What reasonable choices of morphisms are there for the category of Poisson algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:09:54Z http://mathoverflow.net/feeds/question/72931 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72931/what-reasonable-choices-of-morphisms-are-there-for-the-category-of-poisson-algebr What reasonable choices of morphisms are there for the category of Poisson algebras? Qiaochu Yuan 2011-08-15T15:06:52Z 2011-08-15T19:59:18Z <p>The first definition of the category of Poisson algebras that comes to mind is that a morphism between Poisson algebras is an algebra homomorphism that is also a Lie algebra homomorphism with respect to the Poisson bracket. This definition does not seem to be easily compatible with how people actually use Poisson algebras (in particular rings of functions on Poisson manifolds):</p> <ul> <li>A <a href="http://en.wikipedia.org/wiki/Poisson%E2%80%93Lie_group" rel="nofollow">Poisson-Lie group</a> is not a group object in the opposite of the category of Poisson algebras because inversion negates the Poisson bracket.</li> <li>The <a href="http://en.wikipedia.org/wiki/Poisson_manifold#Product_manifold" rel="nofollow">standard choice of bracket</a> on the tensor product of two Poisson algebras is <a href="http://mathoverflow.net/questions/13454/is-a-poisson-group-a-group-object-in-the-category-of-poisson-manifolds" rel="nofollow">not a categorical coproduct</a> (if I have the correct general definition: it's defined by the requirements that it restricts to the given brackets on two Poisson algebras $A, B$ and that every element of $A$ Poisson-commutes with every element of $B$).</li> </ul> <p>This suggests to me that if we used a different choice of morphisms, we might get actual group objects and an actual categorical coproduct. So are there any nice choices that do this?</p> <p>I read somewhere on MO that the correct definition of a morphism between Poisson manifolds is a Lagrangian submanifold of their product. How does this generalize to Poisson algebras? Does it fix the two issues above? (I'm a little more pessimistic about the second issue, so if there's a different general principle that leads to the standard choice, I would be interested in hearing about that as well.) </p> <p><strong>Edit:</strong> The discussion in my <a href="http://mathoverflow.net/questions/66474/the-definition-of-a-group-object-is-wrong" rel="nofollow">previous question</a> about Poisson-Lie groups seems relevant, and perhaps it shows that the above point of view is misguided. Any Poisson algebra $A$ admits an "opposite" $A^{op}$ given by negating the Poisson bracket, and then inversion in a Poisson-Lie group is a "contravariant morphism" rather than a morphism. This suggests to me that it might make more sense to look for a <em>bicategory</em> of Poisson algebras similar to the bimodule bicategory. </p> http://mathoverflow.net/questions/72931/what-reasonable-choices-of-morphisms-are-there-for-the-category-of-poisson-algebr/72936#72936 Answer by Dima Shlyakhtenko for What reasonable choices of morphisms are there for the category of Poisson algebras? Dima Shlyakhtenko 2011-08-15T16:51:34Z 2011-08-15T16:51:34Z <p>I believe your objections could as well be raised for the category of algebras (note that Poisson algebras are in a certain way infinitesimal to algebras, since they encode first-order information about the deformation of the product of an algebra). For example, you will note that the inversion of the group reverses the order of products. There is also no natural notion of tensor products of algebra representations.</p> <p>One way to fix this in the category of algebras is to replace an algebra $M$ by its "enveloping algebra" $M\otimes M^o$ where $M^o$ is the opposite of $M$; in other words, to go from the category of $M$-modules to the category of $M\otimes M^o$ modules (i.e., $M,M$-bimodules). The same idea has been carried out to some extent in the Poisson category. See e.g. the book "Geometric models for noncommutative algebras" by Ana Cannas da Silva, Alan Weinstein. </p> <p>For example, the idea that morphisms between Poisson manifolds are related to Lagrangian submanifolds of a certain symplectic manifold is parallel to the idea from algebra that morphisms can be generalized by considering a bimodule with a preferred vector (the symplectic manifold in question is a bit more complicated than the product of the two Poisson manifolds, since that in general need not have any symplectic structure; it is related to the point that any Poisson manifold can be viewed as the quotient of a symplectic manifold by an action of a groupoid).</p> <p>However, this does not fix the problems you mention above (since, as I wrote, they are somehow the same in the category of algebras).