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\begin{document}
 \title{Discriminant of symplectic involutions}
 \author{Skip Garibaldi}
\address{Department of Mathematics \& Computer Science, Emory University, Atlanta, GA 30322, USA}
\email{skip@member.ams.org}
\urladdr{http://www.mathcs.emory.edu/{\textasciitilde}skip/}

\author{R. Parimala}
\address{Department of Mathematics \& Computer Science, Emory University, Atlanta, GA 30322, USA}
\email{parimala@mathcs.emory.edu}
\urladdr{http://www.mathcs.emory.edu/{\textasciitilde}parimala/}

\author{Jean-Pierre Tignol}
\address{D\'epartement de math\'ematique, chemin du cyclotron, 2, B-1348 Louvain-la-Neuve, Belgium}
\email{jean-pierre.tignol@uclouvain.be}
\urladdr{http://www.math.ucl.ac.be/membres/tignol/}

\subjclass[2000]{16W10 (11E72)}

%\thanks{{\tt Version of \today.}}

\thanks{The third author is partially supported by the Fund for Scientific Research -- FNRS (Belgium). He would like to thank his co-authors and Emory University for their hospitality while the work for this paper was carried out.}

\dedicatory{\`a Jean-Pierre Serre, pour son 80\/$^e$ anniversaire}

\maketitle
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\begin{abstract}
We define an invariant of torsors under adjoint linear algebraic groups of type $C_n$---equivalently, central simple algebras of degree $2n$ with symplectic involution---for $n$ divisible by 4 that takes values in $H^3(k, \mmu2)$.  The invariant is distinct from the few known examples of cohomological invariants of torsors under adjoint groups.  We also prove that the invariant detects whether a central simple algebra of degree 8 with symplectic involution can be decomposed as a tensor product of quaternion algebras with involution.
\end{abstract}


\section{Introduction}

While the Rost invariant is a degree 3 invariant defined for torsors under simply connected simple algebraic groups, there are very few degree 3 invariants known for adjoint groups.  In this paper, we define such an invariant for torsors under adjoint algebraic groups of symplectic type and show that this invariant gives a necessary and sufficient condition for decomposability of symplectic involutions on degree 8 algebras.  This decomposability criterion is analogous to the criteria given for degree 4 algebras with orthogonal involutions in \cite{KPS:16} and for unitary involutions from \cite{KarpQ}.

In the paper \cite{BMT}, the authors defined a ``relative" invariant of symplectic involutions: for $\s, \tau$ symplectic involutions on the same central simple algebra $A$ (defined over a field $F$ of characteristic $\ne 2$ and of degree divisible by 4), they defined
\[
\D_\tau(\s) = (\Nrp(s)) \cdot [A] \quad \in H^3(F, \mmu2)
\]
where $s$ is a $\tau$-symmetric element of $A^\times$ such that $\Int(s) \tau = \s$.  (Alternatively, $\D_\tau$ can be defined by pushing the Rost invariant of the simply connected group $\Sp(A,\tau)$ forward along the natural projection to the adjoint group.)  If the index of $A$ divides half the degree, then one gets an ``absolute" invariant by taking $\tau$ to be hyperbolic.  

Our first result shows that the relative invariant defined in \cite{BMT} leads to an absolute invariant of all symplectic involutions on algebras of a fixed degree, i.e., to a cohomological invariant of the split adjoint group $\psp_{2n}$ of type $C_n$ for $n$ divisible by 4.
Recall that the set $H^1(K, \grp)$ classifies central simple $K$-algebras of degree $2n$ endowed with a symplectic involution \cite[29.22]{KMRT} for each extension $K/F$, and this defines a functor $\mathsf{Fields}_{/F} \ra \mathsf{Sets}$.  The map $K \mapsto H^3(K, \mmu2)$ also defines a functor $\mathsf{Fields}_{/F} \ra \mathsf{Sets}$, and an \emph{invariant} $H^1(*, \psp_{2n}) \ra H^3(*, \mmu2)$ (in the sense of \cite{GMS}) is a morphism of functors, i.e., a map $a_K \!: H^1(K, \psp_{2n}) \ra H^3(K, \mmu2)$ for every extension $K/F$ together with a compatibility condition.  

