Quadratic Jordan algebra

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In mathematics, quadratic Jordan algebras are a generalization of Jordan algebras introduced by Kevin McCrimmon (1966). 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.

Definition

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, aQ(a), satisfying the conditions:

  • Q(1) = id;
  • Q(Q(a)b) = Q(a)Q(b)Q(a) ("fundamental identity");
  • Q(a)R(b,a) = R(a,b)Q(a) ("commutation identity"), where R(a,b)c = (Q(a + c) − Q(a) − Q(c))b.

Further, these properties are required to hold under any extension of scalars.[1]

Elements

An element a is invertible if Q(a) is invertible and there exists b such that Q(b) is the inverse of Q(a) and Q(a)b = a: 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.[2]

Structure

Let B be a subspace of A. Define B to be a quadratic ideal[3] 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.[1] A quadratic Jordan algebra is simple if it contains no non-trivial ideals.[2]

For given b, the image of Q(b) is an inner ideal: we call this the principal inner ideal on b.[2][4]

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

  • T Q(a) = Q(a) T;
  • Q(Ta) = Q(a) T2.

The centroid of a simple algebra is a field: A is central if its centroid is just K.[5]

Examples

Quadratic Jordan algebra from an associative algebra

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) : ba × 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.[2]

Quadratic Jordan algebra from a quadratic form

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) : yq(x,y)xq(x) y .

Then Q defines a quadratic Jordan algebra on A.[6][7]

Quadratic Jordan algebra from a linear Jordan algebra

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):xax 

and define a K-endomorphism of A, called the quadratic representation, by

Q(a)=2L(a)2L(a2).

Then Q defines a quadratic Jordan algebra.

Quadratic Jordan algebra defined by a linear 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.[8] 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 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.[9] In (Jacobson 1969) an elementary proof, due to McCrimmon and Meyberg, is given for Jordan algebras over a field of characteristic not equal to 2.

Koecher's proof

Koecher's arguments apply for finite-dimensional Jordan algebras over the real or complex numbers.[10]

Fundamental identity I

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

::Q(a)1a=a1,Q(a1)=Q(a)1.

Indeed, if Q(a) is invertible it carries R[a] onto itself. On the other hand Q(a)1 = a2, so

(Q(a)1a)a=aQ(a)1a=L(a)Q(a)1a=Q(a)1a2=1.

The Jordan identity

[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

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

Q(a)L(a1)=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.

*If a and b are invertible then so is Q(a)b and it satisfies the inverse identity:
(Q(a)b)1=Q(a1)b1.
  • The quadratic representation satisfies the following fundamental identity:
Q(Q(a)b)=Q(a)Q(b)Q(a).

For c in A and F(a) a function on A with values in End A, let DcF(a) be the derivative at t = 0 of F(a + tc). Then

c=Dc(Q(a)a1)=2Q(a,c)a1+Q(a)Dc(a1),

where Q(a,b) if the polarization of Q

Q(a,c)=12(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)

Q(a,c)a1=(L(a)L(c)+L(c)L(a)L(ac))a1=c.

Hence

c=2c+Q(a)Dc(a1),

so that

::Dc(a1)=Q(a)1c.

Applying Dc to L(a−1)Q(a) = L(a) and acting on b = c−1 yields

(Q(a)b)(Q(a1)b1)=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

(Q(a)b)1=Q(a1)b1.

Taking the derivative Dc in the variable b in the expression above gives

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.

Commutation identity I

As shown above, if a is invertible,

Q(a,b)a1=b.

Taking Dc with a as the variable gives

0=Q(c,b)a1+Q(a,b)Dc(a1)=Q(c,b)a1Q(a,b)Q(a)1c.

Replacing a by a−1 gives, applying Q(a) and using the fundamental identity gives

Q(a)Q(b,c)a=Q(a)Q(a1,b)Q(a)c=12Q(a)[Q(a1+b)Q(a1)Q(b)]Q(a)c=Q(Q(a)b,a)c.

Hence

Q(a)Q(b,c)a=Q(Q(a)b,a)c.

Interchanging b and c gives

Q(a)Q(b,c)a=Q(a,Q(a)c)b.

