Exercise 2

1. Let $J$ be a Jordan algebra. Show that $J$ is special if and only if $\iota :J\to S(J)$ is injective.

\begin{proof} Let $S(J)$ denote the special universal envelope of $J$, and $\iota :J\to S(J)$ be the canonical Jordan homomorphism.

Firstly, assume that $J$ is special. Then there exists an associative algebra $A$ and an injective Jordan homomorphism $\varphi :J\to A^{(+)}$. By the universal property, there exists a unique associative algebra homomorphism $\psi :S(J)\to A$ such that $\varphi =\psi \circ \iota$. Since $\varphi$ is injective, we know $\iota$ is injective.

Now we assume that $\iota :J\to S(J)$ is injective. Note that $S(J{)}^{(+)}$ is a special Jordan algebra. Since $\iota$ is injective, we know $J$ is isomorphic to a subalgebra $\iota (J)\subseteq S(J{)}^{(+)}$ and so $J$ is special. \end{proof}

2. Show that $S(SJ⟨X⟩)=F⟨X⟩$.

\begin{proof} Let $F⟨X⟩$ be the free associative algebra generated by the set $X$, and let $SJ⟨X⟩$ be the free special Jordan algebra generated by $X$. We need to show that the special universal envelope of $SJ⟨X⟩$ is isomorphic to $F⟨X⟩$.

There is a natural inclusion map $j:SJ⟨X⟩\hookrightarrow F⟨X{⟩}^{(+)}$. By the universal property, it can be lifted to a unique associative algebra homomorphism $\Phi :S(SJ⟨X⟩)\to F⟨X⟩$.

On the other hand, consider the canonical map $\iota :SJ⟨X⟩\to S(SJ⟨X⟩)$. Since $X\subset SJ⟨X⟩$, we have a map $\iota :X\hookrightarrow SJ⟨X⟩\to S(SJ⟨X⟩)$. By the universal property of the free associative algebra $F⟨X⟩$, $\iota :X\to S(SJ⟨X⟩)$ can be lifted to $\Psi :F⟨X⟩\to S(SJ⟨X⟩)$.

Now it remains to check $\Phi \circ \Psi =\Psi \circ \Phi =\mathrm{id}$. Since $X$ generated $F⟨X⟩$, it is enough to check

• $\Phi (\Psi (x))=\Phi (\iota (x))=j(x)=x$, and
• $\Psi (\Phi (\iota (x)))=\Psi (x)=\iota (x)$.

Therefore, $S(SJ⟨X⟩)=F⟨X⟩$. \end{proof}

3. Let $J$ be an n-dimensional Jordan algebra. Show that $\dim S(J)\le 2^n$.

\begin{proof} Let $\{e_1,e_2,\dots ,e_n\}$ be a basis for the Jordan algebra $J$.

The special universal envelope $S(J)$ is generated as an associative algebra by these basis elements subject to the relations derived from the Jordan multiplication:

$$xy+yx=2(x\circ y)$$

where $x\circ y$ is the product in $J$. The relation can be rewritten as

$$yx=-xy+2(x\circ y).$$

Since $2(x\circ y)$ is a linear combination of basis elements in $J$, this relation allows us to reorder any product $e_j{e}_i$ (where $j>i$) into a term $e_i{e}_j$ plus terms of lower degree. Also notice that $2x^2=2(x\circ x)$ is a linear combination of basis elements (degree 1). By repeatedly applying the reordering relation and the power reduction relation, any element in $S(J)$ can be written as a linear combination of ordered monomials of the form

$$e_{i_1}{e}_{i_2}\dots e_{i_k}$$

where $1\le i_1<i_2<\cdots <i_k\le n$.

