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%%% Involutions and Trace Forms on Exterior Powers of a Central Simple Algebra
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%%% The n-th exterior power of a central simple algebra of degree 2n is endowed
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\title[Involutions and trace forms on exterior powers]
{Involutions and trace forms on exterior powers of a central simple algebra}
\author[R. S. Garibaldi\and A. Qu\'eguiner-Mathieu \and J.-P. Tignol]
{R. Skip Garibaldi\and Anne Qu\'eguiner-Mathieu \and Jean-Pierre Tignol}

\date{May 15, 2000}
\thanks{The first author would like
to thank ETH Z\"urich for its hospitality while some of the research
for this paper was conducted.
The second and third authors are partially supported by the
European Commission under the TMR contract ERB-FMRX-CT97-0107. The
third author is partially supported by the National Fund for
Scientific Research (Belgium).}

\begin{document}

\begin{abstract}
For $A$ a central simple algebra of degree $2n$, the $n$th exterior power
algebra $\lambda^n A$ is endowed with an involution which provides an
interesting invariant of $A$.  In the case where $A$ is isomorphic to
$Q \otimes B$ for some quaternion algebra $Q$, we describe this involution
quite explicitly in terms of the norm form for $Q$ and the corresponding
involution for $B$.
\end{abstract}

\maketitle

% \section{Introduction}

Associated to every central simple $F$-algebra $A$ and any nonnegative
integer $k\leq\deg A$ is the $k$th exterior power $\la^k A$ of $A$,
which is a
central simple $F$-algebra, of degree $\binom{\deg A}{k}$,
Brauer-equivalent to $A^{\ot k}$,
see \cite[10.A]{KMRT}. It is defined so that when $A$ is
the split algebra $A=\End_F(W)$, this $\la^k \End_F(W)$ is naturally
isomorphic to $\End_F(\wedge^k W)$.
When $A$ has even degree $2n$, the $n$th exterior power $\lambda^n A$
is endowed with a canonical involution $\gamma$ such that
when  $A$ is split, $\gamma$ is adjoint to the
bilinear form $\theta$ defined on $\w^n W$ by the equation
$\theta(x_1\wedge\dots\w x_n,y_1\w \dots\w y_n)e=
x_1\wedge\dots\w x_n\w y_1\w \dots\w y_n$,
where $e$ is any basis of the $1$-dimensional vector space
$\w^{2n}W$.

Besides providing an invariant of $A$, the involution $\gamma$ is of
additional interest because of the even
Clifford algebra. Indeed, any central simple algebra $A$ with hyperbolic
orthogonal involution can be written as $(M_2(B),\sig)$, for some central
simple algebra $B$ of degree $n$ uniquely determined up to isomorphism.
The Clifford algebra $C\bigl(M_{2}(B),\sig\bigl)$ is
itself an algebra with
involution, and has been completely described when $n$ is
odd: in this case, $M_{2}(B)$ is the endomorphism algebra of an
$F$-vector space $V$, and the Clifford algebra
$C\bigl(M_{2}(B),\sigma\bigr)$ is the even Clifford algebra of any
quadratic form on $V$ with adjoint involution $\sigma$.
However, in the case when $n$ is even, $C(M_2(B),\sig)$ has a nontrivial
piece, which is isomorphic to $(\lambda^n M_2(B),\gamma)$.  Please
see \cite{G:clif}
for a precise statement and \cite{QT} for a rational proof.

If $A$ has degree $2n$ for $n$ even, then $\gamma$ is of orthogonal
type, and if moreover $A^{\otimes n}$ is split, then $\lambda^n A$ is
split as well and hence $\gamma$ is adjoint to some quadratic form $q_A$.
This provides a canonical way to associate to $A$ a quadratic form $q_A$
of dimension
$\binom{2n}{n}$, which is uniquely determined up to a scalar factor.
If $A$ is a biquaternion algebra, then $q_A$ is an Albert form for $A$
\cite[6.2]{G:clif}.
Until now, the value of $q_A$ has not been known for
any algebra $A$ of index $\geq 8$.

The main purpose of this paper is to provide a description of this involution
$\gamma$ in the particular case when $A$ admits a decomposition $A=Q\ot B$,
where $Q$ is a quaternion algebra over $F$ (see below for precise statements).
In particular, if $A$ is a tensor product of quaternion algebras,
we get a formula that gives $q_A$ (up to Witt-equivalence) in terms of
the norm forms of the quaternion algebras. In the course of obtaining this
description, we also prove a formula relating the trace forms of the
various exterior powers $\lambda^kB$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Statement of the main results}

We will always assume that our base field $F$ has characteristic
$\ne 2$ and that $A$ is a central simple $F$-algebra of degree $2n$.
We assume moreover that $A$ is isomorphic to a tensor product
$A=Q\ot B$, where $Q$ is a quaternion algebra over $F$, and $B$ is a
central simple $F$-algebra, necessarily of degree $n$.
Note that this is always the case when $n$ is odd.  We write
$\gamma_Q$ for the canonical symplectic involution on $Q$ and $n_Q$
for the norm form.

If $n$ is odd, the main result is the following, proven in Section
\ref{odd.sec}:
\begin{thm} \label{odd.thm}
If $n$ is odd, the algebra with involution
$(\lambda^n(Q\otimes B),\gamma)$ is Witt-equivalent to
$(Q,\gamma_Q)^{\ot n}$.
\end{thm}

Witt-equivalence for central simple algebras is the natural
generalization of Witt-equi\-valence for quadratic forms, see
\cite{DLT} for a definition.

Assume now that $n$ is even, $n=2m$.  Then $\la^n A$ is split and
the involution $\ga$ is orthogonal.
We fix some quadratic form $q_A$ to which $\gamma$ is adjoint.
It is only defined up to similarity.

The algebra $\lambda^m B$ is endowed with a canonical involution
which we denote by $\gamma_m$.
For $k=0,\dots,n$, we let $t_k:\la^kB\ra F$ be the reduced trace
quadratic form
defined by
\begin{neqn} \label{redtr}
t_k(x)=\trd_{\la^k B}(x^2).
\end{neqn}
Moreover, we let $t_m^+$ and $t_m^-$ denote the restrictions
of $t_m$ to the subspaces $\Sym(\lambda^m B,\gamma_m)$ and
$\Skew(\lambda^m B,\gamma_m)$ of elements of $\la^m B$
which are respectively symmetric and skew-symmetric under $\gamma_m$,
so that $t_m=t_m^+\oplus t_m^-$. The forms thus defined are related
by the following equation, proven in~\ref{memorable.pf}:
\begin{thm}
  \label{memorable.thm}
  In the Witt ring of $F$, the following equality holds:
  \[
  \langle2\rangle\cdot
  \sum_{k=0}^{m-1} (-1)^k t_{k}=
  \begin{cases}
    -t_{m}^-&\text{if $m$ is even,}\\
    t_{m}^{+}&\text{if $m$ is odd.}
  \end{cases}
  \]
\end{thm}

The similarity class of $q_{A}$ is determined by the following theorem,
proven in \ref{even.pf}:
\begin{thm} \label{even.thm}
If $n$ is even, $n=2m$, the similarity class of $q_A$ contains the
quadratic form:
\begin{align*}
  t_{m}^{+}-t_{m}^- + n_{Q}\cdot\Bigl(
  t_{m}^- +\sum\nolimits_{
    \substack{0\leq k<m\\
    \text{$k$ even}}}
  \langle2\rangle t_{k}\Bigr)&
  \quad\text{ if $m$ is even,}\\
  t_{m}^- -t_{m}^{+} + n_{Q}\cdot\Bigl(
  \sum\nolimits_{
    \substack{0\leq k<m\\
    \text{$k$ even}}}
  \langle2\rangle t_{k}\Bigr)&
  \quad\text{ if $m$ is odd.}
\end{align*}
\end{thm}

The Witt class of this quadratic form can be described more precisely
under some additional assumptions (see Proposition~\ref{even.prop}
for precise statements).
We just mention here a particular case in which the formula reduces to
be quite nice.

Assume that $m$ is even and $B$ is of exponent at most $2$.
Then $\lambda^m B$ is split, and its canonical involution is
adjoint to a quadratic form $q_B$.
Even though this form is only defined up to a scalar factor, its square
is actually defined up to isometry.
We then have the following, proven in \ref{1.pf}:
\begin{cor}
\label{1.cor}
If $m$ is even (i.e., $\deg B \equiv 0 \mod{4}$) and
$B$ is of exponent at most $2$, then the similarity class of $q_A$
contains a form whose Witt class is
$q_B^2+ n_Q \left(2^{n-2} - \frac12 \binom{n}{m} - \wt{q_B} \right)$.
\end{cor}
Some of the notation needs an explanation.  For a quadratic form $q$
on a vector space $W$ with associated symmetric bilinear form $b$ so
that $q(w) = b(w,w)$, we have an induced quadratic form on $\w^2 W$
which we denote by $\wt{q}$.  For $x_1, x_2, y_1, y_2 \in W$, its associated
symmetric bilinear form $\wt{b}$ is defined by
\[
(\wt{b})(x_1 \wedge x_2, y_1 \wedge y_2) = b(x_1, y_1) b(x_2, y_2) -
b(x_1, y_2) b(x_2, y_1).
\]
Thus if $q = \qform{\alpha_1, \ldots, \alpha_n}$, we have
\[
\wt{q} \simeq \oplus_{1 \le i < j \le n} \qform{\alpha_i \alpha_j}.
\]
In particular, even if $q$ is just defined up to similarity, $\wt{q}$
is well-defined up to isometry.

