12 Infinite-dimensional Jordan Algebras I - i-special Case

Our goal is to consider the classification of prime Jordan algebras, with a particular interest in the infinite-dimensional case.

According to the structure theory, the classification naturally divides into two branches based on $\overline{SJord}$, the variety generated by special Jordan algebras.

• The i-Special Case ($J\in \overline{SJord}$ )
  • This variety is characterized by the set of all polynomial identities that hold for special Jordan algebras (s-identities), such as Glennie’s identity.
  • We show that such algebras fall into
    • the Hermitian type $H(A,\ast )$, or
    • the quadratic type $J(V,g)$.
• The i-Exceptional Case ($J\notin \overline{SJord}$ )
  • 13 Infinite-dimensional Jordan Algebras II - i-exceptional Case considers algebras that fail to satisfy $s$-identites.
  • A key result is that such algebras cannot be infinite-dimensional; they are forms of the 27-dimensional Albert algebra.

This section is organized as follows.

• the Polynomial Identity f and Simple Classification finishes the classification of finite-dimensional simple Jordan algebras.
  • Define $f(x,y,z,t,u,v)=([[x,y{]}^2,x{]}^2,[[z,t{]}^2,z{]}^2,[[u,v{]}^2,u{]}^2{)}_{+}$
  • (capacity ≥ 3) If $f\ne 0$, then $J=H(A,\ast )$
  • If $f=0$, then either
    • (capacity = 1) $J=F$, or
    • (capacity = 2) $J=J(V,g)$.
• Radical and Speciality gives a criteria of special Jordan algebra in the form of $H⟨X⟩/P$.
  • Remark that each Jordan algebra is a quotient of $SJ⟨X⟩$, but here we only consider the quotient of $H⟨X⟩$ for computational convenience.
  • If such Jordan algebra is prime, i.e. $\mathrm{rad}J=0$, then it is special.
  • Furthermore, without proof, a prime Jordan algebra is i-special iff it is special.
  • Therefore, in the following sections, we focus on the classification of Prime Special Jordan Algebras.
• Central Localization of Prime Special PI Jordan Algebras is a reduction from prime PI to finite dimensional simple.
  • It uses the similar method of Posner, that is why we assume PI. (For non-PI case, $J\cong H(A,\ast )$ and they are wild.)
  • We can get finite dimensional simple Jordan algebra from prime PI Jordan algebra via central localization.
  • Therefore, by Classification of JNW, a prime special Jordan PI-algebra $J$ can be classified as
    • $J=H(A,\ast )$
    • $J$ satisfies $f(x,y,z,t,u,v)=([[x,y{]}^2,x{]}^2,[[z,t{]}^2,z{]}^2,[[u,v{]}^2,u{]}^2{)}_{+}=0$, and so $Z^{-1}J$ is a finite-dimensional simple Jordan algebra satisfying $f=0$.
• [[#structure-of-prime-pi-jordan-algebras-satisfying-f0|Structure of Prime PI Jordan Algebras Satisfying $f=0$]] divide the prime Jordan algebra $J$ with $f=0$ into two parts
  • if $[x,y{]}^2\ne 0$, then $J=J(V,f)$;
  • if $[x,y{]}^2=0$, then $J$ is associative.

the Polynomial Identity $f$ and Simple Classification

Let $X$ be a set of variables. Take $x,y\in X$, then $[[x,y{]}^2,x{]}^2\in F⟨X⟩$ is a Jordan element in $SJ⟨X⟩$. There exists a non-associative polynomial $n(x,y)\in F\{X\}$ such that $[[x,y{]}^2,x{]}^2=n(x,y{)}_{+}$. Define $f(x_1,x_2,x_3,x_4,x_5,x_6)=(n(x_1,x_2),n(x_3,x_4),n(x_5,x_6))$.

Recall that we have proved the following theorem.

Transclude of 9-Speciality-and-Exceptionality#^8z1x2m

In this subsection, we consider the later case.

