6 Pierce Decomposition for Jordan Algebras and Idempotents

Definition

Let $A$ be an algebra. Then:

• associative center of $A$ is defined as

  $$N(A)=\{a\in A:(a,A,A)=(A,a,A)=(A,A,a)=0\}.$$

• commutative center of $A$ is defined as

  $$\mathcal{K}(A)=\{a\in A:[a,A]=0\}.$$

• center of $A$ is defined as

  $$Z(A)=N(A)\cap \mathcal{K}(A).$$

Lemma

Let $\mathcal{J}$ be a Jordan algebra, let $I⊲J$ and $e\in I$ be a unity element in $I$. Then $e\in \mathcal{Z}(\mathcal{J})$ and $\mathcal{J}=I\oplus \mathcal{K}$ for some ideal $\mathcal{K}⊲\mathcal{J}$.

\begin{proof} Take $x\in I,a\in \mathcal{J}$ and $e\in I$, then

$$(x,e,a)=(x,e^2,a)=e(x,e,a)+(x,e,a)e=2(x,e,a).$$

Hence $(x,e,a)=0$. Take $a,b\in \mathcal{J}$, then $(e,a,b)=(e^2,a,b)=-(b,a,e)-(b,a,e)=-2(eb,a,e)$ by linearization, and

$$(x,a,e)=(xa)e-x(ae)=xa-x(ae)=(xe)a-x(ae)=(x,e,a)=0.$$

It follows that $(e,a,b)=0$ and so $(e,\mathcal{J},\mathcal{J})=0$.

For any commutative or anti commutative algebra, $(a,b,a)=0$. Linearizing $a\to a+c$, there is $(a,b,c)=-(c,b,a)$. As $(e,\mathcal{J},\mathcal{J})=0$, we have $(\mathcal{J},\mathcal{J},e)=0$.

In commutative algebra we have $(a,b,c)+(b,c,a)+(c,a,b)=0$. Similarly one can check $(a,e,b)=0$ and $(\mathcal{J},e,\mathcal{J})=0$. (exercise.)

Therefore, $e\in \mathcal{Z}(\mathcal{J})$. Define $\mathcal{K}=\{x-xe:x\in \mathcal{J}\}$. We claim $\mathcal{K}⊲\mathcal{J}$. For any $x,y\in \mathcal{J}$, $(x-xe)y=xy-(xe)y=xy-(xy)e\in \mathcal{K}$. Note that $a=ae+(a-ae)\in I+\mathcal{K}$. Furthermore, for $k\in I\cap \mathcal{K}$, there exists $a,b\in \mathcal{J}$ such that $k=ae=b-be$. Hence $k=ae=(ae)e=be-(be)e=be-be=0$ and so $I\cap \mathcal{K}=\{0\}$. Now we finish the proof. \end{proof}

Theorem

Suppose that every simple Jordan algebra contains unity element. Then each finite Jordan algebra $\mathcal{J}$ with $\mathrm{Nil}\mathcal{J}=0$ has unity element and can be written as direct sum of simple algebras.

\begin{proof} Induction in $\dim \mathcal{J}$. If $\mathcal{J}$ is simple, nothing to prove. Let $I⊲\mathcal{J}$ be a non-trivial ideal. It follows that $\mathrm{Nil}I=0$ and $I=I_1\oplus \cdots \oplus I_k$ with $I_j$ simple, by induction hypothesis. Hence $I$ has unity element. By ^fbftxw, $\mathcal{J}=I\oplus K$ for some $K$. Also $\mathrm{Nil}K=0$ and $K=K_1\oplus \cdots \oplus K_{\ell }$ by induction hypothesis. Therefore, $\mathcal{J}=I_1\oplus \cdots \oplus I_k\oplus K_1\oplus \cdots \oplus K_{\ell }$ and $\mathrm{Nil}\mathcal{J}=0$. Now we finish the proof. \end{proof}

Pierce Decomposition for Jordan Algebras

For any idempotent $e$, each $a$ can be written as $a=ea+(1-e)a:=a_0+a_1$, where $e{a}_0=a_0$ and $e{a}_1=0$.

