A Generalization of the First Tits Construction

Abstract

Let F be a field of characteristic, not 2 or 3. The first Tits construction is a well-known tripling process to construct separable cubic Jordan algebras, especially Albert algebras. We generalize the first Tits construction by choosing the scalar employed in the tripling process outside of the base field. This yields a new family of non-associative unital algebras which carry a cubic map, and maps that can be viewed as generalized adjoint and generalized trace maps. These maps display properties often similar to the ones in the classical setup. In particular, the cubic norm map permits some kind of weak Jordan composition law.

1. Introduction

Let F be a field of characteristic, not 2 or 3. Separable cubic Jordan algebras over F play an important role in Jordan theory (where separable means that their trace defines a non-degenerate bilinear form). It is well known that every separable cubic Jordan algebra can be obtained by either a first or a second Tits construction [1] (IX, Section 39). In particular, exceptional simple Jordan algebras, also called Albert algebras, are separable cubic Jordan algebras. The role of Albert algebras in the structure theory of Jordan algebras is similar to the role of octonion algebras in the structure theory of alternative algebras. Moreover, their automorphism group is an exceptional algebraic group of type $F_4$, and their cubic norms have isometry groups of type $E_6$. For some recent developments, see [2,3,4,5,6].

In this paper, we canonically generalize the first Tits construction $J(A,\mu )$. The first Tits construction starts with a separable associative cubic algebra A and uses a scalar $\mu \in F^{\times }$ in its definition. Our construction also starts with A and employs the same algebra multiplication as that used for the classical first Tits construction, but now allows also $\mu \in A^{\times }$.

We obtain a new class of non-associative unital algebras which we again denote by $J(A,\mu )$. They carry a cubic map $N:J(A,\mu )\to A$ that generalizes the classical norm, a map $T:J(A,\mu )\to F$ that generalizes the classical trace, and a map $♯:J(A,\mu )\to J(A,\mu )$ that generalizes the classical adjoint of a Jordan algebra. Starting with a cubic étale algebra E, the algebras obtained this way can be viewed as generalizations of special nine-dimensional Jordan algebras. Starting with a central simple algebra A of degree three, the algebras obtained this way can be viewed as generalizations of Albert algebras.

Cubic Jordan algebras carry a cubic norm that satisfies some Jordan composition law involving the U-operator. Curiously, the cubic map $N:J(A,\mu )\to A$ of our generalized construction still allows some sort of generalized weak Jordan composition law, and some of the known identities of cubic Jordan algebras involving a generalized trace map and adjoint can be at least partially recovered.

We point out that there already exists a canonical non-associative generalization of associative central simple cyclic algebras of degree three, involving skew polynomials: the non-associative cyclic algebras $(K/F,\sigma ,\mu )$, where $K/F$ is a cubic separable field extension or a $C_3$-Galois algebra, and $\mu \in K\setminus F$, were first studied over finite fields [7], and then later over arbitrary base fields and rings [8,9,10,11] and applied in space-time block coding [12,13]. Their “norm maps” reflect some of the properties of the non-associative cyclic algebra $(K/F,\sigma ,\mu )$ and are isometric to the “norm maps” $N:J(K,\mu )\to K$ of the generalized Tits construction $J(K,\mu )$. We show that these algebras are not related, however.

Some obvious questions like if and when the algebras obtained through a generalized first Tits construction are division algebras seem to be very difficult to answer. We will not address these here and only discuss some straightforward implications.

The contents of the paper are as follows: After introducing the terminology in Section 2 and reviewing the classical first Tits construction, we generalize the classical construction in Section 3 and obtain unital non-associative algebras $J(A,\mu )$, where $\mu \in A^{\times }$. The algebras $J(A,\mu )$ carry maps that satisfy some of the same identities we know from the classical setup. If $A\ne F$, then ${Nuc}_l\left(J(A,\mu )\right)={Nuc}_r\left(J(A,\mu )\right)=F$ for all $\mu \in A^{\times }$. If A is a central simple associative division algebra of degree three, then ${Nuc}_m\left(J(A,\mu )\right)=F$ (Theorems 3 and 4). Some necessary conditions on when $J(A,\mu )$ is a division algebra are listed in Theorem 6: If $J(A,\mu )$ is a division algebra, then $\mu \notin N_A\left(A^{\times }\right)$ and A must be a division algebra. If N is anisotropic, then A is a division algebra and $\mu \notin N_A\left(A^{\times }\right)$. If there exist elements $0\ne x=(x_0,x_1,x_2)\in J(A,\mu )$ such that $x^{♯}=0$, we show that either A must have zero divisors, or A is a division algebra and $\mu \in N_A\left(A^{\times }\right)$. Moreover, if A is a division algebra over F and 1, $\mu$, and ${\mu }^2$ are linearly independent over F, then N must be anisotropic.

We investigate in which special cases several classical identities carry over in Section 4.

In Section 5, we compare the algebras obtained from a generalized first Tits construction starting with a cyclic field extension with the algebras ${(K/F,\sigma ,\mu )}^{+}$, where $(K/F,\sigma ,\mu )$ is a non-associative cyclic algebra over F of degree three. If $\mu \in F^{\times }$, then it is well known that these algebras are isomorphic. For $\mu \in K\setminus F$, they are not isomorphic, but their norms are isometric.

This construction was briefly investigated for the first time in Andrew Steele’s PhD thesis [11]. We improved and corrected most of their results, and added many new ones.

2. Preliminaries

2.1. Non-Associative Algebras

Throughout the paper, F is a field of characteristic, not 2 or 3. An algebra over F is an F-vector space A together with an F-bilinear map $A\times A\to A$, $(x,y)\mapsto x·y$, denoted simply by juxtaposition of $xy$, the multiplication of A. An algebra A is unital if there exists an element in A, denoted by 1, such that $1x=x1=x$ for all $x\in A$.

A non-associative algebra $A\ne 0$ is called a division algebra if for any $a\in A$, $a\ne 0$, the left multiplication with a, $L_a\left(x\right)=ax$, and the right multiplication with a, $R_a\left(x\right)=xa$, are bijective. We will only consider unital finite-dimensional algebras, which implies that A is a division algebra if and only if A has no zero divisors. Define $[x,y,z]=\left(xy\right)z-x\left(yz\right)$. The subalgebras ${\mathrm{Nuc}}_l\left(A\right)=\{x\in A|[x,A,A]=0\}$, ${\mathrm{Nuc}}_m\left(A\right)=\{x\in A|[A,x,A]=0\}$, and ${\mathrm{Nuc}}_r\left(A\right)=\{x\in A|[A,A,x]=0\}$ of A are called the left, middle, and right nuclei of A, $\mathrm{Nuc}\left(A\right)=\{x\in A|[x,A,A]=[A,x,A]=[A,A,x]=0\}$ is called the nucleus of A. The center of A is defined as $C\left(A\right)=\{x\in A|xy=yx\mathrm{for}\mathrm{all}y\in A\}\cap \mathrm{Nuc}(A)$ [14]. All algebras we consider will be unital.

A non-associative unital algebra J is called a cubic Jordan algebra over F, if J is a Jordan algebra, i.e., $xy=yx$ and $\left(x^{2}y\right)x=x^2\left(yx\right)$ for all $x,y\in J$, and if its generic minimal polynomial has degree three. Given an associative algebra A over F, its multiplication written simply by juxtaposition, we can define a Jordan algebra over F denoted by $A^{+}$ on the F-vector space underlying the algebra A via $x·y=\frac{1}{2}(xy+yx)$. A Jordan algebra J is called special, if it is a subalgebra of $A^{+}$ for some associative algebra A over F; otherwise, J is exceptional. An exceptional Jordan algebra is called an Albert algebra.

The following easy observation is included for the sake of the reader:

Lemma 1. 

Let A be an associative algebra over F such that $A^{+}$ is a division algebra. Then, A is a divison algebra.

Proof. 

Suppose that $xy=0$ for some $x,y\in A$. Then, $\left(yx\right)·\left(yx\right)=y\left(xy\right)x=0$, and since $A^{+}$ is a division algebra, we obtain $yx=0$. This implies that $x·y=\frac{1}{2}(xy+yx)=0.$ Using again that $A^{+}$ is a division algebra, we deduce that $x=0$ or $y=0$. □

A non-associative cyclic algebra $(K/F,\sigma ,c)$ of degree m over F is an m-dimensional K-vector space $(K/F,\sigma ,c)=K\oplus Kz\oplus K{z}^2\oplus \cdots \oplus K{z}^{m-1},$ with multiplication given by the relations $z^m=c,zl=\sigma \left(l\right)z$ for all $l\in K$. The algebra $(K/F,\sigma ,c)$ is a unital F-central algebra and associative if and only if $c\in F^{\times }$. The algebra $(K/F,\sigma ,c)$ is a division algebra for all $c\in F^{\times }$, such that $c^s\notin N_{K/F}\left(K^{\times }\right)$ for all s which are prime divisors of m, $1\le s\le m-1$. If $c\in K\setminus F$, then $(K/F,\sigma ,c)$ is a division algebra for all $c\in K\setminus F$ such that 1, $c,\cdots ,c^{m-1}$ are linearly independent over F [10]. If m is prime, then $(K/F,\sigma ,c)$ is a division algebra for all $c\in K\setminus F$.

2.2. Cubic Maps

Let V and W be two finite-dimensional vector spaces over F. A trilinear map $M:V\times V\times V\to W$ is called symmetric if $M(x,y,z)$ is invariant under all permutations of its variables. A map $M:V\to W$ is called a cubic map over F, if $M\left(ax\right)=a^{3}M\left(x\right)$ for all $a\in F$, $x\in V$, and if the associated map $M:V\times V\times V\to W$ defined by

$$M(x,y,z)=\frac{1}{6}(M(x+y+z)-M(x+y)-M(x+z)-M(y+z)+M\left(x\right)+M\left(y\right)+M\left(z\right))$$

is a (symmetric) F-trilinear map. We canonically identify symmetric trilinear maps $M:V\times V\times V\to W$ with the corresponding cubic maps $M:V\to W$.

A cubic map $M:V\to F$ is called a cubic form and a trilinear map $M:V\times V\times V\to F$ a trilinear form over F. A cubic map is called non-degenerate if $v=0$ is the only vector such that $M(v,v_2,,v_3)=0$ for all $v_i\in V$. A cubic map M is called anisotropic if $M\left(x\right)=0$ implies that $x=0$; otherwise, it is isotropic. For a non-associative algebra A over F together with a non-degenerate cubic form $M:A\to F$, M is called multiplicative, if $M\left(vw\right)=M\left(v\right)M\left(w\right)$ for all $v,w\in A$.

