The n-Category Café

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:

• André Joyal and Ross Street, Braided monoidal categories, 1986.

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

$$ \beta : A \times A \to B$gives a braided 2-group. <b>Example 2.</b> In particular, if$L$is any lattice in a real inner product space$V$, for any constant$\theta \in \mathbb{R}$we get a bilinear map$$\beta: L \times L \to \mathbb{R}$$given by$$\beta(x,y) = \theta \langle x,y\rangle$$So, we get a 1-parameter family of braided 2-groups with$L$as the group of objects and$\mathbb{R}$as the group of automorphisms of any object. <b>Example 3.</b> We call$L$an <b>integral</b> lattice if$\langle x, y \rangle$is an integer whenever$x, y \in L$. In this case we have$$\beta: L \times L \to \mathbb{Z}$$whenever$\theta \in \mathbb{Z}$. So, we also get a 1-parameter family of braided 2-groups with$L$as the group of objects and$\mathbb{Z}$as the group of automorphisms of any object --- but now the parameter needs to be an integer. <b>Example 4.</b> Alternatively, if$L$is a lattice in a real inner product space$V$, for any constant$\theta \in \mathbb{R}$we get a bilinear map$$\beta: L \times L \to \mathrm{U}(1)$$given by$$\beta(x,y) = e^{i \theta \langle x,y\rangle }$$So, we get a 1-parameter family of braided 2-groups with$L$as the group of objects and$\mathrm{U}(1)$as the group of automorphisms of any object. <b>Example 5.</b> Take the previous example and suppose$L$is integral. In this case$\beta$doesn't change when we replace$\theta$by$\theta + 2 \pi$, so there is really a circle's worth of braidings. Certain nice things happen for special values of$\theta$. When$\theta = 0$,$$\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$$and we get a <b>symmetric 2-group</b> --- that is, a 2-group that's a symmetric monoidal category. So far these examples look a bit general and abstract, so let me give you some examples of these examples. (Category theory can be defined as the branch of math where the examples require examples.) <h3> An example from Lie theory </h3> Suppose$G$is a compact simple Lie group, and let$T \subseteq G$be a maximal torus. The Lie algebra of$T$is abelian: its Lie bracket vanishes, so we might as well think of it as a mere vector space. For this reason I'll call it$V$, instead of the more scary Gothic$\mathfrak{t}$that Lie theorists would prefer. But this vector space$V$is equipped with some other interesting structure: an inner product and a lattice! You see, sitting inside$V$is a lattice$L$consisting of those elements$v$with$$\exp(2 \pi v) = 1 \in T$$Furthermore the Lie algebra of$G$comes with a god-given nondegenerate bilinear form called the <a href = "https://en.wikipedia.org/wiki/Killing_form">Killing form</a>. If we multiply this by a suitable constant and restrict it to$V$, we get an inner product on$V$for which$L$is an integral lattice. Indeed,$L$is usually called the <b>integral lattice of$G$</b>. So, using the trick in Example 4, we get a 1-parameter family of braided 2-groups with$L$as the group of objects and$\mathrm{U}(1)$as the group of automorphisms of any object. <h3> Another example from Lie theory </h3> If you're not familiar with the integral lattice of$G$, you may know about its 'weight lattice'. This is dual to the integral lattice, in a certain sense. Namely, we can start with the integral lattice$L$in$V$and define a new lattice$L^\ast$consisting of all points in the dual vector space$V^\ast$whose pairing with every point in$L$is an integer:$$L^\ast = \{ \ell \in V^\ast : \; \ell(v) \in \mathbb{Z} \; for \; all \; v \in L \}$$Let's call$L^\ast$the <b><a href = "https://en.wikipedia.org/wiki/Weight_$G$, though people usually reserve this term for the case when$G$is simply connected. Why is the weight lattice interesting? For starters, any 1-dimensional unitary representation of the maximal torus$T$is equivalent to one coming from a point in this lattice! If we take$v \in V$then$\exp(2 \pi v) \in T$, and any element$T$arises this way. Given$\ell \in L^$we can thus define a 1-dimensional unitary representation of$T$by$$\exp(2 \pi v) \mapsto \exp(2 \pi i \ell(v)) \in \mathrm{U}(1)$$In fact, we get all the 1d unitary reps of$T$this way. Since every finite-dimensional unitary representation of$T$is a direct sum of 1-dimensional ones, the category$Rep(T)$of finite-dimensional unitary representations of$T$is pretty well described by the weight lattice$L^$. How can we make this precise? Here's one way. There's a category FinVect$[L^\ast]$whose objects are complex vector bundles on$L^\ast$that have <b>finite support</b>: that is, have 0-dimensional fibers outside a finite set. The morphisms are just vector bundle morphisms. This category is monoidal, where the tensor product comes from <a href = "http://ncatlab.org/nlab/show/Day+convolution">Day convolution</a>:$$(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)$$as monoidal categories. (This is pretty obvious if you know your stuff, but if you want a proof, see the section in my paper on <a href = "http://arxiv.org/abs/q-alg/9609018">2-Hilbert spaces</a> where I discuss the categorifed Fourier transform. In case it helps, I should point out that the lattice I'm calling$L^\ast$is actually the Pontryagin dual of$T$. You see, the dual of a lattice in some vector space is really the Pontryagin dual of the vector space mod that lattice.) Now let's build some braided 2-groups as in Example 4. We've got a lattice$L^\ast$in a real inner product space$V^\ast$. This gives a braided 2-group whose objects are points of$L^\ast$, where the automorphisms of any object form the group$\mathrm{U}(1)$, and where the braiding$$\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)$$that is compatible with direct sums. When$\theta = 0$this is the usual braiding on$\mathrm{Rep}(T)$. So, as we move$\theta$away from$0$, we're deforming that braiding. Sitting inside the weight lattice are certain vectors called <a href = "https://en.wikipedia.org/wiki/Root_system">roots</a>. For$\mathrm{E}8$they look like this, if we project them from 8 dimensions down to 2 in a certain nice way: <a href = "https://en.wikipedia.org/wiki/Root_system#E6.2C_E7.2C_E8"> <img src = "http://math.ucr.edu/home/baez/mathematical/500px-E8Petrie_net.png" alt = ""/></a> For any root$r \in L^{\ast}$we have a reflection$$R_r : V^{\ast} \to V^{\ast}$$that maps$r$to$-r$while leaving orthogonal vectors alone. And the wonderful thing is that these reflections preserve the weight lattice! These reflections generate a group called the <b><a href = "https://en.wikipedia.org/wiki/Weyl_group">Weyl group</a></b> of$G$. Since the Weyl group preserves the inner product on$V^{\ast}$and also the weight lattice, it acts on the monoidal category$$\mathrm{FinVect}[L^\ast] \cong \mathrm{Rep}(T)$$in a way that preserves the braiding for any value of the parameter$\theta$. <h3> Puzzles </h3> <b>Puzzle 1.</b> How are the braided 2-groups constructed in Example 3 related to Nora Ganter's <a href = "http://arxiv.org/abs/1406.7046">categorical tori</a>? 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'. <b>Puzzle 2.</b> 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. <b>Puzzle 3.</b> 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}$). <b>Puzzle 4.</b> Among the most interesting integral lattices are the <b>even</b> 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 <b>self-braiding</b>$\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? <b>Puzzle 5.</b> 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?