Let $e$ be idempotent in Jordan algebra $\mathcal{J}$. Apply $a=b=c=e$ to $(ac,b,d)+(ad,b,c)+(cd,b,a)=0$, then we get

$$R_e+2R_e^3-3R_e^2=R_e(2R_e^2-3R_e+1)=R_e(2R_e-1)(R_e-1)=0.$$

This implies that the only possible eigenvalues for the operator $R_e$​ are $0$, $1/2$, and $1$. Hence, we have decomposition of $\mathcal{J}$ as vector space

$$\mathcal{J}={\mathcal{J}}_0(e)\oplus {\mathcal{J}}_{1/2}(e)\oplus {\mathcal{J}}_1(e),$$

where ${\mathcal{J}}_i(e)=\{a\in \mathcal{J}:ea=ia\}$.

Pierce decomposition in Jordan algebra

Let $\mathcal{J}$ be Jordan algebra, and let $e$ be a non-trivial idempotent. We have $\mathcal{J}={\mathcal{J}}_0\oplus {\mathcal{J}}_{1/2}\oplus {\mathcal{J}}_1$, where ${\mathcal{J}}_i$ are defined above. Furthermore, the followings hold.

$${\mathcal{J}}_1^2\subseteq {\mathcal{J}}_1,{\mathcal{J}}_0^2\subseteq {\mathcal{J}}_0,{\mathcal{J}}_i{\mathcal{J}}_{1/2}\subseteq {\mathcal{J}}_{1/2},{\mathcal{J}}_{1/2}^2\subseteq {\mathcal{J}}_0+{\mathcal{J}}_1,{\mathcal{J}}_1{\mathcal{J}}_0=0.$$

\begin{proof} We have decomposition $\mathcal{J}={\mathcal{J}}_0\oplus {\mathcal{J}}_{1/2}\oplus {\mathcal{J}}_1$. Take $a\in {\mathcal{J}}_0\cup {\mathcal{J}}_1$ and $x\in \mathcal{J}$, then use the similar idea here and we have

$$(e,x,a)=(e^2,x,a)=-2(ea,x,e).$$

If $a\in {\mathcal{J}}_0$, then $-2(ea,x,e)=0$ and so $(e,x,a)=0$; if $a\in {\mathcal{J}}_1$, then $-2(ea,x,e)=-2(a,x,e)=2(e,x,a)=0$. Therefore, $(e,\mathcal{J},a)=0$, where $a\in {\mathcal{J}}_0\cup {\mathcal{J}}_1$.

Take $a_i,b_i\in {\mathcal{J}}_i$, then

$$\begin{array}{rl}& (a_1{b}_1)e=a_1(b_{1}e)+(a_1,b_1,e)=a_1{b}_1,\\ & (a_0{b}_0)e=a_0(b_{0}e)+(a_0,b_0,e)=0,\\ & (a_1{b}_0)e=a_1(b_{0}e)+(a_1,b_0,e)=0,\\ & (a_1{b}_0)e=(b_0{a}_1)e=b_0(a_{1}e)+(b_0,a_1,e)=0.\end{array}$$

Finally, let $a\in {\mathcal{J}}_0\cup {\mathcal{J}}_1$ and $b\in {\mathcal{J}}_{1/2}$, then

$$(ab)e=a(be)+(a,b,e)=\frac{1}{2}ab.$$

Set $a=a_0+a_{1/2}+a_1$, and notice that

$$(a,e,e)=(a_{1/2},e,e)=(a_{1/2}e)e-a_{1/2}{e}^2=\frac{1}{4}{a}_{1/2}-\frac{1}{2}{a}_{1/2}=-\frac{1}{4}{a}_{1/2}.$$

Also we have

$$(a_{1/2}{b}_{1/2},e,e)=-(a_{1/2}e,e,b_{1/2})-(b_{1/2}e,a_{1/2})=-\frac{1}{2}(a_{1/2},e,b_{1/2})-\frac{1}{2}(b_{1/2},e,a_{1/2})=0,$$

which deduces that $({\mathcal{J}}_{1/2}{)}^2\cap {\mathcal{J}}_{1/2}=0$ and so ${\mathcal{J}}_{1/2}^2\subseteq {\mathcal{J}}_0+{\mathcal{J}}_1$. \end{proof}

