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I am working over $\mathbb{R}$ (though most of the story goes over any field). I am looking for linear spaces of matrices such that the restriction of the determinant to this spaces can be written (non-trivially) as the power of another polynomial. Let me give some examples where such a phenomenon appears, before asking my question in more details.

  • let $n = 2m$ be an even integer. Then, the restriction of the determinant to $\bigwedge^2 \mathbb{R}^n \subset \mathrm{End}(\mathbb{R}^n)$ can be written as the square of a non-zero polynomial : the pfaffian.

  • let $n = 2m$ be again even. Let $\mathcal{H}_{m}(\mathbb{C})$ be the set of $m \times m$ Hermitian matrices with complex coefficients. Using the matrix representation of $i$ (square root of $-1$) as: $$ i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix},$$ one can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{C})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{C})} \right)^2.$$

  • let $n = 4m$. Let $\mathcal{H}_{m}(\mathbb{H})$ be the set of $m \times m$ Hermitian matrices with quaternionic coefficients. Using the $4 \times 4$ matrix representation of $i,j,k$ (square roots of $-1$ in $\mathbb{H}$), One can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is again easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{H})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{H})} \right)^4.$$

Question : Is there a similar story for $\mathcal{H}_{3}(\mathbb{O})$?

More precisely, one can define a good notion of determinant for Hermitian $3 \times 3$ matrices with octonionic coefficients. I was wondering if there is an embedding $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{S}^{2} \mathbb{R}^{24} \subset \mathrm{End}(\mathbb{R}^{24})$, such that the following hold on $\mathcal{H}_{3}(\mathbb{O})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})} = \left( \mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})} \right)^8?$$

The algebra $\mathcal{H}_{3}(\mathbb{O})$ being non-associative, it can not be embedded as an algebra into a matrix algebra. On the other hand, what I am asking is (probably) considerably weaker : I just want an embedding of vector spaces $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{End}(\mathbb{R}^{24})$ such that the restriction of $\mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})}$ to $\mathcal{H}_{3}(\mathbb{O})$ is the $8$-th power of $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}$.

Of course, since $\mathcal{H}_{3}(\mathbb{O})$ can not be embedded into $\mathrm{End}(\mathbb{R}^{24})$ as an algebra, it will probably be harder to check such an equality of determinants on $\mathcal{H}_{3}(\mathbb{O})$. Indeed, we can't use the representative of the Hermitian comatrix in $\mathrm{End}(\mathbb{R}^{24})$ and just multiply it with the orginal matrix to get $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}.Id_{24}$. But I thought that it was perhaps possible to find another trick for $\mathcal{H}_{3}(\mathbb{O})$?

Thanks for your help.

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  • $\begingroup$ Do you know anything about the $\mathcal{H}_{2}(\mathbb{O})$ case? $\endgroup$
    – MTyson
    Commented Aug 17, 2017 at 22:37
  • $\begingroup$ @MTyson In the case of $\mathcal{H}_{2}(\mathbb{O})$ I think there is something going on. Let $\mathbb{O} \hookrightarrow \mathrm{S}^2 \mathbb{R}^8$ which associates to an octonion its multiplication table. This gives rise to an embedding $\mathcal{H}_{2}(\mathbb{O})$ in $\mathrm{S}^2 \mathbb{R}^{16}$. Using the block computation of a determinant for a $2*2$ block matrix (with the bottom right being a translation), one sees that $\mathrm{det}_{\mathrm{End}(\mathbb{R}^{16})} = (\mathrm{det}_{\mathcal{H}_{2}(\mathbb{O})})^3$ $\endgroup$
    – Libli
    Commented Aug 17, 2017 at 22:48
  • $\begingroup$ What is that first map? The multiplication table should be neither symmetric nor skew-symmetric. I see how given an $\mathbb{R}$-linear map $f:\mathbb{O}\hookrightarrow\mathrm{End}(\mathbb{R}^8)$ satisfying $f(z)f(\bar z)=|z|^2I_8$ one could extend to $\mathcal{H}_2(\mathbb{O})$ with the right determinant relation. That would require 7 anticommuting matrices $e_1,\dots,e_7$ in $\mathrm{End}(\mathbb{R}^8)$ that square to $-I_8$. How are you building these out of the multiplication table? $\endgroup$
    – MTyson
    Commented Aug 18, 2017 at 1:55
  • $\begingroup$ The first suggestion of embedding octonions into matrices is to map octonion $x$ to $8\times 8$ matrix $L_x$ or $R_x$ of left and right multiplication by octonion $x$. Similarly octonion matrix $2\times 2$ can be mapped to block matrix consisting of four blocks $L_{a_{i,j}}$ where $a_{i,j}$ is octonion in original matrix. Do we have formula for determinant of block matrix ? $\endgroup$
    – user21230
    Commented Aug 18, 2017 at 9:32
  • $\begingroup$ See also mathoverflow.net/questions/239954/… $\endgroup$
    – user21230
    Commented Aug 18, 2017 at 10:49

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