3 Enveloping Algebras and Radicals of Algebras

Universal Enveloping Algebra

For any vector space $V$, define

$$T(V)=(F\cdot 1)\oplus V\oplus (V\otimes V)\oplus (V\otimes V\otimes V)+\cdots .$$

Note that for any $v\in V^{\otimes i}$ and $u\in V^{\otimes j}$, there is $vu\in V^{\otimes i+j}$. For any associative algebra $A$, any linear mapping $T:V\to A$ extends uniquely to $T(V)\to A$.

We call associative algebra $\mathcal{U}(\mathcal{J})=T(\mathcal{J})/I$, where

$$I=⟨[a^2,a],a\otimes b\otimes c+c\otimes b\otimes a+(a\circ c)\circ b-a\otimes b\circ c-b\otimes a\circ c-c\otimes a\circ b:a,b,c\in \mathcal{J}⟩.$$

Remark. $I$ is defined by

• ^fgzlin is equivalent to $[R_{a^2},R_a]=0$. $(\mathrm{I})$

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• $d$: $R_a{R}_b{R}_c+R_c{R}_b{R}_a+R_{(a\circ c)\circ b}=R_a{R}_{bc}+R_b{R}_{ac}+R_c{R}_{ba}$ $(\ast \ast )$. $(\mathrm{IV})$

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Let $\mathcal{B}$ be a Jordan algebra which contain $\mathcal{J}$, and let ${\mathrm{Mult}}_{\mathcal{B}}(\mathcal{J})={\mathrm{alg}}_{\mathrm{End}(\mathcal{B})}⟨R_a^{\mathcal{B}}:a\in \mathcal{J}⟩$.

Then ${\mathrm{Mult}}_{\mathcal{B}}(\mathcal{J})$ is a homomorphism image of $\mathcal{U}(\mathcal{J})$

$$\begin{array}{rl}\phi :\mathcal{U}(\mathcal{J})& \to {\mathrm{Mult}}_{\mathcal{B}}(\mathcal{J}),\\ a+I& \mapsto R_a^{\mathcal{B}},\text{ where }a\in \mathcal{J},\\ (a+I)\otimes (b+I)& \mapsto \phi (a+I)\phi (b+I).\end{array}$$

Also we have

$$\begin{array}{rl}\psi :\mathcal{U}(\mathcal{J})& \to \mathrm{Mult}(\mathcal{J}),\\ a+I& \mapsto R_a.\end{array}$$

Exercise. Prove that there exists $\mathcal{B}\supseteq \mathcal{J}$ such that ${\mathrm{Mult}}_{\mathcal{B}}(\mathcal{J})\cong \mathcal{U}(\mathcal{J})$. (Hint: define $\mathcal{B}=\mathcal{J}+\mathcal{M}$ with ${\mathcal{M}}^2=0$. $\mathcal{M}$ is Jordan bimodule iff $\mathcal{B}$ is Jordan algebra.) See here.

Remark. It is a remark on the Split Null Extension.

• The construction $\mathcal{B}=\mathcal{J}+\mathcal{M}$ with the condition ${\mathcal{M}}^2=0$ is a fundamental technique known as the split null extension of the Jordan algebra $\mathcal{J}$ by a Jordan bimodule $\mathcal{M}$.
  • $\mathcal{M}$ is a Jordan bimodule iff $\mathcal{B}$ is a Jordan algebra.
  • It is a effective way to create a larger space for $\mathcal{J}$.
• The parallel construction is $\mathcal{L}=\mathcal{G}+\mathcal{N}$ for Lie algebra $\mathcal{G}$, where $\mathcal{N}$ is a $\mathcal{G}$-module.
  • $\mathcal{N}$ is a $\mathcal{G}$-module iff $\mathcal{L}$ is a Lie algebra.

