Yesterday I discussed the Poisson bracket on a symplectic manifold and some of its basic properties. Naturally enough, someone decided to axiomatize all this.

Poisson manifolds

So, a Poisson manifold is a smooth manifold ${M}$ together with a Lie algebra structure ${\{\cdot, \cdot\}}$ (the Poisson bracket) on the space of smooth functions on ${M}$ such that

$\displaystyle \{ fg, h \} = f\{g, h\} + g \{f, h \}.$

To check that a symplectic manifold is indeed a smooth manifold, we need only recall that there is a vector field ${H_h}$ with ${\{ f, h \} = -H_h f}$, so the above is just the derivation identity. All the other properties of the Poisson bracket on a symplectic manifold were established yesterday.

Now let’s switch to the general Poisson manifold case.

It turns out that the Poisson structure alone is enough to show many similarities. Indeed, the derivation identity above implies that there is a vector field, still denoted by ${H_h}$, with

$\displaystyle H_h f = - \{ f, h \}$

and in particular it follows that ${H_f f = 0}$ by antisymmetry. The set of such vector fields ${H_h}$ will be called, as in the symplectic case, Hamiltonian vector fields. I claim that the identity

$\displaystyle \boxed{ [H_f, H_g] = H_{ \{f,g\}} }$

is still true. In the previous post we proved this directly in the symplectic case, and used it to deduce the Jacobi identity, but this time we will go in the opposite direction. The Jacobi identity implies for ${f,g, h \in C^{\infty}(M)}$:

$\displaystyle 0 = H_f H_g h - H_{\{f,g\}} h + H_g H_h f = (H_fH_g - H_g H_f - H_{ \{f,g\}})h .$

So we get a Lie isomorphism from ${C^{\infty}(M)}$ to the Lie algebra of Hamiltonian vector fields with the usual Lie bracket.

The cosymplectic structure

I now claim that it is possible to take a Poisson structure, and turn it into a cosymplectic structure. That is, we have a ${(0,2)}$ tensor ${\Lambda}$ such that ${\Lambda}$ is an antisymmetric bilinear map on each cotangent space ${T^*_x(M)}$. Note that ${\Lambda}$ is not required to be nondegenerate. This terminology admittedly seems confusing.

Note that a symplectic structure (which is required to be nondegenerate!) induces a cosymplectic structure by the natural isomorphisms ${T_x(M) \simeq T_x^*(M)}$ from duality. Anyway, to construct this structure, we take

$\displaystyle \Lambda( df, dg) := \{ f, g \},$

which at first appears ambiguous, but is in fact seen to be well-defined. What we have to do is prove that if ${f, \tilde{f}}$ are smooth functions with the same differential at the point ${x}$, then

$\displaystyle \{f - \tilde{f}, g \} = 0.$

The analogous identity in the other variable follows by antisymmetry. But this follows because

$\displaystyle \{f - \tilde{f}, g \} = -H_g(f - \tilde{f}) = - d(f-\tilde{f})H_g = 0.$

Morphisms of Poisson manifolds

A Poisson map ${\phi}$ between Poisson manifolds ${M_1, M_2}$ is one that commutes with the Poisson structures. In other words, if ${f,g}$ are smooth on ${M_2}$, then

$\displaystyle \{ f, g \} \circ \phi = \{ f \circ \phi, g \circ \phi \},$

i.e. that the map ${C^{\infty}(M_2) \rightarrow C^{\infty}(M_1)}$ is a Lie algebra homomorphism. Perhaps there is an analogy here to the notion of ${\phi}$-relatedness. Vector fields ${X_1, X_2}$ on ${M_1, M_2}$ respectively are said to be ${\phi}$-related if for every ${p \in M_1}$,

$\displaystyle \phi_*(X_1(p)) = X_2(\phi(p)).$

It is a basic fact that if ${X_1, X_2}$ are respectively ${\phi}$-related to ${Y_1, Y_2}$, then ${[X_1, X_2]}$ is ${\phi}$-related to ${[Y_1, Y_2]}$. Here it is functions that are related by ${\phi}$ and the Poisson bracket that is preserved. (Ok, maybe this was a bit much, but I am a fan of analogies.)

