7 Capacity of Jordan Algebras ≤ 2

Definition

Let $\mathcal{J}$ be a finite-dimensional Jordan algebra with $\mathrm{Nil}\mathcal{J}=0$. Then we call capacity of $\mathcal{J}$, the maximal number of primitive idempotents in decomposition $1=e_1+\cdots +e_k$, where $k$ is capacity of $\mathcal{J}$.

We will consider in classification of simple finite-dimensional Jordan algebra. Since any Jordan algebra $\mathcal{J}$ which is central simple over $F$ ($Z(\mathcal{J})=F$) iff ${\mathcal{J}}_K:=K{\otimes }_F\mathcal{J}$ is central simple over $K$, where $K\supseteq F$ is a field extension, WLOG we can assume that $F$ is algebraically closed.

Invertible Elements

Now we give the definition of invertible elements in Jordan algebra.

Definition

Let $\mathcal{J}$ be a Jordan algebra with unity $1$, we call $a\in \mathcal{J}$ invertible if there exists $b\in \mathcal{J}$ such that $ab=1$ and $a^{2}b=a$. In this case, $b$ is called inverse of $a$, and denoted $a^{-1}$.

Lemma

Let $\mathcal{J}$ be a Jordan algebra with $1$. Then:

• (i) if $b=a^{-1}$, then $b^{-1}=a$;
• (ii) $a$ is invertible iff $U_a$ is invertible；
• (iii) $a,b$ are invertible iff $\{aba\}$ is invertible.

\begin{proof} i) Assume that $b$ is inverse of $a$. Then

$$a^2{b}^2=(a^{2}b)b-(a^2,b,b)=ab+2(an,b,a)=1+2(1,b,a)=1$$

and $b^{2}a=b^2(a^{2}b)=(b^2{a}^2)b=b$. Hence $b$ is invertible and $b^{-1}=a$.

ii) Assume that $a$ is invertible with inverse $b$. Then

$$\begin{array}{rl}& b{U}_a=2(ba)a-b{a}^2=2a-a=a,\\ & b^2{U}_a=2(b^{2}a)a-b^2{a}^2\stackrel{(i)}{=}2ba-1=1.\end{array}$$

It follows that $U_1=2R_1{R}_1-R_1=R_1={\mathrm{id}}_{\mathcal{J}}$ and ${\mathrm{id}}_{\mathcal{J}}=U_1=U_{b^2{U}_a}=U_a{U}_{b^2}{U}_a$. Thus $U_a$ is invertible.

Note that $U_a{U}_b=U_{b{U}_a}{U}_b=U_a{U}_b{U}_a{U}_b$ and $U_a,U_b$ are invertible. So ${\mathrm{id}}_{\mathcal{J}}=U_a{U}_b$ and $U_b=U_{a^{-1}}=(U_a{)}^{-1}$.

Conversely, let $U_a$ is invertible. Define $b=a{U}_a^{-1}\in \mathcal{J}$. Then $b{U}_a=a$. By ^jmr4zb, we have $[U_a,R_a]=0$ and $[U_a^{-1},R_a]=0$. Because $1U_a=1(2R_a^2-R_{a^2})=2a^2-a^2=a^2$, we have

$$ba=a{U}_a^{-1}{R}_a=a{R}_a{U}_a^{-1}=a^2{U}_a^{-1}=1U_a{U}_a^{-1}=1.$$

Then $a=b{U}_a=2(ba)a-b{a}^2=2a-b{a}^2$ yields that $a=b{a}^2$ and so $a$ is invertible.

iii) $\{aba\}$ is invertible iff $U_{\{aba\}}$ is invertible iff $U_{\{aba\}}=U_{b{U}_a}=U_a{U}_b{U}_a$ is invertible iff $U_a,U_b$ invertible iff $a,b$ invertible. \end{proof}

Albert-Jacobson-McCrimmon Theorem

Albert-Jacobson-McCrimmon

Let $\mathcal{J}$ be a Jordan algebra with identity. If the unit element of $\mathcal{J}$ is absolutely primitive idempotent, then the set

$$N=\{n\in \mathcal{J}:n\text{ nilpotent}\}$$

is ideal in $\mathcal{J}$.

