5 Non-degenerated Jordan algebra

For any $a\in \mathcal{J}$, define $U_a=2R_a^2-R_{a^2}$. For any $b\in \mathcal{J}$, $b{U}_a=2(b\circ a)\circ a-b\circ a^2$.

Remark. If $A^{+}=\mathcal{J}$, then $b{U}_a=aba$.

Define $U_{a,b}=\frac{1}{2}(U_{a+b}-U_a-U_b)$ and $\{axb\}=x{U}_{a,b}$ (Jordan triple product).

Exercise. Verify $U_{a,b}=R_a{R}_b+R_b{R}_a-R_{a\circ b}$. See here.

Definition

An element $a\in \mathcal{J}$ is called an absolute zero divisor if ${\mathcal{U}}_a=0$.

We call $\mathcal{J}$ non-degenerated if it only contains trivial absolute zero divisor.

fundamental identities for $U_a$

Let $\mathcal{J}$ be a Jordan algebra. Then we have $U_{a{U}_b}=U_b{U}_a{U}_b$ and $U_{a^2}=U_a^2$ for any $a,b\in \mathcal{J}$.

\begin{proof} (We will prove it for special algebras, and then eventually see that it is enough.)

Note that $x{U}_{a{U}_b}=x{U}_{bab}=babxbab=b(a(bxb)a)b=x{U}_b{U}_a{U}_b$, so we have $U_{a{U}_b}=U_b{U}_a{U}_b$.

Note that $x{U}_{a^2}=a^{2}x{a}^2=a(axa)a=x{U}_a{U}_a$, so we have $U_{a^2}=U_a^2$.

Then by Shirshov-Cohn we finish the proof. \end{proof}

Define $\mathcal{Z}=⟨a\in \mathcal{J}:U_a=0⟩$ as vector space generated by absolute zero divisors.

Theorem

If $\mathcal{J}$ is finite-dimensional Jordan algebra, then $\mathrm{Nil}\mathcal{J}=0$ iff $\mathcal{Z}(\mathcal{J})=0$.

In 1956, Shirshov proved every Jordan algebra from two generators is special.

Remember that for any algebra $A$, $A^{♯}=A\oplus \mathbb{F}\cdot 1$ with $1^1=1$ and $a1=1a=a$ is called unital hull.

Lemma

Let $\mathcal{J}$ be a Jordan algebra, and let $a$ be an absolute zero divisor (AZD). Then for any $b\in {\mathcal{J}}^{♯}$, $a{U}_b$ is an AZD in $\mathcal{J}$.

\begin{proof} Notice that $\mathcal{J}$ is an ideal in ${\mathcal{J}}^{♯}$. For any $a\in \mathcal{J}$ and $b\in {\mathcal{J}}^{♯}$, we have $a{U}_b=2(ab)b-a{b}^2\in \mathcal{J}$. For any $x\in \mathcal{J}$, we have $x{U}_{a{U}_b}=x{U}_b{U}_a{U}_b\subseteq \mathcal{J}{U}_a{U}_b=0$ as $a$ is an AZD. Therefore, $a{U}_b$ is an AZD in $\mathcal{J}$. \end{proof}

Corollary

$\mathcal{Z}(\mathcal{J})$ is an ideal in $\mathcal{J}$.

\begin{proof} We need to prove $a\in \mathcal{Z}(\mathcal{J})$ and $b\in \mathcal{J}$, then $ab\in \mathcal{Z}(\mathcal{J})$. By ^6f8fac, we know $a{U}_{b+1}\in \mathcal{Z}(\mathcal{J})$. Note that

$$a{U}_{b+1}=\{(b+1)a(b+1)\}=\{1a1\}+\{ba1\}+\{1ab\}+\{bab\}.$$

Exercise. Check $\{ba1\}+\{1ab\}=2ab$.

Since $a{U}_b\in \mathcal{Z}(\mathcal{J})$, we know $ab\in \mathcal{Z}(\mathcal{J})$ and so $\mathcal{Z}(\mathcal{J})$ is an ideal in $\mathcal{J}$. \end{proof}

Lemma

$\mathcal{Z}(\mathcal{J})\subseteq \mathrm{Nil}(\mathcal{J})$.

\begin{proof} Let $a\in \mathcal{Z}(\mathcal{J})$. Set $a=a_1+\cdots +a_n$ with $a_i$ being AZD. If $n=1$, then $a=a_1$ and $a^3=a{U}_a=0$. Hence $a$ is nilpotent.

