4 Albert Theorem

The goal of this section is to show that, if $\mathcal{J}$ is a finite-dimensional Jordan algebra, then $\mathcal{J}$ is nil $⟹$ $\mathcal{J}$ is nilpotent. (Albert theorem).

Lemma

Let $\mathcal{J}=\mathcal{B}+Fv$ be a Jordan algebra, where $\mathcal{B}$ is an ideal with ${\mathcal{J}}^2\subseteq \mathcal{B}$. If $\mathcal{U}(\mathcal{B})$ is nilpotent, then $\mathcal{U}(\mathcal{J})$ is nilpotent. Here $\mathcal{U}(\mathcal{J})$ is the universal enveloping algebra of $\mathcal{J}$, and $\mathcal{U}(\mathcal{B})={\mathcal{U}}_{\mathcal{J}}(\mathcal{B})={\mathrm{alg}}_{\mathcal{U}(\mathcal{J})}⟨R_b:b\in \mathcal{B}⟩$.

\begin{proof} We check that

$$(\mathcal{U}(\mathcal{J}){)}^{3m}\subseteq (\mathcal{U}(\mathcal{B}){)}^m+(\mathcal{U}(\mathcal{B}){)}^m\mathcal{U}(\mathcal{J})$$

by induction.

Consider the case of $m=1$. For any $a,b,c\in \mathcal{J}$, there is $R_a{R}_b{R}_c\in \mathcal{U}(\mathcal{J}{)}^3$. If $a,b,c\in \mathcal{B}$, then $R_a{R}_b{R}_c\in (\mathcal{U}(\mathcal{B}){)}^3$. If $a=v$ and $b,c\in \mathcal{B}$, then

$$R_a{R}_b{R}_c=-R_c{R}_b{R}_v-R_{(vc)b}+R_{bc}{R}_v+R_{bv}{R}_c+R_{cv}{R}_b$$

by (IV) and $R_a{R}_b{R}_c\in \mathcal{U}(\mathcal{B})+\mathcal{U}(\mathcal{B}{)}^2+\mathcal{U}(\mathcal{B})\mathcal{U}(\mathcal{J})$. It deduces that

$$\mathcal{U}(\mathcal{J})\mathcal{U}(\mathcal{B}{)}^2\subseteq \mathcal{U}(\mathcal{J})\mathcal{U}(\mathcal{B})\subseteq \mathcal{U}(\mathcal{B})+\mathcal{U}(\mathcal{B})\mathcal{U}(\mathcal{J}).\text{\hspace{1em}(**)}$$

Similarly, we can verify that it holds for $a=b=v,c\in \mathcal{B}$ and $a=b=c=v$. (Exercise)

If $a=c=v,b\in \mathcal{J}$, by (IV) there is

$$R_v{R}_b{R}_v=R_v{R}_{vb}+\frac{1}{2}{R}_{b{v}^2}-\frac{1}{2}{R}_{v^{2}b}.$$

Then for the case of $m+1$, we have

$$\begin{array}{rl}(\mathcal{U}(\mathcal{J}){)}^{3(m+1)}& =(\mathcal{U}(\mathcal{J}){)}^{3m}\cdot (\mathcal{U}(\mathcal{J}){)}^3\\ & \subseteq ((\mathcal{U}(\mathcal{B}){)}^m+(\mathcal{U}(\mathcal{B}){)}^m\mathcal{U}(\mathcal{J}))((\mathcal{U}(\mathcal{B}))+(\mathcal{U}(\mathcal{B}))\mathcal{U}(\mathcal{J}))\\ & \subseteq (\mathcal{U}(\mathcal{B}){)}^{m+1}+(\mathcal{U}(\mathcal{B}){)}^{m+1}\mathcal{U}(\mathcal{J})\end{array}$$

By $(\ast \ast )$ one can check $\mathcal{U}(\mathcal{B}{)}^m\mathcal{U}(\mathcal{J})\mathcal{U}(\mathcal{B})\subseteq \mathcal{U}(\mathcal{B}{)}^m\mathcal{U}(\mathcal{B})\mathcal{U}(\mathcal{J})$ and so the proof of induction is completed.

Therefore, if $\mathcal{U}(\mathcal{B})$ is nilpotent, then $\mathcal{U}(\mathcal{J})$ is nilpotent as well. \end{proof}

Albert theorem

Let $\mathcal{J}$ be a finite-dimensional Jordan algebra. If $\mathcal{J}$ is nil, then $\mathcal{U}(\mathcal{J})$ is nilpotent.

\begin{proof} Let $\mathcal{B}\subseteq \mathcal{J}$ be a maximal subalgebra with respect that $\mathcal{U}(\mathcal{B})$ is nilpotent. Remark that $\mathcal{B}$ exists because $\mathcal{J}$ is finite-dimensional and $0$ is a subalgebra has that property. Then

$$\mathcal{J}\supseteq \mathcal{J}\mathcal{B}\supseteq (\mathcal{J}\mathcal{B})\mathcal{B}\supseteq \cdots \supseteq (((\mathcal{J}\mathcal{B})\mathcal{B})\cdots \mathcal{B})\mathcal{B}=0\text{\hspace{1em}(*)}$$

by $\mathcal{U}(\mathcal{B})$ is nilpotent.

