## SF&PA: What is a Subfactor? March 25, 2008

Posted by Noah Snyder in planar algebras, subfactors.

By a subfactor I will mean a pair of rings A < B, such that (and I apologize in advance if I mess this up, as I said I don’t know any analysis):

• A and B are both Von Neumann algebras
• A and B have trivial centers (that is they are factors)
• A and B are both II_1-factors (that is they have a unique normalized trace)
• A<B has finite index (perhaps I’ll define this later)
• A<B is irreducible (that is the centralizer of A in B is trivial)

Whew! What on earth does all of this mean? And why would a representation theorist care about these conditions?

Well, as we saw before in the Galois theory example, given a pair A<B we get a bi-oidal category C(A<B) which is tensor generated by ${}_A B_B$ and ${}_B B_A$. It turns out that you can use the above conditions to prove a lot of nice algebraic properties about C(A<B). Furthermore, due to a converse theorem of Ocneanu Popa’s given a suitable category C(A<B) it must come from a subfactor. With this (hard!) theorem in hand we can happily ignore the subfactor setup and instead just think about categories that look like C(A<B).

Furthermore, if category theory scares you, tomorrow we’ll be doing the first thing you should do when someone tries to talk to you about categories: translate everything into pictures! At that point you’ll be able to forget both the Subfactor and the category and just start drawing pictures.

The fact that A and B are both Von Neumann algebras has two important consequences. First, all the Hom-spaces in this category are not only vector spaces over $\mathbb{C}$, they also inherit a positive definite Hermitian form. Secondly, the analysis allows one to prove (and this is substantially harder) that C(A<B) is a semisimple category. Morally this isn’t surprising, as positive definite forms are often helpful for proving semi-simplicity, but the proof is far from obvious.

The fact that A and B are factors says that the trivial bimodules ${}_A A_A$ and ${}_B B_B$ are irreducible. The fact that A<B is irreducible says that the generating modules ${}_A B_B$ and ${}_B B_A$ are also irreducible.

Finiteness of the index tells you that there is a good dimension theory for C(A<B). The II_1 property together with finiteness of the index (and maybe irreduciblity?) imply that C(A<B) has a good theory of duals.

Let’s unpack this last bit for a moment. When we were talking about Galois theory, C(L/K) had the property that ${}_L L_K \otimes {}_K L_L$ had a unique copy of the trivial bimodule ${}_L L_L$ (and this “evaluation map” is just given by multiplication in L). So, ${}_K L_L$ behaves like the dual of ${}_L L_K$. However, on the other hand if you took the tensor product in the opposite order it didn’t work out so well. That is, we didn’t have a unique map ${}_K L_K \cong {}_K L_L \otimes_L {}_L L_K \rightarrow {}_K K_K.$ In the Subfactor world, there is a unique A-A bilinear map $B \rightarrow A$ that is called the “conditional expectation.”

To summarize, we’ll call a bi-oidal category C (bi-oidal means that every object comes with two labels A-A, A-B, B-A, or B-B, and that we have two associative tensor products $\otimes_A$ and $\otimes_B$ which are defined when the labels match up) together with a choice of object X in the A-B sector a subfactor category if:

• C is enriched over Hilbert spaces (that is every Hom space has a Hilbert space structure)
• C is semi-simple
• C has a really nice theory of duals. (The right condition here is “spherical” in the sense of Barrett and Westbury, modified appropriately for the bi-oidal setting. Included in this condition is the fact that C has a good notion of dimensions.)
• The trivial objects ${}_A 1_A$ and ${}_B 1_B$ are irreducible
• X is irreducible, furthermore X is a tensor generator in the sense that every object in C occurs as a sub-object of $X \otimes X^* \otimes X \ldots$ or $X^* \otimes X \otimes X^* \ldots$.

One thing I want to emphasize here is that the choice of X is part of the data of a Subfactor category.

1. John Armstrong - March 25, 2008

Some perverse part of me really wants to weaken the definition of a bi-oidal category. What you’ve defined has an associativity

$({}_XL_Y \otimes_Y {}_YM_Z) \otimes_Z {}_ZN_W \cong$
${}_XL_Y \otimes_Y ({}_YM_Z \otimes_Z {}_ZN_W)$

like a strict monoidal category, right? We “should” have weak bi-oidal categories. Can they all be strictified?

For all I know this is already out in the literature. On the other hand, if it’s not…

2. Noah Snyder - March 25, 2008

To spoil a future post, a n-oidal category is a 2-category with n objects. Since 2-categories come in various versions of strictness, so do n-oidal categories.

3. Urs Schreiber - March 26, 2008

I keep forgetting what people tried to teach me about subfactors.

But what it should all amount to is to provide examples for adjunctions in a 2-category which are “special ambidextrous” such that the two monads obtained from them are “special symmetric” Frobenius mondas (as described somewhere here). Give or take a condition, maybe, for the case of subfactors, I don’t recall.

