9 Speciality and Exceptionality

The relationships between the various free algebras discussed in this section can be complex. The following diagram provides a structural roadmap of these subalgebars and their key distinctions.

Error: $SJ⟨X⟩\subseteq F{⟨X⟩}^{+}$ is a subalgebra.

Free Algebra and Special Universal Envelopes

Free Lie and Special Jordan Algebra

Definition

Let $X$ be a set of variables. Denote by $F⟨X⟩$ a free associative algebra generated by $X$.

The subalgebra of $F{⟨X⟩}^{(-)}$ generated by $X$ is the free Lie algebra generated by $X$, denoted $\mathrm{Lie}⟨X⟩$. Its elements are called Lie elements.

• There are at least $2$ ways to check whether given element of $F⟨X⟩$ is Lie element or not.
  • Denote by $[x_{i_1},\cdots ,x_{i_n}]$ the element $[\cdots [[x_{i_1},x_{i_2}],x_{i_3}],\cdots ,x_{i_n}]$. The Dynkin-Specht-Wever criterion guarantees that $f\in F⟨X⟩$ is homogeneous of degree $n$ that $f$ is a Lie element iff $[f]=nf$.
  • The other criterion (Friedrichs’ Criterion) considers the coalgebra structure of $F⟨X⟩$, which is given by a comultiplication $\Delta :F⟨X⟩\to F⟨X⟩\otimes F⟨X⟩$. We say $x$ is primitive, if $\Delta (x)=1\otimes x+x\otimes 1$. For any $f\in F⟨X⟩$, $f$ is a Lie element iff $f$ is primitive.

Analogously, the subalgebra of $F{⟨X⟩}^{+}$ generated by $X$ is the free special Jordan algebra, denoted $SJ⟨X⟩$. Its elements are called Jordan elements.

• Contrary to the case of Lie algebras, $SJ⟨X⟩$ is NOT free Jordan algebra (since it is not closed under homomorphic images) and we do not know criteria to decide if element $f\in F⟨X⟩$ is Jordan element.
• Nevertheless, $SJ⟨X⟩$ has the following universal property of free algebras.

Proposition

Let $\mathcal{J}$ be a special Jordan algebra. Then any mapping $\phi :X\to \mathcal{J}$ can be extended to a homomorphism $\tilde{\phi }:SJ⟨X⟩\to \mathcal{J}$ and $\tilde{\phi }$ is unique, i.e. the diagram commutes.

\begin{proof} Let $\mathcal{J}$ be a special Jordan algebra, then there exists an associative algebra $A$ such that $\mathcal{J}\hookrightarrow A^{+}$. Therefore, we have $X\to \mathcal{J}\hookrightarrow A^{+}$. Since $F⟨X⟩$ is a free associative algebra, there exists $\tilde{\phi }:F⟨X⟩\to A$ associative homomorphism which extends $\phi$. Then $\hat{\phi }:F{⟨X⟩}^{+}\to A^{+}$ is a homomorphism of Jordan algebras and taking the restriction to $SJ⟨X⟩$ as $\hat{\phi }{|}_{SJ⟨X⟩}:=\tilde{\phi }$. We obtain that $\tilde{\phi }:SJ⟨X⟩\to A^{+}$ is a homomorphism of Jordan algebras. Moreover, since $\tilde{\phi }(x)=\phi (x)\in \mathcal{J}$, the map $\tilde{\phi }$ is what we desired. \end{proof}

Remark. The class of special Jordan algebra is closed under direct sums and subalgebras, but they are not closed under homomorphic images (see ^ajsspd) and so not forms a variety of algebras. This class forms what we call quasi-variety, which consists of algebras satisfying quasi-identity. A quasi-identity is a relation of type $f=0⟹g=0$, for example $x^2=0⟹x=0$ define the quasi-variety of algebras without nilpotent elements. Each quasi-variety contains free algebras.

Special Universal Algebra $S(\mathcal{J})$

Let $\mathcal{J}$ be a Jordan algebra. The algebra $S(\mathcal{J})$ with a special injection $i:\mathcal{J}\to S(\mathcal{J}{)}^{(+)}$ is called special universal algebra of $\mathcal{J}$, if for any homomorphism $\phi :\mathcal{J}\to A^{+}$ with $A$ an associative algebra, there exists a unique $\tilde{\phi }:S(\mathcal{J})\to A$ homomorphism of associative algebra such that $\phi =\tilde{\phi }\circ i$.

