In mathematics, quadratic Jordan algebras are a generalization of Jordan algebras introduced by . The fundamental identities of the quadratic representation of a linear Jordan algebra are used as axioms to define a quadratic Jordan algebra over a field of arbitrary characteristic. There is a uniform description of finite-dimensional simple quadratic Jordan algebras, independent of characteristic. If 2 is invertible in the field of coefficients, the theory of quadratic Jordan algebras reduces to that of linear Jordan algebras.
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.
An element a is invertible if is invertible and there exists such that is the inverse of and : such b is unique and we say that b is the inverse of a. A Jordan division algebra is one in which every non-zero element is invertible.
Let B be a subspace of A. Define B to be a quadratic ideal or an inner ideal if the image of Q(b) is contained in B for all b in B; define B to be an outer ideal if B is mapped into itself by every Q(a) for all a in A. An ideal of A is a subspace which is both an inner and an outer ideal. A quadratic Jordan algebra is simple if it contains no non-trivial ideals.
For given b, the image of Q(b) is an inner ideal: we call this the principal inner ideal on b.
The centroid Γ of A is the subset of EndK(A) consisting of endomorphisms T which "commute" with Q in the sense that for all a
The centroid of a simple algebra is a field: A is central if its centroid is just K.
If A is a unital associative algebra over K with multiplication × then a quadratic map Q can be defined from A to EndK(A) by Q(a) : b ↦ a × b × a. This defines a quadratic Jordan algebra structure on A. A quadratic Jordan algebra is special if it is isomorphic to a subalgebra of such an algebra, otherwise exceptional.
Let A be a vector space over K with a quadratic form q and associated symmetric bilinear form q(x,y) = q(x+y) - q(x) - q(y). Let e be a "basepoint" of A, that is, an element with q(e) = 1. Define a linear functional T(y) = q(y,e) and a "reflection" y∗ = T(y)e - y. For each x we define Q(x) by
Q(x) : y ↦ q(x,y∗)x − q(x) y∗ .
Then Q defines a quadratic Jordan algebra on A.
Let A be a unital Jordan algebra over a field K of characteristic not equal to 2. For a in A, let L denote the left multiplication map in the associative enveloping algebra
L(a):x\mapstoax
and define a K-endomorphism of A, called the quadratic representation, by
\displaystyle{Q(a)=2L(a)2-L(a2).}
Then Q defines a quadratic Jordan algebra.
The quadratic identities can be proved in a finite-dimensional Jordan algebra over R or C following Max Koecher, who used an invertible element. They are also easy to prove in a Jordan algebra defined by a unital associative algebra (a "special" Jordan algebra) since in that case Q(a)b = aba.[1] They are valid in any Jordan algebra over a field of characteristic not equal to 2. This was conjectured by Jacobson and proved in : 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.[2] In 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.[3]
An element a in A is called invertible if it is invertible in R[''a''] or C[''a'']. If b denotes the inverse, then power associativity of a shows that L(a) and L(b) commute.
In fact a is invertible if and only if Q(a) is invertible. In that case
Indeed, if Q(a) is invertible it carries R[''a''] onto itself. On the other hand Q(a)1 = a2, so\displaystyle{(Q(a)-1a)a=aQ(a)-1a=L(a)Q(a)-1a=Q(a)-1a2=1.}
The Jordan identity
\displaystyle{[L(a),L(a2)](b)=0}
can be polarized by replacing a by a + tc and taking the coefficient of t. Rewriting this as an operator applied to c yields
\displaystyle{2L(ab)L(a)+L(a2)L(b)=2L(a)L(b)L(a)+L(a2b).}
Taking b = a−1 in this polarized Jordan identity yields
\displaystyle{Q(a)L(a-1)=L(a).}
Replacing a by its inverse, the relation follows if L(a) and L(a−1) are invertible. If not it holds for a + ε1 with ε arbitrarily small and hence also in the limit.
For c in A and F(a) a function on A with values in End A, letDcF(a) be the derivative at t = 0 of F(a + tc). Then| -1 | |
| \displaystyle{c=D | |
| c(Q(a)a |
)=2Q(a,c)a-1
| -1 | |
| +Q(a)D | |
| c(a |
),}
where Q(a,b) if the polarization of Q
\displaystyle{Q(a,c)={1\over2}(Q(a+c)-Q(a)-Q(c))=L(a)L(c)+L(c)L(a)-L(ac).}
Since L(a) commutes with L(a−1)
\displaystyle{Q(a,c)a-1=(L(a)L(c)+L(c)L(a)-L(ac))a-1=c.}
Hence
\displaystyle{c=2c
| -1 | |
| +Q(a)D | |
| c(a |
),}
so that
Applying Dc to L(a−1)Q(a) = L(a) and acting on b = c−1 yields\displaystyle{(Q(a)b)(Q(a-1)b-1)=1.}
On the other hand L(Q(a)b) is invertible on an open dense set where Q(a)b must also be invertible with
\displaystyle{(Q(a)b)-1=Q(a-1)b-1.}
Taking the derivative Dc in the variable b in the expression above gives
\displaystyle{-Q(Q(a)b)-1Q(a)c=-Q(a)-1Q(b)-1c.}
This yields the fundamental identity for a dense set of invertible elements, so it follows in general by continuity. The fundamental identity implies that c = Q(a)b is invertible if a and b are invertible and gives a formula for the inverse of Q(c). Applying it to c gives the inverse identity in full generality.
