# Lectures on Poisson algebras

V. Roubtsov∗, R. Suchánek∙
∗<EMAIL_ADDRESS>∙<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
    1. 1.0.1 Acknowledgment
  2. 2 Motivation
    1. 2.1 Lagrangian and Hamiltonian mechanics
    2. 2.2 Hamiltonian mechanics and Poisson brackets
  3. 3 Poisson algebras
    1. 3.1 Subalgebras and ideals
    2. 3.2 Morphisms and derivations
  4. 4 Hamiltonian derivations and Casimirs
    1. 4.1 Exterior algebra of a commutative algebra
  5. 5 Homology and cohomology
    1. 5.1 Hochschild (co)homology
    2. 5.2 Lichnerowicz-Poisson cohomology
    3. 5.3 Low-dimensional Poisson cohomology
      1. 5.3.1 Compatible Poisson structures
      2. 5.3.2 Interpretation of $HP^{2}(\operatorname{\mathcal{A}})$
    4. 5.4 Poisson homology
    5. 5.5 Duality
  6. 6 Polynomial Poisson Algebras
    1. 6.1 Nambu-Jacobi-Poisson algebras
    2. 6.2 Poisson-Calabi-Yau algebra
      1. 6.2.1 Low-dimensional cohomology of the PCY algebra
    3. 6.3 Dual Poisson complex
  7. 7 Graded Poisson algebras
    1. 7.1 Algebra of differential operators
  8. 8 Intermezzo - Tensor, symmetric and exterior algebras
    1. 8.1 Tensor algebra of a vector space
    2. 8.2 Symmetric algebra of a vector space
    3. 8.3 Exterior algebra of a vector space
    4. 8.4 Poisson structure on a symmetric algebra $S(\operatorname{\mathfrak{g}})$
  9. 9 Universal enveloping and PBW theorem
    1. 9.1 Universal enveloping algebra
    2. 9.2 Poincaré-Birkhoff-Witt (PBW) theorem
    3. 9.3 Universal enveloping and differential operators
  10. 10 Poisson manifolds
    1. 10.1 Poisson structure on the cotangent bundle
    2. 10.2 Poisson manifolds
    3. 10.3 Hamiltonian mapping
    4. 10.4 Poisson bracket on a symplectic manifold
    5. 10.5 Examples of Poisson and Symplectic manifolds
    6. 10.6 Poisson manifolds and Lie theory
    7. 10.7 Symplectic foliation on $\operatorname{\mathfrak{g}}^{*}$
      1. 10.7.1 Coadjoint invariant functions
  11. 11 Differential Calculus on Poisson Manifolds
    1. 11.1 Coordinate-free construction of the Schouten bracket
  12. 12 Modified Double Poisson Brackets
    1. 12.1 Poisson Brackets for General Associative Algebras
      1. 12.1.1 Ginzburg-Voronov lemma
      2. 12.1.2 Representation scheme.
      3. 12.1.3 Moduli Space of Representations
    2. 12.2 Double Poisson Brackets
    3. 12.3 Quadratic double Poisson brackets
    4. 12.4 Examples and classification of low dimensional quadratic double Poisson brackets

##  1 Introduction

The notion of a Poisson algebra was probably introduced in the first time by
A.M. Vinogradov and J. S. Krasil’shchik in 1975 under the name ”canonical
algebra” and by J. Braconnier in his short note ”Algèbres de Poisson” (Comptes
rendus Ac.Sci) in 1977.

It was a natural ”algebraic interpretation” of the notion of Poisson structure
and Poisson brackets appeared in XIX century in the framework of Classical
Mechanics.

Nowadays, Poisson algebras have proved to have a very rich mathematical
structure (see the beautiful short article of Y. Kosmann-Schwarzbach in
”Encyclopedia of Mathematics” (Poisson algebra, Encyclopedia of Mathematics,
http://encyclopediaofmath.org/index.php?title=Poisson_algebra&oldid=51639).

There are many good books and lecture notes devoted to Poisson structures and
much more less sources concerning the algebraic side of the story. We should
mention the extensive book of C. Laurent-Gengoux, A. Pichereau and P.
Vanhaecke ”Poisson structures” which covers also many algebraic aspects and
structures associated with the notion of a Poisson variety [8].

These notes are slightly extended reproduction of the mini-course of the same
title (5 lectures) given by the first author during the virtual (online)
Winter School and Workshop ”Wisla 20-21” and handled and written by the
second.

The virtual character of lectures has imposed few constraints and has defined
a specific choice of material and chosen subjects. The authors had to pass
between Scylla of unknown audience level and tantalizing Charybdis of their
will to exit out of the standard set of content in numerous existed lecture
notes about Poisson structures in algebra and in geometry.

This contradiction can probably explain some strange ”jumps” of a difficulty
level between various chapters of our notes. We hope to come back to these
notes and to extend, to improve, or even, to write a new comprehensive book
covering the subject. Right now we were on a serious time (and the Covid
pandemic) limits pressure and it was also a reason of some non-homogeneous
choice of the lecture notes matters and its details.

There are (almost) no ”new results” in the lectures. The only exception is a
content of the last brief chapter where we provide our classification results
(joint with A. Odesskii and V. Sokolov) on the ”low rank” double quadratic
Poisson brackets. We had kept this chapter almost in it’s specific form of a
computer presentation.

The lectures are based on many various well- and less known sources, which we
had tried to carefully quot in the main body of the text. But we need to
specify most influential: the beautiful paper of A.M. Vinogradov and J.S.
Krasil’shchik [28] ”What is the hamiltonian formalism?”, lectures of P.
Cartier ”Some fundamental techniques in the theory of integrable systems” [9],
a short book of K. H. Bhaskara and K. Viswanath ”Poisson algebras and Poisson
manifolds” [5].

####  1.0.1 Acknowledgment

V. Rubtsov is thankful to the Institut des Hautes Études Scientifiques
Université Paris-Saclay for hospitality during the last stages of this work.
R. Suchánek is grateful to the University of Angers and to the Masaryk
University for their hospitality. The work of R. Suchánek was financially
supported under the project GAČR EXPRO GX19-28628X and by the Barrande
Fellowship program organized by The French Institute in Prague (IFP) and the
Czech Ministry of Education, Youth and Sports (MYES). Authors are also
grateful to the organizers of the Winter School and Workshop Wisla 20-21 for a
wonderful online event, organized during a difficult time of a covid pandemic.

##  2 Motivation

###  2.1 Lagrangian and Hamiltonian mechanics

A fundamental discovery of Lagrange: a _Lagrangian_ function $L\colon
TM\to\operatorname{\mathbb{R}}$

$\displaystyle L=T-V\ ,$

describes the motion of a particle. The equation of motion is described by the
Euler-Lagrange equation

$\displaystyle\frac{d}{dt}(\frac{\operatorname{\partial}L}{\operatorname{\partial}\dot{q^{i}}})=\frac{\operatorname{\partial}L}{\operatorname{\partial}q^{i}}\
.$

Here $(\bar{q})=(q^{1},\ldots,q^{n})$ are coordinates on $M$,
$(\bar{q},\bar{\dot{q}})=(q^{1},\ldots,q^{n},\dot{q}^{1},\ldots,\dot{q}^{n})$
are the induced coordinates on the tangent bundle $TM$. The above Euler-
Lagrange equation follows from the variational principle

$\displaystyle\delta\int L(\bar{q},\bar{\dot{q}})dt=0\ ,$

where $\delta$ (in this context) refers to the variational derivative. Using
the Legendre transformation, one can pass from the tangent bundle to the
cotangent bundle $T^{*}M$ with the coordinates $(\bar{q},\bar{p})$, and
reformulate the Euler-Lagrange equations equivalently by the Hamiltons
equations

$\displaystyle\dot{\bar{q}}=\frac{\operatorname{\partial}H}{\operatorname{\partial}\bar{p}},\
\ \
\dot{\bar{p}}=-\frac{\operatorname{\partial}H}{\operatorname{\partial}\bar{q}}$
(1)

The link between Lagrangian and Hamiltonian is given by

$\displaystyle
H(\bar{q},\bar{p})=<\bar{p},\bar{\dot{q}}>-L(\bar{q},\bar{\dot{q}})\ .$

###  2.2 Hamiltonian mechanics and Poisson brackets

Instead of starting with the Lagrangian $L$ function, one can start with the
Hamiltonian function $H\colon T^{*}M\to\operatorname{\mathbb{R}}$ and build
the mechanics independently of the Lagrangian. We now go to the simpler setup
of flat spaces, i.e. consider $M\cong\operatorname{\mathbb{R}}^{N}$. We will
get back to the setup of smooth manifolds later on.

Let $F,G$ be two differentiable functions on $\operatorname{\mathbb{R}}^{2n}$
with coordinates $(\bar{q},\bar{p})$, where
$\bar{p}=(p_{1},\dots,p_{n}),\bar{q}=(q_{1},\dots,q_{n})$. Introduce a new
functions as the result of the following skew-symmetric operation on $F$ and
$G$.

$\displaystyle\\{F,G\\}:=\sum_{k=1}^{n}(\frac{\operatorname{\partial}F}{\operatorname{\partial}q_{k}}\frac{\operatorname{\partial}G}{\operatorname{\partial}p_{k}}-\frac{\operatorname{\partial}G}{\operatorname{\partial}q_{k}}\frac{\operatorname{\partial}F}{\operatorname{\partial}p_{k}})=\frac{\operatorname{\partial}F}{\operatorname{\partial}\bar{q}}\frac{\operatorname{\partial}G}{\operatorname{\partial}\bar{p}}-\frac{\operatorname{\partial}G}{\operatorname{\partial}\bar{q}}\frac{\operatorname{\partial}F}{\operatorname{\partial}\bar{p}}\in
C^{\infty}(\operatorname{\mathbb{R}}^{2n})\ .$ (2)

This operations allows to write the Hamilton’s equations (1) associated with a
Hamiltonian function $H=H(\bar{q},\bar{p})$ in the following manner (putting
$\bar{q}$ and $\bar{p}$ on an equal footing):

$\displaystyle\dot{\bar{q}}=\\{\bar{q},H\\}\ ,\ \ \
\dot{\bar{p}}=\\{\bar{p},H\\}\ .$ (3)

Indeed we have

$\displaystyle\\{q_{i},H\\}=\sum_{k=1}^{n}(\frac{\operatorname{\partial}q_{i}}{\operatorname{\partial}q_{k}}\frac{\operatorname{\partial}H}{\operatorname{\partial}p_{k}}-\frac{\operatorname{\partial}H}{\operatorname{\partial}q_{k}}\frac{\operatorname{\partial}q_{i}}{\operatorname{\partial}p_{k}})=\frac{\operatorname{\partial}H}{\operatorname{\partial}p_{i}}\
,$
$\displaystyle\\{p_{i},H\\}=\sum_{k=1}^{n}(\frac{\operatorname{\partial}p_{i}}{\operatorname{\partial}q_{k}}\frac{\operatorname{\partial}H}{\operatorname{\partial}p_{k}}-\frac{\operatorname{\partial}H}{\operatorname{\partial}q_{k}}\frac{\operatorname{\partial}p_{i}}{\operatorname{\partial}p_{k}})=-\frac{\operatorname{\partial}H}{\operatorname{\partial}q_{i}}\
,$

since
$\frac{\operatorname{\partial}q_{i}}{\operatorname{\partial}q_{k}}=\frac{\operatorname{\partial}p_{i}}{\operatorname{\partial}p_{k}}=\delta^{i}_{k}$
and
$\frac{\operatorname{\partial}q_{i}}{\operatorname{\partial}p_{k}}=\frac{\operatorname{\partial}p_{i}}{\operatorname{\partial}q_{k}}=0$.

Note that for each solution $\\{(q(t),p(t)\\}$ of the Hamilton system (3) and
for any $F\in C^{\infty}(\operatorname{\mathbb{R}}^{2n})$

$\displaystyle\frac{d}{dt}F(\bar{q}(t),\bar{p}(t))$
$\displaystyle:=\sum_{k=1}^{n}(\frac{\operatorname{\partial}F}{\operatorname{\partial}q_{i}}\dot{q_{i}}+\frac{\operatorname{\partial}F}{\operatorname{\partial}p_{i}}\dot{p_{i}})$
$\displaystyle=\sum_{k=1}^{n}(\frac{\operatorname{\partial}F}{\operatorname{\partial}q_{i}}\frac{\operatorname{\partial}H}{\operatorname{\partial}p_{i}}-\frac{\operatorname{\partial}F}{\operatorname{\partial}p_{i}}\frac{\operatorname{\partial}H}{\operatorname{\partial}q_{i}})$
$\displaystyle=\\{F,H\\}(\bar{q}(t),\bar{p}(t))\ .$

In particular, for the Hamiltonian function $H$,

$\displaystyle\frac{d}{dt}H(\bar{q}(t),\bar{p}(t))=\\{H,H\\}(\bar{q}(t),\bar{p}(t))=0\
,$

which is the _conservation law_ for $H$ (i.e. $H(\bar{q}(t),\bar{p}(t))$ is
constant along the trajectories). D. Poisson had observed that if $\\{F,H\\}$
and $\\{G,H\\}$ vanish, then $\\{\\{F,G\\},H\\}$ vanishes (Poisson bracket of
two constants of motion is again a constant of motion). This statement can be
explained abstractly with the help of the _Jacobi identity_

$\displaystyle\\{\\{F,G\\},H\\}+\\{\\{G,H\\},F\\}+\\{\\{H,F\\},G\\}=0\ ,$ (4)

since if $\\{F,H\\}=\\{G,H\\}=0$, then the above equality gives
$\\{\\{F,G\\},H\\}=0$. Now let $\pi_{ij}$ be arbitrary smooth functions on
$\operatorname{\mathbb{R}}^{d}$, $x_{1},\dots,x_{d}$ cooridnates on
$\operatorname{\mathbb{R}}^{d}$. Define

$\displaystyle\\{F,G\\}:=\sum_{k=1}^{d}\pi_{ij}\frac{\operatorname{\partial}F}{\operatorname{\partial}x_{i}}\frac{\operatorname{\partial}G}{\operatorname{\partial}x_{j}}\
.$ (5)

Then (5) satisfies the Jacobi identity (4) if and only if

$\displaystyle\sum_{k=1}^{d}(\pi_{lk}\frac{\operatorname{\partial}\pi_{ij}}{\operatorname{\partial}x_{l}}+\pi_{li}\frac{\operatorname{\partial}\pi_{jk}}{\operatorname{\partial}x_{l}}+\pi_{lj}\frac{\operatorname{\partial}\pi_{ki}}{\operatorname{\partial}x_{l}})=0\
.$ (6)

Skew-symmetry and the Jacobi identity imply that the operation
$\\{\cdot,\cdot\\}$ defines a _Lie algebra_ structure on
$C^{\infty}(\operatorname{\mathbb{R}}^{d})$. Moreover this structure is
compatible with the product operation $\cdot$ on
$C^{\infty}(\operatorname{\mathbb{R}}^{d}),U\subset\operatorname{\mathbb{R}}^{d}$,
meaning that it satisfies the _Leibniz rule_

$\displaystyle\\{FG,H\\}=F\\{G,H\\}+\\{F,H\\}G\ .$

##  3 Poisson algebras

There are two ways to think about a Poisson structure (given by a Poisson
bracket) on a manifold $M$ (algebraic, smooth, analytic, etc.).

A geometric viewpoint. To each function $H$ (smooth, holomorphic, etc.) on
$M$, one can associate a vector field $X_{H}$, where $H$ is the Hamiltonian in
the mechanical interpretation. The vector field is given by

$\displaystyle X_{H}:=\\{-,H\\}\ .$ (7)

An algebraic viewpoint. Consider a vector space (typically infinite
dimensional) $\operatorname{\mathcal{A}}$, with two algebra structures:

  1. 1.

Structure of an associative and commutative algebra with a unit.

  2. 2.

A Lie algebra structure.

_The Poisson structure:_ commutative structure + Lie algebra structure +
compatibility conditions.

Extracting the algebraic aspects of the above ideas, one is lead to the
following definition.

######  Definition 3.1 (Poisson algebra).

Let $(\operatorname{\mathcal{A}},\cdot)$ be an associative, unital and
commutative $\operatorname{\mathbb{K}}$-algebra
($\operatorname{char}\operatorname{\mathbb{K}}=0$), with the unit denoted $1$.
Let

$\displaystyle\\{-,-\\}\colon\operatorname{\mathcal{A}}\times\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}$

be a Lie bracket defining a $\operatorname{\mathbb{K}}$-Lie algebra structure
on $\operatorname{\mathcal{A}}$. Then $\operatorname{\mathcal{A}}$ is called
_a Poisson algebra_ if the operations $\cdot$ and $\\{-,-\\}$ are compatible
in the following sense

$\displaystyle\\{a\cdot b,c\\}=a\cdot\\{b,c\\}+\\{a,c\\}\cdot b\ .$ (8)

The Lie bracket is then called _a Poisson bracket_.

When the context is clear, we will usually omit the $\cdot$ symbol of the
associative operation in $\operatorname{\mathcal{A}}$.

######  Remark 3.1.

Note that the Lie bracket $\\{-,-\\}$ in the above definition is not
necessarily given by the commutator. In fact, if it is given by the
commutator, i.e. $\\{a,b\\}=a\cdot b-b\cdot a$, then $\\{a,b\\}=0$ for all
$a,b\in\operatorname{\mathcal{A}}$, since the $\cdot$ is assumed to be a
commutative operation on $\operatorname{\mathcal{A}}$.

###  3.1 Subalgebras and ideals

######  Definition 3.2 (Subalgebras).

A vector subspaces
$\operatorname{\mathcal{B}}\subset\operatorname{\mathcal{A}}$ is _a Poisson
subalgebra_ in a Poisson algebra
$(\operatorname{\mathcal{A}},\\{-,-\\},\cdot)$ if
$\operatorname{\mathcal{B}}\cdot\operatorname{\mathcal{B}}subset\operatorname{\mathcal{B}}$
and
$\\{\operatorname{\mathcal{B}},\operatorname{\mathcal{B}}\\}\subset\operatorname{\mathcal{B}}$.

######  Definition 3.3 (Ideals).

A vector subspaces
$\operatorname{\mathcal{J}}\subset\operatorname{\mathcal{A}}$ is _a Poisson
ideal_ of a Poisson algebra $(\operatorname{\mathcal{A}},\\{-,-\\},\cdot)$ if
$\operatorname{\mathcal{J}}\cdot\operatorname{\mathcal{A}}\subset\operatorname{\mathcal{J}}$
and
$\\{\operatorname{\mathcal{J}},\operatorname{\mathcal{A}}\\}\subset\operatorname{\mathcal{J}}$.

######  Observation 3.1.

In the first definition, $\operatorname{\mathcal{B}}$ is itself a Poisson
algebra with operations given by restriction of operations in
$\operatorname{\mathcal{A}}$. The inclusion map
$\iota\colon\operatorname{\mathcal{B}}\subset\operatorname{\mathcal{A}}$ is a
Poisson algebra morphism. In the second definition, the quotient algebra
$\operatorname{\mathcal{A}}/\operatorname{\mathcal{J}}$ inherits a Poisson
bracket from $\operatorname{\mathcal{A}}$ by the requirement that the
canonical projection on the quotient
$p\colon\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}/\operatorname{\mathcal{J}}$
is a Poisson morphism.

######  Remark 3.2.

For a fixed field $\operatorname{\mathbb{K}}$
($\operatorname{char}\operatorname{\mathbb{K}}=0)$,
$\operatorname{\operatorname{Poiss}}_{\operatorname{\mathbb{K}}}$ denotes a
category, whose objects are the Poisson $\operatorname{\mathbb{K}}$-algebras,
and whose morphisms are $\operatorname{\mathbb{K}}$-morphisms of Poisson
algebras. Group object in
$\operatorname{\operatorname{Poiss}}_{\operatorname{\mathbb{K}}}$ is the
Poisson-Lie group $(G,\cdot,\\{-,-\\})$ s.t. the bracket on $G$ is
_multiplicative_ , i.e. the multiplication $\cdot G\times G\to G$ is a Poisson
morphism, where $G\times G$ is considered with a Poisson bracket given by
$\\{-,-\\}_{G\times G}:=\\{-,-\\}_{G}\oplus\\{-,-\\}_{G}$.

