The Conway 2-Groups

August 9, 2019

Posted by John Baez

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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.

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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?

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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

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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

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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

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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: anomalies in conformal field theory

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I find it better to consider the Conway group $Co_0$ from a near horizon geometry perspective. Consider an M2-brane in D=26+1 M-theory. Its near horizon geometry is $AdS_4\times S^{23}$ , corresponding to Poincare symmetry breaking $SO(26,1)\rightarrow SO(2,1)\times SO(24)$ , where the R-symmetry is $SO(24)$ . Let the 24-dimensional transverse space of the M2 be discretized as the Leech lattice. This breaks $SO(24)$ to its maximal finite subgroup $Co_0$ . The boundary of $AdS_4$ can then support a 3-dimensional (S)CFT with Conway group symmetry.

If one further reduces from D=26+1 to D=25+1, say with the M2’s 3-form with a component along the reduced dimension, one can recover $AdS_3\times S^{23}$ . The boundary of $AdS_3$ can then support a 2-dimensional (S)CFT with $Co_0$ symmetry.

How does one get $Co_1$ symmetry? For this reduce from D=26+1 to D=25+1 along an orbifold $S^1/\mathbb{Z}_2$ , thus breaking $Co_0$ down to $Co_1$ . This is what’s needed for D=25+1 string theory with Monster symmetry. The orbifold also kills the massless 24, removing it from the partition function, as required for the FLM construction.

Posted by: Metatron on December 31, 2020 5:05 PM | Permalink | Reply to this

Re: Could someone check this out?

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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?

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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?

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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

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Glad you liked the paper!

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

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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

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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

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By the way, in this blog article:

Braided 2-groups from lattices

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

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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

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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:

Nora Gantner, Categorical tori.

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.

Also, in the conversation about this stuff, Qiaochu Yuan wrote:

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.

So, there are various avenues to explore. I should think about this stuff some more.

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

Re: The Conway 2-Groups

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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

Re: The Conway 2-Groups

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David wrote:

Ok, so none of this actually mentioned the lattices!

Indeed, nothing about building a 3-group from a 4-cocycle has anything to do with lattices. Suppose we’re trying to build $n$ -groups that are only nontrivial at the bottom and top levels. These are the same as connected pointed homotopy types with only $\pi_n$ and $\pi_1$ nontrivial.

It’s a general fact that connected pointed homotopy types with only $\pi_n$ and $\pi_1$ nontrivial are classified by:

an action of $\pi_1$ on $\pi_n$

together with

an element of $H^{n+1}(\pi_1, \pi_n)$ .

Here the cohomology is the group cohomology of $\pi_1$ with coefficients in a module of that group, namely $\pi_n$ . But in the examples I’m talking about, $\pi_1$ is acting trivially on $\pi_n$ , so we’re just talking about group cohomology with coefficients in an abelian group — a special case.

In either case, if you think of an $n$ -group as an $(n+1)$ -groupoid with only one object, then you can think of $\pi_1$ as the group of automorphisms of this object, while $\pi_n$ is the group of $(n+1)$ -morphisms that are automorphisms of the identity on the identity on the … identity of this object. The action of $\pi_1$ on $\pi_n$ comes from ‘whiskering’, while the cocycle describes the weakening of associativity. For example, when $n = 2$ the cocycle describes the associator and the cocycle condition it obeys is the pentagon identity. When $n = 3$ the associator is trivial but it can have a nontrivial ‘pentagonator’, and that’s what the cocycle describes: the cocycle condition it obeys comes from the 3d Stasheff polytope.

The usual place one reads about this stuff is in the theory of Postnikov towers. May’s old book Simplicial Objects in Algebraic Topology develops this theory quite efficiently. But if you want an explanation in terms of higher categories, well, that’s why Mike Shulman and I wrote Lectures on n-categories and cohomology.

Posted by: John Baez on August 23, 2019 9:36 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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Thanks for your reply, this is really beautiful stuff! I’m going to try “type checking” what you say about spaces with just $\pi_1$ and $\pi_n$ .

If I have a group $G$ , an abelian group $A$ , an action of $G$ on $A$ which is a pointed map $B\alpha : B G \to B\text{Aut}_{\text{Type}_\ast}(B A)$ and a cocycle in the cohomology of $G$ in $A$ as a $G$ -module via this action, then I should be able to make an $n$ -type with $G$ as $\pi_1$ and $A$ as $\pi_n$ etc.

