# Postnikov invariants of the Brauer 3-group

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:

• $$\pi_1$$ is the Brauer group of $$k$$.

• $$\pi_2$$ is the Picard group of $$k$$.

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

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 $$\mathbf{Br}(k)$$?

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