%&plain
\magnification=\magstep1
% Here are some notes on spinor actions in the low
% dimensions that may be of some interest.
% Robert Bryant
% 23 April 1998
\font\eu=eufm10
\font\bigbf=cmbx12
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{\hbox{\sevenjap C}}{\hbox{\sevenjap C}}}}
\def\bbO{{\mathchoice{\hbox{\tenjap O}}{\hbox{\tenjap O}}
{\hbox{\sevenjap O}}{\hbox{\sevenjap O}}}}
\def\bbR{ {\mathchoice {\hbox{\tenjap R}} {\hbox{\tenjap R}}
{\hbox{\sevenjap R}} {\hbox{\sevenjap R}}}}
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{\hbox{\sevenjap Z}}{\hbox{\sevenjap Z}}}}
\def\w{{\mathchoice{\,{\scriptstyle\wedge}\,}
{{\scriptstyle\wedge}}
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\def\xb{{\bf x}}\def\yb{{\bf y}}\def\wb{{\bf w}}\def\zb{{\bf z}}
\def\pb{{\bf p}}\def\qb{{\bf q}}\def\ub{{\bf u}}\def\vb{{\bf v}}
\def\ab{{\bf a}}\def\sb{{\bf s}}\def\fb{{\bf f}}
\def\zerob{{\bf 0}}\def\oneb{{\bf 1}}
\def\Cl{{{\rm C}\ell}}\def\Im{{\rm Im}}
\def\bT{{\bf T}}\def\bR{{\bf R}}
\def\la{\langle}\def\ra{\rangle}
\def\euk{\hbox{\eu k}}\def\eur{\hbox{\eu r}}
\def\eug{\hbox{\eu g}}\def\euspin{\hbox{\eu spin}}\def\euso{\hbox{\eu so}}
\def\eusu{\hbox{\eu su}}\def\euh{\hbox{\eu h}}
\def\eugl{\hbox{\eu gl}}\def\eusl{\hbox{\eu sl}}
\def\GL{{\rm GL}}\def\SL{{\rm SL}}\def\End{\mathop{\rm End}}
\def\Spin{{\rm Spin}}\def\SO{{\rm SO}}\def\SU{{\rm SU}}
\def\Sp{{\rm Sp}}\def\Hom{{\rm Hom}}
\def\bfalpha{\hbox{\greekbold\char11}}%bold Greek alpha
\def\bfomega{\hbox{\greekbold\char33}}%bold Greek omega
\def\bfsigma{\hbox{\greekbold\char27}}%bold Greek sigma
\def\bftheta{\hbox{\greekbold\char18}}%bold Greek theta
\def\bfTheta{\hbox{\greekbold\char02}}%bold Greek Theta
\def\bfphi{\hbox{\greekbold\char30}}%bold Greek phi
\def\bfPhi{\hbox{\greekbold\char08}}%bold Greek Phi
\def\bfpsi{\hbox{\greekbold\char32}}%bold Greek psi
\def\bfPsi{\hbox{\greekbold\char09}}%bold Greek Psi
\centerline{\bigbf Remarks on Spinors in Low Dimension}
\bigskip
{\bf 0. Introduction.} The purpose of these notes is to study
the orbit structure of the groups~$\Spin(p,q)$ acting on their spinor
spaces for certain values of~$n=p{+}q$, in particular, the values
$$
(p,q) = (8,0),\ (9,0),\ (10,0),\ \hbox{and}\ (10,1).
$$
though it will turn out in the end that there are a few interesting
things to say about the cases $(p,q) = (10,2)$ and $(9,1)$, as well.
{\bf 1. The Octonions.} Let~$\bbO$ denote the ring of octonions.
Elements of~$\bbO$ will be denoted by bold letters, such as $\xb$, $\yb$,
etc. Thus, $\bbO$ is the unique $\bbR$-algebra of dimension~$8$
with unit~$\oneb\in\bbO$ endowed with a positive definite inner
product~$\la,\ra$ satisfying~$\la \xb\yb,\xb\yb\ra
=\la\xb,\xb\ra\,\la\yb,\yb\ra$ for all~$\xb,\yb\in\bbO$.
As usual, the norm of
an element~$\xb\in\bbO$ is denoted $|\xb|$ and defined as the square root
of~$\la\xb,\xb\ra$. Left and right multiplication
by~$\xb\in\bbO$ define maps~$L_\xb\,,R_\xb:\bbO\to\bbO$ that are isometries
when~$|\xb|=1$.
The conjugate of~$\xb\in\bbO$,
denoted~$\overline\xb$, is defined to be~$\overline\xb =
2\la\xb,\oneb\ra\,\oneb-\xb$. When a symbol is needed, the map of
conjugation will be denoted $C:\bbO\to\bbO$. The identity
$\xb\,\overline\xb = |\xb|^2$ holds, as well as the conjugation
identity~$\overline{\xb\yb}={\overline\yb}\,{\overline\xb}$.
In particular, this implies the useful identities
$C\, L_\xb\, C=R_{\overline\xb}$
and $C\, R_\xb\, C=L_{\overline\xb}$.
\smallskip
The algebra~$\bbO$ is not commutative or associative. However, any
subalgebra of~$\bbO$ that is generated by two elements is associative.
It follows that $\xb\,\bigl({\overline\xb} \yb\bigr)=|\xb|^2\,\yb$ and
that~$(\xb\yb)\xb=\xb(\yb\xb)$ for
all~$\xb,\yb\in\bbO$. Thus,~$R_\xb\, L_\xb
=L_\xb\, R_\xb$ (though, of course,
$R_\xb\, L_\yb\not=L_\yb\, R_\xb$ in general). In
particular, the expression $\xb\yb\xb$ is unambiguously defined.
In addition, there are the {\it Moufang Identities}
$$
\eqalign{
(\xb\yb\xb)\zb &= \xb\bigl(\yb(\xb\zb)\bigr),\cr
\zb(\xb\yb\xb) &= \bigl((\zb\xb)\yb\bigr)\xb,\cr
\xb(\yb\zb)\xb &= (\xb\yb)(\zb\xb),\cr
}
$$
which will be useful below. (See, for example,
{\it Spinors and Calibrations}, by F.~Reese Harvey, for proofs.)
\bigskip
{\bf 2. $\Spin(8)$.}
For~$\xb\in \bbO$, define the linear
map~$m_\xb:\bbO\oplus\bbO\to\bbO\oplus\bbO$ by the formula
$$
m_\xb = \left[\matrix{0&C \, R_\xb\cr -C \, L_\xb&0}\right]\,.
$$
By the above identities, it follows that $(m_\xb)^2 = -|\xb|^2$ and
hence this map induces a representation on the vector
space~$\bbO\oplus\bbO$ of the Clifford algebra generated
by~$\bbO$ with its standard quadratic form.
