## August 9, 2019

### 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.

Posted at August 9, 2019 8:58 AM UTC

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### 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.

Posted by: Arun Debray on August 9, 2019 3:19 PM | Permalink | Reply to this

### Re: anomalies in conformal field theory

My motivation was the following. There is a holomorphic $N=1$ SCFT called $V^{f\natural}$ with automorphism group $Co_1$, first constructed by John Duncan. This $Co_1$ action is anomalous, meaning that there is an obstruction to gauging it. The anomaly comes in two pieces. First, “the” Ramond sector of an SCFT is only well-defined up to isomorphism, and so symmetries of an SCFT act projectively on the Ramond sector, but to gauge a symmetry requires promoting that action to an honest action. This requires lifting from $Co_1$ to its double cover $Co_0$. Second, there is still an $H^4(Co_0; Z)$ valued anomaly, which can be canceled by anomaly-inflow from a 3D Dijkgraaf—Witten model. Equivalently, to trivialize the anomaly requires pulling back from $Co_0$ to the appropriate 3-group.

I claim that of the 24 3-groups, the one that trivializes this anomaly is specifically the one corresponding to our generator of $H^4(Co_0; Z)$, namely the fractional Pontryagin class of the 24-dimensional representation.

Posted by: Theo Johnson-Freyd on August 9, 2019 6:44 PM | Permalink | Reply to this

### Re: anomalies in conformal field theory

Hi Theo, thanks for the explanation. I’m trying to get some of the vocabulary straightened out in my head, so I have a followup question if you don’t mind.

Let $\mathbb{G}$ denote the 3-group extending $\mathrm{Co}_0$ that is specified by the anomaly in $H^4(\mathrm{Co}_0; \mathbb{Z})$. When you write that “to trivialize the anomaly requires pulling back from $\mathrm{Co}_0$ to [$\mathbb{G}$],” is it accurate to say that $\mathbb{G}$ is a nonanomalous symmetry of the theory $V^{f\natural}$?

Posted by: Arun Debray on August 13, 2019 10:05 PM | Permalink | Reply to this

### Re: anomalies in conformal field theory

Would there be a similar anomaly standing in the way if one started with $Co_3$ instead of $Co_1$? I only ask because it’s $Co_3$ that showed up in a problem I’ve worked on, as the symmetry group of the maximal set of equiangular lines in $\mathbb{R}^{23}$.

Posted by: Blake Stacey on August 13, 2019 11:33 PM | Permalink | Reply to this

### Re: Could someone check this out?

Theo: thanks for you explanation! Have you written, or are you writing, a paper where you expand a bit more on these matters? The story surrounding the fact $H^4(\mathrm{Co}_0,\mathbb{Z}) = \mathbb{Z}/24$ seems to appeal to a different audience than the audience who can appreciate the proof of this fact. The proof involves some impressively technical computations. But there’s a lot of nice physics and higher group theory surrounding this fact. It would be fun to see this spelled out, as I was starting to do in my own amateur way.

For example:

Second, there is still an $H^4(Co_0; Z)$ valued anomaly, which can be canceled by anomaly-inflow from a 3D Dijkgraaf–Witten model. Equivalently, to trivialize the anomaly requires pulling back from $Co_0$ to the appropriate 3-group.

I think I get anomaly cancellation: you can sometimes tensor two projective representations and get an honest representation, and this sometimes happens when you glom together two field theories. But how is this equivalent to “pulling back from $Co_0$ to the appropriate 3-group”? I’m not even sure what you’re pulling back, much less how this is equivalent to something else.

Posted by: John Baez on August 10, 2019 4:13 AM | Permalink | Reply to this

### Re: Could someone check this out?

A lot has been written, but it may not be quite what you’re looking for. The closest thing I’ve written is probably my arXiv:1707.08388.

I’ve also been to a number of conferences in the last few years that involve a mix of math, hep-th, and cond-mat people, and that usually include the word “anomaly” in the title. A main theme of those conferences has been “higher groups” and “higher symmetries”. For instance, I saw John Huerta last week at such a conference at Aspen. At a different Aspen workshop, Clay Cordova gave a truly awesome talk about higher groups and anomalies — his work is definitely worth checking out.

Posted by: Theo Johnson-Freyd on August 11, 2019 6:10 AM | Permalink | Reply to this

### Re: Could someone check this out?

The example we’re given at the $n$Lab for higher gauge-theoretic anomalies and their cancellation is the Green-Schwarz mechanism.

