## June 7, 2020

### 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.

Posted at June 7, 2020 6:32 PM UTC

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### 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.

Posted by: Allen Knutson on June 7, 2020 8:20 PM | Permalink | Reply to this

### Re: Elimination

This is a fun thing to think about. Jordan algebras that come from associative algebras by setting $a \circ b = a b + b a$ are called special, and as I see it you’re asking “which identities does $a \circ b$ obey in every special Jordan algebra, aside from those that have been built into the definition of Jordan algebra?”

But this question is rather scary: it’s famous, and it’s hard. Here’s what McCrimmon says in his mammoth tome ironically named A Taste of Jordan Algebras:

These identities are called the special identities or s-identities: they are precisely all Jordan polynomials which vanish on all special Jordan algebras (hence automatically on their homomorphic images as well), but not on all Jordan algebras (they are nonzero elements of the free Jordan algebra).

The first thing to say is that there weren’t supposed to be any s-identities! Remember that Jordan’s goal was to capture the algebraic behavior of hermitian operators in the Jordan axioms. The s-identities are in fact just the algebraic identities involving the Jordan product satisfied by all hermitian matrices (of arbitrary size), and in principle all of these were supposed to have been incorporated into the Jordan axioms!

A nonconstructive proof of the existence of s-identities was first given by A.A. Albert and Lowell J. Paige in 1959 (when it was far too late to change the Jordan axioms), by showing that there must be nonzero Jordan polynomials $f(x, y, z)$ in three variables which vanish on all special algebras (become zero in the free special algebra on three generators). The first explicit s-identities G8 and G9 were discovered by Jacobson’s student Charles M. Glennie in 1963: as we remarked in the Colloquial Survey, these could not possibly have been discovered without the newly-minted notions of the Jordan triple product and U-operators, and indeed even in their U-form no mortal other than Glennie has been able to remember them for more than 15 minutes.

It is known that there are no s-identities of degree $\le 7$, but to this day we do not know exactly what all the s-identities are, or even whether they are finitely generated.

I’m really hoping there is some conceptual way of understanding the Jordan algebra axioms that doesn’t focus on special Jordan algebras.

There’s a subject of Jordan algebras: is this merely the study of what happens when you foolishly ignore the s-identities, or is there a solid idea underlying this subject?

Posted by: John Baez on June 7, 2020 8:57 PM | Permalink | Reply to this

### Re: Jordan Algebras

After schmutzing around trying to make a recursion or induction proof of some kind, I decided to hit the books. This looks like what’s proven just before Eq. (56) in Jacobson (2008). There’s a chain of one notation invented after another, leading to the statement that right multiplications commute:

$[R_{a^k}, R_{a^l}] = 0$

for $k,l \geq 1$.

Posted by: Blake Stacey on June 7, 2020 9:43 PM | Permalink | Reply to this

### Re: Jordan Algebras

Great! And left multiplication is the same as right multiplication in a commutative algebra.

Meanwhile, over on MathOverflow, Gro-Tsen wrote:

This identity (*) is, indeed, true, and is, in fact, a step in one of the standard ways to prove that Jordan algebras are power-associative: see McCrimmon’s 2004 book A Taste of Jordan Algebras, exercise 5.2.2A (question (2)) on page 201.

I don’t know how I overlooked this.

So this is now my favored definition of ‘Jordan algebra’: a commutative power-associative algebra such that for any element $x$, the operations of multiplication by $x^n$ ($n \ge 1$) all commute with each other.

Posted by: John Baez on June 7, 2020 9:55 PM | Permalink | Reply to this

### Re: Jordan Algebras

That, at least, is a definition I have a chance at remembering!

Posted by: Blake Stacey on June 8, 2020 12:19 AM | Permalink | Reply to this

### Re: Jordan Algebras

By a theorem of Kokoris (http://dx.doi.org/10.2307/1969601), over a field of characteristic zero, a semisimple commutative power associative algebra is Jordan.

Posted by: Dan F. on June 9, 2020 10:02 AM | Permalink | Reply to this

### Re: Jordan Algebras

Nice! Is this for finite-dimensional algebras?

Posted by: John Baez on June 10, 2020 6:25 AM | Permalink | Reply to this

### Re: Jordan Algebras

The front page of the linked paper doesn’t mention such a condition.

Posted by: Allen Knutson on June 10, 2020 10:02 PM | Permalink | Reply to this

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