This Clifford algebra is known to be isomorphic
to~$M_{16}(\bbR)$, the algebra of 16-by-16 matrices with real entries, so
this representation must be faithful. By dimension count, this establishes
the isomorphism~$\Cl\bigl(\bbO,\la,\ra\bigr)
=\End_\bbR\bigl(\bbO\oplus\bbO\bigr)$.
\smallskip
The group $\Spin(8)\subset\GL_\bbR(\bbO\oplus\bbO)$ is defined as the
subgroup generated by products of the form~$m_\xb\,m_\yb$
where~$\xb,\yb\in\bbO$ satisfy~$|\xb|=|\yb|=1$.
Such endomorphisms preserve the splitting of $\bbO\oplus\bbO$ into
the two given summands since
$$
m_\xb\,m_\yb
= \left[\matrix{ -L_{\overline\xb} \, L_\yb&0\cr
0&-R_{\overline\xb} \, R_\yb}\right]\,.
$$
In fact, setting $\xb=-\oneb$ in this formula shows that endomorphisms
of the form
$$
\left[\matrix{ L_\ub&0\cr 0&R_\ub}\right],\qquad\hbox{with $|\ub|=1$}
$$
lie in~$\Spin(8)$. In fact, they generate~$\Spin(8)$, since
$m_\xb\,m_\yb$ is clearly a product of two of these when
$|\xb|=|\yb|=1$.
\smallskip
Fixing an identification~$\bbO\simeq\bbR^8$, this defines an embedding
$\Spin(8)\subset\SO(8)\times\SO(8)$, and the projections onto either
of the factors is a group homomorphism. Since neither of these
projections is trivial, since the Lie algebra~$\euso(8)$ is simple, and
since~$\SO(8)$ is connected, it follows that each of these projections
is a surjective homomorphism. Since~$\Spin(8)$ is simply connected
and since the fundamental group of~$\SO(8)$ is~$\bbZ_2$, it follows that
that each of these homomorphisms is a non-trivial double cover of~$\SO(8)$.
Moreover, it follows that the subsets~$\{\ L_\ub\ \vrule\ |\ub|=1\ \}$
and $\{\ R_\ub\ \vrule\ |\ub|=1\ \}$ of~$\SO(8)$ each suffice to
generate~$\SO(8)$.
\smallskip
Let~$H\subset\bigl(\SO(8)\bigr)^3$ be the set of
triples~$(g_1,g_2,g_3)\in\bigl(\SO(8)\bigr)^3$ for which
$$
g_2(\xb\yb) = g_1(\xb)\,g_3(\yb)
$$
for all~$\xb,\yb\in\bbO$. The set~$H$ is closed and is evidently
closed under multiplication and inverse. Hence it is a compact Lie group.
By the third Moufang identity, $H$ contains the subset
$$
\Sigma =
\left\{\ (L_\ub,\,L_\ub{\,}R_\ub,\,R_\ub)\ \vrule\ |\ub|=1
\right\}.
$$
Let~$K\subset H$ be the subgroup generated by~$\Sigma$, and for $i=1,2,3$,
let~$\rho_i:H\to\SO(8)$ be the homomorphism that is
projection onto the $i$-th factor. Since~$\rho_1(K)$ contains
$\{\ L_\ub\ \vrule\ |\ub|=1\ \}$, it follows that $\rho_1(K)=\SO(8)$,
so {\it a fortiori} $\rho_1(H)=\SO(8)$. Similarly, $\rho_3(H)=\SO(8)$.
The kernel of~$\rho_1$ consists of elements~$(I_8,g_2,g_3)$
that satisfy~$g_2(\xb\yb) = \xb\,g_3(\yb)$ for all~$\xb,\yb\in\bbO$.
Setting~$\xb=\oneb$ in this equation yields~$g_2=g_3$,
so that~$g_2(\xb\yb) = \xb\,g_2(\yb)$. Setting~$\yb=\oneb$
in this equation yields~$g_2(\xb) = \xb\,g_2(\oneb)$, i.e., $g_2 = R_\ub$
for $\ub=g_2(\oneb)$. Thus, the elements in the kernel of~$\rho_1$ are
of the form~$(1,R_\ub,R_\ub)$ for some~$\ub$ with~$|\ub|=1$.
However, any such $\ub$ would, by definition,
satisfy~$(\xb\yb)\ub=\xb(\yb\ub)$ for all~$\xb,\yb\in\bbO$,
which is impossible unless~$\ub=\pm\oneb$. Thus, the kernel of~$\rho_1$
is~$\bigl\{(I_8,\pm I_8,\pm I_8)\bigr\}\simeq\bbZ_2$, so that $\rho_1$ is a
2-to-1 homomorphism of~$H$ onto~$\SO(8)$. Similarly, $\rho_3$ is a 2-to-1
homomorphism of~$H$ onto~$\SO(8)$, with
kernel~$\bigl\{(\pm I_8,\pm I_8,I_8)\bigr\}$. Thus, $H$ is either
connected and isomorphic to~$\Spin(8)$ or else disconnected,
with two components.
Now~$K$ is a connected subgroup of~$H$ and the kernel of~$\rho_1$
intersected with~$K$ is either trivial or~$\bbZ_2$. Moreover, the product
homomorphism~$\rho_1{\times}\rho_3:K\to\SO(8){\times}\SO(8)$ maps the
generator~$\Sigma\subset K$ into generators
of~$\Spin(8)\subset\SO(8){\times}\SO(8)$. It follows that
$\rho_1{\times}\rho_3(K)=\Spin(8)$ and hence that~$\rho_1$ and $\rho_3$
must be non-trivial double covers of~$\Spin(8)$ when restricted to~$K$.
In particular, it follows that~$K$ must be all of~$H$ and, moreover, that the
homomorphism~$\rho_1{\times}\rho_3:H\to\Spin(8)$ must be an isomorphism.
It also follows that the homomorphism $\rho_2:H\to\SO(8)$
must be a double cover of~$\SO(8)$ as well.
Henceforth, $H$ will be identified with~$\Spin(8)$ via the
isomorphism~$\rho_1{\times}\rho_3$. Note that the center of $H$
consists of the
elements~$(\varepsilon_1\,I_8,\varepsilon_2\,I_8,\varepsilon_3\,I_8)$
where~${\varepsilon_i}^2=\varepsilon_1\varepsilon_2\varepsilon_3=1$
and is isomorphic to~$\bbZ_2\times\bbZ_2$.
\smallskip
{\it Triality.} For~$(g_1,g_2,g_3)\in H$, the identity
$g_2(\xb\yb) = g_1(\xb)\,g_3(\yb)$ can be conjugated, giving
$$
Cg_2C(\xb\yb)
= \overline{g_2(\overline{\yb}\,\overline{\xb})}
= \overline{g_1(\overline{\yb})\,g_3(\overline{\xb})}
= \overline{g_3(\overline{\xb})}\, \overline{g_1(\overline{\yb})}.
$$
This implies that $\bigl(Cg_3C,Cg_2C,Cg_1C\bigr)$ also lies in~$H$.