\relax From Corollary~\ref{1.cor}, we also get the following, which is proven
in \ref{2.pf}:
\begin{cor}
\label{2.cor}
Let $A_r=Q_1\ot\cdots\ot Q_r$ be a tensor product of $r$ quaternion
$F$-algebras, where $r\geq 3$, and let $T_{A_{r}}$ be the reduced
trace quadratic form on $A_{r}$.
The similarity class of $q_{A_r}$ contains a quadratic form
whose Witt class is
\[
2^{n-1}-\frac{2^{n-2}}{n}\langle 2^r\rangle\cdot T_{A_{r}}=
2^{f(r)}\bigl(2^r-(2-n_{Q_1})\cdots(2-n_{Q_r})\bigr),
\]
where $n=2^{r-1}=\frac{1}{2}\deg A$ and $f(r)=2^{r-1}-r-1$.
\end{cor}

In particular, for $r=3$, we get the quadratic form
$$
4(n_{Q_1}+n_{Q_2}+n_{Q_3})-2(n_{Q_1}n_{Q_2}+n_{Q_1}n_{Q_3}+n_{Q_2}n_{Q_3})
+n_{Q_1}n_{Q_2}n_{Q_3}.
$$
Adrian Wadsworth had casually conjectured a
description of $q_{A_3}$ in \cite[6.8]{G:clif}, and
we now see that his conjecture was not quite correct in that it omitted the
$n_{Q_1}n_{Q_2}n_{Q_3}$ term.

\medskip

As a consequence of Corollary~\ref{2.cor}, we can show that the form $q_{A}$
lies in the $n$th power of the fundamental ideal of the Witt ring $WF$
for many central simple algebras $A$ of degree~$2n$; the
following result is proven in \ref{3.pf}:
\begin{cor} \label{3.cor}
Suppose that $A$ is a central simple algebra of degree $2n\equiv 0\mod
4$ which is isomorphic to matrices over a tensor product of quaternion
algebras.
Then the form $q_A$ lies in $\IF{n}$.
\end{cor}

The first author conjectured \cite[6.6]{G:clif} that $q_A$ lies in
$\IF{n}$ for {\em all} central simple $F$-algebras $A$ of degree $2n
\equiv 0 \mod{4}$ and such that $A^{\otimes 2}$ is split.  Corollary
\ref{3.cor} fails to prove the full conjecture because for every integer
$r \ge 3$ there exists a division algebra $A$ of degree $2^r$ and exponent
2 such that $A$ doesn't decompose as $A' \ot A''$ for any nontrivial
division algebras $A'$ and $A''$ \cite[3.3]{Jacob:indec}, so such an
$A$ doesn't satisfy the hypotheses of Corollary \ref{3.cor}.

If $A$ is a tensor product of two quaternion algebras, the form
$q_{A}$ is an Albert form of $A$, and the Witt index of $q_{A}$
determines the Schur index of $A$, as Albert has shown (see for
instance \cite[(16.5)]{KMRT}). Corollary~\ref{2.cor} shows that one
cannot expect nice results relating the Witt index of $q_{A_{r}}$ and
the Schur index of $A_{r}$ for $r\geq3$.
As pointed out to us by Jan van Geel, the difficulty is that
Merkurjev has constructed in \cite[\S 3]{M:simple} algebras of the
form $A_r$ for $r \ge 3$ (i.e., tensor products of at least 3
quaternion algebras) which are skew fields but whose center, $F$, has
$\IF3 = 0$.  By Corollary~\ref{3.cor}, the forms $q_{A_r}$ are then
hyperbolic.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Description of $\la^n M_2(B)$}
\label{GD.sec}

In order to prove these results, we have to describe the algebra
with involution
$(\la^n(Q\ot B),\ga)$, which we will do by Galois descent.
Hence we first give a description of $\la^n M_2(B)$,
see Theorem~\ref{1.thm} below.

Assume $B=\End_{F}(V)$ for some $n$-dimensional vector space $V$.
For $0 \le k \le n$, we have $\la^k B=\End_F(\wedge^k V)$.
We identify $M_2(B)\simeq\End_F(V\oplus V)$ by
mapping
$\left(
\begin{smallmatrix}
  a&b\\
  c&d
\end{smallmatrix}
\right) \in M_{2}(B)$ to the endomorphism
$$
(x,y)\mapsto\bigl(a(x)+b(y),c(x)+d(y)\bigr).
$$
The distinguished choice of embedding of $B$ in $M_2(B)$ corresponds
with the obvious choice of direct sum decomposition of $V \oplus V$.
(There are many others.)  This gives an identification
$\lambda^n M_{2}(B)=\End_{F}(\wedge^n(V\oplus V))$.
For all
integers $k$, $\ell$, this decomposition determines
$\w^k V \ot \w^\ell V$ as a vector subspace of $\w^{k+\ell} (V \oplus
V)$ by mapping
$(x_{1}\wedge\dots\wedge x_{k})\otimes(y_{1}\wedge\dots
\wedge y_{\ell})$ to
$$
(x_{1},0)\wedge\dots\wedge(x_{k},0)\wedge
(0,y_{1})\wedge\dots\wedge(0,y_{\ell})\in\wedge^{k+\ell}(V\oplus V).
$$
In particular, we have
\begin{neqn} \label{basicid}
\wedge^n(V\oplus V)=
\oplus_{k=0}^n\left(\wedge^kV\otimes\wedge^{n-k}V\right).
\end{neqn}
For each $k$, the space $\wedge^kV\otimes \wedge^{n-k}V$ can be
identified to $\End_{F}(\wedge^kV)$ as follows. Fix a nonzero element
(hence a basis) $e$ of $\w^n V$ and define a
bilinear form
\[
\theta_{k}\colon\,\wedge^kV\times\wedge^{n-k}V\to F
\]
by the equation
\[
\theta_{k}(x_k, x_{n-k}) \,e= x_k \wedge x_{n-k}\
\text{for $x_\ell\in\wedge^\ell V$.}
\]
This form is nonsingular, so it provides the identification mentioned
above
\begin{neqn} \label{1.eq}
\w^k V \ot \w^{n-k} V = \End_F(\w^k V)
\end{neqn}
by sending $x_k \otimes x_{n-k}$ to the map $y \mapsto x_k
\theta_{n-k}(x_{n-k}, y)$.  The product in $\End_F(\w^k V)$ then
corresponds in $\w^k V \otimes \w^{n-k} V$ to
\[
(x_k \otimes x_{n-k}) (y_k \otimes y_{n-k}) = \theta_{n-k}(x_{n-k},
y_k) \, x_k \ot y_{n-k}.
\]
\relax From~\eqref{basicid} and \eqref{1.eq}, we deduce an
identification
of the corresponding endomorphism rings
\[
  \lambda^n M_{2}(B) = \End_{F}(\oplus_{k=0}^n\lambda^kB).
\]
This remains true in the case when $B$ is non split, as we will prove
by Galois descent.
First, we must introduce some maps on $\oplus^n_{k=0} \la^k B$.

Since the bilinear form $\ta_k$ is nonsingular,
for any $f\in \End_F(\wedge^kV)$,
we have a unique element $\gamma_k(f)\in \End_{F}(\wedge^{n-k}V)$
such that
\[
\theta_{k}\left(f(x),y\right)=
\theta_{k}\left(x,\gamma_{k}(f)(y)\right),
\]
for every $x\in\wedge^kV$ and $y\in\wedge^{n-k}V$.
This defines a canonical anti-isomorphism
(not depending on the choice of $e$)
\[
\gamma_{k}\colon\,\End_{F}(\wedge^kV)\to
\End_{F}(\wedge^{n-k}V)
\]
such that
\begin{neqn} \label{gakdef}
\gamma_k(x \ot y) = (-1)^{k (n -k)} y \otimes x
\end{neqn}
for $x$ and $y$ as before.
One may easily verify that
$\gamma_{n-k}\circ\gamma_{k}=\id_{\End_F(\w^k V)}$ for all $k=0,
\ldots, n$.
By Galois descent, the maps $\gamma_k$ are defined even
when $B$ is nonsplit, i.e., we have anti-isomorphisms $\gamma_k
\colon\, \la^k B \ra \la^{n-k} B$ such that $\gamma_k \circ
\gamma_{n-k} = \id_{\la^k B}$ (see~\cite[Exercise~12, p.~147]{KMRT}
for a rational definition).
In the particular case where $n$ is even, by definition of the bilinear
form $\theta_{n/2}$, the map $\gamma_{n/2}$ is actually the canonical
involution
on $\la^{n/2}B$.