Then there exists element $a\in \mathcal{J}$, $a\notin \mathcal{B}$ but $a\mathcal{B}\subseteq \mathcal{B}$. (Otherwise, for any $a\in \mathcal{J}\setminus \mathcal{B}$, $a\mathcal{B}⊈\mathcal{B}$ and then $(\ast )$ does not hold.) There are two possibilities for $a$:

• $a^2\mathcal{B}\subseteq \mathcal{B}$;
• there exists $b\in \mathcal{B}$ such that $a^{2}b\notin \mathcal{B}$.

Case 1: $\mathcal{B}$ is invariant under $R_a$ and $R_{a^2}$.

By ^b3vx4v, $\mathcal{B}$ is invariant under $R_{a^k}$ for all $k⩾1$, i.e. $a^k\mathcal{B}\subseteq \mathcal{B}$. Since $\mathcal{J}$ is nil, we know $a$ is nilpotent and there exists $m$ such that $a^m=0$ while $a^{m-1}\ne 0$.

Set $v=a^{m-1}$ and ${\mathcal{J}}_1=\mathcal{B}+Fv$. Since $\mathcal{B}$ is invariant under $R_{a^k}$, one can check that $\mathcal{B}$ is an ideal in ${\mathcal{J}}_1$, ${\mathcal{J}}_1^2\subseteq \mathcal{B}$, and $\mathcal{U}(\mathcal{B})$ is nilpotent. By ^j7zg8g, $\mathcal{U}({\mathcal{J}}_1)$ is nilpotent, which is contradicts with maximality of $\mathcal{B}$. Hence Case 1 is impossible.

Case 2: $a\mathcal{B}\subseteq \mathcal{B}$ and $a^{2}b\notin \mathcal{B}$ for some $\mathcal{B}$.

We show the following identities holds.

• $(a^2\mathcal{B})\mathcal{B}\subseteq \mathcal{B}$
  • $(a^{2}b)b_1=(a,a,b)b_1+(a(ab))b_1$
  • $(a(ab))b_1\in \mathcal{B}$
  • Since $(a,c,b)=c[R_a,R_b]$ (second linearization), by ^skoiqn we know $(a,c\circ d,b)=(a,c,b)\circ d+c\circ (a,d,b)$
  • Then $(a,a,b)b_1=(a,a{b}_1,b)-a(a,b_1,b)\in \mathcal{B}$
• $(a,a^2\mathcal{B},\mathcal{B})\subseteq \mathcal{B}$
  • $(a,a^{2}b,b_1)=(a(a^{2}b))b_1-a((a^{2}b)b_1)$
  • $(a(a^{2}b))b_1=(a^2(ab))b_1\in \mathcal{B}$
  • $a((a^{2}b)b_1)\in \mathcal{B}$ by i)
• $(a^2\mathcal{B})(a^2\mathcal{B})\subseteq \mathcal{B}$
  • $(a^{2}b)(a^2{b}_1)=(a,a,b)(a^2{b}_1)+(a(ab))(a^2{b}_1)$
  • $(a(ab))(a^2{b}_1)\in \mathcal{B}$ by i)
  • $(a,a,b)(a^2{b}_1)=-(a,a(a^2{b}_1),b)+a(a,a^2{b}_1,b)=(a,a^2(ab),b)+a(a,a^2{b}_1,b)\in \mathcal{B}$ by $(b,cd,a)=c(b,d,a)+(b,c,a)d$

where $(a,b,c)=(ab)c-a(bc)$.

Define $a_1=a^{2}b\notin \mathcal{B}$. Then $a_1\mathcal{B}\subseteq \mathcal{B}$ and $a_1^2\in \mathcal{B}$. Hence $a_1$ satisfies condition of case 1. Thus $\mathcal{B}=\mathcal{J}$ and so $\mathcal{U}(\mathcal{J})$ is nilpotent. \end{proof}

Corollary

Let $\mathcal{J}$ be finite-dimensional and let $a\in \mathcal{J}$ be nilpotent. Then $R_a$ is nilpotent in $\mathcal{U}(\mathcal{J})$.

\begin{proof} Define ${\mathcal{J}}^{\prime }={\mathrm{alg}}_{\mathcal{J}}⟨a⟩=\mathbb{F}a+\mathbb{F}{a}^2+\cdots +\mathbb{F}{a}^n$ where $\mathbb{F}$ is the field. Then ${\mathcal{J}}^{\prime }$ is finite-dimensional and nil, so $\mathcal{U}({\mathcal{J}}^{\prime })$ is nilpotent by ^0ac1ff. It follows that $\mathrm{Mult}{\mathcal{J}}^{\prime }$ is nilpotent and so $R_a$ is nilpotent. \end{proof}

another version of Albert theorem

Every finite-dimensional nil Jordan algebra is nilpotent.

\begin{proof} By ^0ac1ff, we have that $\mathcal{U}(\mathcal{J})$ is nilpotent. It follows that $\mathrm{Mult}(\mathcal{J})$ is nilpotent and so $\mathcal{J}$ is nilpotent by ^e3yi7m. \end{proof}

Corollary

Every finite-dimensional solvable Jordan algebra is nilpotent.

• todo: see

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

  Link to original