Proposition

Let $\mathcal{J}$ be a Jordan algebra. Consider the tensor algebra $T(\mathcal{J})$ generated by a vector space $\mathcal{J}$

$$T(\mathcal{J})=F\cdot 1\oplus \mathcal{J}\oplus \mathcal{J}\oplus \mathcal{J}\otimes \cdots \otimes \mathcal{J}\otimes \cdots \otimes \mathcal{J}\oplus \cdots$$

and an ideal $I$ of $T(\mathcal{J})$ generated by $a\otimes a-a^2$, $a\in \mathcal{J}$. Then $S(\mathcal{J})\cong T(\mathcal{J})/I$.

\begin{proof} Let $i:\mathcal{J}\to T(\mathcal{J})/I$ be a mapping defined by $i(a)=a+I$, for any $a\in J$. Let $\phi :\mathcal{J}\to A^{(+)}$ be a homomorphism of Jordan algebras, with $A$ being an associative algebra. In particular, $\phi$ is linear. Since $T(\mathcal{J})$ is the free associative algebra over $J$, there exists a unique homomorphism of associative algebras $\tilde{\phi }:T(\mathcal{J})\to A$ such that ${\tilde{\phi }|}_J=\phi$. Observe that $I\subset \ker \tilde{\phi }$ since for any $a\in \mathcal{J}$, $\tilde{\phi }\left(a\otimes a-a^2\right)=\tilde{\phi }(a)\tilde{\phi }(a)-\tilde{\phi }\left(a^2\right)=0$. Thus there exists $\bar{\phi }:T(J)/I\to A$ such that $\phi =\bar{\phi }\circ i$. \end{proof}

Exercises.

• Let $J$ be Jordan algebra. Show that $J$ is special Jordan algebra $\iff i:J\to S(J)$ is injective. See here.
• Show that $S(SJ⟨X⟩)=F⟨X⟩$.
• Let $J$ be Jordan algebra of dimension $n$. Show that $\dim S(J)⩽2^n$.

Relation between $SJ⟨X⟩$ and $H⟨X⟩$

Note that $SJ⟨X⟩$ consists of Jordan polynomials. Obviously, $SJ⟨X⟩\ne F⟨X⟩$ because $SJ⟨X⟩$ contains only symmetric elements, whereas $F⟨X⟩$ contains anti-symmetric elements like $[x,y]$.

Therefore, the meaningful comparison is between $SJ⟨X⟩$ and the set of symmetric elements $H⟨X⟩$. While $SJ⟨X⟩\subseteq H⟨X⟩$ is always true, the equality does not hold for $|X|⩾4$.

Definition

Define the free associative algebra $F⟨X⟩$ with $X=\{x_1,\cdots ,x_n,\cdots \}$, and define a involution $\ast$ by $(x_{i_1},\cdots ,x_{i_n}{)}^{\ast }=x_{i_n}\cdots x_{i_2}{x}_{i_1}$ for $x_{i_1},\cdots ,x_{i_n}\in X$.

Denote $H⟨X⟩=H(F⟨X⟩,\ast )=\{f\in F⟨x⟩:f^{\ast }=f\}$. Clearly $SJ⟨X⟩\subseteq H⟨X⟩$.

Example

Note that $X\subseteq H⟨X⟩$ and $H⟨X⟩$ is a Jordan subalgebra of $F{⟨X⟩}^{(+)}$. Furthermore, we have $SJ⟨X⟩⊊H⟨X⟩\subseteq F⟨X⟩$.

\begin{proof} For any $f\in F⟨X⟩$, we denote $\{f\}=f+f^{\ast }$. We claim that $k=\{x_1\cdots x_4\}\in H⟨X⟩$, but $k\notin SJ⟨x_1,\cdots ,x_4⟩$. Otherwise, assume that $k\in SJ⟨x_1,\cdots ,x_4⟩$. Consider the Grassmann algebra $G=\mathrm{alg}⟨1,e_i:e_i{e}_j=-e_j{e}_i,e_i^2=0,i=1,2,\cdots ⟩$. Define $\phi :X\to G,x_i\mapsto e_i$. Then there exists $\tilde{\phi }:F⟨X⟩\to G$, which is also a homomorphism of Jordan algebras $F{⟨X⟩}^{+}\to G^{+}$. Note that

$$\begin{array}{rl}\tilde{\phi }(k)& =\tilde{\phi }\left(x_1{x}_2{x}_3{x}_4+x_4{x}_3{x}_2{x}_1\right)\\ & =\tilde{\phi }\left(x_1\right)\tilde{\phi }\left(x_2\right)\tilde{\phi }\left(x_3\right)\tilde{\phi }\left(x_4\right)+\tilde{\phi }\left(x_4\right)\tilde{\phi }\left(x_3\right)\tilde{\phi }\left(x_2\right)\tilde{\phi }\left(x_1\right)\\ & =e_1{e}_2{e}_3{e}_4+e_4{e}_3{e}_2{e}_1\\ & =2e_1{e}_2{e}_3{e}_4.\end{array}$$