2.3. Associative Cubic Algebras

(cf. for Instance [1,15] (Chapter C.4)) Let A be a unital separable associative algebra over F with cubic norm $N_A:A\to F$. Let $x,y\in A$ and let Z be an indeterminate. The linearization $N_A(x+Zy)=N_A\left(x\right)+Z{N}_A(x;y)+Z^2{N}_A(y;x)+Z^3{N}_A\left(y\right)$ of $N_A$, i.e., the coefficient of Z in the above expansion, is quadratic in x and linear in y, and is denoted by $N_A(x;y)$. Indeed, we have

$$N_A(x+Zx)=N_A\left((1+Z)x\right)={(1+Z)}^3{N}_A\left(x\right)=(1+3Z+3Z^2+Z^3)N_A\left(x\right),$$

so $N_A(1;1)=3N_A\left(1\right)=3$. Linearize $N_A(x;y)$ to obtain a symmetric trilinear map $N_A:A\times A\times A\to F$, $N_A(x,y,z)=N_A(x+z;y)-N_A(x;y)-N_A(z;y).$ We define

$$\begin{array}{cc}\hfill T_A\left(x\right)& =N_A(1;x),\hfill \\ \hfill T_A(x,y)& =T_A\left(x\right)T_A\left(y\right)-N_A(1,x,y),\hfill \\ \hfill S_A\left(x\right)& =N_A(x;1),\hfill \\ \hfill x^{♯}& =x^2-T_A\left(x\right)x+S_A\left(x\right)1,\hfill \end{array}$$

for all $x,y\in A$. We call $x^{♯}$ the adjoint of x, and define the sharp map $♯:A\times A\to A$, $x♯y={(x+y)}^{♯}-x^{♯}-y^{♯}$ as the linearization of the adjoint. We observe that $T_A\left(1\right)=S_A\left(1\right)=3$. Since the trilinear map $N_A(x,y,z)$ is symmetric,

$$T_A(x,y)=T_A(y,x)$$

(1)

for all $x,y\in A$.

The algebra A is called an (associative) cubic algebra (respectively, called an algebra of degree three in [15] (p. 490)), if the following three axioms are satisfied for all $x,y\in A$:

$$\begin{array}{cc}\hfill & x^3-T_A\left(x\right)x^2+S_A\left(x\right)x-N_A\left(x\right)1=0(degree 3 identity),\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill & T_A(x^{♯},y)=N_A(x;y)(trace-sharp formula),\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill & T_A(x,y)=T_A\left(xy\right)(trace-product formula).\hfill \end{array}$$

(4)

For the rest of Section 2.3, we assume that A is a separable cubic algebra over F with cubic norm $N_A:A\to F$. Note that (2) is equivalent to the condition that

$$x{x}^{♯}=x^{♯}x=N_A\left(x\right)1,$$

(5)

and combining (1) with (4) gives

$$T_A\left(xy\right)=T_A\left(yx\right).$$

(6)

An element $x\in A$ is invertible if and only if $N_A\left(x\right)\ne 0$. The inverse of $x\in A$ is $x^{-1}=N_A{\left(x\right)}^{-1}{x}^{♯}$. It can be shown that

$${\left(xy\right)}^{♯}=y^{♯}{x}^{♯}$$

(7)

for all $x,y\in A$. Notice that

$$T_A\left(x^{♯}\right)=T_A(x^{♯},1)=N_A(x;1)=S_A\left(x\right),$$

(8)

using (3) and (4). We also have $S_A\left(x\right)=T_A\left(x^{♯}\right)=T_A\left(x^2\right)-T_A{\left(x\right)}^2+3S_A\left(x\right)$, so

$$2S_A\left(x\right)=T_A{\left(x\right)}^2-T_A\left(x^2\right).$$

(9)

A straightforward calculation shows that

$$x♯y=2(x·y)-T_A\left(x\right)y-T_A\left(y\right)x+(T_A\left(x\right)T_A\left(y\right)-T_A(x·y))1$$

(10)

for all $x,y\in A$. In particular,

$$x·y=\frac{1}{2}(xy+yx)=\frac{1}{2}(x♯y+T_A\left(x\right)y+T_A\left(y\right)x-(T_A\left(x\right)T_A\left(y\right)-T_A(x,y))1)$$

for all $x,y\in A$ and by employing (5) and the adjoint identity in A, we see that the norm $N_A$ satisfies the relation

$$N_A\left(x^{♯}\right)=N_A{\left(x\right)}^2.$$

(11)

$A^{+}$ satisfies the adjoint identity

$${\left(x^{♯}\right)}^{♯}=N_A\left(x\right)x$$

(12)

for all $x\in A$. By (11), we have $N_A\left(x^{♯}\right)1=x^{♯}{\left(x^{♯}\right)}^{♯}=x^{♯}{N}_A\left(x\right)x=N_A{\left(x\right)}^{2}1.$ For $x,y\in A$, we define the operators $U_x:A\to A$, $U_x\left(y\right)=T_A(x,y)x-x^{♯}♯y$ and $U_{x,y}:A\to A$, $U_{x,y}\left(z\right)=U_{x+y}\left(z\right)-U_x\left(z\right)-U_y\left(z\right).$ Then, we have $x·y=\frac{1}{2}{U}_{x,y}\left(1\right)$ for all $x,y\in A$ and

$$xyx=T_A(x,y)x-x^{♯}♯y,$$

(13)

Hence, $U_x\left(y\right)=xyx$ for all $x,y\in A^{\times }$.

Define

$$x\times y=\frac{1}{2}(x♯y),$$

and

$$\overline{x}=\frac{1}{2}(T_A\left(x\right)1-x)$$

for $x,y\in A$. (Note that some literature does not include the factor $\frac{1}{2}$ in the definition of ×, e.g., [16]). By (10), we then have

$$x\times y=x·y-\frac{1}{2}{T}_A\left(x\right)y-\frac{1}{2}{T}_A\left(y\right)x+\frac{1}{2}(T_A\left(x\right)T_A\left(y\right)-T_A(x·y))1$$

for all $x,y\in A$; hence,

$$x\times x=x^2-T_A\left(x\right)x+\frac{1}{2}(T_A{\left(x\right)}^2-T_A\left(x^2\right))=x^{♯},$$

(14)

using (9).

2.4. The First Tits Construction

Let A be a separable cubic associative algebra over F with norm $N_A$, trace $T_A$ and adjoint map ♯. Let $\mu \in F^{\times }$ and define the F-vector space $J=J(A,\mu )=A_0\oplus A_1\oplus A_2$, where $A_i=A$ for $i=0,1,2$. Then, $J(A,\mu )$ together with the multiplication

$$\begin{array}{cc}\hfill & (x_0,x_1,x_2)(y_0,y_1,y_2)\hfill \\ & =(x_0·y_0+\overline{x_1{y}_2}+\overline{y_1{x}_2},\overline{x_0}{y}_1+\overline{y_0}{x}_1+{\mu }^{-1}(x_2\times y_2),x_2\overline{y_0}+y_2\overline{x_0}+\mu (x_1\times y_1))\hfill \end{array}$$

becomes a separable cubic Jordan algebra over F. $J(A,\mu )$ is called a first Tits construction. $A^{+}$ is a subalgebra of $J(A,\mu )$ by canonically identifying it with $A_0$. If A is a cubic etale algebra, then $J(A,\mu )\cong D^{+}$ for with D an associative cyclic algebra D of degree three. If A is a central simple algebra of degree three then $J(A,\mu )$ is an Albert algebra.

We define the cubic norm form $N:J(A,\mu )\to F$, the trace $T:J(A,\mu )\to F$, and the quadratic map $♯:J(A,\mu )\to J(A,\mu )$ (the adjoint) by

$$\begin{array}{cc}\hfill N\left((x_0,x_1,x_2)\right)& =N_A\left(x_0\right)+\mu N_A\left(x_1\right)+{\mu }^{-1}{N}_A\left(x_2\right)-T_A\left(x_0{x}_1{x}_2\right)\hfill \\ \hfill T\left((x_0,x_1,x_2)\right)& =T_A\left(x_0\right),\hfill \\ \hfill {(x_0,x_1,x_2)}^{♯}& =(x_0^{♯}-x_1{x}_2,{\mu }^{-1}{x}_2^{♯}-x_0{x}_1,\mu x_1^{♯}-x_2{x}_0).\hfill \end{array}$$

The intermediate quadratic form $S:J(A,\mu )\to F$, $S\left(x_0\right)=N(x;1)$, linearizes to a map $S:J(A,\mu )\times J(A,\mu )\to F$. The sharp map $♯:J(A,\mu )\times J(A,\mu )\to J(A,\mu )$ is the linearization $x♯y={(x+y)}^{♯}-x^{♯}-y^{♯}$ of the adjoint. For every $x=(x_0,x_1,x_2)\in J(A,\mu )$, we have $x♯1=T\left(x\right)1-x$ and

$$x♯y=(x_0♯y_0-x_1{y}_2-y_1{x}_2,{\mu }^{-1}(x_2♯y_2)-x_0{y}_1-y_0{x}_1,\mu (x_1♯y_1)-x_2{y}_0-y_2{x}_0)$$

for all $x=(x_0,x_1,x_2),y=(y_0,y_1,y_2)\in J(A,\mu )$. We define the trace symmetric bilinear form $T:J(A,\mu )\times J(A,\mu )\to F$, $T(x,y)=T_A\left(x_0{y}_0\right)+T_A\left(x_1{y}_2\right)+T_A\left(x_2{y}_1\right).$ Then, for all $x,y\in J(A,\mu )$, we have

$$T(x,y)=T\left(xy\right).$$

(15)

Remark 1. 

$(N,♯,1)$ is a cubic form with adjoint and base point $(1,0,0)$ on $J(A,\mu )$ which makes $J(A,\mu )$ into a cubic Jordan algebra $J(N,♯,1)$.

3. The Generalized First Tits Construction $\mathbf{J}(\mathbf{A},\mu )$

Let A be a separable associative cubic algebra over F with norm $N_A$, trace $T_A$ and adjoint map ♯.

We now generalize the first Tits construction by choosing the scalar $\mu \in A^{\times }$. Then, the F-vector space $J(A,\mu )=A_0\oplus A_1\oplus A_2$, where again $A_i=A$ for $i=0,1,2$, becomes a unital non-associative algebra over F together with the multiplication given by

$$\begin{array}{cc}\hfill & (x_0,x_1,x_2)(y_0,y_1,y_2)\hfill \\ & =(x_0·y_0+\overline{x_1{y}_2}+\overline{y_1{x}_2},\overline{x_0}{y}_1+\overline{y_0}{x}_1+{\mu }^{-1}(x_2\times y_2),x_2\overline{y_0}+y_2\overline{x_0}+\mu (x_1\times y_1)).\hfill \end{array}$$

The algebra $J(A,\mu )$ is called a generalized first Tits construction. The special Jordan algebra $A^{+}$ is a subalgebra of $J(A,\mu )$ by canonically identifying it with $A_0$. If $\mu \in F^{\times }$, then $J(A,\mu )$ is the first Tits construction from Section 2.4.

We define a (generalized) cubic norm map $N:J(A,\mu )\to A$, a (generalized) trace $T:J(A,\mu )\to F$, and a quadratic map $♯:J(A,\mu )\to J(A,\mu )$ via

$$\begin{array}{cc}\hfill N\left((x_0,x_1,x_2)\right)& =N_A\left(x_0\right)+\mu N_A\left(x_1\right)+{\mu }^{-1}{N}_A\left(x_2\right)-T_A\left(x_0{x}_1{x}_2\right)\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill T\left((x_0,x_1,x_2)\right)& =T_A\left(x_0\right),\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill {(x_0,x_1,x_2)}^{♯}& =(x_0^{♯}-x_1{x}_2,{\mu }^{-1}{x}_2^{♯}-x_0{x}_1,\mu x_1^{♯}-x_2{x}_0).\hfill \end{array}$$

(18)

Put $♯:J(A,\mu )\times J(A,\mu )\to J(A,\mu )$, $x♯y={(x+y)}^{♯}-x^{♯}-y^{♯}$; then, it can be verified by a direct computation that

$$x♯y=(x_0♯y_0-x_1{y}_2-y_1{x}_2,{\mu }^{-1}(x_2♯y_2)-x_0{y}_1-y_0{x}_1,\mu (x_1♯y_1)-x_2{y}_0-y_2{x}_0)$$

for all $x=(x_0,x_1,x_2),y=(y_0,y_1,y_2)\in J(A,\mu )$. We also define a symmetric F-bilinear form $T:J(A,\mu )\times J(A,\mu )\to F$ via $T(x,y)=T_A\left(x_0{y}_0\right)+T_A\left(x_1{y}_2\right)+T_A\left(x_2{y}_1\right).$

The quadratic form $S_A:A\to F$, $S_A\left(x_0\right)=N_A(x;1)$, linearizes to $S_A:A\times A\to F$, and we have $S_A\left(x_0\right)=T_A\left(x_0^{♯}\right)$ for all $x_0\in A$. We extend $S_A$ to $J(A,\mu )$ by defining the quadratic map $S:J(A,a)\to A$, $S\left(x\right)=N(x;1)$. As in the classical case, we obtain:

Theorem 1. 

