5
$\begingroup$

This is a follow-up to The definition of a group object is wrong?. The basic setup is as follows. Let $C$ be a category with finite products, $S : C \to D$ a product-preserving faithful functor, and $G \in C$ be a monoid object. It sometimes happens that $G$ has inverses in the sense that there exists a morphism $S(i) : S(G) \to S(G)$ making $S(G)$ a group object in $D$, but that this morphism does not lift to a morphism in $C$.

Example: A Poisson-Lie group with nontrivial Poisson bracket is not a group object in the category of Poisson manifolds $C$. However, it is a group object in the category $D$ of smooth manifolds; the inverse map negates the Poisson bracket, so does not lift to a Poisson map. (Not an example; see the comments.)

In the above question, Ryan Reich suggested two suitably weak notions of group object (in which the inverse is required to be a morphism) which allow the above example. Here they are, with ad hoc names for the sake of discussion.

A $D$-virtual group object in $C$ is a monoid object $G \in C$ together with a morphism $S(i) : S(G) \to S(G)$ such that $S(G)$ is a group object in $D$ with $S(i)$ as the inverse.

Let $F : C \to C$ be a functor such that $SF \cong S$. An $(F, D)$-virtual group object in $C$ is a monoid object $G \in C$ together with a morphism $i : G \to F(G)$ such that $S(G)$ is a group object in $D$ with $S(i)$ as the inverse.

The first definition is the most general one suggested by the basic setup, but the second definition suffices, at least, for the case of Poisson manifolds, where $F$ negates the Poisson bracket.

Question 1: Is every $D$-virtual group object actually an $(F, D)$-virtual group object for some $F$?

I expect the answer to be "no," but I don't know how I would go about constructing a counterexample. A basic observation is that $S(i)^2 = \text{id}_{S(G)}$, so the types of morphisms that can actually occur in $D$ as inverses are somewhat restricted: while they are "virtual morphisms" (Ben Webster suggested the term heteromorphism) in $C$, they must square to "real morphisms." A sufficiently good theory of heteromorphisms might make it possible to construct $F$ given $C, D, S$.

Various follow-up questions suggest themselves.

Question 2: Given $C, D, S$ as above, is there a unique $F$ such that $C$ admits $(F, D)$-virtual group objects which don't lift to group objects in $C$?

Question 3: Given a category $C$, how can I tell if it admits $D$-virtual group objects for some $D$ which don't lift to group objects in $C$?

Question 4: Can we describe heteromorphisms in a manner internal to $C$? (For example, if $C = \text{Cat}$, it seems a heteromorphism ought to be a contravariant functor. Here we can take $D$ to be the category of graphs, $S : C \to D$ the obvious forgetful functor, and $F$ the opposite category functor. Did we need to do this, or does the notion of contravariant functor naturally fall out of the structure of $C$ in some way?)

$\endgroup$
9
  • $\begingroup$ It seems that "virtual group" is also the name of a concept due to Mackey. Anyone have any other suggestions? I can't think of anything else particularly snappy. $\endgroup$ Jun 1, 2011 at 17:42
  • 2
    $\begingroup$ I think "product" is not the word you want in the first paragraph. For example, given two Poisson manifolds, there is a canonical Poisson structure on the product of underlying manifolds. But the thus-constructed Poisson manifold is not the categorical product of the first two Poisson manifolds in the category of Poisson manifolds. (This is a semiclassical shadow of the fact that the tensor product of commutative rings is their coproduct in commutative rings, but the tensor product of noncommutative rings is not a coproduct.) $\endgroup$ Jun 1, 2011 at 17:50
  • $\begingroup$ Indeed, you can see that the "Poisson product" is not the categorical product because there does not exist a diagonal map. Let $(M,\pi)$ be a Poisson manifold. The "product" $(M,\varpi)\otimes (M,\varpi)$ is the manifold $M\times M$ with Poisson structure $\varpi\oplus0\oplus\varpi$ (identifying $T^{\otimes 2}_{m,n}(M\times N) = T^{\otimes 2}_mM \oplus T_mM\otimes T_nN \oplus T^{\otimes 2}_nN$. A smooth map $f: M\to N$ is poisson if the two Poisson structures are "$f$-related", but the diagonal map $M \to M\times M$ relates $\varpi\oplus0\oplus\varpi$ with $2\pi$, not $\varpi$. $\endgroup$ Jun 1, 2011 at 17:54
  • $\begingroup$ (where $2\pi = 2\varpi$) $\endgroup$ Jun 1, 2011 at 17:57
  • 1
    $\begingroup$ Can't you make topological monoids that are groups but with discontinuous inverse? Like maybe an ordered group with a topology having the sets $(a,+\infty)$ as basis? It seems unlikely that there is a suitable functor $F:Top\to Top$. $\endgroup$ Jun 1, 2011 at 22:52

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.