## January 1, 2015

### Braided 2-Groups from Lattices

#### Posted by John Baez

I’d like to tell you about a cute connection between lattices and braided monoidal categories.

We’ve been looking at a lot of lattices lately, like the $\mathrm{E}_8$ lattice and the Leech lattice. A lattice is an abelian group under addition, so we can try to categorify it and construct a ‘2-group’ with points in the lattice as objects, but also some morphisms. Today I’ll show you that for a lattice in a vector space with an inner product, there’s a nice 1-parameter family of ways to do this, each of which gives a ‘braided 2-group’. Here the commutative law for addition in our lattice:

$a + b = b + a$

is replaced by an isomorphism:

$a + b \cong b + a$

And this has some fun spinoffs. For example: for any compact simple Lie group $G$, the category of representations $Rep(T)$ of any maximal torus $T \subseteq G$, with its usual tensor product, has a 1-parameter family of braidings that are invariant under the action of the Weyl group.

What is this good for? I don’t know! I hope you can help me out. The best clues I have seem to be lurking here:

### Braided 2-groups

Here is the kind of braided monoidal category I’m really interested in right now:

Definition. A braided 2-group is braided monoidal category such that every morphism has an inverse and every object has a ‘weak inverse’. A weak inverse of an object $x$ is an object $x^{-1}$ such that

$x \otimes x^{-1} \cong x^{-1} \otimes x \cong 1$

where $1$ is the unit object — that is, the unit for the tensor product.

We say two braided 2-groups $X$ and $Y$ are equivalent if there’s a braided monoidal functor $f : X \to Y$ that’s an equivalence of categories.

In this paper, Joyal and Street showed how to classify braided 2-groups:

This paper is mainly famous for defining braided monoidal categories, but there’s a lot more in it. They published a closely related paper called ‘Braided tensor categories’ in 1992 — but it left out a lot of fun stuff, so I urge you to read this one.

Here’s the idea behind their classification. First, we take our braided 2-group $X$ and form two abelian groups:

• $A$, the group of isomorphism classes of objects of $X$

• $B$, the group of automorphisms of the unit object $1 \in X$

If we have an automorphism $b: 1 \to 1$ we can ‘translate’ it using the tensor product in $X$ to get an automorphism of any other object, so $B$ becomes, in a canonical way, the group of automorphisms of any object of $X$.

Next, the associator and braiding in $X$ give two functions

• $\alpha: A^3 \to B$

• $\beta: A^2 \to B$

How does this work? We can choose a skeleton of $X$, a full subcategory containing one object from each isomorphism class. This will be a braided 2-group equivalent to $X$, so from now on let’s work with this skeleton and just call it $X$. In the skeleton, isomorphic objects are equal, so we have

$(a \otimes a') \otimes a'' = a \otimes (a' \otimes a'')$

and

$a \otimes a' = a' \otimes a$

and every object $a$ has an object $a^{-1}$ with

$a^{-1} \otimes a = a \otimes a^{-1} = 1$

So, the set of objects now forms an abelian group, which is just $A$.

In this setup we can assume without loss of generality that the ‘unitor’ isomorphisms

$\alpha \otimes 1 \cong 1 \cong 1 \otimes \alpha$

are trivial. We still have an associator

$\alpha(a,a',a'') : (a \otimes a') \otimes a'' \to a \otimes (a' \otimes a'')$

and braiding

$\beta(a,a') : a \otimes a' \to a' \otimes a$

and these can be nontrivial, but now they are automorphisms, so we can think of them as elements of the group $B$. So, we get maps

$\alpha: A^3 \to B$

$\beta: A^2 \to B$

The data $(A,B,\alpha,\beta)$ is enough to reconstruct our braided 2-group $X$ up to equivalence. To reconstruct it, we start by taking the category with one object for each element of $A$ and no morphisms except automorphisms, with the automorphism group of each object being $B$. Then we give this category an associator using $\alpha$ and a braiding using $\beta$.

But suppose someone just hands you a choice of $(A,B,\alpha,\beta)$. Then it needs to obey some equations to give a braided 2-group! The famous pentagon identity for the associator:

says we need

$\alpha(b, c,d) - \alpha(a b, c, d) + \alpha(a, b c, d) - \alpha(b, c, d) = 0$

for all $a,b,c,d \in A$. Here I’m writing the operation in the group $A$ as multiplication and the operation in $B$ as addition. There are also some identities that the associator and unitor must obey, and these say:

$\alpha(a, b, 1) = \alpha(a, 1, c) = \alpha(1, b, c) = 0$

Fans of cohomology will recognize that we’re saying $\alpha$ is a normalized 3-cocycle on $A$ valued in $B$. And if our 2-group weren’t ‘braided’, we’d be done: any such cocycle gives a 2-group!

