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\begin{document}

\title{On the Galois cohomology of Spin(14)}
\author{Markus Rost}
\address{Fakult\"at f\"ur Mathematik,
Universit\"at Bielefeld,
Postfach 100131,
33501 Bielefeld,
Germany}
\email{rost@math.uni-bielefeld.de}
\urladdr{http://www.math.uni-bielefeld.de/\~{}rost}
%% \address{NWF I - Mathematik,
%%   Universit\"at Regensburg,
%%   D-93040 Regensburg, Germany}
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%% \email{markus.rost@mathematik.uni-regensburg.de}
%% \urladdr{http://www.physik.uni-regensburg.de/\~{}rom03516}
%\thanks{}
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\date{March 18, 1999; June 6, 2006}
%\date{\today}
\maketitle

\centerline{\tt Preliminary Notes}

\section*{Note from May/June 2006}

I am very grateful to Skip Garibaldi for comments.  They led to
several corrections and additions.

In the version from 1999 I had claimed without proof
$\ed(\Spin_{13})=6$.  I have now added a new section
(Section~\ref{sec:cohomology-spin13}) containing a proof.

\section*{Abstract}

Let $k$ be a field with $\car k\neq2$.  For $i=6$,~$7$ we define
invariants
\begin{displaymath}
  h_i\colon H^1\bigl(k,\Spin(14)\bigr) \to
  H^i(k,\LZ/2)/(-1)H^{i-1}(k,\LZ/2).
\end{displaymath}

Further we show that the natural map
\begin{displaymath}
  H^1\bigl(k,(G_2\times G_2)\rtimes\mu_8\bigr) \to
  H^1\bigl(k,\Spin(14)\bigr)
\end{displaymath}
is surjective.

One concludes that the essential dimension of $\Spin(14)$ is equal
to~$7$.

Similar considerations are done for~$\Spin(12)$.  We also present the
list of essential dimensions of the split groups $\Spin(n)$ for $n\leq
14$.

\tableofcontents

\section{The Arason invariant}

\subsection{The invariants $e_i$, $i\leq3$}

Let
\begin{displaymath}
  e_i\colon I^i(k)/I^{i+1}(k) \to H^i(k,\LZ/2), \quad\text{$i=0$,
  \dots,~$3$}
\end{displaymath}
be the first invariants on the graded Witt ring given by dimension,
discriminant, the Hasse-Witt invariant, and Arason's invariant,
cf.~\cite{Arason:75,Scharlau:85}.

\subsection{The split groups of type~$D_n$}

We denote by $\SO(n,n)$ the automorphism group of the quadratic form
\begin{displaymath}
  \sum_1^n(x_i^2-y_i^2).
\end{displaymath}
Furthermore, $\Spin(n,n)$ denotes the universal cover of~$\SO(n,n)$
and
\begin{displaymath}
  \PSO(n,n)=\SO(n,n)/\{\pm1\}=\Spin(n,n)/\mu
\end{displaymath}
denotes the corresponding adjoint group.  Here $\mu$ is the center
of~$\Spin(n,n)$.  One has $\mu=\mu_2\times\mu_2$ if $n$ is even and
$\mu=\mu_4$ if $n$ is odd.

If $n$ is odd, every split group of type~$D_n$ is isomorphic to one of
$\Spin(n,n)$, $\SO(n,n)$,~$\PSO(n,n)$.

\subsection{Galois cohomology of $\SO(n,n)$}

The set $H^1\bigl(k,\SO(n,n)\bigr)$ consists of the isomorphism
classes of $2n$-dimensional quadratic forms with trivial discriminant.
We consider $H^1\bigl(k,\SO(n,n)\bigr)$ as a subset of $I^2(k)\subset
W(k)$.

The image of
\begin{displaymath}
  H^1\bigl(k,\SO(n,n)\bigr) \to H^1\bigl(k,\PSO(n,n)\bigr)
\end{displaymath}
consists of the similarity classes of the quadratic forms
in~$H^1\bigl(k,\SO(n,n)\bigr)$.  For $u\in
H^1\bigl(k,\Spin(n,n)\bigr)$ let $q_u$ be the corresponding quadratic
form.

The image of
\begin{displaymath}
  H^1\bigl(k,\Spin(n,n)\bigr) \to H^1\bigl(k,\SO(n,n)\bigr)
\end{displaymath}
consists of those classes in $H^1\bigl(k,\SO(n,n)\bigr)$ with trivial
Hasse-Witt invariant.

\subsection{The invariant $\tilde e_3$ in $K_3^M/2$}

Let $K_n^Mk$ be Milnor's $K$-group~\cite{Milnor:70}.

By Merkurjev's theorem~\cite{Arason:84,Merkurjev:81,Wadsworth:86} the
invariant~$e_2$ is bijective.  Furthermore, Milnor's homomorphism
\begin{displaymath}
  s_3\colon K_3^Mk/2\to I^3(k)/I^4(k)
\end{displaymath}
is bijective
(cf.~\cite{Jacob-Rost:87,Merkurjev-Suslin:86,Milnor:70,Ro:88x}).

Putting things together yields natural maps
\begin{displaymath}
  \tilde e_3\colon H^1\bigl(k,\Spin(n,n)\bigr) \to K_3^Mk/2.
\end{displaymath}
For $u\in H^1\bigl(\Spin(n,n)\bigr)$ the class $\tilde e_3(u)$ depends
alone on~$q_u$.  For $u\in H^1\bigl(\Spin(8,8)\bigr)$ the
corresponding quadratic form~$q_u$ is a $3$-fold Pfister form
(cf.~\cite{Elman-Lam:72,Lam:73,Pfister:95,Scharlau:85}); if
$q_u=\pfister{a,b,c}$, then $\tilde e_3(u)=\{a,b,c\}$.  Furthermore,
the maps~$\tilde e_3$ behave additively with respect to the natural
inclusions
\begin{displaymath}
  \Spin(n,n)\times \Spin(m,m)\to \Spin(n+m,n+m).
\end{displaymath}
These properties determine the family of maps~$\tilde e_3$ uniquely.

\section{Reduced squares}

It has been observed by Serre that for any $n\geq2$ there is a natural
map
\begin{displaymath}
  \OP \colon K^M_n k/2\to K^M_{2n} k/(2K^M_{2n} k + \{-1\}^{n-1}
  K^M_{n+1}k)
\end{displaymath}
characterized by
\begin{displaymath}
  \OP\Bigl(\sum_ix_i\Bigr)=\sum_{i<j}x_ix_j \bmod (2K^M_{2n} k +
  \{-1\}^{n-1} K^M_{n+1}k)
\end{displaymath}
where~$x_i$ are symbols.  (An element $x\in K^M_n k/2$ is called a
symbol if it is of the form $x=\{a_1,\dots,a_n\}$ for some $a_i\in
k^*$.)

To define the operation~$\OP$ one checks that the right hand side of
this formula does not depend on the presentation of an element as a
sum of symbols.  This follows easily from the definition of Milnor's
$K$-theory and the identity $\{a,a\}=\{a,-1\}$, cf.~\cite{Milnor:70}.

