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%%% Authors: G. Berhuy, M. Monsurro, J.-P. Tignol
%%% Title: The discriminant of a symplectic involution
%%% Email: gregory.berhuy@epfl.ch, monsurro@masg1.epfl.ch, tignol@math.ucl.ac.be
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%%% Keywords and phrases: central simple algebras with involution, trace forms
%%% 
%%% Abstract:
%%%      An invariant for symplectic involutions on central simple
%%%      algebras of degree divisible by 4 over fields of characteristic
%%%      different from 2 is defined on the basis of Rost's
%%%      cohomological invariant of degree 3 for torsors under
%%%      symplectic groups. We relate this invariant to trace forms and
%%%      show how its triviality yields a decomposability criterion for
%%%      algebras with symplectic involution.
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%%% Thank you in advance, and best regards,
%%% 
%%%    Jean-Pierre
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\DeclareMathOperator{\Nrd}{Nrd}
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\DeclareMathOperator{\Trd}{Trd}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\Int}{Int}
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\newtheorem{prop}{Proposition}
\newtheorem{thm}[prop]{Theorem}
\newtheorem{lem}[prop]{Lemma}
\newtheorem{corol}[prop]{Corollary}


\theoremstyle{definition}

\newtheorem*{defn}{Definition}
\newtheorem{exemple}[prop]{Example}
\newtheorem{nota}[prop]{Remark}

\title{The discriminant of a symplectic involution}
\author{Gr\'egory Berhuy \and Marina Monsurr\`o \and Jean-Pierre Tignol}
\address{D\'epartement de math\'ematiques\\
\'Ecole Polytechnique F\'ed\'erale de Lausanne\\
CH-1015 Lausanne\\
Switzerland}
\email{gregory.berhuy@epfl.ch, monsurro@masg1.epfl.ch}
\address{Institut de Math\'ematique Pure et Appliqu\'ee\\
Universit\'e catholique de Louvain\\
B-1348 Louvain-la-Neuve\\
Belgium}
\email{tignol@math.ucl.ac.be}

\thanks{The authors gratefully acknowledge support from the TMR
network ``$K$-theory and linear algebraic groups'' (ERB FMRX CT97-0107).
The third author is also grateful to the National Fund for
Scientific Research (Belgium) for partial support.}

\begin{document}

\begin{abstract}
     An invariant for symplectic involutions on central simple
     algebras of degree divisible by~$4$ over fields of characteristic
     different from~$2$ is defined on the basis of Rost's
     cohomological invariant of degree~$3$ for torsors under
     symplectic groups. We relate this invariant to trace forms and
     show how its triviality yields a decomposability criterion for
     algebras with symplectic involution.
\end{abstract}

\maketitle
\section{Introduction and Statement of Results}

In contrast with orthogonal involutions, for which invariants
corresponding to the discriminant and Clifford algebras of quadratic
forms are defined, no ``classical'' invariant is known for symplectic
involutions on central simple algebras, besides the signature (see
\cite[(11.10)]{BoI}). Using the cohomological invariant of degree~$3$
defined by Rost for torsors under simply connected absolutely simple linear
algebraic groups, we introduce an invariant of symplectic involutions
on central simple algeras of degree multiple of $4$ with values in
the third Galois cohomology group of the center with coefficients
$\{\pm1\}$ and give an alternative description in terms of trace
forms. We call this invariant \emph{discriminant} since it is the
first nontrivial invariant, and because it is directly linked to the
discriminant of hermitian forms, see Example~\ref{ind2.ex}.
Even though its definition is elementary, Rost's computation of the
invariants of torsors under symplectic groups is needed to prove that
there is no other cohomological invariant of degree~$3$ and to
establish the relation with trace forms. In the final section, we
prove that symplectic involutions of trivial discriminant on central
simple algebras of degree~$8$ and index~$4$ afford a special type of
decomposition. In a sequel to this paper, the discriminant is used to
give examples of non $R$-trivial adjoint symplectic groups of even
index.

\subsection{Definition of the discriminant}

Throughout this paper, $F$ denotes a field of characteristic
different from $2$. Let $A$ be a finite-dimensional central simple
$F$-algebra, and $\theta\colon\,A\rightarrow A$ be
an anti-automorphism of order $2$. We recall that $\theta$ is
called a \emph{symplectic involution} on $A$ if, after
scalar extension to a splitting field, $\theta$ is adjoint
to  an alternating form, see \cite[(2.5)]{BoI}. From now on,
we suppose the involution $\theta$ is of this type. In this case, the
degree $\deg A$ is necessarily an even integer $n=2m$.

The symplectic group $\Sp(A,\theta)$ is the group scheme over
$F$ defined by
\[
\Sp(A,\theta)(E)=\{x\in A\otimes_{F}E\mid\theta(x)x=1\}
\]
for any commutative $F$-algebra $E$.

Let $\Sym(A,\theta)$ be the $F$-vector space of elements
in $A$ fixed by $\theta$,
\[
\Sym(A,\theta)=\{x\in A\mid \theta(x)=x\}.
\]
We denote by $\Sym(A,\theta)^\times$ the set of units in
$\Sym(A,\theta)$,
\[
\Sym(A,\theta)^\times=\Sym(A,\theta)\cap A^\times.
\]

We recall that the \emph{pfaffian reduced norm} is the
homogeneous polynomial function of degree $m$
\[
\Nrp_{\theta}\colon\,\Sym(A,\theta)\to F
\]
uniquely determined by the following conditions:
\[
\Nrp_{\theta}(1)=1 \qquad\text{and}
\qquad
\Nrp_{\theta}(x)^{2}=\Nrd_{A}(x)
\qquad\text{for $x\in\Sym(A,\theta)$},
\]
see \cite[(2.10)]{BoI}.

The cohomology set $H^{1}\bigl(F,\Sp(A,\theta)\bigr)$ can
be represented as
\begin{equation}
      \label{H1.eq}
      H^{1}\bigl(F,\Sp(A,\theta)\bigr)\simeq \Sym(A,\theta)^\times/{\sim}
\end{equation}
where $\sim$ is the equivalence relation defined by $x\sim y$
if and only if there exists $u\in A^\times$ such that
$y=ux\theta(u)$, see \cite[(29.24)]{BoI}.

Let $\Gm$ be the multiplicative group. The Kummer exact sequence
\[
1\to\mu_{2}\to\Gm\xrightarrow{2}\Gm\to1
\]
allows us to identify the cohomology sets $H^{1}(F,\mu_{2})$ and
$H^{2}(F,\mu_{2})$ respectively
with the quotient $F^\times/F^{\times2}$ and with the $2$-torsion
subgroup of the Brauer group. For all $x\in F^{\times}$, we
denote by $(x)_{2}\in H^{1}(F,\mu_{2})$ the cohomology class
associated to $xF^{\times2}$. Similarly, we denote by
$[A]\in H^{2}(F,\mu_{2})$ the cohomology class associated to
the Brauer class of $A$. We define
\[
\Delta_{\theta}\colon\,\Sym(A,\theta)^\times\to H^{3}(F,\mu_{2})
\]
as the map given by the cup-product
\[
\Delta_{\theta}(s)=\bigl(\Nrp_{\theta}(s)\bigr)_{2}\cup [A].
\]

It follows from the properties of $\Nrp_{\theta}$
(see the proof of Proposition~\ref{elem.prop} below)
that $\Delta_{\theta}$ is well defined on the set
of equivalence classes under the relation $\sim$. The
induced map on the quotient can be interpreted under
the bijection \eqref{H1.eq} as the
Rost invariant of
$H^{1}\bigl(F,\Sp(A,\theta)\bigr)$, see \cite[p.~440]{BoI}.