Also, replacing $\xb$ by $\zb\overline{\yb}$ in the original formula
and multiplying on the right by $\overline{g_3(\yb)}$ shows that
$$
g_2(\zb) \overline{g_3(\yb)} = g_1(\zb\overline{\yb}),
$$
implying that~$\bigl(g_2,g_1,Cg_3C\bigr)$ lies in~$H$ as well. In fact, the
two maps~$\alpha,\beta:H\to H$ defined by
$$
\alpha(g_1,g_2,g_3) = \bigl(Cg_3C,Cg_2C,Cg_1C\bigr),
\quad\hbox{and}\qquad
\beta(g_1,g_2,g_3) = \bigl(g_2,g_1,Cg_3C\bigr)
$$
are outer automorphisms (since they act nontrivially on the center of~$H$)
and generate a group of automorphisms isomorphic to~$S_3$, the symmetric group
on three letters. The automorphism $\tau=\alpha\beta$ is known as the
triality automorphism.
\smallskip
{\it Notation.}
To emphasize the group action, denote~$\bbO\simeq\bbR^8$ by~$V_i$
when regarding it as a representation space of~$\Spin(8)$ via
the representation~$\rho_i$. Thus, octonion multiplication induces
a $\Spin(8)$-equivariant projection
$$
V_1\otimes V_3 \longrightarrow V_2\,.
$$
In the standard notation, it is traditional to identify $V_1$ with
$\bbS_-$ and~$V_3$ with $\bbS_+$ and to refer to $V_2$ as the
`vector representation'. Let~$\rho_i':\euspin(8)\to\euso(8)$ denote
the corresponding Lie algebra homomorphisms, which are, in fact,
isomorphisms. For simplicity of notation, for any~$a\in\euspin(8)$,
the element~$\rho_i'(a)\in\euso(8)$ will be denoted by~$a_i$ when
no confusion can arise.
\smallskip
{\it Orbit structure.} Let~$\SO(\Im\bbO)\simeq\SO(7)$ denote the
subgroup of~$\SO(\bbO)\simeq\SO(8)$ that leaves~$\oneb\in\bbO$ fixed,
and let~$K_i\subset H$ be the preimage of~$\SO(\Im\bbO)$ under the
homomorphism~$\rho_i:H\to\SO(\bbO)$. Then~$K_i$ is a non-trivial
double cover of~$\SO(\Im\bbO)$ and hence is isomorphic to~$\Spin(7)$.
Note, in particular that~$K_1$ contains~$(I_8,-I_8,-I_8)$ and hence
$\rho_3(K_1)\subset\SO(8)$ contains~$-I_8$. This implies that
$\rho_3:K_1\to\SO(V_3)$ is a faithful representation of~$\Spin(7)$ and
hence~$K_1$ acts transitively on the unit sphere in~$V_3$.
\smallskip
In particular, it follows that~$\Spin(8)\subset\SO(V_1)\times\SO(V_3)$
acts transitively on the product of the unit spheres in~$V_1$ and~$V_3$.
Consequently, it follows that the quadratic polynomials
$$
q_1(\xb,\yb) = |\xb|^2\qquad\rm{and}\qquad q_2(\xb,\yb) = |\yb|^2
$$
generate the ring of $\Spin(8)$-invariant polynomials on~$\bbO\oplus\bbO$
and that every point of this space lies on the $\Spin(8)$-orbit of a
unique element of the form~$(a\,\oneb,b\,\oneb)$ for some pair of
real numbers~$a,b\ge0$. For~$ab\not=0$, the stabilizer of such an element
is the 14-dimensional simple group~$G_2$, and this group acts transitively
on the unit sphere in~$\Im\bbO$.
\bigskip
{\bf 2. $\Spin(9)$.} For~$(r,\xb)\in\bbR\oplus\bbO$, define a
$\bbC$-linear map~$m_{(r,\xb)}:\bbC\otimes\bbO^2\to\bbC\otimes\bbO^2$
by the formula
$$
m_{(r,\xb)}
=i \left[\matrix{ r\,I_8& C\,R_\xb\cr C\,L_\xb& -r\,I_8}\right]\,.
$$
Since~$(m_{(r,\xb)})^2$ is $-\left(r^2{+}|\xb|^2\right)$ times the identity
map, this defines a $\bbC$-linear representation
on~$\bbC\otimes\bbO^2$ of the Clifford algebra generated by~$\bbR\oplus\bbO$
endowed with its direct sum inner product. Since this Clifford algebra is
known to be isomorphic to~$M_{16}(\bbC)$, it follows, for dimension reasons,
that this representation is one-to-one and onto, establishing the
isomorphism~$\Cl(\bbR\oplus\bbO,\la,\ra) = \End_\bbC(\bbC\otimes\bbO^2)$.
As usual, $\Spin(9)$ is the subgroup generated by the products of the
form~$m_{(r,\xb)}m_{(s,\yb)}$ where $r^2+|\xb|^2=s^2+|\yb|^2=1$. Note
that these products have real coefficients, and so actually lie in
$\GL_\bbR(\bbO^2)\simeq\GL(16,\bbR)$. In fact, these products are
themselves seen to be products of the products of the special form
$$
p_{(r,\xb)}
= m_{(-1,\zerob)}m_{(r,\xb)}
= \left[\matrix{ r\,I_8& C\,R_\xb\cr -C\,L_\xb& r\,I_8}\right]\,,
\qquad\hbox{where $r^2+|\xb|^2=1$},
$$
so these latter matrices suffice to generate~$\Spin(9)$. By the results
of the previous section, products of an even number of the~$p_{(0,\ub)}$
with~$|\ub|=1$
generate~$\Spin(8)\subset\Spin(9)$.
\smallskip
Since the linear transformations of the form~$p_{(r,\xb)}$ preserve the
quadratic form
$$
q(x,y) = |\xb|^2+|\yb|^2,
$$
it follows that~$\Spin(9)$ is a subgroup of~$\SO(\bbO^2)=\SO(16)$.
