Quadratic Jordan algebra
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
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.
- [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
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
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
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
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
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
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,
As a consequence
Indeed 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
- [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
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
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)
Combining the two previous displayed identities yields
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
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
- ↑ 1.0 1.1 Racine 1973, p. 1
- ↑ 2.0 2.1 2.2 2.3 Racine 1973, p. 2
- ↑ Jacobson 1968, p. 153
- ↑ Jacobson 1968, p. 154
- ↑ Racine 1973, p. 3
- ↑ Jacobson 1968, p. 35
- ↑ Racine 1973, pp. 5–6
- ↑
See:
- Koecher 1999, pp. 72–76
- Faraut & Koranyi 1994, pp. 32–34
- ↑ See:
- Jacobson 1968, pp. 40–47,52
- ↑ See:
- Koecher 1999
- Faraut & Koranyi 1994, pp. 32–35
- ↑ Meyberg 1972, pp. 66–67
- ↑ Meyberg 1972
- ↑ See:
- Meyberg 1972, pp. 85–86
- McCrimmon 2004, pp. 202–203
- ↑ Koecher 1999
- ↑ Loos 2006
References
- Faraut, J.; Koranyi, A. (1994), Analysis on symmetric cones, Oxford Mathematical Monographs, Oxford University Press, ISBN 0198534779
- Jacobson, N. (1968), Structure and representations of Jordan algebras, American Mathematical Society Colloquium Publications, 39, American Mathematical Society, ISBN 978-0-8218-4640-7, https://books.google.com/books?id=aAGWAwAAQBAJ
- Jacobson, N. (1969), Lectures on quadratic Jordan algebras, Tata Institute of Fundamental Research Lectures on Mathematics, 45, Bombay: Tata Institute of Fundamental Research, http://www.math.tifr.res.in/~publ/ln/tifr45.pdf
- Koecher, M. (1999), The Minnesota Notes on Jordan Algebras and Their Applications, Lecture Notes in Mathematics, 1710, Springer, ISBN 3-540-66360-6
- Loos, Ottmar (2006), Jordan pairs, Lecture Notes in Mathematics, 460, Springer, ISBN 978-3-540-37499-2, https://books.google.com/books?id=6Zl8CwAAQBAJ&pg=PR1
- Loos, Ottmar (1977), Bounded symmetric domains and Jordan pairs, Mathematical lectures, University of California, Irvine, archived from the original on 2016-03-03, https://web.archive.org/web/20160303234008/http://molle.fernuni-hagen.de/~loos/jordan/archive/irvine/irvine.pdf
- Macdonald, I. G. (1960), "Jordan algebras with three generators", Proc. London Math. Soc. 10: 395–408, doi:10.1112/plms/s3-10.1.395, http://plms.oxfordjournals.org/content/s3-10/1/395
- McCrimmon, Kevin (1966), "A general theory of Jordan rings", Proc. Natl. Acad. Sci. U.S.A. 56 (4): 1072–9, doi:10.1073/pnas.56.4.1072, PMID 16591377
- McCrimmon, Kevin (1975), "Quadratic methods in nonassociative algebras", Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 1, pp. 325–330, http://www.mathunion.org/ICM/ICM1974.1/Main/icm1974.1.0325.0330.ocr.pdf
- McCrimmon, Kevin (2004), A taste of Jordan algebras, Universitext, Springer-Verlag, doi:10.1007/b97489, Errata, ISBN 978-0-387-95447-9, https://books.google.com/books?isbn=0387954473
- McCrimmon, Kevin (1978), "Jordan algebras and their applications", Bull. Amer. Math. Soc. 84 (4): 612–627, doi:10.1090/s0002-9904-1978-14503-0, https://www.ams.org/journals/bull/1978-84-04/S0002-9904-1978-14503-0/home.html
- Meyberg, K. (1972), Lectures on algebras and triple systems, University of Virginia, http://www.math.uci.edu/~brusso/Meyberg(Reduced2).pdf
- Racine, Michel L. (1973), The arithmetics of quadratic Jordan algebras, Memoirs of the American Mathematical Society, 136, American Mathematical Society, ISBN 978-0-8218-1836-7
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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