Since $\Nrp_{\theta}$ is homogeneous of degree $m$, we
obtain, for $\alpha\in F^\times$ and $s\in\Sym(A,\theta)^\times$,
the following relation
\begin{equation}
      \label{mult.eq}
\Delta_{\theta}(\alpha
s)=\bigl(\alpha^m\Nrp_{\theta}(s)\bigr)_{2}\cup[A]=
\begin{cases}
      \Delta_{\theta}(s)&\text{if $m$ is even,}\\
      \Delta_{\theta}(s)+(\alpha)_{2}\cup[A]&\text{if $m$ is odd.}
\end{cases}
\end{equation}

Therefore, if $m$ is even one can define a relative invariant for
symplectic involutions on $A$ as follows:

\begin{defn}
Let $A$ be a central simple algebra over $F$ of degree
$n=2m\equiv 0\bmod4$. Let $\theta$ and $\sigma$ be symplectic
involutions on $A$. There exists (see \cite[(2.7)]{BoI})
$s\in\Sym(A,\theta)^\times$ such that
\[
\sigma=\Int(s)\circ\theta
\]
where $\Int(s)$ denotes the inner automorphism associated with $s$,
\[
\Int(s)(x)=sxs^{-1}\qquad\text{for $x\in A$}.
\]
The element $s$ is uniquely determined up to multiplication by an
element of $F^\times$.
By \eqref{mult.eq}, it follows that
$\Delta_{\theta}(s)\in H^{3}(F,\mu_{2})$ only depends on $\sigma$,
since $m$ is even. We call this element the \emph{discriminant}
of $\sigma$ with respect to $\theta$ and denote it by
$\Delta_{\theta}(\sigma)$. Thus,
\[
\Delta_{\theta}(\sigma)=\bigl(\Nrp_{\theta}(s)\bigr)_{2}\cup[A]
\in H^{3}(F,\mu_{2}).
\]
\end{defn}

In the case $m=2$, an analogue of this invariant has been
studied in \cite[\S16.B]{BoI}, where it is denoted by
$j_{\theta}(\sigma)$. Theorem~(16.19) of \cite{BoI} shows
that this invariant classifies, up to conjugation, symplectic
involutions on a central simple algebra of degree~$4$.

In section~\ref{elem.sec} we establish the following
elementary result:
\begin{prop}
      \label{elem.prop}
      \begin{enumerate}
         \item[(a)]
The discriminant $\Delta_{\theta}(\sigma)$ only depends on the
conjugacy classes of $\theta$ and $\sigma$, namely, if $u$,
$v\in A^\times$ and
         \[
         \theta'=\Int(u)\circ\theta\circ\Int(u)^{-1},
         \qquad
         \sigma'=\Int(v)\circ\sigma\circ\Int(v)^{-1},
         \]
         then
         \[
         \Delta_{\theta'}(\sigma')=\Delta_{\theta}(\sigma).
         \]
In particular, if $\sigma$ and $\theta$ are conjugate, then
$\Delta_{\theta}(\sigma)=0$.

\item[(b)]
Let $\rho$, $\sigma$ and $\theta$ be symplectic involutions
on $A$; then
\[
\Delta_{\rho}(\sigma)=\Delta_{\rho}(\theta)+\Delta_{\theta}(\sigma)
\qquad\text{and}\qquad
\Delta_{\theta}(\sigma)=\Delta_{\sigma}(\theta).
\]
\end{enumerate}
\end{prop}

If the Schur index $\ind A$ divides
$\frac{1}{2}\deg A$, i.e., if $A\simeq M_{2}(A_{0})$
for some central simple $F$-algebra $A_{0}$,
then $A$ carries hyperbolic symplectic involutions, such as
$\gamma\otimes\theta_{0}$, where $\gamma$ is the (unique) symplectic
involution on $M_{2}(F)$ and $\theta_{0}$ is an arbitrary orthogonal
involution on $A_{0}$.
Since all hyperbolic involutions are pairwise conjugate,
we may set $\Delta=\Delta_{\theta}$ for any hyperbolic symplectic
involution $\theta$.

\begin{exemple}
      \label{ind2.ex}
      Consider the algebra $A=\End_{Q}V$, where $Q$ is a
quaternion division $F$-algebra and $V$ is an $m$-dimensional
$Q$-vector space. Symplectic involutions on $A$ are then adjoint
to hermitian forms on $V$ with respect to the conjugation
involution on $Q$. Suppose that $m$ is even and let $\sigma$
be the involution adjoint to a fixed hermitian form $h$ on
$V$. Let
      \[
      h=\langle\alpha_{1},\ldots,\alpha_{m}\rangle
      \]
be the diagonalization of $h$ relative to some orthogonal
basis $e$ of $V$ ($\alpha_{1}$, \ldots, $\alpha_{m}\in F^\times$), then
      \[
      \Delta(\sigma)=\bigl((-1)^{m/2}\alpha_{1}\cdots\alpha_{m}\bigr)_{2}
      \cup[Q].
      \]
      Indeed, let $\theta$ be the hyperbolic involution adjoint to the
      hermitian form with diagonalization
      $\langle1,-1,\ldots,1,-1\rangle$ relative to the basis $e$.
Then, identifying $A$ with $M_{m}(Q)$ by $e$, we get
      \[
      \sigma=
      \Int\diag(\alpha_{1},-\alpha_{2},\ldots,\alpha_{m-1},-\alpha_{m})
      \circ\theta,
      \]
      and we can compute $\Delta(\sigma)=\Delta_{\theta}(\sigma)$ by
      Lemma~\ref{elem.lem}(e) below.
\end{exemple}


Note that if $V_{0}\subset V$ is the $F$-subspace spanned by $e$,
then $A=(\End_{F}V_{0})\otimes Q$, and we obtain a decomposition
$\sigma=\sigma_{0}\otimes\gamma$ where $\sigma_{0}$ is the
involution adjoint to the bilinear form on $V_{0}$ with
diagonalization $\langle\alpha_{1},\ldots,\alpha_{m}\rangle$
relative to $e$, and $\gamma$ is the canonical (conjugation)
involution on $Q$.

This example can be slightly generalized:

\begin{exemple}
      \label{genex.ex}
      Consider the algebra $A=A_{0}\otimes_{F}Q$
where $Q$ is a quaternion $F$-algebra and $A_{0}$ is a central
simple $F$-algebra. Let $\sigma_{0}$ be an orthogonal involution
on $A_{0}$, $\gamma$ the canonical involution on $Q$, and
      \[
      \sigma=\sigma_{0}\otimes\gamma.
      \]
Suppose that $\ind A_{0}$ divides $\frac{1}{2}\deg A_{0}$. Then
      \begin{equation}
         \label{Discomp.eq}
      \Delta(\sigma)=(\disc\sigma_{0})_{2}\cup[Q],
      \end{equation}
      where $\disc\sigma_{0}\in F^\times/F^{\times2}$ is the
      discriminant of the orthogonal involution $\sigma_{0}$ (see
      \cite[\S7]{BoI}).
Indeed, let $\theta_{0}$ be a hyperbolic orthogonal
involution on $A_{0}$ and let $x_{0}\in\Sym(A_{0}, \theta_{0})^\times$
be such that $\sigma_{0}=\Int(x_{0})\circ\theta_{0}$.
The involution
      $\theta=\theta_{0}\otimes\gamma$ is hyperbolic, and we have
      $\sigma=\Int(x_{0}\otimes1)\circ\theta$, so that
      \[
      \Delta(\sigma)=\bigl(\Nrp_{\theta}(x_{0}\otimes1)\bigr)_{2}\cup
      [A].
      \]
Thus, by Lemma~\ref{elem.lem}(d),
$\Nrp_{\theta}(x_{0}\otimes1)=\Nrd_{A_{0}}(x_{0})$.
Equation~\eqref{Discomp.eq} follows,
since $\disc\sigma_{0}$ is represented by $\Nrd_{A_{0}}(x_{0})$,
and $\bigl(\Nrd_{A_{0}}(x_{0})\bigr)_{2}\cup[A_{0}]=0$.
\end{exemple}

\subsection{Trace forms}
\label{trace.subsec}

Let $A$ be an arbitrary central simple $F$-algebra.
For every involution $\sigma:A\rightarrow A$, the
associated trace form  $T_{\sigma}\colon\,A\to F$ is defined as follows:
\[
T_{\sigma}(x)=\Trd_{A}\bigl(\sigma(x)x\bigr)
\]
where $\Trd_{A}$ denotes the reduced trace. Denote by
$T_{\sigma}^{+}$ the restriction of $T_{\sigma}$ to
$\Sym(A,\sigma)$; this form can also be seen as the
restriction to  $\Sym(A,\sigma)$ of the form
$T_{A}\colon\,A\to F$ defined by
\[
T_{A}(x)=\Trd_{A}(x^{2}).
\]

As is the case with involutions of the other types (see \cite[\S11]{BoI}),
the discriminant of symplectic involutions
can be expressed in terms of trace forms;
indeed we have the following result:

\begin{thm}
      \label{trace.thm}
Let $A$ be a central simple algebra over $F$ and let $\theta$
and $\sigma$ be symplectic involutions on $A$. The class
in the Witt ring  $WF$ of the difference
$T_{\sigma}^{+}-T_{\theta}^{+}$ lies in the third power of the
fundamental ideal, namely
\[
T_{\sigma}^{+}-T_{\theta}^{+}\in I^{3}F.
\]
Moreover, if $e_{3}\colon\, I^{3}F\to H^{3}(F,\mu_{2})$
denotes the Arason invariant, we obtain
\[
e_{3}(T_{\sigma}^{+}-T_{\theta}^{+})=
\begin{cases}
\Delta_{\theta}(\sigma)&\text{if $\deg A\equiv 0\mod4$},\\
0&\text{if $\deg A\equiv 2\mod 4$}.
\end{cases}
\]
\end{thm}

