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When constructing the Brauer group of a field, only the finite-dimensional central simple algebras are considered (because of Artin-Wedderburn's characterization).

But what happens to the infinite-dimensional ones? (I.e., to simple algebras which are infinite-dimensional over their centers).

  • Is there an analogue for the Brauer group?

  • Is there some structure theorem like Artin-Wedderburn's?

  • What can be said about cross products of these algebras?

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3 Answers 3

Perhaps one approach to thinking about this question would be Dixmier-Douady theory (as described in Brylinski's book "Loop Spaces, Characteristic Classes and Geometric Quantization"). This involves considering certain Banach algebras over the complex numbers as infinite dimensional analogs of central simple algebras, and leads to some useful notions generalizing bundles of central simple algebras over topological spaces.

The point of view here starts with the correspondence between bundles of central simple $\mathbb C$-algebras over a topological space $X$ and torsion elements of the second cohomology $H^2(X, \underline{\mathbb C}^*)$. This comes from the identification of isomorphism classes degree $n$ algebra bundles with $H^1(X, \underline{PGL_n})$ and the nonabelian short exact sequence: $$ 1 \to \underline{\mathbb C}^* \to \underline{GL_n} \to \underline{PGL_n} \to 1$$

(here I'm using the underline to denote the corresponding sheaves of continuous functions).

Now, when you do this, you find that those classes which come from $H^1(X, \underline{PGL_n})$ (i.e. from degree $n$ central simple algebra bundles) are necessarily $n$-torsion in $H^2(X, \underline{\mathbb C}^*)$. what Dixmier-Doady theory does is identify all the elements of this $H^2$ as arising from infinite dimensional algebras.

A little more precisely, assuming paracompactness of $X$, on fixes some separable Hilbert space $E$, and defines $\mathcal K$ to be the Banach algebra of compact operators on $E$. It then turns out that one can naturally identify the whole of $H^2(X, \underline{{\mathbb C}^\ast})$ with the set of isomorphism classes of locally trivial Banach algebra bundles with fiber isomorphic to $\mathcal K$. What makes this really work is an analog of the above sequence

$$ 1 \to {\mathbb C}^\ast \to {\mathcal L}^\ast \to {\mathcal L}^\ast/{\mathbb C}^\ast \to 1 $$

where $\mathcal L$ is the Banach algebra of bounded (sup-norm) operators on $E$ (which incidentally happens to have center $\mathbb C$), and one uses an identification $\mathcal L^\ast/\mathbb C^\ast \cong Aut(\mathcal K)$.

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I am very far from an expert, so I hope my answer is useful for you, but I hope also that people who know more than I do will supplement or correct my answer.

Let me fix a commutative ring (or...) $k$. The first thing to recall is that the Brauer group of $k$ is very close to the etale homology group $\mathrm H^2(\operatorname{spec} k, \mathbb G_m)$. For example, when $k$ is a field, then etale homology is group homology for the Galois group, and (writing $k^s$ for the separable closure of $k$) one has $\text{Brauer} = \mathrm H^2(\operatorname{Gal}(k^s/k),(k^s)^\times)$, where $(k^s)^\times$ is the group of units in $k^s$. Grothendieck asked (and at least in some cases answered, but I don't know the history) whether one actually has Brauer=H^2 in general, where "Brauer" is defined by Azumaya algebras and corresponds geometrically to projective bundles over $\operatorname{spec} k$ (modulo proj(vector bundles)). The answer is that one does have an inclusion of Brauer into H^2, but it is not onto in general. First, every Azumaya algebra represents a torsion element in H^2, and in general H^2 has non-torsion part. But in fact there can be torsion in H^2 that also is not in Brauer. To fix these problems, see for example Heinloth and Schröer, "The bigger brauer group and twisted sheaves", 2009. (Sorry, I don't have an arXiv number, but I can email the pdf if you can't find it online.) Perhaps these are the infinite-dimensional algebras you're looking for. If memory serves, I learned this story from Pete L. Clark here on MathOverflow. (If memory has misserved me, I apologize!)

