Exercise 1

1. Let $A$ be an associative algebra. Consider $B=A\oplus A^{op}$, where $A^{op}=\left(A,a^{\cdot op}b:=ba\right)$, an algebra with an involution $(a,b{)}^{\ast }=(b,a)$. Show that the Jordan algebras are isomorphic

$$H(B,\ast )\simeq A^{+}$$

\begin{proof} Note that $H(B,\ast )=\{(a,a):a\in A\}\subseteq A\oplus A^{\mathrm{op}}$. Define $\phi :H(B,\ast )\mapsto A^{+},(a,a)\mapsto a$. It is easy to check $\phi$ is bijective, so it remains to check $\phi$ is a homomorphism. For any $a,b\in A$, we have

$$(a,a)\circ (b,b)=\left(\frac{1}{2}(ab+ba),\frac{1}{2}(a{\cdot }_{\mathrm{op}}b+b{\cdot }_{\mathrm{op}}a)\right)=\left(\frac{1}{2}(ab+ba),\frac{1}{2}(ab+ba)\right).$$

It deduces that $\phi ((a,a)\circ (b,b))=\phi (a)\circ \phi (b)$. Therefore, $H(B,\ast )\simeq A^{+}$. \end{proof}

2. Let $A$ be an associative algebra with involution $\ast$. If $(A,\ast )$ is $\ast$-simple algebra (i.e $A\ne 0$ and there are no non-trivial ideal $I$ in $A$, with $I^{\ast }=I)$, then $H(A,\ast )$ is simple.

\begin{proof}

3. Let $J(V,f)$ be a Jordan algebra of Clifford type. Show that

• a) if $\dim V>1$ and $f$ is non-degenerated then $J(V,f)$ is simple algebra;
• b) $J(V,f)$ is special (Hint: use embedding into the Clifford algebra of $f$).

\begin{proof} a) Assume that $I$ is a non-trivial ideal of $J(V,f)$. If there exists nonzero $v\in V$ such that $v\in I$, then by $f$ non-degenerated there exists $u\in V$ such that $f(u,v)\ne 0$ and so $u\circ v=f(u,v)\in I$ is a nonzero element in $F$. Since $F$ is a field, $f(u,v)\in I$ yields $1\in I$ and so $I=J(V,f)$. If there exists nonzero $\alpha \in F$ such that $\alpha \in I$, then $\alpha \circ {\alpha }^{-1}=1\in I$ and so $I=J(V,f)$.

Take $v+\alpha \in I$, and take $u\in V$ such that $f(u,v)\ne 0$ and $u,v$ linearly independent. (Existence of $u$: if $f(v,v)=0$, then by $f$ non-degenerate $u$ exists; if $f(v,v)\ne 0$ and $f(w,v)=0$, then $v+w$ is what we desired.) Then $(v+\alpha )\circ u=\alpha u+f(u,v)\in I$ and so $\frac{{\alpha }^2}{f(u,v)}u+\alpha \in I$. It deduces that $v-\frac{{\alpha }^2}{f(u,v)}u\in I$ is a non-zero element in $V$ and so $I=J(V,f)$. Therefore, $J(V,f)$ is simple.

b) It suffices to show $J(V,f)\hookrightarrow {\mathrm{Clif}}^{+}(f)$. Recall that $\mathrm{Cliff}(V,f)=C(V,f)$ is an associative algebra with unit element generated by $V$ satisfying relation $v^2=f(v,v)\cdot 1_C$. Then $(u+v{)}^2=f(u+v,u+v)\cdot 1_C$ yields $uv+vu=2f(u,v)\cdot 1_C$.
Define $V\oplus F\cdot 1_C$ as a subalgebra of $C(V,f)$, and define

$$\phi :J(V,f)\mapsto C(V,f{)}^{+},v+\alpha \mapsto v+\alpha \cdot 1_C.$$

Note that $\phi$ is a homomorphism as Jordan algebra because

$$\begin{array}{rl}\phi (v+\alpha )\circ \phi (w+\beta )& =\frac{1}{2}((v+\alpha )(w+\beta )+(w+\beta )(v+\alpha ))\\ & =f(v,w)+\alpha w+\beta v+\alpha \beta =\phi ((v+\alpha )(w+\beta ))\end{array}$$

and $\phi$ is a injective. Hence $J(V,f)\hookrightarrow {\mathrm{Clif}}^{+}(f)$ and so $J(V,f)$ is special. \end{proof}

