Sextonions, Zorn matrices, and $\mathbf {e}_{\mathbf{7} \frac{\mathbf{1}}{\mathbf{2}}}$

Abstract

By exploiting suitably constrained Zorn matrices, we present a new construction of the algebra of sextonions (over the algebraically closed field $\mathbb {C}$). This allows for an explicit construction, in terms of Jordan pairs, of the non-semisimple Lie algebra $\mathbf {e}_{\mathbf{7} \frac{\mathbf{1}}{\mathbf{2}}}$, intermediate between $\mathbf {e_7}$ and $\mathbf {e_8}$, as well as of all Lie algebras occurring in the sextonionic row and column of the extended Freudenthal Magic Square.

A Real forms

We use the notations of [51]. From the treatment in [51], a real form of octonions is obtained by taking $\alpha _0^\pm , \alpha _k^\pm \in \mathbf {R}$. The quaternionic subalgebra generated by $\rho ^\pm , \varepsilon _1^\pm$ is obviously a split form with nilpotent $\varepsilon _1^\pm$.

Another real form is obtained by taking complex coefficients with complex conjugation denoted by ‘$*$’ subject to the conditions:

$$\begin{aligned} \alpha _0^- = (\alpha _0^+)^* , \quad \alpha _1^- = -(\alpha _1^+)^* , \quad \alpha _3^- = (\alpha _2^+)^* \end{aligned}$$

(A.1)

Its quaternionic subalgebra, generated by $1, u_7, iu_1, iu_4$, is also split with nilpotent $u_7 + i u_k, k=1,4$. It is equivalent to the one obtained with all real coefficients, which is generated by $1, u_1, iu_4, iu_7$ upon cyclic permutation of the indices 7, 4, 1.

Let us now restrict the Zorn matrix product [51] to the sextonions and introduce the vectors

$$\begin{aligned} E_1 = (1,0,0) \quad E_2 = (0,1,0) \quad E_3 = (0,0,1) \end{aligned}$$

We get:

$$\begin{aligned} \begin{array}{l} \left[ \begin{array}{ll} \alpha _0^+ &{} A^+ \\ A^- &{} \alpha _0^- \end{array}\right] \left[ \begin{array}{ll} \beta _0^+ &{} B^+ \\ B^- &{} \beta _0^- \end{array}\right] \\ \\ \quad =\left[ \begin{array}{ll} \alpha _0^+ \beta _0^+ - \alpha _1^+ \beta _1^- &{} (\alpha _0^+ \beta _1^+ + \beta _0^- \alpha _1^+) E_1^+ \\ (\alpha _0^- \beta _1^- + \beta _0^+ \alpha _1^-)E_1^- &{} \alpha _0^- \beta _0^- - \alpha _1^- \cdot \beta _1^+ \end{array} \right] \\ \\ \quad \quad +\left[ \begin{array}{ll} 0 &{} (\alpha _0^+ \beta _2^+ + \beta _0^- \alpha _2^+ + \alpha _3^- \beta _1^- - \alpha _1^- \beta _3^-) E_2^+ \\ (\alpha _0^- \beta _3^- + \beta _0^+ \alpha _3^- + \alpha _1^+ \beta _2^+ - \alpha _2^+ \beta _1^+)E_3^- &{} 0\end{array} \right] \end{array} \end{aligned}$$

(A.2)

The algebra generated by $\rho ^\pm , \varepsilon _1^\pm$ is the quaternionic subalgebra. Its divisible real form is obtained by setting: $\alpha _0^- = (\alpha _0^+)^*$ and $\alpha _1^- = (\alpha _1^+)^*$ and it is a real linear span of $1, u_7, u_4, u_1$.

We now show that it is impossible to have e sextonion real algebra that has divisible quaternions as a subalgebra. To this aim, we suppose $\alpha _0^- = (\alpha _0^+)^*$ and $\alpha _1^- = (\alpha _1^+)^*$ and take in (A.2) $\alpha _0^+ = \beta _0^+ = \beta _1^+ = 0$ - which implies $\alpha _0^- = \beta _0^- = \beta _1^- = 0$. The product (A.2) shows that the coefficients of $\varepsilon _2^+$ and $\varepsilon _3^-$ must be complex, hence each coefficient, say $\alpha _2^+$ contains 2 real parameters a and b, and, in order to have a six-dimensional real algebra, $\alpha _3^-$ viewed in $\mathbf {R}_2$ must be a linear transformation T of (a, b), linearity being enforced by the linearity of the algebra.

We loosely write $\alpha _2^+ = T \alpha _3^-$. It is easy to show that $T^2 = Id$, namely T is an involution. By playing with the coefficients in (A.2), we can easily obtain $\alpha _2^+ = - T^2 \alpha _2^+ = - \alpha _2^+$ and similarly for $\alpha _3^-$, a contradiction unless $\alpha _2^+ = \alpha _3^- = 0$.

This ends our proof.