But the hexagon identities for the braiding:

give two more equations:

$\alpha(a, b, c) + \beta(a, b c) + \alpha(b, c, a) = \beta(a,b) + \alpha(b, a, c) + \beta(a,c)$

$\alpha(a,b,c) + \beta(b,c) - \alpha(a,c,b) = \beta(a b, c) - \alpha(c, a, b) - \beta (a,c)$

Long before braided monoidal categories were invented, Eilenberg and Mac Lane called a choice of $\alpha$ and $\beta$ obeying all these equations an abelian 3-cocycle on $A$ with values in $B$. Here’s the easy, obvious part of what Joyal and Street proved:

Theorem (Joyal, Street). Any abelian 3-cocycle on $A$ with values in $B$ gives a braided 2-group with $A$ as its group of objects and $B$ as the group of automorphisms of every object. Moreover, any braided 2-group is equivalent to one of this form.

They went a lot further. For starters, they said when two 2-groups of this form are equivalent. But let’s look at some examples!

### Examples

Example 1. Let’s simplify things by taking associator to be trivial: $\alpha = 0$. Then all the required equations hold automatically except the hexagon identities, which become

$\beta(a, b + c) = \beta(a,b) + \beta(a,c)$

$\beta(a + b, c) = \beta(a,c) + \beta(b,c)$

Now I’m writing the group operation in $A$ as addition, to help you see something nice: these equations just say that $\beta$ is bilinear! So: any bilinear map

$\beta : A \times A \to B$

gives a braided 2-group.

Example 2. In particular, if $L$ is any lattice in a real inner product space $V$, for any constant $\theta \in \mathbb{R}$ we get a bilinear map

$\beta: L \times L \to \mathbb{R}$

given by

$\beta(x,y) = \theta \langle x,y\rangle$

So, we get a 1-parameter family of braided 2-groups with $L$ as the group of objects and $\mathbb{R}$ as the group of automorphisms of any object.

Example 3. We call $L$ an integral lattice if $\langle x, y \rangle$ is an integer whenever $x, y \in L$. In this case we have

$\beta: L \times L \to \mathbb{Z}$

whenever $\theta \in \mathbb{Z}$. So, we also get a 1-parameter family of braided 2-groups with $L$ as the group of objects and $\mathbb{Z}$ as the group of automorphisms of any object — but now the parameter needs to be an integer.

Example 4. Alternatively, if $L$ is a lattice in a real inner product space $V$, for any constant $\theta \in \mathbb{R}$ we get a bilinear map

$\beta: L \times L \to \mathrm{U}(1)$

given by

$\beta(x,y) = e^{i \theta \langle x,y\rangle }$

So, we get a 1-parameter family of braided 2-groups with $L$ as the group of objects and $\mathrm{U}(1)$ as the group of automorphisms of any object.

Example 5. Take the previous example and suppose $L$ is integral. In this case $\beta$ doesn’t change when we replace $\theta$ by $\theta + 2 \pi$, so there is really a circle’s worth of braidings. Certain nice things happen for special values of $\theta$. When $\theta = 0$,

$\beta(x,y) = 1$

for all $x,y \in L$, so the braiding is ‘trivial’. This also happens for a category of line bundles or 1-dimensional representations of a group, and we’ll see this is no coincidence. When $\theta$ is $0$ or $\pi$,

$\beta(x,y) = \beta(y,x) = \pm 1$

so

$\beta(x,y) \beta(y,x) = 1$

and we get a symmetric 2-group — that is, a 2-group that’s a symmetric monoidal category.

So far these examples look a bit general and abstract, so let me give you some examples of these examples. (Category theory can be defined as the branch of math where the examples require examples.)

### An example from Lie theory

Suppose $G$ is a compact simple Lie group, and let $T \subseteq G$ be a maximal torus. The Lie algebra of $T$ is abelian: its Lie bracket vanishes, so we might as well think of it as a mere vector space. For this reason I’ll call it $V$, instead of the more scary Gothic $\mathfrak{t}$ that Lie theorists would prefer.

But this vector space $V$ is equipped with some other interesting structure: an inner product and a lattice!

