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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 kk there is a bicategory with

  • algebras over kk 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 kk its Brauer 3-group, and I’ll denote it as Br(k)\mathbf{Br}(k). It’s discussed on the nnLab: there it’s called the Picard 3-group of kk but denoted as Br(k)\mathbf{Br}(k).

Like any 3-group, we can think of Br(k)\mathbf{Br}(k) as a homotopy type with 3 nontrivial homotopy groups, π 1,π 2,\pi_1, \pi_2, and π 3\pi_3. These groups are wonderful things in themselves:

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 kk.

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

He said:

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

An H\mathrm{H}\mathbb{Z}-module spectrum is the simplest sort of space from a homotopy theory perspective: all its homotopy groups are abelian, π 1\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 H\mathrm{H}\mathbb{Z}-module spectrum might just as well be a chain complex of abelian groups: the homotopy category of H\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 π 1,π 2\pi_1, \pi_2 and π 3\pi_3 for the Brauer 3-group. Suppose for example that kk is a field. Then

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

where KK is the separable closure of kk, G=Gal(K|k)G = \mathrm{Gal}(K|k) is the Galois group, and we’re taking the group cohomology of GG with coefficients in the group of units K *K^\ast, which is a GG-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 H\mathrm{H}\mathbb{Z}-module spectrum whose homotopy groups match those of the Brauer 3-group! Namely:

C 2(G,K *)dC 1(G,K *)dC 0(G,K *)d0d0d 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 H\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 π i\pi_i using Galois cohomology as H 3i(Gal(K|k),K *)H^{3−i}(Gal(K|k),K^*) where KK is the separable closure of kk?

Lurie answered:

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

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

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

Concretely this gives you a formula for π *(Br(k))\pi_*(\mathrm{Br}(k)) in terms of Galois cohomology. It also tells you that Br(k)\mathrm{Br}(k) admits the structure of a topological/simplicial abelian group (since for an Eilenberg Mac Lane space Br(K)\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 GG).

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 *)dC 1(G,K *)dC 0(G,K *)d0d0d 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 xZ 2(G,K *)x \in Z^2(G,K^\ast).
  • morphisms f:xyf \colon x \to y are 1-cochains fC 1(G,K *)f \in C^1(G,K^\ast) with df=yxd f = y - x.
  • 2-morphisms α:fg\alpha \colon f \to g are 0-cochains αC 0(G,K *)\alpha \in C^0(G,K^*) with dα=gfd\alpha = g - f .

and composition is addition. This is how chain complexes turn into nn-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 kk: these turn out to be central simple algebras over kk.
  • 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 xZ 2(G,K *)x \in Z^2(G,K^\ast) should quite explicitly give central simple algebras over kk, because I know a way they give algebra. Given a 2-cochain x:G 2K *x \colon G^2 \to K^\ast we can use it to ‘twist’ the multiplication in the group algebra K[G]K[G] as follows. By definition K[G]K[G] has a basis given by group elements gGg \in G, and we define a twisted multiplication \star by

gh=x(g,h)gh 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 gg and hh in front. This twisted multiplication is associative iff xx 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 KK of a field kk. She took a 2-cocycle xZ 2(Gal(K|k),L *)x \in Z^2(\mathrm{Gal}(K|k), L^*) and used it to twist the group algebra K[G]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 AA over kk that ‘splits’ over KK: in other words, K kAK \otimes_k A is isomorphic to a matrix algebra over KK. Furthermore, she showed that every central simple algebra over kk that splits over KK 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 kk, 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:G 2K *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 xZ 2(G,K *)x \in Z^2(G,K^\ast).
  • morphisms f:xyf \colon x \to y are 1-cochains fC 1(G,K *)f \in C^1(G,K^\ast) with df=yxd f = y - x.
  • 2-morphisms α:fg\alpha \colon f \to g are 0-cochains αC 0(G,K *)\alpha \in C^0(G,K^*) with dα=gfd\alpha = g - f .

to the Brauer 3-group Br(k)\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 kk, or further. We can even do an (,1)(\infty,1)-category version of Brauer theory:

Posted at May 16, 2020 9:33 PM UTC

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2 Comments & 0 Trackbacks

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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