\medskip
{\it The Lie algebra.} Since~$\Spin(9)$ contains $\Spin(8)$, the
containment~$\euspin(8)\subset\euspin(9)$ yields the containment
$$
\left\{\ \pmatrix{a_1&0\cr0&a_3}\ \vrule\ a\in\euspin(8)\ \right\}
\subset\euspin(9)\,.
$$
Moreover, since~$\Spin(9)$ contains the 8-sphere consisting of the
$p_{(r,\xb)}$ with $r^2+|\xb|^2=1$, its Lie algebra must contain
the tangent space to this 8-sphere at $(r,\xb)=(1,\zerob)$, i.e.,
$$
\left\{\ \pmatrix{0&C\,R_\xb\cr-C\,L_\xb&0}\ \vrule\ \xb\in\bbO\ \right\}
\subset\euspin(9)\,.
$$
By dimension count, this implies the equality
$$
\euspin(9) = \left\{\pmatrix{a_1&C\,R_\xb\cr-C\,L_\xb&a_3}\
\vrule\ \xb\in\bbO,\ a\in\euspin(8)\ \right\}\,.
$$
Let~$\rho:\Spin(9)\to\SO(\bbR\oplus\bbO)\simeq\SO(9)$ be the
homomorphism for which the induced map on Lie algebras is
$$
\rho'\left(\pmatrix{a_1&C\,R_\zb\cr-C\,L_\zb&a_3}\right)
= \pmatrix{0&2\,\overline{\zb}^*\cr-2\,\overline{\zb}&a_2}.
$$
where~$\xb^*:\bbO\to\bbR$ is just~$\xb^*(\yb) = \la\xb,\yb\ra$.
(The triality constructions imply that $\rho'$ is, indeed, a Lie algebra
homormophism. Note that, when restricted to~$\Spin(8)$, this becomes
the homomorphism~$\rho_2:\Spin(8)\to\SO(\bbO)=\SO(8)$.) Then~$\rho$
is a double cover of~$\SO(9)$.
\smallskip
Define the squaring map~$\sigma:\bbO^2\to\bbR\oplus\bbO$ by
$$
\sigma\left(\pmatrix{\xb\cr\yb\cr}\right)
= \pmatrix{|\xb|^2-|\yb|^2\cr 2\,\xb\,\yb\cr}.
$$
A short calculation using the Moufang Identities shows that $\sigma$ is
equivariant with respect to~$\rho$, i.e., that~$\sigma\bigl(g\vb\bigr)
=\rho(g)\bigl(\sigma(\vb)\bigr)$ for all~$\vb\in\bbO^2$
and all~$g\in\Spin(9)$. This will be useful below.
\medskip
{\it Orbit structure and stabilizer.} Every point of~$\bbO^2$ lies
on an $\Spin(8)$-orbit of an element of the form~$(a\,\oneb,b\,\oneb)$, for
some pair of real numbers~$a,b\ge0$. Thus, the orbits of~$\Spin(9)$ on
the unit sphere in~$\bbO^2$ are unions of the $\Spin(8)$-orbits of the
elements~$(\cos\theta\,\oneb,\sin\theta\,\oneb)$.
Now, calculation yields
$$
p_{(\cos\phi,\sin\phi\,\oneb)}\pmatrix{\cos\theta\,\oneb\cr\sin\theta\,\oneb}
= \pmatrix{\cos(\theta{-}\phi)\,\oneb\cr\sin(\theta{-}\phi)\,\oneb}.
$$
Since all of the elements~$(\cos\theta\,\oneb,\sin\theta\,\oneb)$ lie on
a single $\Spin(9)$-orbit, it follows that~$\Spin(9)$ acts transitively
on the unit sphere in~$\bbO^2$ and, consequently, that the quadratic
form~$q$ generates the ring of $\Spin(9)$-invariant polynomials
on~$\bbO^2$.
Since the orbit of~$(\oneb,\zerob)\in\bbO^2$ is the 15-sphere and
since $\Spin(9)$ is connected and simply connected, it follows that
the $\Spin(9)$-stabilizer of this element must be connected, simply connected,
and of dimension~$21$. Since~$K_1\subset\Spin(8)\subset\Spin(9)$
lies in this stabilizer and has dimension~$21$, it follows that~$K_1$
must be equal to this stabilizer.
\smallskip
For use in the next two sections, it will be useful to understand the
orbits of~$\Spin(9)$ acting on~$\bbO^2\oplus\bbO^2$ and to understand
the ring of $\Spin(9)$-invariant polynomials on this vector space of
real dimension~$32$. The first observation is that the generic orbit
has codimension~$4$. This can be seen as follows: Since~$\Spin(9)$
acts transitively on the unit sphere in~$\bbO^2$, every element lies
on the ~$\Spin(9)$ orbit of an element of the form
$$
\left(\ \pmatrix{a\,\oneb\cr\zerob\cr},
\ \pmatrix{\xb\cr\yb\cr}\ \right),
$$
where~$a\ge0$. Assuming~$a>0$, the stabilizer in~$\Spin(9)$ of this
first component is~$K_1\simeq\Spin(7)$ and this acts transitively on the
unit sphere in the second $\bbO$-summand of~$\bbO^2$, so that an
element of the above form lies on the orbit of an element of the form
$$
\left(\ \pmatrix{a\,\oneb\cr\zerob\cr},
\ \pmatrix{\xb\cr b\,\oneb\cr}\ \right),
$$
where~$b\ge0$. Assuming~$b>0$, the stabilizer in~$K_1$ of $\oneb$ in
this second $\bbO$-summand is~$G_2$, which acts transitively on the
unit sphere in~$\Im\bbO$ in the first $\bbO$-summand. This implies that
an element of the above form lies on the orbit of an element of the form
$$
\zb = \left(\ \pmatrix{a\,\oneb\cr\zerob\cr},
\ \pmatrix{c\,\oneb + d\,\ub \cr b\,\oneb\cr}\ \right),
$$
for some~$c,d\ge0$ and~$\ub\in\Im\bbO$ some fixed unit imaginary octonion.
Thus, the generic $\Spin(9)$-orbit has codimension at most~$4$. It is
still possible that two elements of the above form with distinct
values of~$a,b,c,d>0$ might lie on the same~$\Spin(9)$-orbit, but this
will be ruled out directly.
To see that these latter elements lie on distinct $\Spin(9)$-orbits,
it will be sufficient to construct $\Spin(9)$-invariant polynomials
on~$\bbO^2\oplus\bbO^2$ that separate these elements. To do so,
write the typical element of~$\bbO^2\oplus\bbO^2$ in the form
$$
(\vb_1,\vb_2) = \left(\ \pmatrix{\xb_1\cr\yb_1\cr},
\ \pmatrix{\xb_2\cr\yb_2\cr}\ \right),
$$
and first consider the quadratic polynomials
$$
\eqalign{
q_{2,0} &= |\xb_1|^2+|\yb_1|^2\cr
q_{1,1} &= \xb_1\cdot\xb_2+\yb_1\cdot\yb_2\cr
q_{0,2} &= |\xb_2|^2+|\yb_2|^2\cr
}
$$
These polynomials are manifestly $\Spin(9)$-invariant and satisfy
$$
q_{2,0}(\zb) = a^2,\qquad q_{1,1}(\zb) = ac,
\qquad q_{0,2}(\zb) = b^2+ c^2 + d^2.
$$
Evidently, these three polynomials span the vector
space of $\Spin(9)$-invariant quadratic polynomials on~$\bbO^2\oplus\bbO^2$.