4. An algebra $A$ is called power associative if $\mathrm{alg}⟨a⟩$ is associative algebra for any $a\in A$. Show that any Jordan algebra is power associative algebra.

\begin{proof} By ^jmr4zb, $\mathrm{Mult}(\mathrm{alg}⟨a⟩)$ is commutative. Hence, for any elements $x,y\in \mathrm{alg}⟨a⟩$, there is $R_x{R}_y=R_y{R}_x$, which deduces that $(z\circ x)\circ y=(z\circ y)\circ x$ for any $z\in \mathrm{alg}⟨a⟩$. Since Jordan algebra is commutative, we have

$$(x\circ z)\circ y=(z\circ x)\circ y=(z\circ y)\circ x=x\circ (z\circ y)$$

and so $\mathrm{alg}⟨a⟩$ is associative. Therefore, any Jordan algebra is power associative algebra.

If $\mathcal{J}=\mathrm{alg}⟨a⟩$, then $\mathrm{Mult}(\mathcal{J})$ is commutative associative algebra.

Link to original

\end{proof}

5. Give an example of an associative non-commutative algebra $A$ such that $A^{+}$ is associative as well.

\begin{proof} Assume that $A$ is an algebra satisfying condition above. Then for any $a,b,c\in A$, there is $(a\circ b)\circ c=a\circ (b\circ c)$, which deduces that

$$bac-bca+cab-acb=[b,[a,c]]=0.$$

Therefore, we have that $[A,A]\subseteq Z(A)$, where $Z(A)$ is the center of $A$.

Let $A=\Lambda ({\mathbb{R}}^2)=\mathrm{span}\{1,e_1,e_2,e_1{e}_2\}$ where $e_1^2=e_2^2=0$ and $e_1{e}_2=-e_2{e}_1$. Notice that $\mathrm{span}\{1,e_1{e}_2\}=Z(A)$ and $[A,A]=\mathrm{span}\{[e_1,e_2]\}=\mathrm{span}\{2e_1{e}_2\}$, then $[A,A]\subseteq Z(A)$ and so $A^{+}$ is associative. Furthermore, as $e_1{e}_2\ne e_2{e}_1$, $A$ is non-commutative. Hence, $A$ is what we desired. \end{proof}

6. Let $A$ be an associative algebra such that $A^{+}$is associative as well. Prove that $A$ is then PI algebra (algebra whose elements satisfy non-trivial polynomial identity). Determine which identity is it.

\begin{proof} By Exercise 5, for any $a,b,c\in A$, the following identity holds

$$bac-bca+cab-acb=[b,[a,c]]=0.$$

Therefore, $A$ is a PI algebra and the non-trivial polynomial identity is $f(a,b,c)=bac-bca+cab-acb$. \end{proof}

7. Let $A$ be commutative associative algebra with a derivation $D$ satisfying $D^3=0$. Define a new multiplication in $A$

$$a\star b=D(a)D(b)$$

Show that $(A,\star )$ is a Jordan algebra.

\begin{proof} To show $(A,\star )$ is a Jordan algebra, it suffices to check $a\star b=b\star a$ and $a^2\star (b\star a)=(a^2\star b)\star a$. Since $A$ is commutative, we have $D(a)D(b)=D(b)D(a)$ and then $a\star b=b\star a$. Note that

$$a^2\star (b\star a)=D(D(a{)}^2)D(D(b)D(a))=2D^2(a)D(a)D(D(a)D(b)).$$

and

$$\begin{array}{rl}(a^2\star b)\star a& =D(2D^2(a)D(a)D(b))D(a)\\ & =2(D^3(a)D(a)D(b)+D^2(a)D(D(a)D(b)))D(a)\\ & =2D^2(a)D(D(a)D(b))D(a).\end{array}$$

Therefore, $a^2\star (b\star a)=(a^2\star b)\star a$ and so $(A,\star )$ is a Jordan algebra. \end{proof}

8. For any Jordan algebra $J$ show that there exists a Jordan algebra $B\supseteq J$ such that the universal multiplicative envelope $\mathcal{U}(J)\simeq {\mathrm{Mult}}_B(J).$

\begin{proof} Define $\mathcal{M}=\mathcal{U}(J)$, then $\mathcal{M}$ is a associative algebra. For any $j\in J$ and $m\in \mathcal{M}$, define $j\circ m=m\circ j=\phi (j)\cdot m$, where $\phi :J\to \mathcal{U}(J)$ is the natural map. Now $\mathcal{M}$ is a Jordan bimodule. Define $B=J+\mathcal{M}$. For any $b_1,b_2\in B$, assume that $b_1=(j_1,m_1),b_2=(j_2,m_2)$, and define

$$b_1\circ b_2=(j_1,m_1)\circ (j_2,m_2):=(j_1\circ j_2,j_1\circ m_2+j_2\circ m_1).$$