\begin{thm} \label{1.thm}
There is a canonical isomorphism
\[
  \Phi\colon\,\lambda^n M_{2}(B)\to
  \End_{F}(\lambda^{0}B\oplus\dots\oplus\lambda^nB).
\]
The canonical involution
$\gamma$ on $\la^n M_2(B)$ induces via $\Phi$ an involution on
$\End_F(\oplus_{k=0}^n \la^k B)$ which is adjoint to the
bilinear form $T$
defined on $\lambda^{0}B\oplus\dots\oplus\lambda^nB$
by
\[
  T(u,v)=
  \begin{cases}
    (-1)^{\ell}\trd_{\lambda^k B}\left(u \gamma_{\ell}(v) \right)&
    \text{if $k+\ell=n$},\\
    0&\text{if $k+\ell\neq n$},
  \end{cases}
\]
for any $u\in\lambda^kB$ and $v\in\lambda^\ell B$.
\end{thm}

\begin{proof}
We prove this by Galois descent.  Fix a separable closure $F_s$ of $F$
and let $\Gamma := \Gal(F_s/F)$ be the absolute Galois group.  We fix
a vector space $V$ over $F$ such that $\dim_F V = \deg B = n$ and let
$V_{s}=V\otimes_{F}F_{s}$. We fix also an $F_{s}$-algebra
isomorphism $\varphi\colon\,B\otimes_{F}F_{s}\iso\End_{F}(V)\ot_{F}F_{s}$.
Every $\sigma\in\Gamma$ acts canonically on $V_{s}$ and
$\End_{F_{s}}(V_{s})=\End_{F}(V)\ot_{F}F_{s}$; we denote again by
$\sigma$ these canonical actions, so that $\sigma(f)=\sigma\circ
f\circ\sigma^{-1}$ for $f\in\End_{F_{s}}(V_{s})$. On the other hand,
the canonical action of $\Gamma$ on $B\ot_{F}F_{s}$ corresponds
under $\varphi$ to some twisted action $*$ on $\End_{F_{s}}(V_{s})$.
Since every $F_{s}$-linear automorphism of $\End_{F_{s}}(V_{s})$ is
inner, we may find $g_{\sigma}\in\GL(V_{s})$ such that
\[
\sigma * f=g_{\sigma}\circ\sigma(f)\circ g_{\sigma}^{-1}=
\Int(g_{\sigma})\circ\sigma(f)
\qquad\text{for all $f\in\End_{F_{s}}(V_{s})$.}
\]
Then $\varphi$ induces an $F$-algebra isomorphism from $B$ onto the
$F$-subalgebra
\[
\bigl\{
f\in\End_{F_{s}}(V_{s})\mid g_{\sigma}\circ\sigma(f)\circ
g_{\sigma}^{-1}=f\text{ for all $\sigma\in\Gamma$}\bigr\}.
\]
The $*$-action of $\Gamma$ on $\End_{F_{s}}(V_{s})$ induces twisted
actions on $\End_{F_{s}}\bigl(\w^n(V_{s}\oplus V_{s})\bigr)$ and on
$\End_{F_{s}}\bigl(\oplus_{k=0}^n\End_{F_{s}}(\w^k V_{s})\bigr)$ such
that the $F$-algebras of $\Gamma$-invariant elements are
$\lambda^n\bigl(M_{2}(B)\bigr)$ and
$\End_{F}(\oplus_{k=0}^n\lambda^kB)$ respectively. To prove the
first assertion of the theorem, we will show that
these actions correspond to each other under the isomorphism
\[
\End_{F_{s}}\bigl(\w^n(V_{s}\oplus V_{s})\bigr)
\iso
\End_{F_{s}}\bigl(\oplus_{k=0}^n\End_{F_{s}}(\w^k V_{s})\bigr)
\]
derived from \eqref{basicid} and \eqref{1.eq}.

For $\sigma\in\Gamma$ and $k=0$, \ldots, $n$, define $\w^k
g_{\sigma}\in \GL(\w^kV_{s})$ by
\[
\w^k g_{\sigma}(x_{1}\w\ldots\w x_{k})=
g_{\sigma}(x_{1})\w\ldots\w g_{\sigma}(x_{k}).
\]
Then $\varphi$ induces an $F$-algebra isomorphism from $\lambda^kB$
onto the $F$-subalgebra
\[
\bigl\{
f\in\End_{F_{s}}(\w^kV_{s})\mid \w^k g_{\sigma}\circ\sigma(f)\circ
(\w^k g_{\sigma})^{-1}=f\text{ for all $\sigma\in\Gamma$}\bigr\},
\]
hence also from $\End_{F}(\oplus_{k=0}^n\lambda^k B)$ to
\begin{multline*}
\Bigl\{
f\in\End_{F_{s}}\bigl(\oplus_{k=0}^n\End_{F_{s}}(\w^k V_{s})\bigr)
\mid\\
\bigl(\oplus_{k}\Int(\w^k g_{\sigma})\bigr)\circ\sigma(f)=
f\circ\bigl(\oplus_{k}\Int(\w^kg_{\sigma})\bigr)
\text{ for all $\sigma\in\Gamma$}\Bigr\}.
\end{multline*}
Similarly, define
$\w^n(g_{\sigma}\oplus g_{\sigma})\in\GL\bigl(
\w^n(V_{s}\oplus V_{s})\bigr)$
by
\[
\w^n(g_{\sigma}\oplus g_{\sigma})\bigl(
(x_{1},y_{1})\w\ldots\w(x_{n},y_{n})\bigr)=
\bigl(g_{\sigma}(x_{1}),g_{\sigma}(y_{1})\bigr)\w\ldots\w
\bigl(g_{\sigma}(x_{n}),g_{\sigma}(y_{n})\bigr),
\]
so that $\lambda^n\bigl(M_{2}(B)\bigr)$ can be identified through
$\varphi$ to
\[
\Bigl\{
f\in\End_{F_{s}}\bigl(\w^n(V_{s}\oplus V_{s})\bigr)\mid
\w^n(g_{\sigma}\oplus g_{\sigma})\circ\sigma(f)=
f\circ\w^n(g_{\sigma}\oplus g_{\sigma})
\text{ for all $\sigma\in\Gamma$}\Bigr\}.
\]
Certainly, $\w^n(g_{\sigma}\oplus g_{\sigma})=
\oplus_{k=0}^n(\w^kg_{\sigma}\otimes\w^{n-k}g_{\sigma})$ under
\eqref{basicid}, and computation shows that
$\w^kg_{\sigma}\otimes\w^{n-k}g_{\sigma}
=(\det g_{\sigma})\Int(\w^kg_{\sigma})$ under \eqref{1.eq}.
Therefore, \eqref{basicid} and \eqref{1.eq} induce an isomorphism of
$F$-algebras
\[
\Phi\colon\,\lambda^n\bigl(M_{2}(B)\bigr)\iso
\End_{F}(\oplus_{k=0}^n\lambda^kB).
\]

To complete the proof of the theorem, we show that the canonical involution
$\gamma$
on $\lambda^n\bigl(M_{2}(B)\bigr)$ corresponds to the adjoint
involution with respect to $T$ under $\Phi$. In order to do so, we
view $\lambda^n\bigl(M_{2}(B)\bigr)$ and
$\End_{F}(\oplus_{k=0}^n\lambda^kB)$ as the fixed subalgebras of
$\End_{F_{s}}\bigl(\w^n(V_{s}\oplus V_{s})\bigr)$ and
$\End_{F_{s}}\bigl(\oplus_{k=0}^n\End_{F_{s}}(\w^k V_{s})\bigr)$, and
show that the canonical involution $\gamma$ on
\linebreak
$\End_{F_{s}}\bigl(\w^n(V_{s}\oplus V_{s})\bigr)$ corresponds to the
adjoint involution with respect to $T$ (extended to $F_{s}$) under the
isomorphism induced by~\eqref{basicid} and \eqref{1.eq}.