A proof of this result is given in section~\ref{trace.sec}
below. For the trace forms $T_{\sigma}$, we have the
following result:

\begin{corol}
      \label{trace.cor}
Keeping the notation of the previous theorem, we have
$T_{\sigma}-T_{\theta}\in I^{4}F$ and
\[
e_{4}(T_{\sigma}-T_{\theta})=
\begin{cases}
(-1)_{2}\cup \Delta_{\theta}(\sigma)&\text{if $\deg A\equiv 0\mod4$},
\\
0&\text{if $\deg A\equiv2\mod4$},
\end{cases}
\]
where $e_{4}\colon\, I^{4}F\to H^{4}(F,\mu_{2})$ denotes the
degree $4$ invariant.
\end{corol}

\begin{proof}
Let $T_{\sigma}^-$ be the restriction of $T_{\sigma}$
(or of $-T_{A}$) to the space of skew-symmetric elements
in $A$. We have
\[
T_{\sigma}=T_{\sigma}^{+}+ T_{\sigma}^-
\qquad\text{and}\qquad
T_{A}=T_{\sigma}^{+}-T_{\sigma}^-,
\]
so that $T_{\sigma}=2T_{\sigma}^{+}-T_{A}$. Similarly,
$T_{\theta}=2T_{\theta}^{+}-T_{A}$, so that
\[
T_{\sigma}-T_{\theta}=2(T_{\sigma}^{+}-T_{\theta}^+),
\]
hence the corollary is a direct consequence of the previous
theorem.
\end{proof}

In the special case where $\theta$ is hyperbolic we get:

\begin{prop}
      \label{hyptrace.prop}
      Suppose $A=M_{2}(A_{0})$ for some central simple $F$-algebra
      $A_{0}$, and let $\theta$ be a hyperbolic symplectic involution on
      $A$. Then $T_{\theta}^{+}$ is Witt-equivalent to
      $\langle2\rangle\cdot T_{A_{0}}$, and $T_{\theta}$
      is hyperbolic. If
      $\deg A\equiv 2\bmod4$, then $A$ is split, hence
      every symplectic
      involution on $A$ is hyperbolic. If $\deg A\equiv0\bmod4$, then,
      for any symplectic involution $\sigma$ on $A$, we have
      $T_{\sigma}\in I^{4}F$ and
      \[
      e_{4}(T_{\sigma})=(-1)_{2}\cup\Delta(\sigma).
      \]
\end{prop}

\subsection{Decomposability of symplectic involutions}
\label{decomp.subsec}

Section~\ref{decomp.sec} below will be devoted to the
relations between the discriminant and the
decomposability of symplectic involutions as tensor
products of involutions defined on subalgebras. Our
main result is concerned with degree $8$ algebras with index
dividing $4$. Such algebras can be written in the form
$A=M_{2}(A_{0})$, where $A_{0}$ is a central simple
algebra of degree $4$, hence they carry hyperbolic
symplectic involutions. The case $\ind A=1$ is
trivial, since every
symplectic involution on a split algebra is hyperbolic,
and is omitted in the following theorem.

\begin{thm}
      \label{decomp0.thm}
      Let $A$ be a central simple $F$-algebra of degree~$8$ having index
      $2$ or $4$. For any symplectic involution $\sigma$ on $A$, there
      is a quaternion subalgebra $Q\subset A$ such that
      \begin{enumerate}
         \item[(1)]
         $Q$ is stable under $\sigma$, and $\sigma$ restricts to the
         conjugation involution $\gamma$ on $Q$,
         \item[(2)]
         the index of the centralizer $A_{0}$ of $Q$ in $A$ is
         $\frac{1}{2}\ind A$.
      \end{enumerate}
      We thus have
      \[
      A=A_{0}\otimes_{F}Q
      \qquad\text{and}\qquad
      \sigma=\sigma_{0}\otimes\gamma ,
      \]
      where $\sigma_{0}$ is the restriction of $\sigma$ to $A_{0}$
      (which is an orthogonal involution on $A_{0}$).
\end{thm}

When $\ind A=2$, this theorem is easily proved and can be readily
generalized to any degree, see Example~\ref{ind2.ex} or
\cite[Proposition~3.4]{BST}.
The case $\ind A=4$ is treated in Section~\ref{decomp.sec}.

Theorem~\ref{decomp0.thm} shows that the discriminant of a
symplectic involution
$\sigma$ on a central simple $F$-algebra of degree~$8$ and index $2$
or $4$ can be computed as in Example~\ref{genex.ex} above. The following
theorem gives a necessary and sufficient condition for the
discriminant to be trivial.

\begin{thm}
      \label{decomp.thm}
Let $A$ be a central simple $F$-algebra of degree $8$ with
index dividing $4$. For any symplectic involution $\sigma$ on $A$,
$\Delta(\sigma)=0$ if and only
if $\sigma$ leaves invariant a subalgebra $A_{1}\subset A$ isomorphic
to $M_{2}(F)$ and restricts to an orthogonal involution
on  $A_{1}$.
\end{thm}

A proof is given in Section~\ref{decomp.sec}.

When the condition of Theorem~\ref{decomp.thm} holds, the
centralizer $A_{1}'$ of $A_{1}$ in $A$ is also invariant
under $\sigma$, and the restriction of $\sigma$ to $A_{1}'$
must be symplectic. As any symplectic involution on a degree
$4$ algebra decomposes as a tensor product of involutions on
quaternion subalgebras (see~\cite[(16.16)]{BoI}), it follows
that $A_{1}'=A_{2}\otimes A_{3}$ for some quaternion subalgebras
$A_{2}$, $A_{3}\subset A_{1}'$ invariant under $\sigma$.
Therefore, Theorem \ref{decomp.thm} can be reformulated as follows:

\begin{corol}
      \label{decomp.cor}
Let $A$ be a central simple $F$-algebra of degree $8$ with
index dividing $4$. Then, for every symplectic involution
$\sigma$ on $A$, we have $\Delta(\sigma)=0$
if and only if there exist quaternion subalgebras $A_{1}$,
$A_{2}$, $A_{3}\subset A$, with $A_1\simeq M_{2}(F)$,
and orthogonal involutions
$\sigma_{1}$ and $\sigma_{2}$ on
$A_{1}$ and $A_{2}$ respectively, such that
\[
A=A_{1}\otimes A_{2}\otimes A_{3}
\qquad\text{and}\qquad
\sigma=\sigma_{1}\otimes\sigma_{2}\otimes\gamma_{3}
\]
where $\gamma_{3}$ denotes the canonical involution on $A_{3}$.
\end{corol}

\section{Discriminants and pfaffian Norms}
\label{elem.sec}

The goal of this section is to prove
Proposition~\ref{elem.prop}. Throughout the section, $A$
denotes a central simple $F$-algebra of degree $n=2m$.

\begin{lem}
      \label{elem.lem}
Let $\sigma$ and $\theta$ be symplectic involutions on $A$ and
let $s\in\Sym(A,\theta)^\times$ be an element such that
$\sigma=\Int(s)\circ\theta$. Then,
\begin{enumerate}
\item[(a)]
for every $x\in\Sym(A,\sigma)\cap\Sym(A,\theta)$,
\[
\Nrp_{\sigma}(x)=\Nrp_{\theta}(x);
\]
\item[(b)]
for every $x\in \Sym(A,\theta)$, the product $sx$ lies in
$\Sym(A,\sigma)$ and
\[
\Nrp_{\sigma}(sx)=\Nrp_{\theta}(s)\Nrp_{\theta}(x);
\]
\item[(c)]
for every $x\in\Sym(A,\theta)^\times$,
\[
\Nrp_{\theta}(x^{-1})=\Nrp_{\theta}(x)^{-1}.
\]
\item[(d)]
Suppose $A=A_{1}\otimes_{F}A_{2}$ for some central simple
$F$-algebras $A_{1}$, $A_{2}\subset A$ of degree $n_{1}=2m_{1}$
and $n_{2}=2m_{2}$ respectively; if $x_{1}\in A_{1}$ and
$x_{2}\in A_{2}$ are such that
$x_{1}\otimes x_{2}\in\Sym(A,\theta)$, then
\[
\Nrp_{\theta}(x_{1}\otimes x_{2})=\Nrd_{A_{1}}(x_{1})^{m_{2}}
\Nrd_{A_{2}}(x_{2})^{m_{1}}.
\]
\item[(e)]
Suppose $A=M_{r}(A_{0})$ and
$\theta\bigl((a_{ij})_{1\leq i,j\leq r}\bigr)=
\bigl(\theta_{0}(a_{ij})\bigr)_{1\leq i,j\leq r}^t$ for some
symplectic involution $\theta_{0}$ on the central simple
$F$-algebra $A_{0}$. For the diagonal matrix
$x=\diag(x_{1},\ldots,x_{r})$ with
$x_{i}\in\Sym(A_{0},\theta_{0})$ for $i=1$, \ldots, $r$, we have
\[
\Nrp_{\theta}(x)=\Nrp_{\theta_{0}}(x_{1})\cdots\Nrp_{\theta_{0}}(x_{r}).
\]
\end{enumerate}
\end{lem}