\begin{specthm} \label{inv}
For every $n$ divisible by $4$, there is a unique invariant
\[
\D \!: H^1(*, \grp) \ra H^3(*, \mmu2)
\]
such that for every extension $K/F$, we have:
\begin{enumerate}
\item If $\As$ is a central simple $K$-algebra with \emph{hyperbolic} symplectic involution, then $\D\As = 0$.
\item If $\s, \tau$ are symplectic involutions on the same $K$-algebra $A$, then
\[
\D_\tau(\s) = \D\As - \D(A, \tau).
\]
\end{enumerate}
\end{specthm}

Property (1) includes the statement that $\D_K$ sends zero to zero for each extension $K/F$, i.e., $\D$ is normalized in the sense of \cite{GMS}.  Property (2) can be replaced by ``$\D$ is nonzero", see \ref{par} below.

No such invariant $\D$ exists in case $n$ is not divisible by 4, see Prop.~\ref{deg3inv} for a precise statement.

\smallskip

We say that $\As$ is \emph{completely decomposable} if there are quaternion algebras $Q_i$ endowed with involutions $\s_i$ of the first kind such that $\As$ is isomorphic to the tensor product $\ot (Q_i, \s_i)$.
For $\s_1, \s_2$ orthogonal, the tensor product $(Q_1, \s_1) \ot (Q_2, \s_2)$ is isomorphic to  a tensor product $(Q'_1, \g_1) \ot (Q'_2, \g_2)$ where $Q'_1, Q'_2$ are quaternion algebras and $\g_1, \g_2$ are the canonical symplectic involutions \cite[5.2]{KPS:16}.  Thus every completely decomposable $\As$ can be written as a tensor product of quaternion algebras with symplectic involutions and at most one quaternion algebra with orthogonal involution.

It follows from results in the literature that $\D$ vanishes on completely decomposable involutions, see Example \ref{dec.eg}.  Our second theorem shows that the converse holds for algebras of degree 8:

\begin{specthm} \label{decomp}
Let $A$ be a central simple algebra of degree $8$ with symplectic involution $\s$.  The element $\D\As$ is zero if and only if $\As$ is completely decomposable.
\end{specthm}

Finally, we address the Pfister Factor Conjecture from \cite{Shapiro}, which asserts: \emph{Let $(Q_i, \s_i)$ be quaternion $F$-algebras endowed with orthogonal involutions for $1 \le i \le n$.  Over every extension $K/F$ such that the tensor product $Q_1 \ot Q_2 \ot \cdots \ot Q_n$ is split, the involution $\s_1 \ot \s_2 \ot \cdots \ot \s_n$ is adjoint to a Pfister form.}  This conjecture is now known in general by \cite{Becher}.  We use our invariant to give an alternate proof for $n = 6$.

\subsection*{Notation and conventions} We assume throughout the paper that the characteristic of $F$ is different from 2.

We sometimes write $\binv$ for the standard (symplectic) involution on a quaternion algebra.  For $g$ an element of a group and $h$ an endomorphism of the group, we write $\Int(g)h$ for the map $x \mapsto g h(x) g^{-1}$.

For an algebraic group $G$ over $F$, we write $H^d(F, G)$ for the Galois cohomology set $H^d(\mathrm{Gal}(\Fs/F), G(\Fs))$ where $\Fs$ denotes a separable closure of $F$.  The group $H^1(F, \mmu2)$ is identified with $F^\times / F^{\times 2}$ by Kummer theory, and we write $(x)$ for the element of $H^1(F, \mmu2)$ corresponding to $xF^{\times 2}$.  The group $H^2(F, \mmu2)$ is identified with the 2-torsion in the Brauer group of $F$, and we write $[A]$ for the element of $H^2(F, \mmu2)$ corresponding to a central simple algebra $A$ of exponent 2.

General background on these topics and on algebras with involution can be found in \cite{KMRT}.  For the Rost invariant, see \cite{GMS}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem \ref{inv}}

This section consists of a proof of Theorem \ref{inv}.