Proposition

Let $A$ be a finite dimensional non nil power-associate algebra. Then such algebra contains idempotent.

\begin{proof} For non-idempotent $a\in A$, $\{a,a^2,\cdots ,a^n,\cdots \}$ is linear dependent by $A$ finite dimensional. Then there exists ${\alpha }_{k+1},\cdots ,{\alpha }_{k+m}$ such that

$$a^k+{\alpha }_{k+1}{a}^{k+1}+\cdots +{\alpha }_{k+m}{a}^{k+m}=0.$$

Hence $a^k=-{\alpha }_{k+1}{a}^{k+1}-\cdots -{\alpha }_{k+m}{a}^{k+m}=a^{k}g(a)$, where $g$ is a polynomial. Then $a^k=a^{k}g(a)=a^k(g(a){)}^k=a^k{g}_1(a)a^k$ with polynomial $g_1$. Thus, $e=a^k{g}_1(a)$ is a non-zero idempotent. \end{proof}

Lemma

If $\mathcal{J}$ is a Jordan algebra and $e$ is an idempotent in $\mathcal{J}$, then $U_e$ is a projection onto ${\mathcal{J}}_1$, i.e., ${\mathcal{J}}_1{U}_e={\mathcal{J}}_1$ and ${\mathcal{J}}_0\oplus {\mathcal{J}}_{1/2}\subseteq \ker U_e$.

\begin{proof} Let $a=b{U}_e\subset \mathcal{J}{U}_e$. There is $b{U}_e=2(be)e-be$, and

$$2((be)e)e=2(be)e+2(be,e,e)=2(be)e-(e,e,b)=2(be)e-eb+e(eb)=3(be)e-eb.$$

Now $ae=(b{U}_e)e=(2(be)e-be)e=2((be)e)e-(be)e=3(be)e-eb-(be)e=2(be)e-eb=b{U}_e=a$. Hence, $a\in {\mathcal{J}}_1$ and $\mathcal{J}{U}_e\subseteq {\mathcal{J}}_1$.

For any $a\in {\mathcal{J}}_1$, we have $a=ae$ and $a{U}_e=2(ae)e-ae=a\subseteq \mathcal{J}{U}_e$, which yields that ${\mathcal{J}}_1\subseteq \mathcal{J}{U}_e$. Obviously, ${\mathcal{J}}_0\subseteq \ker U_e$. For any $b\in {\mathcal{J}}_{1/2}$, we have $b{U}_e=2(be)e-be=0$ and so ${\mathcal{J}}_{1/2}\subseteq \ker U_e$. Now we finish the proof. \end{proof}

Lemma

Let $\mathcal{J}$ be finite-dimensional Jordan with $\mathrm{Nil}\mathcal{J}=0$, and let $e$ be an idempotent. Then $\mathrm{Nil}{\mathcal{J}}_1=0$.

\begin{proof} By ^e4d581, it suffices to show $\mathcal{Z}({\mathcal{J}}_1)=0$. We will show that ${\mathcal{J}}_1$ is non-generated. Take $a\in {\mathcal{J}}_1$ such that $a$ is an AZD in ${\mathcal{J}}_1$. Then ${\mathcal{J}}_1{U}_a=0$.

By ^o05l95, $a{U}_e=a$ and

$$\mathcal{J}{U}_a=\mathcal{J}{U}_{a{U}_e}=(\mathcal{J}{U}_e{U}_a)U_e\subseteq {\mathcal{J}}_1{U}_e=0.$$

Hence $a\in \mathcal{Z}(\mathcal{J})=0$ and so $\mathcal{Z}({\mathcal{J}}_1)=\mathrm{Nil}({\mathcal{J}}_1)=0$. \end{proof}

principal idempotent

An idempotent $e\in \mathcal{J}$ is called principal idempotent if there are no idempotent $f\in \mathcal{J}$ with $ef=0$.