But what does it mean to take cohomology with coefficients in a module (using classifying types)? Since everything should be happening $G$ -equivariantly, it should be happening in the context of a $G$ -torsor $t : B G$ . We have an action of $G$ on $A$ ; can we use this to get an action on $B^{n+1} A$ so that a cocycle could be a map $B G \to B^{n+1} A$ which is $G$ -equivariant? Since $A$ is an abelian group, $B^{n+1}A$ can be constructed as $\|\Sigma^{n} B A \|_{n+1}$ . Since this is a formula “in $B A$ ”, we can define the action on $B^{n+1} A$ by $t \mapsto B^{n+1}\alpha(t) :\equiv \|\Sigma^{n} B A \|_{n+1},$ noting of course that applying this to $G : B G$ yields $B^{n+1} A$ . So a cocycle should assign to any $t : B G$ an element of $B^{n+1}\alpha(t)$ . Of course, if the action is trivial then this is just a function $B G \to B^{n+1} A$ as we’d expect.

Ok, so given an action $B\alpha : B G \to B\text{Aut}_{\text{Type}_\ast}(B A)$ and a cocycle $c : (t : B G) \to B^{n+1}\alpha(t)$ , how do we get the $(n+1)$ -type classified by this data? My guess is it should just come out of the cocycle $c$ , since this depends on the action $\alpha$ anyway. Just like in my last comment, we can take the fiber (a “dependent fiber” this time) to get $(t : B G) \times (c(t) = \text{pt}_{B^{n+1}\alpha(t)})$ which is an $n$ -type. Because the second half is $(n-1)$ -connected, $\pi_1$ of the fiber is $G$ , and because $B G$ is $1$ -truncated, $\pi_n$ of the fiber is $A$ ! Obviously there’s a lot left to check if I was trying to be rigorous, but I think this satisfies my urge to “type check”.

I’ve dipped my toes into your lectures on $n$ -categories and cohomology, but I think it may be about time I took a deep dive. I might try and get a reading series going here at Johns Hopkins!

Posted by: David Jaz Myers on August 26, 2019 7:20 PM | Permalink | Reply to this

Re: The Conway 2-Groups

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David wrote:

I’ve dipped my toes into your lectures on $n$ -categories and cohomology, but I think it may be about time I took a deep dive. I might try and get a reading series going here at Johns Hopkins!

That would be great! I think you’ll find it’s really fun material. It was written before homotopy type theory was invented, so we were just trying to relate two other languages — homotopy theory and higher category theory. But a lot of this material can probably be re-expressed quite nicely in the more ‘type-theoretic’ language you’re talking.

Posted by: John Baez on August 27, 2019 3:50 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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The appearance of Z/24 and the first fractional Pontryagin class reminds me of the table at the end of the examples section on the Whitehead tower nLab page. I don’t feel particularly confident with my understanding of the subject matter here so this may simply be a coincidence, but I would be interested to find out if this has some connection to the J-homomorphism/stable homotopy groups of spheres.

Posted by: Jon Borenstein on August 28, 2019 3:11 PM | Permalink | Reply to this

Re: The Conway 2-Groups

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Many different appearances of the numbers 12 and its multiples are connected, from the fact that

$\lim_{n \to \infty} \pi_{n+3}(S^n) = \mathbb{Z}/24$

to the fact that the abelianization of $\mathrm{SL}(2,\mathbb{Z})$ is $\mathbb{Z}/12$ , to the fact that the discriminant for elliptic curves is a modular form of weight 12, to the fact that $\zeta(-1) = -1/12$ , to the fact that $K_3(\mathbb{Z})$ , the 3rd algebraic $K$ -group of the integers, is $\mathbb{Z}/48$ , to the existence of the 24-cell, as drawn here by Greg Egan:

The existence of 24 even unimodular lattices in 24 dimensions, the Leech lattice, the Mathieu group $\mathrm{M}_{24}$ , and the fact that $H^3(Co_0,\mathbb{Z}) = \mathbb{Z}/24$ , seem harder to connect to the rest of this story… but I’m convinced they’re all manifestations of the same phenomenon.

Posted by: John Baez on August 30, 2019 7:41 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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One answer (I am not sure how serious it is) is that of course it’s going to be difficult to connect these stories together, because the first bunch pertains to complex equiangular lines, and the second to real ones.

Posted by: Blake Stacey on August 30, 2019 11:58 PM | Permalink | Reply to this

Re: The Conway 2-Groups

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Since from Section 1.4 of their paper, this first fractional Pontryagin class is discussed in terms of the obstruction to lifting a spin representation to a string representation, this does seem very much in that terrain.