Since $\Spin(9)$ contains $-1$ times the identity, there are no
$\Spin(9)$-invariant cubic polynomials. A representation-theoretic argument
shows that the $\Spin(9)$-invariant quartic polynomials
on~$\bbO^2\oplus\bbO^2$ form a vector space of dimension~$7$. Six of these
are accounted for by quadratic polynomials in~$q_{2,0}$, $q_{1,1}$,
and $q_{0,2}$, while a seventh can be constructed as follows. Define
$$
q_{2,2} = \sigma(\vb_1)\cdot\sigma(\vb_2)
= \left(|\xb_1|^2-|\yb_1|^2\right)\left(|\xb_2|^2-|\yb_2|^2\right)
+ 4\,(\xb_1\yb_1)\cdot(\xb_2\yb_2).
$$
Using the $\Spin(9)$-equivariance of the squaring map~$\sigma$, it follows
that $q_{2,2}$ is indeed invariant under~$\Spin(9)$.
Note that
$$
q_{2,2}(\zb) = a^2(c^2+d^2-b^2),
$$
so that knowledge
of~$\bigl(q_{2,0}(\zb),q_{1,1}(\zb),q_{0,2}(\zb),q_{2,2}(\zb) \bigr)$
suffices to recover~$a,b,c,d>0$ when these numbers are all non-zero. It
now follows that the simultaneous level sets of these four polynomials are
exactly the $\Spin(9)$-orbits on~$\bbO^2\oplus\bbO^2$. (It seems likely
that these polynomials generate the ring of $\Spin(9)$-invariant polynomials
on~$\bbO^2\oplus\bbO^2$, but such a result will not be needed, so this
problem will not be discussed further.)
\bigskip
{\bf 3. $\Spin(10)$.} Rather than construct the Clifford representation
for an inner product on a vector space of dimension~$10$, it is convenient
to use the fact that~$\Spin(10)$ already appears as a subgroup
of~$\Cl(\bbR\oplus\bbO,\la,\ra)=\End_\bbC(\bbC\otimes\bbO^2)$. In fact, by
the discussion in the last section, $\Spin(10)$ is the connected subgroup of
this latter algebra whose Lie algebra is
$$
\euspin(10)
= \left\{\pmatrix{a_1+ir\,I_8&C\,R_\xb+i\,C\,R_\yb\cr
-C\,L_\xb+i\,C\,L_\yb&a_3-ir\,I_8}\
\vrule\ r\in\bbR,\ \xb,\yb\in\bbO,\ a\in\euspin(8)\ \right\}\,.
$$
Note that~$\euspin(10)$ appears as a subspace of~$\eusu(16)$, so that
$\Spin(10)$ acts on~$\bbC^{16}=\bbC\otimes\bbO^2$ preserving the complex
structure and the quadratic form
$$
q = q_{2,0}+q_{0,2} = |\xb_1|^2+|\yb_1|^2 + |\xb_2|^2+|\yb_2|^2\,,
$$
where, now, the typical element of~$\bbC\otimes\bbO^2$ will be written
as
$$
\zb = \pmatrix{\xb_1+i\,\xb_2\cr \yb_1+i\,\yb_2\cr}.
$$
\smallskip
Note that, because there are no connected Lie groups that lie properly
between~$\Spin(9)$ and $\Spin(10)$, it follows that~$\Spin(10)$ is
generated by~$\Spin(9)$ and the circle subgroup
$$
\bT
=\left\{\pmatrix{e^{ir}\,I_8&0\cr0&e^{-ir}\,I_8}
\ \vrule\ r\in\bbR/2\pi\bbZ\ \right\},
$$
which lies in~$\Spin(10)$, but does not lie in~$\Spin(9)$. In particular,
a polynomial on~$\bbC\otimes\bbO^2$ is~$\Spin(10)$-invariant if and only
if it is both~$\Spin(9)$-invariant and~$\bT$-invariant.
\medskip
{\it Invariant polynomials.}
Among the quadratic polynomials that are $\Spin(9)$-invariant, only the
multiples of~$q = q_{2,0}+q_{0,2}$ are also~$\bT$-invariant. Thus, $q$ spans
the space of $\Spin(10)$-invariant quadratic forms on~$\bbC\otimes\bbO^2$.
In particular, this implies that the action of~$\Spin(10)$
on~$\bbC\otimes\bbO^2$ is irreducible (even as a real vector space).
Among the quartic polynomials that are~$\Spin(9)$-invariant, a short
calculation shows that only linear combinations of~$q^2$ and
$$
\eqalign{
p &= {\ts{1\over2}}\left(q_{2,2} + q_{2,0}\,q_{0,2}-2\,{q_{1,1}}^2\right)\cr
&= |\xb_1|^2|\xb_2|^2+|\yb_1|^2|\yb_2|^2
- \left(\xb_1\cdot\xb_2+\yb_1\cdot\yb_2\right)^2
+ 2\,(\xb_1\yb_1)\cdot(\xb_2\yb_2)\cr
&= |\xb_1\w\xb_2|^2 + |\yb_1\w\yb_2|^2
- 2\,(\xb_1\cdot\xb_2)\,(\yb_1\cdot\yb_2)
+ 2\,(\xb_1\yb_1)\cdot(\xb_2\yb_2) .
}
$$
are invariant under the action of~$\bT$. Thus, it follows that $q^2$~and~$p$
span the space of $\Spin(10)$-invariant quartics. (Note the interesting
feature that, in the latter expression for~$p$, only the final term makes
use of octonion multiplication operations.)
\medskip
{\it Orbits and stabilizers.} Let~$M\subset\bbC\otimes\bbO^2$ be the
$\Spin(10)$-orbit of~$\zb_0=(\oneb+i\,\zerob,\zerob + i\,\zerob)$. The
tangent space to~$M$ at~$\zb_0$ is the set of vectors of the form
$$
\pmatrix{a_1+ir\,I_8&C\,R_\xb+i\,C\,R_\yb\cr
-C\,L_\xb+i\,C\,L_\yb&a_3-ir\,I_8}
\pmatrix{\oneb+i\,\zerob\cr\zerob + i\,\zerob}
= \pmatrix{a_1\oneb+i\,r\oneb\cr -\overline{\xb} + i\,\overline{\yb}}\,.
$$
and the Lie algebra of the $\Spin(10)$-stabilizer of~$\zb_0$ is defined
by the equations~$a_1\oneb = r = \xb = \yb = 0$. Thus, the
identity component of the stabilizer of~$\zb_0$ is~$K_1\simeq\Spin(7)$
and the full stabilizer must lie in the normalizer of~$K_1$ in~$\Spin(10)$.
Evidently, the normalizer of~$K_1$ in~$\Spin(10)$ is~$K_1\cdot\bT$.