You see, sitting inside $V$ is a lattice $L$ consisting of those elements $v$ with

$\exp(2 \pi v) = 1 \in T$

Furthermore the Lie algebra of $G$ comes with a god-given nondegenerate bilinear form called the Killing form. If we multiply this by a suitable constant and restrict it to $V$, we get an inner product on $V$ for which $L$ is an integral lattice. Indeed, $L$ is usually called the integral lattice of $G$.

So, using the trick in Example 4, we get a 1-parameter family of braided 2-groups with $L$ as the group of objects and $\mathrm{U}(1)$ as the group of automorphisms of any object.

### Another example from Lie theory

If you’re not familiar with the integral lattice of $G$, you may know about its ‘weight lattice’. This is dual to the integral lattice, in a certain sense. Namely, we can start with the integral lattice $L$ in $V$ and define a new lattice $L^\ast$ consisting of all points in the dual vector space $V^\ast$ whose pairing with every point in $L$ is an integer:

$L^\ast = \{ \ell \in V^\ast : \; \ell(v) \in \mathbb{Z} \; for \; all \; v \in L \}$

Let’s call $L^\ast$ the weight lattice of $G$, though people usually reserve this term for the case when $G$ is simply connected.

Why is the weight lattice interesting? For starters, any 1-dimensional unitary representation of the maximal torus $T$ is equivalent to one coming from a point in this lattice! If we take $v \in V$ then $\exp(2 \pi v) \in T$, and any element $T$ arises this way. Given $\ell \in L^*$ we can thus define a 1-dimensional unitary representation of $T$ by

$\exp(2 \pi v) \mapsto \exp(2 \pi i \ell(v)) \in \mathrm{U}(1)$

In fact, we get all the 1d unitary reps of $T$ this way.

Since every finite-dimensional unitary representation of $T$ is a direct sum of 1-dimensional ones, the category $Rep(T)$ of finite-dimensional unitary representations of $T$ is pretty well described by the weight lattice $L^*$.

How can we make this precise?

Here’s one way. There’s a category FinVect$[L^\ast]$ whose objects are complex vector bundles on $L^\ast$ that have finite support: that is, have 0-dimensional fibers outside a finite set. The morphisms are just vector bundle morphisms. This category is monoidal, where the tensor product comes from Day convolution:

$(V \otimes W)_x = \bigoplus_{x' + x'' = x} V_{x'} \otimes W_{x''}$

This should remind you of the group algebra of the lattice $L^\ast$. Indeed, it’s just a categorified version of the group algebra where we use finite-dimensional vector spaces instead of numbers as coefficients! Moreover,

$\mathrm{FinVect}[L^\ast] \cong Rep(T)$

as monoidal categories. (This is pretty obvious if you know your stuff, but if you want a proof, see the section in my paper on 2-Hilbert spaces where I discuss the categorifed Fourier transform. In case it helps, I should point out that the lattice I’m calling $L^\ast$ is actually the Pontryagin dual of $T$. You see, the dual of a lattice in some vector space is really the Pontryagin dual of the vector space mod that lattice.)

Now let’s build some braided 2-groups as in Example 4. We’ve got a lattice $L^\ast$ in a real inner product space $V^\ast$. This gives a braided 2-group whose objects are points of $L^\ast$, where the automorphisms of any object form the group $\mathrm{U}(1)$, and where the braiding

$\beta: L^\ast \times L^\ast \to \mathrm{U}(1)$

is given by

$\beta(x,y) = e^{i \theta \langle x,y \rangle }$

for any chosen constant $\theta \in \mathbb{R}$.

We can think of this as a braiding on the category of 1-dimensional unitary representations of $T$. Since every finite-dimensional representation is a direct sum of these, we can extend it uniquely to a braiding on

$\mathrm{FinVect}[L^\ast] \cong \mathrm{Rep}(T)$

that is compatible with direct sums. When $\theta = 0$ this is the usual braiding on $\mathrm{Rep}(T)$. So, as we move $\theta$ away from $0$, we’re deforming that braiding.

Sitting inside the weight lattice are certain vectors called roots. For $\mathrm{E}_8$ they look like this, if we project them from 8 dimensions down to 2 in a certain nice way:

For any root $r \in L^{\ast}$ we have a reflection

$R_r : V^{\ast} \to V^{\ast}$

that maps $r$ to $-r$ while leaving orthogonal vectors alone. And the wonderful thing is that these reflections preserve the weight lattice!

