Question about derivation in Jordan algebra

Question

Let $(G,\circ)$ be a Jordan algebra, then $\sigma:G\to G$ given by $$c\mapsto a\circ(b\circ c)-b\circ(a\circ c),\quad\forall c\in G,$$ is a derivation, where $a$ and $b$ are two fixed elements of $G$.

Answer 1

Let's now actually prove this.

The algebra in question is commutative and satifies $$(xy)x^2=x(yx^2).$$ It is easy to see (if the base field has at least three elements!) that this is equivalent to the identity $$(x,y,zw)+(z,y,wx)+(w,y,xz)=0,$$ where $(a,b,c)=(ab)c-a(bc)$ is the associator map.

If we write $R_x$ the map given by right multiplication by $x$ (and using that it is the same as the left multiplication), this identity is immediately seen to be equivalent to the identity $$R_{xy}R_z-R_xR_yR_z+R_{zy}R_{x}-R_{y(xz)}+R_{xz}R_{y}-R_zR_yR_x=0.$$ If we interchange $x$ and $y$ in this equation, and compute the difference of the two, and arrange things properly, we see that $$[R_z,[R_x,R_y]]+R_{[R_x,R_y](z)}=0$$ If we write $\sigma=[R_x,R_y]$, then this equation is $$[R_z,\sigma]+R_{\sigma(z)}=0,$$ and one can see at once that this last equation means precisely that the map $\sigma$ is a derivation.

Your map if $\sigma$ for $x=a$ and $y=b$.

P.S. I learned this from Schafer's book on non-associative algebras and the technique comes from old papers of Albert.

Answer 2

Just for the record, this is how I checked my claim that this works in special Jordan algebras:

The following Mathematica code makes the operator ** multilinear and defines $\circ$ to be its symmetrization.

Unprotect[NonCommutativeMultiply];
NonCommutativeMultiply[a___, b_Plus, c___] := NonCommutativeMultiply[a, #, c] & /@ b;
NonCommutativeMultiply[a___, \[Lambda]_?NumberQ b_, c___] := \[Lambda] NonCommutativeMultiply[a, b, c];

a_ \[SmallCircle] b_ := (a ** b + b ** a)/2;

This is your map $\sigma$:

\[Sigma][c_] := a\[SmallCircle](b\[SmallCircle]c) - b\[SmallCircle](a\[SmallCircle]c);

And this is the check that it is a derivation:

\[Sigma][x\[SmallCircle]y] - x\[SmallCircle]\[Sigma][y] - \[Sigma][x]\[SmallCircle]y // Simplify

which gives zero.

(All this looks nicer in Mathematica...)