\begin{proof}
(a)
Let $t$ be an indeterminate over $F$. Define
\[
\Prp_{\sigma,x}(t)=\Nrp_{\sigma}(t-x)\in F[t],\qquad
\Prp_{\theta,x}(t)=\Nrp_{\theta}(t-x)\in F[t].
\]
Those polynomials, called \textit{pfaffian characteristic
polynomials} in \cite[p.~19]{BoI}, are monic and satisfy
\[
\Prp_{\sigma,x}^{2}=\Pcrd_{A,x}=\Prp_{\theta,x}^{2},
\]
where $\Pcrd_{A,x}(t)=\Nrd_{A(t)}(t-x)$ is the reduced
characteristic polynomial of $x$. Therefore,
$\Prp_{\sigma,x}(t)=\Prp_{\theta,x}(t)$, and evaluation
at $t=0$ yields $\Nrp_{\sigma}(x)=\Nrp_{\theta}(x)$.

(b)
Straightforward calculations show that $\sigma(sx)=sx$
if $\theta(x)=x$. Let us consider the two sides of the
equality we aim to prove as polynomial functions of
$x$. The squares of the two sides are equal since the
reduced norm is multiplicative, hence they are equal up
to sign. Moreover, they are equal and nonzero for
$x=1$ in view of part~(a). Hence, they are equal for all $x$.

(c)
We apply (b) with $x=s^{-1}$ and use the relation
$\Nrp_{\sigma}(1)=1$.

(d)
By taking the square root on both sides of the equation
\[
\Pcrd_{A,x_{1}\otimes x_{2}}=(\Pcrd_{A_{1},x_{1}})^{n_{2}}
(\Pcrd_{A_{2},x_{2}})^{n_{1}},
\]
we obtain
\[
\Prp_{\theta,x_{1}\otimes x_{2}}=(\Pcrd_{A_{1},x_{1}})^{m_{2}}
(\Pcrd_{A_{2},x_{2}})^{m_{1}}.
\]
The property follows by considering the constant
terms.

(e)
As in the preceding case, the property follows by extracting
the monic square root of each side of the equation
\[
\Pcrd_{A,x}=\Pcrd_{A_{0},x_{1}}\cdots\Pcrd_{A_{0},x_{r}}.
\]
\end{proof}

Proposition~\ref{elem.prop} easily follows from the lemma above.
Indeed, if $\sigma=\Int(s)\circ\theta$, so that
$\Delta_{\theta}(\sigma)=\bigl(\Nrp_{\theta}(s)\bigr)_{2}\cup[A]$,
then $\theta=\Int(s^{-1})\circ\sigma$ and hence
$\Delta_{\sigma}(\theta)=\bigl(\Nrp_{\sigma}(s^{-1})\bigr)_{2}\cup[A]$.
Now Lemma~\ref{elem.lem} shows that
\[
\Nrp_{\sigma}(s^{-1})=\Nrp_{\theta}(s)^{-1},
\]
hence
$\bigl(\Nrp_{\sigma}(s^{-1})\bigr)_{2}=
\bigl(\Nrp_{\theta}(s)\bigr)_{2}$, and so
$\Delta_{\sigma}(\theta)=\Delta_{\theta}(\sigma)$.
If $\rho$ is another symplectic involution, and if
$t\in\Sym(A,\rho)$ is such that $\theta=\Int(t)\circ\rho$,
then $\sigma=\Int(st)\circ\rho$. Part~(b) of Lemma~\ref{elem.lem}
yields
\[
\Nrp_{\sigma}(st)=\Nrp_{\theta}(s)\Nrp_{\theta}(t).
\]
Moreover, part~(a) shows that
$\Nrp_{\sigma}(st)=\Nrp_{\rho}(st)$ and
$\Nrp_{\theta}(t)=\Nrp_{\rho}(t)$. Therefore, the preceding
equality can be written as
\[
\Nrp_{\rho}(st)=\Nrp_{\theta}(s)\Nrp_{\rho}(t).
\]
It follows that
\begin{multline*}
\Delta_{\rho}(\sigma)=\bigl(\Nrp_{\rho}(st)\bigr)_{2}\cup[A]=
\\
\bigl(\Nrp_{\theta}(s)\bigr)_{2}\cup[A]
+\bigl(\Nrp_{\rho}(t)\bigr)_{2}\cup[A]=
\Delta_{\theta}(\sigma)+\Delta_{\rho}(\theta),
\end{multline*}
which completes the proof of part~(b) of
Proposition~\ref{elem.prop}.

Let now $v\in A^\times$ and
$\sigma'=\Int(v)\circ\sigma\circ\Int(v)^{-1}$, so
that $\sigma'=\Int\bigl(vs\theta(v)\bigr)\circ\theta$. Then,
\[
\Delta_{\theta}(\sigma')=\bigl(\Nrp_{\theta}(vs\theta(v))\bigr)_{2}\cup[A].
\]
By \cite[(2.13)]{BoI},
$\Nrp_{\theta}\bigl(vs\theta(v)\bigl)=\Nrd_{A}(v)\Nrp_{\theta}(s)$.
Since $\bigl(\Nrd_{A}(v)\bigr)_{2}\cup[A]=0$, it follows that
\[
\Delta_{\theta}(\sigma')=\Delta_{\theta}(\sigma).
\]
Similarly, if $\theta'$ is a symplectic involution
conjugate to $\theta$, then
$\Delta_{\sigma'}(\theta')=\Delta_{\sigma'}(\theta)$.
Now, part (b) of Proposition~\ref{elem.lem} shows that
$\Delta_{\theta'}(\sigma')=\Delta_{\sigma'}(\theta')$ and
$\Delta_{\theta}(\sigma')=\Delta_{\sigma'}(\theta)$.
Therefore,
\[
\Delta_{\theta'}(\sigma')=\Delta_{\theta}(\sigma').
\]
We already observed that
$\Delta_{\theta}(\sigma')=\Delta_{\theta}(\sigma)$, hence
\[
\Delta_{\theta'}(\sigma')=\Delta_{\theta}(\sigma)
\]
and the proof of Proposition~\ref{elem.lem} is complete.

\section{Discriminant and trace form}
\label{trace.sec}

In this section we prove Theorem \ref{trace.thm}
and Proposition~\ref{hyptrace.prop}.

Let $\CorpsF$ be the category of fields containing $F$ and let
$G$ be any algebraic group over $F$. We consider the
functor
\[
H^{1}(G)\colon\,\CorpsF\to\Ensp
\]
where $\Ensp$ denotes the category of pointed sets,
associating to every  $L\in\CorpsF$ the Galois
cohomology set $H^{1}(L,G)$.

Similarly, the ``cohomological cycle
module''
$M=\bigoplus_{d\geq0}H^d\bigl(\bullet,\QZ(d-1)\bigr)$,
introduced by Rost in \cite{R}, yields a sequence of functors
\[
M_{d}=H^d\bigl(\bullet,\QZ(d-1)\bigr)\colon\,\CorpsF\to\Ensp.
\]
The natural transformations $H^{1}(G)\to M_{d}$ are called
\emph{cohomological invariants of degree $d$} in
Merkurjev's paper \cite{IAG}. These invariants form a group
denoted $\Inv^d(H^{1}(G),M)$. (The group $\Inv^d(H^{1}(G),M)$
is denoted by  $\Inv^d\bigl(G,\QZ(d-1)\bigr)$ in
\cite[\S31.B]{BoI}.)

