I realized that I am very confused about a certain sign in the definition of a Poisson group. I will give some definitions, and then point out my confusion.


Definitions
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### Group objects


Let $\mathcal C$ be a category with Cartesian products. Recall that a *group object* in $\mathcal C$ is an object $G \in \mathcal C$ along with chosen maps $e: 1\to G$ and $m: G\times G \to G$ (choose an initial object $1$ and a particular instance of the categorical product, and they imply all the others), such that (i) the two maps $G^{\times 3} \to G$ agree, (ii) the three natural maps $G\to G$ agree, and (iii) the map $p\_1 \times m: G^{\times 2} \to G^{\times 2}$ is an isomorphism, where $p\_1$ is the "project on the first factor" map $G^{\times 2}\to G$.


You may be used to seeing axiom (iii) presented slightly differently. Namely, if $p\_1 \times m: G^{\times 2} \to G^{\times 2}$ is an isomorphism, then consider the map $i = p\_2 \circ (p\_1\times m)^{-1} \circ (e\times \text{id}) : G = 1\times G \to G$. It satisfies the usual axioms of the inverse map. Conversely, if $i: G\times G$ satisfies the usual axioms, then $p\_1 \times (m \circ (i\times \text{id}))$ is an inverse to $p\_1 \times m$. I learned this alternate presentation from [Chris Schommer-Pries](https://mathoverflow.net/users/184/).


### Poisson manifolds


A *Poisson manifold* is a smooth manifold $M$ along with a smooth bivector field, i.e. a section $\pi \in \Gamma(\wedge^2 TM)$, satisfying an axiom. Recall that if $v\in \Gamma(TM)$, then $v$ defines a (linear) map $C^\infty(M) \to C^\infty(M)$ by differentiating in the direction of $v$. Well, if $\pi \in \Gamma(\wedge^2 TM)$, then it similarly defines a map $C^\infty(M)^{\wedge 2} \to C^\infty(M)$. The axiom states that this map is a Lie brackets, i.e. it satisfies the Jacobi identity.


A *morphism of Poisson manifolds* is a smooth map of manifolds such that the induced map on $C^\infty$ is a Lie algebra homomorphism.


[The category of Poisson manifolds has products](http://en.wikipedia.org/wiki/Poisson_manifold#Product_manifold) (Wikipedia).


### Poisson groups


A *Poisson Group* is a manifold $G$ with a Lie group structure $m : G\times G \to G$ and a Poisson structure $\pi \in \Gamma(\wedge^2 TG)$, such that $m$ is a morphism of Poisson manifolds. Recall that a *Lie group* $G$ is a group object in the category of smooth manifolds.


Recall also that a Lie group is almost entirely controlled by its Lie algebra $\mathfrak g = T\_eG$. Then it is no surprise that the Poisson structure can be described infinitesimally. Indeed, by left-translating, identify $TG = \mathfrak g \times G$. Consider the adjoint action of $G$ on the abelian Lie group $\mathfrak g$. Then we can define a Lie group structure $\mathfrak g^{\wedge 2} \rtimes G$ on $\wedge^2 TG$. Recall that a section $\pi \in \Gamma(\wedge^2 TG)$ is just a manifold map $G \to \wedge^2 TG$ that splits the projection $\wedge^2 TG \to G$. Then a Poisson manifold $(G,\pi)$ is a Poisson group if and only if $\pi : G \to \wedge^2 TG$ is a map of Lie groups.


Thus, a Poisson group structure is precisely the same as a Lie algebra $d\pi : \mathfrak g \to \wedge^2 \mathfrak g \rtimes \mathfrak g$ splitting the obvious projection (here $\wedge^2 \mathfrak g$ is an abelian Lie algebra, and $\mathfrak g$ acts on it via the adjoint action), and such that $(d\pi)^\* : \wedge^2\mathfrak g^\* \to \mathfrak g^\*$ satisfies the Jacobi identity. (Any failure of $G$ to be simply-connected, which might prevent such a map from lifting, also fails in $\wedge^2 TG$, so this really is a one-to-one identification of Poisson group structures on $G$ and "Lie bialgebra" structures on $\mathfrak g$.)


From this perspective, then, it is more or less clear that [the inverse map $i:G\to G$ is not a morphism of Poisson manifolds](http://en.wikipedia.org/wiki/Poisson%E2%80%93Lie_group) (Wikipedia). Indeed, infinitesimally, $di = -1: \mathfrak g \to \mathfrak g$, which takes $d\pi$ to $-d\pi$ (as $d\pi$ has one $\mathfrak g$ on the left and two on the right). Instead, $i$ is an "anti-Poisson map". The monoid $(\mathbb R,\times)$ acts on the category of Poisson manifolds by doing nothing to the underlying smooth manifolds and rescaling the Poisson structures; a smooth map is *anti-Poisson* if it becomes Poisson after twisting by the action of $-1$.


The unit map $e: 1 \to G$, on the other hand, is Poisson; it follows from the axioms of a Poisson group that $\pi(e) = 0$, and the terminal object in the category of Poisson manifolds is $1 = \{\text{pt}\}$ with the trivial Poisson structure. ($C^\infty(\{\text{pt}\}) = \mathbb R$ can only support this Poisson structure.)


My question
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Suppose that $G$ is a Poisson group. Then $p\_1 \times m: G^{\times 2} \to G^{\times 2}$ is a Poisson map, and an isomorphism of smooth manifolds. Thus, I would expect that it is an isomorphism of Poisson manifolds. On the other hand, in the first section above I constructed the inverse map $i : G\to G$ out of this isomorphism and the other structure maps, all of which are Poisson when $G$ is a Poisson group. And yet $i$ is not a Poisson map. So where am I going wrong?



Is a Poisson Group a group object in the category of Poisson Manifolds?
I think the problem is that the product of Poisson manifolds is not actually a categorical product. This is due to the fact (if I remember correctly) that two Poisson maps $f: X \to Y$ and $g: X \to Z$ give a Poisson map $f \times g: X \to Y \times Z$ only when the images of $f^\*$ and $g^\*$ Poisson commute in $C^\infty(X)$. In particular, $p\_1 \times m$ doesn't seem to be Poisson.