\emph{(i):} We first prove uniqueness.  Let $\D, \D'$ be invariants as in the theorem, and consider $\As \in H^1(K, \grp)$.   In the Brauer group, the class of $A$ can be written as a sum $\sum_{i=1}^\ell [Q_i]$ of quaternion algebras, and the \emph{symbol length} of $A$ is defined to be the minimum such $\ell$.  We prove that $\D$ equals $\D'$ by induction on $\ell$.

If $\ell$ is 0 or 1, i.e., if $A$ has index 1 or 2, then applying (1) and (2) to compare $\s$ with the hyperbolic involution $\tau$ on $A$ shows that
\[
\D\As = \D_{\tau}(\s) + 0 = \D'\As.
\]

If $\ell$ is at least 2, then we let $E$ be the function field of the Albert quadric for the product $Q_1 \ot Q_2$.  Recall from, say, \cite[16.4, 16.5]{KMRT} that if $Q_i = (a_i, b_i)$ for $a_i, b_i \in \Fx$, the Albert quadric of $Q_1 \ot Q_2$ is the variety defined by the quadratic polynomial
\[
a_1 X_1^2 + b_1 X_2^2 - a_1 b_1 X_3^2 -a_2 X_4^2 -b_2 X_5^2 + a_2 b_2 X_6^2
\]
and that $Q_1 \ot Q_2 \ot E$ has index 2.
Then $A \ot E$ has symbol length strictly less than $\ell$, hence $\D$ and $\D'$ agree on $\res_{E/K}\As$.  However, the restriction map $H^3(K, \mmu2) \ra H^3(E, \mmu2)$ is injective \cite[5.6]{Ar:th}, so $\D$ and $\D'$ agree on $\As$.

\medskip

\emph{(ii):} 
We prove the existence of $\D$ in the same manner as uniqueness, i.e., by induction on the symbol length $\ell$, supposing  at each stage that $\D$ is defined for every algebra of symbol length $< \ell$ over every extension of $F$ and satisfies (1) and (2) and is functorial.

For $\ell \le 2$, the definition is clear: A central simple algebra $A$ of degree $2n$ that can be written as a sum of at most 2 quaternion algebras has index 1, 2, or 4.  As $n$ is divisible by 4, there is a hyperbolic involution $\tau$ on $A$ and we put 
$\D\As := \D_\tau(\s)$.  This defines an invariant satisfying properties (1) and (2) by the results of \cite{BMT}.

\medskip

\emph{(iii):} Suppose that $\ell$ is at least 3.  Let $X$ be the Albert quadric for the biquaternion algebra $Q_1 \ot Q_2$.  We put:
\[
\delta := \D(\As \ot F(X)) \quad \in H^3(F(X), \mmu2)
\]
where the right side is defined because $A \ot F(X)$ has symbol length strictly less than $\ell$ in the Brauer group of $F(X)$.
We claim that $\delta$ belongs to the image of  $H^3(F, \mmu2) \ra H^3(F(X), \mmu2)$.

We first verify that $\delta$ is unramified.  Let $x$ be a codimension 1 point of $X$, write $F(x)$ for its residue field, and write $K$ for the completion of $F(X)$ at $x$.  By Cohen's Structure Theorem, there is a finite purely inseparable extension $K_1$ of $K$ with residue field $F_1$ such that there is an $F$-embedding $F_1 \injects K_1$ splitting the residue map, see e.g.~\cite[p.~30]{GMS}.  We obtain a commutative diagram \cite[p.~19]{GMS}:
\[
\begin{CD}
H^3(K, \mmu2) @>>> H^2(F(x), \mmu2) \\
@VVV @VVeV \\
H^3(K_1, \mmu2) @>>> H^2(F_1, \mmu2)
\end{CD}
\]
where $e$ denotes the ramification index of $K_1/K$.
The element $\D(\As \ot F(x))$ of $H^3(F(x), \mmu2)$ is well defined (because the symbol length of $A$ is lower over $F(x)$) and has the same image as $\delta$ in $H^3(K_1, \mmu2)$ (via the embedding $F(x) \injects K_1$).
Hence $\delta$ is unramified over $K_1$, i.e., $\delta$ in the upper left corner maps to zero in the lower right.
As $K_1/K$ is a purely inseparable extension, it has odd degree, so the vertical arrows are injections and the residue of $\delta$ in $H^2(F(x), \mmu2)$ is zero.  We have shown that $\delta$ belongs to the subgroup $\Hnr(F(X), \mmu2)$ of $H^3(F(X), \mmu2)$ consisting of unramified classes.  