As shown above, if a is invertible,
\displaystyle{Q(a,b)a-1=b.}
Taking Dc with a as the variable gives
\displaystyle{0=Q(c,b)a-1+
| -1 | |
| Q(a,b)D | |
| c(a |
)=Q(c,b)a-1-Q(a,b)Q(a)-1c.}
Replacing a by a−1 gives, applying Q(a) and using the fundamental identity gives
\displaystyle{Q(a)Q(b,c)a=Q(a)Q(a-1,b)Q(a)c=
| 1 | |
| 2 |
Q(a)[Q(a-1+b)-Q(a-1)-Q(b)]Q(a)c=Q(Q(a)b,a)c.}
Hence
\displaystyle{Q(a)Q(b,c)a=Q(Q(a)b,a)c.}
Interchanging b and c gives
\displaystyle{Q(a)Q(b,c)a=Q(a,Q(a)c)b.}
On the other hand is defined by, so this implies
\displaystyle{Q(a)R(b,a)c=R(a,b)Q(a)c,}
so that for a invertible and hence by continuity for all a
The Jordan identity can be polarized by replacing a by a + tc and taking the coefficient of t. This gives
\displaystyle{c(a2b)+2a(ac)b)=a2(cb)+2(ac)(ab).}
In operator notation this implies
Polarizing in a again gives\displaystyle{c((ad)b)+d((ac)b)+a((dc)b)=(cd)(ab)+(da)(cb)+(ac)(db).}
Written as operators acting on d, this gives
\displaystyle{L(c)L(b)L(a)+L((ac)b)+L(a)L(b)L(c)=L(ab)L(c)+L(cb)L(a)+L(ac)L(b).}
Let
\displaystyle{Q(a)=2L(a)2-L(a2).}
Then Q(a) commutes with L(a) by the Jordan identity.
From the definitions if is the associated symmetric bilinear mapping, then and
\displaystyle{Q(a,b)=L(a)L(b)+L(b)L(a)-L(ab).}
Moreover
IndeedBy the second and first polarized Jordan identities this implies
The polarized version of is
Now with, it follows that\displaystyle{Q(a)R(b,a)=2[Q(a)L(b),L(a)]+2Q(a)L(ab)=2[Q(ab,a),L(a)]+2[L(a),L(b)]Q(a)+2Q(a)L(ab).}
So by the last identity with ab in place of b this implies the commutation identity:
\displaystyle{Q(a)R(b,a)=2[L(a),L(b)]Q(a)+2L(ab)Q(a)=R(a,b)Q(a).}
The identity Q(a)R(b,a) = R(a,b)Q(a) can be strengthened to
Indeed applied to c, the first two terms give\displaystyle{2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.}
Switching b and c then gives
\displaystyle{Q(a)R(b,a)c=2Q(Q(a)b,a)c.}
The identity is proved using the Lie bracket relations
Indeed the polarization in c of the identity gives\displaystyle{Q(c,y)L(x)+L(x)Q(c,y)=Q(yx,c)+Q(cx,y).}
Applying both sides to d, this shows that
\displaystyle{[L(x),R(c,d)]=R(xc,d)-R(c,xd).}
In particular these equations hold for x = ab. On the other hand if T = [''L''(''a''),''L''(''b'')] then D(z) = Tz is a derivation of the Jordan algebra, so that
\displaystyle{[T,R(c,d)]=R(Dc,d)+R(c,Dd).}
The Lie bracket relations follow because R(a,b) = T + L(ab).
Since the Lie bracket on the left hand side is antisymmetric,
As a consequenceIndeed set a = y, b = x, c = z, d = x and make both sides act on y.On the other hand
Indeed this follows by setting x = Q(a)b in\displaystyle{[R(a,b),R(x,y)]a=-R(R(x,y)a,b)a+R(a,R(y,x)b)a.}
Hence, combining these equations with the strengthened commutation identity,
\displaystyle{2Q(Q(a)b)=2R(a,b)Q(Q(a)b,a)-R(b,a)Q(a)R(a,b)+2Q(a)Q(b)Q(a)=2Q(a)Q(b)Q(a).}
Let A be a quadratic Jordan algebra over R or C. Following, a linear Jordan algebra structure can be associated with A such that, if L(a) is Jordan multiplication, then the quadratic structure is given by Q(a) = 2L(a)2 − L(a2).