Remark. Finite dimensional Jordan algebra idempotent $e$ is principal iff ${\mathcal{J}}_0(e)$ is nil. (Proof: $fe=0$ iff $f\in {\mathcal{J}}_0(e)$. $f\in {\mathcal{J}}_0(e)$ is idempotent iff ${\mathcal{J}}_0(e)$ is not nil by ^w804lu. )

Lemma

Let $\mathcal{J}$ be a finite-dimensional non-degenerate Jordan algebra. Then any principal idempotent in $\mathcal{J}$ is unity element in ${\mathcal{J}}_1$.

\begin{proof} Since $\mathcal{J}$ is non-degenerate, we have $\mathrm{Nil}(\mathcal{J})=0$ and so ${\mathcal{J}}_0(e)=\{0\}$. Thus the Pierce decomposition is $\mathcal{J}={\mathcal{J}}_{1/2}\oplus {\mathcal{J}}_1$.

Take $b\in {\mathcal{J}}_{1/2}$, and we claim that $b^2$ is an AZD of ${\mathcal{J}}_1$. For any $a\in {\mathcal{J}}_1$, we have $a{U}_b=2(ab)b-a{b}^2\in {\mathcal{J}}_1$ by ^ei364m. Hence,

$${\mathcal{J}}_1{U}_b={\mathcal{J}}_1{U}_b{U}_e=\mathcal{J}{U}_e{U}_b{U}_e=\mathcal{J}{U}_{b{U}_e}=0.$$

Then ${\mathcal{J}}_1{U}_b=0$ yields that ${\mathcal{J}}_1{U}_b{U}_b={\mathcal{J}}_1{U}_{b^2}=0$. Since $b^2\in {\mathcal{J}}_1$ by ^ei364m, we have $b^2{U}_e=b^2$ and

$${\mathcal{J}}_{1/2}{U}_{b^2}={\mathcal{J}}_{1/2}{U}_{b^2{U}_e}=({\mathcal{J}}_{1/2}{U}_e)U_{b^2}{U}_e=0$$

by ^o05l95. As $\mathcal{J}={\mathcal{J}}_{1/2}\oplus {\mathcal{J}}_1$, we know $b^2$ is an AZD of $\mathcal{J}$. Now we finish the proof the claim.

For any $b_1,b_2\in {\mathcal{J}}_{1/2}$, since $b^2$ is AZD and $\mathcal{J}$ is non-degenerated, we know $2b_1{b}_2=(b_1+b_2{)}^2-b_1^2-b_2^2=0$ and so ${\mathcal{J}}_{1/2}^2=0$. Hence ${\mathcal{J}}_{1/2}$ is nilpotent. Since $\mathcal{J}$ is non-degenerate, ${\mathcal{J}}_{1/2}=0$ and so $\mathcal{J}={\mathcal{J}}_1$. Therefore, $e$ is unity element in ${\mathcal{J}}_1$. \end{proof}

Corollary

Each finite-dimensional non-degenerate Jordan algebra has unity element.

\begin{proof} Since $\mathcal{J}$ is finite-dimensional and non-degenerate, we have $\mathrm{Nil}(\mathcal{J})=0$. By ^w804lu, there exists an idempotent $e\in \mathcal{J}$. If $e$ is not principal, then there exists idempotent $f$ in ${\mathcal{J}}_0(e)$. Set $e_1=e+f$, then $e_1^2=e^2+f^2=e+f=e_1$ and so $e_1$ is an idempotent. Hence ${\mathcal{J}}_0(e_1)⊊{\mathcal{J}}_0(e)$.

We can repeat this procedure if $e_1$ is not principle. However, $\mathcal{J}$ is finite-dimensional, so the procedure terminate. Therefore, there exists principal idempotent $z$ and by ^xxn8qm we finish the proof. \end{proof}

Corollary

Each finite-dimensional Jordan algebra $\mathcal{J}$ with $\mathrm{Nil}\mathcal{J}=0$ has unity element and can be written as direct sum of simple Jordan algebras.

\begin{proof} Since each finite-dimensional simple Jordan algebra is non-degenerated, by ^mprh1v every simple Jordan algebra contains unity element. Then by ^9e5yge, $\mathcal{J}$ can be written as direct sum of simple Jordan algebras. \end{proof}

Describe Simple Finite-dimensional Jordan Algebras

Definition

An idempotent $e$ is called primitive if it can not be decomposed as $e=e_1+e_2$, where $e_i$ are orthogonal idempotents.