If a real representation is $G \to O(n)$ and this is extended to $G \to O$ , is there a name for the postcomposition with the J-homomorphism, $O \to GL_1(\mathbb{S})$ ?

Posted by: David Corfield on August 30, 2019 11:04 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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It’s perhaps worth noting that they use ‘spin representation’ in a non-standard way to mean a map $G \to Spin(n)$ , rather than a linear representation of a Spin group as here, so a kind of map out of $Spin(n)$ .

Posted by: David Corfield on August 31, 2019 7:57 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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David wrote:

It’s perhaps worth noting that they use ‘spin representation’ in a non-standard way to mean a map $G \to Spin(n)$ .

Thanks, I was a bit confused about that. I was more confused about the concept of the Pontryagin class of a representation, since think of Pontryagin classes as cohomology classes that you get on a space whenever you have a real vector bundle over that space.

I should probably think of a finite-dimensional real representation of a topological group $G$ as a homomomorphism $G \to O(n) \hookrightarrow O$ that gives a map $B G \to B O$ , and the Pontryagin classes as cohomology classes on $B O$ which I can pull back to classes on $B G$ , the ‘Pontryagin classes of the representation’.

Or something like that.

This makes me very interested in understanding Pontryagin classes and other characteristic classes of representations, since I know a bunch about group representations and a bunch about characteristic classes and have never thought to connect them in this way.

Each cohomology class on $B O$ , for example, should give a map from the representation ring of $G$ to the cohomology of $B G$ . Organized nicely, maybe these classes taken together give a ring homomorphism from the representation ring of $G$ to the cohomology ring of $B G$ .

All this stuff should have been worked out explicitly in great details, right? — at least when $G$ is a classical group, not so much when it’s a sporadic finite simple group.

Posted by: John Baez on September 1, 2019 6:08 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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The usage of the word ‘representation’ in the paper is more similar to the usage at infinity-representation . I think the characteristic class page along with the Whitehead tower page I linked previously should help explain the rest.

All the replies have been very helpful. Thanks everyone!

Posted by: Jon Borenstein on September 1, 2019 4:00 PM | Permalink | Reply to this

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I’m not very satisfied with how I said this. I’m essentially just reaffirming what you’re saying here. It looks, in fact, like you very likely wrote some of what I’m referencing. Sorry!

Posted by: Jon Borenstein on September 1, 2019 4:29 PM | Permalink | Reply to this

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I should think it was most likely Urs. Characteristic classes are simply formed by composition. See also characteristic class of a linear representation.

Posted by: David Corfield on September 1, 2019 7:40 PM | Permalink | Reply to this

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Somewhat tangentially related, but, inspired by Jack Morava’s AlgTop post we began an entry on the only exotic 2-compact group DI(4). This has an important cohomological relation to the third Conway group, $Co_3$ . Perhaps someone here has insights to add to the entry.

Posted by: David Corfield on August 30, 2019 7:15 AM | Permalink | Reply to this

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Your second link appears broken; I assume it was meant to point to the nLab as well?

Posted by: Blake Stacey on August 31, 2019 4:30 AM | Permalink | Reply to this

Re: The Conway 2-Groups

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Thanks! It’s fixed now.

Posted by: David Corfield on August 31, 2019 7:14 AM | Permalink | Reply to this

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Perhaps the cohomological connection with the Conway group $Co_3$ is the way to make precise the following handwavy analogy.

The Lie group $G_2$ , which is the automorphism group of the octonions, has a finite counterpart, a group $G_2(2)$ of order 12,096 which is the automorphism group of the Cayley integer octonions (called the octavians in Conway and Smith’s book). Up to an overall scaling, these integers form a lattice that is just the $E_8$ lattice, a fact invoked in Wilson’s construction to which the nLab page refers. $G_2(2)$ is very nearly simple; its commutator subgroup $G_2(2)'$ is isomorphic to the finite simple group $PSU(3,3)$ , or $U_3(3)$ in Atlas notation, of order 6,048.

Just as $E_8$ is the really nice lattice in 8 dimensions, so the Leech lattice is the really nice lattice in 24 dimensions. So, there should be a continuous thing whose symmetry group has a discrete counterpart that is (or is closely related to) the symmetry group of the Leech lattice.

