8 Capacity of Jordan Algebras ≥ 3

It is a notes for Jordan Algebra, Lecture Oct 22.pdf

Let $\mathcal{J}$ be a finite-dimensional, simple Jordan algebra of capacity $>2$ and $F=\overline{F}$. Let $e_1,\cdots ,e_n$ be pairwise orthogonal primitive idempotents such that $1=e_1+\cdots +e_n$. By ^x0wce7 we can assume that

$$\mathcal{J}=({\oplus }_{i=1}^{n}F{e}_i)\oplus ({\oplus }_{i<j}{\mathcal{J}}_{ij}).$$

Our main goal is to prove the following theorem.

Proposition

Any finite-dimensional simple Jordan algebra $F=\overline{F}$ with capacity $⩾3$ has a form $H_n(\mathcal{D})$, where $\mathcal{D}$ is a composition algebra over $F$.

If $n⩾4$, then $\mathcal{D}$ is associative; if $n=3$, then $\mathcal{D}$ is alternative.

\begin{proof} By ^f3e9rq, the idempotents are strictly connected. Applying ^g7zbu1, $\mathcal{J}\cong H_n(A)$, where $A$ is associative if $n⩾4$ and is alternative if $n=3$. Since $\mathcal{J}$ is simple, the coordinate algebra $A$ is simple. When $n=3$, $A$ is alternative and simple, then $A$ is an octonion algebra by Kleinfeld’s theorem. When $n=4$, $A$ is associative and simple, which yields that $A=F$ by ^f43ojz.

By the structure theorem of composition algebra, $A$ is a composition algebra. Now we finish the proof. \end{proof}

In fact, we have a general version.

coordinatization theorem

Let $\mathcal{J}$ be a unital Jordan algebra such that

$$1=e_1+\cdots +e_n$$

with $e_1,\dots ,e_n$ strictly connected, pairwise orthogonal idempotents. Then $\mathcal{J}$ is isomorphic to $H_n(A,\ast )$, with $(A,\ast )$ being an alternative algebra (associative if $n⩾4$ ) with involution $\ast$ such that for every $a\in A,a+a^{\ast }$ is an element of $N(A)$ , the center of $A$.

Step 1: the Skeleton

Firstly, we show that idempotents $e_i$ are strictly connected.

Lemma

Let $\mathcal{J}$ be a finite-dimensional simple Jordan algebra with $F=\overline{F}$, and let $1=e_1+\cdots +e_n$. Then ${\mathcal{J}}_{ij}\ne 0$ for all $i,j$ distinct.

\begin{proof} We have proved in ^uh6mw6 that, if $x\in {\mathcal{J}}_{ij}$ with $i\ne j$ and $x{\mathcal{J}}_{ij}=0$, then $x=0$. Hence, if ${\mathcal{J}}_{ij}\ne 0$, then ${\mathcal{J}}_{ij}^2\ne 0$. By ^uh6mw6, $F{e}_i+{\mathcal{J}}_{ij}+F{e}_j$ is a Jordan algebra of Clifford type, and there exists $a_{ij}\in {\mathcal{J}}_{ij}$ such that $a_{ij}^2=\alpha (e_i+e_j)$ with $\alpha \ne 0$.

If $i,j,k$ are distinct, then by linearization we have

$$a_{ij}^2{b}_{jk}=-2(a_{ij}^2{e}_k)b_{jk}+2(a_{ij}^2)(e_k{b}_{jk})=-2(a_{ij}^2,e_k,b_{jk})=4(a_{ij}{b}_{jk},e_k,a_{ij})=2(a_{ij}{b}_{jk})a_{ij}.\text{\hspace{1em}(*)}$$

We claim that if ${\mathcal{J}}_{ij},{\mathcal{J}}_{jk}\ne 0$, then ${\mathcal{J}}_{ik}\ne 0$. Recall that ${\mathcal{J}}_{ij}^2\subseteq {\mathcal{J}}_1(e_i+e_j)$. Hence $a_{ij}^2=\alpha (e_i+e_j)$ and $b_{jk}^2=\beta (e_j+e_k)$. It follows that $2(a_{ij}{b}_{jk}{)}^2\ne 0$ by $(\ast )$ and so ${\mathcal{J}}_{ik}\ne 0$. Now we prove the claim.