Let now $A$ be a central simple $F$-algebra of
degree~$n=2m$ and let $\theta$ be a symplectic involution on $A$.
We take for $G$ the group $\GSp(A,\theta)$ of symplectic
similitudes; this is the algebraic group scheme defined by
\[
\GSp(A,\theta)(E)=\{g\in A\otimes_{F}E\mid\theta(g)g\in E^\times\}
\]
for any commutative $F$-algebra $E$.
The set
$H^{1}\bigl(L,\GSp(A,\theta)\bigr)$ is in one to one
correspondence  with the set of conjugacy classes of
symplectic involutions defined on $A_{L}=A\otimes_{F}L$,
the class of $\theta$ being the distinguished one (see
\cite[(29.23)]{BoI}). The following proposition shows that
for symplectic involutions there is no (nontrivial)
cohomological invariant of degree $1$ or $2$.

\begin{prop}
      \label{invcoh.prop}
Suppose $M=\bigoplus_{d\geq0}H^d\bigl(\bullet,\QZ(d-1)\bigr)$.
If the algebra $A$ is split, we get
$H^{1}\bigl(L,\GSp(A,\theta)\bigr)=1$, for every
$L\in\CorpsF$, so that
\[
\Inv^d\bigl(H^{1}\bigl(\GSp(A,\theta)\bigr),M\bigr)=0
\qquad\text{for all $d$}.
\]
If $A$ is not split, we have
\[
\Inv^d(H^{1}\bigl(\GSp(A,\theta)\bigr),M\bigr)=
  0\qquad\text{for $d=1$, $2$}
\]
and
\[
\Inv^{3}(H^{1}\bigl(\GSp(A,\theta)\bigr),M\bigr)=
\begin{cases}
0&\text{if $\deg A\equiv2\mod4$},\\
\mathbb{Z}/2\mathbb{Z}&\text{if $\deg A\equiv 0\mod 4$}.
\end{cases}
\]
\end{prop}

\begin{proof}
Every symplectic involution on a split algebra
is hyperbolic. Therefore, when $A$ is split,
$H^{1}\bigl(L,\GSp(A,\theta)\bigr)=1$ for all $L\in\CorpsF$.

For the rest of the proof, we may thus assume that $A$ is not split.
Let
\[
\mu\colon\,\GSp(A,\theta)\to\Gm
\]
be the homomorphism which associates to each similitude $g$
its multiplier $\mu(g)=\theta(g)g$.
The cohomology sequence induced by the exact sequence
\[
1\to\Sp(A,\theta)\to\GSp(A,\theta)\xrightarrow{\mu}\Gm\to1
\]
yields for every $L\in\CorpsF$ the exact sequence
\[
L^\times\to H^{1}\bigl(L,\Sp(A,\theta)\bigr)\to
H^{1}\bigl(L,\GSp(A,\theta)\bigr)\to1
\]
since $H^{1}(L,\Gm)=1$ by Hilbert's Theorem~90. Therefore,
for every $d$, we have an exact sequence
\[
0\to
\Inv^d\bigl(H^{1}\bigl(\GSp(A,\theta)\bigr),M\bigr)\to
\Inv^d\bigl(H^{1}\bigl(\Sp(A,\theta)\bigr),M\bigr)\to
\Inv^d(\Gm,M).
\]
For $d=1$ or $2$, we obtain, by \cite[(31.15)]{BoI},
$\Inv^d\bigl(H^{1}\bigl(\Sp(A,\theta)\bigr),M\bigr)=0$ and
hence
\[
\Inv^d\bigl(H^{1}\bigl(\GSp(A,\theta)\bigr),M\bigr)=0.
\]
The group
$\Inv^{3}\bigl(H^{1}\bigl(\Sp(A,\theta)\bigr),M\bigr)$
is of order $2$, the nontrivial element being the Rost
invariant $\Delta_{\theta}$ defined in the introduction.
Equation~\eqref{mult.eq} shows that this invariant
is zero in $\Inv^{3}(\Gm,M)$ if and only if
$\deg A\equiv2\bmod4$.
\end{proof}

When $\deg A\equiv0\bmod4$, the unique nontrivial invariant
of degree $3$ is the discriminant. Our next goal is to give an
explicit description of this invariant in terms of trace forms.

Let $T_{\theta}^{+}\colon\,\Sym(A,\theta)\to F$ be the
quadratic form
\[
T_{\theta}^{+}(x)=\Trd_{A}(\theta(x)x)=\Trd_{A}(x^{2}).
\]
This forms only depends, up to isometry, on the conjugacy
class of $\theta$ since, if
$\theta'=\Int(v)\circ\theta\circ\Int(v)^{-1}$ for some
$v\in A^\times$, then $\Int(v)$ defines an isometry
between $T_{\theta}^{+}$ and $T_{\theta'}^{+}$.
Consider $L\in\CorpsF$. The map sending every symplectic
involution $\sigma \colon\, A_{L}\to A_{L}$ to the discriminant
\[
\disc(T_{\sigma}^{+}-T_{\theta}^{+})\in
L^\times/L^{\times2}=H^{1}(L,\mu_{2})
\]
defines a cohomological invariant
$H^{1}\bigl(\GSp(A,\theta)\bigr)\to H^{1}(\mu_{2})$.
By Proposition~\ref{invcoh.prop}, this invariant is trivial,
hence  $T_{\sigma}^{+}-T_{\theta}^{+}\in I^{2}L$. Similarly,
the map sending every symplectic involution
$\sigma$ to the Witt (--Clifford) invariant
\[
e_{2}(T_{\sigma}^{+}-T_{\theta}^{+})\in H^{2}(L,\mu_{2})
\]
defines a cohomological invariant of degree $2$. Again, by
Proposition~\ref{invcoh.prop}, we get
$e_{2}(T_{\sigma}^{+}-T_{\theta}^{+})=0$, and hence
$T_{\sigma}^{+}-T_{\theta}^{+}\in I^{3}L$ using
Merkurjev's theorem. This proves the first part of Theorem
\ref{trace.thm}. Note that the equality
$e_{2}(T_{\sigma}^{+}-T_{\theta}^{+})=0$ can also be derived
from Qu\'eguiner's explicit calculation of the Hasse invariant of
trace forms in~\cite[p.~307]{Q}.

Consider the map associating to every
symplectic involution $\sigma\colon\, A_{L}\to A_{L}$ the Arason
invariant
\[
e_{3}(T_{\sigma}^{+}-T_{\theta}^{+})\in H^{3}(L,\mu_{2}).
\]
Using Proposition~\ref{invcoh.prop}, we see that this
invariant is trivial either if $\deg A\equiv2\bmod4$ or if
$A$ is split. We claim that it coincides with the
discriminant $\Delta_{\theta}(\sigma)$ if $A$ is nonsplit
and $\deg A\equiv0\bmod4$. To prove this, it suffices to
show that it is nontrivial because, by
Proposition~\ref{invcoh.prop}, there is an
unique nontrivial invariant in
$\Inv^{3}\bigl(H^{1}\bigl(\GSp(A,\theta)\bigr),M\bigr)$.
Therefore, our goal is to find a field $L\in\CorpsF$ and two
symplectic involutions $\sigma_{1}$,
$\sigma_{2}$ on $A_{L}$ such that
$e_{3}(T_{\sigma_{1}}^{+}-T_{\sigma_{2}}^{+})\neq0$. If
$\theta$ is any involution on $A$, the equality
\[
e_{3}(T_{\sigma_{1}}^{+}-T_{\theta}^{+})=
e_{3}(T_{\sigma_{1}}^{+}-T_{\sigma_{2}}^{+})+
e_{3}(T_{\sigma_{2}}^{+}-T_{\theta}^{+})
\]
shows that at least one of the terms
$e_{3}(T_{\sigma_{1}}^{+}-T_{\theta}^{+})$ and
$e_{3}(T_{\sigma_{2}}^{+}-T_{\theta}^{+})$ is nonzero, hence
the invariant
$\sigma\mapsto e_{3}(T_{\sigma}^{+}-T_{\theta}^{+})$
is nontrivial.

\begin{lem}
Let $A$ be a central simple $F$-algebra of exponent $2$. There
exists a field $L\in\CorpsF$ such that $\ind A_{L}=2$.
\end{lem}

\begin{proof}
We may take for $L$ the fonction
field of a suitable generalized Severi--Brauer variety (cf
\cite{Bl}). Alternatively, we may consider
a multiquadratic extension $M/F$ of minimal degree splitting
$A$ (such extensions exist, because $A$ is Brauer--equivalent
to a tensor product of quaternion algebras by a theorem of
Merkurjev) and take for $L$ a proper maximal subfield of
$M$. The algebra $A_{L}$ is not split, because $M$ is of
minimal degree, and it is of index $2$ because it is split
by the quadratic extension $M/L$.
\end{proof}