Lemma

Let $J$ be a simple finite-dimensional Jordan algebra (over $F=\overline{F}$), and $J$ satisfies $f$. Then $J$ is either $F$ or $J=J(V,g)$ is a Jordan algebra of bilinear form.

\begin{proof} Assume that $e_1+\cdots +e_n=1$. For $n⩾3$, $H_3(F)\subseteq H_n(D)=J$. There exists some element $f$ such that $f\ne 0$ in $H_3(F)$, leading to a contradiction. Hence $n<3$.

By 7 Capacity of Jordan Algebras ≤ 2, when $n=1$, $J=F$; when $n=2$, $J=J(V,g)$.
\end{proof}

Radical and Speciality

In this subsection, we aim to prove that

Let $J=H⟨X⟩/P$. If $\mathrm{rad}J=0$, then $J$ is special.

Link to original

Recall that homomorphic image of $H⟨X⟩$ is special, and $SJ⟨X⟩\subseteq H⟨X⟩\subseteq F^{(+)}⟨X⟩$.

Lemma

Let $I⊲H⟨X⟩$. For any $a\in I$, ${\mathrm{id}}_{F⟨X⟩}⟨a^2⟩\cap H⟨X⟩\subseteq I$.

\begin{proof} Take $f\in {\mathrm{id}}_{F⟨X⟩}⟨a^2⟩\cap H⟨X⟩$. Since $f\in {\mathrm{id}}_{F⟨X⟩}⟨a^2⟩$, it can be written as $f={\sum }_i\{u_i{a}^2{v}_i\}$, where $v_i,u_i$ are monomials in $X$.

By ^1k683q, there is $H⟨X⟩=⟨X,\{x_i{x}_j{x}_k{x}_{\ell }\}⟩$. Define $\tilde{f}(t)={\sum }_i\{u_{i}t{v}_i\}\in H⟨X\cup t⟩$, then

$$\tilde{f}(t)=j_1(x_1,\cdots ,x_n,\{x_i{x}_j{x}_k{x}_{\ell }\},\cdots ,t)+j_2(x_1,\cdots ,x_n,\{x_i{x}_j{x}_k{x}_{\ell }\},\cdots ,\{x_i{x}_j{x}_{k}t\}).$$

Take $t=a^2$. By $a^2\in I$, we know $\{x_i{x}_j{x}_k{a}^2\}=j(x_i,x_j,x_l,a^2)+a\circ \{x_{i_1}{x}_{j_2}{x}_{k_1}a\}\in I$ and so $f=\tilde{f}(a^2)\in I$. Now we finish the proof. \end{proof}

Corollary

${\mathrm{id}}_{F⟨X⟩}⟨I^2⟩\cap H⟨X⟩\subseteq I$.

Lemma

Let $J$ be a special Jordan algebra, and let $A$ be an associative algebra. Assume that $J\subseteq A^{(+)}$ and $\mathrm{alg}⟨J⟩=A$. If $I\subseteq J$ and $I^2=0$, then $I\subseteq \mathrm{rad}A$.

\begin{proof} WLOG we may assume that $\mathrm{rad}A=0$ and prove that $I=0$. (Otherwise take $\overline{A}=A/\mathrm{rad}A$ and $\overline{J}=J+\mathrm{rad}A/\mathrm{rad}A$.) Let $a,b\in I$. Then $a\circ b=0$. Consider $c=[a,b]\in A$. For any $x\in J$, $[x,c]=[x,[a,b]]=4(a,x,b{)}_{+}=4(a\circ x)\circ b-4a\circ (x\circ b)\in I^2=0$.

Since $A=\mathrm{alg}⟨J⟩$, we have $[c,A]=0$ and so $c\in Z(A)$. In associative algebra we have

$$[a,b]\circ x=[a\circ x,b]-a\circ [x,b].\text{\hspace{1em}(*)}$$

It follows that

$$\begin{array}{rl}2c^2& =2[a,b{]}^2=[a,b]\circ [a,b]=[a\circ [a,b],b]-a\circ [[a,b],b]\\ & =[[a^2,b],b]-a\circ [[a,b],b]=-4(a^2,b,b{)}_{+}+4a\circ (a,b,b{)}_{+}=0.\end{array}$$

Since $c^2=0$ and $c\in Z(A)$, there is $cA⊲A$. Since $(cA{)}^2=c^2{A}^2=0$, $c$ is annihilates $m$-sequence and $c\in \mathrm{rad}A=0$. It yields that $[a,b]=0$. As $[a,b]=0$ and $a\circ b=0$, we have $ab=0$ for any $a,b\in I$.