Define a relation in $\{1,\cdots ,n\}$ as $i\approx j⟺{\mathcal{J}}_{ij}\ne 0$, and it is a equivalence relation. Let $S_1,\cdots ,S_k$ be equivalence classes, and let $E_i={\sum }_{\sigma \in S_i}{e}_{\sigma }$. Then $1=E_1+\cdots +E_k$ and $\mathcal{J}={\mathcal{J}}_1(E_1)\oplus \cdots \oplus {\mathcal{J}}_1(E_k)$, which is impossible because $\mathcal{J}$ is simple. Thus $\mathcal{J}={\mathcal{J}}_1(E_1)$ and so ${\mathcal{J}}_{ij}\ne 0$ for any $i\ne j$. \end{proof}

Definition

$e_i$ and $e_j$ are called strictly connected if there exists $a_{ij}\in \mathcal{J}$ such that $a_{ij}^2=e_i+e_j$.

Corollary

In $\mathcal{J}$ satisfying ^3lcj22, for any $i\ne j$, $e_i$ and $e_j$ are strictly connected.

\begin{proof} In proof of ^3lcj22, $a_{ij}^2=\alpha (e_i+e_j)$ for some $\alpha \ne 0$. Since $F=\overline{F}$, we can take $\beta$ such that ${\beta }^2=\alpha$, and ${\beta }^{-1}{a}_{ij}$ is what we desired. \end{proof}

Step 2: the Material

Then, we prove that $J_{ij}$ are isomorphic vector spaces for any $i\ne j$ and they are composition algebras.

By ^uh6mw6, $J_1(e_i+e_j)=F{e}_i+J_{ij}+F{e}_j$ is a subalgebra of Clifford type. Let $f_{ij}:J_{ij}\to F$ such that $a_{ij}^2=f_{ij}(a_{ij})(e_i+e_j)$. Then $f$ is quadratic, since $f_{ij}(\alpha a_{ij})={\alpha }^2{f}_{ij}(a_{ij})$ for any $\alpha \in F$ and $a_{ij}\in J_{ij}$.

Let $\overline{f_{ij}}$ be a bilinear form associated to quadratic $f_{ij}$

$$\overline{f_{ij}}(a_{ij},b_{ij})=\frac{1}{2}(f_{ij}(a_{ij}+b_{ij})-f_{ij}(a_{ij})-f_{ij}(b_{ij})).$$

We claim that $\overline{f_{ij}}$ is non-degenerate. If there exists $a_{ij}\in J_{ij}$ such that $\overline{f_{ij}}(a_{ij},J_{ij})=0$, then $a_{ij}{J}_{ij}=0$ and so $F{a}_{ij}⊲J(e_i+e_j)$. It follows that

$$(F{a}_{ij}{)}^2=F{a}_{ij}^2=F{f}_{ij}(a_{ij})(e_i+e_j)=0$$

and $F{a}_{ij}$ is nilpotent. As $J$ is simple, $F{a}_{ij}=0$ and so $a_{ij}=0$. Therefore, $\overline{f_{ij}}$ is non-degenerate.

Lemma

Let $J$ be a finite-dimensional simple Jordan algebra with $F=\overline{F}$ and capacity $n>2$. Let $i,j,k$ be all distinct. Let $f_{ij}$, $f_{ik}$, $f_{jk}$ be quadratic forms for $J_1(e_i+e_j)$, $J_1(e_i+e_k)$ and $J_1(e_j+e_k)$. Then

$$f_{ik}(2a_{ij}{b}_{jk})=f_{ij}(a_{ij})f_{jk}(b_{jk})$$

for any $a_{ij}\in J_{ij}$ and $b_{jk}\in J_{jk}$.

\begin{proof} See Pasted image 20251216230519.png \end{proof}

Lemma

Under the assumption of ^ukbxzg, all subspaces $J_{ij}$ with $i\ne j$ are isomorphic as vector space.

\begin{proof} See Pasted image 20251216230648.png \end{proof}

Composition Algebra $J_{ij}$

Definition

We call an algebra $A$ a composition algebra if $A$ has unity and a quadratic non-degenerate form in $A$ such that $n(xy)=n(x)n(y)$ for all $x,y\in A$.

Remark. Any composition algebra $A$ is quadratic, that is, for all $a\in A$,

$$a\ast a:=a^2=t(a)a-n(a)\cdot 1$$

where $t(a),n(a)\in F$, and has involution $a\to \overline{a}$ such that

$$a+\overline{a}=t(a)\cdot 1\text{ and }a\overline{a}=\overline{a}a=n(a)1.$$

Proposition

Let $J$ be a Jordan algebra satisfying the condition of ^ukbxzg with $e_{ij}$ defined as in ^de81d7. Consider $D=J_{12}$ with the following multiplication

$$a\ast b=8(a{e}_{23})(b{e}_{13})$$

for any $a,b\in D$ and denote $f_{12}$ by $n$. Then $n(a\ast b)=n(a)n(b)$ and $a\ast e_{12}=e_{12}\ast a=a$ for all $a,b\in D$.