Let
\begin{displaymath}
  \alpha_n\colon K_n^Mk/2\to H^n(F,\LZ/2)
\end{displaymath}
be the norm residue homomorphism~\cite{Milnor:70}.  Milnor's
conjecture (cf.~\cite{Voevodsky:96}) asserts that $\alpha_n$ is
bijective.  With Milnor's conjecture, the operations~$P$ give rise to
corresponding maps
\begin{displaymath}
  H^n(k,\LZ/2)\to H^{2n}(k,\LZ/2)/(-1)^{n-1}H^{n+1}(k,\LZ/2).
\end{displaymath}
Combining this with the fact that $(-1)H^{2n-1}(k,\LZ/2)$ is in the
kernel of the natural maps $H^{2n}(k,\LZ/2)\to H^{2n}(k,\LZ/4)$, one
obtains operations
\begin{displaymath}
  H^n(k,\LZ/2)\to H^{2n}(k,\LZ/4).
\end{displaymath}
In the case $n=2$ this operation is nothing else than the Pontryagin
square, cf.~\cite
{Eilenberg-MacLane:54,Eilenberg-MacLane:54a,Whitehead:50,Whitehead:51}.
For $n>2$ I don't know any explanation of the operations~$\OP$ by an
operation defined on the cohomology of topological spaces.

\section{Lambda operations}
\label{sec:lambda-powers}

Let $\GW(k)$ be the Grothendieck (-Witt) ring of quadratic forms
over~$k$.  One defines $\lambda$-operations
\begin{displaymath}
  \lambda^i\colon\GW(k)\to\GW(k)
\end{displaymath}
in the usual fashion (see for instance~\cite{Karoubi:78}):

For a quadratic form $\varphi\colon V\to k$ let
$\lambda^i\varphi\colon\bigwedge^i V\to k$ be its $i$-th exterior
power.  One has $\lambda^0\varphi=\qform1$ and
$\lambda^1\varphi=\varphi$.  The form~$\lambda^2$ is also given by the
Killing form on the Lie algebra~$\so(\varphi)$ (at least if
$\bar\LQ\subset k$).

One forms the formal power series
\begin{displaymath}
  \lambda_t\varphi=\sum_{i\geq0}t^i\lambda^i\varphi.
\end{displaymath}
Then
\begin{displaymath}
  \lambda_t(\varphi\orth\psi)=\lambda_t \varphi\otimes\lambda_t\psi.
\end{displaymath}
The series $\lambda_t$ extends to~$\GW(k)$ by
\begin{displaymath}
  \lambda_t(\varphi-\psi)=\lambda_t \varphi\otimes(\lambda_t\psi)^{-1}
\end{displaymath}
and the operations $\lambda^i$ on~$\GW(k)$ are defined by
\begin{displaymath}
  \lambda_t(x)=\sum_{i\geq0}t^i\lambda^i(x)
\end{displaymath}
for $x\in\GW(k)$.

We are mainly interested in~$\lambda^2$.  Note that
\begin{align*}
  \lambda^0(x)&=1,\\
  \lambda^1(x)&=x,\\
  y^2&=\dim y+2\lambda^2(y),\\
  \lambda^2(x+y)&=\lambda^2(x)+xy+\lambda^2(y),\\
  \lambda^2(x-y)&=\lambda^2(x)-y(x-y)-\lambda^2(y),\\
  \lambda^2(x-y)&=\lambda^2(x)-xy+\dim y+\lambda^2(y),\\
  \lambda^2(\qform a x)&=\lambda^2(x)
\end{align*}
for $x$, $y\in\GW(k)$ and $a\in k^*$.

Let $\GI(k)\subset\GW(k)$ be the fundamental ideal of zero dimensional
virtual quadratic forms.  The projection $\GW(k)\to W(k)$ induces
identifications $\GI^n(k)=I^n(k)$ for $n>0$.  $\GI^n(k)$ is additively
generated by elements of the form
\begin{displaymath}
  \pfister{a_1,\dots,a_n}-\pfister1^n =
  \pfister{a_1,\dots,a_{n-1}}-\qform{a_n}\pfister{a_1,\dots,a_{n-1}}.
\end{displaymath}

\begin{lemma}
  \label{lem:lambda-Pfister}
  Let $\varphi$ be an $n$-fold Pfister form and
  $x=\varphi-\pfister1^n$.  Then
  \begin{displaymath}
    \lambda^2(x)=\pfister{-1}^{n-1}x.
  \end{displaymath}
\end{lemma}

\begin{proof}
  Write $\varphi=\psi\pfister a$ where $\psi$ is an $(n-1)$-fold
  Pfister form and where $a\in k^*$.  Then $x=\psi-\qform a \psi$ and
  one finds
  \begin{align*}
    \lambda^2(x)&= \lambda^2(\psi-\qform a \psi)
    \\
    &=\lambda^2(\psi)-\qform a \psi x -\lambda^2(\psi)
    \\
    &=-\qform a \pfister{-1}^{n-1} x
    \\
    &=\pfister{-1}^{n-1}\qform{-a}x=\pfister{-1}^{n-1}x
  \end{align*}
  Here one uses $\psi^2=\pfister{-1}^{n-1}\psi$, $\qform{-a}x=-\qform
  a x$ if $\dim x=0$, and $\qform{-a}\pfister a=\pfister a$.
\end{proof}

\begin{corollary}
  \label{cor:lambda-pfister}
  Let $\varphi$ be an $n$-fold Pfister form.  Then
  \begin{align*}
    \lambda^2(\varphi)&\simeq\varphi'\pfister{-1}^{n-1},\\
    \lambda^2(\varphi')&\simeq\varphi'(\pfister{-1}^{n-1})'.\qquad\qed
  \end{align*}
\end{corollary}

We define operations
\begin{gather*}
  P'\colon I^n(k)\to I^{2n}(k),\\
  P'(x)=\lambda^2(x)-\pfister{-1}^{n-1}x.
\end{gather*}

It follows from Lemma~\ref{lem:lambda-Pfister} and
$\lambda^2(x+y)=\lambda^2(x)+xy+\lambda^2(y)$ that indeed $P'(x)\in
I^{2n}(k)$.

These operations lift the operations~$P$ to the Witt ring.

\section{Multiplicative transfer}

Let $L/F$ be separable field extension.  In addition to the
restriction map
\begin{displaymath}
  r_{L/F}\colon W(F)\to W(L), \quad[\varphi]\mapsto [\varphi_L]
\end{displaymath}
and the corestriction map
\begin{displaymath}
  c_{L/F}\colon W(L)\to W(F), \quad[\psi]\mapsto [\trace_{L/F}\varphi]
\end{displaymath}
one may define a multiplicative transfer map
\begin{displaymath}
  N_{L/F}\colon W(L)\to W(F).
\end{displaymath}
This map is analogous to the multiplicative transfer in cohomology,
cf.~\cite{Evens:63,Kahn:84,Shapiro:79}.

We are interested in the case~$\degre LF=2$.  Let $\sigma$ denote the
generator of the Galois group.  Then for a quadratic form~$\psi\colon
W\to L$ the form $N_{L/F}(\psi)$ is given by the restriction of
$\psi\otimes{}^\sigma\psi\colon W\otimes{}^\sigma W\to L$ to the
subspace of invariants $(W\otimes{}^\sigma W)^\sigma$.

Suppose $L=F(\sqrt a)$.  One has the following rules
\begin{align*}
  \dim_F\bigl(N_{L/F}(\psi)\bigr)&=(\dim_L\psi)^2,\\
%  N_{L/F}\bigl(r(x)\bigr)&=x^2,\\
  N_{L/F}(\qform\alpha)&=\qform{N_{L/F}(\alpha)},\\
  N_{L/F}(x+y)&=N_{L/F}(x)+c_{L/F}(x\sigma(y))+N_{L/F}(y),\\
  N_{L/F}(x-y)&=N_{L/F}(x)-c_{L/F}(x\sigma(y))+N_{L/F}(y),\\
  \lambda^2\bigl(c_{L/F}(x)\bigr)&= c_{L/F}\bigl(\lambda^2(x)\bigr) +
  aN_{L/F}(x),
  \\
  N_{L/F}(\pfister\alpha)&=\pfister a +
  \begin{cases}
    \pfister{\trace\alpha, -aN_{L/F}(\alpha)}&\text{if
    $\trace\alpha\neq 0$,}
    \\
    0& \text{if $\trace\alpha=0$,}
  \end{cases}
  \\
  N_{L/F}(\pfister{\alpha_1,\dots,\alpha_n})&=\pfister a^n +
  \begin{cases}
    \prod_i\pfister{\trace\alpha_i, -aN_{L/F}(\alpha_i)}&\text{if
    $\trace\alpha_i\neq 0$,}
    \\
    0& \text{else.}
  \end{cases}
\end{align*}

In particular, if $-1$ is a square in~$F$, then
\begin{displaymath}
  N_{L/F}\bigl(I^n(L)\bigr)\subset I^{2n}(F)
\end{displaymath}
for $n\geq2$.