Posted by: Blake Stacey on September 1, 2019 6:18 PM | Permalink | Reply to this

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Interesting. Did you mean that $G_2(2)$ is the automorphism group of the Integral Octonions modulo 2, as this thesis, p. 75 claims? The full symmetry group is much larger, it seems.

Posted by: David Corfield on September 1, 2019 7:54 PM | Permalink | Reply to this

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Wilson’s The Finite Simple Groups (2009) actually defines $G_2(\mathbb{Z})$ , the automorphism group of integral octonions, then defines $G_2(2)$ as the automorphism group of the octonions modulo 2, then proves that $G_2(\mathbb{Z}) \cong G_2(2)$ . See section 4.4, pages 129–134. I don’t think all the reflections one can apply on the $E_8$ lattice, embedded in Euclidean space, play well with octonion multiplication.

Conway and Smith’s On Quaternions and Octonions (2003) discusses the the “octavians” in chapter 10 and proves that their automorphism group has order 12,096 (p. 125). Chapter 11 moves on to discuss reading the octavians modulo $p$ . There’s an embedding from $G_2(\mathbb{Z})$ into $G_2(p)$ that is an isomorphism when $p = 2$ . For $p \gt 2$ , the group $G_2(p)$ is simple.

Posted by: Blake Stacey on September 1, 2019 8:34 PM | Permalink | Reply to this

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I think that in order to relate the algebra to the rotational symmetries of the $E_8$ lattice, you need to talk about isotopies instead of automorphisms. The latter are invertible linear maps that preserve octonion multiplication ( $L(x y) = L(x)L(y)$ ), while the former are triples of linear maps $(L,M,N)$ such that $x y z = 1$ implies $L(x)M(y)N(z) = 1$ . It’s the isotopy group that gives you $Spin(8)$ for the continuous case and a double cover of the rotational symmetries of $E_8$ in the discrete case.

Posted by: Blake Stacey on September 1, 2019 8:51 PM | Permalink | Reply to this

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I have nothing much to say except “Blake’s right”.. but I’ll say it in a very long way.

It’s good to start by understanding the situation over the reals before moving on to the integers or integers mod 2. Using the triality consisting of the three 8-dimensional irreps of $Spin(8)$ , you can build the octonions and the subgroup $G_2 \subseteq Spin(8)$ that acts as algebra automorphisms of the octonions.

Working over the integers in some clever way, I guess $Spin(8)$ gets replaced by the subgroup of $Spin(8)$ that preserves the $E_8$ lattice (also known as the ‘integral octonions’, up to a rescaling), and $G_2$ gets replaced by the subgroup of $G_2$ that preserves the integral octonions. I guess the former group is $W(E_8)$ , the Weyl group of $E_8$ , which has 696,729,600 elements… or maybe a double cover of the even part of this Weyl group, which has just as many elements. The latter group is $G_2(\mathbb{Z})$ , which has just 12,096 elements.

In case all that was as clear as mud, let me say it a different way. $W(E_8)$ consists of all linear transformations of the octonions that preserve the lattice of integral octonions. But the lattice of integral octonions is a ‘nonassociative ring’: you can multiply them as well as add and subtract them. And I believe the group of linear transformations that preserve the lattice of integral octonions and its multiplication is the smaller group $G_2(\mathbb{Z})$ .

Posted by: John Baez on September 2, 2019 6:14 AM | Permalink | Reply to this

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Blake wrote:

Just as $E_8$ is the really nice lattice in 8 dimensions, so the Leech lattice is the really nice lattice in 24 dimensions. So, there should be a continuous thing whose symmetry group has a discrete counterpart that is (or is closely related to) the symmetry group of the Leech lattice.

Greg Egan and I tried to demonstrate that this thing is a certain obvious 24-dimensional subspace of the exceptional Jordan algebra: namely, the space of $3 \times 3$ self-adjoint octonion matrices with vanishing diagonal entries. See Part 10 of this series:

Integer octonions.

though that relies on earlier parts.

Posted by: John Baez on September 2, 2019 6:20 AM | Permalink | Reply to this

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Ah, OK good.

So how to think about $G_3$ (otherwise known as $D I(4)$ or $D W3$ , the solitary exotic 2-compact group). On the one hand, you might think it was trying to play the role of automorphisms of the sedenions:

$DI(0) = 1 = Aut(\mathbb{R})$
$DI(1) = \mathbb{Z}/2 = Aut(\mathbb{C})$
$DI(2) = SO(3) = Aut(\mathbb{H})$
$DI(3) = G_2 = Aut(\mathbb{O})$
$DI(4) = G_3 =??$

On the other hand, people observing that it is 45-dimensional have wondered about a relationship with the $3 \times 3$ skew-hermitian matrices over the octonions, which had they been associative, would have pointed to the Lie algebra of the unitary $3 \times 3$ matrices. These act on the $3 \times 3$ hermitian matrices in the Albert algebra by automorphisms.