######  Definition 3.4 (Prime).

A Poisson ideal $\mathcal{P}\subset\operatorname{\mathcal{A}}$ is _Poisson
prime_ , if for all Poisson ideals
$\operatorname{\mathcal{I}},\operatorname{\mathcal{J}}\subset\operatorname{\mathcal{A}}$

$\displaystyle\operatorname{\mathcal{I}}\operatorname{\mathcal{J}}\subset\mathcal{P}\Rightarrow\operatorname{\mathcal{I}}\subset\mathcal{P}\text{
or }\operatorname{\mathcal{J}}\subset\mathcal{P}$

######  Remark 3.3.

If $\operatorname{\mathcal{A}}$ is _Noetherian_ 111Given an increasing
sequence of ideals in $\operatorname{\mathcal{A}}$,
$\operatorname{\mathcal{I}}_{1}\subseteq\operatorname{\mathcal{I}}_{2}\subseteq\ldots$,
there always exists $k\in\operatorname{\mathbb{N}}$ such that
$\operatorname{\mathcal{I}}_{k}=\operatorname{\mathcal{I}}_{k+n}$ for all
$n\in\operatorname{\mathbb{N}}$., then the above definition is equivalent to
$\mathcal{P}$ being both prime ideal and Poisson ideal.

######  Definition 3.5 (Spectrum).

The _Poisson spectrum of_ $\operatorname{\mathcal{A}}$, denoted
$\operatorname{\operatorname{PSpec}}(\operatorname{\mathcal{A}})$, is the set
of Poisson prime ideals of $\operatorname{\mathcal{A}}$.

######  Definition 3.6 (Maximal ideal).

A maximal ideal $\mathcal{M}\subset\operatorname{\mathcal{A}}$ is a _Poisson
maximal ideal_ if it is a Poisson ideal.

######  Remark 3.4.

A Poisson maximal ideal is not the same thing as a maximal Poisson ideal.

######  Definition 3.7 (Core).

The _Poisson core_ of an ideal
$\operatorname{\mathcal{I}}\subset\operatorname{\mathcal{A}}$, denoted
$\operatorname{\operatorname{PC}}(\operatorname{\mathcal{I}})$, is the largest
Poisson ideal of $\operatorname{\mathcal{A}}$ contained in
$\operatorname{\mathcal{I}}$.

######  Remark 3.5.

If $\operatorname{\mathcal{I}}$ is a prime ideal of
$\operatorname{\mathcal{A}}$, then
$\operatorname{\operatorname{PC}}(\operatorname{\mathcal{I}})$ is Poisson
prime.

######  Definition 3.8 (Primitive ideal).

A Poisson ideal $\mathcal{P}$ of $\operatorname{\mathcal{A}}$ is a _Poisson
primitive_ if $\mathcal{P}$ is the Poisson core of some maximal ideal
$\mathcal{M}$, i.e.
$\operatorname{\operatorname{PC}}(\mathcal{M})=\mathcal{P}$

######  Remark 3.6.

Each Poisson primitive ideal is Poisson prime.

######  Definition 3.9 (Localization).

Let $R$ be a commutative ring and $S\subset R$ a multiplicative
subset222$s_{1},s_{2}\in S\Rightarrow s_{1}s_{2}\in S$. A _localization of $R$
at $S$_, denoted $S^{-1}R$, is the commutative ring of equivalence classes in
$R\times S$ defined by

$\displaystyle(\tilde{r},\tilde{s})\in[r,s]\iff\exists k\in
S:(\tilde{r}s-r\tilde{s})k=0\ .$

One usually denotes the equivalence class $[r,s]$ by $\frac{r}{s}$ or by
$rs^{-1}$ (this reflects the similarity with the construction of the field of
fractions). The ring operations are given by

$\displaystyle\frac{\tilde{r}}{\tilde{s}}+\frac{r}{s}$
$\displaystyle:=\frac{\tilde{r}s+r\tilde{s}}{\tilde{s}s}$
$\displaystyle\frac{\tilde{r}}{\tilde{s}}\cdot\frac{r}{s}$
$\displaystyle:=\frac{\tilde{r}r}{\tilde{s}s}$

######  Exercise 3.1.

Show that the above construction indeed yields a ring structure on the set
$S^{-1}R$. Describe the identity elements with respect to both ring
operations.

If $S\subset\operatorname{\mathcal{A}}$ is a multiplicatively closed subset of
a Poisson algebra $\operatorname{\mathcal{A}}$, then the localization
$S^{-1}\operatorname{\mathcal{A}}$ is also a Poisson algebra with the brackets

$\displaystyle\\{as^{-1},bt^{-1}\\}:=(st\\{a,b\\}-sb\\{a,t\\}-at\\{s,b\\}+ab\\{s,t\\})(s^{2}t^{2})^{-1},$

for all $a,b\in\operatorname{\mathcal{A}}$ and all $s,t,\in S$. For any
Poisson ideal $\mathcal{P}\subset\operatorname{\mathcal{A}}$, the localization
$S^{-1}\mathcal{P}$ is a Poisson ideal in $S^{-1}\operatorname{\mathcal{A}}$.
For two multiplicative sets $S,T\subset\operatorname{\mathcal{A}}$ such that
$S\subset T$, there is a Poisson morphism $\colon
S^{-1}\operatorname{\mathcal{A}}\to T^{-1}\operatorname{\mathcal{A}}$.

######  Exercise 3.2.

Let $s,t\in\operatorname{\mathcal{A}}$ and consider the multiplicative subsets
of $\operatorname{\mathcal{A}}$

$\displaystyle S:=\\{s^{n}|\ n\in\operatorname{\mathbb{N}}\\}$ $\displaystyle
T:=\\{t^{n}|\ n\in\operatorname{\mathbb{N}}\\}\ ,$

where $s,t\in\operatorname{\mathcal{A}}$ are such that $t=su$ for apropriate
$u\in\operatorname{\mathcal{A}}$. Show that the map $\varphi_{s,t}\colon
S^{-1}\operatorname{\mathcal{A}}\to T^{-1}\operatorname{\mathcal{A}}$ defined
by $\varphi_{s,t}(\frac{a}{s^{m}}):=\frac{au^{n}}{t^{n}}$ is independent of
$u$ and is a Poisson morphism.

######  Example 3.1.

Any Poisson maximal ideal is maximal Poisson, but the converse is not true. A
countre-example is the Weyl-Poisson algebra $\operatorname{\mathbb{C}}[x,y]$,
with $\\{x,y\\}=1$. This is a simple Poisson algebra but not a simple
associative algebra (consider the trivial ideal $\\{0\\}$).

###  3.2 Morphisms and derivations

######  Definition 3.10 (Morphisms and isomorphisms).

Let $(\operatorname{\mathcal{A}}_{i},\\{-,-\\}_{i},\cdot_{i})_{i=1,2}$ be two
Poisson algebras over $\operatorname{\mathbb{K}}$ and let
$\varphi\colon\operatorname{\mathcal{A}}_{1}\to\operatorname{\mathcal{A}}_{2}$
be a $\operatorname{\mathbb{K}}$-linear map satisfying for all
$a,b\in\operatorname{\mathcal{A}}_{1}$

  1. 1.

$\varphi(a\cdot_{1}b)=\varphi(a)\cdot_{2}\varphi(b)$,

  2. 2.

$\varphi(\\{a,b\\}_{1})=\\{\varphi(a),\varphi(b)\\}_{2}$

Then $\varphi$ is called _a morphism of Poisson algebras_. If $\varphi$ is a
bijective morphism of Poisson algebras s.t. the inverse map
$\varphi^{-1}\colon\operatorname{\mathcal{A}}_{2}\to\operatorname{\mathcal{A}}_{1}$
is also a Poisson algebra morphism, then $\varphi$ is called _an isomorphism
of Poisson algebras_.

######  Definition 3.11 (Derivation).

A $\operatorname{\mathbb{K}}$-linear map
$\varphi\colon\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}$ is
called _a $\operatorname{\mathbb{K}}$-derivation_ on
$\operatorname{\mathcal{A}}$ if

$\displaystyle\varphi(a\cdot b)=a\varphi(b)+\varphi(a)\cdot b\ .$ (9)

The set of all derivations of $\operatorname{\mathcal{A}}$ is denoted by
$\mathcal{D}(\operatorname{\mathcal{A}})$,
$\mathcal{D}^{1}(\operatorname{\mathcal{A}})$ or
$\mathcal{D}(\operatorname{\mathcal{A}},\operatorname{\mathcal{A}})$. A
bilinear map
$\varphi\colon\operatorname{\mathcal{A}}\times\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}$
is called _a bi-derivation_ if

$\begin{gathered}\varphi(ab,c)=a\varphi(b,c)+b\varphi(a,c),\\\
\varphi(a,bc)=\varphi(a,b)c+b\varphi(a,c)\end{gathered}$ (10)

######  Observation 3.2.

Directly from the definition,
$\mathcal{D}(\operatorname{\mathcal{A}})\subset\operatorname{\operatorname{End}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})=\operatorname{\operatorname{Hom}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}},\operatorname{\mathcal{A}})$.
The $\operatorname{\mathbb{K}}$-linear map, given for every
$b\in\operatorname{\mathcal{A}}$, $a\mapsto\\{a,b\\}$ is a derivation on
$\operatorname{\mathcal{A}}$.

######  Observation 3.3.

The biderivation associated with derivation on $\operatorname{\mathcal{A}}$
given by a Poisson bracket is always skew-symmetric.

From now on, we will often abbreviate the notation and refer to a Poisson
algebra $(\operatorname{\mathcal{A}},\\{-,-\\},\cdot)$ simply by
$\operatorname{\mathcal{A}}$, if the context is clear. Also, all the vector
spaces, algebras (Poisson, Lie), and the corresponding linear maps (morphisms,
derivations), will be considered over a general field
$\operatorname{\mathbb{K}}$, with
$\operatorname{char}\operatorname{\mathbb{K}}=0$, unless stated otherwise.

##  4 Hamiltonian derivations and Casimirs

######  Definition 4.1 (Hamiltonian derivations and Casimirs).

Let $\operatorname{\mathcal{A}}$ be a Poisson algebra and
$a\in\operatorname{\mathcal{A}}$. The derivation
$X_{a}:=\\{-,a\\}\in\mathcal{D}(\operatorname{\mathcal{A}})$ is called a
Hamiltonian derivation with a Hamiltonian (or a Hamilton function) $a$
associated to $X_{a}$. We denote by

$\displaystyle\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}}):=\\{X_{a}\
|\ a\in\operatorname{\mathcal{A}}\\}$ (11)

the set of all Hamiltonian derivations. We have a linear map
$\operatorname{\mathcal{A}}\to\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$,
$a\mapsto X_{a}$. Let $a\in\operatorname{\mathcal{A}}$ be s.t. $X_{a}=0$, then
$a$ is called a _Casimir element_. We denote by

$\displaystyle\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}}):=\\{a\in\operatorname{\mathcal{A}}|X_{a}=0\\}$
(12)

the set of all Casimir elements.

######  Definition 4.2.

The (Poisson) _center of $\operatorname{\mathcal{A}}$_ is
$\operatorname{Z}_{P}(\operatorname{\mathcal{A}}):=\\{a\in\operatorname{\mathcal{A}}\
|\ \\{a,b\\}=0\text{ for all }b\in\operatorname{\mathcal{A}}\\}$. We have
$\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})=\operatorname{Z}_{P}(\operatorname{\mathcal{A}})$.

######  Definition 4.3.

_Augmentation_ of an (associative) $\operatorname{\mathbb{K}}$-algebra
$\operatorname{\mathcal{A}}$ is a $\operatorname{\mathbb{K}}$-algebra
homomorpshism
$\alpha\colon\operatorname{\mathcal{A}}\to\operatorname{\mathbb{K}}$. The pair
$(\operatorname{\mathcal{A}},\alpha)$ is called an _augmented algebra_. The
kernel $\ker\alpha$ is called the augmentation ideal of
$\operatorname{\mathcal{A}}$.

######  Lemma 4.1.

For arbitrary monic polynomial $p(x)\in\operatorname{\operatorname{Cas}}[x]$
and arbitrary $b\in\operatorname{\mathcal{A}}$

$\displaystyle\\{p(x),b\\}=p^{\prime}(x)\\{x,b\\}\ ,$ (13)

where
$p^{\prime}(x)=\frac{\operatorname{\partial}p(x)}{\operatorname{\partial}x}$.

###### Proof.

By induction. Suppose $\tau:=\deg p=0$. Then $p(x)=c_{0}$, where
$c_{0}\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$, and
hence for all $b\in\operatorname{\mathcal{A}}$

$\\{p(x),b\\}=\\{c_{0},b\\}=0=p^{\prime}(x)\\{x,b\\}\ .$

Let the induction hypothesis hold for monic polynomials of $\deg\leq\tau-1$
and suppose $\tau>0$. Then
$p(x)=x^{\tau}+c_{\tau-1}x{\tau-1}+\ldots+c_{1}x+c_{0}$, where all
$c_{i}\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$. We
can rewrite $p(x)=x^{\tau}+c_{\tau-1}q(x)$, where $q(x)$ is monic and $\deg
q=\tau-1$. Then

$\displaystyle\\{p(x),b\\}$
$\displaystyle=\\{xx^{\tau-1},b\\}+c_{\tau-1}\\{q(x),b\\}$
$\displaystyle=x^{\tau-1}\\{x,b\\}+x\\{x^{\tau-1},b\\}+c_{\tau-1}q^{\prime}(x)\\{x,b\\}$
$\displaystyle=(x^{\tau-1}+x(\tau-1)x^{\tau-2}+c_{\tau-1}q^{\prime}(x))\\{x,b\\}$
$\displaystyle=p^{\prime}(x)\\{x,b\\}\ .\qed$

######  Proposition 4.1.

Let $\operatorname{\mathcal{A}}$ be a Poisson algebra.

  1. 1.

$\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$ is a
subalgebra of $(\operatorname{\mathcal{A}})$. Moreover,
$\operatorname{\mathbb{K}}$ can be naturally identified with a subset of
$\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$.

  2. 2.

If $\operatorname{\mathcal{A}}$ is an integral domain (i.e. has no zero
divisors), then
$\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$ is integrally
closed in $\operatorname{\mathcal{A}}$.

  3. 3

Not every representation of $\operatorname{\mathcal{A}}$ on
$\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$ defines an
$\operatorname{\mathcal{A}}$-module structure on
$\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\subset\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$.

  4. 3.

$\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$ is a
$\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$-module.

  5. 4.

The map
$\varphi\colon\operatorname{\mathcal{A}}\to\mathcal{D}(\operatorname{\mathcal{A}})$,
defined by $H\mapsto-X_{H}$ is a morphism of Lie algebras.

  6. 5.

There is a short exact sequence of Lie algebras

$0\to\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})\to\operatorname{\mathcal{A}}\to\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\to
0\ .$

###### Proof.

  * 1.

That $\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$ is a
subalgebra of $\operatorname{\mathcal{A}}$ is obvious. Now we show that
scalars correspond to Casimir elements. Notice that since
$\operatorname{\mathcal{A}}$ is an algebra with the unit, $1$, and because
$\operatorname{\mathbb{K}}$ has an action on $\operatorname{\mathcal{A}}$
(denoted also by juxtaposition), there is a natural injective morphism
$\iota\colon\operatorname{\mathbb{K}}\to\operatorname{\mathcal{A}}$, given by
$k\mapsto k1$. This means that we can identify $\operatorname{\mathbb{K}}$
with
$\iota(\operatorname{\mathbb{K}})=\operatorname{\mathbb{K}}1\subset\operatorname{\mathcal{A}}$.
We will simply write $\iota(k)=k$.333In this notation, $ka$ can be interpreted
equivalently as action of $\operatorname{\mathbb{K}}$ on
$\operatorname{\mathcal{A}}$, or as a multiplication in
$\operatorname{\mathcal{A}}$ after identifying $\operatorname{\mathbb{K}}$
with $\iota(\operatorname{\mathbb{K}})$. Of course we have $k1=1k=k$. Now
consider arbitrary $a\in\operatorname{\mathcal{A}}$ and
$k\in\operatorname{\mathbb{K}}$. Then we have

$\\{a,k1\\}=\\{a,k\\}1+k\\{a,1\\},$

which implies that $k\\{a,1\\}=0$ (otherwise $\\{a,k1\\}\neq\\{a,k\\}$) and
hence $\\{a,1\\}=0$.444since for $a\neq 0:ka=0\Rightarrow k=0$ But if
$\\{a,1\\}=0$ then the $\operatorname{\mathbb{K}}$-bilinearity of $\\{-,-\\}$
implies $k\\{a,1\\}=\\{a,k\\}=0$. So arbitrary
$k\in\operatorname{\mathbb{K}}\cdot 1$ satisfies $\\{a,k\\}=0$ for all
$a\in\operatorname{\mathcal{A}}$, hence
$\operatorname{\mathbb{K}}1\subset\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$.

  * 2.

Suppose $a$ is an integral element over
$\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$, i.e.
$a\in\operatorname{\mathcal{A}}$ and there is a monic polynomial
$p(x)\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})[x]$ s.t.
$p(a)=0$. We need to show that
$a\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$. Without
loss of generality, suppose $p$ is the smallest degree polynomial s.t.
$p(a)=0$. If $\deg p=1$, then $p(x)=x-a$. Hence
$a\in\operatorname{\mathbb{K}}\subset\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$
by the previous statement. Suppose $\tau:=\deg p>1$. Using (13), for any
$b\in\operatorname{\mathcal{A}}$ we have

$0=\\{p(a),b\\}=p^{\prime}(a)\\{a,b\\}\ .$

Now $p^{\prime}(a)\neq 0$, otherwise there exists
$k\in\operatorname{\mathbb{K}}$ s.t. $\frac{p^{\prime}}{k}$ is a monic
polynomial, which satisfies $\frac{p^{\prime}}{k}(a)=0$ and
$\deg(\frac{p^{\prime}}{k})=\deg p^{\prime}=\tau-1$. But this would be a
contradiction with the assumption of $p$ being the smallest degree monic
polynomial with the property $p(a)=0$. Because $\operatorname{\mathcal{A}}$
has no zero divisors, $\\{a,b\\}=0$. As $b\in\operatorname{\mathcal{A}}$ is
arbitrary, this shows that
$a\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$.

  * 3

Consider the left action
$l\colon\operatorname{\mathcal{A}}\to\operatorname{\operatorname{End}}(\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})),x\mapsto
l_{x}$, where $l_{x}(a)$. Suppose $l$ defines an $A$-module structure on
$\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$. Then for any
$x\in\operatorname{\mathcal{A}}$ and arbitrary
$X_{a}\in\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$, we
have
$l_{x}X_{a}\in\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$.
This means there is $d\in\operatorname{\mathcal{A}}$ s.t. for every
$b\in\operatorname{\mathcal{A}}:l_{x}X_{a}(b)=X_{d}(b)$. But this is not true,
in general, since

$l_{x}X_{a}(b)=x\\{b,a\\}=\\{b,xa\\}-\\{b,x\\}a\ ,$

while

$X_{d}(b)=\\{b,d\\}\ .$

In other words,

$\displaystyle(l_{x}X_{a}-X_{d})(b)=0\iff d=xa\text{ and }\\{b,x\\}a=0\ ,$
(14)

for every $a,b,x\in\operatorname{\mathcal{A}}$, which is obviously not always
the case. Notice though, if we assume that
$x\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$, then
$\\{b,x\\}=0$ for every $b\in\operatorname{\mathcal{A}}$ and thus (14) can
always be satisfied. This shows that the action of
$\operatorname{\mathcal{A}}$ on
$\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$, if restricted
to $\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$, defines an
action on $\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$,
i.e. statement 4 follows.

