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Aug 28, 2011 at 2:13 vote accept Qiaochu Yuan
Aug 16, 2011 at 12:22 comment added Theo Johnson-Freyd In Todd Trimble's comment to the main question, he recalls the correct buzzword "coisotropic calculus". The foundational paper seems to be Alan Weinstein, Coisotropic calculus and Poisson groupoids, J. Math. Soc. Japan Vol. 40, No. 4, 1988, projecteuclid.org/euclid.jmsj/1230129807 .
Aug 16, 2011 at 6:29 comment added Dima Shlyakhtenko @Qiaochu: One instance in which all this is relatively easy to understand is the case of integrable Poisson manifolds. As Poisson algebras they have the form $C^\infty(G)^G$ where $G$ is a symplectic groupoid acting on itself by, say, left translation. A module of the Poisson algebra is then a module over the associated Lie algebroid, and, in good cases, the action integrates to that of the symplectic groupoid. Similarly for bimodules. The relative tensor product of bimodules is then the natural notion of a relative fibered product of spaces with groupoid actions.
Aug 16, 2011 at 2:49 comment added Qiaochu Yuan @Theo: I don't have a good sense of what that looks like in the greater generality of Poisson algebras. (Does it still make sense in that generality?)
Aug 15, 2011 at 21:01 comment added Theo Johnson-Freyd Actually, the proof of coisotropy should have something to do with the correct version of "Poisson reduction" akin to symplectic reguction. I'll try to dig up the appropriate Alan Weinstein paper.
Aug 15, 2011 at 21:00 comment added Theo Johnson-Freyd @Qiaochu: At its most basic, the composition is just that of correspondences: if you have $N \to M \times M'$ and $N' \to M' \times M''$, then you can form $N \times_{M'} N' \to M \times M' \times M'' \to M \times M''$. Except that in manifolds, fibered products like this are not well-defined. Spelling it out, what has to happen is that $N \times N' \to M \times M' \times M' \times M''$ must intersect transversally with the diagonal map $M \times M' \times M''\to M \times M' \times M' \times M''$. If the intersection is transverse, then the composition is coisotropic if $N,N'$ are.
Aug 15, 2011 at 20:08 comment added Qiaochu Yuan Thanks! This sounds like a pretty good choice. Can you explain briefly what the composition law is in this category?
Aug 15, 2011 at 19:59 history answered Theo Johnson-Freyd CC BY-SA 3.0