(i) 

[11] (Proposition 5.2.2) For all $x\in J(A,\mu )$, we have $S\left(x\right)=T\left(x^{♯}\right)$ and the linearization $S:J(A,\mu )\times J(A,\mu )\to A$ satisfies

$$S(x,y)=T\left(x\right)T\left(y\right)-T(x,y)$$

for all $y\in J(A,\mu )$.

(ii) 

[11] (Lemma 5.2.3) For all $x,y\in J(A,\mu )$, we have $T(x,Y)=T\left(xy\right)$.

(iii) 

[11] (Lemma 5.2.3) For all $x\in J(A,\mu )$, we have $x♯1=T\left(x\right)1-x$.

Proof. 

(i)

Let $x=(x_0,x_1,x_2),y=(y_0,y_1,y_2)\in J(A,a)$, then

$$\begin{array}{cc}\hfill N(x;y)=& N_A(x_0;y_0)+\mu N_A(x_1;y_1)+{\mu }^{-1}{N}_A(x_2;y_2)\hfill \\ & -T_A\left(x_0{x}_1{y}_2\right)-T_A\left(x_0{y}_1{x}_2\right)-T_A\left(y_0{x}_1{x}_2\right),\hfill \end{array}$$

and since $S\left(x\right)=N(x;1)$ we obtain $S\left(x\right)=N_A(x_0;1)-T_A\left(x_1{x}_2\right)=S_A\left(x_0\right)-T_A\left(x_1{x}_2\right).$ On the other hand,

$$T\left(x^{♯}\right)=T_A(x_0^{♯}-x_1{x}_2)=T_A\left(x_0^{♯}\right)-T_A\left(x_1{x}_2\right)=S_A\left(x_0\right)-T_A\left(x_1{x}_2\right)=S\left(x\right).$$

We have $S_A(x_0,y_0)=T_A\left(x_0\right)T_A\left(y_0\right)-T_A(x_0,y_0)$ for all $x_0,y_0\in A$. Linearizing S gives $S(x,y)=S_A(x_0,y_0)-T_A\left(x_1{y}_2\right)-T_A\left(y_1{x}_2\right)=T_A\left(x_0\right)T_A\left(y_0\right)-T_A(x_0,y_0)-T_A\left(x_1{y}_2\right)-T_A\left(y_1{x}_2\right)=T\left(x\right)T\left(y\right)-T(x,y)$ using the definitions of $T_A\left(x_i\right)$ and $T_A(x_i,y_i)$ and the fact that $T_A(x_0,y_0)=T_A\left(x_0{y}_0\right)$.

(ii)

Let $x=(x_0,x_1,x_2),y=(y_0,y_1,y_2)\in J(A,\mu )$. Since $T_A$ is linear, we obtain

$$\begin{array}{cc}\hfill T\left(xy\right)& =T_A(x_0·y_0)+T_A\left(\overline{x_1{y}_2}\right)+T_A\left(\overline{y_1{x}_2}\right)\hfill \\ & =\frac{1}{2}(T_A\left(x_0{y}_0\right)+T_A\left(y_0{x}_0\right))+\frac{1}{2}(T_A\left(x_1{y}_2\right)T_A\left(1\right)-T_A\left(x_1{y}_2\right))\hfill \\ & +\frac{1}{2}(T_A\left(y_1{x}_2\right)T_A\left(1\right)-T_A\left(y_1{x}_2\right)).\hfill \end{array}$$

By (6) we obtain $T_A\left(x_0{y}_0\right)=T_A\left(y_0{x}_0\right)$ and $T_A\left(y_1{x}_2\right)=T_A\left(x_2{y}_1\right)$. Since we have $T_A\left(1\right)=3$ we obtain $T\left(xy\right)=T_A\left(x_0{y}_0\right)+T_A\left(x_1{y}_2\right)+T_A\left(x_2{y}_1\right)=T(x,y).$

(iii)

Let $x=(x_0,x_1,x_2)\in J(A,\mu )$. By (10), we have $x_0♯1=T_A\left(x_0\right)1-x_0$. Thus, $x♯1=(x_0♯1,-x_1,-x_2)=T\left(x\right)1-x.$

□

Theorem 2. 

Let $\mu \in A^{\times }$, and let $x,y\in J(A,\mu )$. Then,

(i) 

$x^{♯}=x^2-T\left(x\right)x+S\left(x\right)1$,

(ii) 

$S\left(x\right)=T\left(x^{♯}\right)$,

(iii) 

$T(x\times y)=\frac{1}{2}(T\left(x\right)T\left(y\right)-T\left(xy\right))$.

Note that these are relations that also hold for a cubic form with adjoint and base point $(N,♯,1)$ [15,17].

Proof. 

Let $x=(x_0,x_1,x_2),y=(y_0,y_1,y_2)\in J(A,\mu )$.

(i)

We have that $x^2-T\left(x\right)x+S\left(x\right)1={(x_0,x_1,x_2)}^2-T_A\left(x_0\right)(x_0,x_1,x_2)+(S_A\left(x_0\right)1-T_A\left(x_1{x}_2\right)1,0,0)=(x_0^2-T_A\left(x_0\right)x_0+S_A\left(x_0\right)1+2\overline{x_1{x}_2}-T_A\left(x_1{x}_2\right)1,$${\mu }^{-1}{x}_2^{♯}+2\overline{x_0}{x}_1-T_A\left(x_0\right)x_1,\mu x_1^{♯}+2x_2\overline{x_0}-T_A\left(x_0\right)x_2)=(x_0^{♯}-x_1{x}_2,{\mu }^{-1}{x}_2^{♯}-x_0{x}_1,\mu x_1^{♯}-x_2\overline{x_0})=x^{♯}.$

(ii)

As for the classical construction,

$$T\left(x^{♯}\right)=T_A(x_0^{♯}-x_1{x}_2)=T_A\left(x_0^{♯}\right)-T_A\left(x_1{x}_2\right)=S_A\left(x_0\right)-T_A\left(x_1{x}_2\right)=S\left(x\right).$$

(iii)

Since $x\times y=\frac{1}{2}(x♯y)=\frac{1}{2}(x_0♯y_0-x_1{y}_2-y_1{x}_2,{\mu }^{-1}(x_2♯y_2)-x_0{y}_1-y_0{x}_1,\mu (x_1♯y_1)-x_2{y}_0-y_2{x}_0)$, we obtain $T(x\times y)=T_A(x_0\times y_0)-\frac{1}{2}{T}_A\left(x_1{y}_2\right)-\frac{1}{2}{T}_A\left(y_1{x}_2\right)$$=\frac{1}{2}(T_A\left(x_0\right)T_A\left(y_0\right)-T_A\left(x_0{y}_0\right)-T_A\left(x_1{y}_2\right)-T_A\left(y_1{x}_2\right))=\frac{1}{2}(T\left(x\right)T\left(y\right)-T\left(xy\right)).$

□

Define operators $U_x,U_{x,y}:J(A,\mu )\to J(A,\mu )$ via

$$U_x\left(y\right)=T(x,y)x-x^{♯}♯y,U_{x,y}\left(z\right)=U_{x+y}\left(z\right)-U_x\left(z\right)-U_y\left(z\right)$$

for all $z\in J(A,\mu )$.

Proposition 1. 

(cf. [11] (Proposition 5.2.4) without factor $\frac{1}{2}$ because of slightly different terminology) For all $x,y\in J(A,\mu )$, we have $xy=\frac{1}{2}{U}_{x,y}\left(1\right)$.

This generalizes the classical setup. Our proof is different to the one of [11] (Proposition 5.2.4), which also proves this result without the factor $\frac{1}{2}$ because of the slightly different definition of the multiplication.

Proof. 

We find that $U_x\left(1\right)=T(x,1)x-x^{♯}♯1=T\left(x\right)x-T\left(x^{♯}\right)1+x^{♯}$; in the second equality, we have used Theorem 1 and the fact that $T(x,1)=T\left(x\right)$ by Theorem 1. So

$$\begin{array}{cc}\hfill U_{x,y}\left(1\right)& =U_{x+y}\left(1\right)-U_x\left(1\right)-U_y\left(1\right)\hfill \\ & =T(x+y)(x+y)-T\left({(x+y)}^{♯}\right)1+{(x+y)}^{♯}-T\left(x\right)x+T\left(x^{♯}\right)1-x^{♯}\hfill \\ & -T\left(y\right)y+T\left(y^{♯}\right)1-y^{♯}\hfill \\ & =T\left(x\right)y+T\left(y\right)x+x♯y-T(x♯y)1.\hfill \end{array}$$

We look at the first component of $xy$ and $U_{x,y}\left(1\right)$: let $x=(x_0,x_1,x_2)$ and $y=(y_0,y_1,y_2)$. Then, the first component of $U_{x,y}\left(1\right)=T\left(x\right)y+T\left(y\right)x+x♯y-T(x♯y)1$ is

$$T_A\left(x_0\right)y_0+T_A\left(y_0\right)x_0+x_0♯y_0-x_1{y}_2-y_1{x}_2-T_A(x_0♯y_0-x_1{y}_2-y_1{x}_2)1.$$

(19)

Using (10), the linearity of $T_A$ and (6), we obtain—after some simplification—that (19) is equal to

$$2(x_0·y_0)+T_A\left(x_1{y}_2\right)-x_1{y}_2+T_A\left(y_1{x}_2\right)-y_1{x}_2=2(x_0·y_0)+2\overline{x_1{y}_2}+2\overline{y_1{x}_2}.$$

This is equal to 2 times the first component of $xy$. Now, we look at the second component of $xy$ and $U_{x,y}\left(1\right)$: the second component of $U_{x,y}\left(1\right)=T\left(x\right)y+T\left(y\right)x+x♯y-T(x♯y)1$ is

$$T_A\left(x_0\right)y_1+T_A\left(y_0\right)x_1+{\mu }^{-1}(x_2♯y_2)-x_0{y}_1-y_0{x}_1=2\overline{x_0}{y}_1+2\overline{y_0}{x}_1+2{\mu }^{-1}(x_2\times y_2).$$

This is precisely equal to 2 times the second component of $xy$. Finally, the third component of $2xy$ and $U_{x,y}\left(1\right)$ are equal, too. The third component of $U_{x,y}\left(1\right)=T\left(x\right)y+T\left(y\right)x+x♯y-T(x♯y)1$ is

$$T_A\left(x_0\right)y_2+T_A\left(y_0\right)x_2+\mu (x_1♯y_1)-x_2{y}_0-y_2{x}_0=2x_2\overline{y_0}+2y_2\overline{x_0}+2\mu (x_1\times y_1).$$

This is precisely equal to 2 times the third component of $xy$. □

Theorem 3. 

