Simplicity of Lie algebras of Poisson brackets

TL;DR

This paper proves that for a simple Poisson algebra over a field of zero characteristic, the associated Lie algebra formed by Poisson brackets modulo the center is also simple, highlighting a fundamental structural property.

Contribution

It establishes the simplicity of the Lie algebra derived from the Poisson brackets in simple Poisson algebras, extending understanding of their algebraic structure.

Findings

The Lie algebra of Poisson brackets modulo the center is simple.

Simplicity of the Poisson algebra implies simplicity of the associated Lie algebra.

Provides a structural link between Poisson and Lie algebra properties.

Abstract

Let $A$ be an associative commutative algebra with $1$ over a field of zero characteristic, $\{,\} : A \times A \to A$ is a Poisson bracket, $Z = \{ a \in A \mid \{a, A\} = (0) \}.$ We prove that if $A$ is simple as a Poisson algebra then the Lie algebra $\frac{\{A,A\}}{\{A,A\}\cap Z}$ is simple.

Equations

$$\{u\{a,v\},b\}=\{u,\{a,v\}b\}+\{\{a,v\},ub\}.$$

$$\{a\{u,v\},b\}=\{\{au,v\},b\}-\{u,\{a,v\}b\}-\{\{a,v\},ub\}.$$

$$0=ad(a{)}^n{b}_1^n=n!\{a,b_1{\}}^n+\sum _{\genfrac{}{}{0ex}{}{k_1+\cdots +k_n=n}{(k_1,\cdots ,k_n)\ne (1,1,\cdots ,1)}}(ad(a{)}^{k_1}{b}_1)\cdots (ad(a{)}^{k_n}{b}_1)$$

$$ad(u{)}^{3}A=\{u,\{u,\{u,A\}\}\}\subseteq \{U,\{U,\{A,A\}\}\subseteq \{U,U\}=(0).$$

$$ad(u{)}^{4}A=\{u,\{u,\{u,\{u,A\}\}\}\}\subseteq \{U^{[1]},\{U^{[1]},U^{[1]}\}\}\subseteq \{A,U^{[2]}\}=(0).$$

$$ad(u{)}^{4}A=\{u,\{u,\{u,\{u,A\}\}\}\}\subseteq \{U,\{U,U\}\}\subseteq \{A,U^{[1]}\}=(0).$$

Full text

1. Introduction

Let $\displaystyle F$ be a field and let $\displaystyle A$ be an associative commutative $\displaystyle F$- algebra. A bilinear bracket $\displaystyle\{,\}:A\times A\to A$ is called Poisson bracket if

1. (1)

$\displaystyle(A,\{,\})$ is a Lie algebra, 2. (2)

$\displaystyle\{ab,c\}=a\{b,c\}+\{a,c\}b$ for arbitrary elements $\displaystyle a,b,c\in A.$

An algebra $\displaystyle A$ with a Poisson bracket is called a Poisson algebra $\displaystyle(see[4]).$ The conditions (1), (2) imply the Poisson identity 3. (3)

$\displaystyle\{ab,c\}=\{a,bc\}+\{b,ac\}$ that we will use later.

We say that an ideal $\displaystyle I$ of an algebra $\displaystyle A$ is a Poisson ideal if $\displaystyle\{I,A\}\subseteq I.$ A Poisson algebra without nontrivial Poisson ideals is called a simple Poisson algebra. Let $\displaystyle Z=\{a\in A|\{a,A\}=(0)\}.$

2. Main Result

Theorem 1**.**

Let $\displaystyle(A,\{,\})$ be a simple Poisson algebra over a field of zero characteristic, $\displaystyle A\ni 1$. Then the Lie algebra $\displaystyle\nicefrac{{\{A,A\}}}{{\{A,A\}\cap Z}}$ is simple.

This theorem is an analog of the celebrated theorem of I. Herstein [3].

Lemma 1**.**

Let $\displaystyle(A,\{,\})$ be a simple Poisson algebra over a field of zero characteristic, $\displaystyle A\ni 1.$ Then the algebra $\displaystyle A$ does not contain a nonzero nilpotent element.

Proof.

