13 Infinite-dimensional Jordan Algebras II - i-exceptional Case

In this section, we assume that $J$ is i-exceptional, i.e. $J\notin \overline{SJord}$. Then there exists s-identity $f$ such that $f(J)\ne 0$.

Define $\pi :J[X]\to SJ[X]$. Then $\ker \pi =\mathcal{S}$ is the set of all s-identities. Remark that $J\in \overline{SJord}$ iff $\mathcal{S}(J)=\{f(a_1,\cdots ,a_n):f\in \mathcal{S},a_i\in J\}=0$.

Definition

Let $J$ be a Jordan algebra. Define $S_0={\cap }_{I\in \mathcal{I}}I$, where $\mathcal{I}=\{I⊲J:I\ne 0\}$, as the heart of Jordan algebra.

Definition

Let $K\subseteq J$ be a subspace. Then $K$ is called a quadratic ideal if $\forall a\in K$, $J{U}_a\subseteq K$; or equivalently, for $\forall x\in J$, $\{axa\}$ is in $K$.

Definition

We say that Jordan algebra with unity is primitive if there exists quadratic ideal $K\ne J$ of $J$ such that

• $K$ is maximal;
• for any $I⊲J$ and $I\subseteq K$, $I=0$.

Remark. Define $(K:J)=\{a\in K:aJ\subseteq K\}$. Then

• $(K:J)⊲J$;
• $K/(K:J)$ is special.

Therefore, if $J$ is primitive, then $(K:J)=0$ and so $K$ is special.

Lemma

Let $J$ be a primitive Jordan algebra with unit element, $J\notin \overline{SJord}$. Then $S_0=(\mathcal{S}(J){)}^3\ne 0$ is unique minimal bilateral ideal of $J$.

\begin{proof} $\mathrm{Nil}(J)=0$, $S(J)\ne 0$ yields $(\mathcal{S}(J){)}^3\ne 0$. Hence to show that $S(J{)}^3$ is contained in every non-zero ideal of $J$. \end{proof}

Lemma

If $J$ is primitive with unity, then $J$ is prime and $\mathrm{Nil}(J)=0$.

Definition

Take $s\in {\mathcal{S}}_0$, then we define $\mathrm{Spec}(s)=\{\alpha \in F:s-\alpha 1\text{ is not invertible}\}$.

Lemma

For any $s\in {\mathcal{S}}_0$, the set $\mathrm{Spec}(s)\setminus \{0\}$ has at most $2n$ elements.

Lemma

The set $S_0$ is algebraic over $F$.

Lemma

The algebra $S_0$ is simple.

Lemma

$F=\overline{F}$ yields $S_0$ is an Albert algebra.

\begin{proof} Let $e=e_1+\cdots +e_m$ be a decomposition of the identity into pairwise orthogonal, absolutely primitive idempotents. Since $\mathrm{Nil}(J)=0$ and $F=\overline{F}$, we consider the Peirce decomposition

$$J(e)=J_1(e)\oplus J_{1/2}(e)\oplus J_0(e).$$

Given $J$ is primitive with unity, the components $J_{1/2}(e)$ and $J_0(e)$ vanish, leaving $J=J_1(e)$. As shown in the notes, $e$ serves as the unity of $S_0$. According to the Structure Theorem for Jordan algebras, the condition $a_{ij}^2=e_i+e_j$ (for the coordination of the algebra) implies:

$$S_0\cong H_n(\mathbb{D},\gamma )$$

where $\mathbb{D}$ is a composition algebra over $F$. We proceed by case analysis on the coordination algebra $\mathbb{D}$ :

• If $\mathbb{D}$ is associative, then $S_0$ is a special Jordan algebra. However, by the theorem assumption $J\notin \overline{SJord}$, this case is excluded (yielding $S_0=0$ in the non-special context).
• If $n=3$ and $\mathbb{D}=\mathbb{O}$ (the algebra of Octonions), the resulting structure is the 27dimensional exceptional algebra.

Therefore, $S_0$ is an algebra. \end{proof}

Theorem

Let $J$ be primitive Jordan algebra with unit element and $F=\overline{F}$, $\#F>\dim J$, $J\notin \overline{SJord}$. Then $J$ is an Albert algebra.

\begin{proof} We need to show $J=S_0$. $1_{S_0}=e_J\in S_0$, then we prove that $e_J=e\in Z(J)$. For $a\in J$, $a=ae+(a-ae)$. Define $I=\{a-ae:a\in J\}$.

We show that $I$ is an ideal of $J$. Since

$$(a-ae)b=(ae)b-(ab)e\in I,$$

we have $J=Je+I$.

We claim that $Je\cap I=0$. Since $x=ae=b-eb$, $xe=ae=eb-e(eb)=0$. Thus $x=0$ and so $J=Je+I$.

As $J$ is prime, $I=0$ and $J=Je\subseteq S_0$. \end{proof}

From Prime to Primitive

In Zelmanov’s classification theory, a Prime i-exceptional Jordan algebra $J$ is not necessarily Primitive. However, its central localization transforms $J$ into a simple (and thus primitive) algebra. This allows ^domxmn for primitive algebras to be applied to all prime i-exceptional algebras.