Thus we may suppose, after possibly replacing $F$ by a
suitable extension, that $\ind A=2$ i.e.\
that $A$ is Brauer--equivalent to a quaternion division
algebra $Q$ over $F$. Then, denoting by $V$ an $F$-vector
space of dimension $m$, we obtain
\[
A\simeq Q\otimes_{F}\End_{F}V.
\]
For the rest of this section, we fix an isomorphism
identifying $A$ with $Q\otimes \End_{F}V$.
Let $b$ be a symmetric non-degenerate bilinear form on $V$.
The symmetric square $S^{2}V$ and the exterior square
$\bigwedge^{2}V$ are endowed with symmetric bilinear forms
$b^{S^2}$ and $b^{\wedge2}$ respectively, defined by
\[
b^{S^2}(x_{1}\cdot x_{2},y_{1}\cdot y_{2})=
b(x_{1},y_{1})b(x_{2},y_{2})+b(x_{1},y_{2})b(x_{2},y_{1})
\]
and
\[
b^{\wedge2}(x_{1}\wedge x_{2},y_{1}\wedge y_{2})=
b(x_{1},y_{1})b(x_{2},y_{2})-b(x_{1},y_{2})b(x_{2},y_{1}).
\]

\begin{lem}
      \label{trace.lem}
Let $Q=(\alpha,\beta)_{F}$. On
$A=Q\otimes_{F}\End_{F}V$, consider the symplectic
involution $\sigma=\gamma\otimes\ad_{b}$, where $\gamma$
is the quaternion conjugation on $Q$ and $\ad_{b}$ is the (orthogonal)
involution adjoint to $b$. Then, the bilinear form
$B_{\sigma}^{+}(x,y)=\Trd_{A}(xy)$ on $\Sym(A,\sigma)$
(which is the polar form of the quadratic form $T_{\sigma}^{+}$)
decomposes as an orthogonal sum
\[
B_{\sigma}^{+}=b^{S^2}\perp\langle-\alpha,-\beta,\alpha\beta\rangle
\cdot b^{\wedge2}.
\]
\end{lem}


\begin{proof}
Let $\Skew(\End_{F}V,\ad_{b})$ be the $F$-vector space of
endomorphisms $f$ of $V$ such that $\ad_{b}(f)=-f$, and let
$Q^{0}$ be the $F$--vector space of pure quaternions in $Q$.
Straightforward calculation shows that the decomposition
\[
\Sym(A,\sigma)=\bigl(
F\otimes\Sym(\End_{F}V,\ad_{b})\bigr)\oplus
\bigl(Q^{0}\otimes\Skew(\End_{F}V,\ad_{b})\bigr)
\]
is orthogonal with respect to the form $B_{\sigma}^{+}$.
Let $B_{b}^{+}$ and $B_{b}^-$ be the restrictions
of the bilinear trace form $B(f,g)=\tr(fg)$ to
$\Sym(\End_{F}V,\ad_{b})$ and $\Skew(\End_{F}V,\ad_{b})$
respectively. The decomposition above yields
\[
B_{\sigma}^{+}=\langle2\rangle\cdot B_{b}^{+}\perp
\langle2\alpha,2\beta,-2\alpha\beta\rangle\cdot B_{b}^-.
\]
The lemma follows, since, by \cite[(11.4)]{BoI}
$B_{b}^{+}\simeq\frac{1}{2}b^{S^2}$ and
$B_{b}^-\simeq-\frac{1}{2}b^{\wedge2}$.
\end{proof}

If $b=\langle a_{1},\ldots,a_{m}\rangle$ is a diagonalization
of $b$, it is easily verified that
\[
b^{S^2}\simeq m\langle2\rangle\perp\bigl(\perp_{i<j}\langle a_{i}a_{j}
\rangle\bigr)
\qquad\text{and}\qquad
b^{\wedge2}\simeq{}\perp_{i<j}\langle a_{i}a_{j}\rangle
\]
(cf \cite[p.~135]{BoI}). The formula of the preceding lemma
can then be written as
\[
B_{\sigma}^{+}\simeq m\langle 2\rangle\perp\langle1,-\alpha,-\beta,
\alpha\beta\rangle\cdot b^{\wedge2},
\]
hence, in terms of quadratic forms,
\[
T_{\sigma}^{+}=m\langle 2\rangle + n_{Q}\cdot q^{\wedge2}
\]
where $n_{Q}$ denotes the norm form of $Q$ and $q^{\wedge2}$
is the quadratic form defined by $q^{\wedge2}(x):=b^{\wedge2}(x,x)$.


Let $b_{1}$ and $b_{2}$ be two  nonsingular symmetric bilinear forms
on $V$, and let
\[
\sigma_{1}=\gamma\otimes\ad_{b_{1}},
\qquad
\sigma_{2}=\gamma\otimes\ad_{b_{2}}
\]
be the symplectic involutions on $A=Q\otimes\End_{F}V$
constructed as in Lemma~\ref{trace.lem}. Observe that
$T_{\sigma_{1}}^{+}-T_{\sigma_{2}}^{+}=
n_{Q}\cdot(q_{1}^{\wedge2}-q_{2}^{\wedge2})$, hence
\begin{equation}
      \label{etrois.eq}
      e_{3}(T_{\sigma_{1}}^{+}-T_{\sigma_{2}}^{+})=
      [Q]\cup\disc(q_{1}^{\wedge2}-q_{2}^{\wedge2}).
\end{equation}
Explicit calculation shows that
\begin{equation}
      \label{disc.eq}
      \disc(q_{1}^{\wedge2}-q_{2}^{\wedge2})=
      \det b_{1}^{\wedge2}\cdot\det b_{2}^{\wedge 2}=
      (\det b_{1}\cdot\det b_{2})^{m-1}.
\end{equation}
Adjoining an indeterminate to $F$ if necessary, we may assume that
there exists an element $t\in F^\times$
not belonging to $\Nrd(Q)$. By a theorem of Merkurjev, this element
satisfies $[Q]\cup(t)_{2}\neq0$. It is easy to
find two bilinear forms $b_{1}$ and $b_{2}$ on $V$
such that $\det b_{1}\cdot \det b_{2}=tF^{\times2}$.
Since $m$ is even, it follows from \eqref{etrois.eq} and \eqref{disc.eq}
that the corresponding involutions $\sigma_{1}$ and
$\sigma_{2}$ satisfy
\[
e_{3}(T_{\sigma_{1}}^{+}-T_{\sigma_{2}}^{+})\neq0.
\]
This completes the proof of Theorem \ref{trace.thm}.

We now turn to Proposition~\ref{hyptrace.prop} and assume
$A=M_{2}(F)\otimes_{F}A_{0}$. Since all hyperbolic involutions
are conjugate, we may assume moreover
$\theta=\gamma\otimes\theta_{0}$
for some orthogonal involution $\theta_{0}$ on $A_{0}$, where
$\gamma$ is the unique symplectic involution on $M_{2}(F)$
(which is hyperbolic).  As in
Lemma~\ref{trace.lem}, we have an orthogonal decomposition
\[
\Sym(A,\theta)=
\bigl(F\otimes\Sym(A_{0},\theta_{0})\bigr)\oplus
\bigl(\Skew(M_{2}(F),\gamma)\otimes\Skew(A_{0},\theta_{0})\bigr)
\]
which yields
\[
T_{\sigma}^{+}=\langle2\rangle\cdot T_{\theta_{0}}^{+}\perp
\langle2,-2,-2\rangle\cdot T_{\theta_{0}}^-.
\]
Therefore, $T_{\sigma}^{+}$ is Witt-equivalent to
$\langle2\rangle\cdot(T_{\theta_{0}}^{+}-T_{\theta_{0}}^-)
=\langle2\rangle\cdot T_{A_{0}}$. Since the adjoint involution to
$T_{\theta}$ is $\theta\otimes\theta$, by \cite[(11.1)]{BoI},
it is clear that $T_{\theta}$ is hyperbolic when $\theta$ is
hyperbolic. If $\deg A\equiv2\bmod4$, then $\deg A_{0}$ is
odd, hence $A_{0}$ is split.
Therefore, $A$ is also split. The other statements in
Proposition~\ref{hyptrace.prop} follow from
Corollary~\ref{trace.cor}.

