Bimodules, Adjunctions and the Internal Hom, Part II
Posted by Urs Schreiber
I continue where I left off yesterday.
Last time I reviewed some aspects of how every Frobenius algebra object in a monoidal category comes from an adjunction (which is a generalization by Aaron Lauda of a classical result in to arbitrary 2-categories). One important point (for me at least) was that and how the 2-category of bimodules makes an appearance in this process.
There is another way to sort of ‘split an algebra in two adjoined halfs’. This, too, involves categories of modules showing up. It can most probably be related by some general nonsense to Aaron Lauda’s way of telling the story, but I don’t quite see yet how exactly.
What I am talking about is theorem 1 in
V. Ostrik
Module Categories, weak Hopf Algebras and Modular Invariants
math.QA/0111139.
Let again be some monoidal category. I’ll furthermore need to assume that is abelian and has a couple of further properties to be listed later.
In an abelian monoidal category we can take direct products and direct sums of objects and morphisms. Hence any such is like a categorified (semi)ring. From this point of view, it is natural to consider categorified modules for such a categorified (semi)ring.
Accordingly, a module category over is defined to be essentially a module in over the (semi)ring object in .
Hence there is a functor
satisfying the usual properties of a usual left action up to coherent isomorphisms.
A straightforward example for a module category is the category of right modules over algebra objects in .
Let be an algebra object in and let be a right module over in . Then, clearly, for any object the object is again canonically a right module over .
Fixing some algebra object , we have the category of right modules in . By the above process, there is a left action of on and hence is a module category over .
It doesn’t hurt to pause for a moment and digest the different levels at which the notion of a module appears here and how one level gives rise to the other.
Now, the interesting point is that for many interesting cases, the above example for a module category is already essentially the only example, up to equivalence.
More precisely, there is the following theorem:
If is semisimple and rigid and if is semisimple and indecomposable (either guess what this means or look it up in the above paper)…
…then it is equivalent to a category of right modules for some algebra in
Now, we know that two algebra objects and in are (essentially by definition) Morita equivalent iff their module categories are equivalent
Hence it follows (under the above stated assumptions) that specifying a (semisimple indecomposable) module category of a monoidal catgeory is the same as specifying a Morita class of algebra objects in .
Now what does this have to do with ‘splitting in two halfs’?
The answer is in the proof of the above theorem. This works as follows.
Recall that in a category with binary products (like the tensor product in the monoidal category ) the functor
which maps (if it exists) any object to the exponential object which is the internal version of the ‘space of morphisms from to ’ is right-adjoint to the functor
meaning that
Let me here write
and call this object in the internal hom from to .
We are interested in the internal hom between objects , i.e. in an object such that
By construction, the internal hom behaves pretty much like a real space of homomorphism. In particular, there is an associative composition morphism
in .
But this means that given any (nonzero) object , we get an internal algebra object by setting
Even better, fixing that , every becomes an internal right module for this . Hence there is a functor
from our arbitrary module category to a module category which is a category of right -modules in . Proving the above theorem amounts to proving that this functor actually extends to an equivalence of categories.
So there is a close similarity here to the constructions discussed in the previous entry. Recall that there, the internal algebra was realized by two adjoint -modules and as
If we pick then clearly should be nothing but .
It’s sort of obvious – but I haven’t checked it.