Taking any nonzero element $e \in \w^n V_{s}$,
the identification $\w^{2n}(V_{s} \oplus V_{s}) = \w^n V_{s} \ot \w^n V_{s}$
allows us
to write $e \ot e$ for a nonzero element of $\w^{2n}(V_{s}\ot V_{s})$.  Then
$\ga$ is adjoint to the bilinear form
\[
\Theta\colon\,
\wedge^n(V_{s}\oplus V_{s})\times\wedge^n(V_{s}\oplus V_{s})\to F_{s}
\]
given by
\[
\text{$\Theta(x, y)\, e\ot e = x \w y$ for $x, y \in \w^n(V_{s}\oplus V_{s})$}
\]
as was mentioned in the introduction.  Using the identification of
$\w^k V_{s}\otimes \w^{n-k} V_{s}$ as a subspace of $\w^n (V_{s} \oplus
V_{s})$, we
have that for $x_i, y_i \in \w^i V_{s}$,
\begin{nmultline}
\begin{multline}
\Theta(x_k \ot x_{n-k}, y_\ell \ot y_{n - \ell}) = \left\{
\begin{array}{ll}
(-1)^{\ell} \theta_k(x_k, y_\ell) \theta_{n-k}(x_{n-k}, y_{n-\ell})
& \text{if $k + \ell = n$,} \\
0 & \text{if $k + \ell \ne n$.}
\end{array}
\right.
\end{multline}
\end{nmultline}
We translate this into terms involving $B$, using the isomorphism
$\varphi$ to identify $\lambda^kB_{s}:=(\lambda^k B)\otimes_{F}F_{s}$
with $\End_{F_{s}}(\w^k V_{s})$.  In particular, we know
that
\[
\trd_{\la^k B_{s}}(x_k \ot x_{n-k}) = \theta_{n-k}(x_{n-k}, x_k)
\]
for $\trd$ the reduced trace, and that
\[
\theta_k(x_k, x_{n-k}) = (-1)^{k(n-k)} \theta_{n-k}(x_{n-k}, x_k).
\]
So for $x = x_k \ot x_{n-k} \in \la^k B_{s}$
and $y = y_\ell \ot y_{n - \ell} \in \la^\ell B_{s}$,
\begin{neqn}
\Theta(x,y) = \left\{
\begin{array}{ll}
(-1)^{\ell} \trd_{\la^k B_{s}}(\ga_\ell(y) x)
& \text{if $k + \ell = n$,} \\
0 & \text{if $k + \ell \ne n$}.
\end{array}
\right.
\end{neqn}
Of course, in the $k + \ell = n$ case we could just as easily have
taken
\[
\Theta(x,y) = (-1)^{\ell} \trd_{\la^\ell B_{s}}(\gamma_k(x)y).
\]
So, the vector space isomorphism derived from \eqref{basicid} and
\eqref{1.eq} is an isometry of $\Theta$ and $T$, and it follows
that the canonical involution $\gamma$ adjoint to $\Theta$ corresponds
to the adjoint involution to $T$ under $\Phi$.
\end{proof}

For later use, we prove a little bit more about this isomorphism $\Phi$.
Let us consider the elements $e_{1}=
\left(
\begin{smallmatrix}
  1&0\\
  0&0
\end{smallmatrix}
\right)$ and $e_{2}=
\left(
\begin{smallmatrix}
  0&0\\
  0&1
\end{smallmatrix}
\right)\in M_{2}(B)$, and let $t$ be an indeterminate over $F$.
We write $\la^n$ for the map
$M_2(B)\ra\la^n M_2(B)$ defined in \cite[14.3]{KMRT}, which
is a homogeneous polynomial map of degree $n$.
So there exist $\ell_0,\dots,\ell_n\in \la^n M_2(B)$
such that
$$
  \lambda^n(e_{1}+te_{2})=t^n\ell_{0}+t^{n-1}\ell_{1}+\dots+
  t\ell_{n-1}+\ell_{n}.
$$
We then have
\begin{lem}
\label{l.lem}
For $k=0,\dots,n$, the image of $\ell_k$ under $\Phi$
is the projection on $\la^k B$.
Moreover, we have $\gamma(\ell_k)=\ell_{n-k}$.
\end{lem}

\begin{proof}
It is enough to prove it in the split case.
Hence, we may assume $B=\End_F(V)$, and use identification~\eqref{1.eq} of
the previous section.
An element of $\la^k B=\End_F(\wedge^k V)$ can be written
as $(x_1\wedge\dots\wedge x_k)\ot(y_1\wedge\dots\wedge y_{n-k})$,
where $x_1,\dots,x_k,y_1,\dots,y_{n-k}\in V$.
The endomorphism $\la^n(e_1+te_2)$ acts on this element as follows:
\[
\begin{array}{rcl}
\multicolumn{3}{l}{\lambda^n(e_{1}+te_{2})\left(
    (x_{1}\wedge\dots\wedge x_{k})\otimes
    (y_{1}\wedge\dots\wedge y_{n-k})\right)} \\
%&=(e_{1}+te_{2})(x_{1},0)\wedge\dots\wedge(e_{1}+te_{2})(x_{k},0)
%    \wedge(e_{1}+te_{2})(0,y_{1})\wedge\dots\wedge
%    (e_{1}+te_{2})(0,y_{n-k})\\
\hphantom{dummy space}&=&(x_{1},0)\wedge\dots\wedge(x_{k},0)\wedge
    (0,ty_{1})\wedge\dots\wedge(0,ty_{n-k})\\
&=&t^{n-k}(x_{1}\wedge\dots\wedge x_{k})\otimes
    (y_{1}\wedge\dots\wedge y_{n-k}).
\end{array}
\]
Hence, the image under $\ell_i$ of this element is itself if $i=k$ and
$0$ otherwise.
This proves the first assertion of the lemma.
By Theorem~\ref{1.thm}, to prove the second one, one has to check that
for any $u,v\in\la^0B\oplus\dots\oplus\la^n B$, we have
$T(\ell_i(u),v)=T(u,\ell_{n-i}(v))$, which follows easily from the description
of $T$ given in that theorem.
 \end{proof}

\begin{rem}
\label{1.rem}
By the previous lemma, the elements
$\ell_0,\dots,\ell_n\in\la^n M_2(B)$
are orthogonal idempotents.
Hence, the fact that $\ga(\ell_k)=\ell_{n-k}$ for all $k=0,\dots,n$
implies that the involution $\ga$ is hyperbolic if $n$ is odd and
Witt-equivalent to its restriction to $\ell_m\la^nM_2(B)\ell_m$ if $n=2m$.
\end{rem}

We will also use the following:
\begin{lem} \label{g.lem}
For any $b\in F^\times$, consider $g_0 := \left(
  \begin{smallmatrix}
    0&b\\
    1&0
  \end{smallmatrix}
  \right)\in M_2(B)$, and set $g := \la^n(g_0)$.
We have:
\begin{enumerate}
\item for any $u\in\la^k B$, $\Phi(g)(u)=b^{n-k}\gamma_k(u)\in\la^{n-k}B$;

\item $g^2=b^n$ and $\ga(g)=(-1)^n g$;

\item For any $k=0,\dots,n$, $g\ell_k=\ell_{n-k}g$.
\end{enumerate}
\end{lem}

\begin{proof}
Again, it is enough to prove it in the split case.
A direct computation then shows that for any
$x\ot y\in \wedge^kV\ot\wedge^{n-k} V=\la^k B$, we have
\[
g(x\ot y)=(-1)^{k(n-k)}b^{n-k}(y\ot x),
\]
which combined with (\ref{gakdef}) gives (1),
which in turn easily implies (3).  The first part of (2) is because
$\la^n$ restricts to be a group homomorphism on $M_2(B)^\ast$
\cite[14.3]{KMRT}, and the second part then follows since $\ga(g) g =
\nrd_{M_2(B)}(g) = (-b)^n$ by \cite[14.4]{KMRT}.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Description of $\la^n (Q \ot B)$}
\label{QB.sec}

We suppose that $Q=(a,b)_F$ is a quaternion $F$-algebra and $B$ is an
arbitrary central simple $F$-algebra of degree $n$.
We will describe $\la^n(Q\ot B)$ by Galois descent from $K=F(\alpha)$,
where $\alpha\in F_s$ is a fixed square root of $a$.
More precisely, let us identify $Q$ with the $F$-subalgebra of
$M_2(K)$ generated
by
$\left(
  \begin{smallmatrix}
    \alpha&0\\
    0&-\alpha
  \end{smallmatrix}
  \right)$
and
$g_0=\left(
  \begin{smallmatrix}
    0&b\\
    1&0
  \end{smallmatrix}
  \right)$,
i.e.,
\[
Q=\{x\in M_2(K) \mid g_0\bar xg_0^{-1}=x\},
\]
where $\bar{\ }$ denotes the non-trivial automorphism of $K/F$.
We also have
\[
Q\ot B= \{x\in M_2(B_K) \mid g_0\bar xg_0^{-1}=x\},
\]
where $B_K=B\ot_F K$, and $g_0$ is now viewed as an element of
$M_2(B_K)$.

The canonical map $\la^n\colon\, A \ra \la^n A$ restricts to be a group
homomorphism on $A^\ast$ \cite[14.3]{KMRT}.  Moreover, when $\deg
A = 2n$, for $a \in A^\ast$, $\Int(\la^n(a))$ preserves the canonical
involution $\ga$ on $\la^n A$ \cite[14.4]{KMRT}, and so we get a map
\[
\la^n\colon\, \aut(A) \ra \aut(\la^n A, \gamma).
\]
In particular this
holds for $A = M_2(B_K)$.  This induces a map on Galois cohomology
\[
H^1(K/F, \aut(M_2(B_K))) \xrightarrow{H^1(\la^n)} H^1(K/F, \aut(\la^n
M_2(B_K), \gamma)).
\]
The image under this map of the 1-cocycle $\bar{\ } \mapsto
\Int(g_0)$ is the 1-cocycle $\bar{\ } \mapsto
\Int(\lambda^n g_0)$, as in the preceding section. Since the former
1-cocycle corresponds to $Q\otimes B$, the latter corresponds to
$\lambda^n(Q\otimes B)$, so
\begin{neqn} \label{QB.prop}
\la^n (Q \ot B) = \{ x \in \la^n M_2(B_K) \mid g \bar{x} g^{-1} = x \}
\end{neqn}
for $g := \la^n (g_0)$.  We fix this definition of $g$ for the rest of
the paper.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The $n$ odd case}
\label{odd.sec}

This section is essentially the proof of Theorem \ref{odd.thm}.