Define $\hat{I}={\mathrm{id}}_A⟨I⟩$. We prove that $\hat{I}=I+IA=I+AI$. To prove that, it is enough to prove that $I+IA$ and $I+AI$ are ideals in $A$. For any $x\in J$ and $a\in I$, $xa=(xa+ax)-ax\in I+IA$. Since $J$ generates $A$, we have $J(I+IA)\subseteq I+IA$. Thus $I+IA$ is an ideal and so $\hat{I}$ is an ideal of $A$. Since $A$ is semisimple, $\hat{I}=0$ and so $I=0$. \end{proof}

Corollary

Let $J$ be a special Jordan algebra, with $J\subseteq A^{+}$ and $\mathrm{alg}⟨J⟩=A$. Then $\mathrm{rad}J=\mathrm{rad}A\cap J$.

\begin{proof} Define $\overline{A}=A/\mathrm{rad}A$, then $\overline{J}\subseteq {\overline{A}}^{(+)}$ and $\overline{A}=\mathrm{alg}⟨\overline{J}⟩$. Let $\overline{I}⊲\overline{J}$ with $\overline{I}\circ \overline{I}=0$. By ^786164, $\overline{I}\subseteq \mathrm{rad}A=0$ and so $\overline{I}=0$. Hence $\overline{J}$ is semi-prime and $\mathrm{rad}\overline{J}=0$. It follows that $\mathrm{rad}J\subseteq \mathrm{rad}A$ and so $\mathrm{rad}J\subseteq \mathrm{rad}A\cap J$.

The other inclusion: let $a\in \mathrm{rad}A\cap J$, and let $\{a_i:i⩾1\}$ be any $m$-sequence in $J$ which contains $a$. Since ${\mathrm{id}}_J⟨a_i⟩\subseteq {\mathrm{id}}_A⟨a_i⟩$ and $a_{i+1}\in ({\mathrm{id}}_J⟨a_i⟩{)}^2\subseteq ({\mathrm{id}}_A⟨a_i⟩{)}^2$, we know $\{a_i:i⩾1\}$ is a $m$-sequence in $A$ and so $a\in \mathrm{rad}A$. Then it contains $0$ and $a\in \mathrm{rad}J$.

Now we finish the proof. \end{proof}

Theorem

Let $J=H⟨X⟩/P$. If $\mathrm{rad}J=0$, then $J$ is special.

\begin{proof} Define $\widehat{P^2}={\mathrm{id}}_{F⟨X⟩}⟨P^2⟩$. Then $\widehat{P^2}\cap P\subseteq \widehat{P^2}\cap H⟨X⟩\subseteq \widehat{P^2}\cap P$ by ^eh49pd and so $\widehat{P^2}\cap P=\widehat{P^2}\cap H⟨X⟩$.

Note that

$$\frac{\frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}}{\frac{P+\widehat{P^2}}{\widehat{P^2}}}\cong \frac{H⟨X⟩/(H⟨X⟩\cap \widehat{P^2})}{P/(P\cap \widehat{P^2})}\cong \frac{H⟨X⟩}{P}\cong J,$$

Since $\mathrm{rad}J=0$, we have

$$\mathrm{rad}\left(\frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}\right)\subseteq \frac{P+\widehat{P^2}}{\widehat{P^2}}.\text{\hspace{1em}(*)}$$

Let $B=\mathrm{rad}\left(F⟨X⟩/\widehat{P^2}\right)$, then by ^b8ac58 and $(\ast )$

$$B\cap \frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}=\mathrm{rad}\left(\frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}\right)\subseteq (P+\widehat{P^2})/\widehat{P^2}.$$

On the other hand, since $P^2\subseteq \widehat{P^2}$, we have ${\left(\frac{P+\widehat{P^2}}{\widehat{P^2}}\right)}^2=0$ in $F⟨X⟩/\widehat{P^2}$. Thus $\frac{P+\widehat{P^2}}{\widehat{P^2}}\subseteq B$. Combining this with the previous inclusion, we get

$$B\cap \frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}=\frac{P+\widehat{P^2}}{\widehat{P^2}}$$

and

$$J\cong \frac{\frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}}{B\cap \frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}}\cong \frac{\frac{H⟨X⟩+\widehat{P^2}}{\widehat{P^2}}+B}{B}\hookrightarrow {\left(\frac{F⟨X⟩/\widehat{P^2}}{B}\right)}^{+}.$$