Let $\displaystyle N(A)$ be the radical of the algebra $\displaystyle A,$ $\displaystyle N(A)=\{a\in A|\text{ the element }a\text{ is nilpotent}\}.$ It is known (see [2], p.107) that in an algebra over a field of zero characteristic the radical is invariant with respect to all derivations of the algebra $\displaystyle A.$ This implies $\displaystyle\{N(A),A\}\subseteq N(A),$ hence $\displaystyle N(A)$ is a Poisson ideal of $\displaystyle A.$ From the simplicity of the Poisson algebra $\displaystyle A$ it follows that $\displaystyle N(A)=(0)$ or $\displaystyle N(A)=A.$ The second option is not possible since $\displaystyle A\ni 1.$ Hence $\displaystyle N(A)=(0)$, which completes the proof of the lemma. ∎

For a subspace $\displaystyle X\subset A,$ let $\displaystyle id_{A}(X)=X+AX$ be the ideal of $\displaystyle A$ generated by $\displaystyle X.$ Let $\displaystyle U$ be an ideal of the Lie algebra $\displaystyle\{A,A\}.$ Define the descending derived series of ideals: $\displaystyle U^{[0]}=U,U^{[i+1]}=\{U^{[i]},U^{[i]}\}.$ The following lemma is an analog of the Lemma $\displaystyle 1$ from [1].

Lemma 2**.**

$\displaystyle\{id_{A}(U^{[3]}),A\}\subseteq U.$**

Proof.

For an arbitrary ideal $\displaystyle I$ of the Lie algebra $\displaystyle\{A,A\}$ we have $\displaystyle\{\{I,I\},A\}\subseteq\{I,\{I,A\}\}\subseteq I.$ Hence $\displaystyle\{U^{[i+1},A\}\subseteq U^{[i]}$ for $\displaystyle i\geq 0.$

For arbitrary elements $\displaystyle a,b,u,v\in A$ we have $\displaystyle\{au,v\}=a\{u,v\}+u\{a,v\}.$ Hence $\displaystyle\{\{au,v\},b\}=\{a\{u,v\},b\}+\{u\{a,v\},b\}.$ By the Poisson identity (3)

[TABLE]

This implies

[TABLE]

The first and the third summands on the right hand side lie in $\displaystyle\{\{v,A\},A\},$ the second summand lies in $\displaystyle\{u,A\}.$ If $\displaystyle u,v\in U^{[2]}$ then by the remark above $\displaystyle\{v,A\}+\{u,A\}\subseteq U^{[1]}$ and $\displaystyle\{\{v,A\},A\}\subseteq\{U^{[1]},A\}\subseteq U^{[0]}=U,\{id_{A}(U^{[3]}),A\}\subseteq U,$ which completes the proof of the lemma.

∎

From now on we assume that $\displaystyle A$ is a simple Poisson algebra over a field of characteristic $\displaystyle 0$ and $\displaystyle A\ni 1.$

Lemma 3**.**

If $\displaystyle I$ is a nonzero ideal of the algebra $\displaystyle A$ and $\displaystyle\{I,\{A,A\}\}\subseteq I$ then $\displaystyle I=A.$

Proof.

Let $\displaystyle\sqrt{I}$ be the set of all elements of the algebra $\displaystyle A$ that are nilpotent module $\displaystyle I.$ Then $\displaystyle\sqrt{I}\unlhd A$ and $\displaystyle\nicefrac{{\sqrt{I}}}{{I}}$ is the radical of the algebra $\displaystyle\nicefrac{{A}}{{I}}.$

Let $\displaystyle d$ be a derivation of the algebra $\displaystyle A$ such that $\displaystyle d(I)\subseteq I.$ Then $\displaystyle d$ induces a derivation on the factor algebra $\displaystyle\nicefrac{{A}}{{I}}.$ Since the radical of algebra $\displaystyle\nicefrac{{A}}{{I}}$ is invariant under all derivations of $\displaystyle\nicefrac{{A}}{{I}}$ if follows that $\displaystyle d\left(\nicefrac{{\sqrt{I}}}{{I}}\right)\subseteq\nicefrac{{\sqrt{I}}}{{I}}$ and therefore $\displaystyle d(\sqrt{I})\subseteq\sqrt{I}.$

This implies $\displaystyle\{\sqrt{I},\{A,A\}\}\subseteq\sqrt{I}.$ If $\displaystyle\sqrt{I}=A\ni 1$ then $\displaystyle I=A$ as well. From now on considering the ideal $\displaystyle\sqrt{I}$ instead of $\displaystyle I$ we will assume that $\displaystyle\sqrt{I}=I$.

