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, a ↦ Q(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) : 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.[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) : y ↦ q(x,y∗)x − q(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

[math]\displaystyle{ L(a) : x \mapsto a x \ }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a)=2L(a)^2 -L(a^2).} }[/math]

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

::[math]\displaystyle{ \displaystyle{Q(a)^{-1}a=a^{-1},\,\,\, Q(a^{-1})=Q(a)^{-1}.} }[/math]

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

[math]\displaystyle{ \displaystyle{(Q(a)^{-1} a)a=a Q(a)^{-1} a=L(a)Q(a)^{-1}a=Q(a)^{-1}a^2 =1.} }[/math]

The Jordan identity

[math]\displaystyle{ \displaystyle{[L(a),L(a^2)](b)=0} }[/math]

can be polarized by replacing a by a + tc and taking the coefficient of t. Rewriting this as an operator applied to c yields

[math]\displaystyle{ \displaystyle{2L(ab)L(a) + L(a^2)L(b)=2L(a)L(b)L(a) + L(a^2b).} }[/math]

Taking b = a−1 in this polarized Jordan identity yields

[math]\displaystyle{ \displaystyle{Q(a)L(a^{-1})=L(a).} }[/math]

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:
[math]\displaystyle{ \displaystyle{(Q(a)b)^{-1}=Q(a^{-1})b^{-1}.} }[/math]
  • The quadratic representation satisfies the following fundamental identity:
[math]\displaystyle{ \displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a).} }[/math]

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

[math]\displaystyle{ \displaystyle{c=D_c(Q(a)a^{-1})=2Q(a,c)a^{-1} +Q(a)D_c(a^{-1}),} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a,c)={1\over 2} (Q(a+c) -Q(a)-Q(c))=L(a)L(c) + L(c)L(a) - L(ac).} }[/math]

Since L(a) commutes with L(a−1)

[math]\displaystyle{ \displaystyle{Q(a,c)a^{-1}= (L(a)L(c) + L(c)L(a) - L(ac))a^{-1}=c.} }[/math]

Hence

[math]\displaystyle{ \displaystyle{c=2c +Q(a)D_c(a^{-1}),} }[/math]

so that

::[math]\displaystyle{ \displaystyle{D_c(a^{-1})= - Q(a)^{-1}c.} }[/math]

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

[math]\displaystyle{ \displaystyle{(Q(a)b)(Q(a^{-1})b^{-1})=1.} }[/math]

On the other hand L(Q(a)b) is invertible on an open dense set where Q(a)b must also be invertible with

[math]\displaystyle{ \displaystyle{(Q(a)b)^{-1}=Q(a^{-1})b^{-1}.} }[/math]

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

[math]\displaystyle{ \displaystyle{-Q(Q(a)b)^{-1}Q(a)c =-Q(a)^{-1}Q(b)^{-1}c.} }[/math]

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,

[math]\displaystyle{ \displaystyle{Q(a,b)a^{-1} =b.} }[/math]

Taking Dc with a as the variable gives

[math]\displaystyle{ \displaystyle{0=Q(c,b)a^{-1} + Q(a,b)D_c(a^{-1})=Q(c,b)a^{-1} -Q(a,b)Q(a)^{-1}c.} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a)Q(b,c)a= Q(a)Q(a^{-1},b)Q(a)c = \frac{1}{2}Q(a)[Q(a^{-1}+b) -Q(a^{-1}) -Q(b)]Q(a)c = Q(Q(a)b,a)c.} }[/math]

Hence

[math]\displaystyle{ \displaystyle{Q(a)Q(b,c)a=Q(Q(a)b,a)c.} }[/math]

Interchanging b and c gives

[math]\displaystyle{ \displaystyle{Q(a)Q(b,c)a=Q(a,Q(a)c)b.} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a)R(b,a)c=R(a,b)Q(a)c,} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a)R(b,a)=R(a,b)Q(a).} }[/math]

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]

[math]\displaystyle{ \displaystyle{c(a^2b) + 2a (ac)b) = a^2(cb) + 2 (ac)(ab).} }[/math]

In operator notation this implies

::[math]\displaystyle{ \displaystyle{[L(a^2),L(c)]+2[L(ac),L(a)]=0.} }[/math]

Polarizing in a again gives

[math]\displaystyle{ \displaystyle{c((ad)b) + d((ac)b) + a((dc)b)= (cd)(ab)+(da)(cb)+(ac)(db).} }[/math]