If $\mu \in A^{\times }$ and $A\ne F$, then ${Nuc}_l\left(J(A,\mu )\right)={Nuc}_r\left(J(A,\mu )\right)=F$.

Proof. 

Let $(x_0,x_1,x_2)\in {Nuc}_l\left(J(A,\mu )\right)$, then

$$(x_0,x_1,x_2)\left[(0,1,0)(0,0,1)\right]=\left[(x_0,x_1,x_2)(0,1,0)\right](0,0,1)$$

implies that

$$(x_0,x_1,x_2)=(\overline{\overline{x_0}},{\mu }^{-1}\left(\overline{\mu \overline{x_1}}\right),\overline{\overline{x_2}}),$$

that means $x_0=\overline{\overline{x_0}}$ and $x_2=\overline{\overline{x_2}}$. Using the definition of $\overline{\overline{x_0}}$, we obtain $\overline{\overline{x_0}}=\frac{1}{4}(T_A\left(x_0\right)+x_0)$, so $x_0=\frac{1}{4}(T_A\left(x_0\right)+x_0)$. Thus, $x_0=\frac{1}{3}{T}_A\left(x_0\right)\in F$. Furthermore, since $x_2=\overline{\overline{x_2}}$, we find in a similar way that $x_2=\frac{1}{3}{T}_A\left(x_2\right)\in F$. Next, since $x=(x_0,x_1,x_2)\in {Nuc}_l\left(J(A,\mu )\right)$, we have that

$$(x_0,x_1,x_2)\left[(0,0,1)(0,1,0)\right]=\left[(x_0,x_1,x_2)(0,0,1)\right](0,1,0).$$

This implies that

$$(x_0,x_1,x_2)=(\overline{\overline{x_0}},\overline{\overline{x_1}},\mu \left(\overline{{\mu }^{-1}\overline{x_2}}\right)),$$

and so $x_1=\overline{\overline{x_1}}$. We now find in a similar way that $x_1=\frac{1}{3}{T}_A\left(x_1\right)\in F$, thus ${Nuc}_l\left(J(A,\mu )\right)\subseteq \{(x_0,x_1,x_2)\in J|x_0,x_1,x_2\in F\}$. Let $x=(x_0,x_1,x_2)\in {Nuc}_l\left(J(A,\mu )\right)$, and let $a\in A\setminus F$. Then

$$(x_0,x_1,x_2)\left[(0,0,1)(0,a,0)\right]=\left[(x_0,x_1,x_2)(0,0,1)\right](0,a,0)$$

which implies that

$$(x_0·\overline{a},\overline{\overline{a}}{x}_1,x_2\overline{\overline{a}})=(\overline{a\overline{x_0}},\overline{\overline{x_1}}a,\mu ({\mu }^{-1}\overline{x_2}\times a)),$$

and so $\overline{\overline{a}}{x}_1=\overline{\overline{x_1}}a$. Assume towards a contradiction that $x_1\ne 0$. Since $x_1\in F$, this implies that $x_1$ is invertible and $\overline{\overline{x_1}}=x_1$. Thus, the condition $\overline{\overline{a}}{x}_1=\overline{\overline{x_1}}a$ yields $a=\overline{\overline{a}}$, and so $a=\frac{1}{3}{T}_A\left(a\right)\in F$ which is a contradiction. Next, since $(x_0,x_1,x_2)\in {Nuc}_l\left(J(A,\mu )\right)$, we know that

$$(x_0,x_1,x_2)\left[(0,1,0)(0,0,a)\right]=\left[(x_0,x_1,x_2)(0,1,0)\right](0,0,a)$$

which implies that

$$(x_0·\overline{a},\overline{\overline{a}}{x}_1,x_2\overline{\overline{a}})=(\overline{\overline{x_0}a},{\mu }^{-1}(\mu \overline{x_1}\times a),a\overline{\overline{x_2}}),$$

and so $x_2\overline{\overline{a}}=a\overline{\overline{x_2}}$. Assume towards a contradiction that $x_2\ne 0$. Then, since $x_2\in F$, $x_2$ is invertible and $\overline{\overline{x_2}}=x_2$. Thus, the condition $x_2\overline{\overline{a}}=a\overline{\overline{x_2}}$ yields $a=\overline{\overline{a}}$, and so $a=\frac{1}{3}{T}_A\left(a\right)\in F$ which is a contradiction. Therefore, $x=(x_0,0,0)$, $x_{0}1\in F$ which shows that ${Nuc}_l\left(J(A,\mu )\right)=F$.

Let $(x_0,x_1,x_2)\in {Nuc}_r\left(J(A,\mu )\right)$. Then,

$$(0,0,1)\left[(0,1,0)(x_0,x_1,x_2)\right]=\left[(0,0,1)(0,1,0)\right](x_0,x_1,x_2)$$

implies that

$$(\overline{\overline{x_0}},{\mu }^{-1}\left(\overline{\mu \overline{x_1}}\right),\overline{\overline{x_2}})=(x_0,x_1,x_2).$$

Hence, $x_0=\overline{\overline{x_0}}$ and $x_2=\overline{\overline{x_2}}$. Using the definition of $\overline{\overline{x_0}}$, we find that $\overline{\overline{x_0}}=\frac{1}{4}(T_A\left(x_0\right)+x_0)$, so the condition $x_0=\overline{\overline{x_0}}$ gives that $x_0=\frac{1}{4}(T_A\left(x_0\right)+x_0)$. Thus, $x_0=\frac{1}{3}{T}_A\left(x_0\right)\in F$. Furthermore, since $x_2=\overline{\overline{x_2}}$, we find in a similar way that $x_2=\frac{1}{3}{T}_A\left(x_2\right)\in F$. Next, since $x=(x_0,x_1,x_2)\in {Nuc}_r\left(J(A,\mu )\right)$, we have that

$$(0,1,0)\left[(0,0,1)(x_0,x_1,x_2)\right]=\left[(0,1,0)(0,0,1)\right](x_0,x_1,x_2).$$

This implies that

$$(\overline{\overline{x_0}},\overline{\overline{x_1}},\mu \left(\overline{{\mu }^{-1}\overline{x_2}}\right))=(x_0,x_1,x_2),$$

and thus $x_1=\overline{\overline{x_1}}$. We find in a similar way that $x_1=\frac{1}{3}{T}_A\left(x_1\right)\in F$, i.e. ${Nuc}_r\left(J(A,\mu )\right)\subseteq \{(x_0,x_1,x_2)\in J|x_0,x_1,x_2\in F\}$.

Let $x=(x_0,x_1,x_2)\in {Nuc}_r\left(J(A,\mu )\right)$, and let $a\in A\setminus F$. Then, $(0,a,0)\left[(0,0,1)(x_0,x_1,x_2)\right]$$=\left[(0,a,0)(0,0,1)\right](x_0,x_1,x_2)$ which implies that

$$(\overline{a\overline{x_0}},\overline{\overline{x_1}}a,\mu (a\times {\mu }^{-1}\overline{x_2}))=(\overline{a}·x_0,\overline{\overline{a}}{x}_1,x_2\overline{\overline{a}});$$

therefore, $\overline{\overline{a}}{x}_1=\overline{\overline{x_1}}a$. Assume towards a contradiction that $x_1\ne 0$. Then, since $x_1\in F$, $x_1$ is invertible and $\overline{\overline{x_1}}=x_1$. Thus, the condition $\overline{\overline{a}}{x}_1=\overline{\overline{x_1}}a$ yields $a=\overline{\overline{a}}$, and so $a=\frac{1}{3}{T}_A\left(a\right)\in F$ which is a contradiction. Next, since $(x_0,x_1,x_2)\in {Nuc}_r\left(J(A,\mu )\right)$, we know that

$$(0,0,a)\left[(0,1,0)(x_0,x_1,x_2)\right]=\left[(0,0,a)(0,1,0)\right](x_0,x_1,x_2)$$

which implies that

$$(\overline{\overline{x_0}a},{\mu }^{-1}(a\times \mu \overline{x_1}),a\overline{\overline{x_2}})=(\overline{a}·x_0,\overline{\overline{a}}{x}_1,x_2\overline{\overline{a}}),$$

and so $x_2\overline{\overline{a}}=a\overline{\overline{x_2}}$. Assume towards a contradiction that $x_2\ne 0$. Then, since $x_2\in F$, $x_2$ is invertible and $\overline{\overline{x_2}}=x_2$. Thus, the condition $x_2\overline{\overline{a}}=a\overline{\overline{x_2}}$ yields $a=\overline{\overline{a}}$, and so $a=\frac{1}{3}{T}_A\left(a\right)\in F$ which is a contradiction. Therefore, $x=(x_0,0,0)=x_{0}1\in F$ which shows the assertion. □

Theorem 4. 

Let $A\ne F$ be a central simple division algebra of degree three and $\mu \in A^{\times }$. Then, ${Nuc}_m\left(J(A,\mu )\right)=F$.

Proof. 

Let $x=(x_0,x_1,x_2)\in {Nuc}_m\left(J(A,\mu )\right)$, and let $y_0\notin C\left(A\right)$. Then, there exists $z_0\in A$ such that $y_0{z}_0\ne z_0{y}_0$. Since $(x_0,x_1,x_2)\in {Nuc}_m\left(J(A,\mu )\right)$, we know that

$$(y_0,0,0)\left[(x_0,x_1,x_2)(z_0,0,0)\right]=\left[(y_0,0,0)(x_0,x_1,x_2)\right](z_0,0,0)$$

which implies that

$$(y_0·(x_0·z_0),\overline{y_0}\left(\overline{z_0}{x}_1\right),\left(x_2\overline{z_0}\right)\overline{y_0})=((y_0·x_0)·z_0,\overline{z_0}\left(\overline{y_0}{x}_1\right),\left(x_2\overline{y_0}\right)\overline{z_0}).$$

Comparing the second and third components yields

$$\begin{array}{cc}\hfill \overline{y_0}\left(\overline{z_0}{x}_1\right)& =\overline{z_0}\left(\overline{y_0}{x}_1\right),\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill \left(x_2\overline{z_0}\right)\overline{y_0}& =\left(x_2\overline{y_0}\right)\overline{z_0}.\hfill \end{array}$$

(21)

Now, assume towards a contradiction that $x_1\ne 0$. Since A is a division algebra, $x_1$ is invertible. Since A is associative, (20) implies that $\overline{y_0}\overline{z_0}=\overline{z_0}\overline{y_0}$. By definition, this yields

$$(T_A\left(y_0\right)1-y_0)(T_A\left(z_0\right)1-z_0)=(T_A\left(z_0\right)1-z_0)(T_A\left(y_0\right)1-y_0).$$

Hence, $y_0{z}_0=z_0{y}_0$ which is a contradiction. Now, in a similar way we assume towards a contradiction that $x_2\ne 0$. Then, since A is a division algebra, $x_2$ is invertible. Since A is associative, (21) implies again that $\overline{y_0}\overline{z_0}=\overline{z_0}\overline{y_0}$. Hence, $y_0{z}_0=z_0{y}_0$ which is a contradiction. Next, since $x=(x_0,0,0)\in {Nuc}_m\left(J(A,\mu )\right)$, we also have that

$$(0,1,0)\left[(x_0,0,0)(0,0,y_2)\right]=\left[(0,1,0)(x_0,0,0)\right](0,0,y_2)$$

for each $y_2\in A$. This implies $(\overline{y_2\overline{x_0}},0,0)=(\overline{\overline{x_0}{y}_2},0,0),$ and so $\overline{y_2\overline{x_0}}=\overline{\overline{x_0}{y}_2}$. By definition, this means that

$$\frac{1}{2}{T}_A\left(\overline{y_2}{x}_0\right)1-\frac{1}{2}\overline{y_2}{x}_0=\frac{1}{2}{T}_A\left(x_0\overline{y_2}\right)1-\frac{1}{2}{x}_0\overline{y_2}.$$

(22)

We know that $T_A\left(\overline{y_2}{x}_0\right)=T_A\left(x_0\overline{y_2}\right)$ (see (6)), and so (22) gives that $\overline{y_2}{x}_0=x_0\overline{y_2}$. By using the definition of $\overline{y_2}$, this implies that $y_2{x}_0=x_0{y}_2$. Hence, $x_0\in C\left(A\right)$. Therefore, $x=(x_0,0,0)=x_{0}1\in C\left(A\right).$ Since $F\subseteq {Nuc}_m\left(J(A,\mu )\right)$ this implies the assertion if A is a central simple division algebra. □

Theorem 5. 

