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$\DeclareMathOperator\RRe{Re}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sym{Sym}$Let $\mathcal{H}_2(\mathbb{O})$ denote the (10-dimensional) real vector space of octonionic Hermitian matrices of size 2. Recall that a matrix $(a_{ij})$ with octonionic entries is called Hermitian if $a_{ji}=\bar a_{ij}$.

One has a linear imbedding $j\colon \mathcal{H}_2(\mathbb{O})\to \Sym^2(\mathbb{R}^{16})$ to the space of real quadratic forms on $\mathbb{R}^{16}=\mathbb{O}^2$ given by $$(j(A))(\xi)=\sum_{i,j=1}^2\RRe((\bar\xi_iA_{ij})\xi_j),$$ where $\xi=(\xi_1,\xi_2)\in \mathbb{O}^2$.

QUESTION. I am looking for a characterization of the image of $j$.

Ideally I need a description analogous to the complex case as follows. The space $\mathcal{H}_n(\mathbb{C})$ of complex Hermitial matrices is imbedded into the space of real quadratic forms on $\mathbb{R}^{2n}=\mathbb{C}^n$ via the similar map $$j'(A)(\xi)=\sum_{i,j=1}^n \bar\xi_iA_{ij}\xi_j.$$ It is known that its image consists precisely of real quadratic forms invariant under the multiplication $\xi\mapsto z\cdot\xi$ for any complex number with $|z|=1$. However, as far as I can see, this description does not seem to generalize to the octonionic situation.

REMARK. I am aware of a representation theoretical characterization of the image of $j$. There is an action of the group $\Spin(1,9)$ on $\mathbb{R}^{16}=\mathbb{O}^2$ (see e.g. p.29 here https://arxiv.org/pdf/math/0105155v4.pdf ). The symmetric square representation in $\Sym^2(\mathbb{R}^{16})$ is a sum of exactly two non-isomorphic irreducible subspaces (see the answer to this post A representation of Spin(9,1)). One of them is the image of $j$.

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2 Answers 2

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I'm revising my answer because whether the image of the map $j$ that the OP defines is equal to the $10$-dimensional subspace $H$ of $\mathrm{Sym}^2(\mathbb{R}^{16})$ that is invariant under $\mathrm{Spin}(9,1)$ depends on which particular action of $\mathrm{Spin}(9,1)$ on $\mathbb{R}^{16}$ one chooses.

First, independent of the map $j$ there is a characterization of $H$ in terms of the $\mathrm{Spin}(9,1)$-action on $\mathbb{R}^{16}=\mathbb{O}^2$ in other terms that may be what the OP wants. This goes as follows: $\mathrm{Spin}(9,1)$ preserves an $8$-sphere $\Sigma$ of $8$-dimensional subspaces $L\subset\mathbb{O}^2$ such that $\mathbb{O}^2\setminus\{0\}$ is foliated smoothly by the punctured planes $L\setminus\{0\}$ for $L\in\Sigma$. Each of these $8$-planes $L$ carries a 'natural' definite quadratic form defined up to a positive multiple such that, for each $g\in \mathrm{Spin}(9,1)$ and $L\in\Sigma$, the induced isomorphism $g:L\to g(L)$ identifies the two quadratic forms up to a multiple.

Then $H$ consists of the quadratic forms on $\mathbb{O}^2$ that restrict to each $L\in\Sigma$ to be a multiple of its 'natural' quadratic form.