</p> http://mathoverflow.net/questions/72931/what-reasonable-choices-of-morphisms-are-there-for-the-category-of-poisson-algebr/72948#72948 Answer by Theo Johnson-Freyd for What reasonable choices of morphisms are there for the category of Poisson algebras? Theo Johnson-Freyd 2011-08-15T19:59:18Z 2011-08-15T19:59:18Z <p>As you suggest in your question and Todd Trimble mentions in a comment, one interesting choice of morphism between Poisson manifolds is that of a coisotropic correspondence: if $M, M'$ are Poisson manifolds, depending on exactly how you work you either think about coisotropic submanifolds in $\bar M \times M'$, or maps $N \to \bar M \times M'$ with coisotropic image, where $\bar M$ is the same manifold as $M$ but with the opposite Poisson structure (and I give $\bar M \times M'$ the product Poisson structure that you're rightly not fond of). Then it is a straightforward fact that a correspondence $N \subseteq M\times M'$ which is the graph of a smooth map $M \to M'$ is coisotropic in $\bar M \times M'$ iff the map is a Poisson map.</p> <p>Note that this all generalizes the category in which objects are symplectic manifolds and morphisms are Lagrangian correspondences --- then a correspondence that is the graph of a smooth function is the graph of a symplectomorphic open embedding iff it is Lagrangian. It also has just as many bad properties. Notably, only composition between generic morphisms is defined, as in the non generic case some intersections may not be transverse. So to make it into a category requires the same kind of $A_\infty$ work (or Wehrheim-Woodward method, or...). I know that some of Alan Weinstein's recent papers discuss this category.</p> <p>This category generalizes easily to the algebraic case that you ask about. Recall that an ideal in a Poisson algebra is <em>coisotropic</em> if it is a Lie subalgebra for the bracket (not necessarily a Lie ideal!), and that a submanifold of a Poisson manifold is coisotropic iff its vanishing ideal is coisotropic. So what I'm suggesting is that if $P,P'$ are Poisson algebras, and writing $\bar P$ for $P$ with the opposite Poisson structure, then one interesting notion of "morphism" $P \to P'$ is a coisotropic ideal in $\bar P \otimes P'$.</p> <p>Dima Shlyakhtenko has suggested more or less the same category in another answer. There is the following philosophy: Poisson manifolds / algebras are a sort of "infinitesimal" piece of noncommutative algebra, and under this rough relationship coisotropic submanifolds are supposed to correspond to (left, say) modules. Then coisotropic correspondences are roughly the same as bimodules. Recall that from an algebra point of view, bimodules are a fairly natural notion of morphism: they are precisely the left adjoints (say, or right adjoints, or adjunctions) between the corresponding categories of modules. The module theory of an algebra knows a lot about the algebra, including its Hochschild homology and cohomology (and hence its center, its perturbative deformation theory, and so on).</p> <p>Of course, it is far from the case that the tensor product of algebras has much to do with the (co)product in any category. Rather, remembering only the Morita theory of algebras helps to explain what is their tensor product: it is the <em>tensor product</em> in the 2-category of (nice) categories with left-adjoints as morphisms, in the sense of being universal for "bilinear" maps. One can be quite precise about this: the 2-category of algebras and bimodules is a categorification of the 1-category of abelian groups. Actually, if you remember the underlying algebra, then that's the same as remembering its module theory along with the data of a "rank-1 free module", and so this is a categorification of the 1-category of abelian groups with a distinguished element. (Morita theory is like linear maps that ignore the distinguished element.)</p> <p>Incidentally, it is now straightforward to invent the notion of "sesquialgebra", which is an algebra object in the 2-category of algebras and bimodules, or equivalently a closed monoidal category structure on the module theory of said algebra. The same notion in Poisson manifolds is an algebra object in the category of Poisson manifolds and coisotropic correspondences, so this includes the Poisson Lie monoids. Alan Weinstein and collaborators a few years ago tried to write down a good notion of "Hopfish algebra" for controlling when this map would be invertible, but my opinion is that their paper doesn't quite get it right. What you should do is the following. Recall that a functor between monoidal categories is strong-monoidal if it comes equipped with a natural isomorphism between the two ways of composing the functor and the corresponding monoidal structures (and maybe extra data for associativity, etc.). A strong monoidal functor between closed monoidal categories also determines a natural transformation between "inner homs", which need not be a natural iso. If it is, call the monoidal functor "hopfish" or "strongly closed". A <em>bialgebra</em> is a sesquialgebra with a marked right adjoint to Vect (equivalently, a marked "rank 1 free algebra", the image of the 1-dimensional vector space under the corresponding left adjoint) which is strong monoidal; a <em>Hopf algebra</em> is a bialgebra in which the strong monoidal functor is hopfish.</p>