On the other hand, since $k\in SJ⟨X⟩$, there exists a Jordan polynomial such that $k=J(x_1,x_2,x_3,x_4)$, which deduces that

$$\tilde{\phi }(k)=\tilde{\phi }\left(J\left(x_1,x_2,x_3,x_4\right)\right)=J\left(\tilde{\phi }\left(x_1\right),\tilde{\phi }\left(x_2\right),\tilde{\phi }\left(x_3\right),\tilde{\phi }\left(x_4\right)\right)=J\left(e_1,e_2,e_3,e_4\right)=0,$$

which is impossible. Therefore, $k\notin SJ⟨x_1,\cdots ,x_4⟩$ and so $H⟨X⟩⊋SJ⟨X⟩$. \end{proof}

From the proof of ^gk08p4, we know the element in the form of $x_1{x}_2{x}_3{x}_4$ is the gap between $H⟨X⟩$ and $SJ⟨X⟩$.

Cohn' theorem

Jordan algebra $H⟨X⟩$ is generated by $X\cup \{x_{i_1}{x}_{i_2}{x}_{i_3}{x}_{i_4}\}$ with $x_{i_j}\in X$.

\begin{proof} Define $H_0⟨X⟩=⟨X,\{x_{i_1}{x}_{i_2}{x}_{i_3}{x}_{i_4}\}⟩$. It is enough to show that for any $u\in F⟨X⟩$, $\{u\}=u+u^{\ast }\in H_0⟨X⟩$. We prove it by induction on degree of monomial $u$. When $n=2$, $x_1{x}_2=x_1{x}_2+x_2{x}_1=x_1\circ x_2$. When $n=3$, $2\{x_1{x}_2{x}_3\}=2(x_1\circ x_2)\circ x_3+(x_3\circ x_2)\circ x_1-(x_1\circ x_3)\circ x_2$.

Suppose that it is valid for $n-1>4$ and we will show that it is valid for $n$. Define $u\equiv 0$ iff $u\in H_0⟨X⟩$. Note that

$$0\equiv x_1\circ \{x_2\cdots x_n\}=x_1{x}_2\cdots x_n+x_2{x}_3\cdots x_n{x}_1+x_1{x}_n\cdots x_2+x_n\cdots x_2{x}_1=\{x_1\cdots x_n\}+\{x_2{x}_3\cdots x_n{x}_1\}$$

and

$$\{x_1\cdots x_n\}\equiv -\{x_2\cdots x_n{x}_1\}\equiv (-1{)}^n\{x_1\cdots x_n\}.$$

If $n$ is odd, we have finish the proof.

Now suppose that $n$ is even. Then we have

$$\begin{array}{rl}0\equiv & \{x_1{x}_2\}\circ \{x_3\cdots x_n\}\\ =& x_1{x}_2{x}_3\cdots x_n+x_2{x}_1{x}_3\cdots x_n+x_3\cdots x_n{x}_1{x}_2+x_3\cdots x_n{x}_2{x}_1+\\ & x_1{x}_2{x}_n\cdots x_3+x_2{x}_1{x}_n\cdots x_3+x_n\cdots x_3{x}_1{x}_2+x_n\cdots x_3{x}_2{x}_1\\ =& \{x_1\cdots x_n\}+\{x_2{x}_1{x}_3\cdots x_n\}+\{x_1{x}_2{x}_n\cdots x_3\}+\{x_2{x}_1{x}_n\cdots x_3\}.\end{array}$$

Note that $\{x_1\cdots x_n\}\equiv (-1{)}^{n(n-1)/2}\{x_1\cdots x_n\}$, so $\{x_1\cdots x_n\}\in H_0⟨X⟩$ with $4\nmid n$. Finally, if $n=4k$, then $0\equiv \{x_1{x}_2{x}_3{x}_4\}\circ \{x_5\cdots x_n\}\equiv 4\{x_1\cdots x_n\}$.

