SF&PA: One more example April 9, 2008Posted by Noah Snyder in planar algebras, subfactors, Uncategorized.
Sorry for the delay, Scott’s been in town and so I’ve been too busy doing actual research to get much blogging done. This post was also a little delayed because I didn’t understand this example as well as I’d hoped. I still don’t fully grok it so maybe you all can help me out. If you don’t follow this example, don’t worry, we’ll be moving on to pictures in the next post.
Take a finite group G. Let C[G] denote the group algebra and let F(G) denote the algebra of functions on G with pointwise product (that is F(G) has a basis of elements of the form and is 0 unless in which case it is ). Recall that for both C[G] and F(G) there is a notion of tensor product for modules (for the former g acts on a tensor product via , whereas for the latter acts on a tensor product via
We define a bioidal category called the “group Subfactor” as follows
- The A-A objects are C[G]-modules
- The A-B objects are vector spaces (thought of as representations of the trivial group)
- The B-A objects are vector spaces (thought of as representations of the trivial group)
- The B-B objects are F(G)-modules
Now we need to define the four flavors of tensor products. All of these are of the following form: first push things around (using restriction/induction) until they live where the tensor product is supposed to live, then take the tensor product there.
For example if we take an A-B object (i.e. vector space) V and a B-A object (i.e. vector space) W, their tensor product should be a C[G]-module. To do this we first induce V and W up to G getting two C[G]-modules, then we take their tensor product as C[G]-modules. In particular, if V and W are 1-dimensional, then their tensor product is the regular representation. For another example, suppose we want to take the tensor product of an A-A object (that is a representation V) and an A-B object (that is a representation of the trivial). The answer is supposed to be an A-B object, so we first turn V into a vector space by restriction (forgetting the C[G]-module structure) and then we take the tensor product. One last example, we want to take the tensor product of two vector spaces and get a F(G)-module. So first we take their tensor product as Vector spaces, and then we push that up to an F(G)-module by tensoring with the regular.
This whole tensor product process is a bit more confusing than it ought to be. Can anyone figure out what’s really going on here?
In order to make this a Subfactor category, I also need to fix a simple A-B object called X which tensor-generates. That’s easy here since there’s only one such simple: the one-dimensional vector space. It is easy to see that this tensor generates as and arejust the regular representation of the corresponding ring and thus contains all simples.
What is the dimension of X? (I’m going to leave the definition of dimension vague for the moment.) The A-A part of the category is just the representation theory of G, so we know the dimensions of objects there. In particular, the regular representation has dimension #G. Hence, X has dimension .
Since I haven’t told you much about how actual Subfactor people think about Subfactors, let me sketch the construction in this case. Let N be the von Neumann completion of the countable tensor product of 2×2 matrix rings. N is a hyperfinite II_1 factor. Pick your favorite faithful action of G on a set, and use that action to give an outer action of G on N by permuting the tensor factors in N. Now consider the fixed points N^G = M. Because the action is outer you can prove that M is also a factor, and since G is finite the inclusion M<N is finite index. This is the group subfactor.