Nilpotent, Nil and Solvable

Take $\mathcal{B},\mathcal{C}\subseteq \mathcal{A}$ as two subalgebras of algebra $\mathcal{A}$. Define $\mathcal{B}\cdot \mathcal{C}=\{{\sum }_{i\in I,\text{finite}}{b}_i{c}_i:b_i\in \mathcal{B},c_i\in \mathcal{C}\}$, and define

• ${\mathcal{A}}^1=\mathcal{A}$, ${\mathcal{A}}^2=\mathcal{A}\cdot \mathcal{A}$, $\cdots$, ${\mathcal{A}}^n={\sum }_{i,j⩾1,i+j=n}{\mathcal{A}}^i\cdot {\mathcal{A}}^j$
• ${\mathcal{A}}^{⟨1⟩}=\mathcal{A}$, ${\mathcal{A}}^{⟨2⟩}=A^{⟨1⟩}\cdot {\mathcal{A}}^{⟨1⟩}$, $\cdots$, ${\mathcal{A}}^{⟨n⟩}=({\mathcal{A}}^{⟨n-1⟩}{)}^2$

Definition

An algebra $\mathcal{A}$ is called nilpotent if there exists $n>0$ such that ${\mathcal{A}}^n=0$. The least $n$ with ${\mathcal{A}}^n=0$ is called nilpotency index of $\mathcal{A}$.

$\mathcal{A}$ is called solvable if there exists $n$ such that ${\mathcal{A}}^{⟨n⟩}=0$, and the least $n$ with ${\mathcal{A}}^{⟨n⟩}=0$ is called solvability index of $\mathcal{A}$.

$\mathcal{A}$ is a nil algebra if $\mathrm{alg}⟨a⟩$ is nilpotent for any $a\in \mathcal{A}$. Remark that any nilpotent algebra is nil.

Remark. Any nilpotent algebra is solvable, because ${\mathcal{A}}^{⟨n⟩}\subseteq {\mathcal{A}}^{2^n}$ for any $n⩾1$.

Molien-Wedderburn

Let $\mathcal{A}$ be a finite-dimensional associative algebra. Then

• There exists maximal nil ideal $\mathrm{Nil}\mathcal{A}$ which contains all nil ideals of $\mathcal{A}$, which is called nil radical.
• $\mathrm{Nil}\mathcal{A}$ is nilpotent.
• $\mathcal{A}/\mathrm{Nil}\mathcal{A}={\mathcal{A}}_1\oplus \cdots \oplus {\mathcal{A}}_n$ where ${\mathcal{A}}_i$ are simple algebras.
• Any ${\mathcal{A}}_i\cong {\mathrm{M}}_n(\mathcal{D})$, where $\mathcal{D}$ is a division ring.
• For $\mathrm{char}F=0$, $\mathcal{A}\cong S\oplus \mathrm{Nil}\mathcal{A}$ where $S$ is semisimple.

Corollary

Any finite-dimensional nil algebra is nilpotent.

Definition

An algebra $\mathcal{A}$ is called power associative if $\mathrm{alg}⟨a⟩$ is associative for any $a\in \mathcal{A}$.

Exercise. Jordan algebras are power associative. (Hint: if $\mathcal{J}=⟨a⟩$ is Jordan algebra, then $\mathrm{Mult}\mathcal{J}$ is commutative.) See here.

Lemma

Let $\mathcal{A}$ be a power associative algebra, and let $I$ be an ideal of $\mathcal{A}$. If $I$ is nil ideal and $\mathcal{A}/I$ is a nil algebra, then $\mathcal{A}$ is a nil algebra.

\begin{proof} Take $a+I\in \mathcal{A}/I$, then there exists $n>0$ such that $(a+I{)}^n=0+I$ in $\mathcal{A}/I$. Hence $a^n\in I$. Since $I$ is nil, there exists $m$ such that $(a^n{)}^m=0$. Therefore, $\mathcal{A}$ is nil. \end{proof}

Lemma

Let $I,J$ be two nil ideals in power associative ring $\mathcal{A}$. Then $I+J$ is nil.

\begin{proof} Since $I$ is nil, $I/I\cap J$ is nil. By 2nd isomorphism theorem, $I+J/J$ is nil. By ^f572de, $I+J$ is nil. \end{proof}

Corollary

$I_1+\cdots +I_n$ is nil as long as $I_i$ is nil.

