This is a roadmap entry for the field of Poisson geometry.

## Idea

Poisson geometry is, roughly speaking, the study of *Poisson manifolds* and their *deformation quantization*.

A Poisson manifold is foliated by immersed symplectic leaves, thus can be regarded as a degenerate/singular generalisation of a symplectic space. It captures semiclassical aspects - for example the superselection rules - of a quantum system and is hence closely related to the theory of operator algebras and noncommutative geometry. It is also related to higher geometry via diverse approaches such as the dg-geometry (AKSZ-BV) picture, the theory of Lie algebroids and groupoids and an important aspect that is inherited from the ’noncommutative’ side of Poisson manifolds: Morita theory.

The notion of Poisson manifold has a further generalization via Dirac structure where the notion of twisted Poisson manifold is subsumed.

## Basics

### Poisson algebra

Spelled out, the bracket is $\{\cdot,\cdot\}:A\otimes A \to A$, s.t for all $a,b,c\in A$,

- $\{a,b\} = -\{b,a\}$,
- $\{a,\{b,c\}\} + \text{cycl.} = 0$,
- $\{a,bc\} = b\{a,c\} + \{a,b\} c$.

The Leibniz rule is a compatibility relation between the associative and Lie products. The algebra $A$ does not need to be commutative, but it usually is. If $A$ is unital with $1$, from Leibniz rule $\{a,b\} = \{a,b\cdot 1\} = \{a,b\}\cdot 1 + a\cdot\{b,1\}$ we have $\{1,\cdot\} = 0$.

#### Subalgebra, ideal, homomorphism theorem

The notion of Poisson morphism, Poisson subalgebra are obvious. A Poisson ideal $I\subset A$ of a Poisson algebra $A$ is an ideal of the underlying associative algebra of $A$, such that $\{a,i\} \in I$ for all $a\in A,i\in I$.

As usual, For $\phi:A\to B$ Poisson, $\text{Ker}(\phi)$ is a Poisson ideal in $A$, and $\text{Im}\phi$ is a Poisson subalgebra in $B$. There is an exact sequence $$ 0\to \text{Ker}\phi \to A\to \text{Im}\phi \to 0. $$

#### Casimirs, canonical and Hamiltonian endomorphisms

*Casimir*if $\\{c,a\\} = 0$ for all $a\in A$.

The set of all Casimir elements in $A$ is the center of the Lie algebra underlying $A$. This will be denoted by $\text{Cas}(A) = \text{Ker}(f\mapsto {f,\cdot})$.

*canonical*if it is a derivation w.r.t. the associative product and the bracket. This means

- $X(fg) = (Xf)g + f(Xg)$
- $X\{f,g\} = \{Xf,g\} + \{f,Xg\}$

The set of canonical endomorphisms will be denoted by $\text{Can}(A)$. There is a special class of canonical endomorphisms,

Canonical endormorphisms of this form are called *Hamiltonian endomorphisms*. They form a set, denoted by $\text{Ham}(A) = \text{Im}(f \mapsto \{f,\cdot\})$.

Let $\text{Der}(A)$ be the set of derivations of the associative algebra underlying $A$. Obviously we have $$ \text{Ham}(A) \subseteq \text{Can}(A) \subseteq \text{Der}(A). $$ Furthermore, it can be easily verified that $\text{Ham}(A)$ is a Lie ideal in $\text{Can}(A)$ and a Lie subalgebra of $\text{Der}(A)$.

#### Poisson module

- $\{\{f,g\}_A,m\}_M = \{f,\{g,m\}_M\}_M - \{g,\{f,m\}_M\}_M$,
- $\{fg,m\}_M = f\cdot\{g,m\}_M + g\cdot\{f,m\}_M$,
- $\{f,g\cdot\}_M = \{f,g\}_A\cdot m + g\{f,m\}_M$.

### Poisson manifolds

Morphism of Poisson manifolds is defined in a straightforward manner.

Hamiltonian endomorphisms are called Hamiltonian vector fields and canonical endomorphisms are called canonical vector fields in the manifold setting, this is because the map $f\mapsto X_f = \{f,\cdot\}$ takes values in $\text{Der}(C^\infty(M)) = \Gamma(TM)$.

We have a short exact sequence $$ 0\to \text{Cas}(M) \to C^\infty(M)\to \text{Ham}(M) \to 0. $$

Usually the Poisson structure is encoded in a bivector field.

#### The sharp map and distribution

By currying $\Pi: T^\ast M \otimes T^\ast M \to \mathbb{R}$ becomes $\Pi^\sharp: T^\ast M \to TM$ which induces a map on sections $\Pi^\sharp: \Gamma(T^\ast M)=\Omega^1(M) \to \Gamma(TM) : \alpha \mapsto \Pi(\alpha,\cdot)$. This called the *sharp map*.

- On exact 1-forms. from $(\Pi^\sharp df)dg = \Pi(df,dg) = \{f,g\} = X_f g = \iota_{X_f} dg$ and the fact that $X_f$ is uniquely determined by its contractions with exact 1-forms (essentially with the exterior derivative of local coordinates), we have $\Pi^\sharp(df) = X_f$, the Hamiltonian ($X_f = \{f,\cdot\}$).
- The image of the sharp map is pointwisely $\text{Im}\Pi^\sharp = \text{Ham}_x (M)$ the vector subspaces of $T_x M$.

Recall that the assignment of a vector subspaces $S_x \subseteq T_xM$ for every $x\in M$ is called a (generalized) *distribution* (generalized: it is not required that $\text{dim}S_x M$ is constant in $x$). A distribution is *differentiable* if for all $x_0 \in M$ and $X_{x_0} \in S_{x_0}$ there is a neighbourhood $U$ of $x_0$ and a smooth vector field $V\in \Gamma(TU)$, s.t. $V_y \in S_y$ for all $y\in U$ and $V_{x_0} = X_{x_0}$.

*characteristic distribution*.

The number $\text{dim} \text{Im}\Pi^\sharp_x$ is called the *rank* of the Poisson manifold at $x\in M$. There is a map $\rho: M\to \mathbb{Z}: \rho(x) = \text{dim}\text{Im}\Pi^\sharp_x$ which is also called the *rank*. The rank is locally the rank of the matrix $\Pi_{ij}(x) = \Pi(x_i,x_j)$. Actually $\rho: M\to 2\mathbb{Z}$ since the rank of real antisymmetric matrices are even.

*regular*. If for $x_0 M$ there is a neighbourhood $U_{x_0}$ for all $y\in U_{x_0}$ s.t. $\rho(y) = \rho(x_0)$, the point $x_0$ is called a

*regular point*of $M$, and a

*singular point*otherwise.

For a Poisson manifold with singular points, $\text{Im}\Pi^\sharp$ will not be a subbundle of $TM$. In fact $\text{Im}\Pi^\sharp$ is a subbundle iff the rank is constant.

A regular Poisson manifold is the Poisson manifold associated to a symplectic manifold.

The set of regular points of $M$ is open and dense but not necessarily connected. To prove that it is dense one needs to prove that given a singular point $x_0 \in U_{x_0}\subset M$, there is a $y\in U\subset U_{x_0}$ that is regular. Say such $y$ is not regular, then there is $y_1 \in U$ s.t. $\rho(y_1) > \rho(y)$. If $y_1$ is not regular, repeat. Then there is a sequence $\{y_i\}\subset U$ of which $\rho(y_{i+1})>\rho(y_i)$. $M$ is locally compact, so $U$ can be chosen such that the closure $\bar{U}$ is compact, hence there is a converging subsequence of $\{y_i\}$, and the limit point of the subsequence will be regular.