May 16, 2020

The Brauer 3-Group

Posted by John Baez

For reasons ultimately connected to physics I’ve been spending time learning about the Brauer 3-group. For any commutative ring $k$ there is a bicategory with

• algebras over $k$ as objects,
• bimodules as morphisms,
• bimodule homomorphisms as 2-morphisms.

This is a monoidal bicategory, since we can take the tensor product of algebras, and everything else gets along nicely with that.

Given any monoidal bicategory we can take its core: that is, the sub-monoidal bicategory where we only keep invertible objects (invertible up to equivalence), invertible morphisms (invertible up to 2-isomorphism), and invertible 2-morphisms. The core is a monoidal bicategory where everything is invertible in a suitably weakened sense so it’s called a 3-group.

I’ll call the particular 3-group we get from a commutative ring $k$ its Brauer 3-group, and I’ll denote it as $\mathbf{Br}(k)$. It’s discussed on the $n$Lab: there it’s called the Picard 3-group of $k$ but denoted as $\mathbf{Br}(k)$.

Like any 3-group, we can think of $\mathbf{Br}(k)$ as a homotopy type with 3 nontrivial homotopy groups, $\pi_1, \pi_2,$ and $\pi_3$. These groups are wonderful things in themselves:

• $\pi_1$ is the Brauer group of $k$.

• $\pi_2$ is the Picard group of $k$.

• $\pi_3$ is the group of units of $k$.

But in general, a homotopy type contains more information than its homotopy groups. So on MathOverflow I asked if anyone knew the Postnikov invariants of the Brauer 3-group — the extra glue that binds the homotopy groups together. In theory these could give extra information about our commutative ring $k$.

But Jacob Lurie said the Postnikov invariants are trivial in this case.

He said:

$\mathbf{Br}(k)$ is the 0th space of an $\mathrm{H}\mathbb{Z}$-module spectrum, so its Postnikov invariants are trivial (the Postnikov tower is noncanonically split).

An $\mathrm{H}\mathbb{Z}$-module spectrum is the simplest sort of space from a homotopy theory perspective: all its homotopy groups are abelian, $\pi_1$ acts trivially on all the higher homotopy groups, and all the Postnikov invariants — which describe subtler interactions between homotopy groups — are trivial. So it really just amounts to a list of abelian groups.

Put another way, an $\mathrm{H}\mathbb{Z}$-module spectrum might just as well be a chain complex of abelian groups: the homotopy category of $\mathrm{H}\mathbb{Z}$-module spectra is Quillen equivalent to the category of such chain complexes.

This instantly made me think about another description of $\pi_1, \pi_2$ and $\pi_3$ for the Brauer 3-group. Suppose for example that $k$ is a field. Then

$\pi_i \cong H^{3-i}(G,K^\ast)$

where $K$ is the separable closure of $k$, $G = \mathrm{Gal}(K|k)$ is the Galois group, and we’re taking the group cohomology of $G$ with coefficients in the group of units $K^\ast$, which is a $G$-module in the obvious way. This ‘Galois cohomology’ approach to the Brauer 3-group has been on my mind lately.

So, Galois cohomology hands us a chain complex of abelian groups — and thus an $\mathrm{H}\mathbb{Z}$-module spectrum whose homotopy groups match those of the Brauer 3-group! Namely:

$C^2(G,K^\ast) \stackrel{d}{\longleftarrow} C^1(G,K^\ast) \stackrel{d}{\longleftarrow} C^0(G,K^\ast) \stackrel{d}{\longleftarrow} 0 \stackrel{d}{\longleftarrow} 0 \stackrel{d}{\longleftarrow} \cdots$

Note this is a cochain complex turned upside down.

I should have asked if this chain complex, converted into an $\mathrm{H}\mathbb{Z}$-module spectrum, is the Brauer 3-group. (We can think of the Brauer 3-group as a spectrum because it’s the core of a symmetric monoidal bicategory.)

But for some reason I just said

Nice! Is this why you can compute the groups I’m calling $\pi_i$ using Galois cohomology as $H^{3−i}(Gal(K|k),K^*)$ where $K$ is the separable closure of $k$?

They have a common explanation. Assume $k$ a field for simplicity, let $K$ be a separable closure, and let $G = \mathrm{Gal}(K/k)$. Let $\mathrm{Br}(k)$ denote the classifying space for your 2-category (so it’s the loop space space of what you’re denoting by $\mathbf{Br}(k)$). Then $\mathrm{Br}(K)$ carries a continuous action of $G$, and there’s a natural map $e$ from $\mathrm{Br}(k)$ to the (continuous) homotopy fixed points $\mathrm{Br}(K)^{hG}$. The input you need is the following:

1) The map $e$ is a homotopy equivalence (because the construction $k \mapsto \mathrm{Br}(k)$ satisfies etale descent).

2) $\mathrm{Br}(K)$ is an Eilenberg–Mac Lane space $K( K^\ast, 2)$ (since the Picard and Brauer groups of a separably closed field vanish).

Concretely this gives you a formula for $\pi_*(\mathrm{Br}(k))$ in terms of Galois cohomology. It also tells you that $\mathrm{Br}(k)$ admits the structure of a topological/simplicial abelian group (since for an Eilenberg Mac Lane space $\mathrm{Br}(K)$, choosing such a structure is equivalent to choosing a base point, and the structure survives passage to homotopy fixed points when the base point is fixed by $G$).

