### Jordan Algebras

#### Posted by John Baez

I’ve learned a fair amount about Jordan algebras by now, but I still don’t have a clear conceptual understanding of the Jordan algebra axioms, and it’s time to fix that.

A Jordan algebra is a vector space with a commutative bilinear operation $\circ$ that obeys

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

That’s how Wikipedia defines it. This axiom is an affront to my mathematical sense of taste. It looks like a ridiculously restricted version of the associative law, plucked from dozens of variants one could imagine. There has to be a better way to understand what’s going on here!

Here’s one better way. We can define $x^2 = x \circ x$ and then write

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

as

$(x \circ y) \circ x^2 = x \circ (y \circ x^2)$

So far, no big deal. Then, use the commutative law to write this as

$x^2 \circ (x \circ y) = x \circ (x^2 \circ y)$

Let $L_a$ stand for left multiplication by $a$. Then the above equation says

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

Left multiplication by $x$ commutes with left multiplication by $x^2$.

I like this better. But I would like it even better if it were a ‘biased’ version of a more general law

$L_{x^m} L_{x^n} = L_{x^n} L_{x^m}$

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

This more general law *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:

**Puzzle.** Is this more general law *true* in every Jordan algebra?

I don’t know! All I know is that

$L_{x^m} L_{x^n} \, x = L_{x^n} L_{x^m} \, x$

and

$L_{x^2} L_{x} \, y = L_{x} L_{x^2} \, y$

for all elements $x,y$ of a Jordan algebra.

For finite-dimensional formally real Jordan algebras, which are the kind Jordan, von Neumann and Wigner classified in their work on quantum mechanics, I know how to completely dodge the annoying axiom $(x \circ y) \circ (x \circ x) = x \circ (y \circ (x \circ x))$. But now I’m thinking about general Jordan algebras.

## Elimination

Abstractly, this might be a question in “elimination theory”. The algebraic geometry is as follows.

Maybe you believe Jordan algebras “should” come from associative algebras, by taking anticommutator. Then you’re considering the map $[]_+: Hom(V\otimes V,V) \to Hom(Sym^2 V, V)$ taking a multiplication to its anticommutator, plus the variety $A \subseteq Hom(V\otimes V,V)$ consisting of associative products, and want to understand the image $[]_+(A) \subseteq Hom(Sym^2 V, V)$.

For any given dimension of $V$, this is a ring theory question – what are the generators of the pullback along $Fun(Hom(Sym^2 V,V)) \hookrightarrow Fun(Hom(V\otimes V,V))$ of the ideal defining $A$?

I don’t instantly see how to express this question without fixing the dimension of the algebra $V$ (since subspaces aren’t subalgebras), and doubt it’s worth putting on a computer, since even for very small $\dim V$ the dimension of $Hom(V\otimes V,V)$ is a lotta variables. So this is more of a theoretical statement, of what a “conceptual understanding” could amount to.