An idempotent $e$ is called absolutely primitive if any $a\in {\mathcal{J}}_1(e)$ can be written as $a=\alpha e+n$, where $\alpha \in F$ and $n\in \mathcal{J}$ is nilpotent.

Lemma

Any absolutely primitive idempotent is primitive.

\begin{proof} Assume that $e$ is not primitive and $e=f+(e-f)$ where $f,e-f$ are idempotents. Then $ef=f$ and so $f\in {\mathcal{J}}_1(e)$. Since $e$ is absolutely primitive, $f$ can be written as $f=\alpha e+n$ with $\alpha \ne 0$, then $fe=\alpha e+ne=\alpha e+n$ and

$$\alpha e+n=f=f^2={\alpha }^{2}e+2\alpha ne+n^2={\alpha }^{2}e+2\alpha n+n^2.$$

Hence $(\alpha -{\alpha }^2)e=(2\alpha -1)n+n^2\in \mathrm{alg}⟨n⟩$ is nilpotent. So $\alpha -{\alpha }^2=0$ and $\alpha =1$. It follows that $n^2=(1-2\alpha )n=-n$. Since $n$ is nilpotent, $n=0$. Thus $f=e$ and so $e$ is primitive. \end{proof}

Lemma

If $F$ is algebraically closed field and $\mathcal{J}$ is finite-dimensional Jordan algebra, then any primitive idempotent is absolutely primitive.

\begin{proof} Let $e$ be a primitive idempotent. For any $a\in {\mathcal{J}}_1(e)$, as $\mathcal{J}$ is finite-dimensional, there exists polynomial $m_a(x)$ such that $m_a(a)=0$ with minimal degree.

We will prove that $m_a(x)$ is a power of irreducible polynomial. Suppose it is not true. Then $m_a(x)=f_1(x)f_2(x)$ such that $(f_1,f_2)=1$. By Bezout theorem, there exist $h_1$ and $h_2$ such that $f_1{h}_1+f_2{h}_2=1(\ast )$, $\mathrm{deg}{h}_1<\mathrm{deg}{f}_2$ and $\mathrm{deg}{h}_2<\mathrm{deg}{f}_1$. Define

$$\begin{array}{rl}{\phi }_a:F[x]& \to {\mathrm{alg}}_{\mathcal{J}}⟨e,a⟩\subseteq {\mathcal{J}}_1(e)\\ x& \mapsto a,1\mapsto e.\end{array}$$

Applying ${\phi }_a$ to $(\ast )$, we have $f_1(a)h_1(a)+f_2(a)h_2(a)=e$ $(\ast \ast )$, and put $e_1=f_1(a)h_1(a)$, $e_2=f_2(a)h_2(a)$. Note that $e_1{e}_2=f_1(a)f_2(a)h_1(a)h_2(a)=0$. Then we obtain $e_1=e_{1}e=e_1(e_1+e_2)=e_1^2$. Since $\mathrm{deg}{f}_i,\mathrm{deg}{h}_i<\mathrm{deg}{m}_a$, we can assume that $e_i\ne 0$. So $e_i$ are idempotent, leading to a contradiction. Therefore, $m_a(x)=(\mathrm{irr}x{)}^k$ for some integer $k$.

Since $F$ is algebraically closed field, we know $m_a(x)=(x-\alpha {)}^k$ and so

$$0=m_a(a)={\phi }_a(m_a(x))=(a-\alpha e{)}^k.$$

Thus $n=a-\alpha e$ is nilpotent. Now $a$ can be written as $a=\alpha e+n$ where $\alpha \in F$ and $n$ is nilpotent for any $a\in {\mathcal{J}}_1(e)$. Therefore, $e$ is primitive idempotent. \end{proof}

Full Peirce decomposition

Theorem

Let $1=e_1+\cdots +e_n$ with primitive idempotents $e_i$. Then

$$J=\underset{i=1}{\overset{n}{⨁}}{J}_{ii}\oplus \left(\underset{1⩽i<j⩽n}{⨁}{J}_{ij}\right),$$

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

\begin{proof} See here. \end{proof}