# Jordan algebra identities

A Jordan algebra is a vector space with a commutative bilinear operation $$\circ$$ obeying an identity that's often written as

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

I find this identity rather obscure. If we write $$x^2 = x \circ x$$ and use $$L_a$$ to stand for left multiplication by $$a$$, we can rewrite it in a more appealing form:

$$L_{x^2} L_x = L_x L_{x^2} .$$

However, I'd be even happier if this were a special case of a more general identity

$$L_{x^m} L_{x^n} = L_{x^n} L_{x^m} \qquad (\ast)$$

holding for all $$n, m \in \mathbb{N}$$.

This more general identity parses in any Jordan algebra, because any Jordan algebra is power-associative: expressions like $$x \circ \cdots \circ x$$ are independent of how you parenthesize them, so $$x^n$$ is well-defined. But is this more general identity $$(\ast)$$ true in every Jordan algebra?

• Seems not what you want but multilinear form of the Jordan identity does not look so obscure. It is$$[L_{ab},L_c]+[L_{bc},L_a]+[L_{ca},L_b]=0$$(where $[-,-]$ is usual commutator of operators). Another equivalent form:$$\operatorname{as}(ab,d,c)+\operatorname{as}(bc,d,a)+\operatorname{as}(ca,d,b)=0$$where $\operatorname{as}(x,y,z)=(xy)z-x(yz)$ is the associator. – მამუკა ჯიბლაძე Jun 8 at 19:02
• I'm sorry to be foolish, but aren't some of these right multiplications? For example, $L_{x^2}L_x y$ should be $x^2 \circ (x \circ y)$, but we want $(x \circ y) \circ x^2$. – LSpice Jun 8 at 20:56
• A Jordan algebra is commutative so left multiplication is the same as right multiplication. – John Baez Jun 8 at 22:10

For completeness of MathOverflow, I might as well copy the essence of the argument: first prove the identity $$L_{(b\circ d)\circ c} = L_{b\circ d}L_c + L_{c\circ d}L_b + L_{b\circ c}L_d - L_b L_c L_d - L_d L_c L_b$$ by linearizing the Jordan identity (here we use the fact that the characteristic is not two). Apply this to $$b=a^k$$ and $$c=d=a$$, giving $$L_{a^{k+2}} = 2L_{a^{k+1}}L_a + L_{a^2}L_{a^k} - L_{a^k} L_a^2 - L_a^2 L_{a^k}$$ From this it follows by induction that all $$L_{a^k}$$ belong to the algebra generated by $$L_a$$ and $$L_{a^2}$$ and, since these commute, they all commute.
• @DavidRoberts why not, I have no problem with it, but don't see why it should be called the only "true" definition. From another intuition one might find the original definition simpler. Also, what do you think of associativity? $(ab)c=a(bc)$ may sound nice because we're used to it, but what's important is that all bracketings of products $x_1\dots x_n$ lead to the same element. In spite of this I'm happier with the usual short definition. – YCor Jun 8 at 10:47
• [This addresses the edited post, not the current stream of comments] Actually I once checked that the naive definition of Jordan algebra in char 2 (commutative + Jordan identity) is so bad that it does not pass to extension of scalars (for extensions of fields of char 2, probably $F_2\subset F_4$). The correct definition of Jordan ring should be that a ring satisfying all identities with coefficients in $\mathbf{Z}$ satisfied by, say, all complex Jordan algebras. When 2 is not invertible it implies we should add the polarizations among the axioms. – YCor Jun 8 at 19:21