  * 4.

See the proof of statement $3$.

  * 5.

By definition, $\varphi(a)(c)=-X_{a}(c)=-\\{c,a\\}$. We want to show that
$\varphi(\\{a,b\\})=[\varphi(a),\varphi(b)]$, where

$[\varphi(a),\varphi(b)]=[X_{a},X_{b}]:=X_{a}\circ X_{b}-X_{b}\circ X_{a}\ .$

From the Jacobi identity (4), we have for arbitrary
$a,b,c\in\operatorname{\mathcal{A}}$

$-\\{c,\\{a,b\\}\\}=\\{\\{c,b\\},a\\}-\\{\\{c,a\\},b\\}=X_{a}(X_{b}(c))-X_{b}(X_{a}(c))\
.$

Since $\varphi(\\{a,b\\})=-X_{\\{a,b\\}}=-\\{c,\\{a,b\\}\\}$, this shows that
$\varphi$ is a Lie algebra morphism.

  * 6.

The first map is an inclusion,
$\iota\colon\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})\to\operatorname{\mathcal{A}}$,
which is obviously a Lie algebra morphism. The second map is
$\varphi\colon\operatorname{\mathcal{A}}\to\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$,
and is given by $a\mapsto-X_{a}$. The previous statement shows that it is a
Lie algebra morphism. Now consider $a\in\operatorname{\mathcal{A}}$. Then
$X_{a}=0$ iff $\forall b\in\operatorname{\mathcal{A}}:X_{a}(b)=\\{b,a\\}=0$.
Thus $\ker\varphi=\operatorname{\operatorname{im}}\iota$, meaning that the
short sequence

${0}$${\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})}$${\operatorname{\mathcal{A}}}$${\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})}$${0}$$\scriptstyle{0}$$\scriptstyle{\iota}$$\scriptstyle{\varphi}$$\scriptstyle{0}$

is exact. ∎

######  Remark 4.1.

Corollary of the fifth statement of the above proposition is that
$\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\subset\mathcal{D}(\operatorname{\mathcal{A}})$
is a Lie subalgebra.

###  4.1 Exterior algebra of a commutative algebra

_In the following constructions, $\operatorname{\mathcal{A}}$ is always
commutative, associative, unital $\operatorname{\mathbb{K}}$-algebra with the
unit $1$._

Module of Kahler differentials of $\operatorname{\mathcal{A}}$. Let
$\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$
be a $\operatorname{\mathbb{K}}$-vector space generated by elements of the
form $b\operatorname{\underline{\operatorname{d}}}(a)$, for all
$a,b\in\operatorname{\mathcal{A}}$, satisfying relations

$\operatorname{\underline{\operatorname{d}}}(ab)=\operatorname{\underline{\operatorname{d}}}(a)b+a\operatorname{\underline{\operatorname{d}}}(b)\
.$

where
$\operatorname{\underline{\operatorname{d}}}\colon\operatorname{\mathcal{A}}\to\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$
is an $\operatorname{\mathcal{A}}$-linear map called the _universal
derivation_. More precisely, for any derivation
$\operatorname{\partial}\in\mathcal{D}(\operatorname{\mathcal{A}})$, there is
a unique $\operatorname{\mathcal{A}}$-linear map
$\tilde{\operatorname{\partial}}\colon\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})\to\operatorname{\mathcal{A}}$,
such that the following diagram commutes

${A}$${\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})}$${\operatorname{\mathcal{A}}}$$\scriptstyle{\operatorname{\underline{\operatorname{d}}}}$$\scriptstyle{\operatorname{\partial}}$$\scriptstyle{\exists!\tilde{\operatorname{\partial}}}$
(15)

The $\operatorname{\mathcal{A}}$-module structure on
$\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$
is given by multiplication from the left. Using the universal property of
$(\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}}),\operatorname{\underline{\operatorname{d}}})$,
we have a canonical $\operatorname{\mathcal{A}}$-module isomorphism

$\displaystyle\mathcal{D}(\operatorname{\mathcal{A}})\cong\operatorname{\operatorname{Hom}}_{\operatorname{\mathcal{A}}}(\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}}),\operatorname{\mathcal{A}})\
.$ (16)

This $\operatorname{\mathcal{A}}$-isomorphism can be understood as duality
between $\mathcal{D}(\operatorname{\mathcal{A}})$ and
$\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$.
Moreover, it can be extended to polyderivations and exterior forms on the
algebra $\operatorname{\mathcal{A}}$.

Exterior algebra of $\operatorname{\mathcal{A}}$. Let
$\Omega^{\bullet}(\operatorname{\mathcal{A}})$ be the (graded) exterior
algebra of the vector space of Kahler differentials of
$\operatorname{\mathcal{A}}$

$\displaystyle\Omega^{\bullet}(\operatorname{\mathcal{A}}):=\oplus_{k\geq
0}\Omega^{k}(\operatorname{\mathcal{A}}):=\oplus_{k\geq
0}\Lambda^{k}(\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})),$

together with the universal derivative
$\operatorname{\underline{\operatorname{d}}}\colon\operatorname{\mathcal{A}}\to\Omega^{1}(\operatorname{\mathcal{A}})$,
which extends to

$\displaystyle\operatorname{\underline{\operatorname{d}}}\colon\Omega^{k}(\operatorname{\mathcal{A}})\to\Omega^{k+1}(\operatorname{\mathcal{A}})$

for all $k$. The universal derivative satisfies all the properties of the
exterior differential on forms. In particular,
$\operatorname{\underline{\operatorname{d}}}$ is a boundary operator

$\displaystyle\operatorname{\underline{\operatorname{d}}}^{2}=0\ ,$

and acts with respect to the wedge product as

$\displaystyle\operatorname{\underline{\operatorname{d}}}(\omega_{1}\wedge\omega_{2})=\operatorname{\underline{\operatorname{d}}}\omega_{1}\wedge\omega_{2}+(-1)^{k}\omega_{1}\wedge\operatorname{\underline{\operatorname{d}}}\omega_{2}\
,$

where $\omega_{1}\in\Omega^{k}(\operatorname{\mathcal{A}})$ and
$\omega_{2}\in\Omega^{\bullet}(\operatorname{\mathcal{A}})$ are arbitrary. The
$\operatorname{\mathcal{A}}$-duality (16) between
$\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$ and
$\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}$ can be written
as

$\displaystyle\mathcal{D}(\operatorname{\mathcal{A}})\times\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})\to\operatorname{\mathcal{A}}\
,$
$\displaystyle<x,\operatorname{\underline{\operatorname{d}}}a>:=(\operatorname{\underline{\operatorname{d}}}a)(X):=X(a)\in\operatorname{\mathcal{A}}\
,$

and may be extended to the duality

$\Lambda^{k}(\mathcal{D}(\operatorname{\mathcal{A}}))\cong\operatorname{\operatorname{Hom}}_{\operatorname{\mathcal{A}}}(\Omega^{k}(\operatorname{\mathcal{A}}),\operatorname{\mathcal{A}})\
.$

Moreover, one can define the symmetric algebra of derivations as

$S(\mathcal{D}(\operatorname{\mathcal{A}})):=\operatorname{\operatorname{Hom}}_{\operatorname{\mathcal{A}}}(S(\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}}),\operatorname{\mathcal{A}})\
,$

which, as was already shown, is a graded Poisson algebra of degree $1$.

Take $k=2$, then we have

$\displaystyle\Lambda^{2}(\mathcal{D}(\operatorname{\mathcal{A}}))\cong\operatorname{\operatorname{Hom}}_{\operatorname{\mathcal{A}}}(\Omega^{2}(\operatorname{\mathcal{A}}),\operatorname{\mathcal{A}})\
.$ (17)

Suppose now that $\operatorname{\mathcal{A}}$ is a Poisson algebra equipped
with a Poisson bracket

$\displaystyle\\{-,-\\}\colon\operatorname{\mathcal{A}}\times\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}\
.$

Then the above duality (17) provides an element
$\pi\in\Lambda^{2}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$ by
the formula

$\displaystyle<\pi,\operatorname{\underline{\operatorname{d}}}a\wedge\operatorname{\underline{\operatorname{d}}}b>=\\{a,b\\}\
,$ (18)

for all $a,b\in\operatorname{\mathcal{A}}$. Then $\pi$ is called a _Poisson
biderivation_. The Jacobi identity for the Poisson bracket $\\{-,-\\}$ imposes
an important constraint on the biderivation $\pi$, which we will discuss below
(see (22).

##  5 Homology and cohomology

We will discuss two types of (co)homologies that can be defined for an algebra
$\operatorname{\mathcal{A}}$. Firstly we define (co)homology of commutative,
associative, unitary algebras (Hochschild), then for Poisson algebras
(Lichnerowicz-Poisson).

###  5.1 Hochschild (co)homology

We use the following notation $\otimes=\otimes_{\operatorname{\mathbb{K}}}$
and $\operatorname{\mathcal{A}}^{\otimes
k}=\underbrace{\operatorname{\mathcal{A}}\otimes\ldots\otimes\operatorname{\mathcal{A}}}_{\text{$k$-times}}$.
We define $\operatorname{\mathcal{A}}^{\otimes 0}:=\operatorname{\mathbb{K}}$.

Hochschild chain complex. Let $\operatorname{\mathcal{A}}$ be a commutative,
associative, unitary $\operatorname{\mathbb{K}}$-algebra, and let
$\mathcal{M}$ be an $\operatorname{\mathcal{A}}$-bimodule555That is,
$\operatorname{\mathcal{A}}$ acts on $\mathcal{M}$ from left and right.. Then
there is the following chain complex of $\operatorname{\mathcal{A}}$-bimodules
$\mathcal{M}\otimes\operatorname{\mathcal{A}}^{\otimes
k},k\in\operatorname{\mathbb{N}}$

${0}$${\mathcal{M}}$${\mathcal{M}\otimes\operatorname{\mathcal{A}}}$${\mathcal{M}\otimes\operatorname{\mathcal{A}}^{2}}$${\ldots}$$\scriptstyle{0}$$\scriptstyle{d_{1}}$$\scriptstyle{d_{2}}$$\scriptstyle{d_{3}}$
(19)

The boundary operators ${d_{k}},k\in\operatorname{\mathbb{N}},k>0$ are defined
via the ”face maps” $\operatorname{\partial}_{i}$ as

$\displaystyle d_{k}:=\sum_{i=0}^{k}(-1)^{i}\operatorname{\partial}_{i}\ ,$

where the maps $\operatorname{\partial}_{i}$ are

$\displaystyle\operatorname{\partial}_{i}(m\otimes a_{1}\otimes\ldots\otimes
a_{k}):=\begin{cases}ma_{1}\otimes\ldots\otimes a_{k},&\text{if }i=0\\\
m\otimes a_{1}\otimes\ldots\otimes a_{i}a_{i+1}\otimes\ldots\otimes
a_{k},&\text{if }0<i<k\\\ a_{k}m\otimes a_{1}\otimes\ldots\otimes
a_{k-1},&\text{if }i=k\end{cases}$

for all $m\in\mathbf{M}$ and $a_{i}\in\operatorname{\mathcal{A}}$. This is the
_Hochschild chain complex_.

######  Exercise 5.1.

Check that the above defined boundary maps $d_{k}$ are

  1. 1.

$\operatorname{\mathbb{K}}$-multilinear,

  2. 2.

well-defined,

  3. 3.

satisfy ${d_{k}}^{2}=0$

for all $k\in\operatorname{\mathbb{N}}_{0}$.

######  Definition 5.1 (Hochschild homology).

The homology of the chain complex (19), denoted
$HH_{*}(\operatorname{\mathcal{A}},\mathcal{M})$, is called the Hochschild
homology of $\operatorname{\mathcal{A}}$ with coefficients in $\mathcal{M}$.
The $k$-th Hochschild homology $\operatorname{\mathbb{K}}$-module is

$\displaystyle HH_{k}(\operatorname{\mathcal{A}},\mathcal{M}):=\ker
d_{k}/im_{d_{k+1}}\ .$

Hochschild cochain complex. Using the
$\operatorname{\operatorname{Hom}}_{\operatorname{\mathbb{K}}}$-functor, we
can construct the Hochschild cochain complex

${0}$${\mathcal{M}}$${\operatorname{\operatorname{Hom}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}},\mathcal{M})}$${\operatorname{\operatorname{Hom}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}}^{\otimes
2},\mathcal{M})}$${\ldots}$$\scriptstyle{0}$$\scriptstyle{d^{1}}$$\scriptstyle{d^{2}}$$\scriptstyle{d^{3}}$
(20)

The coboundary operators ${d^{k}}$ are defined as

$\displaystyle d^{k}:=\sum_{i=0}^{k}(-1)^{i}\operatorname{\partial}^{i}\ ,$

where the maps $\operatorname{\partial}^{i}$ are defined for
$f\in\operatorname{\operatorname{Hom}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}}^{\otimes
k},\mathcal{M})$

$\displaystyle(\operatorname{\partial}^{i}f)(a_{0}\otimes
a_{1}\otimes\ldots\otimes
a_{k}):=\begin{cases}a_{0}f(a_{1}\otimes\ldots\otimes a_{k}),&\text{if }i=0\\\
f(a_{0}\otimes a_{1}\otimes\ldots\otimes a_{i}a_{i+1}\otimes\ldots\otimes
a_{k}),&\text{if }0<i<k\\\ f(a_{0}\otimes a_{1}\otimes\ldots\otimes
a_{k-1})a_{k},&\text{if }i=k\end{cases}$

where $a_{i}\in\operatorname{\mathcal{A}}$ for all $i$. This is the
_Hochschild cochain complex_.

######  Definition 5.2 (Hochschild cohomology).

The cohomology of the cochain complex (20), denoted
$HH^{*}(\operatorname{\mathcal{A}},\mathcal{M})$, is called the Hochschild
cohomology of $\operatorname{\mathcal{A}}$ with coefficients in $\mathcal{M}$.
The $k$-th Hochschild cohomology $\operatorname{\mathbb{K}}$-module is

$\displaystyle HH^{k}(\operatorname{\mathcal{A}},\mathcal{M}):=\ker
d^{k}/\operatorname{\operatorname{im}}d^{k-1}\ .$

######  Proposition 5.1.

The $0$-th Hochschild homology of $\operatorname{\mathcal{A}}$ with
coefficients in $\mathcal{M}$ satisfy

$\displaystyle HH_{0}(\operatorname{\mathcal{A}},\mathcal{M})$
$\displaystyle=\mathcal{M}/[\mathcal{M},\operatorname{\mathcal{A}}]\ ,$

where $[\operatorname{\mathcal{A}},\mathcal{M}]=\\{am-ma|\
a\in\operatorname{\mathcal{A}},m\in\mathcal{M}\\}$. In particular

$\displaystyle
HH_{0}(\operatorname{\mathcal{A}},\operatorname{\mathcal{A}})=\operatorname{\mathcal{A}}/[\operatorname{\mathcal{A}},\operatorname{\mathcal{A}}]\
.$

The $0$-th Hochschild cohomology of $\operatorname{\mathcal{A}}$ with
coefficients in $\mathcal{M}$ satisfy

$\displaystyle HH^{0}(\mathcal{M},\operatorname{\mathcal{A}})$
$\displaystyle=\\{m\in M|\ am=ma,\ \forall a\in\operatorname{\mathcal{A}}\\}\
.$

In particular,

$\displaystyle
HH^{0}(\operatorname{\mathcal{A}},\operatorname{\mathcal{A}})=\operatorname{Z}(A)\
,$

where $\operatorname{Z}(A)$ is the center of $\operatorname{\mathcal{A}}$.

######  Proposition 5.2.

Let $\operatorname{\mathcal{A}}$ be a commutative
$\operatorname{\mathbb{K}}$-algebra and $\mathcal{M}$ an
$\operatorname{\mathcal{A}}$-bimodule. Then the $1$st Hochschild homology
satisfies

$\displaystyle HH_{1}(\operatorname{\mathcal{A}},\mathcal{M})\cong
M\otimes_{\operatorname{\mathcal{A}}}\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})\
.$

Denote by $\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}},\mathcal{M})$
the space of $\operatorname{\mathbb{K}}$-linear functions
$f\colon\operatorname{\mathcal{A}}\to\mathcal{M}$ such that

$\displaystyle f(ab)=af(b)+f(a)b\ .$

Denote by $\mathcal{PD}(\operatorname{\mathcal{A}},\mathcal{M})$ the space of
$\operatorname{\mathbb{K}}$-linear functions
$f_{m}\colon\operatorname{\mathcal{A}}\to\mathcal{M}$, which are given by

$\displaystyle f_{m}(a)=ma-am\ .$

######  Exercise 5.2.

Check that the above defined $f_{m}$ satisfies
$f_{m}\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}},\mathcal{M})$.

######  Proposition 5.3.

$HH^{1}(\operatorname{\mathcal{A}},\mathcal{M})=\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}},\mathcal{M})/\mathcal{PD}(\operatorname{\mathcal{A}},\mathcal{M})$.

###  5.2 Lichnerowicz-Poisson cohomology

Let $\operatorname{\mathcal{A}}$ be a commutative Poisson algebra with the
Poisson bracket $\\{-,-\\}$ and consider
$\Lambda(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))=\bigoplus_{k\geq
0}\Lambda^{k}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$ defined
as follows. For $k=0$, we define
$\Lambda^{0}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))=\operatorname{\mathcal{A}}$.
An element
$X\in\Lambda^{k}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})),k>0$,
is a multilinear, antisymmetric mapping

$\displaystyle
X\colon\operatorname{\mathcal{A}}^{k}\to\operatorname{\mathcal{A}}$

and the mapping
$\operatorname{\partial}_{X}\colon\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}$
given by

$\displaystyle\operatorname{\partial}_{X}(a):=X(a,a_{1},\ldots,a_{k-1})$

is a derivation.

Schouten-Nijenhuis bracket. Let
$X,Y\in\Lambda(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$ be
decomposable,

$\displaystyle X=x_{1}\wedge\ldots\wedge x_{k}$ $\displaystyle
Y=y_{1}\wedge\ldots\wedge y_{l}\ ,$

where all
$x_{i},y_{j}\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$. The
Schouten-Nijenhuis bracket $[\\![-,-]\\!]$ is given by

$\displaystyle[\\![X,Y]\\!]:=\sum_{i,j}(-1)^{i+j}[x_{i},y_{j}]x_{1}\wedge\ldots\wedge\hat{x_{i}}\wedge\ldots\wedge
x_{k}\wedge y_{1}\wedge\ldots\wedge\hat{y_{j}}\wedge\ldots\wedge y_{l}\ ,$

where $[-,-]$ is the commutator of differential operators, and the above
definition is extended linearly on the whole
$\Lambda(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$.

######  Exercise 5.3.

Show that the Schouten-Nijenhuis bracket $[\\![-,-]\\!]$ satisfies for all
$P,Q\in\Lambda(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$

$\displaystyle[\\![P,Q]\\!]=(-1)^{\deg P\deg Q}[\\![Q,P]\\!]$ (21)

Poisson operator. Let
$\pi\in\Lambda^{2}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$ be
the _Poisson biderivation_ , defined by (18). Then the Schouten-Nijenhuis
bracket gives an element
$[\\![\pi,\pi]\\!]\in\Lambda^{3}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$.

######  Proposition 5.4.

The Jacobi identity for the Poisson bracket (46) implies the triviality of the
$3$-derivation:

$\displaystyle[\\![\pi,\pi]\\!]=0\ .$ (22)

######  Exercise 5.4.

Prove the above proposition.

The equation (22) is sometimes called _Poisson Master equation_. It can be
interpreted as a nilpotency condition (${\delta_{\pi}}^{2}=0$) for the
operator
$\delta_{\pi}\colon\Lambda(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))\to\Lambda(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$,
defined by

$\displaystyle\delta_{\pi}X=[\\![\pi,X]\\!]\ .$

The operator $\delta_{\pi}$ is called _Lichnerowicz-Poisson operator_ and
leads to the following notion of cohomology.