Products of Poisson manifolds

Given Poisson manifolds ${M_1, M_2}$, I claim that there is a unique way to define a Poisson structure on the product manifold ${M = M_1 \times M_2}$ such that the projections ${p_1, p_2}$ are Poisson maps. Moreover, we have injections of Lie algebras

$\displaystyle p_1^*: C^{\infty}(M_1) \rightarrow C^{\infty}(M), \ p_2^*: C^{\infty}(M_2) \rightarrow C^{\infty}(M)$

We want the images to commute. So, we first define the Poisson bracket on ${ C^{\infty}(M_1) \times C^{\infty}(M_2)}$—it is the product Lie algebra. Then we note that given ${x \in M}$ and ${f \in C^{\infty}(M)}$, we can choose ${a \in C^{\infty}(M_1), b \in C^{\infty}(M_2)}$ such that ${a+b}$ and ${f}$ have the same differential at ${x}$. Moreover the differentials of ${a,b}$ at ${x}$ are uniquely determined by this. Thus, we use this near-splitting of ${f}$ into ${a,b}$ to get the full Poisson structure on the product.

The splitting theorem

First, we define the rank of the Poisson structure at a point ${x \in M}$. This is the dimension of the largest subspace of ${T_x^*(M)}$ such that the bilinear form ${\Lambda}$ is nondegenerate.

The splitting theorem, due to Weinstein, describes Poisson manifolds locally:

Theorem 1 (Splitting theorem) Let ${M}$ be a Poisson manifold and ${x \in M}$. Then there is an open neighborhood ${U}$ of ${x}$ Poisson-isomorphic to a product ${S \times N}$ for ${S}$ symplectic and ${N}$ Poisson of rank zero at (the inverse image of) ${x}$.

This will be proved by induction on ${n = \dim M}$. When ${n=1}$, the form ${\Lambda}$ is identically zero, and the result is immediate.

Assume the theorem proved for dimensions smaller than ${n}$. Now there are two cases. One is that the form ${\Lambda}$ vanishes at ${x}$, in which case we can take ${S = \{x\}, N = M}$.

The more interesting one is that ${\Lambda}$ does not vanish on ${T_x^*(M)}$. We consider this next. So we can find a smooth function ${f}$ with

$\displaystyle H_f(x) \neq 0.$

We can straighten out ${H_f}$ by choosing an appropriate coordinate system ${y^1, \dots, y^n}$ so that ${H_f = \frac{\partial}{\partial y^1}}$. Then writing ${g=y^1}$, we have

$\displaystyle \{f,g\} \equiv 1.$

We are going to use ${f,g}$ to obtain a new, better system of local coordinates. First, ${H_f}$ and ${H_g}$ are linearly independent in some neighborhood of ${x}$ because ${H_f f = 0, H_f g = 1}$. Also

$\displaystyle [H_f, H_g]=H_{\{f,g\}} = H_1 = 0,$

so we can choose a set of local coordinates ${x^1, \dots, x^n}$ with

$\displaystyle H_f = \frac{\partial}{\partial x^1}, H_g = \frac{\partial}{\partial x^2}.$

I now claim that ${f,g, x^3, \dots, x^n}$ are also a system of local coordinates. Indeed,

$\displaystyle \frac{\partial g}{\partial x_1} = H_f g \neq 0, \frac{\partial f}{\partial x_2} \neq 0.$

This system of coordinates has the important property that

$\displaystyle \{ f, x^i \} = \{g, x^i \} = 0 \ \mathrm{if} \ i \geq 3$

because these equal respectively ${H_f x^i, H_g x^i}$. Let the corresponding coordinate neighborhood be ${A \times B}$, where ${A}$ corresponds to the ${x^i}$ being fixed and ${B}$ to ${f,g}$. Then ${A,B}$ are open neighborhoods in euclidean spaces.

Now ${A,B}$ have Poisson structures on them; indeed, they are determined by the values on the coordinate system. The “important property” above shows that ${C^{\infty}(A), C^{\infty}(B)}$ commute.

Moreover, on ${A}$ we have the Poisson structure associated to the usual symplectic structure on ${\mathbb{R}^2}$. This is because the brackets on the local coordinates ${f,g}$ determine the whole structure, again.

Now we apply the inductive hypothesis to ${B}$, and take the product of ${A}$ with the symplectic portion of ${B}$ (or some neighborhood thereof) as the ${S}$, and the other factor as ${N}$.

There is a uniqueness theorem as well, which I may or may not get to tomorrow.