Has anyone proved that if you take an "algebra" (actually monoid) object in a monoidal biclosed category that has equalizers and coequalizers, then the category of two-sided modules for that algebra is again a monoidal biclosed category. Mac Lane did this in 1965 when everything is symmetric (the tensor, the algebra and the modules) under the (surely irrelevant) assumption that the original category is also abelian. The fact is certainly true, but writing down the proof would be rather painful. Michael Barr
Hello, Regarding Michael Barr's question: I have this result for bi-closed bicategories, and actually talked about it in Halifax last year. Als Micheal rightly suspects, some of the proofs get tedious, and designing the relevant diagrams is something else, even using xy-pic! I hope that the paper will be finished by March. Best regards, -- J"urgen Koslowski Institut f"ur theoretische Informatik TU Braunschweig Braunschweig, Germany
Has anyone proved that if you take an "algebra" (actually monoid) object in a monoidal biclosed category that has equalizers and coequalizers, then the category of two-sided modules for that algebra is again a monoidal biclosed category. Mac Lane did this in 1965 when everything is symmetric (the tensor, the algebra and the modules) under the (surely irrelevant) assumption that the original category is also abelian. The fact is certainly true, but writing down the proof would be rather painful.
The question made me think - as perhaps it was meant to - of modules over non-commutative rings, and I wondered whether its scope was unnecessarily restricted in considering just 2-sided modules over a single algebra. We also know that there can be a rich and well-behaved structure for bimodules over two algebras: (A,B)-bimod tensor (B,C)-bimod is (A,C)-bimod B (A,B)-bimod hom (A,C)-bimod is (B,C)-bimod A (A,B)-bimod hom (C,B)-bimod is (C,A)-bimod B Does anyone know how to state the question to cover this more general context? But then what works for algebras ought also to work for algebroids, i.e. enriched categories. Steve Vickers.
Date: Wed, 17 Jan 1996 10:00:03 -0500 From: Michael Barr <barr@triples.math.mcgill.ca> - - Date: Wed, 17 Jan 1996 12:22:32 +0000 - From: Steve Vickers <sjv@doc.ic.ac.uk> - - >Has anyone proved that if you take an "algebra" (actually monoid) object - >in a monoidal biclosed category that has equalizers and coequalizers, - >then the category of two-sided modules for that algebra is again a - >monoidal biclosed category. Mac Lane did this in 1965 when everything - >is symmetric (the tensor, the algebra and the modules) under the (surely - >irrelevant) assumption that the original category is also abelian. The - >fact is certainly true, but writing down the proof would be rather - >painful. - - The question made me think - as perhaps it was meant to - of modules over - non-commutative rings, and I wondered whether its scope was unnecessarily - restricted in considering just 2-sided modules over a single algebra. We - also know that there can be a rich and well-behaved structure for bimodules - over two algebras: - - (A,B)-bimod tensor (B,C)-bimod is (A,C)-bimod - B - - (A,B)-bimod hom (A,C)-bimod is (B,C)-bimod - A - - (A,B)-bimod hom (C,B)-bimod is (C,A)-bimod - B - - Does anyone know how to state the question to cover this more general - context? But then what works for algebras ought also to work for - algebroids, i.e. enriched categories. - - Steve Vickers. - - - - Actually, if the ground tensor is not symmetric, you cannot even define the notion of (A,B) bimodule. You can define that of left A, right B bimodule and that seems to be that. Perhaps that is what Steve means when he says (A,B) bimodule. In that case, he is, of course, correct. In fact, in writing this up, I find it less confusing to do the more general situation exactly as described. But I would rather not have to write it up if someone has already done it as it is rather unpleasant. Michael
In response to Michael Barr's question: Theorem: If V is a closed braided monoidal category which is complete and cocomplete then the bicategory V-Mod of V-categories, V-modules (sometimes called V-bimodules, V-distributors or V-profunctors), and V-module morphisms is a monoidal bicategory (meaning the hom of a tricategory with one object). However, in order for V-Mod to be braided, V must be symmetric in which case V-Mod is also symmetric (also called strongly involutory by Baez-Dolan). If you are only interested in monoids in V, just take the subbicategory of one-object V-categories. Of course, then, you can do with less completeness and cocompleteness on V. But yes, the detailed proof of this makes a long, but fairly routine, story. Brian Day and I are working on a paper "Monoidal bicategories and Hopf algebroids" which will contain a discreet amount of detail (along with other things). I have been talking about aspects of the paper in our Category Seminars. Regards, Ross
participants (5)
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categories -
koslowj＠iti.cs.tu-bs.de -
Michael Barr -
Ross Street -
Steve Vickers