I’ve been chewing on this. It’s great, but something in me hungers for a more lowbrow approach. After all, I can take this chain complex

$C^2(G,K^\ast) \stackrel{d}{\longleftarrow} C^1(G,K^\ast) \stackrel{d}{\longleftarrow} C^0(G,K^\ast) \stackrel{d}{\longleftarrow} 0 \stackrel{d}{\longleftarrow} 0 \stackrel{d}{\longleftarrow} \cdots$

and turn it into a 3-group where

• objects are 2-cocycles: elements $x \in Z^2(G,K^\ast)$.
• morphisms $f \colon x \to y$ are 1-cochains $f \in C^1(G,K^\ast)$ with $d f = y - x$.
• 2-morphisms $\alpha \colon f \to g$ are 0-cochains $\alpha \in C^0(G,K^*)$ with $d\alpha = g - f$.

and composition is addition. This is how chain complexes turn into $n$-groups with trivial Postnikov data. So, we can try to see explicitly map these objects, morphisms and 2-morphisms to objects, morphisms and 2-morphisms in the Brauer 3-group, where

• objects are invertible algebras over $k$: these turn out to be central simple algebras over $k$.
• morphisms are invertible bimodules: these turn out to be the ones that give Morita equivalences.
• 2-morphisms are invertible bimodule homomorphisms.

I’d been thinking for a while that 2-cocycles $x \in Z^2(G,K^\ast)$ should quite explicitly give central simple algebras over $k$, because I know a way they give algebra. Given a 2-cochain $x \colon G^2 \to K^\ast$ we can use it to ‘twist’ the multiplication in the group algebra $K[G]$ as follows. By definition $K[G]$ has a basis given by group elements $g \in G$, and we define a twisted multiplication $\star$ by

$g \star h = x(g,h) \, g h$

It’s just like the usual multiplication in the group algebra but with a number that depends on $g$ and $h$ in front. This twisted multiplication is associative iff $x$ is a 2-cocycle!

Then last night I was reading this pleasant document:

and I found that Noether had settled this issue in her Göttingen lectures of 1929-1930. She considered any Galois extension $K$ of a field $k$. She took a 2-cocycle $x \in Z^2(\mathrm{Gal}(K|k), L^*)$ and used it to twist the group algebra $K[G]$ as I’ve just described. But she called the 2-cocycle a ‘factor system’, and there’s even a Wikipedia article with this title. Others call it a ‘Noether system’, and the 2-cocycle condition is sometimes called the ‘Noether equation’ in this context. She called her twisted group algebra a ‘crossed product’.

Noether never published her work on this! But Deuring took notes of her lectures, and Brauer and Hasse got copies — and later, Noether, Brauer and Hasse collaborated on developing these ideas further.

Back in 1929-1930, Noether showed that that for any 2-cocycle, the resulting twisted group algebra is a central simple algebra $A$ over $k$ that ‘splits’ over $K$: in other words, $K \otimes_k A$ is isomorphic to a matrix algebra over $K$. Furthermore, she showed that every central simple algebra over $k$ that splits over $K$ arises this way, up to Morita equivalence! (She called this equivalence relation something like ‘similarity’.)

However, all central simple algebra split over the separable closure of $k$, so the theory simplifies in this case, and that’s what I mainly care about now.

So, we have a concrete map from 2-cocycles $x \colon G^2 \to K^\ast$ to objects of the Brauer 3-group. Noether also showed that two 2-cocycles differ by a coboundary iff their twisted group algebras are Morita equivalent! It should be pretty easy to refine this and get a map from the 3-group where

• objects are 2-cocycles: elements $x \in Z^2(G,K^\ast)$.
• morphisms $f \colon x \to y$ are 1-cochains $f \in C^1(G,K^\ast)$ with $d f = y - x$.
• 2-morphisms $\alpha \colon f \to g$ are 0-cochains $\alpha \in C^0(G,K^*)$ with $d\alpha = g - f$.

to the Brauer 3-group $\mathbf{Br}(k)$. And this should be an equivalence!

Has someone already done this? It seems most of the work has been done already, here and there. But I’d like to see it neatly laid out in one place. It would also be nice to see it all generalized to an arbitrary commutative ring $k$, or further. We can even do an $(\infty,1)$-category version of Brauer theory:

Posted at May 16, 2020 9:33 PM UTC

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Re: The Brauer 3-Group

Somewhat off-topic, but I remember reading about this kind of stuff in This Week’s Finds 209 and being quite inspired by it. I was probably only just starting to think about higher categories back then.

Posted by: Simon Willerton on May 30, 2020 5:48 PM | Permalink | Reply to this

Re: The Brauer 3-Group

Thanks! I’ve just come around back to it from thinking about the 10-fold way, which involves a “super” version of the Brauer 3-group.

I was hoping to find some more interesting structure in the Brauer 3-group of a field, coming from associators and the like, but now I’m fascinated to find that the opposite is true: this 3-group is equivalent to one that’s maximally strict and abelian: it’s basically just a 3-term chain complex.

Posted by: John Baez on May 30, 2020 6:14 PM | Permalink | Reply to this

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