Since only the identity in the subgroup~$\bT$ stabilizes~$\zb_0$,
the full stabilizer of~$\zb_0$ is~$K_1$. Thus, $M$ is
diffeomorphic to~$\Spin(10)/\Spin(7)$, which is a smooth manifold of
dimension~$45-21=24$ that is 2-connected, i.e., $\pi_0(M)=\pi_1(M)=\pi_2(M)=0$.
The normal space to~$M$ at~$\zb_0$ is the orthogonal direct sum of the
line~$\bbR\zb_0$ (which is normal to the unit sphere in~$\bbC\otimes\bbO^2$)
and the subspace of dimension~$7$
$$
N_{\zb_0} =
\left\{\pmatrix{0+i\,\xb\cr\zerob+i\,\zerob}\ \vrule\ \xb\in\Im\bbO\ \right\}.
$$
The stabilizer~$K_1$ acts as~$\SO(7)$ on this
subspace. In particular, it acts transitively on the unit sphere
in~$N_{\zb_0}$, and hence it acts transitively on the space of geodesics
in the unit 31-sphere that meet~$M$ orthogonally at~$\zb_0$. Since~$M$
is itself a $\Spin(10)$-orbit, it follows that~$\Spin(10)$ must
act transitively on the the normal tube of any radius about~$M$ in the
unit 31-sphere. Since, for generic radii, these normal tubes are
hypersurfaces, it follows that the generic $\Spin(10)$-orbit in the
31-sphere must be a hypersurface of dimension~$30$.
Using the fact that such a hypersurface is an $S^6$-bundle over~$M$,
the long exact sequence in homotopy implies that these hypersurface
orbits are also 2-connected, which implies that the $\Spin(10)$-stabilizer
of any point on such a hypersurface must be both connected and
simply connected.
Now, the full group~$\Spin(10)$ must act transitively on the
space of geodesics in the unit 31-sphere that meet~$M$ orthogonally at
any point while every point of the unit 31-sphere lies on some
geodesic that meets~$M$ orthogonally. Thus, fixing some $\ub\in\Im\bbO$
with~$|\ub|=1$, it follows that every element of the 31-sphere lies on
the $\Spin(10)$-orbit of an element of the form
$$
\zb_\theta = \pmatrix{\cos\theta+i\,\sin\theta\,\ub\cr\zerob+i\,\zerob}.
$$
Note that $p(\zb_\theta) = \cos^2\theta\,\sin^2\theta
= {\ts{1\over4}}\,\sin^2(2\theta)$, so it follows that for~$0\le \theta
\le \pi/4$, the elements~$\zb_\theta$ lie on distinct orbits, and that
$0\le p\le {\ts{1\over4}}$, with the endpoints of this interval being
the only critical values of~$p$. While~$M = p^{-1}(0)$ is one critical
orbit, the other critical orbit is~$M^* = p^{-1}({1\over4})$, and consists
of the points of the 31-sphere that are at geodesic
distance~${\sqrt2}/2$ from~$M$. It follows from this that~$M^*$ is also
connected and is a single orbit of~$\Spin(10)$. In particular, the
simultaneous level sets of~$q$ and $p$ are exactly the~$\Spin(10)$-orbits
in~$\bbC\otimes\bbO^2$.
For~$0<\theta<\pi/4$, the nearest point on~$M$ to~$\zb_\theta$ is~$\zb_0$,
so the $\Spin(10)$-stabilizer of~$\zb_\theta$ is a subgroup of~$K_1$
that has already been seen to be both connected and simply connected.
Also, the orbit of~$\zb_\theta$ is a 6-sphere bundle over~$M$. By
dimension count, this stabilizer must be of dimension 15 and must contain
the stabilizer in~$K_1$ of~$\oneb$ and $\ub$, which is~$\Spin(6)$.
Thus, the stabilizer of such a~$\zb_\theta$ is exactly $\Spin(6)\simeq\SU(4)$.
In particular, the stabilizer of any point of the 31-sphere not
on~$M$ or~$M^*$ must be a conjugate of~$\SU(4)$.
Now, the tangent space to~$M^*$ at~$\zb_{\pi/4}$ is the set of vectors
of the form
$$
\pmatrix{a_1+ir\,I_8&C\,R_\xb+i\,C\,R_\yb\cr
-C\,L_\xb+i\,C\,L_\yb&a_3-ir\,I_8}
\pmatrix{\oneb+i\,\ub\cr\zerob + i\,\zerob}
=
\pmatrix{(a_1\oneb-r\,\ub)+i\,(a_1\ub+r\oneb)\cr\noalign{\vskip2pt}
-\overline{(\xb+\yb\ub)}+i\,\overline{(\yb-\xb\ub)}}\,.
$$
Thus, the Lie algebra of the stabilizer~$G$ of~$\zb_{\pi/4}$ is
defined by the relations~$a_1\oneb-r\,\ub=a_1\ub+r\oneb=\yb-\xb\ub=\zerob$.
(Remember that~$\ub^2 = -\oneb$.) It follows that~$a_1\in\euso(8)$ must
belong to the stabilizer of the 2-plane spanned by~$\{\oneb,\ub\}$, so
that~$a_1$ lies in~$\euso(2)\oplus\euso(6)$. Conversely, if~$a_1$ lies
in this subspace, then there exists a unique~$r\in\bbR$ so that
$a_1\oneb-r\,\ub=a_1\ub+r\oneb = 0$. From the matrix representation, it is
clear that the maximal torus in $\euso(2)\oplus\euso(6)$ (which has rank~4)
is a maximal torus in the full stabilizer algebra, which has dimension~$24$.
The root pattern is evident from the matrix representation, implying
that the stabilizer algebra is isomorphic to~$\eusu(5)$.
Now~$M^*$ has dimension~$21$ and is the base of a fibration whose total space
is one of the hypersurface orbits and whose fiber is a 9-sphere.
The 2-connectivity of the hypersurface orbits implies, by the long
exact sequence in homotopy, that~$M^*$ is also 2-connected, which implies
that $M^* = \Spin(10)/G$ where $G$ is both connected and simply connected.
Since its Lie algebra is~$\eusu(5)$, it follows that~$G$ is isomorphic
to~$\SU(5)$.