For any $m_1,m_2\in \mathcal{M}$, there is $(0,m_1)\circ (0,m_2)=(0,0)$ in $B$ and then ${\mathcal{M}}^2=0$. In addition, $\mathcal{M}$ is a Jordan bimodule. So $B=J+\mathcal{M}$ is a Jordan algebra.

Recall that ${\mathrm{Mult}}_B(J)={\mathrm{alg}}_{\mathrm{End}B}⟨R_a^B:a\in J⟩$ and there is a natural map

$$\psi :\mathcal{U}(J)\to {\mathrm{Mult}}_B(J).$$

Take $u\in \ker \psi$, and set $u={\sum }_i{c}_i{a}_{i,1}\cdots a_{i,n_i}$ with $a\in J$. Then $\psi (u)((0,1_M))=(0,{\sum }_i{c}_i{a}_{i,1}\cdots a_{i,n_i})=(0,u)=0$. Hence $u=0$ and so $\psi$ is injective. It is trivial that $\psi$ is surjective. Therefore, $\mathcal{U}(J)\cong {\mathrm{Mult}}_B(J)$. \end{proof}

9. Let $J$ be a Jordan algebra. Show that if $\mathrm{Mult}(J{)}^n=0$ then $J^{2n-1}=0$.

\begin{proof} We aim to show each $x\in J^n$ can be written as $x=a{R}_1\cdots R_s$ where $s⩾n/2$. When $n=1$, it is trivial. Assume that it holds for all $k<n$. To show the case of $n$, it suffices to show for any $a\in J^i$ and $b\in J^j$ with $i+j=n$, $a\circ b$ can be written as $a\circ b=z{R}_1\cdots R_t$ and $t⩾n/2$. Since $a=a_0{R}_{a1}\cdots R_{as}$ and $b=b_0{R}_{b1}\cdots R_{br}$ with $s⩾i/2$ and $r⩾j/2$, there is

$$a\circ b=a_0{R}_{a1}\cdots R_{as}{R}_b.$$

It remains to show $R_b=R_{b1}^{\prime }\cdots R_{br}^{\prime }$ with $r⩾j/2$.

We claim that if $y\in J^k$, then $R_y=R_{y_1}\cdots R_{y_m}$ with $m⩾k/2$. When $k=1$, it is trivial. Assume that the claim holds for all $i<k$. To show the case of $k$, it suffices to show for any $a\in J^i$ and $b\in J^k$ with $i+j=k$, $R_{a\circ b}=R_{z1}\cdots R_{zn}$ with $n>k/2$. Note that

then we have

$$[R_{dc},R_a]+[R_{ad},R_c]+[R_{ac},R_d]=0.$$

It follows that $[R_{ab},R_z]=-[R_{az},R_b]-[R_{bz},R_a]$ with $z\in J^1$. Then $R_{ab}{R}_z-R_z{R}_{ab}=R_1\cdots R_{m^{\prime }}$ with $m^{\prime }⩾(i+j+1)/2=(k+1)/2$.

？？？？

\end{proof}

10. Let $A$ be a finite-dimensional algebra. Show that $A$ contains solvable ideal $\mathrm{Solv}(A)$, which in turn contains all solvable ideals in $A$, moreover $\mathrm{Solv}(A/\mathrm{Solv}A)=0$.

\begin{proof} Firstly, we claim that for two solvable ideals $I,J$ in $A$, $I+J$ is also solvable. Since $(I+J)/I\cong J/I\cap J$ is a quotient of a solvable algebra, we know $(I+J)/I$ is also solvable. Hence, there exists $r$ such that $(I+J{)}^{⟨r⟩}\subseteq I$. Furthermore, there exists $s$ such that $I^{⟨s⟩}=0$ by $I$ solvable, and so $(I+J{)}^{⟨r+s⟩}=0$. Therefore, the sum $I+J$ is solvable.

Let $S$ be the sum of all solvable ideals $I$ in $A$. Since $A$ is finite-dimensional, $A$ satisfies ACC on ideals and so $S$ is a finite sum of solvable ideals. Hence $S={\sum }_{i=1}^k{I}_i$. By induction on the claim above, $S$ is solvable and so $S=\mathrm{Solv}(A)$.