Therefore, $D$ is a composition algebra with quadratic from $n$ and unity $e_{12}$.

\begin{proof} See Pasted image 20251216231814.png. \end{proof}

We have shown that $J_{12}=D$ is a composition algebra with unit element $e_{12}$ and quadratic form $n=f_{12}$. Next, we will show that $t(a)(e_1+e_2)=2a{e}_{12}$ for all $a\in D$.

Let $a\in \mathcal{D}$. Recall that

$$a\ast a=8\left(a{e}_{23}\right)\left(a{e}_{13}\right)$$

then we have

$$\begin{array}{rl}a\ast a& =-4a^2\left(e_{23}{e}_{13}\right)+4\left(\left(a{e}_{23}\right)a\right)e_{13}+4\left(\left(a{e}_{13}\right)a\right)e_{23}+4a\left(a\left(e_{13}{e}_{23}\right)\right)\\ & =2n(a)\left(e_1+e_2\right)e_{12}+2\left(a^2{e}_{23}\right)e_{13}+2\left(a^2{e}_{13}\right)e_{23}+2a\left(a{e}_{12}\right)\\ & =-2n(a)e_{12}+2\left(n(a)\left(e_1+e_2\right)e_{23}\right)e_{13}+2\left(n(a)\left(e_1+e_2\right)e_{13}\right)e_{23}+2\left(a{e}_{12}\right)a\\ & =n(a)\left(-2e_{12}+\frac{1}{2}{e}_{12}+\frac{1}{2}{e}_{12}\right)+2\left(a{e}_{12}\right)a\\ & =2\left(a{e}_{12}\right)a-n(a)e_{12}\end{array}$$

by the linearized Jordan identity $(ac,b,d)+(ad,b,c)+(cd,b,a)=0$ and $a^2=n(a)\left(e_1+e_2\right)$.

Since $D$ is quadratic, it satisfies $a^2-t(a)a+n(a)1=0$. It follows that

$$t(a)\left(e_1+e_2\right)=2a{e}_{12}$$

and

$$\bar{a}=t(a)1_{\mathcal{D}}-a=2\left(a{e}_{12}\right)e_{12}-a\forall a\in \mathcal{D}.$$

Step 3: the Blueprint

Let $A$ be an algebra with involution $a\mapsto \overline{a}$. We may extend this involution to a matrix algebra $M_n(A)$ as $\overline{\{a_{ij}\}}=\{\overline{a_{ji}}\}$ for any $\{a_{ij}\}\in M_n(A)$.

We call a matrix $x\in M_n(A)$ Hermitian if $\overline{x}=x$. The set of Hermitian matrices is denoted by $H(n)A$. In fact, $M_n(A)$ is a subalgebra of $M_n(A{)}^{+}$ since for any $x,y\in H_n(A)$, $x\circ y=\frac{1}{2}(xy+yx)\in H_n(A)$.

For $a\in A$ and $i,j\in 1,\cdots ,n$, define $a[ij]=a{e}_{ij}+\overline{a}{e}_{ji}$, where $e_{ij}$ is the elementary matrix with only $1$ in entry $(i,j)$ while other entries $0$. We have $a[ij]\in H_n(A)$ and $a[ij]=\overline{a}[ji]$.

For each $i=1,\cdots ,n$, set $e_i=e_{ii}$, then $e_1,\cdots ,e_n$ are pairwise orthogonal idempotents $1=e_1+\cdots +e_n$. Then the full Peirce decomposition

$$H_n(A)=\sum _{i=1}^n{H}_n(A{)}_{ii}+\sum _{1⩽i<j⩽n}{H}_n(A{)}_{ij},$$

where $H_n(A{)}_{ii}=\{a[ii]:a\in A\}$ and $H_n(A{)}_{ij}=\{a[ij]:a\in A\}$.

Lemma

Let $i,j,k\in \{1,\cdots ,n\}$ be distinct integers. Then

• $2a[ij]\cdot b[jk]=(ab)[ik]$;
• $2a[ii]\cdot b[ik]=(a+\overline{a})b[ik]$;
• $2a[ij]\cdot b[jj]=a(b+\overline{b})[ij]$;
• $2a[ij]\cdot b[ij]=(a\overline{b})[ii]+(\overline{a}b)[jj]$;
• $2a[rs]\cdot b[k\ell ]=0$ if $\{r,s\}\cap \{k,\ell \}=\varnothing$.