We set $\leven B :=
\oplus_{\substack{0\leq k<n\\ \text{$k$ even}}}\la^k B$.
For $0 \le k \le n$, we let
$t_k$ be the reduced trace quadratic form on $\la^k B$ as in (\ref{redtr}).
We then have the following:

\begin{lem}
When $n = \deg B$ is odd,
the algebra with involution $(\la^n(Q\ot B),\gamma)$
is isomorphic to $(Q,\ga_Q)\ot(C,\sig)$,
where $(C,\sig)$ is isomorphic to
$\End_F(\leven B)$
endowed with the adjoint involution with respect to
$\sum_{\substack{0\leq k<n\\ \text{$k$ even}}}t_k$.
\end{lem}

\begin{proof}
If $i,j\in Q$ satisfy $i^2=a$, $j^2=b$ and $ij=-ji$, then
since $\la^n$ restricts to be a group homomorphism on $(Q \ot B)^\ast$,
$\la^n(i\ot 1)$ and $\la^n(j\ot 1)\in\la^n(Q\ot B)$ anticommute and
satisfy
\[
\begin{array}{cc}
\la^n(i\ot 1)^2=a^n,\ &\ \la^n(j\ot 1)^2=b^n,\\
\ga(\la^n(i\ot 1))=-\la^n(i\ot 1),\ &\
\ga(\la^n(j\ot 1))=-\la^n(j\ot 1).\\
\end{array}
\]
(For the bottom two equations, see \cite[(14.4)]{KMRT}.)
Hence, these two elements generate a copy of $Q$ in $\la^n(Q\ot B)$
on which $\ga$ restricts to be $\ga_Q$ and we have
$(\la^n(Q\ot B),\gamma)\simeq (Q,\ga_Q)\ot(C,\sig)$,
where $C$ is the centralizer of $Q$ in $\la^n(Q\ot B)$ and
$\sig$ denotes the restriction of $\ga$ to $C$ \cite[1.5]{KMRT}.

To describe $C$, we take
$i=\alpha(e_1-e_2)\text{ and }j=g_0$,
as in the beginning of the previous section,
so that $\la^n(j\ot 1)=g$ and
\[
\la^n(i\ot 1)=\alpha^n((-1)^n \ell_0+(-1)^{n-1}\ell_1+\dots+\ell_n)
=-\alpha^n(\ell_{\text{even}}-\ell_{\text{odd}}),
\]
where
$\ell_{\text{even}}=\sum_{\substack{0\leq k\leq n\\ \text{$k$ even}}}\ell_k$
and
$\ell_{\text{odd}}=\sum_{\substack{0\leq k\leq n\\ \text{$k$ odd}}}\ell_k$.

Let us consider the map
$\Psi\colon\,\ell_{\text{even}}\la^n(M_2(B))\ell_{\text{even}}
\ra \la^n(M_2(B_K))$ defined by $\Psi(x)=x+gxg^{-1}$.
Believe it or not, $\Psi$ is an $F$-algebra homomorphism.
Clearly, $\overline{\Psi(x)}=\Psi(x)$ and
since $g^2=b^n$ is central (see Lemma~\ref{g.lem}),
$g\Psi(x)=\Psi(x)g$ for all $x$.
Hence, the image of $\Psi$ is contained in $\la^n(Q\ot B)$ and
is centralized by $g$.
Moreover,
$$\la^n(i\ot 1)\Psi(x)=-\alpha^n(x-gxg^{-1})=\Psi(x)\la^n(i\ot 1).$$
Hence, the image of $\Psi$ also centralizes $\la^n(i\ot 1)$,
and by dimension count it is exactly $C$.

Since $\ga(\Psi(x))=\Psi(g^{-1}\ga(x)g)$,
the involution $\sig$ on $C$ corresponds via $\Psi$ to
$\Int(g^{-1})\circ\ga$ on $\ell_{\text{even}}\la^n(M_2(B))\ell_{\text{even}}$.
Note that if $x\in \ell_{\text{even}}\la^n(M_2(B))\ell_{\text{even}}$, then
$\ga(x)\in \ell_{\text{odd}}\la^n(M_2(B))\ell_{\text{odd}}$ and
$g^{-1}\ga(x)g\in \ell_{\text{even}}\la^n(M_2(B))\ell_{\text{even}}$.
By Theorem~\ref{1.thm}, we get that $(C,\sig)$ is isomorphic
to $\End_F(\la^{\text{even}}B)$ endowed with the involution adjoint
to the quadratic form $T'$ defined by
$T'(u,v)=T(u,\Phi(g)(v))$.
Using the description of $T$ given in Theorem~\ref{1.thm} and
Lemma~\ref{g.lem}(1),
it is easy to check that the $\la^k B$ are pairwise orthogonal for
$T'$ and that $T'$ restricts to be $\qform{(-b)^{n-k}}t_k$ on $\la^k B$.
Thus $T'$ is similar to
$\sum_{\substack{0\leq k<n\\ \text{$k$ even}}}t_k$.
\end{proof}

Let us now prove Theorem~\ref{odd.thm}.
If $n=2m+1$, then the algebra with involution
$(Q,\gamma_Q)^{\ot n}$ is isomorphic to
$(Q,\gamma_Q)\ot (\End_F(Q),\ad_{{n_Q}})^{\ot m}$, where
$\ad_{{n_Q}}$ denotes the adjoint involution with respect to the
quadratic form ${n_Q}$.
Indeed, one may easily check that
$(Q\ot Q,\gamma_Q\ot \gamma_Q)$ is isomorphic to
$\bigl(\End_F(Q),\ad_{T_{(Q, \gamma_Q)}}\bigr)$,
where $T_{(Q, \gamma_Q)}$ is the quadratic form
defined by $T_{(Q, \gamma_Q)}(x)=\trd_Q(x\gamma_Q(x))$.
Since for any $x\in Q$, we have $x\gamma_Q(x)=n_Q(x)\in F$,
$T_{(Q, \gamma_Q)}= \qform{2} n_Q$, and
$(Q^{\ot 2},\gamma_Q^{\ot 2})\simeq (\End_F(Q),\ad_{n_Q})$.
Therefore, to prove Theorem~\ref{odd.thm}, it suffices to show that
the algebras with involution $(Q,\ga_Q)\ot(C,\sig)$ and
$(Q,\ga_Q)\ot(\End_F(Q),\ad_{n_Q})^{\ot m}$ are Witt-equivalent.
We will use the following lemma:

\begin{lem}
Let $(U,q)$ and $(U',q')$ be two quadratic spaces over $F$.
There exists an isomorphism
$$(Q,\ga_Q)\ot(\End_F(U),\ad_q)\simeq (Q,\ga_Q)\ot(\End_F(U'),\ad_{q'})$$
if and only if the quadratic forms
$n_Q\ot q$ and $n_Q\ot q'$ are similar.
\end{lem}

\begin{proof}
Let us write $h$ for the hermitian form $h\colon\,U_Q=U\ot Q\ra (Q,\ga_Q)$
induced by $q$
so that we have
$$(\End_Q(U_Q),\ad_h)=(\End_F(U),\ad_q)\ot (Q,\ga_Q).$$
Its trace form, which is by definition the quadratic form
$$U\ot_F Q\ra F,\ x\mapsto h(x,x)$$ is $q\ot n_Q$.
Similarly, we denote by $h'$ the hermitian form induced by $q'$.
By a theorem of Jacobson \cite[10.1.7]{Sch},
the hermitian modules $(U_Q,h)$ and $(U'_Q,h')$ are isomorphic if
and only if their trace forms are isometric.
Hence, if the quadratic forms $q\ot n_Q$ and $q'\ot n_Q$ are similar,
i.e., $q\ot n_Q\simeq \langle \mu\rangle q'\ot n_Q$ for some $\mu \in
F^\ast$, then
the hermitian forms $h$ and $\langle \mu\rangle h'$ are isomorphic,
which proves
that
$$(Q,\ga_Q)\ot(\End_F(U),\ad_q)\simeq(Q,\ga_Q)\ot(\End_F(U'),\ad_{q'}).$$

Conversely, if the two algebras with involution
$(Q,\ga_Q)\ot(\End_F(U),\ad_q)$ and $(Q,\ga_Q)\ot(\End_F(U'),\ad_{q'})$
are isomorphic, then,
$(Q,\ga_Q)^{\ot 2}\ot(\End_F(U),\ad_q)=(\End_F(Q\ot U),\ad_{n_Q \ot q})$
and
$(Q,\ga_Q)^{\ot 2}\ot(\End_F(U'),\ad_{q'})$ are also isomorphic,
which proves that
$n_Q\ot q$ and $n_Q\ot q'$ are similar.
\end{proof}

These two lemmas reduce the proof of Theorem~\ref{odd.thm} to showing that
the quadratic forms
$n_Q\ot \sum_{\substack{0\leq k<n\\ \text{$k$ even}}}t_k$ and
$n_Q^{\ot (m+1)}$ are Witt-equivalent, up to a scalar factor.