Let $\overline{A}=A/\mathrm{Solv}A$. If ${\overline{I}}^{\prime }$ is a nontrivial solvable ideal of $\overline{A}$, then there exists $I^{\prime }⊋\mathrm{Solv}A$ such that $I^{\prime }/\mathrm{Solv}A$ is solvable. It follows that $I^{\prime }$ is a solvable ideal in $A$ with $I^{\prime }⊋\mathrm{Solv}A$, which is impossible. Hence, $\mathrm{Solv}(A/\mathrm{Solv}A)=0$. \end{proof}

11. Give an example of Jordan algebra $J$ and its ideal $I$ such that $I^2$ is not an ideal of $J$.

\begin{proof} 构造不出来

12. Prove that in any commutative or skew-commutative algebra $A$ the following identities are satisfied for any $a,b,c\in A$

$$(a,b,a)=0(a,b,c)=-(c,b,a).$$

\begin{proof} If $A$ is commutative, then

$$\begin{array}{rl}& (a,b,a)=(ab)a-a(ba)=a(ab)-a(ab)=0\\ & (a,b,c)=(ab)c-a(bc)=c(ba)-(cb)a=-(c,b,a).\end{array}$$

If $A$ is skew-commutative, then

$$\begin{array}{rl}& (a,b,a)=(ab)a-a(ba)=-a(ab)+a(ab)=0\\ & (a,b,c)=(ab)c-a(bc)=-c(ab)+(bc)a=c(ba)-(cb)a=-(c,b,a).\end{array}$$

Now we finish the proof. \end{proof}

13. Show that for any finite-dimensional Jordan algebra $J$ its nil-radical is hereditary, namely for any ideal $I$ of $J$ we have $\mathrm{Nil}I=I\cap \mathrm{Nil}J$.

\begin{proof} Since $\mathrm{Nil}(J/\mathrm{Nil}J)=0$, by ^j6tw1g we know $\mathrm{Nil}(I/I\cap \mathrm{Nil}J)=0$.
As $\mathrm{Nil}I$ is the minimal ideal such that $I/\mathrm{Nil}I$ being semisimple, there is $\mathrm{Nil}I\subseteq I\cap \mathrm{Nil}J$. On the other hand, note that $I\cap \mathrm{Nil}(J)$ is nil, which deduces that $\mathrm{Nil}I\supseteq I\cap \mathrm{Nil}J$. Therefore, we have $\mathrm{Nil}I=I\cap \mathrm{Nil}J$. \end{proof}

14. Verify that in Jordan algebra $J$ we have $U_{a,b}=R_a{R}_b+R_b{R}_a-R_{ab},a,b\in J$.

\begin{proof} Note that

$$x{U}_{a,b}=\frac{1}{2}((a+b)x(a+b)-axa-bxb)=\frac{1}{2}(axb+bxa)$$

and

$$x(R_a{R}_b+R_b{R}_a-R_{ab})=\frac{1}{2}(axb+bxa).$$

Then by Shirshov-Cohn we finish the proof.

Alternating proof. Note that

$$U_{a,b}=(R_{a+b}^2-R_a^2-R_b^2)-\frac{1}{2}(R_{(a+b{)}^2}-R_{a^2}-R_{b^2}).$$

Since $R_{a+b}=R_a+R_b$, we have $R_{a+b}^2=(R_a+R_b{)}^2=R_a^2+R_a{R}_b+R_b{R}_a+R_b^2$. Since $(a+b{)}^2=(a+b)\circ (a+b)=a\circ a+2a\circ b+b\circ b$, we have

$$R_{(a+b{)}^2}=R_{a^2}+2R_{a\circ b}+R_{b^2}=R_{a^2}+2R_{ab}+R_{b^2}.$$

Therefore, $U_{a,b}=R_a{R}_b+R_b{R}_a-R_{ab}$. \end{proof}

15. Show that for special Jordan algebra $\left(A^{+},a\cdot b=\frac{1}{2}(ab+ba)\right)$, $\{abc\}=\frac{1}{2}(abc+cba)$.

\begin{proof} It can be directly verified by Exercise 14 and $\{axb\}=x{U}_{a,b}$. \end{proof}

16. The full Peirce Decomposition: Let $J$ be Jordan algebra with unity element 1. Let $e_1,\dots ,e_n$ be non-zero, pairwise orthogonal idempotents satisfying $e_1+\cdots +e_n=1$. Prove that

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

where $J_{ii}=J_1\left(e_i\right)$ and $J_{ij}=J_{\frac{1}{2}}\left(e_i\right)\cap J_{\frac{1}{2}}\left(e_j\right)$ for $1\le i<j\le n$.

\begin{proof} For any idempotent $e$, one can apply $a=b=c=e$ to $(ac,b,d)+(ad,b,c)+(cd,b,a)=0$, and then 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 ${\mathcal{J}}_i(e)=\{a\in \mathcal{J}:ea=ia\}$.