These reflections generate a group called the Weyl group of $G$. Since the Weyl group preserves the inner product on $V^{\ast}$ and also the weight lattice, it acts on the monoidal category

$\mathrm{FinVect}[L^\ast] \cong \mathrm{Rep}(T)$

in a way that preserves the braiding for any value of the parameter $\theta$.

### Puzzles

Puzzle 1. How are the braided 2-groups constructed in Example 3 related to Nora Ganter’s categorical tori? A 3-cocycle on $\mathbb{Z}^n$ valued in $\mathbb{Z}$ is just what you need to define a gerbe on the $n$-torus. Is an abelian 3-cocycle what you need to define a multiplicative gerbe on the $n$-torus? I believe this is essentially the same as what Ganter would call a ‘2-group extension of the $n$-torus by the circle’.

Puzzle 2. What can we do with this 1-parameter family of braidings on $\mathrm{Rep}(T)$? Perhaps it’s worth noting that any representation of the torus $T$ gives rise to a $G$-equivariant vector bundle on the flag variety $G/T$; the space of holomorphic sections of this bundle forms a representation of $G$, and we can get all the finite-dimensional representations of $G$ this way. This is part of a well-known story. But I haven’t figured out how to do anything exciting with it yet.

Puzzle 3. I’m especially interested in the $\mathrm{E}_8$ weight lattice. The braided 2-groups with this lattice as their group of objects are categorifications of the integral octonions. But what can we do with them? Naturally it makes sense to start with simpler examples like $\mathrm{A}_2$ (ordinary integers in $\mathbb{R}$), $\mathrm{A}_3$ (Eisenstein integers in $\mathbb{C}$), and $\mathrm{F}_4$ (Hurwitz integers in $\mathbb{H}$).

Puzzle 4. Among the most interesting integral lattices are the even ones, where $\langle x, x \rangle$ is even whenever $x$ is in the lattice. For example, the $A, D,$ and $E$ type lattices are even, and so is the Leech lattice. If we construct a braided 2-group from an even lattice as in Example 3, the self-braiding $\beta(x,x)$ is the square of some other natural transformation. What can we do using this fact? Do you know other interesting braided monoidal categories where this happens?

Puzzle 5. How about deforming both the braiding and the associator? Suppose we have abelian groups $A$ and $B$. It’s easy to check that any trilinear map

$\alpha : A \times A \times A \to B$

obeys the pentagon identity

$\alpha(b, c,d) - \alpha(a + b, c, d) + \alpha(a, b + c, d) - \alpha(b, c, d) = 0$

(where I’m writing the group operation in $A$ as addition), and also the other identities we need to get a monoidal category with $A$ as the group of objects and $B$ as the group of automorphisms of any object. That’s nice. I also think any 3-cocycle on $A$ valued in $B$ is cohomologous to a trilinear one; if so we’re not losing anything about assuming $\alpha$ is trilinear. But now suppose we want to give the resulting monoidal category a braiding. Now we want

$\beta : A \times A \to B$

obeying

$\alpha(a, b, c) + \beta(a, b + c) + \alpha(b, c, a) = \beta(a,b) + \alpha(b, a, c) + \beta(a,c)$

and

$\alpha(a,b,c) + \beta(b,c) - \alpha(a,c,b) = \beta(a + b, c) - \alpha(c, a, b) - \beta (a,c)$

How can we find such $\beta$? Suppose we take $\beta$ to be bilinear. Then we need

$\alpha(a, b, c) + \alpha(b, c, a) = \alpha(b, a, c)$

and

$\alpha(a,b,c) - \alpha(a,c,b) = - \alpha(c, a, b)$

But what do these equations really say? Are there any nontrivial solutions? Can we classify them?

Posted at January 1, 2015 1:17 AM UTC

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### Re: Integral Octonions (Part 12)

Re: Puzzle 2, I have been told that there are nice constructions of quantum groups (nice meaning, in particular, not just by writing down generators and relations) starting from more natural objects living in $\text{Rep}(T)$ with one of its nontrivial braidings, but I don’t know where there are details written up.

Re: Puzzle 4, the significance of evenness has to do with the distinction between bilinear forms and quadratic forms: evenness means precisely that the bilinear form $B(x, y)$ is induced from an integral quadratic form $q(x)$ in the sense that $B(x, y) = q(x + y) - q(x) - q(y)$. This whole story has more to do with quadratic forms than bilinear forms.

