Given a 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.

The particular 3-group we get from a commutative ring $k$ could be called its Brauer 3-group and denoted $\mathbf{Br}(k)$. 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, $\mathbf{Br}(k)$ has homotopy groups which I will call $\pi_1, \pi_2, \pi_3$ (though there are choices of where we start numbering). These are well-known things:

My question is whether people have studied, or computed, the Postnikov invariants involving these things. The simplest is the map

$$ a : \pi_1^3 \to \pi_2$$

coming from the associator in the monoidal category of $k$-algebras (with isomorphism classes of bimodules as morphisms). Since the associator obeys the pentagon identity this is a 3-cocycle on $\pi_1$ with values in its module $\pi_2$, so it gives an element of $ H^3(\pi_1, \pi_2)$.

Is this element trivial? If not, what is it?

But in fact $\mathbf{Br}(k)$ is not just a 3-group but also a symmetric monoidal bicategory. So, it's what I call a symmetric 3-group, though some others call it a Picard 2-category. These have a number of other Postnikov invariants:

Has anyone figured out any of these for $\mathbf{Br}(k)$?

  • 3
    $\begingroup$ As you point out, $\mathbf{Br}(k)$ is symmetric, and so the Postnikov invariants are (the destablizations of) stable cohomology operations. This highly constrains the question. $\endgroup$ May 8, 2020 at 22:46
  • 10
    $\begingroup$ Br(k) is the 0th space of an HZ-module spectrum, so its Postnikov invariants are trivial (the Postnikov tower is noncanonically split). $\endgroup$ May 13, 2020 at 20:33
  • 1
    $\begingroup$ Nice! Is this why you can compute the groups I'm calling $\pi_i$ using Galois cohomology as $H^{3-i}(\mathrm{Gal}(K|k), K^\star)$ where $K$ is the separable closure of $k$? $\endgroup$
    – John Baez
    May 14, 2020 at 20:34
  • 4
    $\begingroup$ They have a common explanation. Assume k a field for simplicity, let K be a separable closure, and let G = Gal(K/k). Let Br(k) denote the classifying space for your 2-category (so it's the loop space of what you're denoting by Br(k) ). Then Br(K) carries a continuous action of G, and there's a natural map e from Br(k) to the (continuous) homotopy fixed points Br(K)^hG. The input you need is the following: 1) The map e is a homotopy equivalence (because the construction k -> Br(k) satisfies etale descent). (cont) $\endgroup$ May 15, 2020 at 18:39
  • 4
    $\begingroup$ 2) Br(K) is an Eilenberg-MacLane space K( K^*, 2) (since the Picard and Brauer groups of a separably closed field vanish). Concretely this gives you a formula for pi_*( Br(k) ) in terms of Galois cohomology. It also tells you that Br(k) admits the structure of a topologica/simplicial abelian group (since for an Eilenberg MacLane space 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). $\endgroup$ May 15, 2020 at 18:44

1 Answer 1


Let me see if I understand what Jacob says in the comments. I think his argument can be summarized as: the Brauer 3-group is étale-locally an Eilenberg-MacLane spectrum, hence étale-locally an $\mathbb{Z}$-module spectrum, hence an $\mathbb{Z}$-module spectrum, hence the Postnikov tower splits. Do I have that right?

If so, I want to point out that the situation changes with one tweak, which is to allow superalgebras, super bimodules, etc. When $k = \mathbb{R}$ the super Brauer 3-group has a nontrivial homotopy operation $\pi_1 \to \pi_3$ given by taking the super dimension of the zeroth Hochschild homology of a superalgebra in the super Brauer group, and I computed an example of it taking a nontrivial value here. This implies that the Postnikov tower can't split but I don't know what the $k$-invariants are. I suppose following Jacob we could try to work things out as homotopy fixed points of the $\text{Gal}(\mathbb{C}/\mathbb{R})$-action on the super Brauer 3-group over $\mathbb{C}$ but this is beyond me.

  • 1
    $\begingroup$ The super Morita category is a great source of examples! For another example, the $k$-invariants for $\mathbb C$, see Dan Freed's Vienna notes around (1.42). Another proof, computed in a slightly different way, can be found here (see 4.3.5). $\endgroup$ Sep 8, 2020 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.