By induction, we may deduce to the case when $a=a_n+z$ with $a_n$ being AZD and $z$ being nilpotent. Define $\tilde{\mathcal{J}}=\mathrm{alg}⟨a_n,z⟩$. By Shirshov-Cohn, $\tilde{\mathcal{J}}$ is special. As $z$ is nilpotent, assume that $z^m=0$. Also notice that $a_n^3=0$. Then we have

$$a^{2m+1}=(a_n+z{)}^{2m+1}=\sum _{i+j=2m}{z}^i{a}_n{z}^j+\sum _{i+j=2m-1}{z}^i{a}_n^2{z}^j=0.$$

Therefore, $a\in \mathrm{Nil}\mathcal{J}$ and we finish the proof. \end{proof}

Lemma

Assume that $\mathcal{J}$ is Jordan algebra. If $I$ is an ideal, then $I^3$ is an ideal.

\begin{proof} For any $a,b,c\in I$ and $x\in \mathcal{J}$, we have

$$((ab)c)x=(ab)(cx)+(ab,c,x)=(ab)(cx)+(-ax,c,b)-(bx,c,a)\in I^3.$$

by ^pgknr0. Now we finish the proof. \end{proof}

Theorem

If $\mathcal{J}$ is finite-dimensional Jordan algebra, then $\mathrm{Nil}\mathcal{J}=0$ iff $\mathcal{Z}(\mathcal{J})=0$.

\begin{proof} By ^0rjrnv, one direction is easy. Conversely, assume that $\mathcal{Z}(\mathcal{J})=0$ and $N:=\mathrm{Nil}(\mathcal{J})\ne 0$. Note that

$$N^{⟨1⟩}⊳N^{⟨2⟩}=N^3⊳N^{⟨3⟩}=N^9⊳\cdots ⊳N^{⟨i⟩}=0$$

as $N^{⟨1⟩}=N$ and $N^{⟨i+1⟩}=(N^{⟨i⟩}{)}^3$.

Thus we have $I⊲J$ with $I^3=0$ and $I\ne 0$. Let $0\ne a\in I^2\subset I$, then $a^2\in I^2{I}^2\subseteq I^3=0$ and $a^2=0$. For any $x\in \mathcal{J}$,

$$x{U}_a=2(xa)a-x{a}^2\subseteq I\cdot a\subseteq I^3=0$$

and so $a\in \mathcal{Z}(\mathcal{J})$. Since $\mathcal{Z}(\mathcal{J})=0$, we know $a=0$ and $I^2=0$. However, take $a\in I$, then we have $x{U}_a=2(xa)a-x{a}^2\in I^2=0$ and so $I=0$, leading to a contradiction. Therefore, $\mathcal{Z}(\mathcal{J})=0$ yields that $\mathrm{Nil}(\mathcal{J})=0$. Now we finish the proof. \end{proof}

Corollary

Let $\mathcal{J}$ be a finite-dimensional Jordan algebra with $\mathrm{Nil}(\mathcal{J})=0$. Then for any ideal $I⊲J$, we have $\mathrm{Nil}(I)=0$.

\begin{proof} Suppose $\mathrm{Nil}(\mathcal{I})\ne 0$, then by ^e4d581, $\mathcal{Z}(\mathcal{I})\ne 0$. Take $a\in \mathcal{Z}(\mathcal{I})$, $a\ne 0$, then $\mathcal{I}{U}_a=0$.

From the other side, $\mathcal{Z}(\mathcal{J})=0$ by ^e4d581 and so $\mathcal{J}{U}_a\ne 0$. Choose $x\in \mathcal{J}$ and $b=x{U}_a\ne 0$, then

$$\mathcal{J}{U}_b=\mathcal{J}{U}_{x{U}_a}=(\mathcal{J}{U}_a{U}_x)U_a\subseteq I{U}_a=0$$

and so $b=0$ by $b\in \mathcal{Z}(\mathcal{J})=0$, which is impossible. Therefore, $\mathrm{Nil}(\mathcal{I})=0$. \end{proof}

Corollary

Let $\mathcal{J}$ be a finite-dimensional Jordan algebra, and let $I⊲\mathcal{J}$ be an ideal. Then $\mathrm{Nil}(I)=I\cap \mathrm{Nil}(\mathcal{J})$.

\begin{proof} See here. \end{proof}