2 Linearization of Jordan identities

Let $(\mathcal{J},\circ )$ be a Jordan algebra over a field $k$ with $\mathrm{char}k\ne 2$.

Recall that if $A$ is associative algebra, then $(A^{+},\circ )$ is Jordan, where $a\circ b:=ab+ba$. However, since $a^{\circ 2}=a^2+a^2\ne a^2$, instead of $a\circ b=b\circ a$ we may consider

$$a\cdot b=\frac{1}{2}(ab+ba).$$

Remark that $(A^{+},\circ )$ and $(A^{+},\cdot )$ are isomorphic as Jordan algebra, as we define $\phi :(A^{+},\circ )\to (A^{+},\cdot ),a\mapsto 2a$.

In the remaining parts, we define $a\circ b=(ab+ba)/2$.

Linearization of Jordan identities

Define $(a,b,c)=(a\circ b)\circ c-a\circ (b\circ c)$, then Jordan identity

$$a^2\circ (b\circ a)=(a^2\circ b)\circ a\text{\hspace{1em}(1)}$$

means that $(a^2,b,a)=0$.

Replace $a$ with $a+\lambda c$, then we have

$$0=((a+\lambda c{)}^2,b,a+\lambda c)=\lambda f(a,b,c)+{\lambda }^{2}g(a,b,c),$$

where $f(a,b,c)=2(a\circ c,b,a)+(a^2,b,c)$ and $g(a,b,c)=2(a\circ c,b,c)+(c^2,b,a)$ are both zero. Therefore, in Jordan algebras the following identity is satisfied

$$2(a\circ c,b,a)+(a^2,b,c)=0.\text{\hspace{1em}(2)}$$

Replace $a$ with $a+d$, then we have

$$2(a\circ c,b,a)+2(d\circ c,b,a)+2(a,b,d)+2(d\circ c,b,d)+(a^2,b,c)+2(a\circ d,b,c)+(d^2,b,c)=0.$$

By (2) we have $2(a\circ c,b,a)+(a^2,b,c)=0$ and $2(d\circ c,b,d)+(d^2,b,c)=0$, which deduces that

$$(d\circ c,b,a)+(a\circ c,b,d)+(a\circ d,b,c)=0.\text{\hspace{1em}(3)}$$

It is useful to write it in operator form $L_x:\mathcal{J}\to \mathcal{J}$ and $R_x:\mathcal{J}\to \mathcal{J}$, where $a{L}_x=xa$ and $a{R}_x=ax$.

• $\mathrm{End}(\mathcal{J})$ - associative algebra
• $R(\mathcal{J})$ is an associative algebra generated by $R_x$, $x\in \mathcal{J}$
• $L(\mathcal{J})$ is an associative algebra generated by $L_x$, $x\in \mathcal{J}$
• $\mathrm{Mult}(\mathcal{J})$ - multiplication algebra, is associative algebra generated by $L_x$, $R_x$.

Since in Jordan algebra $\mathcal{J}$ we have $L_x=R_x$, there is $\mathrm{Mult}(\mathcal{J})=L(\mathcal{J})=R(\mathcal{J})$.

In the remaining part, we write $dc$ instead of $d\circ c$. By ^pgknr0, we have

$$((dc)b)a-(dc)(ba)+((ad)b)c-(ad)(bc)+((ac)b)d-(ac)(bd)=0,$$

which deduces that

• $b$: $R_{dc}{R}_a+R_{ad}{R}_c+R_{ac}{R}_d=R_a{R}_{dc}+R_c{R}_{ad}+R_d{R}_{ac}$, i.e. $[R_{dc},R_a]+[R_{ad},R_c]+[R_{ac},R_d]=0$; and
• $d$: $R_a{R}_b{R}_c+R_c{R}_b{R}_a+R_{(a\circ c)\circ b}=R_a{R}_{bc}+R_b{R}_{ac}+R_c{R}_{ba}$ $(\ast \ast )$. $(\mathrm{IV})$

Remark. Using the method above, we can obtain followings.

• ^fgzlin is equivalent to $[R_{a^2},R_a]=0$. $(\mathrm{I})$
• ^mwyshj is equivalent to $[R_{a^2},R_c]+2[R_{ac},R_a]=0$. $(\mathrm{II})$
• ^pgknr0 is equivalent to $[R_{dc},R_a]+[R_{ad},R_c]+[R_{ac},R_d]=0$. $(\mathrm{III})$

Derivation

Definition

For any algebra $A$, we call linear operator $\mathcal{D}:A\to A$ a derivation if it satisfy

$$(ab)\mathcal{D}=(a\mathcal{D})b+a(b\mathcal{D}).$$

Lemma

For any algebra $A$ and linear opeator $\mathcal{D}:A\to A$, $\mathcal{D}$ is a derivation iff $[R_a,\mathcal{D}]=R_{a\mathcal{D}}$ for all $a\in A$.

\begin{proof} Note that $(ab)\mathcal{D}=(a\mathcal{D})b+a(b\mathcal{D})$ iff $R_a\mathcal{D}=R_{a\mathcal{D}}+\mathcal{D}{R}_a$ iff $[R_a,\mathcal{D}]=R_{a\mathcal{D}}$. \end{proof}

Notice that RHS of $(\ast \ast )$ is symmetric in $a$ and $b$. Exchange $a$ and $b$ and substitute $(\ast \ast )$, then we obtain

$$[R_a,R_b]R_c-R_c[R_a,R_b]=R_{(bc)a}-R_{(ac)b}.$$

that is, $R_{(a,c,b)}=[R_c,[R_a,R_b]]$. $(\mathrm{V})$

Corollary

For any Jordan algebra $\mathcal{J}$ and any $a,b\in \mathcal{J}$, $[R_a,R_b]$ is a derivation of $\mathcal{J}$.

\begin{proof} Since $[R_c,[R_a,R_b]]=R_{(a,c,b)}=R_{c[R_a,R_b]}$, by ^70d2f4 $[R_a,R_b]$ is a derivation. \end{proof}

Proposition

For any Jordan algebra $\mathcal{J}=⟨a_i:i\in I⟩$, there is

$$\mathrm{Mult}(\mathcal{J})=⟨a_i,a_{ij}:i,j\in I⟩.$$

\begin{proof} By (IV), we know $R_{(a_i{a}_j)a_k}$ can be generated by $R_{a_i}$, $R_{a_j}$, $R_{a_k}$, $R_{a_i{a}_j}$, $R_{a_i{a}_k}$, $R_{a_j{a}_i}$, and $R_{a_j{a}_k}$. \end{proof}

Corollary

If $\mathcal{J}$ is finitely generated, then $\mathrm{Mult}(\mathcal{J})$ is also finitely generated.

Corollary

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

\begin{proof} Since $⟨R_{a^2},R_a⟩=\mathrm{Mult}(\mathcal{J})$ and $[R_{a^2},R_a]=0$ by Jordan identity, we know $\mathrm{Mult}(\mathcal{J})$ is commutative. \end{proof}