Poisson algebra

Remarks:

• $A$ needs not to be commutative.
• Every associative algebra $A$ can be made into a Poisson algebra by setting $\{f, g\}\equiv 0$.
• When $A$ is unital we get from the assumptions
  
  $$\{f, g\}=\{f, g\cdot 1\}=\{f,g\}\cdot 1+ g\cdot \{f, 1\},$$
  
  
  so $\{f, 1\}=0$ for all $f\in A$.

For Poisson morphism $\varphi\: A\to B$, $\ker \varphi$ is an ideal in $A$, $\im \varphi$ is a subalgebra in $B$, and there is an exact sequence of Poisson algebras

$$0\to \ker\varphi \to A\to \im \varphi\to 0.$$

The set of all Casimir elements in $A$ will be denoted by $\Cas(A)$, and set of canonical endomorphisms by $\Can(A)$.





With $\Der(A)$ we will denote the set of derivations of the associative algebra $A$. We have the following chain of inclusions.

$$\Ham(A)\subseteq \Can(A)\subseteq \Der(A).$$

Let us recall now that $\Der(A)$ is a Lie algebra with respect to the commutator of endomorphisms. $\Can(A)$ is a subalgebra of $\Der(A)$.







Remark: It is a definition of a flat connection when $M$ is the module of sections of a vector bundle. Indeed, when we denote

$$T\: M\to \Hom_{\bK}(A, M),\quad m\mapsto T_m:=\{\cdot, m\}_M$$

then

1. $\Longleftrightarrow T_m(\{f, g\}_A)=\{f, T_m(g)\}_M-\{g, T_m(f)\}_M$ (that is $T_m\in\Der((A, \{\cdot,\cdot\}_A); M)$),
2. $\Longleftrightarrow T_{f\cdot m}=f\cdot T_m(g)+ \{f,g\}_A\cdot m=f\cdot T_m(g)+X_f(g)\cdot m$,
3. $\Longleftrightarrow T_m(fg)=fT_m(g)+gT_m(f)$ (that is $T_m\in\Der((A, \cdot); M)$).

One may ask whether this is a reasonable definition of Poisson module. It is, in a sense, the categorical notion of Poisson bimodule as it verifies the so-called square-zero construction which can be summarized as follows: let $A$ be a Poisson algebra and $M$ Poisson $A$-module; define a Poisson algebra structure on $A\oplus M$ using formulas
$$\begin{align*} (f+m)\cdot(f_1+m_1) &:= ff_1+(f\cdot m_1 + f_1\cdot m),\\ \{f+m, f_1+m_1\} &:= \{f,f_1\}_A+\{f,m_1\}_M-\{f_1,m\}_M. \end{align*}$$