([18] (Chapter IX, Section 12), [15] (Chapter C.5)) For $\mu \in F^{\times }$, $J(A,\mu )$ is a division algebra if and only if $\mu \notin N_A\left(A^{\times }\right)$ and A is a division algebra, if and only if N is anisotropic.

The general situation is much harder to figure out and we were only able to obtain some obvious necessary conditions:

Theorem 6. 

Let $\mu \in A^{\times }$.

(i) 

If $J(A,\mu )$ is a division algebra, then $\mu \notin N_A\left(A^{\times }\right)$ and A is a division algebra.

(ii) 

Let A be a division algebra over F. If 1, μ, ${\mu }^2$ are linearly independent over F then N is anisotropic.

(iii) 

If N is anisotropic then A is a division algebra and $\mu \notin N_A\left(A^{\times }\right)$.

(iv) 

Let $0\ne x=(x_0,x_1,x_2)\in J(A,\mu )$. Then, $x^{♯}=0$ implies that A has zero divisors, or A is a division algebra and $\mu \in N_A\left(A^{\times }\right)$.

Proof. 

(i)

Suppose that $J(A,\mu )$ is a division algebra, then so is $A^{+}$ and thus A (Lemma 1). Assume towards a contradiction that $\mu =N_A\left(x_0\right)1$ for some $x_0\in A^{\times }$. Then, $\mu \in F^{\times }$ and $J(A,\mu )$ is not a division algebra by Theorem 5. Hence, $\mu \notin N_A\left(A\right)1$.

(ii)

Since A is a division algebra, $N_A$ is anisotropic. So, let $N\left((x_0,x_1,x_2)\right)=0$; then, the assumption means that $N\left(x_0\right)=0$, which implies that $x_0=0$. This immediately means that $x_1=x_2=0$, too.

(iii)

If N is anisotropic, then so is $N_A$; so, A is clearly a division algebra. Moreover, $\mu \notin N_A\left(A^{\times }\right)$ by Theorem 5.

(iv)

Let $0\ne x=(x_0,x_1,x_2)\in J(A,\mu )$. Then, $x^{♯}=0$ implies that

$$\begin{array}{cc}\hfill x_0^{♯}& =x_1{x}_2\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill {\mu }^{-1}{x}_2^{♯}& =x_0{x}_1\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill \mu x_1^{♯}& =x_2{x}_0.\hfill \end{array}$$

(25)

We can now multiply (23) (resp. (24), (25)) by $x_0$ (resp. $x_2$, $x_1$) on the right and left to obtain two new equations. Additionally, using the fact that $N_A\left(x_i\right)=x_i{x}_i^{♯}=x_i^{♯}{x}_i$ for all $i=0,1,2$, we obtain the following six equations:

$$\begin{array}{cc}\hfill N_A\left(x_0\right)& =x_1{x}_2{x}_0\hfill \\ \hfill {\mu }^{-1}{N}_A\left(x_2\right)& =x_0{x}_1{x}_2\hfill \\ \hfill \mu N_A\left(x_1\right)& =x_2{x}_0{x}_1\hfill \end{array}\begin{array}{cc}\hfill N_A\left(x_0\right)& =x_0{x}_1{x}_2\hfill \\ \hfill {\mu }^{-1}{N}_A\left(x_2\right)& =x_2{x}_0{x}_1\hfill \\ \hfill \mu N_A\left(x_1\right)& =x_1{x}_2{x}_0.\hfill \end{array}$$

(26)

These imply that $N_A\left(x_0\right)={\mu }^{-1}{N}_A\left(x_2\right)=\mu N_A\left(x_1\right)$. This means that either $N_A\left(x_0\right)={\mu }^{-1}{N}_A\left(x_2\right)=\mu N_A\left(x_1\right)=0$ and so $N_A$ is isotropic, or $N_A\left(x_0\right)={\mu }^{-1}{N}_A\left(x_2\right)=\mu N_A\left(x_1\right)\ne 0$ and $N_A$ is anisotropic. In the later case, $x_0,x_1,x_2$ are all invertible in A, $N_A\left(x_i\right)\ne 0$ for all $i=0,1,2$ and it follows that $\mu \in N_A\left(A^{\times }\right)$. This proves the assertion.

□

In other words: If A is a division algebra and $\mu \notin N_A\left(A^{\times }\right)$, $0\ne x=(x_0,x_1,x_2)\in J(A,\mu )$, then $x^{♯}\ne 0$. Note that (iv) was a substantial part of the classical result that if $\mu \in F^{\times }$, $\mu \notin N_A\left(A^{\times }\right)$ and A is a division algebra, then N is anisotropic. What is missing in order to generalize this result to the generalized first Tits construction is the adjoint identity ${\left(x^{♯}\right)}^{♯}=N\left(x\right)x$. This identity only holds in very special cases—see Lemma 4 below. It would be of course desirable to have conditions on when (or if at all) $J(A,\mu )$ is a division algebra.

4. Some More Identities

Lemma 2. 

Let $x=(x_0,x_1,x_2)$, $y=(y_0,y_1,y_2),z=(z_0,z_1,z_2)\in J(A,\mu )$ be such that one of $x_1,y_1,z_1$ is equal to zero and one of $x_2,y_2,z_2$ is equal to zero. Then, $T(x\times y,z)=T(x,y\times z)$.

Proof. 

We find that

$$\begin{array}{ccc}\hfill T(x\times y,z)& =& \frac{1}{2}{T}_A((x_0♯y_0)z_0-x_1{y}_2{z}_0-y_1{x}_2{z}_0)\hfill \\ & & +\frac{1}{2}{T}_A({\mu }^{-1}(x_2♯y_2)z_2-x_0{y}_1{z}_2-y_0{x}_1{z}_2)\hfill \\ & & +\frac{1}{2}{T}_A(\mu (x_1♯y_1)z_1-x_2{y}_0{z}_1-y_2{x}_0{z}_1)\hfill \end{array}$$

(27)

and

$$\begin{array}{cc}\hfill T(x,y\times z)=& \frac{1}{2}{T}_A(x_0(y_0♯z_0)-x_0{y}_1{z}_2-x_0{z}_1{y}_2)\hfill \\ & +\frac{1}{2}{T}_A(x_1\mu (y_1♯z_1)-x_1{y}_2{z}_0-x_1{z}_2{y}_0)\hfill \\ & +\frac{1}{2}{T}_A(x_2{\mu }^{-1}(y_2♯z_2)-x_2{y}_0{z}_1-x_2{z}_0{y}_1).\hfill \end{array}$$

(28)

Using the definitions, we can show that $T_A\left((x_0♯y_0)z_0\right)=T_A\left(x_0(y_0♯z_0)\right)$. Furthermore, since one of $x_1,y_1,z_1$ is equal to zero, we have that $T_A\left(\mu (x_1♯y_1)z_1\right)=0=T_A\left(x_1\mu (y_1♯z_1)\right)$. Finally, since one of $x_2,y_2,z_2$ is equal to zero, $T_A\left({\mu }^{-1}(x_2♯y_2)z_2\right)=0=T_A\left(x_2{\mu }^{-1}(y_2♯z_2)\right)$. Therefore, applying these equalities and using (6), we deduce that (27) and (28) are equal, so $T(x\times y,z)=T(x,y\times z)$. □

We know that $x{x}^{♯}=x^{♯}x=N\left(x\right)1$ holds for all $x\in J(A,\mu )$ if $\mu \in F^{\times }$. We now show for which $x\in J(A,\mu )$ we still obtain $x{x}^{♯}=x^{♯}x=N\left(x\right)1$:

Lemma 3. 

Let $\mu \in A^{\times }$ and suppose that $x\in J(A,\mu )$, such that one of the following holds:

(i) 

$x=(x_0,0,x_2)\in J(A,\mu )$, $x_0\mu =\mu x_0$ and $N_A\left(x_2\right)=0$.

(ii) 

$x=(x_0,x_1,0)\in J(A,\mu )$, $x_1\mu =\mu x_1$ and $N_A\left(x_1\right)=0$.

Then, we have

$$x{x}^{♯}=x^{♯}x=N\left(x\right)1.$$

(29)

Moreover, assume that one of the following holds:

(iii) 

$x=(x_0,0,x_2)\in J(A,\mu )$, $x_0\mu =\mu x_0$ and $N_A\left(x_2\right)\ne 0$.

(iv) 

$x=(x_0,x_1,0)\in J(A,\mu )$, $x_1\mu =\mu x_1$ and $N_A\left(x_1\right)\ne 0$.

Then, $x{x}^{♯}=x^{♯}x=N\left(x\right)1$ if and only if $\mu \in F^{\times }$.

Proof. 

Let $x=(x_0,x_1,x_2)\in J(A,\mu )$, and let $x{x}^{♯}=(a_0,a_1,a_2)$; then, $x^{♯}=(x_0^{♯}-x_1{x}_2,{\mu }^{-1}{x}_2^{♯}$$-x_0{x}_1,\mu x_1^{♯}-x_2{x}_0).$ Thus, we have

$$\begin{array}{cc}\hfill a_0=& \frac{1}{2}(x_0{x}_0^{♯}-x_0{x}_1{x}_2+x_0^{♯}{x}_0-x_1{x}_2{x}_0)\hfill \\ & +\frac{1}{2}(T_A(x_1\mu x_1^{♯}-x_1{x}_2{x}_0)1-x_1\mu x_1^{♯}+x_1{x}_2{x}_0)\hfill \\ & +\frac{1}{2}(T_A({\mu }^{-1}{x}_2^{♯}{x}_2-x_0{x}_1{x}_2)1-{\mu }^{-1}{x}_2^{♯}{x}_2+x_0{x}_1{x}_2),\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill a_1& =\frac{1}{2}(T_A\left(x_0\right)-x_0)({\mu }^{-1}{x}_2^{♯}-x_0{x}_1)+\frac{1}{2}(T_A\left(x_0^{♯}-x_1{x}_2\right)-x_0^{♯}+x_1{x}_2)x_1\hfill \\ & +{\mu }^{-1}(x_2\times (\mu x_1^{♯}-x_2{x}_0)),\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill a_2& =\frac{1}{2}{x}_2(T_A(x_0^{♯}-x_1{x}_2)-x_0^{♯}+x_1{x}_2)+\frac{1}{2}(\mu x_1^{♯}-x_2{x}_0)(T_A\left(x_0\right)-x_0)\hfill \\ & +\mu (x_1\times ({\mu }^{-1}{x}_2^{♯}-x_0{x}_1))\hfill \end{array}$$

(32)

by the definition of the multiplication on $J(A,\mu )$.