These monomials are in one-to-one correspondence with the subsets of the index set $\{1,\dots ,n\}$. Therefore, the dimension of the vector space $S(J)$ is at most $2^n$. \end{proof}

5. Let $F⟨X⟩$ be a free associative algebra generated by set $X$ and let $SJ⟨X⟩$ be the Jordan subalgebra of $F⟨X{⟩}^{(+)}$ generated by $X$. Show that $xyz+zyx$, $[[x,y],z]$ and $[x,y{]}^2$ are in $SJ⟨X⟩$.

\begin{proof} (i) Recall the standard identity for the Jordan triple product $\{a,b,c\}=(a\circ b)\circ c+(b\circ c)\circ a-(a\circ c)\circ b$. In a special Jordan algebra, this evaluates to $(abc+cba)/2$. Therefore

$$xyz+zyx=2\left((x\circ y)\circ z+(z\circ y)\circ x-(x\circ z)\circ y\right)\in SJ⟨X⟩.$$

(ii) Note that

$$[[x,y],z]=(xy-yx)z-z(xy-yx)=xyz-yxz-zxy+zyx=(xyz+zyx)-(yxz+zxy)$$

and so by (i), $[[x,y],z]\in SJ⟨X⟩$. ence is also in $SJ⟨X⟩$.

(iii) Note that

$$[x,y{]}^2=(xy-yx)(xy-yx)=xyxy-xyyx-yxxy+yxyx,$$

and

$$4(x\circ y{)}^2=(xy+yx{)}^2=xyxy+xyyx+yxxy+yxyx.$$

Then we have

$$4(x\circ y{)}^2-[x,y{]}^2=2(xyyx+yxxy).$$

Since $xyyx=x(y^2)x=U_x(y^2)$, we have $xyyx\in SJ⟨X⟩$ and so for $yxxy$. Now we finish the proof. \end{proof}

12. Let $J$ be a Jordan algebra with the Pierce decomposition $J={\oplus }_{i=1}^n{J}_{ii}+{\oplus }_{i<j}{J}_{ij}$. Prove that $J_{ij}^2{e}_i$ is an ideal of $J_{ii}$.

\begin{proof} Let $J_{ij}^2$ denote the subspace spanned by products $x\circ y$ with $x,y\in J_{ij}$. The expression $J_{ij}^2{e}_i$ denotes the set $\{(x\circ y)\circ e_i:x,y\in J_{ij}\}$. Since $e_i$ is the unit element for $J_{ii}$ and $e_i\circ J_{jj}=\{0\}$, the operation $u\mapsto u\circ e_i$ acts as the projection mapping $J_{ii}+J_{jj}\to J_{ii}$. Recall that $J_{ij}\circ J_{ij}\subseteq J_{ii}+J_{jj}$, then $K=J_{ij}^2{e}_i$ is the projection of $J_{ij}^2$ onto $J_{ii}$.

To show $K$ is an ideal of $J_{ii}$, it suffices to show that for any $a\in J_{ii}$ and any $k\in K$, there is $a\circ k\in K$. Assume that $k=(x\circ y)\circ e_i$ for some $x,y\in J_{ij}$, then we compute $a\circ k=a\circ ((x\circ y)\circ e_i)$. Note that

$$\begin{array}{rl}a\circ k& =a\circ (x\circ y{)}_{ii}\\ & =a\circ (x\circ y{)}_{ii}+0\\ & =a\circ (x\circ y{)}_{ii}+a\circ (x\circ y{)}_{jj}\\ & =a\circ ((x\circ y{)}_{ii}+(x\circ y{)}_{jj})\\ & =a\circ (x\circ y).\end{array}$$

Consider the element $S=(a\circ x)\circ y+(a\circ y)\circ x$. Since $a\in J_{ii}$ and $x,y\in J_{ij}$, we have $a\circ x\in J_{ij}$ and $a\circ y\in J_{ij}$ and so $S\in J_{ij}^2$.

Furthermore, as $\{J_{ij},J_{ii},J_{ij}\}\subseteq J_{jj}$, there is $S-a\circ (x\circ y)\in J_{jj}$ and so we have

$$a\circ (x\circ y)=S\circ e_i\in J_{ij}^2\circ e_i.$$

Now we finish the proof. \end{proof}