On the one hand, we have $n_Q^{\ot (m+1)}=4^m n_Q$, since $n_Q^{\ot 2}=4n_Q$.
On the other hand, since the algebra $B$ is split by an odd-degree
field extension,
Springer's Theorem \cite[VII.2.3]{Lam} shows that
$t_k$ is isometric to the trace form of
$$\la^k(M_n(F))=M_{\binom{n}{k}}(F)$$ which is Witt-equivalent to
$\binom{n}{k}\langle 1\rangle $.
Hence the Witt class of
$n_Q\ot \sum_{\substack{0\leq k<n\\ \text{$k$ even}}}t_k$
is
\[
\sum\nolimits_{\substack{0\leq k<n\\ \text{$k$ even}}}\binom{n}{k}n_Q
=2^{n-1}n_Q=4^mn_Q,
\]
which completes the proof of Theorem \ref{odd.thm}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The $n$ even case}
\label{even.sec}

In this section, we prove Theorems~\ref{memorable.thm},
\ref{even.thm}, and Corollary~\ref{1.cor}.

Assume from now on that $n$ is even and write $n=2m$.
Consider the element of $\la^n(M_2(B_K))$
$$
h=\alpha(1-b^{-m}g)(\ell_{0}+\dots+\ell_{m-1}+
{\textstyle\frac{1}{2}}\ell_{m})+
(1+b^{-m}g)
({\textstyle\frac{1}{2}}\ell_{m}+\ell_{m+1}+\dots+\ell_{n}).
$$
One can check that
$$
h^{-1}=
{\textstyle\frac{1}{2}}\left(
(\alpha^{-1}+b^{-m}g)(\ell_{0}+\dots+\ell_{m})+
(1-b^{-m}g\alpha^{-1})(\ell_{m}+\dots+\ell_{n})\right)
$$
and $g=b^m \, h\overline{h}^{-1}$.

Therefore, it follows from (\ref{QB.prop}) that
$$
\lambda^n(Q\otimes B)=h\lambda^nM_{2}(B)h^{-1}\subset
\lambda^nM_{2}(B)_{K}.
$$
Using the isomorphism $\Phi$ of Theorem~\ref{1.thm} as an
identification, we then have
$$
\lambda^n(Q\otimes B)=
\End_{F}\left(h(\lambda^{0}B)\oplus\dots\oplus h(\lambda^nB)\right),
$$
and the canonical involution on $\lambda^n(Q\otimes B)$ is adjoint to
the restriction of the bilinear form $T_{K}$ to the $F$-subspace
$h(\lambda^{0}B)\oplus\dots\oplus h(\lambda^nB)$.
This restriction is given by the following formula:

\begin{lem}
  \label{5.1.lem}
  The $F$-subspaces $h(\lambda^kB)$ are pairwise orthogonal.
  Moreover, for $u$, $v\in\lambda^kB$ we have
   $$
   T_{K}\left(h(u),h(v)\right)=
   \begin{cases}
   -2a(-1)^kb^{m-k}\trd_{\lambda^kB}(uv)&\text{if $k<m$},\\
   (-1)^m\trd_{\lambda^mB}\left(
   \frac{(1+a)\gamma_{m}(u)+(1-a)u}{2}v\right)&\text{if $k=m$},\\
   2(-1)^kb^{m-k}\trd_{\lambda^kB}(uv)&\text{if $k>m$}.
   \end{cases}
   $$
\end{lem}

\begin{proof}
Using Lemmas~\ref{l.lem} and ~\ref{g.lem}(1), one may easily check that for
any $u\in\la^k B$, we have
$$h(u)=
\begin{cases}
\alpha(u-b^{m-k}\ga_k(u))&\text{if $k<m$},\\
\frac{(1+\alpha)u+(1-\alpha)\ga_k(u)}{2}&\text{if $k=m$},\\
u+b^{m-k}\ga_k(u)&\text{if $k>m$}.\\
\end{cases}$$
The claim then follows from the description of $T$ given in
Theorem~\ref{1.thm} and Lemma \ref{g.lem}(1) by some direct computations.
For instance, if $u,v\in\la^m B$, we get
\[
\begin{array}{rcl}
T_K(h(u),h(v)) &=& T_K\left(\frac{(1+\alpha)u+(1-\alpha)\ga_m(u)}{2},
\frac{(1+\alpha)v+(1-\alpha)\ga_m(v)}{2}\right) \\
&=&(-1)^m\trd_{\la^m B}\left(\frac{(1+\alpha)\ga_m(u)+(1-\alpha)u}{2}
\times\frac{(1+\alpha)v+(1-\alpha)\ga_m(v)}{2}\right) \\
&=&(-1)^m\trd_{\la^m B}\left(\frac{((1+\alpha)^2+(1-\alpha)^2)\ga_m(u)v+
2(1+\alpha)(1-\alpha)uv}{4}\right) \\
&=&(-1)^m\trd_{\lambda^mB}\left(
   \frac{(1+a)\gamma_{m}(u)+(1-a)u}{2}v\right)
\end{array}
\]
\end{proof}

This lemma yields a first description of the similarity class of
$q_{A}$:

\begin{prop}
  \label{temp.prop}
  If $n$ is even, the similarity class of $q_A$ contains the
  quadratic form:
  \[
  \left(\oplus_{0\leq k<m} \qform{2(-1)^kb^{m-k}} \qform{1,-a}
  t_{k}\right) \oplus \qform{(-1)^m} (t_{m}^{+}
  \oplus \qform{-a} t_{m}^-).
  \]
\end{prop}

\begin{proof}
  Since the
  anti-isomorphism $\gamma_{k}$ defines an isometry $t_k\simeq t_{n-k}$,
  the restriction of $T_K$ to $h(\la^k B\oplus\la^{n-k} B)$, for all $k<m$, is
  $$
  \langle 2(-1)^kb^{m-k}\rangle \langle 1,-a\rangle t_k.
  $$
  Moreover, we have
  $$
  \frac{(1+a)\gamma_{m}(u)+(1-a)u}{2}=
  \begin{cases}
    u&\text{if $u\in\Sym(\lambda^mB,\gamma_{m})$},\\
    -au&\text{if $u\in\Skew(\lambda^mB,\gamma_{m})$}.
  \end{cases}
  $$
  Hence, the proposition clearly follows from the lemma.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{borel}{Proof of Theorem~\ref{memorable.thm}}
\label{memorable.pf}

Theorem~\ref{memorable.thm} is a consequence of the preceding results in
the special case
where $Q = (a,b)_F$ is split.  In that case, we may take $b = 1$ so
that the matrix $g_0=\left(
  \begin{smallmatrix}
    0&1\\
    1&0
  \end{smallmatrix}
  \right)$ then decomposes as $g_0=f_0\bar f_0^{-1}$,
where $f_0=\left(
  \begin{smallmatrix}
    1&-\alpha\\
    1&\alpha
  \end{smallmatrix}
  \right)$.
Hence, if we let $f=\la^n f_0$, we have $g=f\bar f^{-1}$. On the other hand,
we also have $g=h\bar h^{-1}$, for $h$ as in the preceding section,
hence $f^{-1}h=\overline{f^{-1}h}$, which means that
$f^{-1}h\in\lambda^n\bigl(M_2(B)\bigr)$. Considering the
isomorphism $\Phi$ of Theorem~\ref{1.thm} as an identification as we did in
the preceding section, we get that
$f^{-1}h\in\End_F(\lambda^0B\oplus\cdots\oplus\lambda^n B)$, hence
\[
h(\la^0 B\oplus\cdots\oplus\la^n B)=f(\la^0 B\oplus\cdots\oplus \la^n B).
\]

To prove Theorem~\ref{memorable.thm}, we compute the restriction of $T_K$ to
this $F$-subspace in two different ways. First, we use
\cite[(14.4)]{KMRT}, which says that
$f$ is a similarity for $T_K$ with similarity factor
$\nrd_{M_2(B_K)}(f_0)=(-2\alpha)^n=2^na^m$.
Hence, for any $u,v\in\la^0B\oplus\dots\oplus\la^nB$, we have
$$T_K(f(u),f(v))=2^na^mT(u,v).$$
By Remark~\ref{1.rem} and Theorem~\ref{1.thm}, the form $T$ is
Witt-equivalent to its restriction
to $\la^m B$, which is isometric to
$\langle (-1)^m\rangle (t_m^+\oplus\langle -1\rangle t_m^-)$.

