$\DeclareMathOperator\RRe{Re}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sym{Sym}$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 $j$.


2 Answers 2


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 $H$.

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.