To see this we’ll pass through the homotopy hypothesis: in the same way that 2-groups are precisely pointed connected homotopy 2-types, braided 2-groups are precisely pointed connected simply connected homotopy 3-types; that is, they describe spaces whose only nontrivial homotopy groups are $\pi_2$ and $\pi_3$. The machinery of Postnikov towers tells us that the extra data needed to describe such a space is a Postnikov invariant living in $H^4(B^2 \pi_2, \pi_3)$, and this is known to be precisely the group of quadratic forms $\pi_2 \to \pi_3$.

The quadratic form $\pi_2 \to \pi_3$ is in fact a homotopy operation corresponding to the Hopf map $S^3 \to S^2$. I believe it refines the bilinear form you wrote down, which as a homotopy operation should correspond to the Whitehead bracket $\pi_2 \times \pi_2 \to \pi_3$, and in particular it contains strictly more information than the Whitehead bracket if $\pi_3$ has any 2-torsion. There is a nice way to visualize all of this by passing it through the cobordism hypothesis, but maybe that’s a digression.

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: Qiaochu Yuan on January 2, 2015 9:47 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks for all that!

While writing this post I started out focused on quadratic forms, since Joyal and Street discuss Eilenberg and Mac Lane’s old work on $H^4(K(\pi_2,2),\pi_3)$, which is what they called ‘the abelian cohomology of $\pi_2$ with coefficients in $\pi_3$’, and they talk about how this winds up being the group of quadratic forms on $\pi_2$ valued in $\pi_3$. But I had difficulty seeing in an explicit way how to get an associator and braiding from a quadratic form. The problem is presumably that the quadratic form encodes the associator and braiding up to equivalence, and one needs to pick a representative. Maybe there’s a nice formula, but I didn’t see it.

On the other hand, there’s an utterly obvious formula for a braiding given a bilinear map — it just is the bilinear map. Since I wanted everything to be very explicit and calculable, I went that way.

On top of this, I’m always a bit confused about bilinear maps versus quadratic forms, since I’ve spent most of my life in characteristic zero where there’s no difference. There’s the recipe you mentioned for getting a bilinear map from a quadratic form

$\beta(x,y) = q(x+y) - q(x) - q(y)$

but also a recipe for getting a quadratic form from a bilinear map

$q(x) = \beta(x,x)$

and these recipes are not inverse to each other, because a factor of 2 pops up. So it seems one should be a bit careful about saying that one contains more information than the other; it seems they each contain information the other does not unless you can divide by 2.

And this business — ways to go back and forth that aren’t inverse to each other — makes me think of adjunctions.

Is there an adjunction between the category of pairs of abelian groups $A, B$ with bilinear map $\beta: A \times A \to B$ and the category of pairs of abelian groups $A, B$ with quadratic form $q: A \to B$?

By what I said in my blog article, a bilinear map $\beta : A \times A \to B$ gives a skeletal braided 2-group with $A$ as objects, $B$ as automorphisms of the identity, and trivial associator. And conversely!

These will correspond to a certain special class of pointed connected simply-connected homotopy 3-types.

Let’s say these are the ones with trivial associator.

On the other hand, by what you said (and Eilenberg and Mac Lane said), a quadratic form $q : A \to B$ gives a pointed connected simply-connected homotopy 3-type with $\pi_2 = A$, $\pi_3 = B$.

These should correspond to all pointed connected simply-connected homotopy 3-types.

Is there an adjunction between the homotopy category of all pointed connected simply-connected homotopy 3-types, and the subcategory of those with trivial associator?

That would ease my confusion.

Posted by: John Baez on January 2, 2015 8:00 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

$q(x) = \beta(x, x)$ is the “wrong” way to pass from a bilinear form to a quadratic form, in the sense that if $\beta(x, y)$ is the Whitehead bracket and $q(x)$ is the Hopf quadratic form then $q(x) \neq \beta(x, x)$, but $\beta(x, y) = q(x + y) - q(x) - q(y)$, which gives $\beta(x, x) = 2 q(x)$. So 1) if you know $q(x)$ you can recover $\beta(x, y)$, but 2) if you know $\beta(x, y)$ you can only recover $q(x)$ up to some ambiguities involving $2$-torsion. So as I said before, $q(x)$ contains strictly more information than $\beta(x, y)$ in general. In this situation $q(x)$ is said to be a quadratic refinement of $\beta(x, y)$.

The factor of $2$ in the relation $\beta(x, x) = 2 q(x)$ really belongs there. Another place where it pops up is in the definition of the Kervaire invariant: this involves a quadratic refinement of the intersection pairing on a framed manifold over $\mathbb{F}_2$, and all of the data in the Kervaire invariant is destroyed if you multiply the quadratic refinement by $2$!