I guess it would be worth getting the story straight about $SO(3)$ , the quaternions, integer quaternions as the $F4$ lattice, etc.

It feels like $G_3$ is hanging on to existence by its finger tips.

Posted by: David Corfield on September 2, 2019 9:22 AM | Permalink | Reply to this

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David wrote:

I guess it would be worth getting the story straight about SO(3), the quaternions, integer quaternions as the F4 lattice, etc.

I think I understand those things, so if you have some questions just ask! You may ask questions I’m not ready for, but that would be good too.

Blake wrote:

In my optimistic moods, I suspect that I could clear up half of my questions with just one good solid afternoon of conversation with the right specialist.

I bet you’re not including all the questions that you’d think of the moment you got your current bunch answered! I’m probably not the right specialist, but again, I wish you’d ask those questions—either here or in a separate guest post.

A point it would be good to understand is why the group that appears in the cohomology discussion is $Co_3$ , as opposed to $Co_1$ . Why is holding a vector of square norm 6 in place the important thing?

That’s too tough for me, since I don’t know much about $G_3$ . Are you sure $Co_3$ is really the crux here? All I read is that “ $BG_3$ receives a map from $BCo_3$ ” — a fairly weak relationship if that’s all we’ve got to go on.

I would want to start with some fairly explicit construction of $G_3$ , if possible, and examine it for clues.

Given that $G_3$ is a creature of characteristic 2, one thing I can’t help thinking about is how everything gets eerily nice in characteristic 2, since all sign problems go away.

For example: the Cayley-Dickson construction, which goes from the reals to the complex numbers to the quaternions to the octonions, only loses properties like commutativity and associativity because of sign problems. If we do it starting with $\mathbb{F}_2$ we get fields forever — which however are just the usual fields $\mathbb{F}_{q}$ where $q$ is a power of 2, nothing exotic.

Another example: the Weyl group of $E_8$ looks extremely exotic, but if you take its even part (consisting of guys that are a product of an even number of reflections), and then mod out by the center (which is just $\{\pm 1$ ), you get a version of $SO(8,\mathbb{F}_2)$ . So the Weyl group of $E_8$ is really not much worse than a rotation group defined over $\mathbb{F}_2$ .

When I say “a version of”, that’s because the usual way of picking out $SO(n)$ inside $O(n)$ doesn’t work over $\mathbb{F}_2$ , since saying “determinant 1” doesn’t add anything in a world where $1 = -1$ . So okay, there is some trickiness here:

John Baez, The Weyl group of E8 versus $O_8^+(2)$ .

But I think a lot of “exotic” math looks less exotic when we view it through the eyes of characteristic 2. I call this “the power of two”, and I’d like to write something about someday.

Posted by: John Baez on September 4, 2019 8:28 AM | Permalink | Reply to this

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I guess one strategy to unpick the exceptional is to start from the appearance of a single exotic thing, and for some reason there’s only one exotic 2-compact group, $DI(4)$ or $G_3$ .

$DI(4) = G_3$ corresponds to the finite $\mathbb{Z}_2$ -reflection group which is number 24 on the Shepard-Todd list. It is the only irreducible finite complex reflection group which is realizable over $\mathbb{Z}_2$ but not $\mathbb{Z}$ .

$Co_3$ does appear to be important in the story, since the mod 2 cohomology of $B DI(4)$ is finitely generated over its mod 2 cohomology.

There are more leads in

Michael Aschbacher, Andrew Chermak, A group-theoretic approach to a family of 2-local finite groups constructed by Levi and Oliver, (paper)

One of the achievements of [LO02] is to suggest that the single “sporadic” object $Co_3$ in the category of groups is a member of an infinite family of exceptional objects in the category of 2-local groups. But in addition, [LO02] establishes a special relationship between the $\mathcal{G}_{Sol}(q)$ s and the exotic 2-adic finite loop space $DI(4)$ of Dwyer and Wilkerson.

Whether one learns anything by understanding how earlier $DI(n)$ are the automorphism groups of the real normed division algebras, I don’t know.

