# Determinants of octonionic hermitian matrices

For quaternionic hermitian matrices (i.e. quaternionic square matrices $(a_{ij})$ satisfying $a_{ji}=\bar a_{ij}$) there is a nice notion of (Moore) determinant which can be defined as follows.

First, for any quaternionic $n\times n$ matrix $A$ one can define its realization $A^{\mathbb{R}}$ which is a real $4n\times 4n$ matrix as follows. Fix an $\mathbb{R}$-linear identification $\mathbb{H}^n\simeq \mathbb{R}^{4n}$. Consider the $\mathbb{R}$-linear map $\mathbb{H}^n\to \mathbb{H}^n$ given by $q\mapsto Aq$. Its matrix (of size $4n$) under the above identification $\mathbb{H}^n\simeq \mathbb{R}^{4n}$ will be denoted by $A^{\mathbb{R}}$. The following result is well known (unfortunately I do not know who is its author in this form):

Theorem. There exists a real polynomial $\det_M$, called Moore determinant, on the space of quaternionic hermitian matrices of size $n$ which is uniquely characterized by the following two properties:

1) for any quaternionic hermitian matrix $A$ one has $\det(A^{\mathbb{R}})=(\det_M(A))^4$;

2) $\det_M(I)=1.$

The Moore determinant has many nice properties similar to the properties of the usual determinant on real symmetric and complex hermitian matrices, e.g. the Sylvester criterion of positive definiteness holds in terms for this determinant. For more properties see Section 1 in http://arxiv.org/abs/math/0104209 for example.

Question. Let us consider octonionic hermitian $n\times n$ matrices instead of quaternionic ones. In the same way as previously one can define the realization of such a matrix which has now size $8n$. Is it true that $\det A^{\mathbb{R}}$ is the 8th power of some other polynomial on the space of octonionic hermitian $n\times n$ matrices?

Remark. For octonionic hermitian matrices of size 2 or 3 I am aware of a nice notion of determinant which is a polynomial in its entries and does satisfy Sylvester criterion of positive definiteness, however I have not checked whether that determinant provides also a positive answer to my question above.

• mathoverflow.net/questions/72751/… – Carlo Beenakker May 30 '16 at 5:09
• I don't understand how you can realize octonionic matrix n x n as real matrix of dimension 8n ? For quaternions it is clear because each quaternion $q$ can be represented as matrix 4x4 - in two ways: 1) as left multiplication by $q$; 2) by right multiplication by $\bar q$. – user21230 Jun 7 '16 at 21:01
• @MarekMitros: Let $A$ be octonionic $n\times n$ matrix. Consider the map $\mathbb{O}^n\to \mathbb{O}^n$ given by $x\mapsto Ax$. It is $\mathbb{R}$-linear map. Let us choose $\mathbb{R}$-linear isomorphism $\mathbb{O}^n\simeq \mathbb{R}^{8n}$. This induces $\mathbb{R}$-linear map $\mathbb{R}^{8n}\to \mathbb{R}^{8n}$. Its matrix is called realization of $A$. – makt Jun 8 '16 at 7:52
• I see. In such case, say for symmetric hermitian 2x2 matrix $$\begin{pmatrix} r & a \\ \bar a & s \end{pmatrix}$$ the realization matrix is $$\begin{pmatrix} L_r & L_a \\ L_\bar a & L_s \end{pmatrix}$$ which is $$\begin{pmatrix} rI & L_a \\ L_\bar a & sI \end{pmatrix}$$ for octonion a and real r,s; $L_a$ is matrix 8x8 of left multiplication by octonion $a$. What is the nice notion of determinant in this case ? – user21230 Jun 8 '16 at 10:36
• @MarekMitros: $$\det\left[\begin{array}{cc} r&a\\ \bar a&s\end{array}\right]=rs-a\bar a.$$ – makt Jun 8 '16 at 11:55