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\begin{document}

\title[14-dimensional quadratic forms]
  {On 14-dimensional quadratic forms, their spinors, and the
  difference of two octonion algebras}
\author{Markus Rost}
\address{NWF I - Mathematik,
  Universit\"at Regensburg,
  D-93040 Regensburg, Germany}
%\curraddr{}
\email{markus.rost@mathematik.uni-regensburg.de}
\urladdr{http://www.physik.uni-regensburg.de/\~{}rom03516}
%\thanks{}
%\subjclass{}
\date{\today}
\date{March 9, 1999}
\maketitle

This text originates from an E-mail to in May 1994.  It has been
revised on September 18, 1996 and March 9, 1999 (but still may contain
some inaccuracies).

\dots

Let me give a short report on some thoughts on low dimensional
quadratic forms. ``low'' means here up to dimension~$14$.

\section{Introduction.}

One may start with the following: Consider a quadratic form~$q$ of
dimension~$d$ and with trivial $e^1$- and $e^2$-invariants.  Then $q$
comes from a torsor $\tilde q\in H^1\bigl(\Spin(d)\bigr)$.  Consider a
fundamental spinor representation~$S$ of $\Spin(d)$ and let $S(\tilde
q)$ be the corresponding ``twisted'' representation.

It turns out that if $d$ is smaller or equal to~$14$, then the
dimension of $S$ is smaller than the dimension of the group
$\Spin(d)$.  These dimensions are ``exceptional'' in this sense.  The
significance is as follows: If you take any nontrivial vector $s\in
S(\tilde q)$, it will have a nontrivial isotropy group, and one may
hope that this gives an interesting reduction of the torsor $\tilde
q\in H^1\bigl(\Spin(d)\bigr)$ to $H^1$ of a subgroup of~$\Spin(d)$.

This is indeed the case: The isotropy group of a generic spinor $s\in
S$ has the $1$-component

\begin{itemize}
  \item
  $G_2 \times G_2$ (for $d=14$),
  \item
  $\SL(6)$ (for $d=12$),
  \item
  semi-direct product of an additive group with $\Spin(7)$ (for
  $d=10$),
  \item
  $\Spin(7)$ (for $d=8$, $9$),
  \item
  $G_2$ (for $d=7$).
\end{itemize}

Moreover if $d=10$, then the ($1$-component of the) isotropy group of
a generic vector in $S\oplus S$, (i.e., the intersection of the
isotropy group of two generic~$s$) is~$G_2$.

Let me note that for $d=9$ the embedding of the isotropy group
$\Spin(7)\to\Spin(9)$ is not the usual one: it is in $\Spin(8)$
twisted by a triality, so that the center of $\Spin(7)$ does not map
to the center of~$\Spin(9)$.  Similarly for $\Spin(7)\to\Spin(10)$ in
the case $d=10$.

The computations (together with computations of the normalizers, e.g.,
in the $14$-dimensional case the normalizer is $(G_2\times G_2)\rtimes
\mu_8$) yield the following:

\begin{itemize}
  \item
  for a $14$-dimensional form~$q$ in~$I^3$ there is a quadratic
  extension $L=F(\sqrt a)$ and a $3$-fold Pfister form $p$ over~$L$,
  such that $q$ is the trace of $\sqrt ap'$, where $p'$ is the pure
  subform of $p$.  (This was not known to me before and I don't know
  any other proof than the one described below.)
  \item
  a $12$-dimensional form in $I^3$ is the trace of a $6$-dimensional
  hermitian form with trivial determinant.  (A known fact.)
  \item
  a $10$-dimensional form in $I^3$ is isotropic.  (A well-known fact.)
  \item etc.
\end{itemize}

I will discuss here only the cases $d=10$, $14$. The case $d=12$ is of
different nature and will not be considered.

\section{A construction of ``the difference of two composition
algebras''}

The goal of the following considerations is to give a certain
description of the ($64$-dimensional) spinor representation of
$\Spin(14)$.  The advantage of this description is, that it makes it
easy to determine the dimension of the main orbit and to compute its
isotropy group.

I assume $\car(F)\neq2$, or even $\car(F)=0$.

Let $C$ be a composition algebra over a field~$F$ with norm form~$N$
and unit~$e$ and let
\begin{displaymath}
  C = e F \oplus V
\end{displaymath}
be the orthogonal decomposition.

Let $D$ be another composition algebra over~$F$ with norm form~$M$ and
unit~$f$ and let
\begin{displaymath}
  D = f F \oplus W
\end{displaymath}
be the orthogonal decomposition.