Now we finish the proof. \end{proof}

Corollary

If $|X|⩽3$, then $SJ⟨X⟩=H⟨X⟩$.

Remark. Shirshov gave direct proof $H⟨X⟩=SJ⟨X⟩$, who gave a algorithm how to write symmetric word in these variables on Jordan elements.

Remark. Although $SJ⟨X⟩=H⟨X⟩$ holds when $|X|=3$, its homomorphism image can be exceptional.

Let $I={\mathrm{id}}_{SJ⟨x,y,z⟩}⟨x\circ y⟩$. Then $SJ⟨x,y,z⟩/I$ is an exceptional Jordan algebra.

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Speciality and Exceptionality Criteria

Method 1: Cohn Criterion

Cohn criterion

If $J$ is a special Jordan algebra, $I⊲J$, denote $\hat{I}={\mathrm{id}}_{S(J)}⟨I⟩$. Then $J/I$ is special iff $\hat{I}\cap J=I$.

\begin{proof} Suppose $J/I$ is special, then there exists associative $A$ such that $J/I\hookrightarrow A^{+}$. Then by definition of special universal algebra, there exists $\tilde{\phi }:S(J)\to A$ such that $\tilde{\phi }\circ i=\phi$, where $i:J\hookrightarrow S(J)$. Note that $\ker \phi =\ker \tilde{\phi }=I$, so $I\subseteq \hat{I}\cap J$ is obvious. Additionally, $\hat{I}\subseteq \ker \tilde{\phi }=I$. Therefore, $\hat{I}\cap J=I$.

Conversely, assume that $\hat{I}\cap J=I$. Consider $A=S(J)\cap \hat{I}$, and define $\phi :S(J)\to A$. Note that $\ker \phi {|}_J=J\cap \ker \phi =J\cap \hat{I}=I$. So $J/I\cong \mathrm{im}\phi {|}_J\subseteq A^{+}$ and the proof is complete. \end{proof}

Corollary

Let $J=SJ⟨X⟩/I$, and let $I⊲SJ⟨X⟩$. If $J$ is special, then $S(J)=F⟨X⟩/\hat{I}$ with $\hat{I}={\mathrm{id}}_{S(J)}(I)$.

By ^6uo6n2, if $SJ⟨x⟩/I$ is special, then $\hat{I}\cap SJ⟨X⟩=I$. If $I\ne \hat{I}\cap SJ⟨X⟩$, then there exists $k\in \hat{I}\cap SJ⟨X⟩$ such that $k\notin I$. We call such element Cohn element.

Method 2: Shirshov-Cohn theorem

Let $T\{V\}$ be the free non-associative algebra over vector space $V$. Then $T\{V\}=F\cdot 1\oplus V\oplus V\otimes V\oplus \cdots \oplus V^{\otimes n}$, where $V^{\otimes n}={\oplus }_{i+j=n}{V}^{\otimes i}\otimes V^{\otimes j}$. Recall that $T\{V\}$ has universal property: for any $F$-algebra $A$ and linear map $\phi :V\to A$, there exists the unique homomorphism between $F$-algebras

$$\tilde{\phi }:T\{V\}\to A$$

such that $\tilde{\phi }{|}_V=\phi$.

Remark. When $V=X$,

• $T\{V\}=F\{X\}$ is the free non-associative algebra on $X$
• $F⟨X⟩$ is the free associative algebra on $X$
• $F[X]$ is the free commutative associative algebra on $X$

Definition

Define $I_{\mathrm{Jord}}=\mathrm{id}⟨[a,b],(a^2,b,a),a,b\in F\{X\}⟩$. Then $\mathrm{Jord}[X]:=F\{X\}/I_{\mathrm{Jord}}$ is the free Jordan algebra over $X$.

Shirshov theorem

The free Jordan algebra with two generators $\mathrm{Jord}[x,y]$ is special.

Remark. By ^xrk076, $\mathrm{Jord}[x,y]\simeq SJ⟨x,y⟩$. When $|X|>3$, there is no information on $SJ⟨X⟩$ and $\mathrm{Jord}[X]$.

