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?

rightmultiplications? For example, $L_{x^2}L_x y$ should be $x^2 \circ (x \circ y)$, but we want $(x \circ y) \circ x^2$. $\endgroup$ – LSpice Jun 8 '20 at 20:56