Put
\begin{displaymath}
  R = C\otimes D
\end{displaymath}
and
\begin{displaymath}
  L =\bigwedge^2(V\oplus W)=\bigwedge^2 V \oplus \bigwedge^2 W \oplus
  V\otimes W.
\end{displaymath}
I define a map
\begin{displaymath}
  \Theta\colon L \longrightarrow \End (R)
\end{displaymath}
by the following formulas:
\begin{align*}
  \Theta(v\wedge\bar v)(c\otimes d) & = [v\cdot (\bar v\cdot c)-\bar
  v\cdot (v\cdot c)] \otimes d,
  \\
  \Theta(w\wedge\bar w)(c\otimes d) & = c \otimes [w\cdot (\bar w\cdot
  d)-\bar w\cdot (w\cdot d)],
  \\
  \Theta (v\wedge w)(c\otimes d) &= (v\cdot c) \otimes (w\cdot d).
\end{align*}
Here $x\cdot y$ is the product in the composition algebras.

\begin{lemma}
  The map\/ $\Theta$ is injective and its image is a Lie subalgebra.
\end{lemma}

\begin{proof}
  This follows from standard rules for composition algebras.
\end{proof}

Let
\begin{displaymath}
  p\colon V \longrightarrow F,\quad p(v)=N(v)
\end{displaymath}
be the restriction of~$N$.  I identify $\bigwedge^2 V$ with the Lie
algebra~$\so(p)$.  (In the natural way as subspaces of the even
Clifford algebra of~$p$.)

Let
\begin{displaymath}
  J\colon \bigwedge^2 V\longrightarrow V,\quad J(v\wedge w)=v\cdot
  w-w\cdot v.
\end{displaymath}

\begin{itemize}
  \item
  Suppose dim $C=4$. Then $C$ is a quaternion algebra and $J$ is
  bijective.
  \item
  Suppose dim $ C=8$. Then $C$ is a octonion algebra. In this case $J$
  is surjective and its kernel is the Lie algebra of $G_2$.
\end{itemize}

Let
\begin{displaymath}
  q\colon W \longrightarrow F,\quad q(v)=M(v)
\end{displaymath}
be the restriction of~$M$.  I identify $\bigwedge^2 W$ with the Lie
algebra $\so(q)$.

Let
\begin{displaymath}
  U = V\oplus W
\end{displaymath}
and consider the quadratic form
\begin{gather*}
  [p-q]\colon U \longrightarrow F,\\ [p-q](v,w)=p(v)-q(w).
\end{gather*}
Note that $L=\bigwedge^2 U$.  I identify $L$ with the Lie algebra of
$\so([p-q])$.  In this way the inclusion
\begin{displaymath}
  \bigwedge^2V \oplus \bigwedge^2 W \longrightarrow L
\end{displaymath}
is identified with the inclusion
\begin{displaymath}
  \so(p) \oplus \so(q) \longrightarrow \so([p-q]).
\end{displaymath}

\begin{lemma}
  With these identifications (with the Lie algebra structures
  eventually altered by a scalar factor), the map\/~$\Theta $ is a
  homomorphism of Lie algebras.
\end{lemma}

\begin{proof}
  By carefully comparing both Lie brackets.
\end{proof}

\section{The main orbit}

We have now established $R$ as $\Spin([p-q])$-module.  The tangent
space of the $\Spin([p-q])$-orbit of some vector $r$ of~$R$ is
$\Theta(L)(r)$.

We now assume that $\dim C$, $\dim D$ are equal to~$4$ or to~$8$.
This means that the maps $J$ are in both cases surjective.

Let us consider the specific element $r=e\otimes f$. One finds for the
tangent space of the orbit through $r$:
\begin{displaymath}
  \Theta (L)(e\otimes f) = e\otimes W \oplus V\otimes f \oplus
  V\otimes W.
\end{displaymath}

This is a subspace of codimension~$1$, transversal to~$r$.  So, if we
extend the group by scalar multiplication, we see that the orbit
of~$r$ is open.  Therefore $r$ generates a generic line and the orbit
of~$r$ has codimension~$1$.

Let us compute the Lie algebra of the isotropy group of~$r$.  This is
the sum of the Lie algebras of the isotropy groups of $e$
and~$f$. These are the kernels of the maps~$J$ on $\so(p)=
\bigwedge^2V$, and on $\so(q)=\bigwedge^2 W$. These are trivial if
$C$, $D$ have dimension~$4$ and are the Lie algebras of~$G_2$ if $C$,
$D$ have dimension~$8$.

Since the main orbit has codimension~$1$ it follows that there is an
invariant form
\begin{displaymath}
  \psi \colon R \longrightarrow F
\end{displaymath}
on~$R$ such that the varieties
\begin{displaymath}
  \{\, \psi = \text{nonzero constant}\,\}
\end{displaymath}
are the main orbits.