(i)

If $x_1=0$ and $N_A\left(x_2\right)=0$, using the fact that $x_i{x}_i^{♯}=x_i^{♯}{x}_i=N_A\left(x_i\right)1$ for all $i=0,1,2$ (see (5)), (30) simplifies to $a_0=N_A\left(x_0\right)1=N\left(x\right)$. Since we have $x_0{\mu }^{-1}={\mu }^{-1}{x}_0$, (31) gives that $a_1=\frac{1}{2}{\mu }^{-1}(T_A\left(x_0\right)-x_0)x_2^{♯}-{\mu }^{-1}(x_2\times \left(x_2{x}_0\right)).$ By (10),

$$x_2^{♯}♯x_0=x_2^{♯}{x}_0+x_0{x}_2^{♯}-T_A\left(x_2^{♯}\right)x_0-T_A\left(x_0\right)x_2^{♯}+(T_A\left(x_2^{♯}\right)T_A\left(x_0\right)-T_A\left(x_2^{♯}{x}_0\right))1.$$

(33)

Using the fact that $T_A\left(x_2^{♯}\right)=S_A\left(x_2\right)$ (by (8)) on the right-hand side of (33), we further obtain after some simplification that

$$x_2^{♯}♯x_0=x_2^2{x}_0-T_A\left(x_2\right)x_2{x}_0+x_0{x}_2^{♯}-T_A\left(x_0\right)x_2^{♯}-T_A\left(x_2^2{x}_0\right)1+T_A\left(x_2{x}_0\right)1.$$

(34)

Now, combining (13) with (34) yields $T_A\left(x_0\right)x_2^{♯}-x_0{x}_2^{♯}=x_2^2{x}_0+x_2{x}_0{x}_2-T_A\left(x_2\right)x_2{x}_0-T_A\left(x_2{x}_0\right)x_2+(T_A\left(x_2\right)T_A\left(x_2{x}_0\right)-T_A\left(x_2^2{x}_0\right))1=2(x_2\times \left(x_2{x}_0\right))$, so $x_2\times \left(x_2{x}_0\right)$$=\frac{1}{2}(T_A\left(x_0\right)x_2^{♯}-x_0{x}_2^{♯})$. Hence, (4) implies $a_1=0$. For $a_2$, (32) yields $a_2=\frac{1}{2}{x}_2(T_A\left(x_0^{♯}\right)-x_0^{♯})-\frac{1}{2}{x}_2(T_A\left(x_0\right)x_0-x_0^2).$ Then, using the definition of $x_0^{♯}$ and the fact that $2S_A\left(x_0\right)=T_A{\left(x_0\right)}^2-T_A\left(x_0^2\right)$, we find that $T_A\left(x_0^{♯}\right)-x_0^{♯}=T_A\left(x_0\right)x_0-x_0^2$. Therefore, $a_2=0$.

(ii)

In this case, we have $x_2=0$, $x_1\mu =\mu x_1$ and $N_A\left(x_1\right)=0$. So, (30) simplifies to $a_0=N_A\left(x_0\right)1=N\left(x\right)$. For $a_1$, (31) simplifies to $a_1=-\frac{1}{2}(T_A\left(x_0\right)x_0-x_0^2)x_1+\frac{1}{2}(T_A\left(x_0^{♯}\right)-x_0^{♯})x_1.$ Then, in a similar way to how we found $a_2$ in (i), we find here that $a_1=0$. For $a_2$, (32) simplifies to $a_2=\frac{1}{2}\mu x_1^{♯}(T_A\left(x_0\right)-x_0)-\mu (x_1\times \left(x_0{x}_1\right)).$ We now find in a similar way to how we found $a_1$ in (i) that $a_2=0$.

To prove that the claimed equivalence holds assuming (iii) or (iv), we only need to show the forward direction since we know from the classical first Tits construction that the reverse direction holds:

(iii)

Here, (30) yields $a_0=N_A\left(x_0\right)1+\frac{1}{2}(T_A\left({\mu }^{-1}\right)N_A\left(x_2\right)-{\mu }^{-1}{N}_A\left(x_2\right));$ thus, $x{x}^{♯}=N\left(x\right)1=(N_A\left(x_0\right)1+{\mu }^{-1}{N}_A\left(x_2\right))1$ gives that $N_A\left(x_0\right)1+\frac{1}{2}(T_A\left({\mu }^{-1}\right)$$N_A\left(x_2\right)-{\mu }^{-1}{N}_A\left(x_2\right))=a_0=N_A\left(x_0\right)1+{\mu }^{-1}{N}_A\left(x_2\right).$ Therefore, we have ${\mu }^{-1}=\frac{1}{3}{T}_A\left({\mu }^{-1}\right)\in F^{\times }$, so $\mu \in F^{\times }$.

(iv)

In this case, (30) yields $a_0=N_A\left(x_0\right)1+\frac{1}{2}(T_A\left(\mu \right)N_A\left(x_1\right)-\mu N_A\left(x_1\right));$ thus, $x{x}^{♯}=N\left(x\right)1=(N_A\left(x_0\right)1+\mu N_A\left(x_1\right))1$ yields $N_A\left(x_0\right)1+\frac{1}{2}(T_A\left(\mu \right)N_A\left(x_1\right)-\mu N_A\left(x_1\right))=a_0=N_A\left(x_0\right)1+\mu N_A\left(x_1\right).$ Therefore, we obtain $\mu =\frac{1}{3}{T}_A\left(\mu \right)\in F^{\times }$. The proof that $x^{♯}x=N\left(x\right)1$ is performed similarly.

□

Corollary 1. 

Let $\mu \in A^{\times }$. Suppose that $x\in J(A,\mu )$ satisfies $N\left(x\right)\ne 0$, and assume that one of the following holds:

(i) 

$x=(x_0,x_1,0)\in J(A,\mu )$, $x_1\mu =\mu x_1$ and $N_A\left(x_1\right)=0$.

(ii) 

$x=(x_0,0,x_2)\in J(A,\mu )$, $x_0\mu =\mu x_0$ and $N_A\left(x_2\right)=0$.

Then, x is invertible in $J(A,\mu )$ with $x^{-1}=N{\left(x\right)}^{-1}{x}^{♯}$.

Proof. 

Let $\mu \in A^{\times }$ and suppose that $x\in J(A,\mu )$ satisfies (i) or (ii); then, $x{x}^{♯}=x^{♯}x=N\left(x\right)1.$ Since $F=C\left(J\right(A,\mu \left)\right)$ this yields the assertion. □

In particular, if N is anisotropic, then every $0\ne xx=(x_0,x_1,0)\in J(A,\mu )$ in (i) or (ii) is of the type $x=(x_0,0,0)\in J(A,\mu )$, i.e., lies in A; so, this result then becomes trivial.

Corollary 2. 

Let $\mu \in A^{\times }$ and suppose that $x\in J(A,\mu )$, such that one of the following holds:

(i) 

$x=(x_0,0,x_2)\in J(A,\mu )$, $x_0\mu =\mu x_0$ and $N_A\left(x_2\right)=0$.

(ii) 

$x=(x_0,x_1,0)\in J(A,\mu )$, $x_1\mu =\mu x_1$ and $N_A\left(x_1\right)=0$.

Then, we have

$$x^3-T\left(x\right)x^2+S\left(x\right)x-N\left(x\right)1=0.$$

Proof. 

Using the fact that $x^{♯}=x^2-T\left(x\right)x+S\left(x\right)1$ from Theorem 2 (i), we have that $x^3-T\left(x\right)x^2+S\left(x\right)x-N\left(x\right)1=0$ if and only if $x{x}^{♯}=x^{♯}x=N\left(x\right)1$. Thus, the result now follows as a consequence of Lemma 3. □

Theorem 7. 

The identity $x{x}^{♯}=x^{♯}x=N\left(x\right)1$ holds for all $x\in J(A,\mu )$ if and only if $\mu \in F^{\times }$.

Proof. 

If $\mu \in F^{\times }$, then $x{x}^{♯}=x^{♯}x=(N\left(x\right),0,0)$ for all $x\in J(A,\mu )$. Conversely, suppose that $x{x}^{♯}=x^{♯}x=(N\left(x\right),0,0)$ holds for all $x\in J(A,\mu )$. Take $x=(0,1,0)$. Then, $x^{♯}=(0,0,\mu )$, and so

$$x{x}^{♯}=(\overline{\mu },0,0)=(\frac{1}{2}(T_A\left(\mu \right)1-\mu ),0,0).$$

We also know that by definition, $N\left(x\right)=\mu N_A\left(1\right)=\mu$, so the condition $x{x}^{♯}=(N\left(x\right),0,0)$ gives that $\mu =\frac{1}{2}(T_A\left(\mu \right)1-\mu )$. Hence $\mu =\frac{1}{3}{T}_A\left(\mu \right)1\in F^{\times }.$ □

We know that the adjoint identity ${\left(x^{♯}\right)}^{♯}=N\left(x\right)x$ holds for all $x\in J(A,\mu )$, if $\mu \in F^{\times }$ [15] (Chapter C.4). In the general construction, it holds only in very special cases:

Lemma 4. 

Let $\mu \in A^{\times }$ and suppose that $x\in J(A,\mu )$, such that one of the following holds:

(i) 

$x=(0,x_1,0)\in J(A,\mu )$ and $N_A\left(x_1\right)=0$.

(ii) 

$x=(x_0,x_1,0)\in J(A,\mu )$ and $N_A\left(x_1\right)=0$ and $x_1\mu =\mu x_1$.

(iii) 

$x=(x_0,0,x_2)\in J(A,\mu )$ and $N_A\left(x_2\right)=0$ and $x_0\mu =\mu x_0$.

Then, we have ${\left(x^{♯}\right)}^{♯}=N\left(x\right)x$.

Moreover, if one of the following holds:

(iv) 

$x=(x_0,x_1,0)\in J(A,\mu )$, $N_A\left(x_1\right)\ne 0$, $x_0\mu =\mu x_0$ and $x_1\mu =\mu x_1$.

(v) 

$x=(x_0,0,x_2)\in J(A,\mu )$, $N_A\left(x_2\right)\ne 0$, $x_0\mu =\mu x_0$ and $x_2\mu =\mu x_2$.

Then, ${\left(x^{♯}\right)}^{♯}=N\left(x\right)x$ for all $x\in J(A,\mu )$ if and only if $N_A\left(\mu \right)={\mu }^3$.

Proof. 