McCrimmon

Let $z\in \mathcal{J}$ is nilpotent, then there exists $w\in \mathcal{J}$ such that $1-4z=(1-2w{)}^2$.

\begin{proof} Assume that $\mathrm{alg}⟨z⟩=⟨z,\cdots ,z^k⟩$ and take $w\in ⟨z,\cdots ,z^k⟩$. Then $w={\sum }_{j=1}^k{\alpha }_j{z}^j$, where ${\alpha }_j\in F$. It follows that

$$(1-2w{)}^2=1-4\sum _{j=1}^k{\alpha }_j{z}^j+4\sum _{i,j=1}^k{\alpha }_i{\alpha }_j{z}^{i+j}=1-4z$$

if ${\alpha }_1=1$, ${\alpha }_2-{\alpha }_1^2=0$, $\cdots$, ${\alpha }_t={\sum }_{i+j}{\alpha }_i{\alpha }_j=0$ hold. Such ${\alpha }_1,\cdots ,{\alpha }_k$ exists and so we finish the proof. \end{proof}

Now we are ready to show ^cvblbw.

\begin{proof} Since $1$ is an absolutely primitive idempotent, we have ${\mathcal{J}}_1(1)=\mathcal{J}$ and each $a\in \mathcal{J}$ can be written as $a=\alpha \cdot 1+n$ where $n$ is nilpotent.

It is enough to prove that $N$ is subspace in $\mathcal{J}$. If it holds, then

$$2n_1{n}_2=(n_1+n_2{)}^2-n_1^2-n_2^2\in N$$

for any $n_1,n_2\in N$; and

$$an=(\alpha \cdot 1+n_1)n=\alpha n+n_{1}n\in N$$

for any $a\in J$ with $a=\alpha \cdot 1+n_1$.

So it remains to show $N$ is a subspace. Suppose $n_1,n_2\in N$ and $n_1+n_2=\alpha \cdot 1+n$ with $\alpha \ne 0$. WLOG $n_1+n_2=1-4n$ for some $n\in N$. By ^2addo4, there exists $w$ such that $(1-2w{)}^2=1-4n$. Using the proof of ^2addo4, we know $w$ is invertible. Hence $1-2w$ is invertible (assume $a=1-2w$, then $(2w{)}^m=(1-a{)}^m=0$, $1=a(p(a))$ and so $a$ is invertible).

It follows $U_{1-2w}$ invertible from $1-2w$ invertible by ^3mr10v. Note that

$$1U_{1-2w}=2(1-2w{)}^2-(1-2w{)}^2=1-4n=n_1+n_2$$

and so $1=n_1{U}_t+n_2{U}_t$ with $t=(1-2w{)}^{-1}\in \mathcal{J}$.

We claim that $n_1{U}_t$ is nilpotent. Note that $n_1$ is nilpotent, thus non-invertible. Since $t$ is invertible, the map $U_t$ is invertible. If $n_1{U}_t$ is invertible, then $n_1$ is invertible, which is impossible as $n_1$ is nilpotent. Hence $n_1$ is non-invertible. Since $1$ is absolutely primitive, every element in $\mathcal{J}$ is either invertible or nilpotent (because $1$ is absolutely primitive), and it yields that $n_1{U}_t$ is nilpotent.

Thus $1-n_1{U}_t=n_2{U}_t$ is both invertible and nilpotent, leading to a contradiction. Therefore, $n_1+n_2$ is nilpotent and so $N$ is a subspace. Now we finish the proof. \end{proof}

Simple Finite-dimensional Jordan algebra

Recall that

where ${\mathcal{J}}_{ii}={\mathcal{J}}_1(e_i)$ and ${\mathcal{J}}_{ij}={\mathcal{J}}_{1/2}(e_i)\cap {\mathcal{J}}_{1/2}(e_j)$.