Firstly the axiom Q(a)R(b,a) = R(a,b)Q(a) can be strengthened to
\displaystyle{Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).}
Indeed applied to c, the first two terms give
\displaystyle{2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.}
Switching b and c then gives
\displaystyle{Q(a)R(b,a)c=2Q(Q(a)b,a)c.}
Now let
| \displaystyle{L(a)= | 1 |
| 2 |
R(a,1).}
Replacing b by a and a by 1 in the identity above gives
\displaystyle{R(a,1)=R(1,a)=2Q(a,1).}
In particular
\displaystyle{L(a)=Q(a,1),L(1)=Q(1,1)=I.}
If furthermore a is invertible then
\displaystyle{R(a,b)=2Q(Q(a)b,a)Q(a)-1=2Q(a)Q(b,a-1).}
Similarly if b is invertible
\displaystyle{R(a,b)=2Q(a,b-1)Q(b).}
The Jordan product is given by
\displaystyle{a\circb=L(a)b=
| 1 | |
| 2 |
R(a,1)b=Q(a,b)1,}
so that
\displaystyle{a\circb=b\circa.}
The formula above shows that 1 is an identity. Defining a2 by a∘a = Q(a)1, the only remaining condition to be verified is the Jordan identity
\displaystyle{[L(a),L(a2)]=0.}
In the fundamental identity
\displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a),}
Replace a by a + t, set b = 1 and compare the coefficients of t2 on both sides:
\displaystyle{Q(a)=2Q(a,1)2-Q(a2,1)=2L(a)2-L(a2).}
Setting b = 1 in the second axiom gives
\displaystyle{Q(a)L(a)=L(a)Q(a),}
and therefore L(a) must commute with L(a2).
In a unital linear Jordan algebra the shift identity asserts that
Following, it can be established as a direct consequence of polarized forms of the fundamental identity and the commutation or homotopy identity. It is also a consequence of Macdonald's theorem since it is an operator identity involving only two variables.[4]For a in a unital linear Jordan algebra A the quadratic representation is given by
\displaystyle{Q(a)=2L(a)2-L(a2),}
so the corresponding symmetric bilinear mapping is
\displaystyle{Q(a,b)=L(a)L(b)+L(b)L(a)-L(ab).}
The other operators are given by the formula
| \displaystyle{ | 1 |
| 2 |
R(a,b)=L(a)L(b)-L(b)L(a)+L(ab),}
so that
\displaystyle{Q(a,b)=2L(a)L(b)-
| 1 | |
| 2 |
R(a,b).}
The commutation or homotopy identity
\displaystyle{R(a,b)Q(a)=Q(a)R(b,a)=2Q(Q(a)b,a),}
can be polarized in a. Replacing a by a + t1 and taking the coefficient of t gives
The fundamental identity\displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a)}
can be polarized in a. Replacing a by a +t1 and taking the coefficients of t gives (interchanging a and b)
Combining the two previous displayed identities yieldsReplacing a by a +t1 in the fundamental identity and taking the coefficient of t2 gives\displaystyle{2Q(Q(a)b,b)-L(a)Q(b)L(a)=Q(a)Q(b)+Q(b)Q(a)-Q(ab).}
Since the right hand side is symmetric this implies
These identities can be used to prove the shift identity:\displaystyle{R(Q(a)b,b)=R(a,Q(b)a).}
It is equivalent to the identity
\displaystyle{2L(Q(a)b)L(b)-Q(Q(a)b,b)=2L(a)L(Q(b)a)-Q(a,Q(b)a).}
By the previous displayed identity this is equivalent to
\displaystyle{[L(Q(a)b)+L(b)Q(a)]L(b)=L(a)[L(Q(b)a)+Q(b)L(a)].}
On the other hand, the bracketed terms can be simplified by the third displayed identity. It implies that both sides are equal to .
For finite-dimensional unital Jordan algebras, the shift identity can be seen more directly using mutations. Let a and b be invertible, and let be the Jordan multiplication in Ab. Then. Moreover.on the other hand and similarly with a and b interchanged. Hence
| 2,a | |
| \displaystyle{L | |
| a(b |
)Q(b)
| 2,b | |
| =Q(b)L | |
| b(a |
).}
Thus
\displaystyle{R(Q(b)a,a)Q(b)=Q(b)R(Q(a)b,b)=R(b,Q(a)b)Q(b),}
so the shift identity follows by cancelling Q(b). A density argument allows the invertibility assumption to be dropped.
A linear unital Jordan algebra gives rise to a quadratic mapping Q and associated mapping R satisfying the fundamental identity, the commutation of homotopy identity and the shift identity. A Jordan pair consists of two vector space and two quadratic mappings from to . These determine bilinear mappings from to by the formula where . Omitting ± subscripts, these must satisfy
the fundamental identity
\displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a),}
the commutation or homotopy identity
\displaystyle{R(a,b)Q(a)=Q(a)R(b,a)=2Q(Q(a)b,a),}
and the shift identity
\displaystyle{R(Q(a)b,b)=R(a,Q(b)a).}
A unital Jordan algebra A defines a Jordan pair by taking V± = A with its quadratic structure maps Q and R.