Posted by: David Corfield on September 4, 2019 10:02 AM | Permalink | Reply to this

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Puzzle 1: The free modular lattice on 3 generators has 28 nontrivial elements. Is this the same 28 as the number of bitangents to a quartic, which we already know to be the number of antisymmetric elements in the 3-qubit (dimension 8) Pauli group?

Puzzle 2: This is in the vein of the Ashbacher and Chermak paper that David linked above. They note that their work shows a context in which the Conway group $Co_3$ and the pariah groups $O'N$ and $J_1$ “can live together in harmony.” Potential relations between pariahs and subgroups of the Monster are interesting, and so: What’s up with the Hall–Janko and Rudvalis groups? The former expresses the symmetries of the icosian Leech lattice. It is also what we blunder into when we try to express the symmetries of a conjugate pair of Hoggar-type SICs in a graph structure. Each vector in one Hoggar-type SIC is orthogonal to exactly 28 of the 64 vectors in its conjugate SIC, and its overlaps with the other 36 vectors have constant magnitude. The Hall–Janko group arises when we play around with the structure-preserving permutations of those 36 nonorthogonal pairs. If we instead consider permutations of the 28 orthogonal pairs, we seem to run into the Rudvalis group. Wilson’s book has this to say:

The Rudvalis group has order 145 926 144 000 $= 2^{14}.3^{3}.5^{3}.7.13.29$ , and its smallest representations have degree 28. These are actually representations of the double cover $2 \cdot Ru$ over the complex numbers, or over any field of odd characteristic containing a square root of $-1$ , but they also give rise to representations of the simple group $Ru$ over the field $\mathbb{F}_2$ of order 2.

Wilson then describes in some detail the 28-dimensional complex representations of $2 \cdot Ru$ , using a basis in which a nonsplit extension $2^6 \cdot G_2(2)$ appears as the monomial subgroup. There is also a characterization of $Ru$ by O’Nan:

Let $G$ be a finite simple group having an elementary abelian subgroup $E$ of order 64 such that $E$ is a Sylow 2-subgroup of the centralizer of $E$ in $G$ and the quotient of the normalizer of $E$ in $G$ by the centralizer is isomorphic to the group $G_2(2)$ or its commutator subgroup $G_2(2)'$ . Then $G$ is isomorphic to the Rudvalis group.

So, characterizing this group that is sporadic among the sporadics — floating off to the side, not contained in the Monster — hinges upon the symmetries of the integral octonions. And the 28 appears to be the same 28 we saw before; that is, it’s the 28 pairs of cube roots of the identity in the integer octonions mod 2, so it’s a set of 28 that is naturally shuffled by $G_2(2)'$ . See section 10 of Duncan (2008).

Puzzle 3: We mentioned the Monster group, so let’s go all in! The discrete affine plane on 9 points appears in the study of the Monster, as explained in Miyamoto (1995) and Gannon (2004). This same finite geometry arises from the Hesse SIC in $\mathbb{C}^3$ . I would say that this is a coincidence — there are only so many small structures to go around — but I’d like to know for sure.

All of this runs right up against my own frontier of ignorance. Lots of it is in papers that have the property where, when I read them, I understand the words (eventually) but I don’t see why you would define a thing that way and try characterizing it in that manner. Reading the words is one thing, but “hearing the music” is another.

Posted by: Blake Stacey on September 4, 2019 5:57 PM | Permalink | Reply to this

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To make a “yes” answer to Puzzle 1 seem a bit more plausible, it would be convenient if we could draw a line between the idea of a lattice of subspaces and quantum theory, particularly the three-qubit Hilbert space on which those Pauli operators act. Fortunately, the quantum logic people have argued for a good long while that the lattice of closed subspaces of a Hilbert space, ordered by inclusion, can be thought of as a lattice of propositions pertaining to a quantum system. If the system in question is a set of three qubits, then we’d be talking about the lattice of subspaces of $\mathbb{C}^8$ . To make this look exactly like the setting of the other problem, we would have to do quantum mechanics over real vector space (“rebits” instead of qubits), but that’s OK. Dedekind’s lattice, the free modular lattice on 3 generators with the top and bottom elements removed, can be constructed as a sublattice of the lattice of subspaces of $\mathbb{R}^8$ , so it’s a “sublogic” of the logic of three rebits.

The 3 subspace problem asks us to classify the triples of subspaces of a finite-dimensional vector space, up to linear transformations. Translating to quantum logic, that means classifying triples of propositions about a quantum system. Each such triple corresponds to a representation of the $D_4$ quiver.