Posted by: David Corfield on August 11, 2019 6:20 PM | Permalink | Reply to this

### Re: they may not all be inequivalent as 3-groups

I think the only isomorphisms between the 24 different 3-groups are that you can act by -1 on the central Z (thereby switching the sign of the extension class). In general, there might also be isomorphisms coming from the action of outer automorphism group of G on $H^4(G;Z)$, but $Co_0$ has no outer automorphisms.

Posted by: Theo Johnson-Freyd on August 9, 2019 6:51 PM | Permalink | Reply to this

### Re: they may not all be inequivalent as 3-groups

Nice! I was hoping it worked like that. So I guess there are

$\left|\frac{\mathbb{Z}/24}{\mathbb{Z}/2}\right| = 13$

of them, where $\mathbb{Z}/2$ acts as automorphisms of $\mathbb{Z}/24$ by flipping the sign. Unlucky!

It reminds me ever so slightly about when some people say there are 28 smooth structures on the 7-sphere, other people like to point out that if we allow orientation-reversing maps there are just about half that many. I guess there are

$\left| \frac{\mathbb{Z}/28}{\mathbb{Z}/2}\right| = 15$

where $\mathbb{Z}/2$ acts on $\mathbb{Z}/28$ by flipping the sign: two of these spheres have orientation-reversing automorphisms.

Posted by: John Baez on August 10, 2019 3:30 AM | Permalink | Reply to this

### Re: we can also start with not just Co 0 but with the whole semidirect product Co 0 ⋉ Λ

I think $\Lambda \rtimes Co_0$ deserves the name “affine Conway group”.

It has 3-group extensions pulled back from $Co_0$, and also some that look like a 2-group extension of $Co_0$, but with coefficients linear in $\Lambda$. Derek Holt has computed $H^3(Co_0; \Lambda) = Z/2$. After using the Bockstein to relate this to $H^2(Co_0; R^24/\Lambda)$, you should get the (nonsplit!) extension $(R^24/\Lambda) \cdot Co_0$ which arises as the automorphism group of the Leech lattice CFT (used famously in the construction of the Monster).

Posted by: Theo Johnson-Freyd on August 9, 2019 7:04 PM | Permalink | Reply to this

### Re: The Conway 2-Groups

By the way, in this blog article:

I showed how to get from any lattice $L \subset \mathbb{R}^n$ a 1-parameter family of braided 2-groups having $L$ as their group of 1-morphisms and a circle’s worth of 2-morphisms from any object to itself. We can apply these ideas to the Leech lattice and get a 1-parameter family of braided 2-groups $\Lambda_\theta$ extending the Leech lattice.

Because the construction is completely natural, $\mathrm{Co}_0$ acts on all these braided 2-groups. So, we get some 2-groups $\mathrm{Co}_0 \ltimes \Lambda_\theta$ extending the affine Conway group. I don’t know if these are good for anything.

Posted by: John Baez on August 10, 2019 9:13 AM | Permalink | Reply to this

### Re: The Conway 2-Groups

John wrote:

Because the construction is completely natural, Co0 acts on all these braided 2-groups. So, we get some 2-groups $Co_0\ltimes\Lambda_\theta$ extending the affine Conway group

Is it obvious that $Co_0$ acts (by group homomorphisms) on each individual $\Lambda_\theta$, or just on the space of them? I remember Ganter explaining, in terms of categorical tori, a story similar to the “classical” story from VOAs and “loops” that the $\Co_0$-action takes some extension-data cocycle to a cohomologous one, but there is an ambiguity in choosing the cocycle…

Posted by: Theo Johnson-Freyd on August 11, 2019 5:59 AM | Permalink | Reply to this

### Re: The Conway 2-Groups

Yes, $\mathrm{Co}_0$ acts on each individual braided 2-group $\Lambda_\theta$. The construction of $\Lambda_\theta$ is really simple. If we shift viewpoints a bit from my article and think of a braided 2-group as a special sort of braided monoidal category, it goes like this. The objects of $\Lambda_\theta$ are just points in the Leech lattice $\Lambda$. The tensor product of objects is addition. All morphisms are endomorphisms and the endomorphisms of any $v \in \Lambda$ form the circle group $\mathrm{U}(1)$. The tensor product of morphisms is the simplest thing it can be. The associator is trivial. The braiding is not: the braiding of two objects $v, w \in \Lambda$ is

$e^{i \theta v \cdot w} \in \mathrm{U}(1)$

The only mildly interesting thing here is that the hexagon identity for the braiding follows from the bilinearity of the dot product. This idea goes back to Joyal and Street’s unpublished paper Braided monoidal categories; it’s part of the material that was omitted from the published version, which had a slightly different title. In fact it goes back to earlier work of Eilenberg and Mac Lane on classifying 2-connected homotopy 3-types, as Joyal and Street explain.