For each idempotent $e_i$, there is a composition $\mathcal{J}={\mathcal{J}}_0(e_i)\oplus {\mathcal{J}}_{1/2}(e_i)\oplus {\mathcal{J}}_1(e_i)$. Since $e_1,\cdots ,e_n$ are pairwise orthogonal, there is $[R_{e_i},R_{e_j}]=0$ and so $\mathcal{J}$ can be decomposed as

$$\mathcal{J}=V_1\oplus \cdots \oplus V_m$$

where $V_i$ is a eigenvector spaces of $R_{e_1},\cdots ,R_{e_n}$ and $m=3^n$. Take $a\in V_j$ with $a{e}_i={\lambda }_{i}a$, then

$$a=a(e_1+\cdots +e_n)=\sum _{i=1}^{n}a{e}_i=\left(\sum _{i=1}^n{\lambda }_i\right)a,$$

where ${\lambda }_i\in \{0,1/2,1\}$. It follows that either $a{e}_t=a$ for some $s$, or $a=a{e}_s+a{e}_r=1/2a+1/2a$. Hence $V_j={\mathcal{J}}_1(e_t)$ or ${\mathcal{J}}_{1/2}(e_r)\cap {\mathcal{J}}_{1/2}(e_s)$. Now we finish the proof. \end{proof}

17. Show that Peirce decomposition components satisfy the following relations:

$$J_{ii}^2\subseteq J_{ii},J_{ij}^2\subseteq J_{ii}+J_{jj},J_{ii}{J}_{ij}\subseteq J_{ij},J_{ii}{J}_{jj}=0i\ne j,$$ $$J_{ij}{J}_{jk}\subseteq J_{ik}(i,j,k\text{ distinct}),J_{ij}{J}_{kl}=J_{ii}{J}_{kl}(i,j,k,l\text{ distinct}).$$

\begin{proof} Take $a\in {\mathcal{J}}_0\cup {\mathcal{J}}_1$ and $x\in \mathcal{J}$, then we have

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

by linearization. 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 $a\in {\mathcal{J}}_0\cup {\mathcal{J}}_1$.

To show $J_{ii}^2\subseteq J_{ii}$, for any $a,b\in J_{ii}=J_1(e_i)$, we have $e_{i}a=a$ and $e_{i}b=b$. Then $e(ab)=(ab)e=a(be)+(a,b,e)=ab+(a,b,e)=ab$ and so $J_{ii}^2\subseteq J_{ii}$.

To show $J_{ij}^2\subseteq J_{ii}+J_{jj}$, since $J_{1/2}^2(e_i)\subseteq J_0(e_i)+J_1(e_i)$ (see ^ei364m), we have

$$\begin{array}{rl}{J}_{ij}^2& \subseteq (J_{1/2}(e_i{)}^2\cap J_{1/2}(e_j{)}^2)\subseteq (J_0(e_i)+J_1(e_i))\cap (J_0(e_j)\cap J_1(e_j))\\ & \subseteq (J_0(e_i)\cap J_1(e_j))\oplus (J_0(e_j)\cap J_1(e_i))\subseteq J_{ii}+J_{jj}.\end{array}$$

To show $J_{ii}{J}_{ij}\subseteq J_{ij}$, by ^ei364m we know $J_1(e_i)(J_{1/2}(e_i))\subseteq J_{1/2}(e_i)$ and $J_0(e_i)(J_{1/2}(e_i))\subseteq J_{1/2}(e_i)$. Then $J_{ii}{J}_{ij}\subseteq J_{1/2}(e_i)\cap J_{1/2}(e_j)=J_{ij}$.

To show $J_{ii}{J}_{jj}=0$, take $a\in J_{ii}$ and $b\in J_{jj}$. Since $a\in J_1(e_i)$, $a\in J_0(e_j)$. By ^ei364m, we have $J_1{J}_0=0$ and so $ab\in J_0(e_j)J_1(e_j)=0$.

