# Quadratic forms on $\mathbb{R}^{16}$ coming from octonions

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

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 $$H$$.

• I think $H=Im(j)$. A proof is given below. Am I wrong? If I am correct your answer would be the final one.
– 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.

• 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.
– makt
Mar 27 at 16:28