MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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.

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 (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 (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

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?

share|cite|improve this question
I'll just note that discussion of this at tea several years ago spawned a theory of categories with structure reversing maps, termed "heteromorphisms." – Ben Webster Jan 30 '10 at 3:49
Your multiplication map is backwards. – S. Carnahan Aug 8 '10 at 13:32
@Scott: Oops, fixed. – Theo Johnson-Freyd Aug 8 '10 at 18:13
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.