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?)

  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Jun 1 '11 at 17:42
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    $\begingroup$ 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.) $\endgroup$ – Theo Johnson-Freyd Jun 1 '11 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 $\varpi$. $\endgroup$ – Theo Johnson-Freyd Jun 1 '11 at 17:54
  • $\begingroup$ (where $2\pi = 2\varpi$) $\endgroup$ – Theo Johnson-Freyd Jun 1 '11 at 17:57
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    $\begingroup$ 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 $F:Top\to Top$. $\endgroup$ – Tom Goodwillie Jun 1 '11 at 22:52

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