This is not a proof of anything, but it does hint that these topics might belong in the same chapter of some book still to be written.

Posted by: Blake Stacey on September 9, 2019 10:35 PM | Permalink | Reply to this

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Regarding the 28 bitangents to a quartic, John McKay has speculated some sporadic connections, see here.

Posted by: David Corfield on September 10, 2019 9:52 AM | Permalink | Reply to this

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The 28 bitangents to a smooth plane quartic curve $Q$ in the projective plane $\mathbb{C}\mathrm{P}^2$ somehow give lines in the branched double cover $M \to \mathbb{C}\mathrm{P}^2$ that has $Q$ as branch points. I don’t know how — maybe someone here can explain it. But we get a configuration of 28 lines this way, and the symmetry group of this configuration is the even part of the Weyl group of $\mathrm{E}_7$ .

Starting from another direction, the smallest nontrivial representation of the compact form of $\mathrm{E}_7$ is 56-dimensional. Its weights, which are 56 vectors in $\mathbb{R}^7$ , form the vertices of a beautiful polytope called the $3_{21}$ Gosset polytope.

These 56 vertices come in opposite pairs, so we get 28 lines through the origin on $\mathbb{R}^7$ . The Weyl group of $\mathrm{E}_7$ acts as symmetries of this lines, preserving the angles between them. But these lines all go through the origin: this is not the configuration of 28 lines I was just talking about!

Here’s another way to get into this circle of ideas. When you pack balls in the densest known way in 7 dimensions, each one touches 56 others at the vertices of the $3_{21}$ Gosset polytope. The symmetries of this bunch of balls, fixing the central one, form the $\mathrm{E}_7$ Weyl group.

So all this is very nice, but as you can see, the stuff about bitangents to a quartic is the part I understand least. The guy who understands this really well is Laurent Manivel, and I recommend section 4 here:

Laurent Manivel, Configurations of lines and models of Lie algebras.

though he takes the stuff about bitangents almost for granted. I had started a series of blog posts about this amazing paper, but quit before getting anywhere.

I don’t think this ‘28’ is particularly related to the 28 elements of the free modular lattice on 3 generators. That 28 should be connected to the fact that $SO(8)$ is 28-dimensional, but I haven’t even been able to figure out that connection!

Posted by: John Baez on September 10, 2019 10:59 AM | Permalink | Reply to this

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If it weren’t for the presence of an eight-dimensional space underlying both problems, I’d be willing to go ahead and say that this is a case where “ $28 \neq 28$ ”; the fact that one problem is about carving up $\mathbb{R}^8$ while the other concerns linear operators on $\mathbb{C}^8$ made it just plausible enough that there could be a commonality. Plausible enough for me to occasionally think about it on train rides and such. It seems just conceivable that generators of $SO(8)$ could be labeled by nonincident (point,line) pairs in the Fano plane, which would reduce puzzle 1 to the older unsolved problem. A concrete connection would be a labeling of the elements in the Dedekind lattice by points in $\mathbb{F}_2^6$ , such that a point corresponds to a Dedekind lattice element when

$x_0 x_1 + x_2 x_3 + x_4 x_5 = 1\ mod\ 2.$

A concrete argument for a no answer might be along the lines that the natural way to shuffle one set of 28 is not the natural way to permute the other.

Posted by: Blake Stacey on September 11, 2019 9:31 PM | Permalink | Reply to this

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Puzzle 4: Is there a set of 27 equiangular lines in $\mathbb{O}^3$ that we can construct in a similarly appealing way to the icosahedron and the Hesse SIC, by starting with a fairly nice “fiducial vector” and applying a straightforward set of transformations? Cohn, Kumar and Minton (2016) give a nonconstructive proof that a set saturating the Gerzon bound in $\mathbb{O}^3$ exists, along with a numerical solution, but that numerical solution doesn’t look like the decimal-place version of a really pretty exact solution in any obvious way. (Their set of mutually unbiased bases in $\mathbb{O}^3$ does look like a generalization of a familiar set thereof in $\mathbb{C}^3$ .) An equiangular set of 27 lines would provide a map from the set of density matrices for an “octonionic qutrit” to the probability simplex in $\mathbb{R}^{27}$ , yielding a convex body that would be a higher-dimensional analogue of the Bloch ball. The extreme points of this Bloch body, the images of the “pure states”, might form an interesting variety.