\section{Discriminant and decomposability of involutions}
\label{decomp.sec}

Our first goal in this section is to give a proof of
Theorem~\ref{decomp0.thm}. As observed in
Section~\ref{decomp.subsec},
the theorem is easy if $\ind A=2$. Therefore, we assume
$\ind A=4$. We may then represent $A$ as
\[
A=\End_{D}V
\]
where $D$ is a division algebra of degree~$4$ and $V$ is a
$2$-dimensional $D$-vector space. Let $\theta_{0}$ be an arbitrary
symplectic involution on $D$. The involution $\sigma$ is adjoint to a
hermitian form $h$ on $V$ (with respect to $\theta_{0}$). Using an
orthogonal basis of $V$ relative to $h$, we may identify
\[
A=M_{2}(D)
\qquad\text{and}\qquad
\sigma=\Int\diag(u_{1},u_{2})\circ\hat\theta_{0}
\]
for some $u_{1}$, $u_{2}\in\Sym(D,\theta_{0})^\times$, where
\[
\hat\theta_{0}\bigl((a_{ij})_{1\leq i,j\leq 2}\bigr)=
\bigl(\theta_{0}(a_{ij})\bigr)_{1\leq i,j\leq2}^t,
\]
i.e., $\hat\theta_{0}=t\otimes\theta$ on $A=M_{2}(F)\otimes_{F}D$.
Substituting $\Int(u_{1})\circ\theta_{0}$ for $\theta_{0}$, we may
assume $u_{1}=1$. By \cite[(16.16)]{BoI}, we may find a decomposition
of $D$ into a tensor product of quaternion subalgebras stable under
$\theta_{0}$,
\[
D=Q_{1}\otimes_{F}Q,\qquad\theta_{0}=\theta_{1}\otimes\gamma
\]
where $\theta_{1}$ is an orthogonal involution on $Q_{1}$ and
$\gamma$ is the canonical involution on $Q$. Moreover, we may
assume $u_{2}\in Q_{1}$. Then
\[
\sigma=\Int\diag(1,u_{2})\circ (t\otimes\theta_{1}\otimes\gamma)=
\sigma_{0}\otimes\gamma
\]
with $\sigma_{0}=\Int\diag(1,u_{2})\circ t\otimes\theta_{1}$
on $M_{2}(F)\otimes Q_{1}$. Theorem~\ref{decomp0.thm} is thus
proved.
Note that the quaternion algebra $Q$ is not uniquely
determined by \cite[(16.16)]{BoI}.

Let us now prove Theorem \ref{decomp.thm}, starting with
the following general remark:

\begin{lem}
      \label{decomp.lem}
Let $\sigma$ and $\theta$ be two symplectic involutions on a
central simple algebra $A$ having degree a multiple of $4$.
Assume that the algebra $A$ decomposes as
$A=A_{1}\otimes A_{2}$ for some subalgebras $A_{1}$,
$A_{2}\subset A$ and, furthermore, that there exist
involutions $\sigma_{1}$, $\theta_{1}$ on $A_1$ and
$\sigma_{2}$, $\theta_{2}$ on $A_2$ such that
$\sigma=\sigma_{1}\otimes\sigma_{2}$ and
$\theta=\theta_{1}\otimes\theta_{2}$. Moreover assume
$\deg A_{1}\equiv 2\bmod4$ and $\deg A_{2}\equiv0\bmod4$.
Then, if $\sigma_{1}$ and $\theta_{1}$ are of orthogonal type,
$\Delta_{\theta}(\sigma)=0$.
\end{lem}

\begin{proof}
Consider $u_{1}\in\Sym(A_{1},\theta_{1})^\times$ and
$u_{2}\in\Sym(A_{2},\theta_{2})^\times$ such that
\[
\sigma_{1}=\Int(u_{1})\circ\theta_{1}
\qquad\text{and}\qquad
\sigma_{2}=\Int(u_{2})\circ\theta_{2},
\]
hence
\[
\sigma=\Int(u_{1}\otimes u_{2})\circ\theta.
\]
Then $\Delta_{\theta}(\sigma)=
\bigl(\Nrp_{\theta}(u_{1}\otimes u_{2})\bigr)_{2}\cup [A]$,
and, by Lemma~\ref{elem.lem}(d),
\begin{equation}
\label{Nrpu1u2.eq}
\Nrp_{\theta}(u_{1}\otimes u_{2})=
\Nrd_{A_{1}}(u_{1})^{\frac{1}{2}\deg A_{2}}
\Nrd_{A_{2}}(u_{2})^{\frac{1}{2}\deg A_{1}}.
\end{equation}
Since $\deg A_{2}\equiv0\bmod4$, the first factor is a square.
Moreover, since $\sigma_{1}$ and $\theta_{1}$ are of orthogonal
type, $\sigma_{2}$ and $\theta_{2}$ must be of symplectic type.
Hence,
\[
\Nrd_{A_{2}}(u_{2})=\Nrp_{\theta_{2}}(u_{2})^{2}.
\]
Therefore, equation \eqref{Nrpu1u2.eq} shows that
$\Nrp_{\theta}(u_{1}\otimes u_{2})\in F^{\times2}$, so that
$\Delta_{\theta}(\sigma)=0$.
\end{proof}

Even in the case $\deg A=8$, the converse of Lemma~\ref{decomp.lem}
does not hold, since there are examples of
algebras with involution which do not contain any invariant quaternion
subalgebra on which the restriction of the involution is of orthogonal
type. Indeed, arguing as in the proof of Corollary
\ref{decomp.cor} (see section~\ref{decomp.subsec}), we see that if
$\theta$ is a symplectic involution on a central
simple algebra $A$ of degree~$8$, and if $A_{1}\subset A$ is a quaternion
subalgebra on which $\theta$ restricts to an orthogonal
involution $\theta_{1}$, then $A$ and $\theta$ decompose in
the form
\[
A=A_{1}\otimes A_{2}\otimes A_{3}
\qquad\text{and}\qquad
\theta=\theta_{1}\otimes \theta_{2}\otimes\gamma_{3},
\]
where $A_{1}$, $A_{2}$ and $A_{3}$ are quaternion algebras,
$\theta_{1}$ and $\theta_{2}$ are orthogonal involutions on
$A_{1}$ and $A_{2}$ respectively, and $\gamma_{3}$ is the
canonical involution on $A_{3}$.
This implies, in particular, that the signature of
$\theta$ with respect to every ordering of the field $F$ is
either $0$ or $8$. For example, if $F=\mathbb{R}$ is the field
of real numbers and $\theta$ is the involution adjoint to the
hermitian form $\langle1,1,1,-1\rangle$ on the usual quaternion
algebra $\mathbb{H}$, then $\sgn\theta=4$, and so
$A=M_{4}(\mathbb{H})$ has no quaternion subalgebras on which
$\theta$ restricts to an orthogonal involution. Therefore, even
though $\Delta_{\theta}(\theta)=0$, there is no decomposition as in
Lemma~\ref{decomp.lem}. (See Example~\ref{nodec.exmple} for a subtler
example where the converse of Lemma~\ref{decomp.lem} fails.)

Returning to the proof of Theorem~\ref{decomp.thm}, we
suppose until the end of this section that $A$ is a central
simple $F$-algebra of degree $8$, with index dividing $4$.
Let $\sigma$ be a symplectic involution on $A$, and suppose
$A_{1}\simeq M_{2}(F)$ is an invariant subalgebra on which the
restriction of $\sigma$ is an orthogonal involution.
In this situation, we have
a decomposition $A=A_{1}\otimes A_{1}'$, where $A_{1}'$
denotes the centralizer of $A_{1}$, and
$\sigma=\sigma_{1}\otimes\sigma_{1}'$, where $\sigma_{1}$ and
$\sigma_{1}'$ are the restrictions of $\sigma$ to $A_{1}$ and
$A_{1}'$ respectively. As $A_{1}\simeq M_{2}(F)$, we can find
a hyperbolic symplectic involution $\theta_{1}$ on $A_{1}$ and
set
\[
\theta=\theta_{1}\otimes\sigma_{1}'.
\]
The involution $\theta$ is hyperbolic of symplectic type and,
by Lemma \ref{decomp.lem}, we have
\[
\Delta(\sigma)=\Delta_{\theta}(\sigma)=0.
\]

Conversely, let $\sigma$ be a symplectic involution on $A$
such that $\Delta(\sigma)=0$. To prove that $\sigma$ leaves invariant a
subalgebra of $A$ isomorphic to $M_{2}(F)$ on which it
restricts to an orthogonal involution, we consider
separately various cases, depending on the index of $A$.
If $A$ is split, every symplectic involution is hyperbolic and
the property is a consequence of \cite[Theorem~2.2]{BST}.
If $\ind A=2$, we can always represent $A$ in the form
\[
A=\End_{Q}V
\]
where $Q$ is a quaternion algebra and $V$ is a $4$-dimensional
vector space over $Q$. The involution $\sigma$ is then
adjoint to a hermitian form $h$ on $V$ (with respect to the canonical
involution $\gamma$ on $Q$). Let $e$ be an orthogonal basis for $h$.
Since $h$ is determined by $\sigma$ up to a factor in $F^\times$, we
may assume that the diagonalization of $h$ with respect to the basis
$e$ is
$\langle1,\alpha_{1},\alpha_{2},\alpha_{3}\rangle$ with
$\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}\in F^\times$.
Let $V_{0}\subset V$ be the $F$-subspace with basis $e$.
We have $V=V_{0}\otimes_{F}Q$ and
\[
A=(\End_{F}V_{0})\otimes_{F}Q,
\qquad
\sigma=\sigma_{0}\otimes\gamma
\]
where $\sigma_{0}$ is the orthogonal involution on
$\End_{F}V_{0}$ adjoint  to the bilinear symmetric form
$\langle1,\alpha_{1},\alpha_{2},\alpha_{3}\rangle$.
As in Example~\ref{ind2.ex},
$\Delta(\sigma)=(\alpha_{1}\alpha_{2}\alpha_{3})_{2}\cup[Q]$.
Therefore, the condition $\Delta(\sigma)=0$ implies, by a theorem
of Merkurjev, that
$\alpha_{1}\alpha_{2}\alpha_{3}\in\Nrd_{Q}(Q^\times)$.
Changing basis if necessary, we may assume that
$\alpha_{3}=\alpha_{1}\alpha_{2}$. Then
\[
\langle1,\alpha_{1},\alpha_{2},\alpha_{3}\rangle=
\langle1,\alpha_{1}\rangle\otimes\langle1,\alpha_{2}\rangle .
\]
This implies $\sigma_{0}=\sigma_{1}\otimes\sigma_{2}$ on
$\End_{F}V_{0}\simeq M_{2}(F)\otimes M_{2}(F)$, where
$\sigma_{1}$ and $\sigma_{2}$ are the involutions adjoint to
the bilinear forms $\langle1,\alpha_{1}\rangle$ and
$\langle1,\alpha_{2}\rangle$, respectively. This proves the
theorem in this case.