Cohn theorem

Homomorphic image of $SJ⟨x,y⟩$ is special.

\begin{proof} Take $I⊲SJ⟨x,y⟩$ and $\hat{I}={\mathrm{id}}_{F⟨x,y⟩}⟨I⟩$. We will prove that there are no Cohn elements ($k\in \hat{I}\cap SJ⟨x,y⟩$ and $k\notin I$). Assume that $k={\sum }_n{u}_n{i}_n{v}_n$ with $u_n,v_n\in F⟨x,y⟩$ and $i_n\in I$, then

$$k^{\ast }=\sum _n{v}_n^{\ast }{i}_n{u}_n^{\ast }=\sum _n{u}_n{i}_n{v}_n=k.$$

It follows that $k=1/2({\sum }_n{u}_n{i}_n{v}_n+v_n^{\ast }{i}_n{u}_n^{\ast })$.

For any $u,v\in F⟨x,y⟩$, consider in $F⟨x,y,z⟩$ the element $r=uzv+v^{\ast }z{u}^{\ast }$, where $z\in I$. Since $r^{\ast }=r$, the element $r\in H⟨x,y,z⟩=SJ⟨x,y,z⟩$ by ^0xovb9. Then there exists $j(x,y,z)\in SJ⟨x,y,z⟩$ such that $r=j(x,y,z)$.

For any $i\in I$, $uiv+v^{\ast }i{u}^{\ast }=j(x,y,i)\in I$ as $I$ is an ideal, and so $k\in I$. By ^6uo6n2, $SJ⟨x,y⟩/I$ is special. \end{proof}

Shirshov-Cohn theorem

Any two generated Jordan algebra is special.

\begin{proof} Any Jordan algebra with two generators is a homomorphic image of $\mathrm{Jord}[x,y]=SJ⟨x,y⟩$ by ^xrk076. \end{proof}

Method 3: Glennie identity

Remark

However, $\mathrm{Jord}[x,y,z]\ne SJ⟨x,y,z⟩$, whose proof is as follows.

\begin{proof} There is Glennie identity of multi-degree $(3,3,2)$. ${\mathfrak{W}}_8$ is a scalar multiple of Glennie identity discovered by Shirshov.

Define $f=[[x,y{]}^3,z]=[x,y{]}^{3}z-z[x,y{]}^3$ with $[x,y{]}^{\ast }=-[x,y]$. Then $f^{\ast }=f$ and so $f=j(x,y,z)$.

In any associative algebra, $[a,b^2]=[a,b]\circ b=2b\circ [a,b]$. Hence $j(x,y,z^2)=[[x,y{]}^3,z^2]=2z\cdot [[x,z{]}^3,z]$.

Define $Sh(x,y,z)=j(x,y,z^2)-2z\cdot j(x,y,z)$. For any special Jordan algebra $J$ and $x,y,z\in J$, we have $Sh(x,y,z)=0$.

On the other hand, $Sh(x,y,z)\ne 0$ for exceptional Jordan algebra, as the following example shows.

Exercise. For $H_3(\mathbb{O})$, define

$$x=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right],y=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right],z=\left[\begin{array}{ccc}0& a& c\\ \overline{a}& 0& \overline{b}\\ \overline{c}& \overline{b}& 0\end{array}\right]$$

with $a,b,c\in \mathbb{O}$. There is $Sh(x,y,z)\ne 0$. \end{proof}

Construction of Exceptional Algebras

Example

Let $I={\mathrm{id}}_{SJ⟨x,y,z⟩}⟨x\circ y⟩$. Then $SJ⟨x,y,z⟩/I$ is an exceptional Jordan algebra.

\begin{proof} Consider $k=\{(x\circ y)xyz\}\in F⟨x,y,z⟩$, then $k\in \hat{I}={\mathrm{id}}_{F⟨x,y,z⟩}⟨I⟩$. Since $|X|=3$, by ^0xovb9 we have $k\in H⟨X⟩=SJ⟨x,y,z⟩$. Thus it is enough to show that $k$ is a Cohn’s element.

Suppose $k\in I$, then there exists $j(t,x,y,z)\in SJ⟨t,x,y,z⟩$ such that all monomials contain $t$ and $j(x\circ y,x,y,z)=k$. Since $k$ is homogeneous and $j$ is homogeneous, we know ${\mathrm{deg}}_{t}j=1$ or $2$.

