$3 \times 3$ Analogue of Zorn's Vector Matrices

Question

A Zorn vector matrix is a $2 \times 2$ matrix whose diagonal elements are scalars and whose off-diagonal elements are 3-vectors. I read about them on this Wikipedia page for split-octonions. The product of two Zorn matrices is given by

$\begin{pmatrix} \alpha & \mathbf a \\ \mathbf b & \beta \end{pmatrix} \begin{pmatrix} \alpha' & \mathbf a' \\ \mathbf b' & \beta' \end{pmatrix} = \begin{pmatrix} \alpha \alpha' + \mathbf a \cdot \mathbf b' & \alpha \mathbf a' + \beta' \mathbf a + \mathbf b \times \mathbf b' \\ \alpha' \mathbf b + \beta \mathbf b' + \mathbf a' \times \mathbf a & \beta \beta' + \mathbf a' \cdot \mathbf b \end{pmatrix}$

Zorn matrices were developed as a representation of the octonions; as with the octonions, multiplication of Zorn matrices is non-associative. The determinant of a Zorn matrix is given by

$\det \begin{pmatrix} \alpha & \mathbf a \\ \mathbf b & \beta \end{pmatrix} = \alpha \beta - \mathbf a \cdot \mathbf b$

and satisfies $(\det AB) = (\det A)(\det B)$ for all Zorn matrices $A$ and $B$. If the off-diagonal elements of $A$ and $B$ are all parallel vectors (say, only the first component is nonzero), then the cross products vanish and Zorn multiplication behaves like ordinary matrix multiplication.

Is there a $3 \times 3$ analogue of the Zorn matrix algebra that preserves the form of multiplication for $2 \times 2$ blocks and has a norm that reduces to the $2 \times 2$ determinant defined above?

1. $\begin{pmatrix} \alpha & \mathbf a & \mathbf 0 \\ \mathbf b & \beta & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} \begin{pmatrix} \alpha' & \mathbf a' & \mathbf 0 \\ \mathbf b' & \beta' & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} = \begin{pmatrix} \alpha \alpha' + \mathbf a \cdot \mathbf b' & \alpha \mathbf a' + \beta' \mathbf a + \mathbf b \times \mathbf b' & \mathbf 0 \\ \alpha' \mathbf b + \beta \mathbf b' + \mathbf a' \times \mathbf a & \beta \beta' + \mathbf a' \cdot \mathbf b & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} $
2. $\det \begin{pmatrix} \alpha & \mathbf a & \mathbf 0 \\ \mathbf b & \beta & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} = \det \begin{pmatrix} \alpha & \mathbf 0 & \mathbf a \\ \mathbf 0 & 1 & \mathbf 0 \\ \mathbf b & \mathbf 0 & \beta \end{pmatrix} = \det \begin{pmatrix} 1 & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \alpha & \mathbf a \\ \mathbf 0 & \mathbf b & \beta \end{pmatrix} = \alpha \beta - \mathbf a \cdot \mathbf b$
3. $\det AB = (\det A)(\det B)$ for all $A,B$ of the form $\begin{pmatrix} \alpha & \mathbf a & \mathbf c \\ \mathbf b & \beta & \mathbf e \\ \mathbf d & \mathbf f & \gamma \end{pmatrix}$

Thanks!

Answer 1

There is no analogue, or at least not a natural one.

Forms permitting composition were classified in Schafer (1970). The classification implies that if a unital algebra over a field of characteristic not dividing $6$ admits a nondegenerate form $N$ of degree $3$ such that $N(xy)=N(x)N(y)$ for all $x,y$, then it must have dimension $1,2,3,5$ or $9$ over that field (note that this generalizes Hurwitz's theorem that composition algebras only exist in dimensions $1,2,4,8$).

Since a $3\times3$ analogue of the Zorn vector-matrix algebra would have dimension $6\cdot 3 + 3 = 21$, and you require that $\det(AB) = \det(A) \det(B)$ for all $A, B$ which implies that $\det$ would be a form (presumably of degree $3$) admitting composition, it follows that any possible examples would be either degenerate or in characteristic $2$ or $3$. In particular there is no such algebra over $\mathbb{R}$ or $\mathbb{C}$ where $\det$ is a degree-$3$ homogeneous polynomial function of the vector-matrix's entries, like in the $2\times 2$ case.