Finally, suppose $\ind A=4$. As in the proof of
Theorem~\ref{decomp0.thm}, given at the beginning of this
section, we may then represent $A$ as
$
A=\End_{D}V
$
where $D$ is a division algebra of degree~$4$ and $V$ is a
$2$-dimensional $D$-vector space. For the rest of the proof,
we use the same notation as in the proof of
Theorem~\ref{decomp0.thm}. We may thus assume
$A=M_{2}(D)=M_{2}(F)\otimes_{F}D$ and
$\sigma=\Int\diag(1,u_{2})\circ\hat\theta_{0}$ for some
symplectic involution $\theta_{0}$ on $D$ and
$\hat\theta_{0}=t\otimes\theta_{0}$.
The involution
$\theta=\Int\diag(1,-1)\circ\hat\theta_{0}$ is hyperbolic,
and $\sigma=\Int\diag(1,-u_{2})\circ\theta$. By
Lemma~\ref{elem.lem}(e), we have
\[
\Nrp_{\theta}\bigl(\diag(1,-u_{2})\bigr)=
\Nrp_{\theta_{0}}(-u_{2})=\Nrp_{\theta_{0}}(u_{2}),
\]
hence
\[
\Delta(\sigma)=\Delta_{\theta}(\sigma)=
\bigl(\Nrp_{\theta_{0}}(u_{2})\bigr)_{2}\cup[D].
\]
Therefore, the condition $\Delta(\sigma)=0$ implies by
\cite[(16.19)]{BoI}
that the involution $\Int(u_{2})\circ\theta_{0}$ on $D$ is
conjugate to $\theta_{0}$. We may then find $v\in D^\times$
such that
\[
\Int(u_{2})\circ\theta_{0}=\Int(v)\circ\theta_{0}\circ\Int(v)^{-1}
=\Int\bigl(v\theta_{0}(v)\bigr)\circ\theta_{0},
\]
hence $u_{2}=v\theta_{0}(v)\lambda$ for some
$\lambda\in F^\times$.
The involution $\theta$ is conjugate to
\[
\Int\diag(1,v)^{-1}\circ\theta\circ\Int\diag(1,v)
=\Int\diag(1,\lambda)\circ\hat\theta_{0},
\]
which restricts to $\Int\diag(1,\lambda)\circ t$ on
$M_{2}(F)\subset M_{2}(D)$. Therefore, $\theta$ leaves the
subalgebra $\diag(1,v)M_{2}(F) \diag(1,v^{-1})$ invariant and
restricts to an orthogonal involution on that subalgebra.
The proof of Theorem~\ref{decomp.thm} is thus complete.

\begin{exemple}
     \label{nodec.exmple}
     The following is another example where the converse of
     Lemma~\ref{decomp.lem} fails. Consider three quadratic
     extensions $K_{1}$, $K_{2}$, $K_{3}$ of a field $k$,
     \[
     K_{i}=k(\sqrt{a_{i}})
     \qquad\text{for some $a_{i}\in k$},
     \]
     such that $K_{1}\otimes_{k}K_{2}\otimes_{k}K_{3}$ is a field, and
     let $F=k(x_{1},x_{2},x_{3})$ be the field of rational fractions
     in three indeterminates over $k$. For $i=1$, $2$, $3$, consider
     $K_{i}$ as a subfield of the quaternion algebra
     $A_{i}=(a_{i},x_{i})_{F}$. On the tensor product
     \[
     A=A_{1}\otimes_{F}A_{2}\otimes_{F}A_{3},
     \]
     consider the symplectic involution
     \[
     \theta=\theta_{1}\otimes\theta_{2}\otimes\gamma_{3},
     \]
     where $\gamma_{3}$ is the conjugation involution on $A_{3}$ and
     $\theta_{1}$ (resp.~$\theta_{2}$) is an orthogonal involution on
     $A_{1}$ (resp.~$A_{2}$) which is the identity on $K_{1}$
     (resp.~$K_{2}$).

     Let $\lambda\in N_{K_{1}/k}(K_{1}^\times)\cap
     N_{K_{2}/k}(K_{2}^\times)$. By a well-known property of
     biquadratic extensions (see for instance \cite[2.13]{EL}), we may
     find $u\in K_{1}\otimes_{k}K_{2}$ and $v\in k^\times$ such that
     \[
     \lambda=v^{2}N_{K_{1}\otimes_{k}K_{2}/k}(u).
     \]
     Viewing $u=u\otimes1$ in $A$, we let
     \[
     \sigma=\Int(u)\circ\theta.
     \]
     By Lemma~\ref{elem.lem}(d), we have
     $\Nrp_{\theta}(u)=\Nrd_{A_{1}\otimes_{F}A_{2}}(u)=
     N_{K_{1}\otimes_{k}K_{2}/k}(u)$, hence
     \[
     \Delta_{\theta}(\sigma)=(\lambda)\cup[A].
     \]
     Since $\lambda$ is a norm from $K_{1}$ and $K_{2}$, hence a
     reduced norm from $A_{1}$ and $A_{2}$, it follows that
     \[
     \Delta_{\theta}(\sigma)=(\lambda)\cup[A_{3}].
     \]
     Therefore, $\Delta_{\theta}(\sigma)=0$ if and only if
     $\lambda\in N_{K_{1}/k}(K_{1}^\times)\cap
     N_{K_{2}/k}(K_{2}^\times)\cap N_{K_{3}/k}(K_{3}^\times)$.

     Now, suppose $\sigma=\Int(w)\circ\sigma'\circ\Int(w)^{-1}$ for
     some involution $\sigma'$ leaving $A_{1}$ invariant. Then
     $\sigma'=\Int\bigl(w^{-1}u\theta(w)^{-1}\bigr)\circ\theta$, and the
     proof of Lemma~\ref{decomp.lem} shows that
     $\Nrp_{\theta}\bigl(w^{-1}u\theta(w)^{-1}\bigr)\in F^{\times2}$.
     Since $\Nrp_{\theta}\bigl(w^{-1}u\theta(w)^{-1}\bigr)=
     \Nrp_{\theta}(u)\Nrd_{A}(w)^{-1}=\lambda v^{-2}\Nrd_{A}(w)^{-1}$,
     it follows that
     \[
     \lambda\in F^{\times2}\cdot\Nrd_{A}(A^\times).
     \]
     By Proposition~9 of \cite{CDTWY}, we then have
     \[
     \lambda\in k^{\times2}\cdot N_{M/k}(M^\times)
     \qquad\text{with $M=K_{1}\otimes_{k}K_{2}\otimes_{k}K_{3}$.}
     \]
     Therefore, examples of triquadratic extensions
     $M=K_{1}\otimes_{k}K_{2}\otimes_{k}K_{3}/k$ such that
     \[
     N_{K_{1}/k}(K_{1}^\times)\cap N_{K_{2}/k}(K_{2}^\times)
     \cap N_{K_{3}/k}(K_{3}^\times)\neq k^{\times2}\cdot
     N_{M/k}(M^\times)
     \]
     yield examples of involutions $\sigma$ for which
     $\Delta_{\theta}(\sigma)=0$ even though $\sigma$ is not conjugate
     to an involution leaving $A_{1}$ invariant. Triquadratic
     extensions of this type were constructed in \cite{STW} (see also
     \cite[Proposition~3]{G}).
\end{exemple}

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\end{document}
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