3-groups, the one object has 8,315,553,613,086,720,000 automorphisms. These 1-morphisms compose in a manner given by the group $\mathrm{Co}_0$. The only 2-morphisms are identities, but each 2-morphism has $\mathbb{Z}$’s worth of automorphisms, so there are infinitely many 3-morphisms.

In a general 3-group the composition of 1-morphisms is associative only up to an ‘associator’, which obeys the pentagon equation up to a ‘pentagonator’, which obeys an equation coming from the 3d associahedron. In the Conway 3-groups, the associator is trivial because there are only identity 2-morphisms. The 24 fundamentally different integral 4-cocycles on $\mathrm{Co}_0$ give us 24 choices for the pentagonator. The cocycle condition means these pentagonators obey the equation coming from the associahedron.

We can play various games with these data. For example, any group $G$ has a ‘Bockstein homomorphism’ relating its 4th cohomology with $\mathbb{Z}$ coefficients and its 3rd cohomology with $\mathrm{U}(1)$ coefficients:

$\beta \colon H^3(G, \mathrm{U}(1)) \to H^4(G, \mathbb{Z})$

and for some reason I think this is an isomorphism when $G$ is finite. Could someone check this out? I think it’s because $H^n(G,\mathbb{R})$ vanishes for $n ≥ 0$, and I vaguely seem to recall that’s because of some group averaging argument, but it’s been a long time since I’ve thought about this stuff.

If so, we also get a bunch of **Conway 2-groups**: 24 different ways of extending $\mathrm{Co}_0$ to a 2-group. These are one-object weak 2-groupoids where the morphisms form the group $\mathrm{Co}_0$, each morphism has a circle’s worth of automorphisms, and while composition of morphisms is associative there is still an associator given by some 3-cocycle $\alpha \colon \mathrm{Co}_0^3 \to \mathrm{U}(1)$.

For either the 2-groups or the 3-groups, we can also start with not just $\mathrm{Co}_0$ but with the whole semidirect product $\mathrm{Co}_0 \ltimes \Lambda$, which is the full symmetry group of the Leech lattice including translations.

Perhaps the 2-groups are a bit more intuitive. Here we start with the Leech lattice symmetries, but then decree that they compose associatively *only up to a phase*, which is the $\mathrm{U}(1)$-valued 3-cocycle. There are, nicely, 24 consistent ways to do this!

There’s a lot more to say, but maybe this will get the ball rolling.

## Re: Could someone check this out?

Yes, it is true that $\beta\colon H^3(G; \mathrm{U}_1)\to \mathrm{H}^4(G; \mathbb{Z})$ is an isomorphism.

Since $G$ is finite, the map $\pi\colon E G\to B G$ is a $\#G$-fold covering map. Therefore there is a Gysin map $\pi_!\colon H^\ast(E G; A)\to H^\ast(B G; A)$, and $\pi_!\circ\pi^\ast\colon H^\ast(B G; A)\to H^\ast(B G; A)$ is multiplication by $\#G$. So if $\#G$ is invertible in the ring $A$, then $\pi_!$ is an isomorphism, so $H^n(G;\mathbb{R})$ vanishes for $n$ positive.

Also, if you don’t mind me asking a pretty speculative question: if I remember correctly, Johnson-Freyd and Treumann’s paper was motivated by a question about anomalies in conformal field theory. Is there any relationship between these higher Conway groups and some sort of anomaly vanishing? There have been some papers about higher-group symmetries in quantum field theory recently (e.g. previous n-Category Café discussion here), so it seems plausible to me that there is a connection.