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

There are two generalizations of usual groups: groupoids, where the multiplication operation becomes "partial", and hypergroups, for which the result of multiplying two elements is a probability measure rather than a single element. I am interested in the structure combining both these features (it might have been called "hypergroupoid", but apparently this term is already in use): multiplication is partial and obeys the same rules as in a groupoid, but its result is a probability measure on morphisms with the appropriate source and target objects. Any references?

share|improve this question

1 Answer 1

(This is more of an extended comment, but contains ideas that may help answer the question)

This might link up with informal ideas that Tim Porter has expressed (only in online forums, not in any formal setting that I know of). He posits the idea that instead of asking for existence fillers in horns for simplicial sets, so as to get Kan complexes, what if there is a probability distribution on the set of possible fillers? To recover your situation, take the nerve of the group considered as a one-object groupoid and this is a reduced Kan complex (reduced = one 0-simplex). That there is a group multiplication means that inner horns have unique fillers. (inverses means unique fillers for outer horns) It may be possible to find a mixture of these two ideas that recovers the definition of a hypergroup.

Then simply relax the condition that your simplicial set is reduced. It is well known that Kan complexes with unique fillers (and poss some other conditions I am too lazy to check) are nerves of groupoids. So this should give the concept you are looking for.

If there were references I'm fairly sure Tim would have mentioned them, but then again, maybe not. If you do find some I'm sure he'd like to hear from you.

share|improve this answer

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.