quadratic Jordan algebra

In mathematics, quadratic Jordan algebras are a generalization of Jordan algebras introduced by {{harvs|txt|first=Kevin|last=McCrimmon|authorlink=Kevin McCrimmon|year=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:

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

Further, these properties are required to hold under any extension of scalars.{{harvnb|Racine|1973|p=1}}

=Elements=

An element a is invertible if {{math|Q(a)}} is invertible and there exists {{math|b}} such that {{math|Q(b)}} is the inverse of {{math|Q(a)}} and {{math|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.

=Structure=

Let B be a subspace of A. Define B to be a quadratic ideal{{harvnb|Jacobson|1968|p=153}} 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.{{harvnb|Racine|1973|p=2}}

For given b, the image of Q(b) is an inner ideal: we call this the principal inner ideal on b.{{harvnb|Jacobson|1968|p=154}}

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.{{harvnb|Racine|1973|p=3}}

=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.

==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.{{harvnb|Jacobson|1968|p=35}}{{harvnb|Racine|1973|pp=5–6}}

==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) : x \mapsto a x \

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

:\displaystyle{Q(a)=2L(a)^2 -L(a^2).}

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.

See:

  • {{harvnb|Koecher|1999|pp=72–76}}
  • {{harvnb|Faraut|Koranyi|1994|pp=32–34}} They are valid in any Jordan algebra over a field of characteristic not equal to 2. This was conjectured by Jacobson and proved in {{harvtxt|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.See:
  • {{harvnb|Jacobson|1968|pp=40–47,52}}

In {{harvtxt|Jacobson|1969|pp=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 proof=

Koecher's arguments apply for finite-dimensional Jordan algebras over the real or complex numbers.See:

  • {{harvnb|Koecher|1999}}
  • {{harvnb|Faraut|Koranyi|1994|pp=32–35}}

==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

{{quote box|align=left|::\displaystyle{Q(a)^{-1}a=a^{-1},\,\,\, Q(a^{-1})=Q(a)^{-1}.}}}

{{Clear}}

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

:\displaystyle{(Q(a)^{-1} a)a=a Q(a)^{-1} a=L(a)Q(a)^{-1}a=Q(a)^{-1}a^2 =1.}

The Jordan identity

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

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.

{{quote box|align=left|*If a and b are invertible then so is Q(a)b and it satisfies the inverse identity:

::\displaystyle{(Q(a)b)^{-1}=Q(a^{-1})b^{-1}.}

  • The quadratic representation satisfies the following fundamental identity:

::\displaystyle{Q(Q(a)b)=Q(a)Q(b)Q(a).}

}}

{{Clear}}

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

:\displaystyle{c=D_c(Q(a)a^{-1})=2Q(a,c)a^{-1} +Q(a)D_c(a^{-1}),}

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

:\displaystyle{Q(a,c)={1\over 2} (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 +Q(a)D_c(a^{-1}),}

so that

{{quote box|align=left|::\displaystyle{D_c(a^{-1})= - Q(a)^{-1}c.}}}

{{Clear}}

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)^{-1}Q(a)c =-Q(a)^{-1}Q(b)^{-1}c.}

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,

:\displaystyle{Q(a,b)a^{-1} =b.}

Taking Dc with a as the variable gives

:\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.}

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 = \frac{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 {{math|1=R(x,y)}} is defined by {{math|1=R(x,y)z = 2 Q(x,z)y}}, 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

{{quote box|align=left|

::\displaystyle{Q(a)R(b,a)=R(a,b)Q(a).}

}}

{{Clear}}

=Mccrimmon–Meyberg proof=

==Commutation identity II==

The Jordan identity {{math|1=a(a2b) = a2(ab)}} can be polarized by replacing a by a + tc and taking the coefficient of t. This gives{{harvnb|Meyberg|1972|pp=66–67}}