4. Urs Schreiber - March 26, 2008

We “should” have weak bi-oidal categories. Can they all be strictified?

Yes. Every bicategory is biequivalent to a strict 2-category.

But beware, this does not go through for morphisms: if you start with an arbitrary bifunctor between two bicategories, then pass to the strictification of these, the induced bifunctor between the strict 2-categories may not be strict itself. See Steve Lack Bicat is not triequivalent to Gray.

By the way, I would tend to discourage using the non-standard “n-oidal”-category for n-object bicategory. Unless you really care about the subtle different between the morphisms between k-tuply monoidal n-categories and k-fold degenerate k+n categories. Some people do, see for instance the remarks in section 5.6 of Baez, Shulman: Lectures on n-categories and cohomology. For more see Cheng, Gurski: The periodic table of n-categories for low dimensions I: degenerate categories and degenerate bicategories.

But I doubt that it is thins kind of issue you want to focus on here. So why not say n-object bicategory?

It’s useful, too, because it correctly keeps track of the dimension. It is not a coincidence that 2d CFT requires 2-categories. And it is not a coincidence that these holographically live in certain 3-categories (which can be found from the fact that a modular tensor category is braided monoidal, hence really a one-object one-1-morphism 3-category).

5. Urs Schreiber - March 26, 2008

We “should” have weak bi-oidal categories. Can they all be strictified?

I had posted a detailed reply twice, but it doesn’t seem to get through. [moderator: sorry about that, our spam filter overreacted] Let’s see if the short answer does:

Yes, every bicategory is biequivalent to a strict bicategory. And “n-oidel categories” are just a non-standard term for n-object bicategories.

6. John Armstrong - March 26, 2008

And so what happens when we add more structure? n-oidal bicategories are some kind of tricategory, and not all of those can be strictified. How much can these special ones be strictified?

7. Urs Schreiber - March 26, 2008

Every tricategory is triequivalent to a “Gray category”, namely a category enriched over strict 2-categories equipped with the Gray tensor product.

The proof of this works essentially by applying the fact that each bicategory is equivalent to a strict 2-category Hom-wise.

As far as I am aware it is not known how this pattern continues. One would in this vein want to know what “semistrict tetracategories” are such that every tetracategory is tetraequivalent to such a semistrict one, with the semistrict one being as strict as possible.

We had some discussion of this point in the recent thread “Infinity groups with specified composites” on the n-Category Cafe, but unfortunately the site is down right this moment, so I can’t give you the link.

8. Ben Webster - March 26, 2008

Note to commenters: if you have commented before, and you write a comment which doesn’t appear, please email one of us. It almost certainly got caught in the spam filter (usually because it had too many links).

9. Urs Schreiber - March 26, 2008

Note to commenters:

Oh, sorry, I should have waited for a little longer for the comments to appear (did wait quite a bit though). Maybe you should remove the first and the last of the three equivalent comments now. Sorry again.

10. Chris Schommer-Pries - March 26, 2008

I have to say I am wholly opposed to the term “bi-oidal”. I can’t say it with a straight face. Not to mention it is confusing and unnecessary.

On the other hand I can appreciate what Noah’s trying to do here. Many of our readers will know what a monoidal category is, especially the example where the tensor product is tensoring over a ring. I think that Noah doesn’t want to scare people by talking about “bicategories with 2 objects”. He wants to explain this subfactor stuff in a way more people can relate to.

11. Noah Snyder - March 26, 2008

Aside from not scaring people with 2-categories, I also didn’t want to spend a lot of time initially discussing stuff like “we will write horizontal composition of 2-morphisms as tensor product of morphisms.” That stuff’s kinda scary, whereas everyone already knows what bimodules look like and already knows what sort of structure categories of bimodules have. I do intend to write a full post on this at a later point, but not in the introduction.

12. Noah Snyder - March 26, 2008

3 is basically right. What you’re getting at there is what I mean by “a good theory of duals.” On top of the two compatible adjunctions, you’re also going to want a spherical condition (which says that the “left trace” equals the “right trace”). Then the only other thing you need is that the 2-morphism spaces are Hilbert spaces.

And a choice of generating morphism! Part of the data of a subfactor is a choice of 1-morphism $X \in Hom(A,B)$.

13. Urs Schreiber - March 26, 2008

There is a big story to be told about these ambidextrous adjunctions. They are to local trivializations as Frobenius algebras are to gluing data/descent data. I talk about that for instance here.

But the partuclar subfactor realization of this general fact has not entirely revealed its meaning to me until now. Jens Fjelstad, my collaborator on this program kept emphasizing subfactors to me, though. They should be related to boundary conditions for conformal nets of local operator algebras, which is why they are factors.

14. Ben Webster - March 26, 2008

Urs,

My point wasn’t that you should have waited. If you write a comment and it doesn’t appear instantly, it’s caught in spam, where we may find it, or may not. It will be found and posted much faster if you tell us it is caught in the spam filter.

Anyways, it’s the spam filter who should be apologizing.

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