######  Definition 5.3 (Lichnerowicz-Poisson cohomology).

The cohomology of the chain complex

${\ldots}$${\Lambda^{k-1}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))}$${\Lambda^{k}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))}$${\Lambda^{k+1}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))}$${\ldots\
,}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{\delta_{\pi}}$

is called the Lichnerowicz-Poisson cohomology, denoted
$HP^{*}(\operatorname{\mathcal{A}})$.

In terms of the Poisson bracket, the Lichnerowicz-Poisson operator can be
rewritten as

$\displaystyle\left(\delta_{\pi}X\right)(a_{0},a_{1},\ldots,a_{k})$
$\displaystyle=\sum_{i}(-1)^{i}\\{a_{i},X(a_{0},a_{1},\ldots,\hat{a_{i}},\ldots,a_{k})\\}$
(23) $\displaystyle\ \ \ +\sum_{0\leq
i<j}X(\\{a_{i},a_{j}\\},a_{1},\ldots,\hat{a_{i}},\ldots,\hat{a_{j}},\ldots,a_{k})\
.$ (24)

###  5.3 Low-dimensional Poisson cohomology

$k=0$: the operator
$\delta_{\pi}\colon\operatorname{\mathcal{A}}\to\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))\cong\Lambda^{1}(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))$
acts as $a\mapsto\operatorname{\partial}_{a}=\\{-,a\\}$, where
$a\in\operatorname{\mathcal{A}}$ can be seen as a Hamiltonian for a
Hamiltonian derivation $\\{-,a\\}$. We have

$\displaystyle
HP^{0}\cong\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$

since $\delta_{\pi}(a)(b)=0$ for all $b\in\operatorname{\mathcal{A}}$ if and
only if $a\in\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$.
Elements of
$HP^{0}\cong\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$ are
called Hamiltonians with zero dynamics.

$k=1$: $1$-coboundary is a derivation
$X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$ which is a
Hamiltonian derivation $\operatorname{\partial}_{a}$ for some element
$a\in\operatorname{\mathcal{A}}$. $1$-cocycle is an element
$X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$ s.t.

$\displaystyle\delta_{\pi}(X)=0\Rightarrow[\\![\pi,X]\\!]=0\ .$

Derivations $X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$ which
satisfies $\delta_{\pi}(X)=0$ are called _Poisson, or canonical, derivations_.
The set of all such elements is denoted
$\operatorname{Can}(\operatorname{\mathcal{A}})$. The equation
$[\\![\pi,X]\\!]=0$ represents a conservation of the Poisson structure along
$X$. If we denote $L_{X}\pi=[\\![\pi,X]\\!]$, one can easily show that

$\displaystyle<L_{X}\pi,\operatorname{\underline{\operatorname{d}}}a\wedge\operatorname{\underline{\operatorname{d}}}b>=L_{X}<\pi,\operatorname{\underline{\operatorname{d}}}a\wedge\operatorname{\underline{\operatorname{d}}}b>\
.$

######  Theorem 5.1 (basic theorem of classical mechanics).

$\displaystyle
HP^{1}(\operatorname{\mathcal{A}})\cong\operatorname{Can}(\operatorname{\mathcal{A}})/\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\
,$

where $\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$ is given
by (11).

######  Example 5.1 (Hamiltonian derivations on polynomial algebra [28]).

Consider $\operatorname{\mathcal{A}}=\operatorname{\mathbb{K}}[x_{1},x_{2}]$
with the Poisson bracket $\\{x_{1},x_{2}\\}=x_{2}$. Then
$\operatorname{\partial}\in\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\subset\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$
if

$\displaystyle\operatorname{\partial}=\alpha_{1}\frac{\operatorname{\partial}}{\operatorname{\partial}x_{1}}+\alpha_{2}\frac{\operatorname{\partial}}{\operatorname{\partial}x_{2}}$
(25)

and exists $h\in\operatorname{\mathcal{A}}$ such that

$\displaystyle\operatorname{\partial}=\operatorname{\partial}_{h}=\\{-,h\\}\
.$

We will compute the coefficients $\alpha_{i}$ so that the above equation
holds. We have

$\displaystyle\\{x_{1},h\\}=\frac{\operatorname{\partial}x_{1}}{\operatorname{\partial}x_{1}}\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{2}}-\frac{\operatorname{\partial}x_{1}}{\operatorname{\partial}x_{2}}\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{1}}=\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{2}}\
.$

Similarly we get

$\displaystyle\\{x_{2},h\\}=-\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{1}}\
.$

For $f\in\operatorname{\mathcal{A}}$, the bracket with $h$ is

$\displaystyle\\{f,h\\}=\frac{\operatorname{\partial}f}{\operatorname{\partial}x_{1}}\\{x_{1},h\\}+\frac{\operatorname{\partial}f}{\operatorname{\partial}x_{2}}\\{x_{2},h\\}=\frac{\operatorname{\partial}f_{1}}{\operatorname{\partial}x_{1}}\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{2}}-\frac{\operatorname{\partial}f_{1}}{\operatorname{\partial}x_{2}}\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{1}}\
.$

From (25) we have

$\displaystyle\operatorname{\partial}(f)=\alpha_{1}\frac{\operatorname{\partial}f}{\operatorname{\partial}x_{1}}+\alpha_{2}\frac{\operatorname{\partial}f}{\operatorname{\partial}x_{2}}\
,$

so to have $\operatorname{\partial}=\operatorname{\partial}_{h}$, one must
have

$\displaystyle\alpha_{1}=\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{2}}\
,$
$\displaystyle\alpha_{2}=-\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{1}}\
.$

This means that the set of Hamiltonian derivations on
$\operatorname{\mathcal{A}}$ is given by

$\displaystyle\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})=\\{\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{2}}\frac{\operatorname{\partial}}{\operatorname{\partial}x_{1}}-\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{1}}\frac{\operatorname{\partial}}{\operatorname{\partial}x_{2}}|\
h\in\operatorname{\mathcal{A}}\\}\ .$

######  Exercise 5.5.

Consider $\operatorname{\mathcal{A}}$ from the previous example. Show that

$\displaystyle\operatorname{Can}(\operatorname{\mathcal{A}})/\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\cong\operatorname{\mathbb{K}}$

Indication: check that if
$\operatorname{\partial}=\alpha_{1}\frac{\operatorname{\partial}}{\operatorname{\partial}x_{1}}+\alpha_{2}\frac{\operatorname{\partial}}{\operatorname{\partial}x_{2}}\in\operatorname{Can}(\operatorname{\mathcal{A}})$,
then

$\displaystyle\alpha_{1}$
$\displaystyle=c+x_{2}\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{2}}$
$\displaystyle\alpha_{2}$
$\displaystyle=-x_{2}\frac{\operatorname{\partial}h}{\operatorname{\partial}x_{1}}$

where $c\in\operatorname{\mathbb{K}}$.

######  Example 5.2.

Consider the set

$\operatorname{\operatorname{Sym}}(\operatorname{\operatorname{Hom}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{D}}^{p}(\operatorname{\mathcal{A}}),\operatorname{\mathcal{A}}))=\\{\varphi\colon\operatorname{\mathcal{D}}^{p}(\operatorname{\mathcal{A}})\to\operatorname{\mathcal{A}}\
|\ \varphi\text{ multilinear}\\}\ .$

An element of the above set is not only $\operatorname{\mathbb{K}}$-linear in
all $p$-arguments but also a $\operatorname{\mathbb{K}}$-derivation in every
argument. Recall that $\mathcal{D}(\operatorname{\mathcal{A}})$ is the
$\operatorname{\mathcal{A}}$-module of $\operatorname{\mathbb{K}}$-derivations
of $\operatorname{\mathcal{A}}$.

####  5.3.1 Compatible Poisson structures

As a useful application of Lichnerowicz-Poisson cohomology, consider the
following description of _compatible Poisson structures_.

######  Definition 5.4.

Two Poisson structures given by biderivations $\pi$ and $\theta$ are
compatible if for all $\lambda\in\operatorname{\mathbb{K}}$,
$\pi+\lambda\theta$ is again a Poisson structure.

As an immediate consequence of the above definition, we have the following
proposition.

######  Proposition 5.5.

If two Poisson structures
$\pi,\theta\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$
are compatible, then

$\displaystyle[\\![\pi,\theta]\\!]=0$

Equivalently, if $\pi$ and $\theta$ are compatible, then they are closed with
respect to coderivations in the following sense

$\displaystyle\delta_{\pi}\theta=\delta_{\theta}\pi=0\ .$

######  Exercise 5.6.

Prove the proposition 5.5.

Notice that if $[\\![\pi,\theta]\\!]=0$ then for all
$\lambda\in\operatorname{\mathbb{K}}$

$\displaystyle[\\![\pi+\lambda\theta,\pi+\lambda\theta]\\!]=0\ .$

Let us fix a biderivation $\pi$. Consider the _Lie derivative_ of $\pi$ along
$X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$

$\displaystyle
L_{X}\pi:=[\\![\pi,X]\\!]\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)\
.$

Let $\pi$ be a Poisson biderivation (i.e. it corresponds to a Poisson
structure). Then $L_{X}\pi$ is called an _infinitesimal deformation of the
Poisson structure_. Now consider a deformation of $\pi$ of the form

$\displaystyle\pi\mapsto\pi+L_{X}\pi\ .$

By assumption, we have $[\\![\pi,\pi]\\!]=0$ since $\pi$ yields a Poisson
structure. If we assume that the deformation also yields a Poisson structure,
then the graded antisymmetry of the bracket (see (21) yields

$\displaystyle
0=[\\![\pi+L_{X}\pi,\pi+L_{X}\pi]\\!]=2[\\![\pi,L_{X}\pi]\\!]+[\\![L_{X}\pi,L_{X}\pi]\\!]\
.$

Let us denote $\gamma:=L_{X}\pi$ and recall that
$[\\![\theta,X]\\!]=\delta_{\theta}(X)$. Then the above can be written as

$\displaystyle\delta_{\pi}\gamma+\frac{1}{2}[\\![\gamma,\gamma]\\!]=0\ .$

This equation is called _Maurer-Cartan equation_ (for a differential graded
algebra
$\Lambda\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$).

####  5.3.2 Interpretation of $HP^{2}(\operatorname{\mathcal{A}})$

Let
$\pi\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$
and consider the image of
$\delta_{\pi}\colon\Lambda^{1}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)\to\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$,
given by

$\displaystyle{B_{\pi}}^{2}=\\{\theta\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)|\
\theta=[\\![\pi,X]\\!]\text{ for some
$X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$}\\}\ ,$

The kernel of
$\delta_{\pi}\colon\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)\to\Lambda^{3}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$
is

$\displaystyle{Z_{\pi}}^{2}=\\{\theta\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)|\
[\\![\pi,\theta]\\!]=0\\!]\\}\ .$

By definition

$\displaystyle HP^{2}(\operatorname{\mathcal{A}})={Z_{\pi}}^{2}/{B_{\pi}}^{2}\
.$

We have the following proposition, which yields an interpretation of the
second Lichnerowicz-Poisson cohomology.

######  Proposition 5.6 ([18]).

If
$\pi\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$
is a Poisson biderivation ( $[\\![\pi,\pi]\\!]=0$) and
$HP^{2}(\operatorname{\mathcal{A}})=0$, then the set of all structures
compatible with $\pi$

$\displaystyle\operatorname{Comm}(\pi):=\\{\theta\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)|\
[\\![\theta,\theta]\\!]=[\\![\pi,\theta]\\!]=0\\}$ (26)

is a set of infinitesimal deformations of $\pi$ along
$X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$. That is,
$\theta=L_{X}\pi$ such that

$\displaystyle{L_{X}}^{2}(\theta)=L_{Y}\pi$

for some $Y\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$, where
${L_{X}}^{2}=L_{X}\circ L_{X}$.

###### Proof.

Let $\theta\in\operatorname{Comm}(\pi)$, then $\delta_{\pi}\theta=0$ so

$\displaystyle\operatorname{Comm}(\pi)\subset\ker\delta_{\pi}\ .$

If $\theta=\delta_{\pi}X$ then $0=[\theta]\in
HP^{2}(\operatorname{\mathcal{A}})$. Suppose
$\theta\in\operatorname{Comm}(\pi)$ and
$HP^{2}(\operatorname{\mathcal{A}})=0$. Then
$\theta=\delta_{\pi}X=[\\![\pi,X]\\!]=L_{X}\pi$. But at the same time
$[\\![\theta,\theta]\\!]=[\\![L_{X}\pi,L_{X}\pi]\\!]=0$. This is equivalent to
$(L_{X}\circ L_{X})(\theta)\in\ker\delta_{\pi}$ since

$\displaystyle
L_{X}\left(L_{X}\pi\right)=L_{X}\left([\\![\pi,X]\\!]\right)=[\\![[\\![\pi,X]\\!],X]\\!]=[\\![\theta,X]\\!]\
,$

and

$\displaystyle\delta_{\pi}\left([\\![\theta,X]\\!]\right)=\delta_{\pi}\left(\delta_{\theta}X\right)=-\delta_{\theta}\left(\delta_{\pi}X\right)=[\\![\delta_{\pi}X,\delta_{\pi}X]\\!]=0\
.$

Hence if $HP^{2}(\operatorname{\mathcal{A}})=0$ then there is
$Y\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$ such that
${L_{X}}^{2}\theta=L_{Y}(\pi)=\delta_{\pi}(Y)$. ∎

From the above, we obtain a mapping

$\displaystyle\tau\colon\operatorname{Comm}(\pi)\to
HP^{2}(\operatorname{\mathcal{A}})$

and

$\displaystyle\tau(\theta)=0\iff\theta=\delta_{\pi}(X)$

for some $X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$. The
image of $\tau$ is described by the following proposition.

######  Proposition 5.7.

Let $\operatorname{\partial}\in HP^{2}(\operatorname{\mathcal{A}})$. Then

$\displaystyle\operatorname{\partial}\in\operatorname{\operatorname{im}}\tau\iff\operatorname{\partial}=\Delta+\delta_{\pi}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)\
,$

where the representative
$\Delta\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$
of the class $\operatorname{\partial}$ satisfies

  1. 1.

$[\\![\operatorname{\partial},\operatorname{\partial}]\\!]=[\\![\Delta,\Delta]\\!]=\operatorname{\partial}_{\pi}(\alpha)$,
where
$\alpha\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$,

  2. 2.

there is $X\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$ such
that $\alpha+2L_{X}(\Delta)-{L_{X}}^{2}(\pi)=\ker\delta_{\pi}$.

###### Proof.

Let
$\Delta\in\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)$
be such that

$\displaystyle\operatorname{\partial}=\Delta+\delta_{\pi}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)\in
HP^{2}(\operatorname{\mathcal{A}})\ .$

If $\operatorname{\partial}\in\operatorname{\operatorname{im}}\tau$, then
there is a $\theta\in\operatorname{Comm}(\pi)$ (see (26)) such that

$\displaystyle\tau(\theta)=\Delta+\delta_{\pi}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)\
.$

Because

$\displaystyle
0=[\\![\operatorname{\partial},\operatorname{\partial}]\\!]\Rightarrow[\\![\Delta,\Delta]\\!]=\operatorname{\partial}_{\pi}(\alpha)\in
HP^{3}(\operatorname{\mathcal{A}})$

and $\tau[\\![\theta,\theta]\\!]=0$, we have

$\displaystyle[\\![\tau(\theta),\tau(\theta)]\\!]$
$\displaystyle=[\\![\Delta+\delta_{\pi}(X),\Delta+\delta_{\pi}(X)]\\!]$
$\displaystyle=[\\![\Delta,\Delta]\\!]+[\\![\Delta,\delta_{\pi}(X)]\\!]+[\\![\delta_{\pi}(X),\Delta]\\!]+[\\![\delta_{\pi}(X),\delta_{\pi}(X)]\\!]$
$\displaystyle=\delta_{\pi}(\alpha)-[\\![\Delta,L_{X}\pi]\\!]-[\\![L_{X}\pi,\Delta]\\!]-\delta_{\pi}{L_{X}}^{2}(\pi)$
$\displaystyle=\delta_{\pi}\left(\alpha-{L_{X}}^{2}(\pi)\right)-2[\\![L_{X}\pi,\Delta]\\!]$
$\displaystyle=\delta_{\pi}\left(\alpha-{L_{X}}^{2}(\pi)\right)-2[\\![[\\![\pi,X]\\!],\Delta]\\!]$
$\displaystyle=\delta_{\pi}\left(\alpha+2L_{X}(\Delta)-{L_{X}}^{2}(\pi)\right)=0\
,$

since $[\\![\tau(\theta),\tau(\theta)]\\!]=0$. ∎

###  5.4 Poisson homology

The following construction is due to Brylinski [7] and [17]. Let
$\operatorname{\mathcal{A}}$ be a commutative
$\operatorname{\mathbb{K}}$-algebra and
$\Omega(\operatorname{\mathcal{A}})=\Lambda(\Omega_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}})$
be the exterior algebra of $\operatorname{\mathcal{A}}$. The boundary morphism
$d_{\pi}\colon\Omega^{k}(\operatorname{\mathcal{A}})\to\Omega^{k-1}(\operatorname{\mathcal{A}})$
is defined by the ”homotopy-like” formula

$\displaystyle
d_{\pi}\omega=(i_{\pi}\circ\operatorname{\underline{\operatorname{d}}})(\omega)-(\operatorname{\underline{\operatorname{d}}}\circ
i_{\pi})(\omega)\ ,$ (27)

where $i_{\pi}$ is the contraction operator with respecet to $\pi$, i.e
.$i_{\pi}:=<\pi,->$. It is straightforward to check that this operator
satisfies $d_{\pi}\circ d_{\pi}=0$. In coordinates, we can describe $d_{\pi}$
on a decomposable
$\omega=a_{0},\operatorname{\underline{\operatorname{d}}}a_{1}\wedge\ldots\wedge
a_{k}$ by the general formula

$\displaystyle d_{\pi}\omega$
$\displaystyle=\sum_{l=1}^{k+1}(-1)^{l+1}\\{a_{0},a_{i}\\}\wedge\operatorname{\underline{\operatorname{d}}}a_{1}\wedge\ldots\wedge\hat{\operatorname{\underline{\operatorname{d}}}a}_{l}\wedge\ldots\wedge\operatorname{\underline{\operatorname{d}}}a_{k+1}$
$\displaystyle\ \ \
-\sum_{i<j}(-1)^{i+j}\operatorname{\underline{\operatorname{d}}}\left(\\{a_{i},a_{j}\\}\right)\wedge\operatorname{\underline{\operatorname{d}}}a_{1}\wedge\ldots\wedge\hat{\operatorname{\underline{\operatorname{d}}}a}_{i}\wedge\ldots\wedge\hat{\operatorname{\underline{\operatorname{d}}}a}_{j}\wedge\ldots\wedge\operatorname{\underline{\operatorname{d}}}a_{k+1}\
.$

By definition,

$\displaystyle d_{\pi}(a_{0}da_{1})=i_{\pi}(da_{0}\wedge
da_{1})=\\{a_{0},a_{1}\\}\ .$

Note that the Jacobi identity for the triple $a_{0},a_{1},a_{2}$

$\displaystyle\\{a_{0},\\{a_{1},a_{2}\\}\\}+\\{a_{1},\\{a_{2},a_{O}\\}\\}+\\{a_{2},\\{a_{0},a_{1}\\}\\}=0$

implies $(d_{\pi})^{2}=0$.