:\displaystyle{c(a^2b) + 2a (ac)b) = a^2(cb) + 2 (ac)(ab).}

In operator notation this implies

{{quote box|align=left|::\displaystyle{[L(a^2),L(c)]+2[L(ac),L(a)]=0.}}}

{{Clear}}

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).}

{{Clear}}

Replacing c by b and b by a gives

{{quote box|align=left|::\displaystyle{L(b)L(a)^2 +L(a(ab))+L(a)^2L(b)=L(a^2)L(b)+2L(ab)L(a).}}}

{{Clear}}

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

:\displaystyle{Q(a)=2L(a)^2 - L(a^2).}

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

From the definitions if {{math|1=Q(a,b) = ½ (Q(a = b) − Q(a) − Q(b))}} is the associated symmetric bilinear mapping, then {{math|1=Q(a,a) = Q(a)}} and

:\displaystyle{Q(a,b)=L(a)L(b)+L(b)L(a)-L(ab).}

Moreover

{{quote box|align=left|:\displaystyle{2Q(ab,a)=L(b)Q(a) +Q(a)L(b).}}}

{{Clear}}

Indeed

:{{math|1=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

:{{math|1=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 {{math|1=[Q(a),L(a)] = 0}} is

{{quote box|align=left|::\displaystyle{2[Q(a,b),L(a)]=2[L(b),Q(a)].}}}

{{Clear}}

Now with {{math|1=R(a,b) = 2[L(a),L(b)] + 2L(ab)}}, 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

{{quote box|align=left|::\displaystyle{Q(a)R(b,a)=R(a,b)Q(a)=2Q(Q(a)b,a).}}}

{{Clear}}

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.}

==Fundamental identity II==

The identity {{math|1=Q(Q(a)b) = Q(a)Q(b)Q(a)}} is proved using the Lie bracket relations{{harvnb|Meyberg|1972}}

{{quote box|align=left|

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

{{Clear}}

Indeed the polarization in c of the identity {{math|1=Q(c)L(x) + L(x)Q(c) = 2Q(cx,c)}} 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,

{{quote box|align=left|::\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).}}}

{{Clear}}

As a consequence

{{quote box|align=left|:\displaystyle{R(y,Q(x)y)=R(y,x)^2 - 2Q(y)Q(x).}}}

{{Clear}}

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

On the other hand

{{quote box|align=left|::\displaystyle{2Q(Q(a)b)=2R(a,b)Q(Q(a)b,a)-Q(a)R(b,Q(a)b).}}}

{{Clear}}

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).}

Linear Jordan algebra defined by a quadratic Jordan algebra

Let A be a quadratic Jordan algebra over R or C. Following {{harvtxt|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

:\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)=\frac{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\circ b = L(a)b=\frac{1}{2}R(a,1)b=Q(a,b)1,}

so that

:\displaystyle{a\circ b = b \circ a.}

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

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

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).

Shift identity

In a unital linear Jordan algebra the shift identity asserts that

{{quote box|align=left|:\displaystyle{R(Q(a)b,b)=R(a,Q(b)a).}}}

{{Clear}}

Following {{harvtxt|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.See:

  • {{harvnb|Meyberg|1972|pp=85–86}}
  • {{harvnb|McCrimmon|2004|pp=202–203}}

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

:\displaystyle{Q(a)=2L(a)^2 - L(a^2),}

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{\frac{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)-\frac{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

{{quote box|align=left|:\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).}}}

{{Clear}}

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)

{{quote box|align=left|:\displaystyle{2Q(ab,a)= Q(a)L(b) + L(b)Q(a).}}}

{{Clear}}

Combining the two previous displayed identities yields

{{quote box|align=left|:\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).}}}

{{Clear}}

Replacing 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

{{quote box|align=left|:\displaystyle{2Q(Q(a)b,b)-2Q(Q(b)a,a)= L(a)Q(b)L(a) - L(b)Q(a)L(b).}}}

{{Clear}}

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 {{math|1=½ L(a)R(b,a)L(b)}}.