\bigskip
{\bf 4. $\Spin(10,1)$.} To construct the spinor representation
of~$\Spin(10,1)$, it will be easiest to construct the Lie algebra
representation by extending the Lie algebra representation of~$\Spin(10)$
that was constructed in~\S3. It is convenient to identify~$\bbC\otimes\bbO^2$
with~$\bbO^4$ explicitly via the identification
$$
\zb = \pmatrix{\xb_1+i\,\xb_2\cr \yb_1+i\,\yb_2\cr}
= \pmatrix{\xb_1\cr\yb_1\cr\xb_2\cr\yb_2\cr}.
$$
Via this identification, $\euspin(10)$ becomes the subspace
$$
\euspin(10)
= \left\{\pmatrix{
a_1 & C\,R_\xb & -r\,I_8 & -C\,R_\yb\cr
-C\,L_\xb & a_3 & -C\,L_\yb & r\,I_8 \cr
r\,I_8 & C\,R_\yb & a_1 & C\,R_\xb\cr
C\,L_\yb & -r\,I_8 & -C\,L_\xb & a_3 \cr}\
\vrule\ r\in\bbR,\ \xb,\yb\in\bbO,\ a\in\euspin(8)\ \right\}\,.
$$
Consider the one-parameter subgroup~$\bR\subset\SL_\bbR(\bbO^4)$ defined by
$$
\bR
=\left\{\pmatrix{t\,I_{16}&0\cr0&t^{-1}\,I_{16}}
\ \vrule\ t\in\bbR^+\ \right\}.
$$
It has a Lie algebra~$\eur\subset\eusl(\bbO^4)$. Evidently, the
the subspace~$\left[\euspin(10),\eur\right]$ consists of matrices of
the form
$$
\pmatrix{
0_8 & 0_8 & r\,I_8 & C\,R_\yb\cr
0_8 & 0_8 & C\,L_\yb & -r\,I_8 \cr
r\,I_8 & C\,R_\yb & 0_8 & 0_8 \cr
C\,L_\yb & -r\,I_8 & 0_8 & 0_8 \cr},
\qquad r\in\bbR,\ \yb\in\bbO\,.
$$
Let~$\eug = \euspin(10) \oplus\eur\oplus\left[\euspin(10),\eur\right]$.
Explicitly,
$$
\eug
= \left\{\pmatrix{
a_1 +x\,I_8 & C\,R_\xb & y\,I_8 & C\,R_\yb\cr
-C\,L_\xb & a_3 +x\,I_8 & C\,L_\yb & -y\,I_8 \cr
z\,I_8 & C\,R_\zb & a_1 -x\,I_8 & C\,R_\xb\cr
C\,L_\zb & -z\,I_8 & -C\,L_\xb & a_3 -x\,I_8 \cr}\
\vrule\
\matrix{x,y,z\in\bbR,\cr\noalign{\vskip2pt} \xb,\yb,\zb\in\bbO,\cr
\noalign{\vskip2pt}\ a\in\euspin(8)}\ \right\}\,.
$$
Compuation using the Moufang Identities shows that $\eug$ is closed
under Lie bracket and hence is a Lie algebra of (real) dimension~$55$
that contains~$\euspin(10)$. The induced representation
of~$\Spin(10)$ on~$\eug/\euspin(10)$ evidently restricts to~$\Spin(9)$
to preserve the splitting corresponding to the
sum~$\eur\oplus\left[\euspin(10),\eur\right]\simeq\bbR\oplus\bbR^9$ and
acts as the standard (irreducible) representation on the ~$\bbR^9$ summand.
It follows that~$\Spin(10)$ must act via its standard (irreducible, ten
dimensional) representation on~$\eug/\euspin(10)$. Since the trace of the
square of a non-zero element in the
subspace~$\eur\oplus\left[\euspin(10),\eur\right]$ is positive, $\eug$ is
semisimple of non-compact type. It follows that $\eug$ is isomorphic
to~$\euso(10,1)$ and hence is the Lie algebra of a representation
of~$\Spin(10,1)$. This representation must be faithful since it is faithful
on the maximal compact subgroup~$\Spin(10)$.
Thus, define~$\Spin(10,1)$ to be the (connected) subgroup
of~$\SL_\bbR(\bbO^4)$ that is generated by~$\Spin(10)$ and the
subgroup~$\bR$. Its Lie algebra~$\eug$ will henceforth be written
as~$\euspin(10,1)$.
\medskip
{\it Invariant Polynomials and Orbits.} Consider the
$\Spin(10)$-invariant polynomial
$$
p = |\xb_1|^2|\xb_2|^2+|\yb_1|^2|\yb_2|^2
- \left(\xb_1\cdot\xb_2+\yb_1\cdot\yb_2\right)^2
+ 2\,(\xb_1\yb_1)\cdot(\xb_2\yb_2)\,,
$$
Evidently, $p$ is invariant under~$\bR$ and is therefore invariant under
$\Spin(10,1)$. In particular, it follows that the orbits of~$\Spin(10,1)$
on~$\bbO^4\simeq\bbR^{32}$ must lie in the level sets of~$p$.
Also from the previous section, it is known that every element of~$\bbO^4$
lies on the $\Spin(10)$-orbit of exactly one of the elements
$$
\zb_{a,b} = \pmatrix{a\,\oneb\cr\zerob\cr b\,\ub\cr\zerob}
\qquad\hbox{ where $0\le b\le a$.}
$$
and where~$\ub\in\Im\bbO$ is a fixed unit imaginary octonion.
However, all of the elements of the form
$$
\pmatrix{at\,\oneb\cr\zerob\cr (b/t)\,\ub\cr\zerob}
\qquad\hbox{(where $0< t$)}
$$
lie on the same~$\bR$-orbit and, hence, on the same
$\Spin(10,1)$-orbit. Since $p(\zb_{a,b}) = a^2b^2$, it now follows that
each of the nonzero level sets of~$p$ is a single $\Spin(10,1)$-orbit while
the zero level set is the union of the origin and a single
$\Spin(10,1)$-orbit, say, the orbit of~$\zb_{1,0}$.
Moreover, it follows that $p$ generates the ring of $\Spin(10,1)$-invariant
polynomials on~$\bbO^4$.
\medskip
{\it Stabilizers.} Multiplication by positive scalars acts transitively
on the non-zero level sets of~$p$, so they are all diffeomorphic. In
fact, each such level set is contractible to the $\Spin(10)$-invariant
locus where~$q$ reaches its minimum on this level set and this is a
manifold of dimension~$21$ that is diffeomorphic to~$M^*$. In particular,
it follows that each of the non-zero level sets of~$p$ is 2-connected,
so that the stabilizer in~$\Spin(10,1)$ of a point on such a level set
must be connected and simply connected.
If~$\zb\in\bbO^4$ has $p(\zb)\not=0$, then the $\Spin(10,1)$-orbit of~$\zb$
has dimension 31 and so its stabilizer in~$\Spin(10,1)$ must be of
dimension~$55{-}31=24$. Moreover all of these stabilizers must be
conjugate in~$\Spin(10,1)$. Since the $\Spin(10)$-stabilizer of the
point~$\zb_{1,1}$ is already known to be~$\SU(5)$, which has dimension~$24$,
it follows that this must be the~$\Spin(10,1)$-stabilizer as well.