\section{The invariants $h_6$ and~$h_7$}

For this section it is assumed for simplicity that $\sqrt{-1}\in k$.

We define
\begin{gather*}
  h_6\colon H^1\bigl(k,\Spin(7,7)\bigr) \to H^6(k,\LZ/2),
  \\
  h_6(u) = \alpha_6\circ\OP\circ \tilde e_3(u).
\end{gather*}
The invariant~$h_6(u)$ depends only on~$q_u$.

By the remarks of Section~\ref{sec:lambda-powers} one can lift this
invariant to $I^6(k)$.

In some cases the invariant~$h_6$ can be described explicitly.  For a
Pfister form~$\varphi$ one denotes by $\varphi'$ its pure subform (one
has $\varphi=\qform1\orth\varphi'$).  Let $a_i$, $b_i$, $c\in k^*$,
$i=1$, $2$, $3$, and put
\begin{equation}
  \label{eq:diff-pfister}
  q=c(\pfister{a_1,a_2,a_3}'\orth-\pfister{b_1,b_2,b_3}')
\end{equation}
Then $q=q_u$ for some $u\in H^1\bigl(k,\Spin(7,7)\bigr)$ and for any
such~$u$ one finds
\begin{equation}
  \label{eq:diff-pfister-h6}
  h_6(u)=(a_1,a_2,a_3,b_1,b_2,b_3).
\end{equation}

\begin{lemma}
  \label{lem:isotr-h6}
  Let $u\in H^1\bigl(k,\Spin(7,7)\bigr)$.  If $q_u$ is isotropic, then
  $h_6(u)=0$.
\end{lemma}

\begin{proof}
  If $q_u$ is isotropic, the $q_u$ has a
  representation~(\ref{eq:diff-pfister}) with $a_1=b_1$, see
  \cite[Satz 14, Zusatz]{Pfister:66} or~\cite{Ro:94x}.  The claim
  follows from (\ref{eq:diff-pfister-h6}).
\end{proof}

\begin{proposition}
  \label{prop:h7-well-def}
  Let $u\in H^1\bigl(k,\Spin(7,7)\bigr)$ and let $c$ be a nonzero
  value of~$q_u$.  The element
  \begin{displaymath}
    h_6(u)\cup (c)\in H^7(k,\LZ/2)
  \end{displaymath}
  does not depend on the choice of~$c$.
\end{proposition}

\begin{proof}[Proof (Variant 1)]
  Write $q=q_u$.  If $q$ is isotropic, then $h_6(u)=0$ by
  Lemma~\ref{lem:isotr-h6}.  We may therefore assume that $q$ is
  anisotropic.  Let $c=q(v)$ and $c'=q(v')$ be two values of~$q$ with
  $v$, $v'$ linearly independent.  Then $c/c'$ is a norm from the
  quadratic extension~$L$ splitting the $2$-dimensional subform
  $q|(vk+v'k)$.  Say $c/c'=N_{L/k}(\lambda)$.  Then
  \begin{align*}
    h_6(u)\cup (c)-h_6(u)\cup (c')&=h_6(u)\cup (c/c')
    \\
    &=h_6(u)\cup N_{L/k}\bigl((\lambda)\bigr)
    \\
    &=N_{L/k}\bigl(h_6(u_L)\cup(\lambda)\bigr)
    \\
    &=N_{L/k}\bigl(0\cup(\lambda)\bigr)=0
  \end{align*}
  since $q_L$ is isotropic and by Lemma~\ref{lem:isotr-h6}.
\end{proof}
\begin{proof}[Proof (Variant 2)]
  Write $q=q_u$ as $q\colon V \to k$.  Then any $x=[v]\in\LP V $
  determines an element $$q(x)\in\kappa(x)/\bigl(\kappa(x)^*\bigr)
  ^2.$$ Let $\xi\in\LP V $ be the generic point and consider
  \begin{displaymath}
    \omega=h_6(u)\cup \bigl(q(\xi)\bigr) \in H^7\bigl(k(\LP V)
    ,\LZ/2\bigr).
  \end{displaymath}

  The element~$\omega$ is unramified on~$\LP V $, except possibly at
  the divisor
  \begin{displaymath}
    Z=\{q=0\}\subset\LP V
  \end{displaymath}
  Here the residue is a multiple of (in fact, equal to)
  \begin{displaymath}
    h_6(u)_{k(Z)}\in H^6\bigl(k(Z),\LZ/2\bigr)
  \end{displaymath}
  But the quadratic form~$q_{k(Z)}$ is isotropic, whence
  $h_6(u)_{k(Z)}=0$ by Lemma~\ref{lem:isotr-h6}.  Hence $\omega$ is
  unramified everywhere on~$\LP V $ and therefore $\omega=
  (\omega_0)_{k(\LP V )}$ for some $\omega_0\in H^7(k,\LZ/2)$.  The
  claim follows by specialization.
\end{proof}

Proposition~\ref{prop:h7-well-def} gives rise to an invariant
\begin{gather*}
  h_7\colon H^1\bigl(k,\Spin(7,7)\bigr) \to H^7(k,\LZ/2),
  \\
  h_7(u) = h_6(u)\cup \bigl(q_u(v)\bigr)
\end{gather*}
where $q_u(v)$ is any nonzero value of~$q_u$.

As for~$h_6$, the invariant~$h_7(u)$ depends only on~$q_u$.  If
$q_u=q$ with $q$ as in~(\ref{eq:diff-pfister}), then
\begin{displaymath}
  h_7(u)=(a_1,a_2,a_3,b_1,b_2,b_3,c).
\end{displaymath}
This computation shows that the invariant~$h_7$ is non-trivial.

In the next two statements (Proposition~\ref{prop:symbol},
Lemma~\ref{lem:killing}) we assume that~$k$ contains the algebraic
closure of $\LQ$.  This assumption is made to be sure that we can
neglect some universal constants arising in decompositions of Killing
forms and of~$\lambda^2(q)$.  I have not tried to figure out the best
possible conditions.

\begin{proposition}
  \label{prop:symbol}
  Assume $\bar\LQ\subset k$.  Any value of $h_6$ and of~$h_7$ is a
  symbol.
\end{proposition}

\begin{proof}
  It suffices to consider~$h_6$.  Let $u\in
  H^1\bigl(k,\Spin(7,7)\bigr)$ and write $q=q_u$.  Then $h_6(u)$ is
  represented by $92$-dimensional form
  \begin{displaymath}
    \lambda^2q\orth\qform1.
  \end{displaymath}
  The form $\lambda^2q$ is also given by the Killing form on~$\so(q)$.

  We may assume that $u$ is induced from an element~$x\in
  H^1\bigl(k,(G_2 \times G_2) \rtimes \mu_8\bigr)$, see
  Corollary~\ref{14-spinor-reduction}.  Let $\dg\subset\so(q)$ be the
  Lie algebra of type $G_2+G_2$ corresponding to~$x$.  Its Killing
  form is the trace of the Killing form of a Lie algebra of type~$G_2$
  over some quadratic extension.  In view of the next Lemma, this form
  is hyperbolic.