Written as operators acting on d, this gives

[math]\displaystyle{ \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).} }[/math]

Replacing c by b and b by a gives

::[math]\displaystyle{ \displaystyle{L(b)L(a)^2 +L(a(ab))+L(a)^2L(b)=L(a^2)L(b)+2L(ab)L(a).} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a)=2L(a)^2 - L(a^2).} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a,b)=L(a)L(b)+L(b)L(a)-L(ab).} }[/math]

Moreover

:[math]\displaystyle{ \displaystyle{2Q(ab,a)=L(b)Q(a) +Q(a)L(b).} }[/math]

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

::[math]\displaystyle{ \displaystyle{2[Q(a,b),L(a)]=2[L(b),Q(a)].} }[/math]

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

[math]\displaystyle{ \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).} }[/math]

So by the last identity with ab in place of b this implies the commutation identity:

[math]\displaystyle{ \displaystyle{Q(a)R(b,a)=2[L(a),L(b)]Q(a)+2L(ab)Q(a)=R(a,b)Q(a).} }[/math]

The identity Q(a)R(b,a) = R(a,b)Q(a) can be strengthened to

::[math]\displaystyle{ \displaystyle{Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).} }[/math]

Indeed applied to c, the first two terms give

[math]\displaystyle{ \displaystyle{2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.} }[/math]

Switching b and c then gives

[math]\displaystyle{ \displaystyle{Q(a)R(b,a)c=2Q(Q(a)b,a)c.} }[/math]

Fundamental identity II

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

[math]\displaystyle{ \displaystyle{[R(a,b),R(c,d)]=R(R(a,b)c,d) - R(c,R(b,a)d).} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(c,y)L(x) +L(x)Q(c,y) = Q(yx,c) + Q(cx,y).} }[/math]

Applying both sides to d, this shows that

[math]\displaystyle{ \displaystyle{[L(x),R(c,d)]= R(xc,d) - R(c,xd).} }[/math]

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

[math]\displaystyle{ \displaystyle{[T,R(c,d)] = R(Dc,d) + R(c,Dd).} }[/math]

The Lie bracket relations follow because R(a,b) = T + L(ab).

Since the Lie bracket on the left hand side is antisymmetric,

::[math]\displaystyle{ \displaystyle{R(R(a,b)c,d) - R(c,R(b,a)d)=- R(R(c,d)a,b) +R(a,R(d,c)b).} }[/math]

As a consequence

:[math]\displaystyle{ \displaystyle{R(y,Q(x)y)=R(y,x)^2 - 2Q(y)Q(x).} }[/math]

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

On the other hand

::[math]\displaystyle{ \displaystyle{2Q(Q(a)b)=2R(a,b)Q(Q(a)b,a)-Q(a)R(b,Q(a)b).} }[/math]

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

[math]\displaystyle{ \displaystyle{[R(a,b),R(x,y)]a=- R(R(x,y)a,b)a +R(a,R(y,x)b)a.} }[/math]

Hence, combining these equations with the strengthened commutation identity,

[math]\displaystyle{ \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).} }[/math]

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)2 − L(a2).

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

[math]\displaystyle{ \displaystyle{Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).} }[/math]

Indeed applied to c, the first two terms give

[math]\displaystyle{ \displaystyle{2Q(a)Q(b,c)a=2Q(Q(a)c,a)b.} }[/math]

Switching b and c then gives

[math]\displaystyle{ \displaystyle{Q(a)R(b,a)c=2Q(Q(a)b,a)c.} }[/math]

Now let

[math]\displaystyle{ \displaystyle{L(a)=\frac{1}{2}R(a,1).} }[/math]

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

[math]\displaystyle{ \displaystyle{R(a,1)=R(1,a)=2Q(a,1).} }[/math]

In particular

[math]\displaystyle{ \displaystyle{L(a)=Q(a,1),\,\,\,L(1)=Q(1,1)=I.} }[/math]

If furthermore a is invertible then

[math]\displaystyle{ \displaystyle{R(a,b)=2Q(Q(a)b,a)Q(a)^{-1}= 2Q(a)Q(b,a^{-1}).} }[/math]

Similarly if 'b is invertible

[math]\displaystyle{ \displaystyle{R(a,b)= 2Q(a,b^{-1})Q(b).} }[/math]