The $\Spin(10,1)$-orbit consisting of nonzero vectors in the zero locus
of~$p$ is just the deleted cone on~$M$, and so has dimension~$25$. Since it
is contractible to~$M$, it is 2-connected, so that the stabilizer
in~$\Spin(10,1)$ of a point in this orbit must be connected and simply
connected and of dimension~$55{-}25=30$. In fact, the Lie algebra of
this stabilizer is just
$$
\left\{\pmatrix{
a_1& 0 & y\,I_8 & C\,R_\yb\cr
0 &a_3& C\,L_\yb & -y\,I_8 \cr
0 & 0 &a_1& 0 \cr
0 & 0 & 0 &a_3\cr}\
\vrule\
\matrix{y\in\bbR,\cr\noalign{\vskip2pt} \yb\in\bbO,\cr
\noalign{\vskip2pt}\ a\in\euk_1}\ \right\}
$$
where~$\euk_1$ is the Lie algebra of~$K_1\subset\Spin(8)$. Thus, the
stabilizer is a semi-direct product of~$\Spin(7)$ with a copy of~$\bbR^9$.
\bigskip
{\bf 5. $\Spin(10,2)$.} It might be tempting to conjecture that
$\Spin(10,1)$ could be defined directly as the stabilizer of~$p$. However,
this is not the case, as the stabilizer of~$p$ is larger.
One can see this directly by looking at the alternative expression
$$
p = |\xb_1\w\xb_2|^2 + |\yb_1\w\yb_2|^2
- 2\,(\xb_1\cdot\xb_2)\,(\yb_1\cdot\yb_2)
+ 2\,(\xb_1\yb_1)\cdot(\xb_2\yb_2)\ .
$$
which makes it evident that~$p$ is invariant under the 6-dimensional
Lie group
$$
G = \left\{\pmatrix{
a\,I_8 & 0 & b\,I_8 & 0\cr
0 & a'\,I_8 & 0 & b'\,I_8 \cr
c\,I_8 & 0 & d\,I_8 & 0\cr
0 & c'\,I_8 & 0 & d'\,I_8 }\
\vrule\ ad{-}bc=a'd'{-}b'c'=\pm1\ \right\}\,.
$$
Since $G$ does not lie in~$\Spin(10,1)$, the invariance group of~$p$ must
be properly larger than~$\Spin(10,1)$.
In particular, consider the $G$-subgroup~$\bR'\simeq\bbR^+$ consisting of
matrices of the form
$$
\pmatrix{
t\,I_8 & 0 & 0 & 0\cr
0 & t^{-1}\,I_8 & 0 & 0 \cr
0 & 0 & t^{-1}\,I_8 & 0\cr
0 & 0 & 0 & t\,I_8 }\qquad \hbox{where $t>0$,}
$$
which is not a subgroup of~$\Spin(10,1)$. Let~$\eur'$ denote its Lie algebra.
Calculation shows that
$$
\eur'\oplus\left[\euspin(10,1),\eur'\right]
= \left\{\pmatrix{
w\,I_8 & C\,R_\wb & u\,I_8 & 0\cr
C\,L_\wb & -w\,I_8 & 0 & u\,I_8 \cr
v\,I_8 & 0 & -w\,I_8 & -C\,R_\wb\cr
0 & v\,I_8 & -C\,L_\wb & w\,I_8 \cr}\
\vrule\
\matrix{u,v,w\in\bbR,\cr\noalign{\vskip2pt} \wb\in\bbO}\ \right\}\,.
$$
and that the
sum~$\euspin(10,1)\oplus\eur'\oplus\left[\euspin(10,1),\eur'\right]$
is closed under Lie bracket. Thus, this defines a Lie algebra of
dimension~$66$ that lies in the stabilizer of~$p$.
The details of further analysis will be omitted, but by using arguments
similar to those used in previous sections, one sees that this algebra
is isomorphic to~$\euso(10,2)$ and that the connected Lie subgroup of
$\GL_\bbR(\bbO^4)$ whose Lie algebra is this one is simply connected,
so that it this group is~$\Spin(10,2)$. Henceforth, this algebra will
be denoted~$\euspin(10,2)$. Thus,
$$
\euspin(10,2)
= \left\{\pmatrix{
a_1 +x\,I_8 & C\,R_\wb & y\,I_8 & C\,R_\yb\cr
C\,L_\xb & a_3 +w\,I_8 & C\,L_\yb & u\,I_8 \cr
z\,I_8 & C\,R_\zb & a_1 -x\,I_8 & -C\,R_\xb\cr
C\,L_\zb & v\,I_8 & -C\,L_\wb & a_3 -w\,I_8 \cr}\
\vrule\
\matrix{u,v,w,x,y,z\in\bbR,\cr\noalign{\vskip2pt} \wb,\xb,\yb,\zb\in\bbO,\cr
\noalign{\vskip2pt}\ a\in\euspin(8)}\ \right\}\,.
$$
Moreover, representation theoretic methods show that the only
connected proper subgroup of~$\SL_\bbR(\bbO^4)$ that properly
contains~$\Spin(10,2)$ is~$\Sp(16,\bbR)$, the symplectic group
preserving the symplectic $2$-form~$\Omega$ defined by
$$
\Omega = d\xb_1\,\hat{\cdot}\,d\xb_2 + d\yb_1\,\hat{\cdot}\,d\yb_2\,.
$$
Of course, $\Sp(16,\bbR)$ does not stabilize any nonzero polynomials.
It follows that~$\Spin(10,2)$ is the identity component of the stabilizer
of~$p$, and hence that the stabilizer of~$p$ must lie in the
normalizer of~$\Spin(10,2)$ in~$\GL_\bbR(\bbO^4)$. However, this
normalizer is just~$\bbR^+{\cdot}I_{32}\times\Spin(10,2)$ and the only
element in~$\bbR^+{\cdot}I_{32}$ that stabilizes~$p$ is the identity
element. It follows that $\Spin(10,2)$ is the stabilizer of~$p$.
\bigskip
{\bf 6. $\Spin(9,1)$.} As a final note, inspection reveals that the
subalgebra
$$
\euspin(9,1) =
\left\{\pmatrix{
a_1 +x\,I_8 & C\,R_\wb \cr
C\,L_\xb & a_3 -x\,I_8 }\
\vrule\
\matrix{ x\in\bbR,\cr\noalign{\vskip2pt} \wb,\xb\in\bbO,\cr
\noalign{\vskip2pt}\ a\in\euspin(8)}\ \right\}
\subset\eusl(16,\bbR)\,,
$$
which contains~$\euspin(9)$, is actually the Lie algebra of a faithful
representation of~$\Spin(9,1)$ on~$\bbR^{16}\simeq\bbO^2$. This action
of~$\Spin(9,1)$ has the interesting feature that it has only two orbits:
The origin and the set of all non-zero vectors. This follows because the
compact group~$\Spin(9)\subset\Spin(9,1)$ already acts transitively on
the unit spheres, but the larger group does not even preserve the
quadratic form.
\bye