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

One way to categorify the non-negative integers is to consider the category FinSet whose objects are finite sets and whose morphisms are functions. The isomorphism classes of objects in FinSet can be labeled "sets of cardinality 0, sets of cardinality 1," and so forth, so are a natural way of talking about the non-negative integers.

What I want is a good definition of a category FinSSet (where SSet stands for "super set") whose isomorphism classes can be thought of as "sets of cardinality k" for all integers k in a natural way. A natural candidate for the set of objects is the set of "Z2-graded sets," i.e. pairs of sets (S0, S1). What we want is to define the cardinality of such a set as card(S0) - card(S1), so we want the isomorphism classes of objects to only depend on this number.

Unfortunately, I can't find a definition of the morphisms that actually accomplishes this. What should it be? For what it's worth, I've read "From Finite Sets to Feynman Diagrams" and I think John Baez gives up a little too early on the integers.

share|improve this question

4 Answers 4

up vote 5 down vote accepted

Not sure if you've seen this already, but it looks like Baez talks about this in one of his "This Week's Finds" columns, where he shows that the morphisms are given by tangles.

share|improve this answer
    
Perfect! I've read that column but I somehow managed to forget the point of that construction. –  Qiaochu Yuan Oct 30 '09 at 23:35
1  
Hmm. The categorification you get if you want the morphisms to be invertible is actually extremely boring. That's unfortunate. –  Qiaochu Yuan Oct 31 '09 at 3:00
1  
Do you mean that it's boring because every morphism (tangle) ends up being an isomorphism? If so, I wonder if tangles can be generalized so that this is not the case. (For example, one might relax the requirement that each point pairs with exactly one other point.) It would be nice to get the usual FinSet category in the cases where S1 is empty. –  Ari Oct 31 '09 at 15:09

Loeb has a paper called "Sets with a negative number of elements" (MR1144345), but I'm not convinced by it. My favorite categorification of the integers is Schanuel's paper "Negative sets have Euler characteristic and dimension" (MR1173024).

share|improve this answer
1  
I've read a lot of papers that cite Schanuel's, but I can't seem to actually track it down. Do you have a link? –  Qiaochu Yuan Oct 31 '09 at 0:21
1  
Sure. If you're at a university that pays for access to Springer archives, the paper is at: springerlink.com/content/h166711076633m7x/… otherwise, you can find my e-mail address by googling me at Berkeley (I don't want to post it lest more spammers find it) --- send me a note and I'll send you the paper. –  Theo Johnson-Freyd Oct 31 '09 at 3:55

For many years I am planning to try the following construction but never had enough time/energy to look at it carefully.

Take the object classifier $\mathscr E:=Set^{set_{fin}}$, and let $I$ be the generic object (the embedding of finite sets into sets). There is a triple of adjoint endofunctors $I\times(\_)\dashv(\_)^I\dashv\nabla_I(\_)$, induced by $(\_)+1:set_{fin}\to set_{fin}$; in particular these give an essential geometric endomorphism $E:\mathscr E\to\mathscr E$.

We now can either look at the $E$-completion or at the $E$-localization of $\mathscr E$, i. e. either at the colimit or at the limit of $$ \cdots\to\mathscr E\xrightarrow[]{E}\mathscr E\xrightarrow[]{E}\mathscr E\to\cdots; $$ what we get must contain an additive inverse of $(\_)+1$ in one sense or other...

share|improve this answer

You could also try some example similar to the loops on the circle — I don't know if it should be called categorification, perhaps you want to take a more advanced category, but I think the idea should be realizable.

share|improve this answer
    
Can you explain in more detail? –  მამუკა ჯიბლაძე Jan 7 at 6:53

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.