  Therefore the $92$-dimensional form~$\lambda^2q\orth\qform1$
  contains a $28$-dimensional hyperbolic subform.  Thus $h_6(u)$ is
  represented by a~$92-28=64$-dimensional quadratic form, which
  therefore must be a multiple of a $6$-fold Pfister form.
\end{proof}

This proof indicates that one may represent $h_6(u)$ by a form on the
spinor representation~$S$, cf.~below.  In fact there is a natural way
to represent $h_6(u)$ as $N_{L/k}(\psi)$ on~$S$, where $\psi/L$ is the
$3$-fold Pfister form corresponding to a reduction~$x\in
H^1\bigl(k,(G_2 \times G_2) \rtimes \mu_8\bigr)$ of~$u$,
cf.~\cite{Ro:94x}.

\begin{lemma}
  \label{lem:killing}
  Assume $\bar\LQ\subset k$.  Let $\dg$ be a Lie algebra of type~$G_2$
  and let $\varphi$ be the associated $3$-fold Pfister form.  Then the
  Killing form on~$\dg$ is hyperbolic.
\end{lemma}

\begin{proof}
  Let $V$ be the $7$-dimensional representation of~$\dg$.  Then
  \begin{displaymath}
    \dg\orth V=\bigwedge^2V.
  \end{displaymath}
  Let further $\psi$ denote the Killing form on~$\dg$ and let
  $\varphi$ be the associated $3$-fold Pfister form.  Then
  \begin{displaymath}
    \psi\orth\varphi'=\lambda^2(\varphi')=\varphi'\pfister{-1,-1}'
  \end{displaymath}
  by Corollary~\ref{cor:lambda-pfister}.  The claim follows.
\end{proof}

Our considerations in the construction of the invariants $h_6$, $h_7$
may be also applied to the group $\SO(6)$.  This leads to invariants
$$H^1\bigl(k,\SO(6)\bigr)\to H^i(k,\LZ/2)$$ for $i=4$,~$5$, given by
\begin{align*}
  c(\pfister{a_1,a_2}'\orth\pfister{b_1,b_2}')&\mapsto
  (a_1,a_2,b_1,b_2),
  \\
  c(\pfister{a_1,a_2}'\orth\pfister{b_1,b_2}')&\mapsto
  (a_1,a_2,b_1,b_2,c).
  \\
  \intertext{The latter coincides with the invariant}
  \qform{a_1,a_2,a_3,a_4,a_5,a_6}&\mapsto (a_1,a_2,a_3,a_4,a_5)
\end{align*}
defined by Serre.

\section{A reduction lemma}

Let $G$ be an algebraic group over~$k$ and let $i\colon H\subset G$ be
a subgroup.  For $x\in H^1(k,G)$ we denote by $P_x$ a corresponding
$G$-torsor.

\begin{lemma}
  \label{lem:simple-reduction}
  Let $x\in H^1(k,G)$.  Then $x$ is in the image of
  \begin{displaymath}
    i_*\colon H^1(k,H)\to H^1(k,G)
  \end{displaymath}
  if and only if the variety $P_x/H$ has a $k$-rational point.
\end{lemma}

\begin{proof}
  Indeed, if $x=i_*(y)$, then $P_x\simeq P_y\times_HG$ and $P_x/H$ has
  the $k$-rational point given by $[P_y,1]\bmod H$.

  Conversely, if $z\in P_x/H$ is $k$-rational, then the fiber of~$z$
  under $P\to P_x/H$ is an $H$-torsor~$Q$ with $Q\times_HG\simeq P_x$.
\end{proof}

This simple lemma is the basis of many structure theorems on quadratic
forms and algebras.  It applies usually when there is a ``small''
representation of~$G$, i.e., a representation $G\to \GL(V)$ with $\dim
V<\dim G$.

A fairly simple example is given by $G=\On(n)$ and
$H=\On(n-1)\times\mu_2$:  Let $x\in H^1\bigl(k,\On(n)\bigr)$; if
$q_x\colon V\to k$ is the corresponding quadratic form, then $P_x/H$
is naturally isomorphic to $U=\LP V\setminus\{q_x=0\}$.  Since $U$ has
a rational point, it follows that $x$ has a reduction to~$H$.

Her majesty~$E_8$ does not have a small representation.

\section{$14$-dimensional spinors}

Let $\Spin(7,7)\to\GL(S)$ be one of the spinor representations ($\dim
S=64$) and let $\PSO(7,7)\to\PGL(S)$ be the induced homomorphism.  We
denote $G=\PSO(7,7)$ and define $H\subset G$ as the image of
\begin{displaymath}
  (G_2\times G_2)\rtimes \LZ/2 \to \PSO(7,7)
\end{displaymath}
given by
\begin{displaymath}
  (g,h)\epsilon^n\mapsto
  \begin{pmatrix}
    \rho(g)&0\\0&\rho(h)
  \end{pmatrix}
  \begin{pmatrix}
    0&1\\1&0
  \end{pmatrix}^n
\end{displaymath}
where $\rho\colon G_2\to \Spin(7)$ is the standard representation.

We need the following fact, see
\cite{Gatti-Viniberghi:78,Igusa:87,Popov:78,Ro:94x}.

\begin{proposition}
  The action of~$G$ on\/~$\LP S$ has an open and dense orbit~$U$.  If
  $k$ is algebraically closed, then the isotropy group $H_u$ of~$u\in
  U$ is conjugate to~$H$.  In particular, $U=G/H$.
\end{proposition}

Now let $x\in H^1(k,G)$.  Then $X_x=P_x\times_G\LP S$ is a
Brauer-Severi variety whose Brauer class coincides with the Tits class
$t(x)\in H^2(k,\mu_4)$ of~$x$.  Further, the variety $U_x=P_x\times_G
U=P_x/H$ is a dense open subscheme of $X_x$.  It follows that $P_x/H$
has $k$-rational points if and only if $t(x)=0$ (to be sure, let us
assume that~$k$ is infinite).  Lemma~\ref{lem:simple-reduction} shows
\begin{corollary}
  An element $x\in H^1(k,G)$ has an $H$-reduction if and only if
  $t(x)=0$.\qed
\end{corollary}

Let $\widetilde H$ be the preimage of~$H$ under
$\Spin(7,7)\to\PSO(7,7)$.  One finds (see~\cite{Ro:94x})
\begin{displaymath}
  \widetilde H=(G_2\times G_2)\rtimes \mu_8
\end{displaymath}
where~$\mu_8\subset \Spin(7,7)$ is the normalizer of $G_2\times G_2$.
\begin{corollary}
  \label{14-spinor-reduction}
  The homomorphism
  \begin{displaymath}
    H^1\bigl(k,(G_2 \times G_2) \rtimes \mu_8\bigr) \to
    H^1\bigl(k,\Spin(7,7)\bigr)
  \end{displaymath}
  is surjective.
\end{corollary}

\begin{proof}
  This follows from a diagram chase in
  \begin{displaymath}
    \begin{cd}[0.8]
      \\
      H^1(k,\mu_4)@>>> H^1\bigl(k,(G_2 \times G_2) \rtimes
      \mu_8\bigr)@>>> H^1\bigl(k,(G_2 \times G_2) \rtimes
      \LZ/2\bigr)@>>> H^2(k,\mu_4)
      \\
      @|@VVV@VVV@|
      \\
      H^1(k,\mu_4)@>>> H^1\bigl(k,\Spin(7,7)\bigr)@>>>
      H^1\bigl(k,\PSO(7,7)\bigr)@>t>> H^2(k,\mu_4)
    \end{cd}
  \end{displaymath}
\end{proof}

It can be shown that there exist a field~$k$ and $x\in
H^1\bigl(k,\Spin(7,7)\bigr)$ such that $x$ has no reduction to the
subgroup $(G_2 \times G_2) \times \mu_4$.  This means that the
appearing forms of $G_2\times G_2$ are necessarily of type
$R_{\ell/k}(G_2)$ with $\ell/k$ a quadratic \emph{field} extension.
Examples have been provided in \cite{Hoffmann-Tignol:98} using residue
arguments and in \cite{Izhboldin-Karpenko:99} using computations of
the $K$-theory of certain homogeneous varieties.

