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This is probably straightforward, but I'm having trouble writing down a precise statement. "Everyone knows" that the cobordism category $\text{2Cob}$ (all manifolds compact and oriented) is the free symmetric monoidal category on a commutative Frobenius object. What is the analogous statement for $\text{1Cob}$?

It looks something like the free symmetric monoidal category on an object with a (left and right) dual, but I'm not sure if I'm interpreting the orientation on points correctly.

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How is it symmetric monoidal? i don't think it there is a cobordism from 2 points to one point that is not a "projection". – Sean Tilson Feb 6 '11 at 4:53
Sean, it's symmetric monoidal under disjoint union. The cobordism you're talking about is part of "Frobenius" for the 2cob story, not "symmetric monoidal." – Gabriel C. Drummond-Cole Feb 6 '11 at 5:04
thanks GCDC! (blank space) – Sean Tilson Feb 6 '11 at 5:18
I would have said "the free symmetric monoidal 1-category on a dualizable object", but you seem to think that this isn't quite right? How's your site-specific google-fu? Have you tried searching the TWF archives? See also . For the unoriented category, you might say it's "the free symmetric monoidal 1-category on a self-dual object". As you move up the ladder, you'll also have to start thinking about other questions, like framed versus oriented versus ... – Theo Johnson-Freyd Feb 6 '11 at 6:20
@Theo: ah. I wasn't sure enough about all the definitions on that page, but I think I know what "stable" means now. – Qiaochu Yuan Feb 6 '11 at 12:05

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up vote 4 down vote accepted

Qiaochu, yes, but you don't need to say "left and right dual" because a left dual is a right dual in a symmetric monoidal category. It would be enough to say "with a left dual", or say it as Theo did.

An equivalent way of describing it is "the free compact closed category generated by a single object". For some information on free compact closed categories, there is an old paper by Kelly and LaPlaza, Coherence for compact closed categories, Journal of Pure and Applied Algebra 19 (1980), pp. 193–213. The description in terms of 1-cobordisms is implicit in that paper.

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Thanks. Could you clear up a minor issue about orientations? At first I thought it would be sensible to regard the positively oriented point as an object and the negatively oriented point as its dual, but the straight-line cobordism appears to only go from a negatively oriented point to a positively oriented point. In other words, there doesn't seem to be an identity morphism. Or am I misinterpreting orientation here? – Qiaochu Yuan Feb 6 '11 at 12:22
You might want to think of morphisms $S \to T$ as in bijection with morphisms $I \to -S + T$, where $I$ is the unit (represented by the empty 0-manifold). Then all the components or edges of the latter morphism do go from $-$ to $+$. The general rule is that an edge between points on opposite sides of the morphism go from $+$ to $+$ or $-$ to $-$, and go between points of opposite orientation if they are on the same side of the morphism. – Todd Trimble Feb 6 '11 at 13:34
@Todd: ah, I see. That makes sense. – Qiaochu Yuan Feb 6 '11 at 14:42

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