Suppose ${\mathrm{deg}}_{t}j=2$, then $j(t,z)=\alpha t^2\circ z+\beta (t\circ z)\circ t$. Since $j(t,z)=\{(x\circ y)xyz\}$, such $\alpha ,\beta$ do not exist. Hence ${\mathrm{deg}}_{t}j=1$ and ${\mathrm{deg}}_{x}j={\mathrm{deg}}_{y}j=1$. The multi-degree of $j(t,x,y,z)$ is $(1,1,1,1)$. As $j\in SJ⟨t,x,y,z⟩$, there is $j^{\ast }=j$ and $j$ is a linear combination of tetrads, that is,

$$j(t,x,y,z)={\alpha }_1\{txyz\}+{\alpha }_2\{tyxz\}+{\alpha }_3\{xtyz\}+{\alpha }_4\{xytz\}+{\alpha }_5\{ytxz\}+{\alpha }_6\{yxtz\}.$$

We can compute that ${\alpha }_1=1$, ${\alpha }_5={\alpha }_6=0$, and $-{\alpha }_2={\alpha }_4=-{\alpha }_3=0$. So $j=\{txyz\}$. However, it is impossible, because tetrad does not belong to $SJ⟨t,x,y,z⟩$. Therefore, $k\notin I$ and so $SJ⟨x,y,z⟩/I$ is exceptional. \end{proof}

This example can be generalized as the following theorem, which will be proved later.

For big enough $|X|$ and nonzero $n\in N_t$, $SJ⟨X⟩/\mathrm{id}⟨n⟩$ is an exceptional Jordan algebra.

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Structure Theory of Special Jordan Algebras

Tetrad Eater

Definition

$f\in F⟨X⟩$ is called a tetrad eater, if $\{abcf\}\in SJ⟨X⟩$ for all $a,b,c\in SJ⟨X⟩$. Note that

$$SJ⟨X⟩⊊H⟨X⟩=⟨X,\{x_{i_1}{x}_{i_2}{x}_{i_3}{x}_{i_4}\}⟩.$$

For $u,v\in F⟨X⟩$, we say $u\equiv v$ iff $u-v\in SJ⟨X⟩$.

In ^1k683q, we prove that $\{\}$ is skew-symmetric under transposition.

Exercise 6. For any $x,y,z,t\in X$, $\{xyzt\}\equiv -\{yxzt\}\equiv \{yxtz\}$.

Exercise 7. For any $x,y,z,t\in X$, $\{x^{2}yzt\}\equiv x\circ \{xyzt\}$.

Exercise 8. For any $x,y,z,t\in X$, $\{xyzt\}\equiv \frac{1}{4}[x,y]\circ [z,t]$.

Lemma. All these exercises can be generalized for any $x,y,z,t\in SJ⟨X⟩$.

Corollary

Let $x,y,z\in SJ⟨X⟩$. Then we have

• (1) $\{xyzt\}\equiv 1/4[x,y]\circ [z,t]$;
• (2) $[x,y]\circ [x,z]\equiv 0$.

Denote by $(x,y,z{)}_{+}$ associator of $x,y,z$ with respect to $x\circ y$ Jordan product. That is, for any $a,b,c\in SJ⟨X⟩$,

$$0\equiv (a,b,c{)}_{+}=(a\circ b)\circ c-a\circ (b\circ c)=[[c,a],b]..\text{\hspace{1em}(3)}$$

Lemma

For all $x,y,z,t\in SJ⟨X⟩$, there is $\{[x,y{]}^{2}xzt\}=0$.

\begin{proof} Take $v=[x,y]$. Then

$$\{[x,y{]}^{2}xzt\}=-\{v^{2}zxt\}\stackrel{(1)}{\equiv }-\frac{1}{4}[v^2,z]\circ [x,t]\equiv -\frac{1}{4}([v,z]\circ v)\circ [x,t]\equiv -\frac{1}{4}([v,z]\circ (v\circ [x,t])+([v,z],v,[x,t]{)}_{+}).$$

Define $j=[v,z]\in SJ⟨X⟩$. Then

$$([v,z],v,[x,t]{)}_{+}=(j,v,[x,t]{)}_{+}\stackrel{(2)}{\equiv }[[[v,t],j],v]=[j_1,[x,y]]\stackrel{(3)}{\equiv }0$$

where $j_1=[[v,t],j]\in SJ⟨X⟩$. \end{proof}

Proposition

Let $x,y\in SJ⟨X⟩$ such that $\{xyab\}\equiv 0$ for all $a,b,c\in SJ⟨X⟩$. Then $[x,y{]}^2$ is a tetrad eater.