Proposition

Let $\mathcal{A}$ be power-associative algebra. Then there exists maximal nil ideal $\mathrm{Nil}\mathcal{A}$ and $\mathrm{Nil}(\mathcal{A}/\mathrm{Nil}\mathcal{A})=0$.

\begin{proof} Define $N$ as the sum of all nil ideals $I$ in $\mathcal{A}$.

Claim that $N$ is in fact the nil radical $\mathrm{Nil}\mathcal{A}$. Note that $N$ is nil since any $a\in N$ can be written as a finite sum of element of $I_1+\cdots +I_n$, which is nil by ^1b3e09. By definition of $N$, it is the maximal nil ideal.

Suppose $I$ is some ideal in $A/\mathrm{Nil}\mathcal{A}$. By correspondence theorem we have an ideal $J⩽\mathcal{A}$ of $\mathcal{A}$ such that $J/\mathrm{Nil}\mathcal{A}=I$. If $J/\mathrm{Nil}\mathcal{A}$ is nil, then by ^f572de $J$ is nil and so $J⩽\mathrm{Nil}\mathcal{A}$. Then $I=0$ and so $\mathrm{Nil}(\mathcal{A}/N)=0$. \end{proof}

As a consequence we showed that the nil radical exists for any associative algebra $\mathcal{A}$ or Jordan algebra $\mathcal{J}$.

Lemma

Let $\mathcal{A}$ be a commutative algebra. Then $\mathcal{A}$ is nilpotent iff $\mathrm{Mult}(\mathcal{A})$ is nilpotent.

\begin{proof} "→" Note that

$$a{R}_{b_1}{R}_{b_2}\cdots R_{b_{n-1}}=(((a{b}_1)b_2)\cdots )b_{n_1}\in {\mathcal{A}}^n=0$$

for some $n⩾1$. It deduces that $(\mathrm{Mult}(A){)}^{n-1}=0$. Hence $\mathrm{Mult}(\mathcal{A})$ is nilpotent.

"←" Let us show that any element of ${\mathcal{A}}^{2^n}$ can be written as sum of $a_1{R}_{b_1}\cdots R_{b_{n-1}}$. We prove it by induction on $n$. When $n=1$, each $ab\in {\mathcal{A}}^2$ can be written as $a{R}_b$. Suppose it for $k=n-1$ and consider element of ${\mathcal{A}}^{2^n}$. Take $a_1\in {\mathcal{A}}^{2^{n-1}}$ and $a_2\in {\mathcal{A}}^{2^{n-1}}\subseteq \mathcal{A}$, then by induction hypothesis $a_1=a_1^{\prime }{R}_{b_1}\cdots R_{b_{n-1}}$, there is

$$a_1{a}_2=a_1^{\prime }{R}_{b_1}\cdots R_{b_{n-1}}{a}_2=a_1^{\prime }{R}_{b_1}\cdots R_{b_{n-1}}{R}_{a_2}$$

and we finish the proof of the induction. Therefore, if $(\mathrm{Mult}(A){)}^n=0$, then $A^{2^n}=0$. \end{proof}

Remark. In fact, for Jordan algebra $\mathcal{J}$, there is ${\mathcal{J}}^n=0$ iff $(\mathrm{Mult}(\mathcal{J}){)}^{n-1}=0$.

Corollary

For a Jordan algebra $\mathcal{J}$, the radical of Jordan algebra is $\mathrm{Nil}\mathcal{J}=\{a\in \mathcal{J}:a\text{ is nil}\}$. Then $\mathrm{Nil}(\mathcal{J}/\mathrm{Nil}\mathcal{J})=0$.