## January 3, 2006

### Bimodules, Adjunctions and the Internal Hom, Part II

#### Posted by urs 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 $\mathrm{Cat}$ 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 $C$ be some monoidal category. I’ll furthermore need to assume that $C$ 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 $C$ 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 $M$ over $C$ is defined to be essentially a module in $\mathrm{Cat}$ over the (semi)ring object $C$ in $\mathrm{Cat}$.

Hence there is a functor

(1)$l:C×M\to M$

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 $C$.

Let $A$ be an algebra object in $C$ and let $m$ be a right module over $A$ in $C$. Then, clearly, for any object $V\in C$ the object $V\otimes m$ is again canonically a right module over $A$.

Fixing some algebra object $A$, we have the category ${\mathrm{Mod}}_{A}\left(C\right)$ of right $A$ modules in $C$. By the above process, there is a left action of $C$ on ${\mathrm{Mod}}_{A}\left(C\right)$ and hence ${\mathrm{Mod}}_{A}\left(C\right)$ is a module category over $C$.

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 $C$ is semisimple and rigid and if $M$ 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 ${\mathrm{Mod}}_{A}\left(C\right)$ for some algebra $A$ in $C$

(2)$M\simeq {\mathrm{Mod}}_{A}\left(C\right)\phantom{\rule{thinmathspace}{0ex}}.$

Now, we know that two algebra objects $A$ and $A\prime$ in $C$ are (essentially by definition) Morita equivalent iff their module categories are equivalent

(3)$A\stackrel{\mathrm{Morita}}{\simeq }A\prime \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⇔\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\mathrm{Mod}}_{A}\left(C\right)\simeq {\mathrm{Mod}}_{A\prime }\left(C\right)\phantom{\rule{thinmathspace}{0ex}}.$

Hence it follows (under the above stated assumptions) that specifying a (semisimple indecomposable) module category $M$ of a monoidal catgeory $C$ is the same as specifying a Morita class of algebra objects in $C$.

Now what does this have to do with ‘splitting $A$ 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 $C$) the functor

(4)$\mathrm{IHom}\left(Y,Z\right):Z↦{Z}^{Y}$

which maps (if it exists) any object $Z$ to the exponential object which is the internal version of the ‘space of morphisms from $Y$ to $Z$’ is right-adjoint to the functor

(5)$--\otimes Y:X↦X\otimes Y$

meaning that

(6)$\mathrm{Hom}\left(X\otimes Y,Z\right)\simeq \mathrm{Hom}\left(X,{Z}^{Y}\right)\phantom{\rule{thinmathspace}{0ex}}.$

Let me here write

(7)$\mathrm{IHom}\left(Y,Z\right):={Z}^{Y}$

and call this object in $C$ the internal hom from $Y$ to $Z$.

We are interested in the internal hom $\mathrm{IHom}\left({m}_{1},{m}_{2}\right)\in C$ between objects ${m}_{1},{m}_{2}\in M$, i.e. in an object $\mathrm{IHom}\left({m}_{1},{m}_{2}\right)\in C$ such that

(8)${\mathrm{Hom}}_{C}\left(X\otimes {m}_{1},{m}_{2}\right)\simeq {\mathrm{Hom}}_{C}\left(X,\mathrm{IHom}\left({m}_{1},{m}_{2}\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

By construction, the internal hom $\mathrm{IHom}\left({m}_{1},{m}_{2}\right)$ behaves pretty much like a real space of homomorphism. In particular, there is an associative composition morphism

(9)$\mathrm{IHom}\left({m}_{2},{m}_{3}\right)\otimes \mathrm{IHom}\left({m}_{1},{m}_{2}\right)\to \mathrm{IHom}\left({m}_{1},{m}_{3}\right)$

in $C$.

But this means that given any (nonzero) object $m\in M$, we get an internal algebra object $A\in C$ by setting

(10)$A=\mathrm{IHom}\left(m,m\right)\phantom{\rule{thinmathspace}{0ex}}.$

Even better, fixing that $m\in M$, every $\mathrm{IHom}\left(m,\stackrel{˜}{m}\right)$ becomes an internal right module for this $A$. Hence there is a functor

(11)$\begin{array}{ccccc}F& :& M& \to & {\mathrm{Mod}}_{A}\left(C\right)\\ & & \stackrel{˜}{m}& ↦& \mathrm{IHom}\left(m,\stackrel{˜}{m}\right)\end{array}$

from our arbitrary module category to a module category which is a category of right $A$-modules in $C$. 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 $A$ was realized by two adjoint $A$-modules ${}_{k}{L}_{A}$ and ${}_{A}{R}_{k}$ as

(12)$A\simeq {}_{k}{L}_{A}{\otimes }_{A}{}_{A}{R}_{k}\phantom{\rule{thinmathspace}{0ex}}.$

If we pick $m={L}_{A}$ then clearly $\mathrm{IHom}\left(m,m\right)$ should be nothing but ${}_{k}{L}_{A}{\otimes }_{A}{}_{A}{R}_{k}$.

It’s sort of obvious – but I haven’t checked it.

Posted at January 3, 2006 3:17 PM UTC

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