\section{The essential dimension of $\Spin(14)$}

We denote by $\ed(G)$ the essential dimension of~$G$,
see~\cite{Reichstein:99}.

\begin{proposition}
  $\ed\bigl(\Spin(14)\bigr)=7$.
\end{proposition}

\begin{proof}
  $\ed\bigl(\Spin(14)\bigr)\geq 7$ follows from the non-triviality of
  the invariant~$h_7$.

  It remains to show $\ed\bigl(\Spin(14)\bigr)\leq 7$.  By
  Corollary~\ref{14-spinor-reduction} it suffices show $\ed(\widetilde
  H)\leq 7$.  To describe any $\widetilde H$-torsor one needs one
  parameter to describe a class $(a)\in H^1(k,\mu_8)=k^*/(k^*)^8$ and
  $3\cdot 2$ parameters to describe an octonion algebra $$O(a_1+\sqrt
  ab_1,a_2+\sqrt ab_2,a_3+\sqrt ab_3)$$ over $k(\sqrt a)$.
\end{proof}

\section{On the cohomology of $\Spin(12)$}

We briefly sketch a proof of $\ed\bigl(\Spin(6,6)\bigr)=6$.

We define $H\subset \SO(6,6)$ as the image of
\begin{displaymath}
  \SL(6)\rtimes \LZ/2 \to \SO(6,6)
\end{displaymath}
given by
\begin{displaymath}
  g\epsilon^n\mapsto
  \begin{pmatrix}
    g&0\\0&(g^t)^{-1}
  \end{pmatrix}
  \begin{pmatrix}
    0&1\\1&0
  \end{pmatrix}^n.
\end{displaymath}
Here we understand coordinates $(x,y)$ with respect to the quadratic
form $\sum_ix_iy_i$.  The preimage $\widetilde H$ of~$H$ in
$\Spin(6,6)$ is
\begin{displaymath}
  \widetilde H=\SL(6)\rtimes \mu_4.
\end{displaymath}
By the mentioned theorem of Pfister (\cite[Satz 14,
Zusatz]{Pfister:66} or~\cite{Ro:94x}), any $\Spin(6,6)$-torsor admits
an $\widetilde H$-reduction.  Since any hermitian form can be
diagonalized, the map
\begin{displaymath}
  H^1\bigl(k,\SO(6)\times\mu_4\bigr) \to H^1(k,\widetilde H)
\end{displaymath}
is surjective.  Hence
\begin{corollary}
  $\ed\bigl(\Spin(6,6)\bigr) \leq 6$.
\end{corollary}

We define invariants in $H^5(\LZ/2)$, $H^6(\LZ/2)$ by a variant of the
previous method.  It is based on the following facts:

\begin{lemma}
  \label{lem:pfister-kernel}
  Let $a\in k^*$.  Then the kernel of
  \begin{displaymath}
    W(k)\to W(k),\quad x\mapsto\pfister ax
  \end{displaymath}
  is generated by $2$-dimensional forms of the form
  $\pfister{N_{\ell/k}(\alpha)}$ with $\alpha\in\ell^*$, $\ell=k(\sqrt
  a)$.
\end{lemma}

\begin{proof}
  Well known\dots
\end{proof}

\begin{lemma}
  \label{lem:chain-2}
  Let $a$, $b\in k^*$ and let $x$, $y\in W(k)$.  If
  \begin{displaymath}
    \pfister ax=\pfister by,
  \end{displaymath}
  then there exist $z\in W(k)$ with
  \begin{displaymath}
    \pfister ax=\pfister az=\pfister bz=\pfister by.
  \end{displaymath}
  Moreover, any such $z$ may be written as a sum of $2$-dimensional
  forms of the form $\pfister{N_{\ell/k}(\alpha)}$ with $\alpha\in
  \ell^*$, $\ell=k(\sqrt{ab})$.
\end{lemma}

\begin{proof}
  Let $\varphi$ be a quadratic form representing~$x$, let $K=k(\sqrt
  b)$, and suppose that $\pfister a\varphi_K$ is split.

  Since $\pfister a\varphi_K$ is isotropic, one has $\pfister
  a\varphi=\pfister a(c\pfister d+\varphi')$ such that $\pfister
  {a,d}_K$ is isotropic.  To see this, let
  $\varphi=\qform{a_1,\ldots,a_n}$ and let
  \begin{gather*}
    q\colon V=L^n\to k
    \\
    q(\lambda_1,\ldots,\lambda_n)=\sum_i a_i N_{L/k}(\lambda_i)
  \end{gather*}
  with $L=k(\sqrt a)$.  Note that $q=\pfister a\varphi$ and that
  $q(\lambda v)=N_{L/k}(\lambda)q(v)$ for $\lambda\in L$.  If $q_K$ is
  isotropic, there exists a $2$-dimensional $L$-submodule~$W$ of~$V$
  such that $q|W$ is isotropic over~$K$.  Next note that
  $q|W=c\pfister {a,d}$ for some $c$,~$d$.

  We may assume $\varphi=c\pfister
  d\orth\varphi'$.  There exists~$e$ such that
  \begin{displaymath}
    \pfister {a,d}=\pfister {a,e}=\pfister {b,e}
  \end{displaymath}
  Then $\pfister
  a\varphi=c\pfister {a,e}+\pfister a\varphi'$.  The claim follows by
  induction on~$\dim \varphi$ and Lemma~\ref{lem:pfister-kernel}.
\end{proof}

Let $I_2(k)\subset I(k)$ be the subset of elements which are split
over \emph{some} quadratic extension.  One defines an operation
\begin{gather*}
  Q\colon I_2(k)\to I_2(k),\\
  Q(\pfister ax)=\pfister a\lambda^2(x).
\end{gather*}
This map is well defined by Lemma~\ref{lem:pfister-kernel} and
Lemma~\ref{lem:chain-2}.

We assume that $-1$ is a square.  Let $u\in
H^1\bigl(k,\Spin(6,6)\bigr)$.  Then
\begin{displaymath}
  q_u=a\pfister b(\pfister{c,d}'-\pfister{e,f}')
\end{displaymath}
and
\begin{displaymath}
  Q(q_u)=\pfister {b,c,d,e,f}
\end{displaymath}
Hence we an invariant
\begin{displaymath}
  k_5\colon H^1\bigl(k,\Spin(6,6)\bigr) \to H^5(k,\LZ/2).
\end{displaymath}

If $q_u$ is isotropic, then $q_u=a\pfister{b,c',d'}\orth\qform{1,-1}$.
This shows $Q(q_u)=0$.  By the same argument as in the proof of
Proposition~\ref{prop:h7-well-def} we get an invariant
\begin{gather*}
  k_6\colon H^1\bigl(k,\Spin(6,6)\bigr) \to H^6(k,\LZ/2),
  \\
  k_6(u) = k_5(u)\cup \bigl(q_u(v)\bigr)
\end{gather*}
where $q_u(v)$ is any nonzero value of~$q_u$.