There is a story I have told myself, but I don't know a reference or let alone a precise statement. Let me be imprecise about some list of "nice categorical properties", and wrap everything up into the word "green". ("green" may include words like "presentable" or "abelian" or ....) Then for example the category $K = \text{Mod}_k$ of right $k$-modules is a green category. Because $k$ is commutative, $K$ has a natural structure as symmetric-monoidal category. For any algebra $A$ over $k$, one can also consider the green category $\text{Mod}_A$ of right $A$-modules. Any such module is also a left $k$-module, and so there is an operation of $k$-modules on $A$-modules, and this makes $\text{Mod}_A$ into some categorified version of left $K$ module. In general, we can consider a 2-category whose objects are green categories equipped with the structure of left $K$-module; I will call this 2-category $_K\text{MOD}$.

Then since $K$ is symmetric monoidal, $_K\text{MOD}$ carries a natural structure as symmetric monoidal 2-category. After you've decided what "green" means, a natural question to ask is: What are the "units" in $_K\text{MOD}$? For comparison, the group of units in $K = \text{Mod}_k$ are precisely the line bundles over $\operatorname{spec} k$, i.e. the Picard group $\mathrm H^1(\operatorname{spec} k,\mathbb G_m)$. So this is a sort of "2-Picard group", and the "0-Picard group" is $\mathbb G_m(k) = \mathrm{GL}_1(k) = \mathrm H^0(\operatorname{spec} k,\mathbb G_m)$.

When "green" means "is the representation theory of an algebra", probably this "2-Picard group" is the classically-defined Brauer group on the nose. What you would like to believe (but as I said, I have no theorem to this effect) is that for the correct definition of "green", the 2-Picard group is precisely $\mathrm H^2(\operatorname{spec} k,\mathbb G_m)$. Then the answer to your question is that rather than looking for infinite-dimensional algebras that extend the Brauer group, what you should be asking for are representation theories that behave as if they were representations of an "infinite-dimensional" central simple algebra. And you should believe that the story continues for the correct categorification of "green".

But then I think that even for the correct definition of "green", over a field you don't get anything new. That's because we already have Brauer = H^2 on the nose, and so you should believe that all reasonable definitions of "green" have the same group of units. Most generally, there is a weakening of "unit" to "dualizable object" and probably when $k$ is a field, all reasonable definitions of "green" give the same class of dualizable objects in $_K\text{Mod}$. But I'm starting to digress far from your original question, and so I will stop.

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I only know a little bit about this, so for your sake I hope someone more knowledgeable comes along...

As for your first question:

It is not hard to see that the tensor product of any two central simple algebras over a field is again a central simple algebra over that field: see e.g. Theorem 79 of these notes. So for any field $K$ and any infinite cardinal number $\kappa$, the set of isomorphism classes of central simple $K$-algebras of cardinality at most $\kappa$ forms a commutative monoid. I don't know what can be said about the structure of these monoids.

However in this case it's not clear to me that there is some natural equivalence relation to impose here such that the quotient under this relation forms a group (as in the finite-dimensional case). One perspective on Brauer equivalence of finite dimensional CSAs is that it is a special case of Morita equivalence. I would tend to doubt that considering infinite dimensional central simple algebras under Morita equivalence will give a group: perhaps some expert can confirm this. Failing this, it seems that the burden is on you to come up with some natural equivalence relation and explain why it generalizes the Brauer group in the classical case. On the other hand I know people who like to think of categorical analogues of the Brauer group, so possibly this is a fruitful way to go here.

As to your second question:

No, definitely not. Among other things, Wedderburn's Theorem shows that any Artinian simple algebra is "almost" a division algebra (in fact is Morita equivalent to a division algebra). But the class of non-Artinian simple algebras over a field $K$ is far larger than the class of division algebras over a field $K$, and there is no reasonable classification of one in terms of the other. For some examples of this, see Chapter 1 in the notes linked to above, and then see Lam's First Course on non-commutative algebra for much more in this direction.

I have nothing to offer on your third question.

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