For arbitrary element $\displaystyle a\in A,u\in I$ we have $\displaystyle\{u,\{u,a^{2}\}\}=2\left(\{u,a\}^{2}+\{u,\{u,a\}\}a\right),$ $\displaystyle\{u,\{u,A\}\}\subseteq\{u,\{A,A\}\}\subseteq I.$ Hence $\displaystyle\{u,a\}^{2}\in I,\{u,a\}\in\sqrt{I}=I.$ We proved that $\displaystyle\{I,A\}\subseteq I,$ i.e $\displaystyle I$ is a Poisson ideal of $\displaystyle A.$ Since the Poisson algebra$\displaystyle\left(A,\{,\}\right)$ is simple we conclude that $\displaystyle I=A.$ This completes the proof of lemma. ∎

Lemma 4**.**

If $\displaystyle U$ is a proper ideal of the Lie algebra $\displaystyle\{A,A\}$ then $\displaystyle U^{[3]}=0.$

Proof.

Let $\displaystyle I=id_{A}\left(U^{[3]}\right).$ Clearly $\displaystyle\{I,\{A,A\}\}\subseteq I.$ If $\displaystyle U^{[3]}\neq 0$ then by Lemma 3 we have $\displaystyle I=A.$ Hence by Lemma 2 $\displaystyle\{A,A\}\subseteq U.$ This completes the proof of the lemma. ∎

For an element $\displaystyle a\in A$ consider the adjoint operator$\displaystyle ad(a):A\to A,x\to\{a,x\}.$

Lemma 5**.**

Let $\displaystyle a\in A,n\geq 1$ and suppose that $\displaystyle ad(a)^{n}=0.$ Then $\displaystyle ad(a)=0.$

Proof.

We will prove that $\displaystyle ad(a)^{n}=0,n\geq 2,$ implies $\displaystyle ad(a)^{n-1}=0.$ Indeed, let $\displaystyle b\in A.$ Consider the element $\displaystyle b_{1}=ad(a)^{n-2}b,$ we have $\displaystyle\{a,\{a,b_{1}\}\}=ad(a)^{n}b=0.$ Now,

[TABLE]

In each summaned on the right hand side for at least one $\displaystyle i,\,1\leq i\leq n,$ we have $\displaystyle k_{i}\geq 2.$ Hence $\displaystyle ad(a)^{k_{i}}b_{1}=0.$ Since the characteristic of the ground field is $\displaystyle 0$ we conclude that $\displaystyle\{a,b_{1}\}^{n}=0.$ By Lemma $\displaystyle 1$ it implies that $\displaystyle\{a,b_{1}\}=ad(a)^{n-1}b_{1}=0,$ which completes the proof of the lemma. ∎

Lemma 6**.**

Let $\displaystyle U$ be an abelian ideal of the algebra $\displaystyle\{A,A\}.$ Then $\displaystyle\{U,A\}=(0).$

Proof.

For an arbitrary element $\displaystyle u\in U$ we have

[TABLE]

By Lemma $\displaystyle 5$ $\displaystyle ad(u)A=\{u,A\}=(0),$ which completes the proof of the lemma. ∎

Lemma 7**.**

Let $\displaystyle U$ be an ideal of the algebra $\displaystyle\{A,A\}$ such that $\displaystyle U^{[3]}=(0).$ Then $\displaystyle\{U,A\}=(0).$

Proof.

The ideal $\displaystyle U^{[2]}$ of the Lie algebra $\displaystyle\{A,A\}$ is abelian. Hence by Lemma $\displaystyle 6$ $\displaystyle\{U^{[2]},A\}=(0).$ Let $\displaystyle u\in U^{[1]}.$ we have

[TABLE]

Hence from Lemma $\displaystyle 5$ it follows that $\displaystyle\{U^{[1]},A\}=(0).$ Again choosing an arbitrary element $\displaystyle u\in U$ we get

[TABLE]

Hence by Lemma $\displaystyle 5$ $\displaystyle\{U,A\}=(0),$ which completes the proof of the lemma. ∎

Proof of Theorem 1.

Now combining Lemmas $\displaystyle 4$ and $\displaystyle 7$ we get the assertion of the theorem. ∎