\section{On the case $\dim C=\dim D=8$}

Consider the above construction for split $C$ and $D$ of
dimension~$8$.  Then we are given a representation
\begin{displaymath}
  \text{split-}\Spin(14)\to\End(R).
\end{displaymath}

Now given a $14$-dimensional quadratic form $h\colon X\longrightarrow
F$ with trivial $e^1$- and $e^2$-invariants, we may twist this
representation to a representation
\begin{displaymath}
  \Spin(h)\longrightarrow \End(R).
\end{displaymath}
leaving invariant a certain form $\rho\colon R\longrightarrow F$, with
$\rho$ the twist of~$\psi$.

If you take a vector $r$ in~$R$ such that $\rho(r)$ is nontrivial,
then there comes along a decomposition $X=Y\oplus Z$, such that $h$
restricted to~$Y$ and $h$ restricted to~$Z$ are similar to the pure
subform of $8$-dimensional composition algebras $C$ and~$D$,
respectively.  Moreover $R=C\otimes D$.

Well, this is not quite true: The decomposition $X=Y\oplus Z$ might be
defined only after passing to a quadratic extension~$E/F$, whose
Galois group interchanges the two factors. In fact, there are elements
in the normalizer of $G_2\times G_2$ which interchange the two
factors.

To make this more precise we compute the normalizer of $G_2\times
G_2$:

\begin{lemma}
  The normalizer of\/ $G_2\times G_2$ in $\Spin(7,7)$ is $(G_2\times
  G_2)\rtimes \mu_8$.  Here a generator $\omega\in\mu_8$ acts on
  $G_2\times G_2$ via $(g,h)\mapsto(h,g)$.
\end{lemma}
\begin{proof}
  Every automorphism of $G_2$ is inner and $G_2\to\GL(7)$ is an
  irreducible representation.  Therefore the normalizer of $G_2\times
  G_2$ in $\GL(14)$ is
  \begin{displaymath}
    \bigl((G_2\times\Gm)\times(G_2\times\Gm)\bigr)\rtimes\LZ/2.
  \end{displaymath}
  Intersecting this group with $\SO(7,7)$ yields the normalizer of
  $G_2\times G_2$ in $\SO(7,7)$:
  \begin{displaymath}
    (G_2\times G_2)\rtimes\mu_4
  \end{displaymath}
  where $\mu_4$ is considered as a subgroup of $\SO(7,7)$ via
  \begin{displaymath}
    \zeta\mapsto \tau(\zeta):=\zeta
    \begin{pmatrix}
      0&1\\1&0
    \end{pmatrix}.
  \end{displaymath}
  The extension $\mu_2\to\Spin(7,7)\to\SO(7,7)$ is nontrivial on the
  subgroup $\tau(\mu_4)\subset \SO(7,7)$, whence the claim.

  To be more specific, we describe $\mu_8$ as a subgroup of
  $\Spin(7,7)$ explicitly.  Let $v_1$, \dots, $v_7$, $w_1$, \dots,
  $w_7$ be an orthogonal basis with $\langle v_i, v_i\rangle=1$ and
  $\langle w_i, w_i\rangle=-1$.  Consider the element
  \begin{displaymath}
    \omega = \prod_{i=1}^7 \left( \frac{1+\zeta v_iw_i}{\sqrt 2}
    \right) \in \Spin(7,7)\subset C_0(7,7)
  \end{displaymath}
  where $\zeta$ is a primitive $4$-th root of unity.  Then $\omega$ is
  of order~$8$ and its image in $\SO(7,7)$ is~$\tau(\zeta)$.
\end{proof}

One concludes that the map
\begin{displaymath}
  H^1\bigl(F,(G_2 \times G_2) \rtimes \mu_8\bigr) \to
  H^1\bigl(F,\Spin(7,7)\bigr)
\end{displaymath}
is surjective.

This gives the mentioned result on $14$-dimensional quadratic forms.

\section{On the cases $\dim D=4$, $\dim C=4$, $8$.}

\subsection{The case $\dim C=8$ and $\dim D=4$}
Here the isotropy group is $G_2$ with normalizer $G_2\times
\Spin(3)\cdot \mu_4$.

The associativity of~$D$ shows that the group action is compatible
with the right $D$-module structure of $R=C\otimes D$.  Moreover
$R=S\oplus S$, where $S$ is a fundamental spinor representation
of~$\Spin([p-q])$.

Suppose that $D$ is split, $D=\End(K)$ where $K$ is a $2$-dimensional
vector space. Then $S$ is the tensor product over $D$ of $R$ with $K$
so that $S=C$ $\otimes$ $K$. A computation similar as above shows that
the $\Spin([p-q])$-group action on~$S$ has a dense orbit.

\subsection{The case $\dim C=\dim D=4$}
Here the isotropy group is trivial.  If $\dim D=\dim C=4$, then $R$ is
an algebra and the group action is given by multiplication from the
left with $\Spin([p-q])=\SL(R)$.  In this case the form $\psi$ is the
reduced norm on~$R$.

\end{document}

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