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Let $k$ be a field, and let $\alpha \in \mathrm{Br}(k)$. Let $A$ be an Azumaya algebra representing $\alpha$. Then the category $A$–$\mathrm{mod}$ depends only on $\alpha$.

I would like to know whether there's a way to describe which $k$-linear categories arise this way. Thus I'd like to know if there's a way to define the Brauer group of a field $k$ as classifying certain kinds of $k$-linear categories. I'd also like to know if there's a good description of the sum of elements of the Brauer group in terms of categories (is it some sort of tensor product of categories?).

The one condition I can come up with is that it should be a semisimple abelian category over $k$ for which the endomorphism algebra of the unit object is $k$.

An even more bold hope is to express the invariant map $\mathrm{Br}(\mathbb{Q}_p) \to \mathbb{Q}/\mathbb{Z}$ in terms of this category.

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    $\begingroup$ If $A$ isn't commutative there won't be a tensor product. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 30, 2020 at 6:20
  • $\begingroup$ Vector spaces over $k$. This category does not depend on $\alpha$ $\endgroup$
    – Bugs Bunny
    Commented Jun 30, 2020 at 14:47

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$k$-linear cocomplete categories admit a "tensor product over $\text{Mod}(k)$" (thinking of them as module categories over $\text{Mod}(k)$) and the only thing you need to know about it to answer this question is that the tensor product of $\text{Mod}(A)$ and $\text{Mod}(B)$ is $\text{Mod}(A \otimes_k B)$. Azumaya algebras then give rise to invertible categories with respect to this tensor product and I believe every invertible such category has this form although I don't know how to prove it. More details about the tensor product in this blog post.

A more explicit characterization isn't hard but also isn't particularly enlightening - over a field, Azumaya algebras are central simple algebras, so you are looking for $k$-linear cocomplete categories where

  • there's exactly one isomorphism class of simple object
  • every object is a direct sum of copies of this simple object, and
  • the (categorical) center (endomorphisms of the identity functor) is $k$.

But the invertibility characterization should hold for $k$ any commutative ring, and is in my opinion very conceptually clean; it tells us that the Brauer group is a kind of "higher Picard group" classifying a categorified version of line bundles. Some difficulties here in general with the difference between the Brauer group and the cohomological Brauer group.

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    $\begingroup$ Fix a commutative ring $k$. I'd suspect that any element $\alpha \in H^2(k,\mathbf{G}_m)$ would give rise to an invertible category: look at the "degree 1" subcategory of quasicoherent sheaves on the corresponding $\mathbf{G}_m$-gerbe over $k$. For this to come from an Azumaya algebra, I think you want $\alpha$ to be a torsion element. This is automatic if $k$ is a field (or a regular ring), but not in general. $\endgroup$
    – Anonymous
    Commented Jun 30, 2020 at 16:14
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    $\begingroup$ Are you suggesting that even when the ring is not regular, it might be isomorphic to the cohomological Brauer group? (but not the group defined in terms of Azumaya algebras) $\endgroup$
    – David Corwin
    Commented Jun 30, 2020 at 23:20
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    $\begingroup$ I wouldn't be surprised if the statement "if $\mathcal{C}$ is invertible over $\mathrm{Mod}(k)$ then $\mathcal{C} \cong \mathrm{Mod}(A)$ for an Azumaya algebra $A$" will depend on "functional analytic" questions about how completed you want your categories to be. These are answered already by what you want "$\mathrm{Mod}$" to mean: all modules, finitely generated modules, finitely generated projective modules, ...? $\endgroup$
    – Theo Johnson-Freyd
    Commented Jul 1, 2020 at 4:17
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    $\begingroup$ You probably mean the first option, and so will work with locally (finitely?) presentable categories. But if you will allow me to work with the last option, then the natural tensor product is only completed for direct sums and idempotents; then Tillmann showed that $\mathcal{C}$ is 2-dualizable iff it is $\mathrm{mod}(A)$ for a finite-dimensional separably algebra, and so invertible iff it $A$ is Azumaya. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jul 1, 2020 at 4:19
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    $\begingroup$ Actually, these categories do know each other: Tillmann's result kicks in as soon as you demand that $\mathcal{C}$ have enough compact projective objects. Whether dualizability/invertiblity in $\{$locally presentable $k$-linear categories$\}$ implies enough compact projectives is, so far as I know, still open, but it is expected to hold. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jul 1, 2020 at 4:22

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