Let $x=(x_0,x_1,x_2)\in J(A,\mu )$ and ${\left(x^{♯}\right)}^{♯}=(a_0,a_1,a_2)$. By definition, $x^{♯}=(x_0^{♯}-x_1{x}_2,{\mu }^{-1}{x}_2^{♯}-x_0{x}_1,\mu x_1^{♯}-x_2{x}_0),$ so $a_0={(x_0^{♯}-x_1{x}_2)}^{♯}-({\mu }^{-1}{x}_2^{♯}-x_0{x}_1)(\mu x_1^{♯}-x_2{x}_0)$. Now, using (9) and (10), it is easy to show that

$$\begin{array}{cc}\hfill {(x_0^{♯}-x_1{x}_2)}^{♯}& ={(x_0^{♯}-x_1{x}_2)}^2-T_A(x_0^{♯}-x_1{x}_2)(x_0^{♯}-x_1{x}_2)+S_A(x_0^{♯}-x_1{x}_2)\hfill \\ & ={\left(x_0^{♯}\right)}^{♯}-x_0^{♯}♯\left(x_1{x}_2\right)+{\left(x_1{x}_2\right)}^{♯}.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill a_0& ={(x_0^{♯}-x_1{x}_2)}^{♯}-({\mu }^{-1}{x}_2^{♯}-x_0{x}_1)(\mu x_1^{♯}-x_2{x}_0)\hfill \\ & ={\left(x_0^{♯}\right)}^{♯}-x_0^{♯}♯\left(x_1{x}_2\right)+{\left(x_1{x}_2\right)}^{♯}-{\mu }^{-1}{x}_2^{♯}\mu x_1^{♯}+{\mu }^{-1}{x}_2^{♯}{x}_2{x}_0+x_0{x}_1\mu x_1^{♯}-x_0{x}_1{x}_2{x}_0.\hfill \end{array}$$

(35)

Similarly, we find that

$$\begin{array}{cc}\hfill a_1& ={\mu }^{-1}{(\mu x_1^{♯}-x_2{x}_0)}^{♯}-(x_0^{♯}-x_1{x}_2)({\mu }^{-1}{x}_2^{♯}-x_0{x}_1)\hfill \\ & ={\mu }^{-1}({\left(\mu x_1^{♯}\right)}^{♯}-\left(\mu x_1^{♯}\right)♯\left(x_2{x}_0\right)+{\left(x_2{x}_0\right)}^{♯})\hfill \\ & -x_0^{♯}{\mu }^{-1}{x}_2^{♯}+x_0^{♯}{x}_0{x}_1+x_1{x}_2{\mu }^{-1}{x}_2^{♯}-x_1{x}_2{x}_0{x}_1\hfill \end{array}$$

(36)

and

$$\begin{array}{cc}\hfill a_2& =\mu {({\mu }^{-1}{x}_2^{♯}-x_0{x}_1)}^{♯}-(\mu x_1^{♯}-x_2{x}_0)(x_0^{♯}-x_1{x}_2)\hfill \\ & =\mu ({\left({\mu }^{-1}{x}_2^{♯}\right)}^{♯}-\left({\mu }^{-1}{x}_2^{♯}\right)♯\left(x_0{x}_1\right)+{\left(x_0{x}_1\right)}^{♯})\hfill \\ & -\mu x_1^{♯}{x}_0^{♯}+\mu x_1^{♯}{x}_1{x}_2+x_2{x}_0{x}_0^{♯}-x_2{x}_0{x}_1{x}_2.\hfill \end{array}$$

(37)

(i)

Here, $x_0=x_2=0$; therefore, (35) implies $a_0=0$ and (36) gives that $a_1={\mu }^{-1}{\left(\mu x_1^{♯}\right)}^{♯}={\mu }^{-1}{N}_A\left(x_1\right)x_1{\mu }^{♯}=0=N\left(x\right)x_1.$ Finally, (37) gives that $a_2=0$ as required.

(ii)

Since $x_2=0$, $x_1\mu =\mu x_1$ and $N_A\left(x_1\right)=0$, we find by (35) that $a_0={\left(x_0^{♯}\right)}^{♯}+x_0\mu x_1{x}_1^{♯}=N_A\left(x_0\right)x_0+x_0\mu N_A\left(x_1\right)=N\left(x\right)x_0.$ Now, (36) gives that $a_1={\mu }^{-1}{\left(\mu x_1^{♯}\right)}^{♯}+x_0^{♯}{x}_0{x}_1={\mu }^{-1}{N}_A\left(x_1\right)x_1{\mu }^{♯}+N_A\left(x_0\right)x_1=N\left(x\right)x_1,$ and by (37), we obtain $a_2=\mu {\left(x_0{x}_1\right)}^{♯}-\mu x_1^{♯}{x}_0^{♯}=0=N\left(x\right)0.$

(iii)

Since $x_1=0$ and $N_A\left(x_2\right)=0$, (35) yields $a_0={\left(x_0^{♯}\right)}^{♯}+{\mu }^{-1}{x}_2^{♯}{x}_2{x}_0=N_A\left(x_0\right)x_0+{\mu }^{-1}{N}_A\left(x_2\right)x_0=N\left(x\right)x_0.$ Now, since $x_0{\mu }^{-1}={\mu }^{-1}{x}_0$, we have that $x_0^{♯}{\mu }^{-1}={\mu }^{-1}{x}_0^{♯}$, so (36) gives that $a_1={\mu }^{-1}{\left(x_2{x}_0\right)}^{♯}-{\mu }^{-1}{x}_0^{♯}{x}_2^{♯}=0=N\left(x\right)0.$ Finally, (37) gives that $a_2=\mu {\left({\mu }^{-1}{x}_2^{♯}\right)}^{♯}+x_2{x}_0{x}_0^{♯}=\mu N_A\left(x_2\right)x_2{\left({\mu }^{-1}\right)}^{♯}+N_A\left(x_0\right)x_2=N\left(x\right)x_2.$

(iv)

Since $x_2=0$, $x_0\mu =\mu x_0$ and $x_1\mu =\mu x_1$, (35) yields $a_0={\left(x_0^{♯}\right)}^{♯}+\mu x_0{x}_1{x}_1^{♯}=N_A\left(x_0\right)x_0+\mu N_A\left(x_1\right)x_0=N\left(x\right)x_0.$ Now, (37) gives that $a_2=\mu {\left(x_0{x}_1\right)}^{♯}-\mu x_1^{♯}{x}_0^{♯}=0$. Finally, (36) gives that $a_1={\mu }^{-1}{\left(\mu x_1^{♯}\right)}^{♯}+x_0^{♯}{x}_0{x}_1={\mu }^{-1}{N}_A\left(x_1\right)x_1{\mu }^{♯}+N_A\left(x_0\right)x_1.$ Thus, $a_1=N\left(x\right)x_1$ if and only if ${\mu }^{-1}{N}_A\left(x_1\right)x_1{\mu }^{♯}+N_A\left(x_0\right)x_1=N\left(x\right)x_1$, which occurs if and only if ${\mu }^{-1}{N}_A\left(x_1\right)x_1{\mu }^{♯}=\mu N_A\left(x_1\right)x_1$. Since $N_A\left(x_1\right)\ne 0$ and $x_1{\mu }^{♯}={\mu }^{♯}{x}_1$, this occurs if and only if ${\mu }^{♯}{x}_1={\mu }^2{x}_1$. Finally, $N_A\left(x_1\right)\ne 0$ implies that $x_1$ is invertible, so ${\mu }^{♯}{x}_1={\mu }^2{x}_1$ if and only if $N_A\left(\mu \right)=\mu {\mu }^{♯}={\mu }^3$.

(v)

Since $x_1=0$, (35) yields $a_0={\left(x_0^{♯}\right)}^{♯}+{\mu }^{-1}{x}_2^{♯}{x}_2{x}_0=N\left(x\right)x_0.$ Furthermore, since $x_0$ commutes with $\mu$, $x_0^{♯}$ commutes with $\mu$. So, $x_0^{♯}{\mu }^{-1}={\mu }^{-1}{x}_0^{♯}$. Hence, (36) gives that $a_1={\mu }^{-1}{\left(x_2{x}_0\right)}^{♯}-{\mu }^{-1}{x}_0^{♯}{x}_2^{♯}=0=N\left(x\right)0.$ Finally, (37) yields $a_2=\mu {\left({\mu }^{-1}{x}_2^{♯}\right)}^{♯}+x_2{x}_0{x}_0^{♯}=\mu N_A\left(x_2\right)x_2{\left({\mu }^{-1}\right)}^{♯}+N_A\left(x_0\right)x_2.$ Thus, $a_2=N\left(x\right)x_2$ if and only if $\mu N_A\left(x_2\right)x_2{\left({\mu }^{-1}\right)}^{♯}+N_A\left(x_0\right)x_2=N\left(x\right)x_2$, which occurs if and only if $\mu N_A\left(x_2\right)x_2{\left({\mu }^{-1}\right)}^{♯}$$={\mu }^{-1}{N}_A\left(x_2\right)x_2$. Since $N_A\left(x_2\right)\ne 0$ and $x_2{\left({\mu }^{-1}\right)}^{♯}={\left({\mu }^{-1}\right)}^{♯}{x}_2$, this occurs if and only if ${\left({\mu }^{-1}\right)}^{♯}{x}_2={\mu }^{-2}{x}_2$. Finally, $N_A\left(x_2\right)\ne 0$ implies that $x_2$ is invertible, so ${\left({\mu }^{-1}\right)}^{♯}{x}_2={\mu }^{-2}{x}_2$ if and only if $N_A\left({\mu }^{-1}\right)={\mu }^{-1}{\left({\mu }^{-1}\right)}^{♯}={\mu }^{-3}$. This is equivalent to $N_A\left(\mu \right)={\mu }^3$.

□

Proposition 2. 

Let A be a central simple algebra over F. Then, ${\left(x^{♯}\right)}^{♯}=N\left(x\right)$ for all $x\in J(A,\mu )$ if and only if $\mu \in F^{\times }$.

Proof. 

Let $\mu \in F^{\times }$ then by Lemma 4, the adjoint identity holds for all $x\in J(A,\mu )$. Suppose now that the adjoint identity holds for all $x\in J(A,\mu )$. Let $x=(x_0,1,0)\in J(A,\mu )$ for some $x_0\in A$. Then, $x^{♯}=(x_0^{♯},-x_0,\mu )$ and so

$${\left(x^{♯}\right)}^{♯}=({\left(x_0^{♯}\right)}^{♯}+x_0\mu ,{\mu }^{-1}{\mu }^{♯}+{\left(x_0^{♯}\right)}^{♯}{x}_0,0).$$

(38)

Furthermore, $N\left(x\right)=N_A\left(x_0\right)1+\mu$. Since the adjoint identity holds by assumption, we see that by using (38),

$$({\left(x_0^{♯}\right)}^{♯}+x_0\mu ,{\mu }^{-1}{\mu }^{♯}+{\left(x_0^{♯}\right)}^{♯}{x}_0,0)=(N_A\left(x_0\right)x_0+\mu x_0,N_A\left(x_0\right)+\mu ,0).$$

(39)

We know that ${\left(x_0^{♯}\right)}^{♯}=N_A\left(x_0\right)x_0$ for all $x_0\in A$ by (12), and so by comparing the first components of (39), we find that $x_0\mu =\mu x_0$ for all $x_0\in A$. Hence, $\mu \in C\left(A\right)$, and since A is a central simple algebra by assumption, $\mu \in F^{\times }$. □

If $\mu \in F^{\times }$, then the norm N permits Jordan composition, i.e. $N\left(U_{x}y\right)=N_A{\left(x\right)}^{2}N\left(y\right)$ for all $x,y\in J(A,a)$. The following result is a corrected version of [11] (Theorem 5.2.5), and a weak generalization of the Jordan composition for $\mu \in A^{\times }\setminus F$:

Theorem 8. 

Let $x=(x_0,0,0)\in A$, $y=(y_0,y_1,y_2)\in J(A,\mu )$ and suppose that one of the following holds:

(i) 

$T_A\left(y_0{y}_1{y}_2\right)=T_A\left(N_A\left(y_0\right)x_0^{♯}{y}_1{y}_2{x}_0^{♯}\right).$

(ii) 

$y_0{y}_1{y}_2=N_A\left(y_0\right)x_0^{♯}{y}_1{y}_2{x}_0^{♯}$.