All of this follows from the description of $\mathrm{Spin}(9,1)\subset\mathrm{SL}(\mathbb{R}^{16})=\mathrm{SL}(\mathbb{O}^{2})$ given at the end of my Notes on spinors in low dimension, where the Lie algebra of $\mathrm{Spin}(9,1)$ is described as $$ {\frak{spin}}(9,1) = \left\{\pmatrix{ a_1 +x\,I_8 & C\,R_{\bf w} \cr C\,L_{\bf x} & a_3 -x\,I_8 }\ :\ \matrix{ x\in\mathbb{R},\cr {\bf w},{\bf x}\in\mathbb{O},\cr \ a\in{\frak{spin}}(8)}\ \right\} \subset{\frak{sl}}(16,\mathbb{R})\,, $$ where $C$ denotes conjugations in the octonions, $R_{\bf w}$ denotes right multiplication by ${\bf w}$ in the octonions, $L_{\bf x}$ denotes left multiplication by ${\bf x}$ in the octonions, and the assignments $a\mapsto a_k$ for $k=1,3$ are Lie algebra isomorphisms $\mathfrak{spin}(8)\to \mathfrak{so}(8)$ described in the Notes referenced above.

For example, the subspaces $L\in\Sigma\simeq S^8 = \mathbb{OP}^1$ are of the form either $L_\infty = \{\ (q,0)\ | \ q\in\mathbb{O\ }\}$ or of the form $L_w = \{\ (wq, \bar q)\ |\ q\in\mathbb{O}\ \}$ for some $w\in\mathbb{O}$. Alternatively, they can be described as either of the form $L'_\infty = \{\ (0,q)\ | \ q\in\mathbb{O\ }\}$ or of the form $L'_x = \{\ (\bar q, qx)\ |\ q\in\mathbb{O}\ \}$ for some $x\in\mathbb{O}$. (Note that $L_w = L'_x$ when $w\bar x = 1$ while $L_0 = L'_\infty$ and $L_\infty = L'_0$.) [Here, I am writing elements of $\mathbb{O}^2$ as pairs $(p,q)$ to save space, but they really should be written in the form $\textstyle\begin{pmatrix}p\\ q\end{pmatrix}$ to be consistent with the matrix notation used above for ${\frak{spin}}(9,1)$.]

Now, with this identification of $\mathbb{R}^{16}$ with $\mathbb{O}^2$, $H$ is not equal to the image of $j$, as is easy to check.

On the other hand, if we take the conjugate of $\mathrm{Spin}(9,1)$ under the matrix $$ \pmatrix{I_8 & 0 \cr 0 & C } $$ (which does not belong to $\mathrm{Spin}(9,1)$), then, for the $H$ associated to this conjugate subgroup, the image of $j$ is equal to $H$.

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    $\begingroup$ I think $H=Im(j)$. A proof is given below. Am I wrong? If I am correct your answer would be the final one. $\endgroup$
    – makt
    Mar 27 at 13:13
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This is not an answer but a long comment on the Robert Bryant's answer. I think that $H=Im(j)$.

One has to show that $(jA)[q,\xi q]=(jA)[1,\xi q]$ for any $\xi,q\in \mathbb{O}$ with $|q|=1$. Using that $A_{11},A_{22}\in \mathbb{R}$ and $Re((ab)c)=Re(a(bc))$ for any octnonions $a,b,c$ one has \begin{eqnarray} (jA)[q,\xi q]=\\ \bar qA_{11}q+\overline{(\xi q)}A_{22}(\xi q)+2Re(\bar qA_{12}(\xi q))=\\ A_{11}+A_{22}|\xi|^2+2Re(((\xi q)\bar q) A_{12})=\\ A_{11}+A_{22}|\xi|^2+2Re(\xi(q\bar q) A_{12})=\\ A_{11}+A_{22}|\xi|^2+2Re( A_{12}\xi)=\\ (jA)[1,\xi]. \end{eqnarray} QED.

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  • $\begingroup$ I think we are talking about two different (but conjugate) imbeddings of $Spin(9,1)$ into $GL_{16}(\mathbb{R})$. They correspond to two different imbeddings of $\mathbb{O}\mathbb{P}^1$ into the Grassmannian $Gr_8(\mathbb{R}^{16})$. The conjugation is apparently given by the map $(x_1,x_2)\mapsto (x_2,\bar x_1)$. Thus in your normalization the map $j$ should be modified. $\endgroup$
    – makt
    Mar 27 at 16:28

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