Link to original

By ^73xp32, we know $e_i$ are absolutely primitive. By ^cvblbw, $J_1(e_i)$ can be written as $J_1(e_i)=F{e}_i+N_i$, where $N_i$ is nilpotent ideal in $J_1(e_i)$. Note that if $e$ is an idempotent and $\mathrm{Nil}(J)=0$, then $\mathrm{Nil}(J_1(e))=0$. Hence, $N_i=0$ and $J_1(e_i)=F{e}_i$ for all $i$.

Theorem

For finite-dimensional simple Jordan algebra with $\overline{F}=F$,

$$\mathcal{J}=\sum _{i=1}^n{F}_{e_i}+\sum _{i<j}{\mathcal{J}}_{ij}.\text{\hspace{1em}(*)}$$

Such $\mathcal{J}$ with Pierce decomposition $(\ast )$ is called reduced.

Capacity 1: Field

Corollary

Any simple finite-dimensional Jordan algebra of capacity $1$ is isomorphic to a field.

\begin{proof} When $n=1$, $(\ast )$ yields that $\mathcal{J}=F_{e_1}$ and we finish the proof. \end{proof}

Capacity 2: Clifford Type

Proposition

Any simple finite-dimensional Jordan algebra $\mathcal{J}$ of capacity $2$ is isomorphic to Jordan algebra of Clifford type.

\begin{proof} Since the capacity is $2$, $\mathcal{J}=F{e}_1+F{e}_2+{\mathcal{J}}_{12}$ with $1=e_1+e_2$. Define $V={\mathcal{J}}_{12}+F(e_1-e_2)$.

For $u,v\in V$, we claim that $uv=\alpha 1$. Since $(e_1-e_2{)}^2=e_1^2+e_2^2=e_1+e_2=1$ and $u(e_1-e_2)=0$ for any $u\in {\mathcal{J}}_{12}$, it suffices to consider the case where $u,v\in {\mathcal{J}}_{12}$. Since $2uv=(u+v{)}^2-u^2-v^2$, it is enough to prove that $v^2=\alpha 1$.
Note that ${\mathcal{J}}_{12}^2\subseteq F{e}_1+F{e}_2$. Take $v\in {\mathcal{J}}_{12}$, then $v^2=\alpha e_1+\beta e_2$ for some $\alpha ,\beta$. It remains to show $\alpha =\beta$. Because

$$(\alpha e_1)v=(v^2{e}_1)v=v^2(e_{2}v)=(v^2{e}_2)v=(\beta e_2)v,$$

there is $\frac{1}{2}(\alpha -\beta )v=0$ and so $\alpha =\beta$. Now we prove the claim.

Define $f:V\times V\to F$, $f(v,u)=vu\in F$. Then $f$ is symmetric. If $f$ is non-degenerate, then $\mathcal{J}=F\cdot 1+V=\mathcal{J}(V,f)$, see here.

Therefore, it is enough to show $f$ is non-degenerate. If there exists $x$ such that $f(x,v)=0$ for all $v$, then $x=y+\alpha (e_1-e_2)$ for some $y\in {\mathcal{J}}_{12}$ and $\alpha \in F$. Note that

$$0=x(e_1-e_2)=(y+\alpha (e_1-e_2))(e_1-e_2)=\alpha ,$$

then we have $x=y\in {\mathcal{J}}_{12}$. Define $I=Fx$. Since $x{\mathcal{J}}_{12}=0$ and $x{e}_1=x{e}_2=x/2\in I$, we know $⟨Fx⟩⊲J$ is an ideal. As $\mathcal{J}$ is simple, $Fx=\mathcal{J}$ or $0$. Since $x\in {\mathcal{J}}_{12}$, $Fx\ne \mathcal{J}$ and so $Fx=0$. Therefore, $f$ is non-degenerate and we finish the proof. \end{proof}

Capacity $⩾3$

See 8 Capacity of Jordan Algebras ≥ 3.