A third place where it pops up is in the “correct” definition of Clifford algebras over a base ring where you can’t divide by $2$. In this context it’s important to decide once and for all whether Clifford algebras take as input bilinear forms or quadratic forms. I think the correct definition involves quadratic forms, and the defining relation should be $v^2 = q(v)$, which in particular gives $v w + w v = \beta(v, w)$ where $\beta(v, w) = q(x + y) - q(x) - q(y)$. Here one should think of $v, w \in V$ as having odd degree and hence of $v w + w v$ as a supercommutator.

The reason is that the Clifford algebra construction can be thought of as a slight variant of the construction of the universal enveloping algebra, but of a graded Lie algebra rather than an ordinary Lie algebra. Whatever a graded Lie algebra is over, say, $\mathbb{Z}$, at the very least, graded derivations of a graded algebra should be an example. But if $D : A \to A$ is such a graded derivation of odd degree, then the supercommutator $[D, D]$ naturally admits a quadratic refinement, namely $D^2$, which is also a graded derivation. In other words, graded Lie algebras ought to have as part of their data a quadratic refinement of the supercommutator on odd elements. For example, the Whitehead bracket defines a graded Lie algebra structure on the homotopy groups of a space in this refined sense (at least I know this is true for the Whitehead bracket $\pi_2 \times \pi_2 \to \pi_3$ and I believe it’s true in general).

Posted by: Qiaochu Yuan on January 2, 2015 10:17 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

You know, I think the way I’m connecting the braiding to a bilinear form is a bit different than the Whitehead bracket.

Usually when we have two objects $x, y$ in a braided 2-group and we want to get an automorphism we take the ‘double braiding’

$B_{x,y} B_{y, x} : x \otimes y \to x \otimes y$

If we call the group of isomorphism classes of objects $\pi_2$ and call the group of automorphisms of any object $\pi_3$, this gives a bilinear map

$[\cdot, \cdot]: \pi_2 \times \pi_2 \to \pi_3$

which is the Whitehead product. If we turn this into a quadratic form in the obvious way:

$\begin{array}{ccc} \pi_2 &\to& \pi_3 \\ x &\mapsto & [x,x] \end{array}$

the resulting quadratic form encodes the process of sending any object $x$ to the automorphism

$B_{x,x}^2 : x \otimes x \to x \otimes x$

This is not as informative as the automorphism

$B_{x,x} :x \otimes x \to x \otimes x$

However, in my post, I was getting a bilinear map in a different way! Every braided 2-group is equivalent to a skeletal one, in which $x \otimes y$ and $y \otimes x$ are equal, because they’re isomorphic. Then, any pair of objects gives an automorphism

$B_{x,y} : x \otimes y \to y \otimes x = x \otimes y$

and this gives me a bilinear map

$\beta: \pi_2 \times \pi_2 \to \pi_3$

This is not the Whitehead product. Indeed, it’s a bit like ‘half’ the Whitehead product, since

$[x, x] = 2 \; \beta(x,x)$

And if we turn my bilinear map into a quadratic form in the obvious way:

$\begin{array}{rcc} q: \pi_2 &\to& \pi_3 \\ x &\mapsto & \beta(x,x) \end{array}$

we see that this quadratic form encodes the process of sending any object $x$ to the automorphism

$B_{x,x} : x \otimes x \to x \otimes x$

This quadratic form, $q$, is more informative than the last one: it’s what you’re calling the Hopf quadratic form, the one that completely describes our braided 2-group after we know $\pi_2$ and $\pi_3$. Note that

$[x,y] = q(x+y) - q(x) - q(y)$

so this quadratic form contains all the information in the Whitehead product… but not vice versa.

We’re left with the small puzzle of how I managed to cook up a bilinear form that acts a bit like ‘half’ the Whitehead product, in the limited sense that

$[x, x] = 2 \; \beta(x,x)$

We certainly do not have

$[x,y] = 2 \; \beta(x,y)$

for all $x, y$. That would be good to be true: the Whitehead product is not always divisible by 2.

The answer to this puzzle is that…

… well, I’ll let people tackle it if they want, but this is one I know the answer to, so I’ll give it away if nobody else does.