\begin{proof} By ^nswau2 we know $\{[x,y{]}^{2}xzt\}\equiv 0$. Linearizing in $y$, i.e. $y=y+u$, we get

$$\{[x,y][x,u]xzt\}\equiv 0.\text{\hspace{1em}(*)}$$

Similarly, by linearizing in $x$, there is

$$\{[x,y{]}^{2}uzt\}+\{[x,y][u,y]xzt\}+\{[u,y][x,y]xzt\}\equiv 0.\text{\hspace{1em}(**)}$$

Linearizing $(\ast )$ in $x$, we obtain

$$\{([x,y]\circ [x,u])wzt\}+\{([x,y]\circ [w,u])xzt\}+\{[w,y]\circ [x,u]xzt\}\equiv 0,$$

which follows that

$$\begin{array}{rl}\{[x,y{]}^{2}uzt\}& \stackrel{(\ast \ast )}{\equiv }-\{[x,y]\circ [u,y]xzt\}\equiv \{([x,y]\circ [u,y])zxt\}\\ & \equiv -\{[y,x]\circ [z,u]yxt\}+\{([z,x]\circ [y,u]yxt)\}\\ & \equiv -\{[y,x]\circ [z,u]-[z,x]\circ [y,u]yxt\}\equiv 0.\end{array}$$

Now we finish the proof. \end{proof}

Corollary

For all $x,y\in SJ⟨X⟩$, the element $[[x,y{]}^2,x{]}^2$ is tetrad eater.

\begin{proof} Let $X_0=[x,y{]}^2$, and let $Y_0=x$. For any $z,t\in SJ⟨X⟩$, there is $\{[x,y{]}^{2}xzt\equiv 0\}$ by ^nswau2, which deduces that $\{X_0{Y}_{0}ab\}\equiv 0$ for any $a,b\in SJ⟨X⟩$. Hence by ^kymkph, $[[x,y{]}^2,x{]}^2$ is tetrad eater. \end{proof}

Definition

Let $N_t=\{f\in SJ⟨X⟩:\{fabc\}\in SJ⟨X⟩,\forall a,b\in SJ⟨X⟩\}$ be the set of tetrad eaters.

Theorem

For big enough $|X|$ and nonzero $n\in N_t$, $SJ⟨X⟩/\mathrm{id}⟨n⟩$ is an exceptional Jordan algebra.

\begin{proof} Take $x,y,z$ such that $x,y,z\notin \mathrm{id}⟨n⟩$. Then $k=\{nxyz\}\in SJ⟨X⟩\cap \hat{I}$, where $I={\mathrm{id}}_{S⟨J⟩}⟨n⟩$ and $\hat{I}={\mathrm{id}}_{F⟨X⟩}⟨I⟩$.

We claim that $k\notin I$. Otherwise $k=j(n,x,y,z)$ is a Jordan element. Since $x,y,z$ do not appear in $n$, $j$ has form $\alpha \{txyz\}$ for some $\alpha \in F$. We know that $\{txyz\}$ is not Jordan, contradiction.

By ^6uo6n2, $SJ⟨X⟩/I$ is exceptional. \end{proof}

Remark. For simple special Jordan algebras, the ideal generated by tetrad eaters is either the entire algebra (standard type) or zero (PI type)—an “all-or-nothing” property. The quotient construction $SJ⟨X⟩/\mathrm{id}⟨n⟩$ forces an intermediate state that violates this rigidity, resulting in an exceptional algebra.

the Main theorem

Any special Jordan algebra is either $H(A,\ast )$ for $(A,\ast )$ for some $\ast$-simple associative $A$, or PI Jordan algebra.

Define $(a,b,c{)}_{+}$ as associator with respect to $a\circ b$ in associative algebra, that is, $(a,b,c{)}_{+}=[b,[a,c]]$. Define $(a,x,b):=x{\mathcal{D}}_{a,b}$, where ${\mathcal{D}}_{a,b}$ is a derivation of $\mathcal{J}$.

The set $N_t$ satisfies the following closure properties.

Proposition

• If $n\in N_t$, then $(a,n,b{)}_{+}\in N_t$ for any $a,b\in SJ⟨X⟩$.
• For any $m,n\in N_t$ and any $a\in SJ⟨X⟩$, there is $n\circ m\in N_t$ and $(n,a,m{)}_{+}\in N_t$.
• For any $n_1,n_2,n_3\in N_t$ and $m=(n_1,n_2,n_3{)}_{+}$, we have ${\mathrm{id}}_{SJ⟨X⟩}⟨m⟩\subseteq N_t$.

By this proposition, one can construct an ideal contained in $N_t$.