Therefore, $J$ is special. \end{proof}

Remark. Although ^vtiqhh is stated for algebras in the form of $H⟨X⟩/P$, the conclusion holds generally for $\overline{SJord}$. For any i-special Jordan algebra $J$ (i.e., $J\in \overline{SJord}$), if $J$ is prime, then $J$ is special. Specifically, Cohn’s examples, which are i-special but not special, are known to be not prime.

Central Localization of Prime Special PI Jordan Algebras

For associative prime PI-algebra, its central localization is central simple.

Posner

Let $R$ be an associative prime PI-algebra over $F$ with center $Z$ and let $S=Z\setminus \{0\}$. Then $S^{-1}R$ is $S^{-1}Z$-algebra, which is central simple and of finite dimension.

In fact, for a prime special PI Jordan algebra, we can use the generalized Posner theorem and get a finite-dimensional simple Jordan algebra via central Localization. Then by Classification of JNW, we have the following conclusions.

• If $f(J)\ne 0$, then $J$ is of Hermitian type (^5pmm0t);
• If $f(J)=0$, then $Z^{-1}J$ is a field or quadratic (^boupap).

Theorem

Let $A$ be an associative algebra with involution. Assume that $J=H(A,\ast )$ is a prime PI Jordan algebra. Then $Z=Z(J)\ne 0$, $Z^{-1}J:=\{z^{-1}a:0\ne z\in Z,a\in J\}$ is a simple algebra over field $Z^{-1}Z$, and dimension of $Z^{-1}J$ over $Z^{-1}Z$ is finite.

Lemma

Let $J$ be a prime Jordan algebra generated by three elements. Suppose $J\in SJord$ and $J$ satisfies $f=0$, where

$$f(x_1,x_2,x_3,x_4,x_5,x_6)=(n(x_1,x_2),n(x_3,x_4),n(x_5,x_6)).$$

Then $Z=Z(J)\ne 0$. Moreover, $Z^{-1}J$ is of finite dimension over $Z^{-1}Z$ and $Z^{-1}J\cong Z^{-1}Z$ or $Z^{-1}J$ is a Clifford algebra.

\begin{proof} Since $J$ is generated by three elements (variables in $f$), we can write $J\cong SJ⟨x,y,z⟩/P$.

By Cohn’s Theorem, for $|X|⩽3$, the free special Jordan algebra coincides with the symmetric elements of the free associative algebra, i.e., $SJ⟨x,y,z⟩=H⟨x,y,z⟩$. Thus, $J\cong H⟨X⟩/P$.

Since $J$ is prime, we have $\mathrm{rad}J=0$. By ^vtiqhh, we conclude that $J$ is special.

Since $J$ is special and prime, it satisfies the condition of the analog of Posner’s Theorem. This implies that $Z^{-1}J$ is a finite-dimensional central simple algebra over the field $K=Z^{-1}Z$.

Since $J$ satisfies the identity $f=0$, its localization $Z^{-1}J$ also satisfies $f=0$. Recall the classification of finite-dimensional simple Jordan algebras: if a simple algebra satisfies $f=0$ (i.e., degree $<3$), it must be either:

• Degree 1: $Z^{-1}J\cong K=Z^{-1}Z$; or
• Degree 2: $Z^{-1}J$ is a Jordan algebra of a bilinear form (Clifford type).

This completes the proof. \end{proof}

Structure of Prime PI Jordan Algebras Satisfying $f=0$

Let $\overline{SJord}$ be the variety generated by special Jordan algebras. The following lemma is a “local global principle”.

