3 typo

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 Jodan 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.

2 added 142 characters in body

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 Jodan 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.

1

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 Jodan 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.