To show $J_{ij}{J}_{jk}\subseteq J_{ik}$, it suffices to show $ab\in J_{1/2}(e_i)$ and $ab\in J_{1/2}(e_j)$ for any $a\in J_{ij}$ and $b\in J_{jk}$. Note that $a\in J_{ij}$ yields $a\in J_0(e_k)$. Then by ^ei364m, $a\in J_0(e_k)$ and $b\in J_{1/2}(e_k)$ imply that $ab\in J_{1/2}(e_k)$. Similarly, $b\in J_0(e_i)$ yields $ab\in J_{1/2}(e_i)$. Therefore, $ab\in J_{ik}$ and we finish the proof.

To show $J_{ij}{J}_{kl}=J_{ii}{J}_{kl}$, note that $J_{kl}\subseteq J_0(e_i)$ and so RHS is zero. So it suffices to show $J_{ij}{J}_{kl}=0$. Since $a\in J_0(e_k)\cap J_0(e_l)$ and $b\in J_0(e_i)\cap J_0(e_j)$, by ^ei364m we know $ab\in J_{1/2}(e_i)\cap J_{1/2}(e_j)\cap J_{1/2}(e_k)\cap J_{1/2}(e_l)=0$. Hence LHS and RHS are both zero and we finish the proof. \end{proof}

18. Suppose that $J$ has the full Peirce decomposition. Show that $J_{ij}^2{e}_i$ is an ideal in $J_{ii}$.

\begin{proof} In Exercise 17, we have proved that $(e_i,\mathcal{J},a)=0$, where $a\in {\mathcal{J}}_0(e_i)\cup {\mathcal{J}}_1(e_i)$.

Since $J_{ij}^2\subseteq J_i+J_j$ by Exercise 17, $J_{ij}^2{e}_i\subseteq J_{ii}$. For any $x\in J_{ii}$ and $y=a\circ b\in J_{ij}^2$ with $a,b\in J_{ij}$, we have

$$\begin{array}{rl}x\circ (y\circ e_i)& =(x\circ y)\circ e_i-(x,y,e_i)=(x\circ y)\circ e_i+(e_i,y,x)=(x\circ y)\circ e_i\\ & =((x\circ a)\circ b-(x,a,b))\circ e_i=((x\circ a)\circ b)\circ e_i-(x,a,b)\circ e_i.\end{array}$$

Since $(x\circ a)\circ b\in J_{ij}^2$, it suffices to show $(x,a,b)\circ e_i\in J_{ij}^2{e}_i$.

Recall that in ^pgknr0 there is $(d\circ c,b,a)+(a\circ c,b,d)+(a\circ d,b,c)=0$, then we have

$$(x,a,b)=-2(e_{j}b,a,x)=2(e_{j}x,a,b)+2(bx,a,e_j)=2(bx,a,e_j)$$

because $J_{ii}{J}_{jj}=0$. Note that

$$(bx,a,e_j)=(bx\circ a)\circ e_j-bx\circ (a\circ e_j)=(bx\circ a)\circ e_j-\frac{1}{2}bx\circ a.$$

Since $(bx\circ a)\circ e_j\in (J_{ii}+J_{jj})\circ e_j\subseteq J_{jj}$ by Exercise 17 and $bx\circ a\in J_{ij}^2$, there is

$$(x,a,b)\circ e_i=2((bx\circ a)\circ e_j)\circ e_i-(bx\circ a)\circ e_i=-(bx\circ a)\circ e_i\in J_{ij}^2{e}_i$$

and so we finish the proof. \end{proof}

**19. Consider polynomial algebra $F\left[x_1,\dots ,x_n\right]$ with a derivations $D(f)={\sum }_{i=1}^n\frac{\partial f}{\partial x_i}$ and its subspace $J=\{f:f\text{ homogeneous},\mathrm{deg}f=2\}$.

