Homework 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}

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 $B=J\oplus \mathcal{M}$. 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 $B=J\oplus \mathcal{M}$ 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

$$\psi :\mathcal{U}(J)\to {\mathrm{Mult}}_B(J),a\mapsto R_a^B\text{ for all }a\in J.$$

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. Since ${\mathrm{Mult}}_B(J)$ is generated by $R_a^B$ for all $a\in J$, we know $\psi$ is surjective. Therefore, $\mathcal{U}(J)\cong {\mathrm{Mult}}_B(J)$. \end{proof}