###  5.5 Duality

Let $\pi$ be a _symplectic_ (or _non-degenerate_) Poisson structure on
$\operatorname{\mathcal{A}}$, meaning that the Hamiltonian map

$\displaystyle\Gamma_{\pi}\colon\Omega^{1}_{\operatorname{\mathcal{A}}/\operatorname{\mathbb{K}}}\to\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$

defined by

$\displaystyle\Gamma_{\pi}(\alpha)$
$\displaystyle=<\pi,\alpha>\in\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})$

is an isomorphism, and there exists the inverse
${\Gamma_{\pi}}^{-1}\colon\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\to\Omega^{1}(\operatorname{\mathcal{A}})$.
The inverse is given by

$\displaystyle{{\Gamma}_{\pi}}^{-1}(\operatorname{\partial})$
$\displaystyle=\alpha_{\operatorname{\partial}}\text{ such that
}<\pi,\alpha_{\operatorname{\partial}}>=\operatorname{\partial}\ .$

In this case, one can check that

$\displaystyle\delta_{\pi}=\Gamma_{\pi}\circ\operatorname{\underline{\operatorname{d}}}\circ{\Gamma_{\pi}}^{-1}\
,$

where $\delta_{\pi}$ is the Lichnerowicz-Poisson operator. The above equation
is symbolical. It can be described more precisely by the following commutative
diagram

${\Omega^{1}(\operatorname{\mathcal{A}})}$${\Lambda^{1}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)}$${\Omega^{2}(\operatorname{\mathcal{A}})}$${\Lambda^{2}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)}$$\scriptstyle{\operatorname{\underline{\operatorname{d}}}}$$\scriptstyle{\Gamma_{\pi}}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{\Lambda^{2}\Gamma_{\pi}}$

which holds for any Poisson biderivation $\pi$.

######  Proposition 5.8.

For general $k$, the following diagram commutes

${\Omega^{k}(\operatorname{\mathcal{A}})}$${\Lambda^{k}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)}$${\Omega^{k+1}(\operatorname{\mathcal{A}})}$${\Lambda^{k+1}\left(\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}})\right)}$$\scriptstyle{\operatorname{\underline{\operatorname{d}}}}$$\scriptstyle{\Lambda^{k}\Gamma_{\pi}}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{\Lambda^{k+1}\Gamma_{\pi}}$

###### Proof.

To check the commutativity of the diagram, consider decomposable
$\omega=a_{0}\operatorname{\underline{\operatorname{d}}}a_{1}\wedge\ldots\wedge\operatorname{\underline{\operatorname{d}}}a_{k}\in\Omega^{k}(\operatorname{\mathcal{A}})$.
Then

$\displaystyle\Lambda^{k+1}\Gamma_{\pi}(\operatorname{\underline{\operatorname{d}}}\omega)=\Lambda^{k+1}\Gamma_{\pi}(\operatorname{\underline{\operatorname{d}}}a_{0}\wedge\operatorname{\underline{\operatorname{d}}}a_{1}\wedge\ldots\wedge\operatorname{\underline{\operatorname{d}}}a_{k})=\operatorname{\partial}_{a_{0}}\wedge\operatorname{\partial}_{a_{1}}\wedge\ldots\wedge\operatorname{\partial}_{a_{k}}\
,$

and

$\displaystyle\delta_{\pi}(\Lambda^{k}\Gamma_{\pi}(\omega))$
$\displaystyle=\delta_{\pi}(a_{0}\operatorname{\partial}_{a_{1}}\wedge\ldots\wedge\operatorname{\partial}_{a_{k}})$
$\displaystyle=-[\\![a_{0}\operatorname{\partial}_{a_{1}}\wedge\ldots\wedge\operatorname{\partial}_{a_{k}},\pi]\\!]$
$\displaystyle=-[\\![a_{0},\pi]\\!]\wedge\operatorname{\partial}_{a_{1}}\wedge\ldots\wedge\operatorname{\partial}_{a_{k}}$
$\displaystyle=\operatorname{\partial}_{a_{0}}\wedge\operatorname{\partial}_{a_{1}}\wedge\ldots\wedge\operatorname{\partial}_{a_{k}}\
.$

Since the above can be extended linearly, the diagram commutes. ∎

######  Remark 5.1.

If $\operatorname{\mathcal{A}}=C^{\infty}(M)$ is the algebra of smooth
function on a smooth manifold $M$, then it was shown in [7] that

$\displaystyle HP_{k}(\operatorname{\mathcal{A}})\cong
HP^{2n-k}(\operatorname{\mathcal{A}})\ .$

Moreover, by the proposition 5.8, the latter group is isomorphic with the de
Rham cohomology

$\displaystyle HP^{2n-k}(\operatorname{\mathcal{A}})=H_{DR}^{2n-k}(M)\ .$

In the following lemma, we use the notion of a _graded commutator_ in a graded
algebra

$\displaystyle[\operatorname{\partial}_{1},\operatorname{\partial}_{2}]:=\operatorname{\partial}_{1}\circ\operatorname{\partial}_{2}-(-1)^{\deg\operatorname{\partial}_{1}\deg\operatorname{\partial}_{2}}\operatorname{\partial}_{2}\circ\operatorname{\partial}_{1}\
.$

We have $\deg
d_{\pi}=-1,\deg\operatorname{\underline{\operatorname{d}}}=1,\deg
i_{\pi}=-2,L_{\pi}=1$

######  Lemma 5.1.

The operator
$d_{\pi}\colon\Omega^{k}(\operatorname{\mathcal{A}})\to\Omega^{k-1}(\operatorname{\mathcal{A}})$
commutes in graded sense with $\operatorname{\underline{\operatorname{d}}}$
and $i_{\pi}$, i.e.

$\displaystyle[d_{\pi},\operatorname{\underline{\operatorname{d}}}]$
$\displaystyle=d_{\pi}\circ\operatorname{\underline{\operatorname{d}}}+\operatorname{\underline{\operatorname{d}}}\circ
d_{\pi}=0\ ,$ $\displaystyle[d_{\pi},i_{\pi}]$ $\displaystyle=d_{\pi}\circ
i_{\pi}-i_{\pi}\circ d_{\pi}=0\ .$

###### Proof.

To prove the first equation, we use the definition of $d_{\pi}$ (see (27)) and
that
$\operatorname{\underline{\operatorname{d}}}\circ\operatorname{\underline{\operatorname{d}}}=0$.
Then

$\displaystyle(i_{\pi}\circ\operatorname{\underline{\operatorname{d}}}-\operatorname{\underline{\operatorname{d}}}\circ
i_{\pi})\circ\operatorname{\underline{\operatorname{d}}}+\operatorname{\underline{\operatorname{d}}}\circ(i_{\pi}\circ\operatorname{\underline{\operatorname{d}}}-\operatorname{\underline{\operatorname{d}}}\circ
i_{\pi})=-\operatorname{\underline{\operatorname{d}}}\circ
i_{\pi}\circ\operatorname{\underline{\operatorname{d}}}+\operatorname{\underline{\operatorname{d}}}\circ
i_{\pi}\circ\operatorname{\underline{\operatorname{d}}}=0$

Hence $[d_{\pi},\operatorname{\underline{\operatorname{d}}}]=0$. Similarly for
the second equation

$\displaystyle[d_{\pi},i_{\pi}]=[L_{\pi},i_{\pi}]=i_{[\pi,\pi]}=0\ .$

∎

######  Remark 5.2.

When $\operatorname{\mathcal{A}}=C^{\infty}(M)$, where $M$ is a smooth Poisson
manifold, we will see all the above, and more general, identities in the later
section as well.

######  Example 5.3 ($0$th Poisson homology).

For the kernel of
$d^{0}_{\pi}\colon\Omega^{0}(\operatorname{\mathcal{A}})\cong\operatorname{\mathcal{A}}\to
0$ we obvously have $\ker d^{0}_{\pi}\cong\operatorname{\mathcal{A}}$. For the
image of
$d^{1}_{\pi}\colon\Omega^{1}(\operatorname{\mathcal{A}})\to\Omega^{0}(\operatorname{\mathcal{A}})\cong\operatorname{\mathcal{A}}$
we have
$\operatorname{\operatorname{im}}d^{1}_{\pi}=\\{\operatorname{\mathcal{A}},\operatorname{\mathcal{A}}\\}$.
This is because $\Omega^{1}(\operatorname{\mathcal{A}})$ consists of element
$a_{0}da_{1}$, where $a_{0},a_{1}\in\operatorname{\mathcal{A}}$ are arbitrary,
and

$\displaystyle
d^{1}_{\pi}(a_{0}da_{1})=\\{a_{0},a_{1}\\}\in\\{\operatorname{\mathcal{A}},\operatorname{\mathcal{A}}\\}\
.$

Hence

$\displaystyle HP_{0}(\operatorname{\mathcal{A}})\cong
A/\\{\operatorname{\mathcal{A}},\operatorname{\mathcal{A}}\\}$

There is no simple interpretation for higher homology groups, $k>0$.

##  6 Polynomial Poisson Algebras

Let $\operatorname{\mathcal{A}}=\operatorname{\mathbb{C}}[x_{1},\ldots,x_{n}]$
be the polynomial algebra over complex numbers. If $x_{i}$ is a generator of
$\operatorname{\mathcal{A}}$, then
$\operatorname{\partial}_{i}:=\frac{\operatorname{\partial}}{\operatorname{\partial}x_{i}}\in\mathcal{D}(\operatorname{\mathcal{B}})$.
The vector operator
$\nabla:=(\operatorname{\partial}_{1},\ldots,\operatorname{\partial}_{n})$ is
called _gradient_ One can define the _Jacobian matrix_ of $n$ elements
$f_{1},\ldots,f_{n}\in\operatorname{\mathcal{A}}$ as

$\displaystyle\operatorname{\operatorname{Jac}}(f_{1},\ldots,f_{n}):=\left(\frac{\operatorname{\partial}f_{i}}{\operatorname{\partial}x_{j}}\right).$

The determinant $\det(\operatorname{\operatorname{Jac}}(f_{1},\ldots,f_{n}))$
is called _Jacobian_. It is clear the the $i$-th row of
$\operatorname{\operatorname{Jac}}(f_{1},\ldots,f_{n})$ is $\nabla(f_{i})$.

###  6.1 Nambu-Jacobi-Poisson algebras

.

Now we fix $f_{1},\ldots,f_{n-2}\in\operatorname{\mathcal{A}}$ and define the
following bilinear operation
$\\{-,-\\}\colon\operatorname{\mathcal{A}}\times\operatorname{\mathcal{A}}\to\operatorname{\mathcal{A}}$,
which yields a Poisson algebra structure on the polynomial algebra
$\operatorname{\mathcal{A}}$, called _Nambu-Poisson-Jacobi structure_.

######  Definition 6.1.

The _Nambu-Jacobi-Poisson bracket_ of $F,G\in\operatorname{\mathcal{A}}$ is

$\displaystyle\\{F,G\\}:=\det\operatorname{\operatorname{Jac}}(F,G,f_{1},\ldots,f_{n-2})\in\operatorname{\mathcal{A}}\
.$ (28)

When
$\operatorname{\mathcal{A}}=\operatorname{\mathbb{C}}[x_{1},x_{2},x_{3}]$,
there is only one $f$ determining the Nambu-Jacobi-Poisson bracket, and we
will denote the bracket by $\\{-,-\\}_{f}$. We proceed with the following
elementary lemma.

######  Lemma 6.1.

For $1\leq i\leq n$ and
$f_{1},\ldots,f_{i-1},f_{i+1},\ldots,f_{n}\in\operatorname{\mathcal{B}}$, the
operation $D\colon\operatorname{\mathcal{B}}\to\operatorname{\mathcal{B}}$
given by
$D(g):=\operatorname{\operatorname{Jac}}(f_{1},\ldots,f_{i-1},g,f_{i+1},\ldots,f_{n})$
is a derivation of $\operatorname{\mathcal{B}}$.

######  Theorem 6.1.

The bracket (28) is a Poisson bracket on $\operatorname{\mathcal{A}}$.

###### Proof.

To prove this theorem for any $n$, we observe that skew symmetry is evident
from the skew-symmetry of $\det$, the Leibniz rule follows from the lemma 6.1,
so the only non-trivial statement is, as usual, the Jacobi identity. But this
follows from the Fundamental identity for the Nambu bracket
$\\{f_{1},\ldots,f_{n}\\}:=\operatorname{\operatorname{Jac}}(f_{1},\ldots,f_{n})$
(see (29) below). ∎

An interesting property of the algebraic bracket structure on the polynomial
algebra is given by the following theorem.

######  Proposition 6.1.

If $f_{1},\ldots,f_{n-2}$ are algebraically dependent over
$\operatorname{\mathbb{C}}$, then $\\{-,-\\}=0$.

###### Proof.

The proof is left as an easy exercise for the reader. ∎

######  Remark 6.1.

If we consider $F,G$ as rational functions, the result of bracket (28) is
still a polynomial. Consider a quotient by (convenient) ideal
$\operatorname{\mathcal{B}}:=\operatorname{\mathbb{C}}[x_{1},\ldots,x_{n}]/<p_{1},\ldots,p_{k}>$
where $p_{i}\in\operatorname{\mathcal{A}}$. Then the theorem 6.1 still ohlds,
i.e. the bracket (28) yields a Poisson algebra structure on
$\operatorname{\mathcal{B}}$. This is valid in an even more general setup of
power series rings [3].

The Nambu-Jacobi-Poisson bracket is a special case of a $(n-m)$-ary operation

$\displaystyle\\{F_{1},\ldots,F_{n-m}\\}:=\lambda\det\operatorname{\operatorname{Jac}}(F_{1},\ldots,F_{n-m},f_{1},\ldots,f_{m})\in\operatorname{\mathcal{A}}\
,$

where $\lambda,F_{i}\in\operatorname{\mathcal{A}}$ for all $i$. The above
multibracket
$\\{-,\ldots,-\\}\colon\operatorname{\mathcal{A}}^{\otimes(n-m)}\to\operatorname{\mathcal{A}}$
satisfies for every permutation $\sigma$ the antisymmetry condition

$\displaystyle\\{F_{1},\ldots,F_{n-m}\\}=(-1)^{\sigma}\\{F_{\sigma(1)},\ldots,F_{\sigma(n-m)}\
.$

It also satisfies the Leibniz rule in every argument

$\displaystyle\\{hF_{1},\ldots,F_{n-m}\\}=F_{1}\\{h,\ldots,F_{n-m}\\}+h\\{F_{1},\ldots,F_{n-m}\\}\
,$

and the _Fundamental identity_

$\displaystyle\\{F_{1},\ldots,F_{n-m-1},\\{G_{1},\ldots,G_{n-m}\\}\\}$
$\displaystyle=$ (29)
$\displaystyle=\sum_{k}\\{G_{1},\ldots,G_{k-1},\\{F_{1},\ldots,F_{n-m-1},G_{k}\\},G_{k+1},\ldots,G_{n-m}\\}\
,$ (30)

which is a generalization of the Jacobi identity (the Jacobi identity and
Nambu-Jacobi-Poisson bracket is restored in the case $n-m=2$). For more
details about the Nambu structures and their generalizations, see [3].

Consider now a $(n-2)\times 2$ matrix over $\operatorname{\mathcal{A}}$

$\displaystyle M=\begin{pmatrix}a_{1,1}&\ldots&a_{1,n}\\\ \vdots&&\vdots\\\
a_{n-2,1}&\ldots&0a_{n-2,n}\end{pmatrix}.$

Suppose $i\neq j$ and denote by $\hat{M}_{ij}$ the matrix given by deleting
the $i$-th and $j$-th column (thus the result being $(n-2)\times(n-2)$ matrix.
If we choose

$\displaystyle
M=\begin{pmatrix}\frac{\operatorname{\partial}f_{1}}{\operatorname{\partial}x_{1}}&\ldots&\frac{\operatorname{\partial}f_{1}}{\operatorname{\partial}x_{n}}\\\
\vdots&&\vdots\\\
\frac{\operatorname{\partial}f_{n-2}}{\operatorname{\partial}x_{1}}&\ldots&\frac{\operatorname{\partial}f_{n-2}}{\operatorname{\partial}x_{n}}\end{pmatrix},$

then for the generators of $\operatorname{\mathcal{A}}$ we have

$\displaystyle\\{x_{i},x_{j}\\}=(-1)^{i+j-1}\det\hat{M}_{ij}$

and we define $\\{x_{i},x_{i}\\}=0$.

######  Example 6.1.

Let $n=3,f=\frac{1}{3}({x_{1}}^{3}+{x_{2}}^{3}+{x_{3}}^{3})+\tau
x_{1}x_{2}x_{3}$, where $\tau\in\operatorname{\mathbb{C}}$. Then
$\\{x_{i},x_{j}\\}=\operatorname{\operatorname{Jac}}(x_{i},x_{j},f)=zx_{i}x_{j}+{x_{k}}^{2}$,
where $(i,j,k)$ is a permutation of $\\{1,2,3\\}$.

######  Example 6.2 (Sklyanin elliptic Poisson brackets).

Let $n=4$ and consider
$f_{1}=q_{1}(x_{1},x_{2},x_{3},x_{4}),f_{2}=q_{2}(x_{1},x_{2},x_{3},x_{4})$,
where $q_{1},q_{2}$ are _quadratic_ polynomials. Choosing
$q_{1}={x_{1}}^{2}+{x_{2}}^{2}+{x_{3}}^{2}+{x_{4}}^{2}$ and
$q_{2}=\alpha{x_{2}}^{2}+\beta{x_{3}}^{2}+\gamma{x_{4}}^{2}$ such that
$\alpha\beta\gamma+\alpha+\beta+\gamma\neq 0$, we obtain the original
Sklyanin-Poisson structure [24, 3, 25].

###  6.2 Poisson-Calabi-Yau algebra

This is a Jacobian algebra
$\operatorname{\mathcal{A}}=\operatorname{\mathbb{C}}[x_{1},x_{2},x_{3}]$ with
$f=-{x_{1}}^{2}x_{3}$ [4]. The Nambu-Jacobi-Poisson bracket (28) is

$\displaystyle\\{x_{1},x_{2}\\}_{f}=-{x_{1}}^{2},$
$\displaystyle\\{x_{2},x_{3}\\}_{f}=-2x_{1}x_{3},$
$\displaystyle\\{x_{1},x_{3}\\}_{f}=0\ .$

It is interesting that the algebra
$\Omega^{1}(\operatorname{\mathcal{A}})=\operatorname{\mathcal{A}}<dx_{1},dx_{2},dx_{3}>$
(think of a free algebra over $A$) is also a Poisson algebra with

$\displaystyle\\{dx_{1},dx_{2}\\}_{\Omega}$
$\displaystyle=d\\{x_{1},x_{2}\\}_{f}=-2x_{1}dx_{1}\ ,$
$\displaystyle\\{dx_{2},dx_{3}\\}_{\Omega}$
$\displaystyle=d\\{x_{2},x_{3}\\}_{f}=-2x_{3}dx_{1}-2x_{1}dx_{3}\ ,$
$\displaystyle\\{dx_{1},dx_{3}\\}_{\Omega}$
$\displaystyle=d\\{x_{1},x_{3}\\}_{f}=0\ .$

There is a corresponding sequence [19]

${0}$${\Omega^{0}(\operatorname{\mathcal{A}})}$${\Omega^{1}(\operatorname{\mathcal{A}})}$${\Omega^{2}(\operatorname{\mathcal{A}})}$${\Omega^{3}(\operatorname{\mathcal{A}})}$${0}$${0}$${\Lambda^{0}((\mathcal{D}(\operatorname{\mathcal{A}}))}$${\Lambda^{1}((\mathcal{D}(\operatorname{\mathcal{A}}))}$${\Lambda^{2}((\mathcal{D}(\operatorname{\mathcal{A}}))}$${\Lambda^{3}((\mathcal{D}(\operatorname{\mathcal{A}}))}$${0}$$\scriptstyle{\operatorname{\underline{\operatorname{d}}}}$$\scriptstyle{\cong}$$\scriptstyle{\operatorname{\underline{\operatorname{d}}}}$$\scriptstyle{\cong}$$\scriptstyle{\operatorname{\underline{\operatorname{d}}}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$

where $\operatorname{\underline{\operatorname{d}}}$ is the universal
derivative (see (15)) and $d_{\pi}$ is given by (27). The isomorphism are
coming from the duality (16).

####  6.2.1 Low-dimensional cohomology of the PCY algebra

.