$A$ is called alternative if $(a,a,b)=0$ and $(a,b,b)=0$ for all $a,b\in A$.

An algebra

Theorem

Let $n⩾3$. If $H_n(A)$ is a Jordan algebra, then $A$ is an alternative algebra (associative, if $n⩾4$). For any $a\in A$, $a+\overline{a}$ is in the center of $A$, that is $a+\overline{a}\in N(A)$.

\begin{proof} See Pasted image 20251218020342.png. \end{proof}

Step 4: Assembly

coordinatization theorem

Let $\mathcal{J}$ be a unital Jordan algebra with $1=e_1+\cdots +e_n$ with $e_1,\cdots ,e_n$ strictly connected ($e_{ij}\in {\mathcal{J}}_{ij}$. $e_{ij}^2=e_i+e_j$) pairwise orthogonal idempotents. Then $\mathcal{J}$ is isomorphic to $H_n(A,\ast )$ with $(A,\ast )$ being associative if $n⩾4$ or alternative if $n=3$.

Remark. Let $H_n(A)$ be a Jordan algebra. For $n⩾4$, $A$ is associative. For $n=3$, $A$ is alternative.

Lemma

If $\mathcal{J}$ satisfies condition of ^3lcj22, then there exists $e_{ij}\in {\mathcal{J}}_{ij}$ such that

• $e_{ij}=e_i+e_j$;
• $2e_{ij}{e}_{jk}=e_{ik}$ for each $k\ne i,j$.

\begin{proof} See Pasted image 20251216223351.png. \end{proof}

Let $\mathcal{J}$ be a finite-dimensional simple Jordan algebra with capacity $>2$ and $F=\bar{F}$. Let $e_1,\dots ,e_n$ be primitive, pairwise orthogonal idempotents such that $1=e_1+\cdots +e_n$. Let $e_{ij}\in {\mathcal{J}}_{ij}$ such that $e_{ij}^2=e_i+e_j$ and $2e_{ij}{e}_{jk}=e_{ik}$ (for distinct $i,j,k$ ). Then by ^1bnoml, $\mathcal{D}={\mathcal{J}}_{12}$ is a composition algebra.

We will show that $\mathcal{J}$ is isomorphic to $H_n(\mathcal{D})$. For any element $a\in \mathcal{D}={\mathcal{J}}_{12}$, we define $a_{ij}\in {\mathcal{J}}_{ij}$ such that the mapping $a[ij]\to a_{ij}$ determines an isomorphism. And for this we show that $a_{ij}\in \mathcal{J}$ satisfy the same relations as $a[ij]\in H_n(\mathcal{D})$.

We define $a_{ij}$ as follows. For each $a\in \mathcal{D}$ and $i,j=1,\dots ,n$ :

• For $j\ne 1:a_{1j}=2a{e}_{2j}{a}_{j1}={\bar{a}}_{1j}$ (Conjugate)
• For $i\ne 2:a_{i2}=2a{e}_{1i}{a}_{2i}={\bar{a}}_{i2}$
• Diagonal elements: $a_{ii}=\mathrm{tr}(a)e_i$ for $i=1,\dots ,n$
• For $i,j⩾2$ distinct: $a_{ij}=2a_{i2}{e}_{2j}=-4\left(a{e}_{1i}\right)e_{2j}$

Then we have the following lemma.

Lemma

The constructed elements satisfy the following relations (matching Matrix Algebra rules):

• For any $i\ne j:a_{ji}={\bar{a}}_{ij}$ (Hermitian symmetry)
• For $i,j,k$ distinct: $2a_{ij}{b}_{jk}=(a\ast b{)}_{ik}$ (Transitivity)
• Action of diagonal: $2a_{ii}{b}_{ij}=\mathrm{tr}(a)b_{ij}2a_{ij}{b}_{jj}=\mathrm{tr}(b)a_{ij}$
• Norm relation: $2a_{ij}{b}_{ji}=\mathrm{tr}(a\ast b)\left(e_i+e_j\right)=(a\ast b{)}_{ii}+(b\ast a{)}_{jj}$

Finally, we can prove the following theorem.

Theorem

Let $\mathcal{J}$ be a finite-dimensional simple reduced Jordan algebra of capacity $n⩾3$, with $n$ idempotents pairwise orthogonal, strictly connected. Then $\mathcal{J}\cong H_n(A,\ast )$, with $A$ being a composition algebra, which is associative if $n>3$.

\begin{proof} See Pasted image 20251218220048.png. \end{proof}