• Prove that $J$ is Jordan algebra with respect to the product $f\star g=D(f)D(g)$.
• Prove that $J$ is finite-dimensional and simple.
• Which simple finite-dimensional Jordan algebra from classification theorem $(J,\star )$ corresponds to?

\begin{proof} i) Note that $f\star g\in J$ and $f\star g=g\star f$ for any $f,g\in J$, so it suffices to show $(f^2,g,f)=0$. Since $f\in J$, $\mathrm{deg}f=2$ and then $D^3(f)=0$. Hence, we can compute that

$$\begin{array}{rl}(f^2,g,f)& =(f^2\star g)\star f-f^2\star (g\star f)\\ & =D(D((D(f){)}^2)D(g))D(f)-D((D(f){)}^2)D(D(f)D(g))\\ & =D(2D^2(f)D(f)D(g))D(f)-2D^2(f)D(f)D(D(f)D(g))\\ & =2D^2(f)D(D(f)D(g))D(f)-2D^2(f)D(f)D(D(f)D(g))\\ & =0.\end{array}$$

Therefore, $J$ is a Jordan algebra.

ii) Define $I=\{f:D^2(f)=0\}$. Then $x_1^2-x_2^2\in I$ and $x_1^2\notin I$, so $J$ is not simple. \end{proof}

20. Prove that simple Jordan algebra $J(V,f)$ over an algebraically closed field is an algebra of capacity 2.

\begin{proof} If $v+\alpha$ is an idempotent of $J(V,f)$, then

$$v+\alpha =(v+\alpha )\circ (v+\alpha )=2\alpha v+f(v,v)+{\alpha }^2$$

and so $f(v,v)=1/4$, $\alpha =1/2$.

Take $v$ such that $f(v,v)=1/4$ ( the existence of $v$ is guaranteed by $F$ being algebraically closed field), then $e_1:=1/2+v$ and $e_2:=1/2-v$ are two idempotents in $J(V,f)$. Since $e_1{e}_2=0$, they are orthogonal. It remains to show $e_1$ and $e_2$ are primitive. Otherwise, $e_1$ can be written as $e_1=e_{11}+e_{12}$ where $e_{11}$ and $e_{12}$ are orthogonal idempotents. Then $e_{11}=v_{11}+1/2$ and $e_{12}=v_{12}+1/2$, where $f(v_{11},v_{11})=f(v_{12},v_{12})=1/4$. It deduces that $e_1=v_{11}+v_{12}+1$, which is impossible. So $e_1$ is primitive. Similarly, we can prove that $e_2$ is also primitive. Therefore, $J(V,f)$ is of capacity $2$. \end{proof}

21. Prove that $J(V,f)$ contains non-trivial idempotent if and only if there exists $x\in V$ with $f(x,x)=1$.

\begin{proof} If there exists $x\in V$ such that $f(x,x)=1$, then define $e=(x+1)/2$. One can verify that

$$e^2=\frac{1}{4}(f(x,x)+2x+1)=(x+1)/2=e$$

and so $e$ is a non-trivial idempotent.

Conversely, if $J(V,f)$ contains a non-trivial idempotent $e$. By Exercise 20, an idempotent $e$ can be written as $e=v+1/2$ with $f(v,v)=1/4$. Then $x:=2v$ is what we desired. \end{proof}

22. Prove that every commutative nilpotent algebra of dimension three is Jordan.

\begin{proof} Let $A$ be a commutative nilpotent algebra of dimension three. Note that $A\supset A^2\supset A^3\supset \cdots$ is strictly decreased and $\dim A=3$, then $A^4=0$. It yields that $(a^2,b,a)=(a^{2}b)a-a^2(ba)\in A^4$ is zero for any $a,b\in A$. Therefore, $A$ is Jordan. \end{proof}

23. An algebra $A$ is called locally nilpotent if every finitely generated subalgebra of $A$ is nilpotent. Prove that simple algebra is not locally nilpotent.

\begin{proof} Assume that $A$ is a locally nilpotent simple algebra. Take $0\ne a\in A$, then the ideal generated by $a$ equals $A$, i.e., there exist $x_i,y_i$ such that ${\sum }_{i=1}^n{x}_{i}a{y}_i=a$. Define $B=⟨x_i,y_i,a:1⩽i⩽n⟩$, then $B$ is nilpotent and there is $n$ such that $B^n=0$. Note that

$$a=\sum _{i=1}^n{x}_{i}a{y}_i=\sum _{i=1}^n{x}_i\left(\sum _{i=1}^n{x}_{i}a{y}_i\right)y_i=\cdots =\sum _{i=1}^n{x}_i\left(\sum _{i=1}^n\left(x_i\left(\sum _{i=1}^n{x}_{i}a{y}_i\right)y_i\right)\cdots \right)y_i\in B^n=0,$$

which is impossible. Therefore, a simple algebra is not locally nilpotent. \end{proof}

24. Prove that for octonion algebra $\mathbb{O},{\mathbb{O}}^{+}$is isomorphic to a Jordan algebra of a symmetric nondegenerate bilinear form.

\begin{proof} Define $\mathbb{O}=\{1,e_1,\cdots ,e_7\}$ with $e_i^2=-1$, $e_i{e}_{i+1}=e_{i+3}(\mathrm{mod}7)$ and $e_i{e}_j=-e_j{e}_i$.