Posted by: John Baez on January 2, 2015 11:44 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Oh, I see. That makes sense. Based on the formulas you’ve written down it looks a bit like what you’re doing is at least analogous to taking two odd derivations $D, E$ and directly writing down their product $D E$ rather than their (super)commutator $[D, E] = D E + E D$. Unlike the supercommutator, the product isn’t guaranteed to be another derivation in general, but it might be in some special cases; even if it is, though, it’s not part of the graded Lie algebra structure on derivations, but is extra structure coming from a particular choice of representation of the graded Lie algebra as derivations on a particular algebra. This is all by analogy though.

Posted by: Qiaochu Yuan on January 3, 2015 5:43 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I will answer my latest puzzle before I forget to do so.

The question was: given a braided 2-group (or equivalently, a pointed connected simply-connected homotopy 3-type), how am I managing to find a bilinear map

$\beta: \pi_2 \times \pi_2 \to \pi_3$

such that $x \mapsto \beta(x,x)$ is the Hopf map from $\pi_2 \to \pi_3$? This seems odd given that the well-known Whitehead product

$[-, -] : \pi_2 \times \pi_2 \to \pi_3$

only gives twice the Hopf map:

$[x,x] = 2 \; \beta(x,x)$

The answer is that my bilinear map, unlike the Whitehead product, is not a symmetric bilinear map. Moreover, defining it required an arbitrary choice!

I started with a braided 2-group with:

• $\pi_2$ as its group of isomorphism classes of objects

• $\pi_3$ as the group of automorphisms of any object

and chose an equivalence between this 2-group and a skeleton of this 2-group, thus making the skeleton into a braided 2-group. Why the italics? When you’re a category theorist, it’s tremendously traumatic to do something that depends on an arbitrary choice. And that turns out to be the key to the puzzle.

In the skeleton, isomorphic objects are equal, so we have

$B_{x,y} : x \otimes y \to y \otimes x = x \otimes y$

Thus, the braiding $B_{x,y}$ becomes an automorphism, and thus an element of $\pi_3$. It thus defines a map

$\beta: \pi_2 \times \pi_2 \to \pi_3$

For convenience I assumed the associator in the skeleton was trivial; given this, $\beta$ is bilinear. But I don’t think we need this in what follows.

What’s really important is this: $\beta$ depends on the choice I made! A different choice would give a different choice of $\beta$, say $\beta'$. By Section 7 of Joyal–Street you can see that

$\beta'(x,y) - \beta(x,y) = k(x,y) - k(y,x)$

for some map $k : \pi_2 \times \pi_2 \to \pi_3$.

So, only the ‘antisymmetric part’ of $\beta$ depends on the choice made.

That’s a bit vague, so let me be precise. First, $\beta(x,x)$ does not depend on the choice made, and this is the Hopf map. Second, $\beta(x,y) + \beta(y,x)$ does not depend on the choice made, and this is the Whitehead product:

$\beta(x,y) + \beta(y,x) = [x,y]$

I thank Qiaochu for his probing comments, which helped me realize what’s going on here.

Posted by: John Baez on January 3, 2015 11:42 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

It would be great to extend this to Freudenthal triple systems and their quartic forms.

Freudenthal triple systems by root system methods Fred W. Helenius

Posted by: Metatron on January 3, 2015 3:37 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I think I can do something like that — thanks!

Posted by: John Baez on January 3, 2015 10:59 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Concerning Puzzle 1: The following things are the same:

1. a Lie 2-group with $A$ the group of iso classes of objects and $B$ the group of automorphisms of 1
2. a multiplicative $B$-gerbe over $A$
3. a central 2-group extension of $A$ by $B$ in the sense of Schommer-Pries.

Up to isomorphism, all these things are classified by Segal’s and Brylinski’s smooth Lie group cohomology $H^3(B A,B)$, which is $H^4(B A,\mathbb{Z})$ if $B=U(1)$.

When the 2-group of (1) is braided, it should be equivalent under (2) to a multiplicative $B$-2-gerbe over $B A$ (the classifying space of $A$), and under (3) to a “central” 3-group extension of $B A$ by $B$, whatever that means.

If $A=\mathbb{Z}^n$ so that $BA=T^n$, then we have a multiplicative $B$-2-gerbe over $T^n$. If further $B=\mathbb{Z}$, then a $B$-2-gerbe is the same as a $U(1)$-gerbe, so that we finally have a multiplicative gerbe over $T^n$, just as you wrote. Cool!

### Re: Integral Octonions (Part 12)

Hi, Konrad! Thanks for tackling Puzzle 1!

