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 $\mathbf{Br}(k)$?

| cite | improve this question | | | | |
  • 2
    $\begingroup$ 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 May 8 at 22:46
  • 7
    $\begingroup$ 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 May 13 at 20:33
  • $\begingroup$ 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 at 20:34
  • 2
    $\begingroup$ 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 May 15 at 18:39
  • 2
    $\begingroup$ 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 May 15 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.