If $q_u=a\pfister b(\pfister{c,d}'-\pfister{e,f}')$, then
\begin{displaymath}
  k_6(u)=(a,b,c,d,e,f).
\end{displaymath}
This shows that $k_6$ is nontrivial.

\begin{corollary}
  $\ed\bigl(\Spin(6,6)\bigr) \geq 6$.
\end{corollary}

\section{On the cohomology of $\Spin(13)$}
\label{sec:cohomology-spin13}

Let
\begin{gather*}
  q\colon V=k^{13}\to k
  \\
  q(x_1,\ldots,x_{13})=(x_1^2+x_2^2+x_3^2)-(x_4^2+x_5^2+x_6^2)
  \\
  {}-(x_7^2+x_8^2+x_9^2)+(x_{10}^2+x_{11}^2+x_{12}^2)-x_{13}^2
\end{gather*}
An element of $H^1(k,\SO(q))$ is given by a $13$-dimensional quadratic
form~$q'$ with
\begin{displaymath}
  q'\orth\qform1\in H^1(k,\SO(7,7))\subset I^2\subset W(k)
\end{displaymath}

Let $G$ be the subgroup of $\SO(q)$ generated by (matrix notation with
respect to $k^{13}=k^3\times k^3\times k^3\times k^3\times k$)
\begin{displaymath}
  U(g,h)=\begin{pmatrix}
  g&0&0&0&0
  \\
  0&g&0&0&0
  \\
  0&0&h&0&0
  \\
  0&0&0&h&0
  \\
  0&0&0&0&1
  \\
\end{pmatrix}
\end{displaymath}
and
\begin{displaymath}
  V(\alpha,\beta)=
  \begin{pmatrix}
    1&0&0&0&0
    \\
    0&\alpha&0&0&0
    \\
    0&0&1&0&0
    \\
    0&0&0&\beta &0
    \\
    0&0&0&0&\alpha\beta
    \\
  \end{pmatrix}
  ,\quad W(\eta)=
  \begin{pmatrix}
    0&0&\eta&0&0
    \\
    0&0&0&\eta&0
    \\
    \eta&0&0&0&0
    \\
    0&\eta&0&0&0
    \\
    0&0&0&0&1
  \end{pmatrix}
\end{displaymath}
with $g$, $h\in \SO(3)$, $\alpha$, $\beta \in\mu_2$ and
$\eta\in\mu_4$.  One has
\begin{displaymath}
  G=\bigl(\SO(3)\times\mu_2\bigr)^2\rtimes\mu_4\subset\SO(q)
\end{displaymath}
with $\mu_4$ acting via the projection $\mu_4\to\mu_2=\LZ/2$ by
permutation of the factors.

We consider the commutative diagram
\begin{equation}
  \label{eq:diagram}
  \begin{cd}
    1@>>>\mu_2@>>>\widetilde G@>\pi_G>>G@>>>1
    \\
    @.@|@VV\tilde jV@VVjV
    \\
    1@>>>\mu_2@>>>\Spin(q)@>\pi>>\SO(q)@>>>1
  \end{cd}
\end{equation}
where $\widetilde G\subset \Spin(q)$ is the preimage of~$G$ under the
projection $\pi\colon \Spin(q)\to \SO(q)$ and $\tilde j$,~$j$ are the
inclusions.

We describe the image
\begin{displaymath}
  J=j_*\bigl(H^1(k,G)\bigr)\subset H^1(k,\SO(q))
\end{displaymath}
\begin{lemma}
  \label{lem:G13lem}
  The set~$J$ consists exactly of the (isomorphism classes of)
  quadratic forms~$q'$ of the type
  \begin{equation}
    \label{eq:q13=dec}
    q'=\tilde q\orth\qform{-\det(\tilde q)}
  \end{equation}
  with
  \begin{equation}
    \label{eq:q13=trace}
    \tilde q=(T_{K/k})_*\bigl(\qform s\qform{1,-\lambda}
    \qform{-\mu_1,-\mu_2,\mu_1\mu_2} \bigr)
  \end{equation}
  with $K=k[s]/(s^2-b)$ for some $b\in k^\times$ and $\lambda$,
  $\mu_1$, $\mu_2\in K^\times$.
\end{lemma}
\begin{proof}
  Note that $G\subset\SO(q)$ leaves the subspace
  $V'=k^{12}\times\{0\}\subset V$ invariant.  Let
  \begin{gather*}
    \ell\colon G\to \On(q|V')
    \\
    \ell(g)=j(g)|V'
  \end{gather*}
  Then
  \begin{displaymath}
    j(g)=\bigl(\ell(g),\det(\ell(g))\bigr) \in \On(q|V')\times
    \On(1)\subset \On(q)
  \end{displaymath}
  This yields the decomposition~\eqref{eq:q13=dec}.

  It remains to show that $\ell_*\bigl(H^1(k,G)\bigr)\subset
  H^1(\On(q|V'))$ consists of the forms~$\tilde q$ as
  in~\eqref{eq:q13=trace}.

  Elements of $H^1(k,G)$ are the isomorphism classes of triples
  $(K',\varphi,\varphi_1)$, where $K'=k[t]/(t^4-b)$ is a Galois
  $\mu_4$-algebra and where $\varphi$, $\varphi_1$ are quadratic forms
  over the quadratic subextension $K=k[s]\subset K'$, $s=t^2$ with
  $\varphi$ of rank~$3$ and determinant~$1$ and with $\varphi_1$ of
  rank~$1$.  Let
  \begin{displaymath}
    H=\bigl(\On(1)\times \On(1)\times \On(1)\bigr)\cap\SO(3)\simeq
    \mu_2\times \mu_2
  \end{displaymath}
  and
  \begin{displaymath}
    G'=(H\times \mu_2)^2\rtimes\mu_4\subset G
  \end{displaymath}
  Since quadratic forms (over~$K$) can be diagonalized, it follows
  that $H^1(k,G')\to H^1(k,G)$ is surjective.

  The claim follows from Corollary~\ref{cor:tracecomp} below.
\end{proof}

\begin{lemma}
  \label{lem:J-lem}
  Let
  \begin{displaymath}
    G''=(\mu_2)^2\rtimes\mu_4
  \end{displaymath}
  generated by $\mu_4$ and elements $\alpha$, $\beta$ with the
  relations
  \begin{displaymath}
    \alpha^2=\beta^2=(\alpha\beta)^2=1,\quad \zeta\alpha\zeta\inv =
    \beta, \quad \zeta \beta \zeta\inv =\alpha
  \end{displaymath}
  for a generator $\zeta$ of~$\mu_4$.

  Let
  \begin{gather*}
    q_0\colon k^2\to k
    \\
    q_0(x,y)=x^2-y^2
  \end{gather*}
  and let
  \begin{displaymath}
    \varphi\colon G''\to\On(q_0)
  \end{displaymath}
  be the homomorphism with
  \begin{displaymath}
    \varphi(\alpha)=
    \begin{pmatrix}
      -1&0\\0&1
    \end{pmatrix}
    ,\quad \varphi(\beta)=
    \begin{pmatrix}
      1&0\\0&-1
    \end{pmatrix}
    ,\quad \varphi(\eta)=
    \begin{pmatrix}
      0&\eta\\\eta&0
    \end{pmatrix}
  \end{displaymath}
  with $\alpha$, $\beta \in\mu_2$ and $\eta\in\mu_4$.