(iii) 

$y_i=0$ for some $i=0,1,2$.

Then, $N\left(U_x\left(y\right)\right)=N{\left(x\right)}^{2}N\left(y\right)$.

Proof. 

Using the definitions, we see that $T(x,y)=T_A\left(x_0{y}_0\right)$ and $x^{♯}♯y=(x_0^{♯}♯y_0,-x_0^{♯}{y}_1,-y_2{x}_0^{♯})$. So $U_x\left(y\right)=T(x,y)x-x^{♯}♯y=(U_{x_0}\left(y_0\right),x_0^{♯}{y}_1,y_2{x}_0^{♯}).$ This yields

$$\begin{array}{cc}\hfill N\left(U_x\left(y\right)\right)& =N_A\left(U_{x_0}\left(y_0\right)\right)1+\mu N_A\left(x_0^{♯}{y}_1\right)+{\mu }^{-1}{N}_A\left(y_2{x}_0^{♯}\right)-T_A\left(U_{x_0}\left(y_0\right)x_0^{♯}{y}_1{y}_2{x}_0^{♯}\right)1\hfill \\ \hfill & =N_A{\left(x_0\right)}^2(N_A\left(y_0\right)1+\mu N_A\left(y_1\right)+{\mu }^{-1}{N}_A\left(y_2\right)-T_A\left(N_A\left(y_0\right)x_0^{♯}{y}_1{y}_2{x}_0^{♯}\right)1)\hfill \\ \hfill & =N{\left(x\right)}^2(N\left(y\right)+T_A\left(y_0{y}_1{y}_2\right)1-T_A\left(N_A\left(y_0\right)x_0^{♯}{y}_1{y}_2{x}_0^{♯}\right)1),\hfill \end{array}$$

where in the second equality we have used the fact that $N_A\left(x_0^{♯}\right)=N_A{\left(x_0\right)}^2$, and that $N_A\left(U_{x_0}\left(y_0\right)\right)=N_A{\left(x_0\right)}^2{N}_A\left(y_0\right)$. Therefore, $N\left(U_x\left(y\right)\right)=N{\left(x\right)}^{2}N\left(y\right)$, if and only if $T_A\left(y_0{y}_1{y}_2\right)1=T_A\left(N_A\left(y_0\right)x_0^{♯}{y}_1{y}_2{x}_0^{♯}\right).$

(ii) and (iii) are examples where this is the case. □

Remark 2. 

Let $f:J(A,\mu )\to J(A,\mu )$ be an automorphism. Then,

$$f\left((x_0,x_1,x_2)\right)=f\left((x_0,0,0)\right)+f\left((0,x_1,0)\right)+f\left((0,0,x_2)\right).$$

(40)

Now, for each $x\in A$, we have $f\left((0,\overline{x},0)\right)=f\left((x,0,0)\right)f((0,1,0),$ and $f\left((0,0,\overline{x})\right)=f\left((x,0,0)\right)f\left((0,0,1)\right)$. On the other hand, by using the definition of $\overline{x}$,

$$\begin{array}{cc}\hfill f\left((0,\overline{x},0)\right)& =\frac{1}{2}{Tr}_A\left(x\right)f\left((0,1,0)\right)-\frac{1}{2}f\left((0,x,0)\right),\hfill \\ \hfill f\left((0,0,\overline{x})\right)& =\frac{1}{2}{Tr}_A\left(x\right)f\left((0,0,1)\right)-\frac{1}{2}f\left((0,0,x)\right).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill f\left(\right(0,x,0\left)\right)& =f\left((0,1,0)\right)({Tr}_A\left(x\right)-2f\left((x,0,0)\right)),\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill f\left(\right(0,0,x\left)\right)& =f\left((0,0,1)\right)({Tr}_A\left(x\right)-2f\left((x,0,0)\right)).\hfill \end{array}$$

(42)

So, by (40)–(42), we see that any automorphism of $J(A,\mu )$ is determined by its restriction on $A^{+}$, and its value on $(0,1,0)$ and $(0,0,1)$. Let $f:J(A,\mu )\to J(A,\mu )$ be an automorphism that fixes $A^{+}$; then, ${f|}_{A^{+}}=\tau$ is either an automorphism or an anti-automorphism of A. Moreover, clearly $f\left(\right(1,0,0\left)\right)=(1,0,0)$, so

$$f\left((x_0,x_1,x_2)\right)=(\tau \left(x_0\right),0,0)+(\tau \left(x_1\right),0,0)f\left((0,1,0)\right)+(\tau \left(x_2\right),0,0)f\left((0,0,1)\right).$$

Calculation to try gain some deeper understanding on the automorphisms are tedious and did not lead us anywhere so far.

5. The Nine-Dimensional Non-Associative Algebras $\mathbf{J}(\mathbf{K},\mathbf{\mu })$

Let $K/F$ be a separable cubic field extension with $\mathrm{Gal}(K/F)=\langle \sigma \rangle$, norm $N_K$, and trace $T_K$. For all $x_0\in K$, we have $x_0^{♯}=\sigma \left(x_0\right){\sigma }^2\left(x_0\right)$ and $\overline{x_0}=\frac{1}{2}(\sigma \left(x_0\right)+{\sigma }^2\left(x_0\right))$. Assume $\mu \in K^{\times }$.

Let us compare the first Tits construction $J(K,\mu )$ with the algebra $D^{+}$ for a (perhaps non-associative) cyclic algebra $D=(K/F,\sigma ,\mu )$ over F of degree three. Consider D as a left K-vector space with basis $\{1,z,z^2\}.$ Write $R_x$ for the matrix of right multiplication by $x=x_0+x_{1}z+x_2{z}^2,$$x_i\in K$, with respect to the basis $\{1,z,z^2\}$, then the cubic map $N_D:D\to K$, $N_D\left(x\right)=\det \left(R_x\right)$ (which is the reduced norm of the central simple algebra D if $\mu \in F^{\times }$), is given by

$$N_D\left(x\right)=N_K\left(x_0\right)+\mu N_K\left(x_1\right)+{\mu }^2{N}_K\left(x_2\right)-\mu T_K\left(x_0\sigma \left(x_1\right){\sigma }^2\left(x_2\right)\right).$$

If $N_D$ is anisotropic then D is a division algebra over F. If $\mu \in K\setminus F$, we obtain $N_D\left(lx\right)=N_K\left(l\right)N_D\left(x\right)$ for all $x\in D$, $l\in K$ [11] (Propositions 4.2.2 and 4.2.3).

On the other hand, $J(K,\mu )$ is a nine-dimensional non-associative unital algebra over F with multiplication

$$xy=(x_0·y_0+\overline{x_1{y}_2}+\overline{x_2{y}_1},\overline{x_0}{y}_1+\overline{y_0}{x}_1+{\mu }^{-1}(x_2\times y_2),\overline{x_0}{y}_2+\overline{y_0}{x}_2+\mu (x_1\times y_1))$$

for $x=(x_0,x_1,x_2),y=(y_0,y_1,y_2)\in J(K,\mu )$, cubic norm map

$$N\left((x_0,x_1,x_2)\right)=N_K\left(x_0\right)+\mu N_K\left(x_1\right)+{\mu }^{-1}{N}_K\left(x_2\right)-T_K\left(x_0{x}_1{x}_2\right),$$

and trace $T\left(x\right)=T_K\left(x_0\right)$. Moreover, we have

$$x^{♯}=(\sigma \left(x_0\right){\sigma }^2\left(x_0\right)-x_1{x}_2,{\mu }^{-1}\sigma \left(x_2\right){\sigma }^2\left(x_2\right)-x_0{x}_1,\mu \sigma \left(x_1\right){\sigma }^2\left(x_1\right)-x_2{x}_0).$$

If $\mu \in F^{\times }$, then $D=(K/F,\sigma ,\mu )$ is an associative cyclic algebra over F of degree three and $J(K,\mu )\cong D^{+}$ is a special cubic Jordan algebra. It is well known that the isomorphism $G:D^{+}={(K/F,\sigma ,\mu )}^{+}\to J(K,\mu )$ is given by

$$x_0+x_{1}z+x_2{z}^2\mapsto (x_0,\sigma \left(x_1\right),\mu {\sigma }^2\left(x_2\right)).$$

However, if $\mu \in K\setminus F$, then the map $G:D^{+}\to J(K,\mu )$ is not an algebra isomorphism between ${(K/F,\sigma ,\mu )}^{+}$ and $J(K,\mu )$, where now $(K/F,\sigma ,\mu )$ is a non-associative cyclic algebra, since $\sigma \left(\mu \right)\ne \mu$. However, for $\mu \in K\setminus F$, the map $G:D^{+}\to J(K,\mu )$ still yields an isometry of norms, since

$$\begin{array}{cc}\hfill N((x_0,\sigma \left(x_1\right),\mu {\sigma }^2\left(x_2\right))& =N_K\left(x_0\right)+\mu N_K\left(x_1\right)+{\mu }^{-1}{N}_K\left(x_2\right)-\mu T_K\left(x_0\sigma \left(x_1\right)\sigma \left(x_2\right)\right)\hfill \\ & =N_D\left((x_0,x_1,x_2)\right);\hfill \end{array}$$

hence, the norms of the two nonisomorphic non-associative algebras $D^{+}={(K/F,\sigma ,\mu )}^{+}$ and $J(K,\mu )$ are isometric.

6. Conclusions

We looked at the following canonical question: “what happens if we choose the element $\mu$ that is used in the first Tits construction $J(A,\mu )$ in $A^{\times }$ instead of in $F^{\times }$?” We showed that the basic ingredients for an interesting theory are in place: our new algebras $J(A,\mu )$ carry maps that can be understood as generalizations of the classical norms and traces, and that behave surprisingly similar to the norms and traces of their classical counterparts; we have a function N on $J(A,\mu )$ that extends the cubic norm of A (however, it has values in A), a trace function $T:J(A,\mu )\to F$, and a quadratic map $♯:J(A,\mu )\to J(A,\mu )$. Operations like $x♯y$ can easily be defined. Some of the main identities from the classical setup hold (Theorems 1 and 2), some others hold only for some elements, e.g., Lemmas 2 and 3, Corollaries 1 and 2, but not in general, and some hold if—and only if—$\mu \in F^{\times }$ (Proposition 2, Theorem 7), i.e., they hold only in the classical case.

It seems a hard problem to check when the algebras $J(A,\mu )$ are division algebras. It would also be interesting to compute their automorphisms; however, we expect the automorphism group to be “small”. Here is one indication as to why this is the case: For Albert algebras over fields F of characteristic not 2 or 3, we know that the similarities of their norms are given either by scalar multiplications or the U operators [4]. Using Theorem 8 (iii), we see that for $J(A,\mu )$ with $\mu \in A^{\times }\setminus F$, scalar multiplications still give similarities; the U-operators, however, do not.

Even partial results on automorphisms or similarities could give an insight on what is happening in this general context, and it would be interesting to address questions of whether there are inner automorphisms, whether there are cubic subfields fixed by automorphisms like in the classical case [2], etc.

The fact that two nonisomorphic algebras $D^{+}={(K/F,\sigma ,\mu )}^{+}$ and $J(K,\mu )$ have isometric norms is an example of how rich the structure theory for non-associative algebras really is (Section 4).

This is an exploratory paper, but our results show that the algebras $J(A,\mu )$ obtained via a generalized first Tits construction merit a closer look. As one referee pointed out, they also show the weaknesses of the language that we have at our disposal, which describes highly non-associative structures.

7. Materials and Methods

We used classical methods from algebra.