Second, the restriction of $T_K$ to $h(\la^0 B\oplus\cdots\oplus\la^n B)$
has been computed in
Lemma~\ref{5.1.lem} and the proof of
Proposition~\ref{temp.prop}. Comparing the results, we get that the
quadratic forms
\[
\bigl(\oplus_{0\leq k<m} \langle  2(-1)^k\rangle\langle 1,-a\rangle
  t_{k}\bigr)\oplus\langle (-1)^m\rangle (t_{m}^{+}
\oplus\langle -a\rangle t_{m}^-)
\]
and
\[
\langle 2^na^m\rangle \langle (-1)^m
\rangle (t_m^+\oplus\langle -1\rangle t_m^-)
\]
are Witt-equivalent.
If $m$ is even, we get that the following equality holds in the Witt ring:
\[
\left(\sum\nolimits_{0\leq k<m}\langle  2(-1)^k\rangle\langle 1,-a\rangle
  t_{k}\right)+t_m^++\langle -a\rangle t_m^-=t_m^+-t_m^-,
\]
from which we deduce
\[
\qform{1,-a} \left( \left( \sum\nolimits_{0\leq k<m} \qform{2(-1)^k}
  t_{k} \right) +t_m^{-} \right) = 0.
\]
To finish the proof, we may assume $a$ is an indeterminate over
the base field $F$.
The previous equality then implies that the quadratic form
\[
\left( \bigoplus\nolimits_{0\leq k<m} \qform{2(-1)^k} t_{k}\right)
\oplus t_m^{-}
\]
is hyperbolic, which proves the theorem
in this case.
A similar argument finishes the proof for the $m$ odd case.

\begin{rem}
Let $t_{(\la^m B, \gamma_m)}\colon\, \la^m B \ra F$ be the quadratic
form
\[
t_{(\la^m B, \gamma_m)}(x) = \trd_{\la^m B}(\gamma_m(x) x).
\]
Using Theorem~\ref{memorable.thm}, together with the facts that $t_{n-k} =
t_k$, $t_{(\la^m B, \gamma_m)} = t_m^{+} - t_m^{-}$, and that $2q
\simeq 2\qform{2} q$ for an arbitrary quadratic form $q$ since
$2\qform{2} = 2\qform{1}$, we obtain the following memorable formula:
\[
\sum^n_{k = 0} (-1)^k t_k = t_{(\la^m B, \gamma_m)}
\qquad\text{in $WF$}.
\]
\end{rem}
\end{borel}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{borel}{Proof of Theorem~\ref{even.thm}}
\label{even.pf}
Consider first the case where $m$ is even. In that case,
Theorem~\ref{memorable.thm} yields
\[
\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}}\langle 2\rangle
t_k+t_m^-
=\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ odd}}}\langle 2\rangle t_k.
\]
Substituting in the formula given in Proposition~\ref{temp.prop}, we get that
the similarity class of $q_A$ contains a quadratic form whose Witt class
is
\begin{multline*}
\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}} \qform{2,-2a} t_k
+\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}} \qform{-2b,2ab} t_k
+\langle -a,-b,ab\rangle t_m^{-} + t_m^{+} \\
=\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}} \qform{2} n_Q t_k
+t_m^+-t_m^-+n_Q t_m^-.
\end{multline*}
Now, suppose $m$ is odd. Multiplying by $\langle a\rangle$ the
quadratic form given in Proposition~\ref{temp.prop} does not change
its similarity class, and shows that the similarity class of $q_{A}$
contains a quadratic form whose Witt class is
\[
\langle1,-a\rangle\cdot\Bigl(t_{m}^{+}+\sum\nolimits_{0\leq k<m}
\langle 2(-b)^{k+1}\rangle t_{k}\Bigr) +t_{m}^- -t_{m}^{+}.
\]
Substituting for $t_{m}^{+}$ the formula of
Theorem~\ref{memorable.thm} simplifies the expression in brackets to
$\langle1,-b\rangle\cdot\Bigl(\sum\nolimits_{\substack{0\leq k<m\\
\text{$k$ even}}}
\langle2\rangle t_{k}\Bigr)$ and completes the proof.
\end{borel}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{borel}{Proof of Corollary~\ref{1.cor}}
\label{1.pf}
Let us assume that $B$ is of exponent at most $2$.
Then, for any even $k$, the algebra $\la^k B$ is split.
Hence, its trace form $t_k$ is Witt-equivalent to
$\binom{n}{k}$.
Since $m$ is even, $\la^m B$ is also split,
and its canonical involution $\ga_m$
is adjoint to a quadratic form $q_B$. This form is only defined up to a
scalar factor, but its square is defined up to isometry.
Now \cite[11.4]{KMRT} gives relationships between $q_B$ and the forms
$t_m^+$ and $t_m^-$:
\[
\text{$t_m^{+} - t_m^{-} \simeq q^2_B$ and $-t_m^{-} \simeq
\qform{1/2} \wt{q_B}$.}
\]
Hence, by Theorem~\ref{even.thm}, the similarity class of $q_A$ contains
a form whose Witt class is
\[
q_B^2 + n_Q
\Bigl( \qform{-2} (\wt{q_B})
+
\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}}
{\textstyle\binom{n}{k}}\langle 2\rangle \Bigr).
\]
One may easily check that, since
$\qform{2,2} \simeq \qform{1,1}$ and $q_B$
is even-dimensional, $q_B^2\simeq \qform{2} q_B^2$. Since we are
concerned only with the similarity class of $q_{A}$, we may therefore
forget the factors $\langle 2\rangle$ throughout.
Moreover, since $m$ is even,
$\sum_{\substack{0\leq k<m\\ \text{$k$ even}}}\binom{n}{k}
=2^{n-2}-\frac{1}{2}\binom{n}{m}$, and Corollary~\ref{1.cor} follows.
\end{borel}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Another approach to the $n$ even case}
\label{result.section}

Let us decompose $B=B_0\ot B_1$, where $\deg B_0=2m_0$ is a power of $2$
and $\deg B_1=m_1$ is odd.
We have $m=m_0m_1$, and $m$ is even if and only if $m_0> 1$.
We write $T_0$ for the trace form of $B_0$.  Under the assumption that
$B_0^{\otimes 2}$ is split (which is automatic if $m$ is odd), we will
give a different characterization of $q_A$ for $A = Q \otimes B$ than
the one in Theorem \ref{even.thm}.  Corollaries \ref{2.cor} and
\ref{3.cor} will follow from this.

\begin{prop} \label{even.prop} \label{odd.prop}
Suppose that $B_0^{\otimes 2}$ is split. Then the similarity class of $q_A$
contains a form whose Witt class
is
\[
2^{n-1}+\frac{2^{n-3}}{m_0}T_0(n_Q-2) \text{\ if $m$ is even}
\]
and
\[
2^{n-2} (n_Q - n_{B_{0}}) \text{\ if $m$ is odd}.
\]
(Note that $B_{0}$ is a quaternion algebra if $m$ is odd.)
\end{prop}

This result is already known for $m$ odd: If $A$ is a biquaternion algebra
it is \cite[6.2]{G:clif}, and in general it follows from
\cite[6.4]{G:clif} by a straightforward computation, using the fact
that for any integer $k\geq 1$, one has $n_Q^k=2^{2(k-1)}n_Q$.
However, the results from \cite{G:clif} make use of Clifford algebras,
which seems a long
way to go.  So we include a direct proof, at least for $m\geq 3$.

We start with a lemma.
\begin{lem}
Suppose that $B_0^{\ot 2}$ is split.
Then the quadratic form $t_k$ is Witt-equivalent to
$\binom{n}{k}$ if $k$ is even and
$\frac{1}{2m_0}\binom{n}{k}T_0$ if $k$ is odd.
Moreover, we have:
\[
t_m^-
=\frac{2^{n-3}}{m_0} \qform{2}
T_0- \Bigl( 2^{n-2}-{\textstyle\frac{1}{2}\binom{n}{m}} \Bigr) \qform{2}
\ \text{if $m$ is even,}
\]
and
\[
t_m^+=2^{n-2} \qform{2}
- \Bigl( 2^{n-3}-{\textstyle\frac{1}{4}\binom{n}{m}} \Bigr) \qform{2} T_0
\ \text{ if $m$ is odd.}
\]
\end{lem}

This lemma actually specifies $t_m^{+}$ and $t_m^{-}$ whatever the
parity of $m$ since in both cases $t_m = t_m^{+} + t_m^{-}$, and $t_m$
is known.

\begin{proof}
Since $B_1$ is split by an odd-degree field extension, Springer's Theorem
shows that $t_k$ is isometric to the trace form of
$\la^k\bigl(B_0\ot M_{m_1}(F)\bigr)$.
If $k$ is even, this algebra is split, and the result is clear.
If $k$ is odd, the algebra is Brauer-equivalent to $B_0$, hence isomorphic to
$M_p(F)\ot B_0$, where $p=\frac{1}{2m_0}\binom{n}{k}$.
The form of $t_k$ for $k$ odd then follows from the fact that the trace form
of a tensor product of central simple algebras is isometric to
the product of the trace forms of each factor.