Lemma

Let $J\in \overline{SJord}$ satisfying $f=0$. Then $\mathrm{Nil}(J)=\{a\in J:a,ax\text{ are nilpontent for all }x\in J\}$. 

\begin{proof} Let $\tilde{N}=\{a\in J:a,ax\text{ are nilpontent for all }x\in J\}$, then clearly $\mathrm{Nil}(J)\subseteq \tilde{N}$.  To show the other inclusion $\tilde{N}\subseteq \mathrm{Nil}(J)$, it is enough to show that for any $a,b\in \tilde{N}$ and $x,y\in J$, 

$$a+b\in \mathrm{Nil}(\mathrm{alg}⟨a,b,x⟩),(ax)y\in \mathrm{Nil}(\mathrm{alg}⟨a,x,y⟩).$$

It is equivalent to showing that $\tilde{N}$ is an ideal in $J$. 

We will assume that $J=\mathrm{alg}⟨r,s,t⟩$ (finitely generated).  Consider the quotient by the nil radical

$$\overline{J}=J/\mathrm{Nil}(J)\hookrightarrow \prod _{\alpha }{J}_{\alpha },$$

where each $J_{\alpha }$ is a prime Jordan algebra. Since $J$ satisfies $f=0$, each component $J_{\alpha }$ also satisfies $f=0$. 

By ^boupap, for each $\alpha$, the center $Z_{\alpha }=Z(J_{\alpha })\ne 0$, and the localization ${\mathcal{J}}_{\alpha }:=Z_{\alpha }^{-1}{J}_{\alpha }$ is a simple finite-dimensional algebra over the field $K_{\alpha }=Z_{\alpha }^{-1}{Z}_{\alpha }$. Specifically, ${\mathcal{J}}_{\alpha }$ is either the field $K_{\alpha }$ itself or a Clifford algebra.

Suppose there exist $a,b\in \tilde{N}$ such that $a+b\notin \mathrm{Nil}(J)$. Then there exists some $\alpha$ and an epimorphism ${\phi }_{\alpha }:J\to J_{\alpha }$ such that the image $\overline{a+b}\ne 0$ (and is not nilpotent) in $J_{\alpha }$.

However, consider the images of $a$ and $b$ in the central localization ${\mathcal{J}}_{\alpha }$. Since $a\in \tilde{N}$, $a$ is “locally nilpotent” in $J$. This property is preserved by homomorphism and localization. In a finite-dimensional simple Jordan algebra (which is either a field or Clifford type), the only elements satisfying the definition of $\tilde{N}$ are the zero elements (since the radical is 0). Therefore, ${\phi }_{\alpha }(a)=0$ and ${\phi }_{\alpha }(b)=0$ in ${\mathcal{J}}_{\alpha }$.

It follows that ${\phi }_{\alpha }(a+b)={\phi }_{\alpha }(a)+{\phi }_{\alpha }(b)=0$ in ${\mathcal{J}}_{\alpha }$. Since $J_{\alpha }$ is prime and embeds into ${\mathcal{J}}_{\alpha }$ (up to central torsion, but prime algebras are torsion-free over center), this implies ${\phi }_{\alpha }(a+b)=0$ in $J_{\alpha }$ (or is nilpotent), which contradicts the assumption that $a+b\notin \mathrm{Nil}(J)$.

Thus $a+b\in \mathrm{Nil}(J)$, and $\tilde{N}$ is an ideal. \end{proof}

Now we classify the elements in $\overline{SJord}$ with $f=0$. Define $[x,y{]}^2=-2(x^2,y,y{)}_{+}+4x(x,y,y{)}_{+}$.

• If $J$ is a prime algebra which satisfies $f=0$ and $[x,y{]}^2\ne 0$, then $Z^{-1}J\cong J(V,f)$. ($Z^{-1}J$ is of finite dimension over $Z^{-1}Z$.)
• If $f\equiv 0$, then $J$ is described in the following theorem.

Theorem

Let $J$ be a non-degenerate Jordan algebra which satisfies $[x,y{]}^2=0$. Then $J$ is associative and thus a domain.

Therefore, if $J$ is simple and $J\in \overline{SJord}$, then one of the followings hold.

• $J=H(A,\ast )$, for some simple associative $A$;
• $J$ is PI algebra;
  • if $J$ does not satisfy $[x,y{]}^2=0$, then $J=J(V,f)$;
  • if $J$ satisfies $[x,y{]}^2=0$, then $J$ is associative.