  Let $\xi\in H^1(k,G'')$ and write the corresponding Galois
  $G''$-algebra as
  \begin{displaymath}
    E_\xi=k[t,s,x,y]/(t^4-b,s-t^2,x^2-u-sv,y^2-u+sv)
  \end{displaymath}
  with $b$, $u$, $v\in k$, $b\neq 0$, $u^2-bv^2\neq 0$.  Here the
  action of $G''$ is given by
  \begin{gather*}
    \zeta(t)=\zeta t,\quad \zeta(s)=-s,\quad \zeta(x)=y,\quad
    \zeta(y)=x
    \\
    \alpha(t)=t,\quad \alpha(s)=s,\quad \alpha(x)=-x,\quad \alpha(y)=y
    \\
    \beta(t)=t,\quad \beta(s)=s,\quad \beta(x)=x,\quad \beta(y)=-y
  \end{gather*}

  Then the associated quadratic form $q_\xi=\varphi_*(\xi)\in
  H^1(k,\On(q_0))$ is given by
  \begin{displaymath}
    q_\xi=(T_{K/k})_*\bigl(\qform s\qform{u+sv}\bigr)
  \end{displaymath}
  with $K=k[s]\subset E_\xi$.
\end{lemma}
\begin{proof}
  One has (more or less by definition)
  \begin{displaymath}
    q_u=(q_0\tensor_k E)| (k^2\tensor_k E)^{G''}
  \end{displaymath}
  with $G''$ acting on $k^2$ via $\On(q_0)$ and on~$E$ as Galois
  algebra, respectively.

  The claim follows from the following explicit computation (for a
  related consideration see Garibaldi's Lens notes from May 2006,
  Example 16.5):

  One finds that $(k^2\tensor_k E)^{G''}$ is the free $k$-module with
  basis
  \begin{displaymath}
    X=(xt,-yt),\qquad Y=(xt^3,yt^3)=(xts,yts)
  \end{displaymath}
  For $c$, $d\in k$ one has with $\lambda=x^2=u+sv$ and
  $\bar\lambda=y^2=u-sv$
  \begin{align*}
    q_0(cX+dY)&=\bigl(xt(c+ds)\bigr)^2-\bigl(yt(-c+ds)\bigr)^2
    \\
    &=\lambda s (c+ds)^2+\bar \lambda (-s) (c-ds)^2
    \\
    &=T_{K/k}\bigl(\lambda s (c+ds)^2\bigr)
  \end{align*}
\end{proof}
\begin{corollary}
  \label{cor:tracecomp}
  Let $n$, $m\geq 0$, let $U=(\mu_2)^n$ and let
  \begin{displaymath}
    \Phi\colon U\to \On(1)^m\subset\On(m)
  \end{displaymath}
  be some homomorphism.  Let
  \begin{displaymath}
    G''=U^2\rtimes\mu_4
  \end{displaymath}
  with $\mu_4$ acting via the projection $\mu_4\to\mu_2=\LZ/2$ by
  permutation of the factors and let
  \begin{gather*}
    \varphi\colon G''\to \On(m,m)
    \\
    \varphi(u_1,u_2)=
    \begin{pmatrix}
      \Phi(u_1)&0\\0&\Phi(u_2)
    \end{pmatrix}
    \\
    \varphi(\zeta)=
    \begin{pmatrix}
      0&\zeta\\\zeta&0
    \end{pmatrix}
  \end{gather*}
  for $(u_1,u_2)\in U^2$ and a generator~$\zeta$ of~$\mu_4$.

  Let $\xi\in H^1(k,G'')$ and write the corresponding Galois
  $G''$-algebra as
  \begin{displaymath}
    E_\xi=
    k[t,s,x_i,y_i;i=1,\ldots,n]/(t^4-b,s-t^2,x_i^2-u_i-sv_i,y_i^2-u_i+sv_i)
  \end{displaymath}
  with $b$, $u_i$, $v_i\in k$, $b\neq 0$, $u_i^2-bv_i^2\neq 0$ (with
  obvious $G''$ action, see Lemma~\ref{lem:J-lem}).

  Then the associated quadratic form $q_\xi=\varphi_*(\xi)\in
  H^1(k,\On(m,m))$ is given by
  \begin{displaymath}
    q_\xi=(T_{K/k})_*\bigl(\qform s\qform {\mu_1,\ldots,\mu_m}\bigr)
  \end{displaymath}
  with $K=k[s]\subset E_\xi$ and with
  \begin{displaymath}
    \mu_j=\prod_{i=1}^n\lambda_i^{\Phi_{ij}}\in K,\quad j=1,\ldots,m
  \end{displaymath}
  where
  \begin{displaymath}
    \lambda_i=u_i+sv_i
  \end{displaymath}
  and where $\Phi_{ij}=0$, $1$ is defined by
  \begin{displaymath}
    \Phi(\alpha_1,\ldots,\alpha_n)=
    \bigl(\prod_{i=1}^n\alpha_i^{\Phi_{ij}} \bigr)_{j=1,\ldots,m}
  \end{displaymath}
\end{corollary}
\begin{proof}
  One easily reduces to the case $m=1$, $n=1$ and $\Phi=\id$, which is
  treated in Lemma~\ref{lem:J-lem}.
\end{proof}
\begin{proposition}
  \label{prop:G13sur}
  The natural map $\tilde j_*\colon H^1(\widetilde G)\to
  H^1(\Spin(q))$ is surjective.
\end{proposition}
\begin{proof}
  Let $u\in H^1(k,\Spin(q))$ and let $q_u\in H^1(k,\SO(q))$ be the
  associated quadratic form.  Then
  \begin{displaymath}
    q_u\orth\qform1\in I^3
  \end{displaymath}
  By the results on $14$-dimensional forms in~$I^3$ one has
  \begin{displaymath}
    q_u\orth\qform1=(T_{K/k})_*(\qform s\varphi')
  \end{displaymath}
  with $K=k[s]/(s^2-b)$ for some~$b\in k^\times$ and with~$\varphi$ a
  $3$-fold Pfister form over~$K$ (and with
  $\varphi=\qform1\orth\varphi'$).  Since the left hand side
  represents~$1$, there exists a value~$-\lambda$ of~$\varphi'$ with
  $T_{K/k}(-s\lambda)=1$.  As for any (invertible) value~$-\lambda$
  of~$\varphi'$, one has $\varphi=\pfister{\lambda,\mu_1,\mu_2}$ for
  some $\mu_1$, $\mu_2\in K^\times$.  Note that
  \begin{displaymath}
    (T_{K/k})_*(\qform{-s\lambda})=\qform{1,-N_{K/k}(\lambda)}
  \end{displaymath}
  Thus
  \begin{displaymath}
    q_u=(T_{K/k})_*\bigl(\qform s\pfister\lambda
    \pfister{\mu_1,\mu_2}' \bigr)\orth\qform{-N_{K/k}(\lambda)}
  \end{displaymath}
  By Lemma~\ref{lem:G13lem} it follows that $q_u\in J$.  A diagram
  chase (see diagram~\eqref{eq:diagram}) involving the coboundary maps
  $H^1(k,G)$, $H^1(k,\SO(q))\to H^2(k,\mu_2)$ shows that there exists
  $\tilde u\in H^1(k,\widetilde G)$ such that $\tilde j(\tilde u)$,
  $u\in H^1(k,\Spin(q))$ have the same image in $H^1(k,\SO(q))$.
  Another diagram chase shows that we can arrange $\tilde j(\tilde
  u)=u$.
\end{proof}

We next compute $\widetilde G\subset \Spin(q)\subset C(q)$ inside the
Clifford algebra.  Let $e_1$, \dots, $e_{13}$ be the standard base
of~$V$.

Let $\zeta$ be a primitive $4$-th root of unity.