The Jordan product is given by

[math]\displaystyle{ \displaystyle{a\circ b = L(a)b=\frac{1}{2}R(a,1)b=Q(a,b)1,} }[/math]

so that

[math]\displaystyle{ \displaystyle{a\circ b = b \circ a.} }[/math]

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

[math]\displaystyle{ \displaystyle{[L(a),L(a^2)]=0.} }[/math]

In the fundamental identity

[math]\displaystyle{ \displaystyle{Q(Q(a)b)= Q(a)Q(b)Q(a),} }[/math]

Replace a by a + t, set b = 1 and compare the coefficients of t2 on both sides:

[math]\displaystyle{ \displaystyle{Q(a)=2Q(a,1)^2 - Q(a^2,1)= 2L(a)^2 - L(a^2).} }[/math]

Setting b = 1 in the second axiom gives

[math]\displaystyle{ \displaystyle{Q(a)L(a)=L(a)Q(a),} }[/math]

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

Shift identity

In a unital linear Jordan algebra the shift identity asserts that

:[math]\displaystyle{ \displaystyle{R(Q(a)b,b)=R(a,Q(b)a).} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(a)=2L(a)^2 - L(a^2),} }[/math]

so the corresponding symmetric bilinear mapping is

[math]\displaystyle{ \displaystyle{Q(a,b)=L(a)L(b)+L(b)L(a) -L(ab).} }[/math]

The other operators are given by the formula

[math]\displaystyle{ \displaystyle{\frac{1}{2}R(a,b)=L(a)L(b) -L(b)L(a) +L(ab),} }[/math]

so that

[math]\displaystyle{ \displaystyle{Q(a,b)= 2L(a)L(b)-\frac{1}{2}R(a,b).} }[/math]

The commutation or homotopy identity

[math]\displaystyle{ \displaystyle{R(a,b)Q(a)=Q(a)R(b,a)=2Q(Q(a)b,a),} }[/math]

can be polarized in a. Replacing a by a + t1 and taking the coefficient of t gives

:[math]\displaystyle{ \displaystyle{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).} }[/math]

The fundamental identity

[math]\displaystyle{ \displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a)} }[/math]

can be polarized in a. Replacing a by a +t1 and taking the coefficients of t gives (interchanging a and b)

:[math]\displaystyle{ \displaystyle{2Q(ab,a)= Q(a)L(b) + L(b)Q(a).} }[/math]

Combining the two previous displayed identities yields

:[math]\displaystyle{ \displaystyle{ 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).} }[/math]

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

[math]\displaystyle{ \displaystyle{2Q(Q(a)b,b)-L(a)Q(b)L(a)= Q(a)Q(b)+Q(b)Q(a) -Q(ab).} }[/math]

Since the right hand side is symmetric this implies

:[math]\displaystyle{ \displaystyle{2Q(Q(a)b,b)-2Q(Q(b)a,a)= L(a)Q(b)L(a) - L(b)Q(a)L(b).} }[/math]

These identities can be used to prove the shift identity:

[math]\displaystyle{ \displaystyle{R(Q(a)b,b)=R(a,Q(b)a).} }[/math]

It is equivalent to the identity

[math]\displaystyle{ \displaystyle{2L(Q(a)b)L(b)- Q(Q(a)b,b) = 2L(a)L(Q(b)a) -Q(a,Q(b)a).} }[/math]

By the previous displayed identity this is equivalent to

[math]\displaystyle{ \displaystyle{[L(Q(a)b)+L(b)Q(a)]L(b)= L(a)[L(Q(b)a) + Q(b)L(a)].} }[/math]

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)2 − Lb(a2,b) and similarly with a and b interchanged. Hence

[math]\displaystyle{ \displaystyle{L_a(b^{2,a})Q(b) =Q(b)L_b(a^{2,b}).} }[/math]

Thus

[math]\displaystyle{ \displaystyle{R(Q(b)a,a)Q(b)=Q(b)R(Q(a)b,b)=R(b,Q(a)b)Q(b),} }[/math]

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

[math]\displaystyle{ \displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a),} }[/math]

the commutation or homotopy identity

[math]\displaystyle{ \displaystyle{R(a,b)Q(a)=Q(a)R(b,a)=2Q(Q(a)b,a),} }[/math]

and the shift identity

[math]\displaystyle{ \displaystyle{R(Q(a)b,b)=R(a,Q(b)a).} }[/math]

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 



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