Since my next post on Scott’s talk concerns the construction of a new subfactor, I wanted to give another attempt at explaining what a subfactor is. In particular, a subfactor is just a finite-dimensional simple algebra over C!

Now, I know what you’re thinking, doesn’t Artin-Wedderburn say that finite dimensional algebras over C are just matrix algebras? Yes, but those are just the finite dimensional algebras in the category of vector spaces! What if you had some other C-linear tensor category and a finite dimensional simple algebra object in that category?

Let me start with an example (very closely related to Scott Carnahan’s pirate post).

Fix your favorite finite group G, and look at the category of G-modules. The ring of functions C(G) can be thought of as a G-module, and the pointwise multiplication map is a morphism of G-modules. Hence C(G) is an algebra object in C[G]. Furthermore it’s simple (in the Killing form sense: it has a nondegenerate inner-product given by tr(xy) where tr ~~pulls out the value of the function on 1~~ averages over the group).

Any time you have an algebra object A in some tensor category C, you can look at the left A-modules, the right A-modules, and the A-A bimodules. The tensor product in the tensor category and the tensor product over A together turn these (together with the original category) into what I call a bi-oidal category. It’s like a monoidal tensor category, but with two different tensor products. (A more standard name for a bi-oidal categry is a 2-category with exactly two objects.)

Ok, what does all this have to do with subfactors? Well suppose I have an inclusion of von Neuman algebras with trivial centers B<A. Then I can look at the category of B-B bimodules. If the subfactor has “finite Jones index,” then A is automatically a simple finite dimensional algebra object in this category!

Furthermore, under the right circumstances you can go backwards and realize any simple algebra object in a tensor category as an inclusion of vN algebras with trival centers. I’ll have more to say about this in the future, but not for a month or two.

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Fix your favorite finite group G, and look at the category of G-modules. The ring of functions C(G) can be thought of as a G-module, and the pointwise multiplication map is a morphism of G-modules. Hence C(G) is an algebra object in C[G]. Furthermore it’s simple (in the Killing form sense: it has a nondegenerate inner-product given by tr(xy) where tr pulls out the value of the function on 1).Well, something here is wrong. In the (commutative) ring of functions C(G), it’s certainly true that pointwise multiplication is a morphism of G-modules. But the trace that evaluates at the identity is _not_ nondegenerate; indeed, if x(1) = 0, then (xy)(1) = (x(1))(y(1)) = 0.

Probably the trace you want is not evaluation at 1 but something like the integral of the function over G. Oh, yeah, your G is finite, so that integral makes perfect sense.

You’re exactly right, thanks!