Notice that nothing I just said takes advantage of $\Lambda$ being even, or for that matter integral! Perhaps a more interesting avenue is the work of Nora Ganter:

Abstract. We give explicit and elementary constructions of the categorical extensions of a torus by the circle and discuss an application to loop group extensions. Examples include maximal tori of simple and simply connected compact Lie groups and the tori associated to the Leech and Niemeyer lattices. We obtain the extraspecial 2-groups as the isomorphism classes of categorical fixed points under an involution action.

A lattice equipped with a quadratic form gives you slightly more than a braided monoidal category: I think taking vector bundles should give you a braided monoidal category with duals and a ribbon structure, and while the braiding can only see the bilinear form the ribbon structure can see the quadratic form. Or at least that’s what Teleman says in these TFT notes.

Posted by: John Baez on August 11, 2019 7:15 AM | Permalink | Reply to this

### Re: The Conway 2-Groups

This is a very tantalizing read, but unfortunately I’m not sure I’m quite ready to understand it yet. I’m going to write out my “informal type checking” here, with the hope that some more knowledgeable denizen of the cafe can correct and prod me where I go astray.

The cohomology of a group $G$ in an abelian group $A$ is the cohomology of the classifying space ${B}G$ in $A$, or the sets of maps $H^n(G;\, A) = ||{B}G \to {B}^n A||_0$

The unit of the shape modality which sends a type to its homotopy type gives us a map $U(1) \to {B}\mathbb{Z}$, and I think we can deloop this by abstract nonsense (that I am happy to go into). Is pushing forward by the resulting maps ${B}^{n}U(1) \to {B}^{n+1}\mathbb{Z}$ the Bockstein homomorphism associated to the exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to U(1) \to 0$? Maybe it helps to note that we have a fiber sequence ${pt} \to \mathbb{Z} \to \mathbb{R} \to U(1) \to {B}\mathbb{Z}.$

I don’t quite understand how a cocyle ${B}G \to {B}^4 \mathbb{Z}$ gives a way of extending $G$ into a $3$-group. An identification in ${B}G$ (an element of $G$) gets turned into an identification in ${B}^4\mathbb{Z}$, which is trivial, and simiarly a witness to composition gets turned into a 2-cell, which is trivial, and then a witness to associativity gets turned into a 3-cell, which is trivial. But a pentagonator gets sent to some 4-cell, or an integer! And then an associahedron gets sent to a 5-cell, or equality between sums of these integers.

If I had to guess how to turn a cocyle ${B}G \to {B}^4 \mathbb{Z}$ into a 3-group, I would try its fiber first. This is a sum of 3-types indexed by a 1-type, so it is a 3-type. Assuming the cocyle is pointed (so a cocycle in reduced cohomology?), we can also fiddle around to show that the fiber is a pointed, 0-connected type. This makes it a 3-group! If we take another fiber, we’ll get $B^3\mathbb{Z}$, so this is probably why it’s called an “extension”. But does it match the description John gave above?

The elements of the fiber are $G$-torsors equipped with an identification of their pushforward with the point of ${B}^4\mathbb{Z}$. At the low levels, this identification won’t really matter since the low levels of ${B}^4 \mathbb{Z}$ are trivial. This means than an identification in the fiber is an identification in $\mathbb{B}G$ – an element of $G$ – and a commuting triangle of the pushforwards of their respective identifications – which doesn’t say much, since we are still low enough that things are trivial. A 2-cell is therefore an identification in $G$, which is trivial, together with some 3-dimensional shape witnessing the coherence of the pushforwards – still trivial. Finally, a 3-cell in the fiber is something trivial on the ${B}G$ side, but a 4-cell in ${B}^4 \mathbb{Z}$, or an integer! So it seems to work out.

So a similar story now makes sense for the 3-cocyles valued in $U(1)$. Excellent.

Ok, so none of this actually mentioned the lattices! Luckily, I have some more blog posts to read and help me out.

Posted by: David Jaz Myers on August 22, 2019 5:24 PM | Permalink | Reply to this

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