Considering the above sequence, we firstly notice that
$\Lambda^{0}((\mathcal{D}(\operatorname{\mathcal{A}}))\cong\operatorname{\mathcal{A}}$
and
$\Lambda^{1}((\mathcal{D}(\operatorname{\mathcal{A}}))\cong\mathcal{D}(\operatorname{\mathcal{A}})$,
so that the first map amounts to mapping
$\operatorname{\mathcal{A}}\to\mathcal{D}(\operatorname{\mathcal{A}})$, and
$\delta_{\pi}(a)=[\\![\pi,a]\\!]=\operatorname{\partial}_{a}\in\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})$
is a Hamiltonian derivation. Thus $\delta_{\pi}(a)=0$ iff $a$ is a Casimir
element (see def. (12)) of $\operatorname{\mathcal{A}}$ and we have

$\displaystyle
HP^{0}(\operatorname{\mathcal{A}})\cong\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})=<{x_{1}}^{2}x_{3}>_{\operatorname{\mathcal{A}}}\
,$

where $\operatorname{\operatorname{Cas}}(\operatorname{\mathcal{A}})$ are the
Cassimir elements of $\operatorname{\mathcal{A}}$. For the first cohomology,
consider $\operatorname{\partial}\in\mathcal{D}(\operatorname{\mathcal{A}})$,
we have
$\delta_{\pi}(\operatorname{\partial})=[\\![\pi,\operatorname{\partial}]\\!]=-[\\![\operatorname{\partial},\pi]\\!]=\mathcal{L}_{\operatorname{\partial}}\pi$.
So we see that $\operatorname{\partial}\in\ker\delta_{\pi}$ iff
$\mathcal{L}_{\operatorname{\partial}}\pi=0$, meaning that $\pi$ is invariant
with respect to $\operatorname{\partial}$. We have already met these operators
in the section in which we computed the low-dimensional Poisson cohomology for
more general algebras: the set of such operators is denoted
$\operatorname{Can}(\operatorname{\mathcal{A}})$ and $\operatorname{\partial}$
is called Poisson canonical. The first cohomology is

$\displaystyle
HP^{1}(\operatorname{\mathcal{A}})=\operatorname{Can}(\operatorname{\mathcal{A}})/\operatorname{\operatorname{Ham}}(\operatorname{\mathcal{A}})\
.$

###  6.3 Dual Poisson complex

Consider the chain complex

${\Lambda^{3}((\operatorname{\mathcal{D}}(\operatorname{\mathcal{A}}))}$${\Lambda^{2}((\mathcal{D}(\operatorname{\mathcal{A}}))}$${\Lambda^{1}((\mathcal{D}(\operatorname{\mathcal{A}}))}$${\Lambda^{0}((\mathcal{D}(\operatorname{\mathcal{A}}))\
.}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$

This complex was introduced by Brylinski in [7]. It has highly non-trivial
(Poisson) homology. For example, the lowest homology is

$\displaystyle
HP_{0}(\operatorname{\mathcal{A}})\cong\operatorname{\mathcal{A}}/<\\{a,b\\}_{f}|\
a,b\in\operatorname{\mathcal{A}}>_{\operatorname{\mathcal{A}}}\ ,$

since
$\Lambda^{0}((\mathcal{D}(\operatorname{\mathcal{A}}))=\operatorname{\mathcal{A}}$
and the image of
$\delta_{\pi}\colon\Lambda^{1}((\mathcal{D}(\operatorname{\mathcal{A}}))\to\Lambda^{0}((\mathcal{D}(\operatorname{\mathcal{A}}))$
is given by $\delta_{\pi}(adb)=\\{a,b\\}_{f}$. Following the definition of the
Nambu-Jacobi-Poisson bracket on $\operatorname{\mathcal{A}}$ we get

$\displaystyle\\{a,b\\}_{f}=\det\begin{pmatrix}\operatorname{\partial}_{1}a&\operatorname{\partial}_{2}a&\operatorname{\partial}_{3}a\\\
\operatorname{\partial}_{1}b&\operatorname{\partial}_{2}b&\operatorname{\partial}_{3}b\\\
\operatorname{\partial}_{1}f&\operatorname{\partial}_{2}f&\operatorname{\partial}_{3}f\end{pmatrix}$

Writing $\nabla
a=(\operatorname{\partial}_{1}a,\operatorname{\partial}_{2}a,\operatorname{\partial}_{3}a)$,
we can express $\\{a,b\\}_{f}$ as

$\displaystyle\\{a,b\\}_{f}=\nabla f\cdot(\nabla a\times\nabla b)\ ,$

where $\cdot$ is the dot product and $\times$ is the vector product. More
details on Poisson (co)homology of the Dual Poisson complex can be found in
[8].

######  Exercise 6.1.

Describe all differentials in the complex above in terms of vector alanysis
operations: $\nabla,\operatorname{curl},\times,(-,-),\operatorname{div}$.

Generalized SPDUNR Poisson algebra. The following is the generalized Sklyanin-
Painlevé-Dubrovin-Ugaglia-Nelson-Regge Poisson algebra:
$\operatorname{\mathcal{A}}_{f}=(\operatorname{\mathbb{C}}[x_{1},x_{1},x_{3}],\\{-,-\\}_{f})$,
where $\\{-,-\\}_{f}$ is the Jacobian Poisson-Nambu structure on
$\operatorname{\mathbb{C}}^{3}$ (see (28)), and
$F,G\in\operatorname{\mathbb{C}}[x_{1},x_{1},x_{3}]$. Let $M_{f}$ be the zero
locus of

$\displaystyle
f=x_{1}x_{1}x_{3}+\sum_{i=1}^{3}a_{i}{x_{i}}^{3}-\sum_{i=1}^{3}\epsilon_{i}{x_{i}}^{2}+\sum_{i=1}^{3}c_{i}x_{i}+\omega\
,$

where $\epsilon_{i}\in\\{0,1\\}$ and
$a_{i},c_{i},\omega\in\operatorname{\mathbb{C}}$. The bracket is given by

$\displaystyle\\{f,x_{i}\\}_{f}:=0,\text{for }i=1,2,3\ ,$

and

$\displaystyle\\{x_{1},x_{2}\\}_{f}:=x_{1}x_{2}+3a_{3}{x_{3}}^{2}-2\epsilon_{3}x_{3}+b_{3}\
,$

the result being cyclic in $(1,2,3)$ for other $x_{i},x_{j}$. For a generic
set of constraints, the bracket is nowhere vanishing on $M_{f}$.

##  7 Graded Poisson algebras

Let $\operatorname{\mathcal{A}}$ be a Poisson algebra. We shall suppose that
$\operatorname{\mathcal{A}}$ is an associative _gradded algebra_ , that is,
$\operatorname{\mathcal{A}}$ contains a set of vector subspaces
$(\operatorname{\mathcal{A}}^{k})_{k\in\operatorname{\mathbb{N}}_{0}}$ s.t.
$A=\bigoplus_{k\in\operatorname{\mathbb{N}}_{0}}\operatorname{\mathcal{A}}^{k}$
and
$\operatorname{\mathcal{A}}^{k}\cdot\operatorname{\mathcal{A}}^{l}\subset\operatorname{\mathcal{A}}^{k+l}$
for all $k,l\in\operatorname{\mathbb{N}}_{0}$. Moreover, we assume
$\operatorname{\mathcal{A}}_{0}:=\operatorname{\mathbb{K}}$.

######  Definition 7.1.

Let $d\in\operatorname{\mathbb{N}}_{0}$ be arbitrary.
$\operatorname{\mathcal{A}}$ is called a _graded Poisson algebra of degree
$d$_ if $\forall
a\in\operatorname{\mathcal{A}}^{k},b\in\operatorname{\mathcal{A}}^{l}:\\{a,b\\}\in\operatorname{\mathcal{A}}^{k+l-d}$
(for $n<0$ define $A^{n}=0$).

The graded Poisson algebras can be constructed from non-commutative,
associative, unital algebras $\operatorname{\mathcal{U}}$, which are
_filtered_ :

$\operatorname{\mathcal{U}}=\bigcup_{k\in\operatorname{\mathbb{N}}_{0}}\operatorname{\mathcal{U}}_{k},\text{
where
}1\in\operatorname{\mathbb{K}}=\operatorname{\mathcal{U}}_{0}\subset\operatorname{\mathcal{U}}_{1}\subset\dots\subset\operatorname{\mathcal{U}}_{k}\subset\dots.$

To every such algebra $\operatorname{\mathcal{U}}$, we can define the
associated graded algebra
$\operatorname{\mathcal{S}}=\operatorname{\operatorname{gr}}(\operatorname{\mathcal{U}})$,
$\operatorname{\mathcal{S}}=\bigoplus_{k\in\operatorname{\mathbb{N}}_{0}}\operatorname{\mathcal{S}}_{k}$,
where
$\operatorname{\mathcal{S}}_{k}:=\operatorname{\mathcal{U}}_{k}/U_{k-1}$,
$k\geq 1$, and
$\operatorname{\mathcal{S}}_{0}:=\operatorname{\mathcal{U}}_{0}=\operatorname{\mathbb{K}}$.
We denote by
$\operatorname{\operatorname{gr}}_{k}\colon\operatorname{\mathcal{U}}_{k}\to\operatorname{\mathcal{S}}_{k}$
the canonical projections. Then $\operatorname{\mathcal{U}}$ is a graded
Poisson algebra of degree $d\geq 1$ if $uv-
vu\in\operatorname{\mathcal{U}}_{k+l-d}$, for all
$u\in\operatorname{\mathcal{U}}_{k},v\in\operatorname{\mathcal{U}}_{l}$
(define $\operatorname{\mathcal{U}}_{k}=0$ for $k<0$).

######  Proposition 7.1.

The associated graded algebra $\operatorname{\mathcal{S}}$ given by the graded
algebra $\operatorname{\mathcal{U}}$ of degree $d$ is a graded Poisson algebra
of degree $d$.

###### Proof.

Since the canonical projection
$\operatorname{\operatorname{gr}}_{k}\colon\operatorname{\mathcal{U}}_{k}\to
S_{k}$ is a surjection, to each $a\in\operatorname{\mathcal{S}}$ exists
$k\in\operatorname{\mathbb{N}}_{0}$ so that
$a\in\operatorname{\mathcal{S}}_{k}$, thus there exists
$u\in\operatorname{\mathcal{U}}_{k}$ s.t.
$a=\operatorname{\operatorname{gr}}_{k}(u)$. Let
$b\in\operatorname{\mathcal{S}}_{l}$ and $v\in\operatorname{\mathcal{U}}_{l}$
s.t. $b=gr_{l}(v)$. The product in $\operatorname{\mathcal{S}}$ is defined by

$ab=gr_{k+l}(uv)\in\operatorname{\mathcal{S}}_{k+l}\ .$

The product is well-defined, because if we pick different representatives, say
$\tilde{a}=a+x,x\in\operatorname{\mathcal{U}}_{k-1}$ and
$\tilde{b}=b+y,y\in\operatorname{\mathcal{U}}_{l-1}$, then

$\tilde{a}\tilde{b}=(a+x)(b+y)=ab+\underbrace{ay+xb+xy}_{\in\operatorname{\mathcal{U}}_{k+l-1}}=ab\
.$

The unit in $\operatorname{\mathcal{S}}$ is the same as in
$\operatorname{\mathcal{U}}$, and the associativity of
$\operatorname{\mathcal{S}}$ is also inherited from the associativity of
$\operatorname{\mathcal{U}}$. So we see that $\operatorname{\mathcal{S}}$ is a
graded algebra. The Lie bracket is

$\\{a,b\\}=\\{\operatorname{\operatorname{gr}}_{k}(u),\operatorname{\operatorname{gr}}_{l}(v)\\}:=gr_{k+l-d}(uv-
vu)$

and is of degree $d$. By a similar argument as for the product, $\\{-,-\\}$ is
well-defined on $\operatorname{\mathcal{S}}$. Since
$ab\in\operatorname{\mathcal{S}}_{k+l}$ and
$ba\in\operatorname{\mathcal{S}}_{l+k}=\operatorname{\mathcal{S}}_{k+l}$,
there exist $u,u^{\prime}\in\operatorname{\mathcal{U}}_{k+l}$ s.t.
$ab=\operatorname{\operatorname{gr}}_{k+l}(u),ba=\operatorname{\operatorname{gr}}_{k+l}(u^{\prime})$
and thus $ab-ba=\operatorname{\operatorname{gr}}_{k+l}(u-u^{\prime})=0$. This
shows that the product in $\operatorname{\mathcal{S}}$ is commutative, hence
$\operatorname{\mathcal{S}}$ is Poisson graded of degree $d$. ∎

Module structures on $\operatorname{\mathcal{A}}$. Let
$\operatorname{\mathcal{A}}$ be a commutative, associative
$\operatorname{\mathbb{K}}$-algebra with the unit $1$.
$\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})=\operatorname{\operatorname{Hom}}(\operatorname{\mathcal{A}},\operatorname{\mathcal{A}})$
has two $\operatorname{\mathcal{A}}$-module structures, _left_ and _right_ :
$\forall a,x\in\operatorname{\mathcal{A}}$

$\displaystyle l_{a}\varphi(x):=a\varphi(x),$ $\displaystyle
r_{a}\varphi(x):=\varphi(ax)\ .$ (31)

######  Lemma 7.1.

For arbitrary $a\in A$, denote

$\displaystyle\delta_{a}:=r_{a}-l_{a}.$ (32)

Then $\delta_{a}$ satisfies the Leibniz rule

$\delta_{a}(\varphi\circ\psi)=\delta_{a}\varphi\circ\psi+\varphi\circ\delta_{a}\psi\
,$

that is,
$\delta_{a}\in\mathcal{D}(\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}}))\subset\operatorname{\operatorname{Hom}}(\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}}),\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}}))$.

###### Proof.

For arbitrary
$\varphi,\psi\in\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$
and $a,u\in\operatorname{\mathcal{A}}$, using the definition (32), we have

$\delta_{a}(\varphi\circ\psi)(u)=\delta_{a}(\varphi(\psi(u)))=\varphi(\psi(au))-a\varphi(\psi(u))\
.$

On the other hand

$\displaystyle(\delta_{a}\varphi\circ\psi+\varphi\circ\delta_{a}\psi)(u)$
$\displaystyle=\delta_{a}\varphi(\psi(u))+\varphi(\delta_{a}\psi(u))$
$\displaystyle=\varphi(a\psi(u))-a\varphi(\psi(u))+\varphi(\psi(au)-a\psi(u))$
$\displaystyle=\varphi(\psi(au))-a\varphi(\psi(u))\ .\qed$

######  Lemma 7.2.

$[\delta_{a},\delta_{b}]=0$ for all $a,b\in\operatorname{\mathcal{A}}$, where
$\delta_{a}$ is given by (32).

###### Proof.

The proof is a straightforward computation and we leave it to the reader as an
exercise. ∎

###  7.1 Algebra of differential operators

######  Definition 7.2.

For all $k\in\operatorname{\mathbb{N}}_{0}$, we define

$\operatorname{\operatorname{Diff}}_{k}(\operatorname{\mathcal{A}}):=\bigcap_{\underset{0\leq
i\leq
k}{a_{i}\in\operatorname{\mathcal{A}}}}\ker(\delta_{a_{0}}\circ\dots\delta_{a_{i}})$

and

$\operatorname{\operatorname{Diff}}_{*}(\operatorname{\mathcal{A}}):=\bigcup_{k\geq
0}\operatorname{\operatorname{Diff}}_{k}(\operatorname{\mathcal{A}})\ ,$

which is an abelian group under the addition $+$. Then
$\operatorname{\operatorname{Diff}}_{*}(\operatorname{\mathcal{A}})$ inherits
the two $\operatorname{\mathcal{A}}$-module structures (31) of
$\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$. We will write
$\operatorname{\operatorname{Diff}}_{*}^{(+)}(\operatorname{\mathcal{A}})$ to
emphasize the bimodule structure. The elements of
$\operatorname{\operatorname{Diff}}_{k}(\operatorname{\mathcal{A}})$ will be
called diffferential operators of order $\leq k$ on a commutative algebra
$\operatorname{\mathcal{A}}$.

######  Observation 7.1.

Directly from the above definition we have that for all
$k\in\operatorname{\mathbb{N}}_{0}:\operatorname{\operatorname{Diff}}_{k-1}(\operatorname{\mathcal{A}})\subset\operatorname{\operatorname{Diff}}_{k}(\operatorname{\mathcal{A}})$.

######  Remark 7.1.

One can generalize the above definition to the case $\varphi\colon P\to Q$,
where $P$ and $Q$ are projective, finitely generated
$\operatorname{\mathcal{A}}$-modules ($0\leq k$) and $\varphi$ is an
$\operatorname{\mathcal{A}}$-module homomorphism. Then

$\operatorname{\operatorname{Diff}}_{k}(P,Q):=\\{\varphi\colon P\to Q\ |\
\delta_{a_{0}}\circ\dots\circ\delta_{a_{k}}(\varphi)=0\text{ for all
}a_{0},\dots,a_{k}\in\operatorname{\mathcal{A}}\\}\ .$

In this notation,
$\operatorname{\operatorname{Diff}}_{k}(\operatorname{\mathcal{A}})=\operatorname{\operatorname{Diff}}_{k}(\operatorname{\mathcal{A}},\operatorname{\mathcal{A}})$.

######  Example 7.1 (Lie algebra structure on
$\operatorname{\operatorname{Diff}}_{*}^{(+)}(\operatorname{\mathcal{A}})$).

Consider
$\operatorname{\operatorname{End}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$,
equipped with a Lie algebra structure given by the commutator

$[\varphi,\psi]=\varphi\circ\psi-\psi\circ\varphi\ .$

Using the derivation property from lemma 7.1, we have

$\displaystyle\delta_{a}[\varphi,\psi]=[\delta_{a}\varphi,\psi]+[\varphi,\delta_{a}\psi]$
(33)

for all $a\in\operatorname{\mathcal{A}}$ and
$\varphi\in\operatorname{\operatorname{End}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$.
Hence $\delta_{a}$ acts as a derivation on the commutator. Suppose that
$\varphi,\psi\in\operatorname{\operatorname{Diff}}_{1}^{(+)}(\operatorname{\mathcal{A}})$,
then from (33) we get

$\displaystyle\delta_{b}\circ\delta_{a}[\varphi,\psi]=[\delta_{a}\varphi,\delta_{b}\psi]+[\delta_{b}\varphi,\delta_{a}\psi]\
,$

which does not have to vanish. Hence
$[\varphi,\psi]\notin\operatorname{\operatorname{Diff}}_{1}^{(+)}(\operatorname{\mathcal{A}})$,
meaning that
$\operatorname{\operatorname{Diff}}_{1}^{(+)}(\operatorname{\mathcal{A}})$ is
not a Lie algebra. Applying $\delta$ once again yields

$\displaystyle\delta_{c}\circ\delta_{b}\circ\delta_{a}[\varphi,\psi]=0\ .$

Thus
$[\varphi,\psi]\in\operatorname{\operatorname{Diff}}_{2}^{(+)}(\operatorname{\mathcal{A}})$.
Proceeding in a similar fashion one can show that the composition of
$\varphi\in\operatorname{\operatorname{Diff}}_{k}^{(+)}(\operatorname{\mathcal{A}})$
and
$\psi\in\operatorname{\operatorname{Diff}}_{l}^{(+)}(\operatorname{\mathcal{A}})$
is of order $\leq k+l$, that is
$\varphi\circ\psi\in\operatorname{\operatorname{Diff}}_{k+l}^{(+)}(\operatorname{\mathcal{A}})$
and the filtered bimodule
$\operatorname{\operatorname{Diff}}_{*}^{(+)}(\operatorname{\mathcal{A}})$ is
a Lie algebra.

