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I found myself "naturally" dealing with an object of this form:

X is a complex vector space, with a "product" (a,b) → {aba} which is quadratic in the first variable, linear in the second, and satisfies some associativity conditions. These conditions are actually complicated, but more or less say that {aba} looks like the product (aba) in an alternative algebra Y containing X as a subspace.

For example, the main "associativity condition" I am interested in is: {a{b{aca}b}a}={{aba}c{aba}}


  1. Symmetric matrices
  2. Octonions, or indeed any alternative algebra
  3. Let J belong to GL(n,ℂ), with tJ=-J and J²=-Id, and W={w∈M(n×n,ℂ)|JtwJ=-w}

all with the standard product {aba}=aba.

All of these examples are Jordan algebras, with respect to the symmetrized product a∘b=½(ab+ba), but I cannot see any direct link between the Jordan product and my product.

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Unless you can define the product more precisely, it is very hard to answer this question. There is a bewildering array of "triple systems" in the literature, but it's hard to say if that is what you have. –  José Figueroa-O'Farrill Feb 22 '10 at 12:46
One point is that I don't really have a triple system, as {abc} is defined only for a=c. (e.g. when X&sub;Y, an alternative algebra, aba needs not to be in X) The main "associativity condition" I am interested in is {a{b{aca}b}a}={{aba}c{aba}} (Question edited) –  user175348 Feb 22 '10 at 13:01
I meant "when $X\subset Y$, an alternative algebra, abc needs not to be in X" –  user175348 Feb 22 '10 at 13:07

2 Answers 2

up vote 25 down vote accepted

In a Jordan algebra with product $\cdot$, a triple product is defined by $$\{abc\}=(a\cdot b)\cdot c+(b\cdot c)\cdot a-(a\cdot c)\cdot b.$$ In a special Jordan algebra (constructed by symmetrising an associative product) one has $\{aba\}=aba$, and it is easy to show that in such algebras one always has the identity $$\{\{aba\}c\{aba\}\}=\{a\{b\{aca\}b\}a\}.$$ Now, there is an amazing general theorem of Macdonald's that states that any identity in three variables which is of degree at most one in one of them and which is valid in special Jordan algebras actually holds in all Jordan algebras. This is proved in Jacobson's breath-taking Structure and representations of Jordan algebras.

So your identity holds in all Jordan algebras. As a consequence, from the information you give it is more or less impossible to distinguish your structure from Jordan algebras, as far as I can see.

By the way, in his book, Jacobson notes that McCrimmon has developed the theory of Jordan algebras based exclusively on the composition $(a,b)\mapsto aba$, and gives [McCrimmon, Kevin. A general theory of Jordan rings. Proc. Nat. Acad. Sci. U.S.A. 56 1966 1072--1079. MR0202783 (34 #2643)] as reference. I do not have access to the paper, though. The paper can be gotten from this link Andrea provided in a comment below.

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This is a wonderful answer. –  Harry Gindi Feb 22 '10 at 14:17
The article on PNAS is open access, and available here: pnas.org/content/56/4/1072.full.pdf –  user175348 Feb 22 '10 at 14:23
Great answer. Everything fits together so nicely now! –  user175348 Feb 22 '10 at 14:40
Time for a shameless plug: The book Jordan Operator Algebras by Erling Størmer and myself (Pitman, 1984) contains a (hopefully more accessible) proof of Macdonald's theorem. A scanned pdf of the book is available under a CC license from my homepage: math.ntnu.no/~hanche/joa –  Harald Hanche-Olsen Feb 22 '10 at 15:11

How hard/lengthy would it be to give a concrete example of your product?

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In the three examples, the product {aba} is just the ordinary product aba. –  user175348 Feb 22 '10 at 12:43
I edited the question to clarify. –  user175348 Feb 22 '10 at 12:45

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