Then for $e_i,e_j\in {\mathbb{O}}^{+}$, there is

$$e_i\circ e_j=\frac{1}{2}(e_i{e}_j+e_j{e}_i)=0$$

and $e_i\circ e_i=-1$.

Let $V=\mathrm{span}\{e_1,\cdots ,e_7\}$. If $f(e_i,e_j)=0$ for all $i\ne j$ and $f(e_i,e_i)=-1$, then $f$ is a symmetric non-degenerate bilinear form and one can verify that ${\mathbb{O}}^{+}\cong J(V,f)$. \end{proof}

25. Let $J$ be finite dimensional Jordan algebra over a field $F$ and $R_a:x\to xa$ be the operator of right multiplication in $J$. Show that

• (i) if $a$ is nilpotent $\mathrm{tr}\left(R_a\right)=0$,
• (ii) if $\mathrm{char}F=0$ and $e$ is an idempotent in $J$ then $\mathrm{tr}{R}_e\ne 0$.

\begin{proof} i) If $a$ is nilpotent, then $R_a^m=0$ for some integer $m$. Assume that the corresponding matrix of $R_a$ is $A$, then $A^m=0$ and so the characteristic polynomial of $A$ is $f(x)=x^t$ for some $t\mid m$. Hence, all eigenvalues of $A$ are $0$ and $\mathrm{tr}(R_a)=0$.

ii) Since $e$ is an idempotent in $J$, eigenvalues of $R_e$ are $\{0,1/2,1\}$. As all eigenvalues are non-negative and $e$ is an eigenvector with eigenvalue $1$, $\mathrm{tr}(R_e)\ne 0$. \end{proof}

26. Describe all Jordan algebras of dimension two up to an isomorphism.

\begin{proof} Let $J=\mathrm{span}\{e_1,e_2\}$ be a Jordan algebra of dimension $2$.

Assume that $J$ is nilpotent. Then using similar argument to that of Exercise 22, $J^3=J^4=0$ and so each commutative nilpotent algebra is Jordan. If $J^2=0$, then $e_1\circ e_1=e_1\circ e_2=e_2\circ e_2=0$. If $J^2\ne 0$, then $\dim J^2=1$. Since $J$ is nilpotent, it does not have idempotent. So $\{e_1,e_1\circ e_1\}$ is linear independent, WLOG $e_1\circ e_1=e_2$. It deduces that $e_1\circ e_2\in J^3=0$ and $e_2\circ e_2\in J^4=0$. Therefore, if $J=\mathrm{span}\{e_1,e_2\}$ is nilpotent, there are two Jordan algebras:

• $e_i\circ e_j=0$ for all $i,j\in \{1,2\}$;
• $e_1\circ e_1=e_2$, $e_1\circ e_2=e_2\circ e_2=0$.

Now assume that $J$ is not nilpotent. Then by ^w804lu and Exercise 4, there exists an idempotent $e\in J$. Take $f\in J$ such that $J=\mathrm{span}\{e,f\}$. Thus $J$ can be decomposed as $J=J_0(e)\oplus J_{1/2}(e)\oplus J_1(e)$. Take $f\in J$ such that $J=\mathrm{span}\{e,f\}$. Since $\dim J_1(e)⩾1$ and $\dim J=2$, there are three possibilities.

• $e\circ e=e$, $e\circ f=0$, $f\circ f\in J_0^2\subseteq J_0$ and so $f\circ f=0$ or $f$.
• $e\circ e=e$, $e\circ f=1/2f$, $f\circ f\in J_{1/2}^2\subseteq J_0+J_1=J_1$ and so $f\circ f=ae$. By Jordan identity, $a=0$.
• $e\circ e=e$, $e\circ f=f$, $f\circ f\in J_1$ and so $f\circ f=ae+bf$ for any $a,b\in F$. Define $f^{\prime }=f-b/2e$, then $f^{\prime }\circ f^{\prime }=\alpha e$. When $\alpha =0$, we get bace to the first case.

Therefore, there are $6$ possibilities up to isomorphism. \end{proof}