######  Remark 7.2.

The $\operatorname{\mathcal{A}}$-bimodule
$\operatorname{\operatorname{Diff}}_{*}^{(+)}(P,P)$ is also filtered, since

$\delta(\varphi\circ\psi)=\delta(\varphi)\circ\psi+\varphi\circ\delta\psi$

and so the composition of a differential operator of degree $\leq k$ with a
differential operator of degree $\leq l$ results in a differential operator of
degree $\leq k+l$

$\operatorname{\operatorname{Diff}}_{k}^{(+)}(P,P)\otimes_{\operatorname{\mathbb{K}}}\operatorname{\operatorname{Diff}}_{l}^{(+)}(P,P)\to\operatorname{\operatorname{Diff}}_{k+l}^{(+)}(P,P)\
.$

######  Example 7.2.

Let
$\varphi\in\operatorname{\operatorname{End}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathcal{A}})$,
and recall that $\operatorname{\mathcal{A}}$ is associative, commutative and
unital algebra over $\operatorname{\mathbb{K}}$.

  * •

$\operatorname{\operatorname{Diff}}_{0}(\operatorname{\mathcal{A}})=\underset{a\in\operatorname{\mathcal{A}}}{\cap}\ker\delta_{a}$.
Using the definition (32), we have

$\displaystyle\delta_{a}\varphi(u)=\varphi(au)-a\varphi(u)\ .$

So
$\varphi\in\operatorname{\operatorname{Diff}}_{0}(\operatorname{\mathcal{A}})$
iff $\varphi(au)=a\varphi(u)$ for all $a,u\in\operatorname{\mathcal{A}}$.
Choosing $u=1$ and writing $a=a1$ gives $\varphi(a1)=a\varphi(1)$, meaning
that $\varphi$ is completely determined by its value on the unit element of
$\operatorname{\mathcal{A}}$, which gives

$\operatorname{\operatorname{Diff}}_{0}(\operatorname{\mathcal{A}})=\operatorname{\operatorname{End}}_{\operatorname{\mathcal{A}}}(\operatorname{\mathcal{A}})=\operatorname{\mathcal{A}}\
.$

Note that the above case is rather special. If we consider the case of
$\operatorname{\mathcal{A}}$-modules $P,Q$, then we get

$\operatorname{\operatorname{Diff}}_{0}(P,Q)=\operatorname{\operatorname{End}}_{\operatorname{\mathcal{A}}}(P,Q)\
.$

  * •

$\operatorname{\operatorname{Diff}}_{1}(\operatorname{\mathcal{A}})=\underset{a,b\in\operatorname{\mathcal{A}}}{\cap}\ker\delta_{b}\circ\delta_{a}$,
where

$\displaystyle\delta_{b}\circ\delta_{a}\varphi(u)=\varphi(bau)-b\varphi(au)-a\varphi(bu)+ba\varphi(u)\
.$ (34)

  * •

$\operatorname{\operatorname{Diff}}_{2}(\operatorname{\mathcal{A}})=\underset{a,b,c\in\operatorname{\mathcal{A}}}{\cap}\ker\delta_{c}\circ\delta_{b}\circ\delta_{a}$,
where

$\displaystyle\delta_{c}\circ\delta_{b}\circ\delta_{a}\varphi(u)=\varphi(bau)-b\varphi(au)-a\varphi(bu)+ba\varphi(u)\
.$

The goal of the following example is to demonstrate that the above given
algebraic definition of differential operators on a commutative algebra
$\operatorname{\mathcal{A}}$ fits with the standard picture of differential
operators on functions.

######  Example 7.3.

Let $\operatorname{\mathcal{A}}=C^{\infty}(\operatorname{\mathbb{R}})$ be the
algebra of smooth functions of one real variable, the algebra binary operation
given by multiplication of functions. Take
$\varphi=\operatorname{\partial}_{x}:=\frac{\operatorname{\partial}}{\operatorname{\partial}x}$.
Then

$\displaystyle\delta_{a}\operatorname{\partial}_{x}(u)=0\iff\operatorname{\partial}_{x}(au)=a\operatorname{\partial}_{x}u\
.$ (35)

which is not the case for all $a,u\in C^{\infty}(\operatorname{\mathbb{R}})$.
As expected (since
$\operatorname{\partial}_{x}\notin\operatorname{\mathcal{A}}$), the above
implies
$\operatorname{\partial}_{x}\notin\operatorname{\operatorname{Diff}}_{0}(\operatorname{\mathcal{A}})$.
On the other hand, using (34),

$\displaystyle\delta_{b}\circ\delta_{a}\varphi(u)=\operatorname{\partial}_{x}(bau)-b\operatorname{\partial}_{x}(au)-a\operatorname{\partial}_{x}(bu)+ba\operatorname{\partial}_{x}u=0$

is satisified for all $a,b,u\in C^{\infty}(\operatorname{\mathbb{R}})$, thus
$\operatorname{\partial}_{x}\in\operatorname{\operatorname{Diff}}_{1}(C^{\infty}(\operatorname{\mathbb{R}}))$.
Similarly, if we take $f\operatorname{\partial}_{x}$, where $f\in
C^{\infty}(\operatorname{\mathbb{R}})$, then

$\displaystyle\delta_{b}\circ\delta_{a}f\operatorname{\partial}_{x}(u)=f\operatorname{\partial}_{x}(bau)-bf\operatorname{\partial}_{x}(au)-af\operatorname{\partial}_{x}(bu)+baf\operatorname{\partial}_{x}u=0\
,$

and so
$f\operatorname{\partial}_{x}\in\operatorname{\operatorname{Diff}}_{1}(C^{\infty}(\operatorname{\mathbb{R}}))$.
Finally, consider
$\operatorname{\partial}^{2}_{x}:=\operatorname{\partial}_{x}\circ\operatorname{\partial}_{x}$.
Then

$\displaystyle\delta_{b}\circ\delta_{a}\operatorname{\partial}^{2}_{x}=\delta_{b}(\delta_{a}\operatorname{\partial}_{x}\circ\operatorname{\partial}_{x}+\operatorname{\partial}_{x}\circ\delta_{a}\operatorname{\partial}_{x})=\delta_{a}\operatorname{\partial}_{x}\circ\delta_{b}\operatorname{\partial}_{x}+\delta_{b}\operatorname{\partial}_{x}\circ\delta_{a}\operatorname{\partial}_{x}\
,$

which is not vanishing for all $a,b\in C^{\infty}(\operatorname{\mathbb{R}})$.
Thus
$\operatorname{\partial}^{2}_{x}\notin\operatorname{\operatorname{Diff}}_{1}(C^{\infty}(\operatorname{\mathbb{R}}))$.
We can easily check that

$\delta_{c}\delta_{b}\circ\delta_{a}f\operatorname{\partial}^{2}_{x}=0\ ,$

so
$f\operatorname{\partial}^{2}_{x}\in\operatorname{\operatorname{Diff}}_{2}(C^{\infty}(\operatorname{\mathbb{R}}))$.
One can show that
$f\operatorname{\partial}^{i}_{x}\in\operatorname{\operatorname{Diff}}_{i}(C^{\infty}(\operatorname{\mathbb{R}}))$,
for all $i\in\operatorname{\mathbb{N}}_{0}$ and $f\in
C^{\infty}(\operatorname{\mathbb{R}})$, where
$\operatorname{\partial}^{i}_{x}:=\underbrace{\operatorname{\partial}_{x}\circ\ldots\circ\operatorname{\partial}_{x}}_{\text{$i$-times}}$.
Since we have the sequence of inclusions

$\operatorname{\operatorname{Diff}}_{0}{\operatorname{\mathcal{A}}}\xhookrightarrow{}\operatorname{\operatorname{Diff}}_{1}{\operatorname{\mathcal{A}}}\xhookrightarrow{}\ldots\xhookrightarrow{}\operatorname{\operatorname{Diff}}_{k}{\operatorname{\mathcal{A}}}\xhookrightarrow{}\ldots\
.$

Altogether we get

$D_{k}:=\sum_{i=0}^{k}f_{i}\operatorname{\partial}^{i}_{x}\in\operatorname{\operatorname{Diff}}_{k}(C^{\infty}(\operatorname{\mathbb{R}}))\
.$

######  Definition 7.3.

Consider the factor space

$\displaystyle\operatorname{\operatorname{Smbl}}_{k}(\operatorname{\mathcal{A}}):=\operatorname{\operatorname{Diff}}_{k}^{(+)}(\operatorname{\mathcal{A}})/\operatorname{\operatorname{Diff}}_{k-1}^{(+)}(\operatorname{\mathcal{A}})\
.$

The _symbols algebra of $\operatorname{\mathcal{A}}$_ is

$\displaystyle\operatorname{\operatorname{Smbl}}_{*}(\operatorname{\mathcal{A}}):=\bigoplus_{k\in\operatorname{\mathbb{N}}_{0}}\operatorname{\operatorname{Smbl}}_{k}(\operatorname{\mathcal{A}})\
,$

with the graded algebra structure

$\displaystyle\operatorname{\operatorname{Smbl}}_{k}\cdot\operatorname{\operatorname{Smbl}}_{l}\subset\operatorname{\operatorname{Smbl}}_{k+l}$
$\displaystyle\\{\operatorname{\operatorname{Smbl}}_{k},\operatorname{\operatorname{Smbl}}_{l}\\}\subset\operatorname{\operatorname{Smbl}}_{k+l-1}\
.$

######  Remark 7.3.

Recall that to any filtered algebra, we can associate a graded commutative
algebra. This is precisely the case of the symbol algebra
$\operatorname{\operatorname{Smbl}}_{*}(\operatorname{\mathcal{A}})$ with
respect to the filtered algebra
$\operatorname{\operatorname{Diff}}_{*}^{(+)}(\operatorname{\mathcal{A}})$.

######  Proposition 7.2.

The symbol algebra
$\operatorname{\operatorname{Smbl}}_{*}(\operatorname{\mathcal{A}})$ is a
Poisson graded algebra of degree $1$.

######  Exercise 7.1.

  1. 1.

$\operatorname{\operatorname{Diff}}_{k}^{(+)}(\operatorname{\mathcal{A}})$ is
an $\operatorname{\mathcal{A}}$-submodule in
$\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$.

  2. 2.

$\operatorname{\operatorname{Diff}}_{*}(\operatorname{\mathcal{A}})$ is a
filtered subalgebra in
$\operatorname{\operatorname{End}}(\operatorname{\mathcal{A}})$.

  3. 3.

$\operatorname{\operatorname{Diff}}_{0}(\operatorname{\mathcal{A}})=\operatorname{\mathcal{A}}$
and
$\operatorname{\operatorname{Diff}}_{1}(\operatorname{\mathcal{A}})=\mathcal{D}(\operatorname{\mathcal{A}})\oplus\operatorname{\mathcal{A}}$
.

  4. 4.

$\operatorname{\operatorname{Smbl}}_{0}(\operatorname{\mathcal{A}})=\operatorname{\mathcal{A}}$
and
$\operatorname{\operatorname{Smbl}}_{1}(\operatorname{\mathcal{A}})=\mathcal{D}(\operatorname{\mathcal{A}})$.

##  8 Intermezzo - Tensor, symmetric and exterior algebras

###  8.1 Tensor algebra of a vector space

Tensor algebra of a $\operatorname{\mathbb{K}}$ vector space, denoted by
$T(V)$ is

$\displaystyle T(V):=\bigoplus_{k=0}^{\infty}T^{k}(V)\ ,$

where

$\displaystyle T^{k}(V):=\otimes^{k}V=V\otimes\ldots\otimes V\ .$

The space $T^{k}(V)$ consists of $\operatorname{\mathbb{K}}$-multilinear
mappings

$\displaystyle\tau\colon V^{*}\times\ldots\times
V^{*}\to\operatorname{\mathbb{K}}\ ,$

where $V^{*}$ is the vector space dual to $V$. by definition,
$T^{0}(V)=\operatorname{\mathbb{K}}$. Also, $T^{1}(V)=V$. The algebra product
in $T(V)$ is given by the canonical isomorphism defined by the tensor product

$\displaystyle T^{k}(V)\otimes T^{l}(V)\to T^{k+l}(V)\ .$

The tensor algebra satisfies the following universal property: for every
$\operatorname{\mathbb{K}}$-algebra $\operatorname{\mathcal{A}}$ and arbitrary
linear map $\psi\colon V\to\operatorname{\mathcal{A}}$, there is a uniquely
given linear map $\tilde{\psi}\colon T(V)\to\operatorname{\mathcal{A}}$ such
that the following diagram commutes

${V}$${T(V)}$${\operatorname{\mathcal{A}}}$$\scriptstyle{\iota}$$\scriptstyle{\psi}$$\scriptstyle{\exists!\tilde{\psi}}$

That is $\tilde{\psi}\circ\iota=\psi$, where $\iota$ is the canonical
embedding of $V$ into $T(V)$.

###  8.2 Symmetric algebra of a vector space

Let $V$ be a $\operatorname{\mathbb{K}}$-vector space, $T(V)$ its tensor
algebra. Consider

$\displaystyle\operatorname{\mathcal{J}}:=<u\otimes v-v\otimes u\ |\ u,v\in
V>\ ,$ (36)

which is a two-sided ideal in $T(V)$. Then the qoutient algebra
$S(V):=T(V)/\operatorname{\mathcal{J}}$, called the _symmetric algebra of
$V$_, is an associative and commutative algebra, satisfying the following
universal property. For any associative, commutative and unital algebra
$\operatorname{\mathcal{A}}$ and every linear mapping $\psi\colon
V\to\operatorname{\mathcal{A}}$, there is precisely one unital algebra
homomorphism $\tilde{\psi}$ such that the following diagram commutes

${V}$${S(V)}$${\operatorname{\operatorname{End}}(V)}$$\scriptstyle{\iota}$$\scriptstyle{\psi}$$\scriptstyle{\exists!\tilde{\psi}}$

where $\epsilon$ is the canonical embedding of $V$ into $S(V)$, given by the
composition of the canonical embedding $V\to T(V)$ and the canonical quotient
projection $T(V)\to S(V)$.

######  Remark 8.1.

Let us mention some useful properties about the above defined algebras.

  * •

$S(V)$ is a free, associative, commutative and unital algebra on $n=\dim V$
generators. The product is given as follows. To avoid confusion, we will
denote the classes in $S(V)$ with bracket notation. Let $[s]\in
S^{k}(V),[t]\in S^{l}(V)$. Then the product is

$[s][t]:=[s\otimes t]\in S^{k+l}(V)\ .$

This product is well defined. To check this, consider different
representatives of the equivalence classes $[\tilde{s}]=[s],[\tilde{t}]=[t]$.
This means there exist $j_{s},j_{t}\in\operatorname{\mathcal{J}}$ such that
$s=\tilde{s}+j_{s}$ and $t=\tilde{t}+j_{t}$. Then

$\displaystyle[s\otimes
t]=[(\tilde{s}+j_{s})\otimes\tilde{t}+j_{t})]=[\tilde{s}\otimes\tilde{t}+\tilde{s}\otimes
j_{t}+j_{s}\otimes\tilde{t}+j_{s}\otimes j_{t}]=[\tilde{s}\otimes\tilde{t}]\
,$

where the last equality follows from the fact that $\tilde{s}\otimes
j_{t}+j_{s}\otimes\tilde{t}+j_{s}\otimes j_{t}\in\operatorname{\mathcal{J}}$,
since $\operatorname{\mathcal{J}}$ is a two-sided ideal in $T(V)$.

  * •

The ideal $\operatorname{\mathcal{J}}$ is homogeneous, meaning that the factor
algebra $S(V)$ inherits the grading of $T(V)$. Thus we have
$S(V)=\oplus_{k=0}^{\infty}S^{k}(V)$, where
$S^{k}(V):=T^{k}(V)/\operatorname{\mathcal{J}}^{k}$, where
$\operatorname{\mathcal{J}}^{k}=\operatorname{\mathcal{J}}\cap T^{k}(V)$.

  * •

Consider the canonical projection $\operatorname{\operatorname{pr}}_{k}\colon
T^{k}(V)\to S^{k}(V)$. We can restrict $\operatorname{\operatorname{pr}}_{k}$
to the linear subspace of symmetric $k$-tensors

$\displaystyle\operatorname{\operatorname{pr}}_{k}|_{\operatorname{\operatorname{Sym}}_{k}(V)}\colon\operatorname{\operatorname{Sym}}_{k}(V)\to
S^{k}(V)\ .$

Because we always assume $\operatorname{char}\operatorname{\mathbb{K}}=O$, the
above map can be inverted, yielding a graded vector space isomorphism

$\operatorname{\operatorname{Sym}}(V)=\oplus_{k=0}^{\infty}\operatorname{\operatorname{Sym}}_{k}(V)\cong\oplus_{k=0}^{\infty}S^{k}(V)=S(V)$

Although the space of symmetric tensors and the symmetric algebra are
isomorphic as a graded $\operatorname{\mathbb{K}}$-vector spaces, it does not
make sense to speak about isomorphism of algebras, since
$\operatorname{\operatorname{Sym}}(V)$ does not posses algebra structure in
the sense that the tensor product of two symmetric tensors does not have to be
a symmetric tensor. In $\operatorname{char}\operatorname{\mathbb{K}}>0$, we
even lose the graded vector spaces isomorphism.

  * •

$\operatorname{\operatorname{Sym}}(V)$ is a linear subspace of $T(V)$.

  * •

$S(V)$ is a not a subalgebra of $T(V)$.

Symmetrization. Let $\operatorname{\operatorname{s}}_{k}\colon
T^{k}(V)\to\operatorname{\operatorname{Sym}}^{k}(V)$ be the symmetrization
map, given for arbitrary $t\in T^{k}(V)$ by

$\displaystyle\operatorname{\operatorname{s}}_{k}(t):=\frac{1}{k!}\sum_{\sigma\in\mathfrak{S}_{k}}\sigma\cdot
t\ ,$ (37)

where $\mathfrak{S}_{k}$ is the permutation group on $k$-elements and
$\sigma\cdot t$ is the action of the permutation group on $k$-tensors, given
on $t=t_{1}\otimes\ldots\otimes t_{k}$ by $\sigma\cdot
t:=t_{\sigma(1)}\otimes\ldots\otimes t_{\sigma(k)}$ (and we extend $\cdot$ on
general $t\in S_{k}$ by $\operatorname{\mathbb{K}}$-linearity).

######  Proposition 8.1.

The ideal (36) is a graded ideal
$\operatorname{\mathcal{J}}=\oplus_{k=0}^{\infty}\operatorname{\mathcal{J}}_{k}$,
where $\operatorname{\mathcal{J}}_{k}:=J\cap T^{k}(V)$. Moreover,
$\ker\operatorname{\operatorname{s}}_{k}=\operatorname{\mathcal{J}}_{k}$ and
$T^{k}(V)=\operatorname{\mathcal{J}}_{k}\oplus\operatorname{\operatorname{Sym}}^{k}(V)\cong\operatorname{\mathcal{J}}_{k}\oplus
S^{k}(V)$

Let $\\{e_{1},\ldots,e_{n}\\}$ be a basis of $V$. The map (37) gives an
isomorphism

$S(V)\cong\operatorname{\mathbb{K}}[x_{1},\ldots,x_{n}]$

such that

$\operatorname{\operatorname{s}}_{k}(e_{i_{1}}\otimes\ldots\otimes
e_{i_{k}})=x_{i_{1}}\dots x_{i_{k}}\ .$

The special case of PBW theorem says that the monoms
$\\{e_{1}^{i_{1}},\ldots,e_{n}^{i_{1}}\\}$ form a basis of $S(V)$ as a
$\operatorname{\mathbb{K}}$-vector space and
$S(V)\cong\operatorname{\mathbb{K}}[e_{1},\ldots,e_{n}]$.

######  Example 8.1.

Let $\operatorname{\mathfrak{g}}$ be a finite dimensional Lie algebra over
$\operatorname{\mathbb{K}}=\operatorname{\mathbb{R}}$ or
$\operatorname{\mathbb{K}}=\operatorname{\mathbb{C}}$. The symmetric algebra
$S(\operatorname{\mathfrak{g}}^{*})$ is isomorphic to
$\operatorname{\mathbb{K}}[\operatorname{\mathfrak{g}}^{*}]$, the polynomial
algebra with $\dim\operatorname{\mathfrak{g}}^{*}$ variables. The bracket on
$\operatorname{\mathfrak{g}}^{*}$ makes
$\operatorname{\mathbb{K}}[\operatorname{\mathfrak{g}}^{*}]$ a (commutive)
Poisson algebra.