We have $m=m_0m_1$, and $m$ is odd if and only if $m_0=1$.
Recall that
\[
\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}}
{\textstyle\binom{n}{k}}=
\begin{cases}2^{n-2}&\text{if $m$ is odd},\\
2^{n-2}-\frac{1}{2}\binom{n}{m}&\text{if $m$ is even},\\
\end{cases}
\]
and
\[
\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ odd}}}
{\textstyle\binom{n}{k}}=
\begin{cases}2^{n-2}-\frac{1}{2}\binom{n}{m}&\text{if $m$ is odd},\\
2^{n-2}&\text{if $m$ is even}.\\
\end{cases}
\]
The second part of the lemma then follows from
Theorem~\ref{memorable.thm}
by a direct computation.
\end{proof}

Let us now prove Proposition~\ref{even.prop}.
Assume first that $m$ is even. The preceding lemma yields
\[
t_{m}^- +\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}}
\langle 2\rangle t_{k}=\frac{2^{n-3}}{m_{0}}\langle 2\rangle T_{0}
\]
and
\[
t_{m}^{+}-t_{m}^-={\textstyle\binom{n}{m}}-2t_{m}^-=
2^{n-1}\langle 2\rangle -\frac{2^{n-2}}{m_{0}}\langle 2\rangle T_{0}
+{\textstyle\binom{n}{m}}\langle 1,-2\rangle.
\]
Since $\binom{n}{m}$ is even, the last term on the right side
vanishes, hence the quadratic form given by Theorem~\ref{even.thm}
is
\[
\langle 2\rangle \Bigl(2^{n-1}-\frac{2^{n-2}}{m_{0}}T_{0}
+\frac{2^{n-3}}{m_{0}}n_{Q}T_{0}\Bigr).
\]
This finishes the $m$ even case.

Assume now that $m$ is odd.
Then, $B_0$ is a quaternion algebra, and $T_{0}=\langle 2\rangle
(2-n_{B_{0}})$. The preceding lemma yields
\[
\sum\nolimits_{\substack{0\leq k<m\\ \text{$k$ even}}}
\langle 2\rangle t_{k}=2^{n-2}\langle 2\rangle=2^{n-2}
\]
and
\[
t_{m}^- -t_{m}^{+}={\textstyle\frac{1}{2}\binom{n}{m}}T_{0}-2t_{m}^{+}
=
{\textstyle\frac{1}{2}\binom{n}{m}}T_{0}-2^{n-1}\langle 2\rangle +
\Bigl(2^{n-2}-{\textstyle\frac{1}{2}\binom{n}{m}}\Bigr)\langle
2\rangle T_{0}.
\]
Since $m$ is odd, $\frac{1}{2}\binom{n}{m}$ is even, hence
$\frac{1}{2}\binom{n}{m}\langle 2\rangle = \frac{1}{2}\binom{n}{m}$
and the right side simplifies to yield
\[
t_{m}^- -t_{m}^{+}=-2^{n-2}n_{B_{0}}.
\]
Therefore, the quadratic form given by Theorem~\ref{even.thm} is
$2^{n-2}(n_{Q}-n_{B_{0}})$, and the proof of
Proposition~\ref{even.prop} is complete.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{borel}{Proof of Corollary~\ref{2.cor}} \label{2.pf}
Corollary~\ref{2.cor} can be proved by induction, using the formula
given in Corollary~\ref{1.cor},
but it can also be directly deduced from Proposition~\ref{even.prop}.
Indeed, let us assume $A=A_r=Q_1\ot\dots\ot Q_r$ is a product of
$r\geq 3$ quaternion algebras.
We let $B=Q_2\ot\dots\ot Q_r$.
Its degree $n=2^{r-1}$ is a power of $2$, and since $r\geq 3$,
$m=2^{r-2}$ is even.
In the notation from earlier in this previous section, we have
$B_0=B$ and $B_0^{\ot 2}$ is split.
Hence, we may apply Proposition~\ref{even.prop}.
The form $T_0$ is the trace form of $B$, that is the tensor
product of the trace forms of the quaternion algebras
$Q_i$ for $i=2,\ldots,r$.
Hence, we have $T_0=\langle 2^{r-1}\rangle (2-n_{Q_2})\cdots(2-n_{Q_r})$,
and Proposition~\ref{even.prop} tells us that the similarity class of
$q_A$ contains a form whose Witt class is
\[
\begin{array}{rcl}
\multicolumn{3}{l}{2^{n-1}+\frac{2^{n-3}}{2^{r-2}}\langle 2^{r-1}\rangle
(n_{Q_1}-2)(2-n_{Q_2})\cdots(2-n_{Q_r})} \\
\hphantom{Dummy space} &=& 2^{n-1}\langle 2^{r-1}\rangle -2^{n-r-1}\langle
2^{r-1}\rangle
(2-n_{Q_1})(2-n_{Q_2})\cdots(2-n_{Q_r}) \\
&=&\langle 2^{r-1}\rangle 2^{n-r-1}\bigl(2^r-(2-n_{Q_1})
\cdots(2-n_{Q_r})\bigr),
\end{array}
\]
which proves the corollary.
\end{borel}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{borel}{Proof of Corollary \ref{3.cor}} \label{3.pf}
Let us now consider a central simple algebra $A$ as in the statement
of Corollary \ref{3.cor}.  Then $A$ is isomorphic to $M_k(A_r)$, where
$A_r=Q_1\ot\cdots\ot Q_r$ is a product of $r$ quaternion algebras.
If $A$ is split then $q_A$ is hyperbolic and the result is clear, so
we may assume that $r \ne 0$.  Because $\deg A \equiv 0 \mod{4}$ by
hypothesis, we may further assume that $r \ne 1$ (so that $r \ge 2$),
with perhaps some of the $Q_i$ being split.

We first treat the $k=1$ case.
If $r=2$, then $A$ is biquaternion algebra and $q_A$ is an Albert
form, which lies in $\IF2$.
If $r \geq 3$, then by Corollary~\ref{2.cor} we have to prove that
\[
2^{n-1}-2^{n-r-1}(2-n_{Q_1})\cdots(2-n_{Q_r})
\]
lies in $\IF{n}$.
When we expand this product, the terms of the form $2^{n-1}$ cancel,
and we are left with a sum of terms of the form
$\pm 2^{n-\ell-1}n_{Q_{i_1}}\cdots n_{Q_{i_\ell}}$, where $\ell\geq 1$.
Since for any $i$ the form $n_{Q_i}$ lies in $\IF2$,
$2^{n-\ell-1}n_{Q_{i_1}}\cdots n_{Q_{i_\ell}}$
belongs to $\IF{n-\ell-1+2\ell} = \IF{n+\ell-1}$, and hence to $\IF{n}$.

Now suppose that $k\geq 2$.
Since $r\geq 2$, we have $\deg(A_r)\equiv 0\mod 4$ and we can apply
\cite[6.3(1)]{G:clif}.
Hence, the similarity class of $q_A$ contains a form which is Witt-equivalent
to $q_{A_r}^{\ot k}$.
Since the result holds for $A_r$ by the $k=1$ case, we are
done.
\end{borel}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{8}

\bibitem{DLT}
I.~Dejaiffe, D.~Lewis, and J.-P. Tignol, \emph{Witt equivalence of central
  simple algebras with involution}, preprint, 1998.

\bibitem{G:clif}
R.S. Garibaldi, \emph{Clifford algebras of hyperbolic involutions}, to appear
  in Math.~Z.

\bibitem{Jacob:indec}
B.~Jacob, \emph{Indecomposable division algebras of prime exponent}, J. Reine
  Angew. Math. \textbf{413} (1991), 181--197.

\bibitem{KMRT}
M.-A. Knus, A.S. Merkurjev, M.~Rost, and J.-P. Tignol, \emph{The book of
  involutions}, Colloquium Publications, vol.~44, AMS, Providence, RI, 1998.

\bibitem{Lam}
T.Y.\ Lam, \emph{The algebraic theory of quadratic forms}, Benjamin, Reading,
  MA, 1973.

\bibitem{M:simple}
A.S. Merkurjev, \emph{Simple algebras and quadratic forms}, Math. USSR Izv.
  \textbf{38} (1992), no.~1, 215--221.

\bibitem{QT}
A.~Qu{\'e}guiner and J.-P. Tignol, \emph{Clifford and discriminant algebras},
  in preparation.

\bibitem{Sch}
W.~Scharlau, \emph{Quadratic and hermitian forms}, Grund.\ math.\ Wiss., vol.
  270, Springer, Berlin-New York, 1985.

\end{thebibliography}

\bigskip

\noindent \parbox{2.8in}{R.~Skip Garibaldi\\%
e-mail: {\tt skip@member.ams.org} \\%
web: {\tt http://www.math.ucla.edu/\~{}skip/}\\ %
UCLA\\%
Dept.~of Mathematics\\%
Los Angeles, CA 90095-1555\\%
USA}

\vspace{0.5in}

\noindent \parbox{2.8in}{Anne Qu\'eguiner-Mathieu\\%
e-mail: {\tt queguin@math.univ-paris13.fr} \\%
UMR CNRS 7539\\
D\'epartement de Math\'ematiques\\
Universit\'e Paris 13\\
F-93430 Villetaneuse\\
France}

\vspace{0.5in}

\noindent\parbox{3.2in}{Jean-Pierre Tignol\\%
e-mail: {\tt tignol@agel.ucl.ac.be} \\%
web: {\tt http://www.math.ucl.ac.be/tignol/}\\ %
D\'epartment de Math\'ematiques\\
Universit\'e Catholique de Louvain\\
B-1348 Louvain-la-Neuve\\
Belgium}

\end{document}