On the other hand R(x,y) is defined by R(x,y)z = 2 Q(x,z)y, so this implies

Q(a)R(b,a)c=R(a,b)Q(a)c,

so that for a invertible and hence by continuity for all a

Q(a)R(b,a)=R(a,b)Q(a).

Mccrimmon–Meyberg proof

Commutation identity II

The Jordan identity a(a2b) = a2(ab) can be polarized by replacing a by a + tc and taking the coefficient of t. This gives[11]

c(a2b)+2a(ac)b)=a2(cb)+2(ac)(ab).

In operator notation this implies

::[L(a2),L(c)]+2[L(ac),L(a)]=0.

Polarizing in a again gives

c((ad)b)+d((ac)b)+a((dc)b)=(cd)(ab)+(da)(cb)+(ac)(db).

Written as operators acting on d, this gives

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).

Replacing c by b and b by a gives

::L(b)L(a)2+L(a(ab))+L(a)2L(b)=L(a2)L(b)+2L(ab)L(a).

Also, since the right hand side is symmetric in b and 'c, interchanging b and c on the left and subtracting , it follows that the commutators [L(b),L(c)] are derivations of the Jordan algebra.

Let

Q(a)=2L(a)2L(a2).

Then Q(a) commutes with L(a) by the Jordan identity.

From the definitions if Q(a,b) = ½ (Q(a = b) − Q(a) − Q(b)) is the associated symmetric bilinear mapping, then Q(a,a) = Q(a) and

Q(a,b)=L(a)L(b)+L(b)L(a)L(ab).

Moreover

:2Q(ab,a)=L(b)Q(a)+Q(a)L(b).

Indeed

2Q(ab,a) − L(b)Q(a) − Q(a)L(b) = 2L(ab)L(a) + 2L(a)L(ab) − 2L(a(ab)) − 2L(a)2L(b) − 2L(b)L(a)2 + L(a2)L(b) + L(b)L(a2).

By the second and first polarized Jordan identities this implies

2Q(ab,a) − L(b)Q(a) − Q(a)L(b) = 2[L(a),L(ab)] + [L(b),L(a2)] = 0.

The polarized version of [Q(a),L(a)] = 0 is

::2[Q(a,b),L(a)]=2[L(b),Q(a)].

Now with R(a,b) = 2[L(a),L(b)] + 2L(ab), it follows that

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:

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

::Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).

Indeed applied to c, the first two terms give

2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.

Switching b and c then gives

Q(a)R(b,a)c=2Q(Q(a)b,a)c.

Fundamental identity II

The identity Q(Q(a)b) = Q(a)Q(b)Q(a) is proved using the Lie bracket relations[12]

[R(a,b),R(c,d)]=R(R(a,b)c,d)R(c,R(b,a)d).

Indeed the polarization in c of the identity Q(c)L(x) + L(x)Q(c) = 2Q(cx,c) gives

Q(c,y)L(x)+L(x)Q(c,y)=Q(yx,c)+Q(cx,y).

Applying both sides to d, this shows that

[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

[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,

::R(R(a,b)c,d)R(c,R(b,a)d)=R(R(c,d)a,b)+R(a,R(d,c)b).

As a consequence

:R(y,Q(x)y)=R(y,x)22Q(y)Q(x).

Indeed set a = y, b = x, c = z, d = x and make both sides act on y.

On the other hand

::2Q(Q(a)b)=2R(a,b)Q(Q(a)b,a)Q(a)R(b,Q(a)b).

Indeed this follows by setting x = Q(a)b in

[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,

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).

Linear Jordan algebra defined by a quadratic Jordan algebra

Let A be a quadratic Jordan algebra over R or C. Following (Jacobson 1969), 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)2L(a2).

Firstly the axiom Q(a)R(b,a) = R (a,b)Q(a) can be strengthened to

Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).

Indeed applied to c, the first two terms give

2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.

Switching b and c then gives

Q(a)R(b,a)c=2Q(Q(a)b,a)c.

Now let

L(a)=12R(a,1).

Replacing b by a and a by 1 in the identity above gives

R(a,1)=R(1,a)=2Q(a,1).

In particular

L(a)=Q(a,1),L(1)=Q(1,1)=I.