A simpler version of this construciton is
$\operatorname{\mathcal{A}}=S(V^{*})=\operatorname{\mathbb{K}}[V^{*}]$, where
$V$ is a finite dimensional vector space equipped with a skew-symmetric
bilinear form $B\colon V\times V\to\operatorname{\mathbb{K}}$, which provides
a Poisson brackets on $\operatorname{\mathbb{K}}[V^{*}]$. For instance, let
$\dim V=2$. Consider $X,Y\in V$ linearly independent, so that $V=<X,Y>$. Then
$\operatorname{\mathbb{K}}[V^{*}]\cong\operatorname{\mathbb{K}}[X,Y]$ and
$\\{X,Y\\}=B(X,Y):=1$. In this case, the pair $(V,B)$ is a symplectic plane.

###  8.3 Exterior algebra of a vector space

Let $V$ be a $n$-dimensional $\operatorname{\mathbb{K}}$-vector space. The
_exterior algebra_ of $V$, denoted $\Lambda(V)$, is defined as a graded
subspace in the tensor algebra $T(V)$, formed by completely antisymmetric
tensors. Recall that $t\in T^{k}(V)$ is _completely antisymmetric_ (or
_alternating_ , or _completely skew-symmetric_) if

$\displaystyle\sigma\cdot t:=t_{\sigma(1)}\otimes\ldots\otimes
t_{\sigma(k)}=\operatorname{\operatorname{sgn}}(\sigma)t_{1}\otimes\ldots\otimes
t_{k}$

for all $k$-permutations $\sigma$ (and we extend $\cdot$ on general $t\in
S_{k}$ by $\operatorname{\mathbb{K}}$-linearity). Then $\Lambda^{k}(V)\subset
T^{k}(V)$ is formed by all such $k$-tensors. Note that $\Lambda^{1}(V)=V$ and
for $k=0$ we define $\Lambda^{0}(V)=\operatorname{\mathbb{K}}$. The whole
algebra is given by the $\operatorname{\mathbb{K}}$-vector space

$\Lambda(V):=\oplus_{k\geq 0}\Lambda^{k}(V)\ ,$

with the product $\wedge$, called _wedge product_ ,

$\Lambda^{k}(V)\otimes\Lambda^{l}(V)\to\Lambda^{k+l}(V)$

given by

$\displaystyle\omega_{1}\wedge\omega_{2}:=\frac{(k+l)!}{k!l!}\operatorname{\operatorname{Alt}}(\omega_{1}\otimes\omega_{2})\
,$ (38)

where

$\displaystyle\operatorname{\operatorname{Alt}}\colon
T^{k}(V)\to\Lambda^{k}(V)\ ,$

is a projection on the subspace of alternating tensors, called _alternating
map_ , defined for arbitrary $t\in T^{k}(V)$ as

$\displaystyle\operatorname{\operatorname{Alt}}(t):=\frac{1}{k!}\sum_{\sigma\in\mathfrak{S}_{k}}\operatorname{\operatorname{sgn}}(\sigma)\sigma\cdot
t\ .$

######  Example 8.2.

Consider $t\in T^{2}(V)$, given by $t=v\otimes w-w\otimes v$. Then
$\operatorname{\operatorname{Alt}}(t)=\frac{1}{2}(v\otimes w-w\otimes
v-w\otimes v+v\otimes w=t$, i.e. $t$ is already an element of
$\Lambda^{2}(V)$.

The wedge product satisfies

$\displaystyle\omega_{1}\wedge\omega_{2}$
$\displaystyle=(-1)^{pq}\omega_{2}\wedge\omega_{1},$
$\displaystyle\omega_{1}\wedge(\omega_{2}\wedge\omega_{3})$
$\displaystyle=(\omega_{1}\wedge\omega_{2})\wedge\omega_{3}$

meaning that $\Lambda(V)$ is a graded, associative
$\operatorname{\mathbb{K}}$-algebra with the unit $1$.

Similarly as in the case of symmetric algebra, we can define $\Lambda(V)$ as a
quotient of the tensor algebra. Consider a two-sided ideal
$\operatorname{\mathcal{J}}\subset T(V)$, given by

$\operatorname{\mathcal{J}}:=<t_{1}\otimes t_{2}+t_{2}\otimes t_{1}\ |\
t_{1},t_{2}\in T(V)>$

Then we define $\Lambda(V):=T(V)/\operatorname{\mathcal{J}}$. Moreover,
$\Lambda(V)$ satisfies the following universal property. For every
associative, unital algebra $\operatorname{\mathcal{A}}$ and any
$\operatorname{\mathbb{K}}$-linear map $\varphi\colon
V\to\operatorname{\mathcal{A}}$, such that $j(v)^{2}=0$ for all $v\in V$,
there is a unique algebra homorphism
$\tilde{\varphi}\colon\Lambda(V)\to\operatorname{\mathcal{A}}$, such that the
following diagram commutes

${V}$${\Lambda(V)}$${\operatorname{\mathcal{A}}}$$\scriptstyle{\iota}$$\scriptstyle{\varphi}$$\scriptstyle{\exists!\tilde{\varphi}}$

Using the universal property, one can show that the two above construction of
$\Lambda(V)$, either as a subspace of alternating tensors or the quotient
algebra, are isomorphic (in a unique way).

###  8.4 Poisson structure on a symmetric algebra
$S(\operatorname{\mathfrak{g}})$

######  Theorem 8.1.

$S(\operatorname{\mathfrak{g}})$ is a graded Poisson algebra of degree $1$ or
of degree $2$.

###### Proof.

(Degree 1.) Let $\operatorname{\mathfrak{g}}$ be a
$\operatorname{\mathbb{K}}$-Lie algebra. We will show that the Lie bracket
extends in a unique way to a Poisson bracket of degree $1$ on
$S(\operatorname{\mathfrak{g}})$. Let $g\in\operatorname{\mathfrak{g}}$ be
arbitrary. Consider the endomorphism666Notice that using the action of
$\operatorname{\mathfrak{g}}$ on itself given by multiplication from the left,
we can identify elements in $\operatorname{\mathfrak{g}}$ as endomorphisms of
$\operatorname{\mathfrak{g}}$, $h\mapsto l_{h}$. Then we have
$\delta_{g}(l_{h})(u)=h(gu)-gh(u)=(ad_{g}h)u$, thus
$\delta_{g}|_{\operatorname{\mathfrak{g}}}\equiv\operatorname{\operatorname{ad}}_{g}$.
$\operatorname{\operatorname{ad}}_{g}\equiv\delta_{g}\in\operatorname{\operatorname{End}}_{\operatorname{\mathbb{K}}}(\operatorname{\mathfrak{g}})$,
given by

$\displaystyle\operatorname{\operatorname{ad}}_{g}(h)=[g,h]\
,h\in\operatorname{\mathfrak{g}}\ .$

Then $\operatorname{\operatorname{ad}}_{g}$ can be extended in a unique way to
the whole $S(\operatorname{\mathfrak{g}})$ to a derivation on
$S(\operatorname{\mathfrak{g}})$ as follows. Consider a _decomposable_ $t\in
S(\operatorname{\mathfrak{g}})$, i.e. $t$ can be written as
$t=t_{1}\cdot\ldots\cdot t_{k}$, where $\cdot$ denotes the (symmetric) product
in $S(\operatorname{\mathfrak{g}})$ and
$t_{1},\ldots,t_{k}\in\operatorname{\mathfrak{g}}$. Then

$\displaystyle\operatorname{\operatorname{ad}}_{g}(t):=\sum_{i=1}^{k}t_{1}\cdot\ldots\cdot
t_{i-1}\cdot ad_{x}(t_{i})\cdot t_{i+1}\cdot\ldots\cdot t_{k}\ .$

A general element of $S(\operatorname{\mathfrak{g}})$ is a
$\operatorname{\mathbb{K}}$-linear combination of decomposables, so the above
definition of the bracket can be extended linearly to the whole
$S(\operatorname{\mathfrak{g}})$. This extension is also denoted
$\operatorname{\operatorname{ad}}_{g}$. For arbitrary $t,u\in
S(\operatorname{\mathfrak{g}})$ we have

$\displaystyle\operatorname{\operatorname{ad}}_{g}(t\cdot
u)=\operatorname{\operatorname{ad}}_{g}(t)\cdot
u+t\cdot\operatorname{\operatorname{ad}}_{g}(u)\ .$

(Degree 2.) It is sufficient to define a skew-symmetric bilinear form

$\displaystyle\Omega\colon\operatorname{\mathfrak{g}}\times\operatorname{\mathfrak{g}}\to\operatorname{\mathbb{K}}\
.$

Then there is a unique extension of
$\Omega(g,-)\colon\operatorname{\mathfrak{g}}\to\operatorname{\mathbb{K}}$ to
a derivation

$\operatorname{\partial}\colon
S(\operatorname{\mathfrak{g}})\to\mathcal{D}(S(\operatorname{\mathfrak{g}}))\
,$

which can be extended to a degree $2$ derivation on
$S(\operatorname{\mathfrak{g}})$. ∎

######  Example 8.3.

Let $\Omega$ be non-degenerate, skew-symmetric $2$-form on
$\operatorname{\mathfrak{g}}=\operatorname{\mathbb{R}}^{2m}$, that is, we
consider $\operatorname{\mathfrak{g}}$ to be abelian (the Lie bracket is
zero). Then there is a basis, called _symplectic basis_ , $e_{1},\ldots,e_{n}$
($n=2m$), such that the matrix of $\Omega$ is written in this basis as

$\begin{pmatrix}0&I_{m}\\\ -I_{m}&0\end{pmatrix}\ ,$

where $I_{n}$ is the identity $m\times m$ matrix. Then for each
$i\in\\{1,\ldots,m\\}$, the brackets of basis elements satisfy
$\\{e_{i},e_{i+m}\\}=-\\{e_{i+m},e_{i}\\}=1$ and all other combinations of
basis elements have zero brackets. Using this choice of basis, we can identify
$S(\operatorname{\mathfrak{g}})\cong\operatorname{\mathbb{K}}[x_{1},\ldots,x_{2m}]$.
Note that this is a degree 2 bracket on $\operatorname{\mathfrak{g}}$, and
differs from the Lie bracket $[-,-]_{\operatorname{\mathfrak{g}}}$, which we
assumed is trivial. Let $m=1$, so
$\operatorname{\mathfrak{g}}=\operatorname{\mathbb{R}}^{2}$ and we consider
coordinates $x,y$. Consider $P,Q\in\operatorname{\mathbb{K}}[x,y]$ given by

$\displaystyle P(x,y)=\sum_{i,j}a_{ij}x^{i}y^{j}\ ,$ $\displaystyle
Q(x,y)=\sum_{p,q}b_{pq}x^{p}y^{q}\ .$

The bracket is

$\\{P,Q\\}=\sum_{i,j,p,q}a_{ij}b_{pq}\\{x^{i}y^{j},x^{p}y^{q}\\}=\sum_{i,j,p,q}a_{ij}b_{pq}(jp-
iq)x^{i+p-1}y^{q+j-1}.$

If we pick $P=xy,Q=4x^{2}$ and choose the bracket as

$\\{P,Q\\}=\frac{\operatorname{\partial}P}{\operatorname{\partial}x}\frac{\operatorname{\partial}Q}{\operatorname{\partial}y}-\frac{\operatorname{\partial}Q}{\operatorname{\partial}x}\frac{\operatorname{\partial}P}{\operatorname{\partial}y}=-8x^{2}\
,$

with the general formula being

$\\{P,Q\\}=\sum_{i=1}^{m}\frac{\operatorname{\partial}P}{\operatorname{\partial}x_{i}}\frac{\operatorname{\partial}Q}{\operatorname{\partial}x_{i+m}}-\frac{\operatorname{\partial}Q}{\operatorname{\partial}x_{i+m}}\frac{\operatorname{\partial}P}{\operatorname{\partial}x_{i}}$

then this gives a bracket of degree $2$ on
$\operatorname{\mathbb{K}}[x_{1},\ldots,x_{2m}]$. The corresponding degree $2$
filtered Poisson algebra is
$\operatorname{\mathcal{A}}_{2m}(\operatorname{\mathfrak{g}})$. The algebra
$\operatorname{\mathcal{A}}_{2m}(\operatorname{\mathfrak{g}})$ has $2m$
generators $(x_{i},y_{i})$, $1\leq i\leq m$ and relations

$\displaystyle x_{i}y_{i}-y_{i}x_{i}=1\ ,1\leq i\leq m$ $\displaystyle
x_{i}y_{j}-y_{j}x_{i}=y_{i}y_{j}-y_{j}y_{i}=x_{i}x_{j}-x_{j}x_{i}=0\ ,i\neq j\
.$

$\operatorname{\mathcal{A}}_{2m}(\operatorname{\mathfrak{g}})$ is called the
$2m$-th Weyl algebra.

Let us conclude this section with the following remark, which puts the above
algebraic constructions in the context of smooth manifolds.

######  Remark 8.2.

The constructions of tensor algebra, symmetric algebra, and exterior algebra
can be extended to the case of smooth manifolds by considering the tangent (or
cotangent) space at a given point. This leads to the notion of _tensor fields_
as sections of the bundle $\otimes TM\to M$, _symmetric tensor fields_ as
sections of the bundle $S(TM)\to M$, and _differential forms_ as sections of
the bundle $\Lambda(TM)\to M$. We will speak more about these objects in
chapter concerning the differential calculus on Poisson manifolds.

##  9 Universal enveloping and PBW theorem

Let $\operatorname{\mathfrak{g}}$ be a finite dimensional Lie algebra over
$\operatorname{\mathbb{K}}$. We assume that there is an associative algebra
$\operatorname{\mathcal{A}}$ such that $\operatorname{\mathfrak{g}}$ can be
embedded in $\operatorname{\mathcal{A}}$ (so that one can multiply elements of
$\operatorname{\mathfrak{g}}$), and the Lie bracket of
$\operatorname{\mathfrak{g}}$ is given by the commutator in
$\operatorname{\mathcal{A}}$, i.e. for all $x,y\in\operatorname{\mathfrak{g}}$

$\displaystyle[x,y]=xy-yx\ ,$

where the product on the right-hans side is the product in
$\operatorname{\mathcal{A}}$. Let us denote by $\\{e_{i}\\}_{1\leq i\leq n}$ a
basis of $\operatorname{\mathfrak{g}}$ as a vector space. The Lie structure is
defined by the _structure constants_ $c_{ij}^{k}$

$\displaystyle[e_{i},e_{j}]=c_{ij}^{k}e_{k}.$

We consider
$S\subset\operatorname{\mathfrak{g}}\times\operatorname{\mathfrak{g}}$, a set
of pairs, defined as follows

$S=\\{(e_{i}e_{j},e_{j}e_{i}+\sum_{k=1}^{n}c_{ij}^{k}e_{k})\in\operatorname{\mathfrak{g}}\times\operatorname{\mathfrak{g}}\
|\ i>j\\}\ .$

It is easy to verify that there is no ambiguity in the presentation of the
triple product $e_{i}e_{j}e_{k}$ with $i>j>k$, since

$\displaystyle(e_{i}e_{j})e_{k}$
$\displaystyle=(\sum_{l=1}^{n}c_{ij}^{l}e_{l}+e_{j}e_{i})e_{k}=\sum_{l=1}^{n}c_{ij}^{l}e_{l}e_{k}+e_{j}e_{i}e_{k}\
,$ $\displaystyle e_{i}(e_{j}e_{k})$
$\displaystyle=e_{i}(\sum_{m=1}^{n}c_{jk}^{m}e_{m}+e_{k}e_{j})=\sum_{m=1}^{n}c_{jk}^{m}e_{i}e_{m}+e_{i}e_{k}e_{j}\
.$

Thus the difference vanishes due to the Jacobi identity (4)

$(e_{i}e_{j})e_{k}-e_{i}(e_{j}e_{k})=\sum_{l=1}^{n}\sum_{m=1}^{n}(c_{ij}^{m}c_{km}^{l}+c_{jk}^{m}c_{im}^{l}+c_{ki}^{m}c_{jm}^{l})e_{l}=0\
.$

###  9.1 Universal enveloping algebra

Let $\operatorname{\mathfrak{g}}$ be a $n$-dimensional ($n<\infty$) Lie
algebra, $\\{e_{i}\\}_{1\leq i\leq n}$ a basis of
$\operatorname{\mathfrak{g}}$. Let
$T(\operatorname{\mathfrak{g}})=\oplus_{i}T^{i}(\operatorname{\mathfrak{g}})$
be the tensor algebra of the underlying vector space of
$\operatorname{\mathfrak{g}}$.

######  Proposition 9.1.

$T(\operatorname{\mathfrak{g}})$ is an associative algebra with respect to the
tensor product $\otimes\colon T^{p}(\operatorname{\mathfrak{g}})\times
T^{q}(\operatorname{\mathfrak{g}})\to T^{p+q}(\operatorname{\mathfrak{g}})$

$\displaystyle(x_{1}\otimes\dots x_{p},y_{1}\otimes\dots y_{q})\mapsto
x_{1}\otimes\dots x_{p}\otimes y_{1}\otimes\dots y_{q}\ .$

######  Remark 9.1.

$T(\operatorname{\mathfrak{g}})$ is generated as
$T(\operatorname{\mathfrak{g}})=\operatorname{\mathbb{K}}<e_{1},\dots,e_{n}>$
(in the algebra sense). Put differently, the tensor algebra is isomorphic (as
a $\operatorname{\mathbb{K}}$-algebra) to a free, associative, non-commutative
$\operatorname{\mathbb{K}}$-algebra on $n$-generators.

Consider a subspace $\operatorname{\mathcal{I}}\subset
T(\operatorname{\mathfrak{g}})$, generated as

$\displaystyle\operatorname{\mathcal{I}}:=<t_{1}\otimes([x,y]-x\otimes
y+y\otimes x)\otimes t_{2}\ |\ t_{1},t_{2}\in
T(\operatorname{\mathfrak{g}}),x,y\in\operatorname{\mathfrak{g}}>\ .$ (39)

Note that $[x,y]-x\otimes y+y\otimes x\in\operatorname{\mathfrak{g}}\oplus
T^{2}(\operatorname{\mathfrak{g}})$.

######  Proposition 9.2.

$\operatorname{\mathcal{I}}$ is a (two-sided) ideal in
$T(\operatorname{\mathfrak{g}})$, i.e. $\forall
j\in\operatorname{\mathcal{I}},t\in T(\operatorname{\mathfrak{g}}):j\otimes
t,t\otimes j\in\operatorname{\mathcal{I}}$.

###### Proof.

The proof is obvious from the form of generators of
$\operatorname{\mathcal{I}}$. ∎

######  Remark 9.2.

Let $\operatorname{\mathcal{A}}$ be an associative algebra. We will denote by
$\operatorname{\mathcal{A}}_{\operatorname{\operatorname{Lie}}}$ the
corresponding Lie algebra
$\operatorname{\mathcal{A}}_{\operatorname{\operatorname{Lie}}}$, with the Lie
bracket given by the commutator $[x,y]:=xy-yx$, for all
$x,y\in\operatorname{\mathcal{A}}$.

######  Definition 9.1 (universal enveloping algebra).

Let $\operatorname{\mathfrak{g}}$ be a Lie algebra. The factor space

$\displaystyle\operatorname{\mathcal{U}}(\operatorname{\mathfrak{g}}):=T(\operatorname{\mathfrak{g}})/\operatorname{\mathcal{I}}\
,$

where $\operatorname{\mathcal{I}}$ is given by (39), is called the universal
enveloping algebra of $\operatorname{\mathfrak{g}}$.

######  Proposition 9.3.