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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 $\mathbf{Br}(k)$?

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    $\begingroup$ 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. $\endgroup$
    – Theo Johnson-Freyd
    Commented May 8, 2020 at 22:46
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    $\begingroup$ Br(k) is the 0th space of an HZ-module spectrum, so its Postnikov invariants are trivial (the Postnikov tower is noncanonically split). $\endgroup$
    – Jacob Lurie
    Commented May 13, 2020 at 20:33
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    $\begingroup$ 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
    Commented May 14, 2020 at 20:34
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    $\begingroup$ 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) $\endgroup$
    – Jacob Lurie
    Commented May 15, 2020 at 18:39
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    $\begingroup$ 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). $\endgroup$
    – Jacob Lurie
    Commented May 15, 2020 at 18:44

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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.

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    $\begingroup$ 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). $\endgroup$
    – Arun Debray
    Commented Sep 8, 2020 at 0:50

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