Fitting the Leech lattice into the traceless part of the exceptional Jordan algebra, as mentioned above, means that each point in the Leech lattice is a “Hamiltonian” for a three-level octonionic quantum system. This sounds a bit like a toy version of a vertex operator algebra construction.

Posted by: Blake Stacey on September 14, 2019 4:14 PM | Permalink | Reply to this

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Yes, the space of pure octonionic qutrits is the Cayley plane, $\mathbb{OP}^2$ . In the context of magic supergravity, the qutrits are 1/2-BPS extremal black hole states, i.e., quantum black holes with vanishing entropy. I wrote about this back in 2013: arXiv:1307.1554 [hep-th].

When the 24-dimensional state space $\mathbb{O}^3$ is integral (yielding the Leech lattice), the mapping to the Cayley plane gives subclasses of the rank-one operators, which are 1/2-BPS states—forming in total 196560=720+11520+184320 states. $F_4$ (in the U-duality group $E_6$ ) being the isometry group of the Cayley plane preserves rank (thus 1/2-BPS property) and Wilson has shown that $Co_0$ is generated by 3x3 matrices that are easily shown to be unitary, hence of type $F_4$ .

The Monster VOA only sees an orbifold of such states, thus 98280=196560/2 states. To see the bigger picture here, one must go beyond D=26 and clarify why the Monster acts on 299+1, as well as 98304. Just as the $E_8$ lattice can be constructed from the D=11 graviton and M-branes, the Leech lattice can be constructed from gauge/gravity data as well. This is discussed in a forthcoming paper.

Posted by: Metatron on July 26, 2020 3:13 PM | Permalink | Reply to this

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If I am reading Manogue and Schray (1993) correctly, then we can label a set of 28 generators for $SO(8)$ in the following way. 7 correspond to the unit imaginary octonions $e_1$ through $e_7$ . Then, for each of the $e_i$ , there are three pairs of unit imaginary octonions $(e_j, e_k)$ that multiply to $e_i$ . These pairs give the other 21 generators. Considering the Fano plane, we have four generators for each point: one for the point itself, and one for each line through that point.

We can associate each generator with a line in $\mathbb{R}^7$ in the following way. First, we label each point in the Fano plane by the lines which meet there. For example,

$(1, 1, 1, 0, 0, 0, 0)$

stands for the point at which the first three lines coincide. There are seven such vectors, any two of which coincide at a single entry (because any two points in the Fano plane have exactly one line between them), and so any two of these vectors have the same inner product. Next, we pick one of the three lines through our chosen point, and we mark it by flipping the sign of that entry. For example,

$(1, -1, 1, 0, 0, 0, 0).$

There are three ways to do this for each point in the Fano plane, and so we get 28 vectors in all. Note that the magnitude of the inner product is constant for all pairs. If the vectors correspond to different Fano points, then their supports overlap at only a single entry, and so the inner product is $\pm 1$ . If they correspond to the same Fano point, then the inner product is $1 - 1 + 1 = 1$ .

So, the 28 of $SO(8)$ seems to be tied in with the other 28’s: The same procedure counts generators of the group and equiangular lines in $\mathbb{R}^7$ . (Counting — a kind of math I understand, some of the time.)

Posted by: Blake Stacey on September 30, 2019 6:10 PM | Permalink | Reply to this

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Thanks, John, for reminding me about Part 10, which I know I had read but which I’d apparently forgotten the substance of. (I find the study of exceptional objects to be full of suggestive analogies and vague intimations, and it’s hard to remember what is what. In my optimistic moods, I suspect that I could clear up half of my questions with just one good solid afternoon of conversation with the right specialist.)

A point it would be good to understand is why the group that appears in the cohomology discussion is $Co_3$ , as opposed to $Co_1$ . Why is holding a vector of square norm 6 in place the important thing?

Posted by: Blake Stacey on September 2, 2019 5:01 PM | Permalink | Reply to this

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In my optimistic moods, I suspect that I could clear up half of my questions with just one good solid afternoon of conversation with the right specialist.

You are evidently much further on than I am. I’ve only dabbled. It seems such a thicket. And yet there should be an order to it all.

I wonder whether physics might provide the key. It seems we can arrive at M-theoretic exceptional structure by invariant central extension from a superpoint, see here.

The pattern breaking down at $(9,1)$ is clearly related to the octonions getting up to their tricks.

Posted by: David Corfield on September 3, 2019 9:11 AM | Permalink | Reply to this

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August 9, 2019