Actually I now see Nora Ganter had similar things to say in her paper Categorical tori, which by the way acknowledges you. She gets a multiplicative $\mathrm{U}(1)$ gerbe over a torus $T$ from a lattice $L$ with a bilinear form $\beta : L \times L \to \mathbb{Z}$. She also thinks of this as a (central) 2-group extension of $T$. And changing her notation to match mine here, she shows:

Proposition 2.7 For any bilinear form $\beta: L \times L \to \mathbb{Z}$, the bilinear form $\beta^t$ given by $\beta^t(x,y) = \beta(y,x)$ gives an equivalent 2-group extension of $T = L \otimes_{\mathbb{Z}} \mathrm{U}(1)$.

Corollary 2.8 i) If $\beta: L \times L \to \mathbb{Z}$ is an even symmetric bilinear form on a lattice $L$, then the multiplicative bundle gerbe associated to $\beta$ possesses a square root.

(ii) If $\beta$ is a skew symmetric bilinear form on a lattice $L$, then $\beta$ yields the trivial 2-group extension of the torus $T = L \otimes_{\mathbb{Z}} \mathrm{U}(1)$.

Corollary 2.9 Let $\beta: L \times L \to \mathbb{Z}$ be a bilinear form on a lattice $L$. Then up to equivalence over $T$, the 2-group we obtain from $\beta$ only depends on the even symmetric bilinear form $\beta + \beta^t$.

All this leads to further questions:

What are some fun things you would do with a multiplicative gerbe over a torus?

I imagine that if I were doing string theory on a spacetime of the form $M \times T^n$ it would be interesting to have a nontrivial gerbe with connection over $T^n$. People love to talk about torus compactifications of string theory coming from integral lattices like $\mathrm{E}_8 \times \mathrm{E}_8$, $\mathrm{D}_{16}^+$ and the Leech lattice, as well as many others. Do the gerbes over tori coming from integral lattices in the way we’re discussing come equipped with a connection in some canonical way? Have people studied this already? Have they used the multiplicativity of the gerbes?

Posted by: John Baez on January 4, 2015 11:17 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Good questions. First of all, the mutliplicative gerbes of Nora’s construction are have all trivial underlying gerbes. The bilinear form is used to construct the multiplication, see Construction 2.3 in Nora’s paper. That multiplication is given by the Poincaré line bundle over $T \times T$ associated to the bilinear form, and the Poincaré line bundle does indeed have a canonical connection. As such, it induces a multiplicative connection on the multiplicative gerbe , this is explained in my paper arxiv:0804.4835, Example 1.4 (b). So, yes, all the multiplicative gerbes of Nora’s construction come equipped with multiplicative connections.

If these connections have been used, I don’t know. Thomas Nikolaus and I have a project that tries to relate them to T-duality, but we have not finally succeded.

Connections on mutliplicative gerbes, in general, have the following meaning. As a gerbe with connection (forgetting the multiplicative structure), it is a B-field for string theory on oriented surfaces with target space the torus. The multiplicative structure means that this string theory lives on the boundary of a Chern-Simons theory with the torus as its gauge group, and the characteristic class of the multiplicative gerbe as its level.

This is explained here:

• Bundle gerbes for Chern-Simons and Wess-Zumino-Witten theories, arxiv:0410013
• Polyakov-Wiegmann Formula and Multiplicative Gerbes, arxiv:0908.1130

### Re: Integral Octonions (Part 12)

By the way, my favorite section in Nora Ganter’s paper is the part on ‘Mathieu, Conway and Weyl 2-groups’. Even though I imagine a lot is left for future papers, this hints at what I believe will become a very fun subject: exceptional higher algebraic structures. The first ones will be based on known exceptional structures — e.g., 2-groups or $n$-groups extending the sporadic finite simple groups. But someday we may start getting structures whose ‘exceptionalness’ is only visible in the higher parts. And with luck, these will shed new light on the exceptional structures we already know.

A simple but very nice example is the Mathieu groupoid, $M_{13}$.

Posted by: John Baez on January 4, 2015 11:37 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

just wondering about octonions and particles - maybe the exceptional groups have to cope with both normal particles of the standard model, and also with ‘dark’ particles - one might break the 480 tables into 40 particles by factoring out spin and color and signature reversal - giving 10 particles in each of 4 generations instead of 4 particles per generation. The extra 6 need to be ‘dark’ and not really fermions or bosons.

Posted by: Joel Rice on January 15, 2015 4:03 PM | Permalink | Reply to this

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