If furthermore a is invertible then

R(a,b)=2Q(Q(a)b,a)Q(a)1=2Q(a)Q(b,a1).

Similarly if 'b is invertible

R(a,b)=2Q(a,b1)Q(b).

The Jordan product is given by

ab=L(a)b=12R(a,1)b=Q(a,b)1,

so that

ab=ba.

The formula above shows that 1 is an identity. Defining a2 by aa = Q(a)1, the only remaining condition to be verified is the Jordan identity

[L(a),L(a2)]=0.

In the fundamental identity

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:

Q(a)=2Q(a,1)2Q(a2,1)=2L(a)2L(a2).

Setting b = 1 in the second axiom gives

Q(a)L(a)=L(a)Q(a),

and therefore L(a) must commute with L(a2).

Shift identity

In a unital linear Jordan algebra the shift identity asserts that

:R(Q(a)b,b)=R(a,Q(b)a).

Following (Meyberg 1972), 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.[13]

For a in a unital linear Jordan algebra A the quadratic representation is given by

Q(a)=2L(a)2L(a2),

so the corresponding symmetric bilinear mapping is

Q(a,b)=L(a)L(b)+L(b)L(a)L(ab).

The other operators are given by the formula

12R(a,b)=L(a)L(b)L(b)L(a)+L(ab),

so that

Q(a,b)=2L(a)L(b)12R(a,b).

The commutation or homotopy identity

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

:L(b)Q(a)+R(a,b)L(a)=Q(a)L(b)+L(a)R(b,a)=L(Q(a)b)+2Q(ab,a).

The fundamental identity

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)

:2Q(ab,a)=Q(a)L(b)+L(b)Q(a).

Combining the two previous displayed identities yields

:L(Q(a)b)+L(b)Q(a)=L(a)R(b,a),L(Q(b)a)+Q(b)L(a)=R(b,a)L(b).

Replacing a by a +t1 in the fundamental identity and taking the coefficient of t2 gives

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

:2Q(Q(a)b,b)2Q(Q(b)a,a)=L(a)Q(b)L(a)L(b)Q(a)L(b).

These identities can be used to prove the shift identity:

R(Q(a)b,b)=R(a,Q(b)a).

It is equivalent to the identity

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

[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 ½ L(a)R(b,a)L(b).

For finite-dimensional unital Jordan algebras, the shift identity can be seen more directly using mutations.[14] Let a and b be invertible, and let Lb(a)=R(a,b) be the Jordan multiplication in Ab. Then Q(b)Lb(a) = La(b)Q(b). Moreover Q(b)Qb(a) = Q(b)Q(a)Q(b) =Qa(b)Q(b). on the other hand Qb(a)=2Lb(a)2Lb(a2,b) and similarly with a and b interchanged. Hence

La(b2,a)Q(b)=Q(b)Lb(a2,b).

Thus

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.

Jordan pairs

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 (V+,V) consists of two vector space V± and two quadratic mappings Q± from V± to V. These determine bilinear mappings R± from V± × V to V± by the formula R(a,b)c = 2Q(a,c)b where 2Q(a,c) = Q(a + c) − Q(a) − Q(c). Omitting ± subscripts, these must satisfy[15]

the fundamental identity

Q(Q(a)b)=Q(a)Q(b)Q(a),

the commutation or homotopy identity

R(a,b)Q(a)=Q(a)R(b,a)=2Q(Q(a)b,a),

and the shift identity

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.

See also

Notes

  1. 1.0 1.1 Racine 1973, p. 1
  2. 2.0 2.1 2.2 2.3 Racine 1973, p. 2
  3. Jacobson 1968, p. 153
  4. Jacobson 1968, p. 154
  5. Racine 1973, p. 3
  6. Jacobson 1968, p. 35
  7. Racine 1973, pp. 5–6
  8. See:
  9. See:
  10. See:
  11. Meyberg 1972, pp. 66–67
  12. Meyberg 1972
  13. See:
  14. Koecher 1999
  15. Loos 2006

References

Further reading

  • Faulkner, John R. (1970), Octonion Planes Defined by Quadratic Jordan Algebras, Memoirs of the American Mathematical Society, 104, American Mathematical Society, ISBN 0-8218-5888-2