For finite-dimensional unital Jordan algebras, the shift identity can be seen more directly using mutations.{{harvnb|Koecher|1999}} Let a and b be invertible, and let

{{math|1=Lb(a)=R(a,b)}} be the Jordan multiplication in Ab. Then

{{math|1=Q(b)Lb(a) = La(b)Q(b)}}. Moreover

{{math|1=Q(b)Qb(a) = Q(b)Q(a)Q(b) =Qa(b)Q(b)}}.

on the other hand {{math|1=Qb(a)=2Lb(a)2Lb(a2,b)}} and similarly with a and b interchanged. Hence

:\displaystyle{L_a(b^{2,a})Q(b) =Q(b)L_b(a^{2,b}).}

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.

Jordan pairs

{{see also|Jordan pair}}

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 {{math|1=(V+,V)}} consists of two vector space {{math|1=V±}} and two quadratic mappings {{math|1=Q±}} from {{math|1=V±}} to {{math|1=V}}. These determine bilinear mappings {{math|1=R±}} from {{math|1=V± × V}} to {{math|1=V±}} by the formula {{math|1=R(a,b)c = 2Q(a,c)b}} where {{math|1=2Q(a,c) = Q(a + c) − Q(a) − Q(c)}}. Omitting ± subscripts, these must satisfy{{harvnb|Loos|2006}}

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.

See also

Notes

{{reflist|2}}

References

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  • {{citation|mr=0202783|last=McCrimmon|first= Kevin|title=A general theory of Jordan rings|journal=Proc. Natl. Acad. Sci. U.S.A. |volume=56|year= 1966|issue=4|pages= 1072–9|jstor=57792|doi=10.1073/pnas.56.4.1072 | zbl=0139.25502 |pmc=220000 |pmid=16591377|doi-access=free}}
  • {{citation|last=McCrimmon|first= Kevin|chapter=Quadratic methods in nonassociative algebras|title= Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 1|pages= 325–330|year=1975|chapter-url=http://www.mathunion.org/ICM/ICM1974.1/Main/icm1974.1.0325.0330.ocr.pdf}}
  • {{Citation | last1=McCrimmon | first1=Kevin | title=A taste of Jordan algebras | url=https://books.google.com/books?isbn=0387954473 | publisher=Springer-Verlag | series=Universitext | isbn=978-0-387-95447-9 | doi=10.1007/b97489 | id=[http://www.math.virginia.edu/Faculty/McCrimmon/ Errata] | year=2004 | mr=2014924 | zbl=1044.17001}}
  • {{citation|last=McCrimmon|first= Kevin|title= Jordan algebras and their applications|journal= Bull. Amer. Math. Soc. |volume=84| year=1978|issue= 4|pages= 612–627|url=https://www.ams.org/journals/bull/1978-84-04/S0002-9904-1978-14503-0/home.html|doi=10.1090/s0002-9904-1978-14503-0|doi-access=free}}
  • {{citation|last=Meyberg|first= K.|title=Lectures on algebras and triple systems|publisher=University of Virginia|year= 1972|url=http://www.math.uci.edu/~brusso/Meyberg(Reduced2).pdf}}
  • {{citation | last=Racine | first=Michel L. | title=The arithmetics of quadratic Jordan algebras | series=Memoirs of the American Mathematical Society |volume=136 | isbn=978-0-8218-1836-7 | year=1973 | publisher=American Mathematical Society | zbl=0348.17009 }}

Further reading

  • {{citation | title=Octonion Planes Defined by Quadratic Jordan Algebras | volume=104 | series=Memoirs of the American Mathematical Society | first=John R. | last=Faulkner | publisher=American Mathematical Society | year=1970 | isbn=0-8218-5888-2 | zbl=0206.23301 }}

Category:Non-associative algebras