For $v$, $w\in V$ with $q(v)=1$, $q(w)=-1$ and $v\perp w$ let
\begin{displaymath}
  \omega(v,w)= \frac{1+\zeta wv}{\sqrt 2}
\end{displaymath}
Then $\omega(v,w)\omega(w,v)=1$ and therefore $\omega(v,w)\in
\Spin(q)$.  Moreover $\omega(v,w)^2=\zeta wv$ and $\omega(v,w)^4=-1$.
Furthermore $\omega(v,w)v=v\omega(v,w)\inv$ and
$\omega(v,w)w=w\omega(v,w)\inv$.  Also
$\omega(v,w)v\omega(v,w)\inv=\zeta w$ and
$\omega(v,w)w\omega(v,w)\inv=\zeta v$.

Consider the element
\begin{displaymath}
  \omega = \omega(e_1,e_7)\omega(e_2,e_8)\omega(e_3,e_9)
  \omega(e_{10},e_4)\omega(e_{11},e_5)\omega(e_{12},e_6)\in \Spin(q)
\end{displaymath}
Its image in $\SO(q)$ is~$\pi(\omega)=W(\zeta)$.  Moreover
\begin{displaymath}
  \omega^4=1
\end{displaymath}

Next let
\begin{displaymath}
  \tilde \alpha=e_4e_5e_6e_{13},\quad \tilde \beta=-\zeta
  e_{10}e_{11}e_{12}e_{13}
\end{displaymath}
Both elements are in $\Spin(q)$ and $\pi(\tilde\alpha)=V(-1,1)$ and
$\pi(\tilde\beta)=V(1,-1)$.  Moreover
\begin{align*}
  \tilde\alpha^2&=1
  \\
  \tilde\beta^2&=1
  \\
  \tilde\alpha\tilde\beta&=-\tilde\beta \tilde\alpha
  \\
  (\tilde\alpha\tilde\beta)^2&=-1
  \\
  \omega\tilde\alpha\omega\inv&=\tilde\beta
  \\
  \omega\tilde\beta\omega\inv&=-\tilde\alpha
  \\
  \omega\tilde\alpha\tilde\beta\omega\inv&=\tilde\alpha\tilde\beta
  \\
  \tilde\alpha\omega\tilde\alpha\inv&=\tilde\alpha\tilde\beta\omega
  \\
  \tilde\alpha \omega^2 \tilde\alpha\inv &= -\omega^2
\end{align*}

Let $H$ be the subgroup generated by $\omega$ and~$\tilde\alpha$.
Then $\tilde\beta\in H$ and
\begin{displaymath}
  H=(\mu_4\times\mu_4)\rtimes\mu_2
\end{displaymath}
with the $\mu_2$ generated by~$\tilde\alpha$ and $\mu_4\times\mu_4$
generated by $\omega$ and~$\tilde\alpha\omega\tilde\alpha\inv$.

Note further that the diagonal embedding $\SO(3)\to\SO(3,3)$ lifts to
$\Spin(3,3)$.  Thus the connected component of~$G$ lifts (uniquely)
to~$\Spin(q)$.  This yields:
\begin{lemma}
  One has
  \begin{displaymath}
    \widetilde G\simeq \bigl(\SO(3)\bigr)^2\rtimes_\varphi H
  \end{displaymath}
  where $H$ acts by permutation of the factors via $\varphi\colon
  H\to\LZ/2$, $\varphi(\tilde\alpha)=0$, $\varphi(\omega)=1$.
\end{lemma}

(I was surprised about the simple structure of~$H$.  There ought to be
a better approach to the subgroup $\widetilde G$ of $\Spin(13)$ than
just by a computation starting from~$G$.)

%% \begin{gather*}
%%   \On(3)  \times \LZ/2 \to \On(3,3)
%%   \\
%%     g\epsilon^n\mapsto
%%   \begin{pmatrix}
%%     g&0\\0&g
%%   \end{pmatrix}
%%   \begin{pmatrix}
%%     1&0\\0&-1
%%   \end{pmatrix}^n
%% \end{gather*}
%% \begin{displaymath}
%%   H=(\SO(3)  \times \mu_2)^2  \rtimes \mu_4
%% \end{displaymath}

\begin{proposition}
  \label{prop:edG13}
  $\ed(\widetilde G)\leq 6$
\end{proposition}

\begin{proof}
  Elements of $H^1(k,H)$ are given by Galois $H$-algebras which can be
  written as
  \begin{displaymath}
    L=k[z,x,y]/(z^2-a,x^4-u-sv,y^4-u+sv)
  \end{displaymath}
  with $a$, $u$, $v\in k$, $a\neq 0$, $u^2-av^2\neq 0$.  For the
  generic case we may assume $v\neq 0$ and replace $s$ by $sv$ and $a$
  by $av^2$.  Then $v=1$.  Therefore $H$-torsors are parameterized by
  $a$ and $u$ and we have $\ed(H)\leq 2$.

  Thus an element of $H^1(k,\widetilde G)$ is given by a Galois
  $H$-algebra
  \begin{displaymath}
    L=k[z,x,y]/(z^2-a,x^4-u-s,y^4-u+s)
  \end{displaymath}
  and a quadratic form of rank~$3$ and determinant~$1$
  over~$K=k[t]\subset L$ with $t=(xy)^2$ and $t^2=u^2-a$.  Thus $\ed(
  \widetilde G)\leq \ed(H)+2\cdot 2$.
\end{proof}

\begin{corollary}
  \label{cor:edspin13}
  $\ed(\Spin(q))\leq 6$
\end{corollary}
\begin{proof}
  This is clear from Proposition~\ref{prop:G13sur} and
  Proposition~\ref{prop:edG13}.
\end{proof}

\section{The essential dimension of split $\Spin(n)$ for $n\leq 14$}

Let $\Spin_n$ denote a split form of~$\Spin(n)$.
\begin{theorem}
  \begin{align*}
    \ed(\Spin_n)&=0 \quad\text{for $n\leq 6$},\\
    \ed(\Spin_7)&=4,\\
    \ed(\Spin_8)&=5,\\
    \ed(\Spin_9)&=5,\\
    \ed(\Spin_{10})&=4,\\
    \ed(\Spin_{11})&=5,\\
    \ed(\Spin_{12})&=6,\\
    \ed(\Spin_{13})&=6,\\
    \ed(\Spin_{14})&=7.
  \end{align*}
\end{theorem}

\begin{proof}
  (Sketch) The cases $n=12$, $14$ have been just considered.  It is
  not difficult to extend our considerations to the case $n=11$.

  As for $n=13$:  By corollary~\ref{cor:edspin13} one has
  $\ed(\Spin_{13})\leq 6$.  The invariant $h_6$ restricted to
  $\Spin_{13}$ is nontrivial, for example for
  \begin{displaymath}
    q\perp \qform 1 =
    b_1\bigl(\pfister{a_1,a_2,a_3}'-\pfister{b_1,b_2,b_3}'\bigr)
  \end{displaymath}
  Hence $\ed(\Spin_{13})\geq 6$.

  For $n=7$,~$10$ one uses that any $\Spin_n$-torsor admits a
  reduction to $G_2\times\mu_2$ resp.\ to~$G_2\times\mu_4$.  For
  $n=8$,~$9$ one may use the fact that
  \begin{displaymath}
    \Spin_8\to\Spin_9\to F_4
  \end{displaymath}
  induce surjections on~$H^1$ at the prime~$2$ and Serre's
  $H^5(\LZ/2)$-invariant for~$F_4$, cf.~\cite[III.~Annexe,
  \S~3.4]{Serre:94} or~\cite[III.~Appendix~2, 3.4]{Serre:97}
  and~\cite[\S~40]{Knus-Merkurjev-Rost-Tignol:98},~\cite{Ro:91x}.  For
  $n\leq 6$ note that any $n$-dimensional quadratic form with trivial
  $e_1$-, $e_2$-invariants is split.
\end{proof}

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