Corollary

For all $x_1,x_2,x_3,x_4,x_5,x_6\in X$, there is

$${\mathrm{id}}_{SJ⟨X⟩}⟨([[x_1,x_2{]}^2,x_1],[[x_3,x_4{]}^2,x_3],[[x_5,x_6{]}^2,x_5]{)}_{+}⟩\subseteq N_t.$$

A simple special Jordan algebra $J$ falls into one of two types:

• $P=J$: $J\cong H(A,\ast )$ for a $\ast$-simple associative algebra $A$ with involution $\ast$;
• $P=0$: $J$ satisfies the polynomial identity $[[x_1,x_2{]}^2,x_1],[[x_3,x_4{]}^2,x_3],[[x_5,x_6{]}^2,x_5]=0$;

where $P=\sum \{I⊲SJ⟨X⟩:I\subseteq N_t\}$.

Definition

We call Jordan algebra $\mathcal{J}$ a PI Jordan algebra if $\mathcal{J}$ satisfies polynomial identity $f$ such that $f{|}_{SJ⟨X⟩}=0$.

Theorem

Let $J$ be a special simple Jordan algebra. Then $J=H(A,\ast )$, where $A$ is a $\ast$-simple associative algebra with involution, or $J$ satisfies polynomial identity $f(x,y,z,t,u,v)=0$, where

$$f(x,y,z,t,u,v)=([[x,y{]}^2,x{]}^2,[[z,t{]}^2,z{]}^2,[[u,v{]}^2,u{]}^2{)}_{+}.$$

\begin{proof} Let $P=\sum \{I⊲SJ⟨X⟩:I\subseteq N_t\}$. We show that $P(J)=\{g(a_1,\cdots ,a_n):g\in P,a_1,\cdots ,a_n\in J\}$ is an ideal in $J$.

$P(J)$ is a subspace in $J$. For any $a\in P(J)$ and $b\in J$, take $g\in P$ and $a_1,\cdots ,a_n\in J$ such that $a=g(a_1,\cdots ,a_n)$. Consider $\pi :SJ⟨X⟩\to J$ such that $\pi (x_i)=a_i$ for $i=1,\cdots ,n$, $\pi (x_{n+1})=b$ and $\pi (x_m)=0$ for $m>n+1$. Then $g_1:=g(x_1,\cdots ,x_n)x_{n+1}\in P$ and $a\circ b=g(a_1,\cdots ,a_n)b\in P(J)$.

Since $J$ is simple, either $P(J)=J$ or $P(J)=0$. If $P(J)=0$, then the element of $J$ satisfy $f$ by ^9fb994 and so $J$ is a PI Jordan algebra. If $P(J)=J$, then $J$ is special. It follows that $J=SJ⟨X⟩/I$ and by ^6uo6n2 $\hat{I}\cap SJ⟨X⟩=I$. By universal special enveloping, we have

$$J=SJ⟨X⟩/\hat{I}\cap SJ⟨X⟩\cong SJ⟨X⟩+\hat{I}/\hat{I}\subseteq H⟨X⟩+\hat{I}/\hat{I}\subseteq F⟨X⟩/\hat{I}\cong S(J).$$

Define $\pi :F⟨X⟩\to F⟨X⟩/\hat{I}=S(J)$. Note that $SJ⟨X⟩\subseteq H⟨X⟩\subseteq F⟨X⟩$, and $J\subseteq H(S(J),\ast )\subseteq S(J)$. Since $P(J)=J$, we have that $\{JJJJ\}=J$. By ^1k683q, $H(S(J),\ast )={\mathrm{alg}}_{F⟨X⟩}⟨J\cup \{JJJJ\}⟩=J$.

In general, $S(J)$ is not $\ast$-simple. Define $\mathcal{M}=\{I{}_{\ast }⊲S(J):I\cap J=0\}$, where ${}_{\ast }⊲$ means stable by $\ast$. Then $M\ne \varnothing$ as $0\in \mathcal{M}$. By Zorn’s lemma, it has maximal element $K$. Set $A=S(J)/K$, then $A$ is still an algebra with involution since $K{}_{\ast }⊲S(J)$. Since $J\cap K=0$, $J\cong J/J\cap K\cong J_K/K\hookrightarrow A$ and so $J\subseteq A$. Let $\phi :S(J)\to A=S(J)/K$ be the canonical projection, then $\phi (H(S(J),\ast ))=H(A,\ast )$. Hence $J\cong H(A,\ast )$, and $A$ is $\ast$-simple by the maximality of $K$. \end{proof}