Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

One of the most important observations in the representation theory of algebras is that if one has a subalgebra $A\subset B$, then these is a functor $B\otimes_A -\colon A\operatorname{-pmod}\to B\operatorname{-pmod}$ called induction.

This is a special case of a notion for representations of a category. (The representations of a category are the functors from that category into vector spaces.) In this case, one replaces the category with its idempotent completion (which doesn't change representations, since vector spaces are idempotent complete), which we'll assume for now is compactly generated. In this case, the induction of a representation $V$ is $$\operatorname{Ind} V(Y)=V(X)\otimes_{\operatorname{End}_A(X)}\operatorname{Hom}_B(X,Y).$$


Now, I can just as well talk about representations of 2-categories, which are 2-functors to the 2-category of categories (with whatever additions you like; additive and enriched over a field would be good choices).

Has anyone written up the basics of induction in this situation?

I don't strictly need to state things this way for what I want to do, but it would be better to give credit if someone else has done it and to have things "work out of the box."

share|improve this question
add comment

1 Answer

What's going on here formally in the category case is that the inclusion $A \subset B$ extends to a map $f: A \to [B^{\operatorname{op}}, \mathcal{V}]$ (where $\mathcal{V}$ is the category you're enriching over) via the Yoneda embedding. The Yoneda embedding is the free cocompletion of a category (i.e., the completion under weighted colimits), so this map extends to a cocontinuous functor $\widehat{f}: [A^{\operatorname{op}}, \mathcal{V}] \to [B^{\operatorname{op}}, \mathcal{V}]$, given by the weighted colimit formula you indicated. This functor is right adjoint to $\operatorname{Hom}(f-, -): [B^{\operatorname{op}}, \mathcal{V}] \to [A^{\operatorname{op}}, \mathcal{V}]$, which is just the restriction functor in this case. (What I've said works more generally for any such $f$, not just one induced by an inclusion of categories.)

Everything I have said should extend more or less identically to the setting of bicategories. A pseudofunctor between bicategories gives you an induction pseudofunctor between the corresponding presheaf categories by a weighted bicolimit formula analogous to the weighted colimit formula in the 1-categorical setting. This should be right biadjoint to the obvious restriction pseudofunctor, although I haven't checked the details (and I don't know if this is written up anywhere).

The situation you are interested in seems a bit more delicate, because you want to work in the setting where your hom-categories are enriched. To formulate all of this properly, one would need a notion of "weak enrichment," which generalizes the theory of bicategories. I've heard that there are people working on this, but I don't know of any available sources yet.

Note that while in principle every "weakly $\mathcal{V}$-$\operatorname{Cat}$-enriched category" may be equivalent to a category enriched in the underlying 1-category $\mathcal{V}$-$\operatorname{Cat}$ (and indeed, you may only be interested in the latter kinds of categories), the $\mathcal{V}$-$\operatorname{Cat}$-enriched setting probably isn't flexible to properly handle the kind of "representations" you would like to handle.

Edit: I really just wanted to fix the spelling of "principle" above, but as long as I'm bumping a two-year-old question, I might as well link to a recent paper by Garner and Shulman that develops the theory of bicategories enriched in a monoidal bicategory. In particular, they show that the bicategory of (certain) modules over an enriched bicategory is the free cocompletion under (certain) weighted colimits, which allows you to describe induction as a weighted colimit as in the 1-categorical setting.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.