There is a uniform description of finite-dimensional simple quadratic Jordan algebras, independent of characteristic.
A quadratic Jordan algebra consists of a vector space A over a field K with a distinguished element 1 and a quadratic map of A into the K-endomorphisms of A, a ↦ Q(a), satisfying the conditions: Further, these properties are required to hold under any extension of scalars.
The quadratic identities can be proved in a finite-dimensional Jordan algebra over R or C following Max Koecher, who used an invertible element.
This was conjectured by Jacobson and proved in Macdonald (1960): Macdonald showed that if a polynomial identity in three variables, linear in the third, is valid in any special Jordan algebra, then it holds in all Jordan algebras.
19–21) an elementary proof, due to McCrimmon and Meyberg, is given for Jordan algebras over a field of characteristic not equal to 2.
Koecher's arguments apply for finite-dimensional Jordan algebras over the real or complex numbers.
On the other hand Q(a)1 = a2, so The Jordan identity can be polarized by replacing a by a + tc and taking the coefficient of t. Rewriting this as an operator applied to c yields Taking b = a−1 in this polarized Jordan identity yields Replacing a by its inverse, the relation follows if L(a) and L(a−1) are invertible.
Then where Q(a,b) if the polarization of Q Since L(a) commutes with L(a−1) Hence so that Applying Dc to L(a−1)Q(a) = L(a) and acting on b = c−1 yields On the other hand L(Q(a)b) is invertible on an open dense set where Q(a)b must also be invertible with Taking the derivative Dc in the variable b in the expression above gives This yields the fundamental identity for a dense set of invertible elements, so it follows in general by continuity.
On the other hand if T = [L(a),L(b)] then D(z) = Tz is a derivation of the Jordan algebra, so that The Lie bracket relations follow because R(a,b) = T + L(ab).
Defining a2 by a∘a = Q(a)1, the only remaining condition to be verified is the Jordan identity In the fundamental identity Replace a by a + t, set b = 1 and compare the coefficients of t2 on both sides: Setting b = 1 in the second axiom gives and therefore L(a) must commute with L(a2).
[13] For a in a unital linear Jordan algebra A the quadratic representation is given by so the corresponding symmetric bilinear mapping is The other operators are given by the formula so that The commutation or homotopy identity can be polarized in a.
For finite-dimensional unital Jordan algebras, the shift identity can be seen more directly using mutations.