Let now $L$ be a generic splitting field for $Q_\ell$.
We claim that $X$ remains anisotropic over $L$ (because $\ell \ne 1, 2$).  Otherwise, by \cite[XIII.2.13]{Lam}, $Q_1 \ot Q_2$ is isomorphic to $H \ot Q_\ell$ for some quaternion $F$-algebra $H$.  But in that case the Brauer class of $A$ equals $[H] + \sum_{i=3}^{\ell - 1} [Q_i]$, contradicting  the definition of $\ell$.

By \cite{Kahn:lower}, we have a diagram with exact rows:
\begin{equation} \label{LX}
\begin{CD}
0 @>>> H^3(F, \mmu2) @>\res_{F(X)/F}>> \Hnr(F(X), \mmu2) @>>> \Zm2 @>>> 0\\
@. @VVV @VVV @| \\
0 @>>> H^3(L, \mmu2) @>\res_{L(X)/L}>> \Hnr(L(X), \mmu2) @>>> \Zm2 @>>> 0
\end{CD}
\end{equation}
Since $X$ is anisotropic over $F$ and $L$, the diagram commutes by the explicit description of an 
 element of $\Hnr(F(X), \mmu2)$ mapping to the nonzero element in $\Zm2$ from \cite[p.~249]{Kahn:lower}.
 The symbol lengths of $A$ over $F(X)$ and over $L$ are at most $\ell -1$, so we have:
\[
 \D\left( [ \As \ot_F F(X) ] \ot_{F(X)} L(X) \right)
= \D\left( [\As \ot_F L] \ot_L L(X) \right),
\]
that is,
\[
\res_{L(X)/F(X)}(\delta) = \D\left( \As \ot L(X) \right) = \res_{L(X)/L} \D( \As \ot L).
\]
A diagram chase shows that $\delta$ is in the image of  the restriction map $H^3(F, \mmu2) \ra \Hnr(F(X), \mmu2)$.  We define $\D\As \in H^3(F, \mmu2)$ to be the unique element that maps to $\delta$.

\medskip

\emph{(iv):} Using the method of \emph{(i)}, one can verify properties (1) and (2) and functoriality for $\D$ relative to algebras of symbol length at most $\ell$ by induction.   This completes the proof of existence of $\D$, hence of Theorem \ref{inv}.$\hfill\qed$

\begin{rmk} \label{biquat}
In part \emph{(iii)} of the proof above, the condition that $\ell\geq3$ is necessary to prove that $\delta$ lies in the image of $H^3(F,\mu_2)$. By way of contrast,
consider the case where $\As$ is a central simple $F$-algebra of degree 4 with symplectic involution; this case is not covered by Theorem \ref{inv}.  The algebra $A$ is a tensor product of quaternion algebras $Q_1 \ot Q_2$.  Over the function field $F(X)$ of the corresponding Albert quadric, the algebra is isomorphic to 2-by-2 matrices over a quaternion algebra $H$ and $\s$ is adjoint to a hermitian form $\qform{1, -c}$ with respect to the canonical involution on $H$.  By \cite[Example 3]{BMT}, the relative invariant with respect to the hyperbolic form $\tau$ on $M_2(H)$ is 
\[
\D_\tau(\s) = (c) \cdot [H] \quad \in H^3(F(X), \mmu2).
\]
This is an unramified class that is not in the image of $H^3(F, \mmu2)$ \cite[p.~249]{Kahn:lower}, i.e., it represents the nonzero element in the cokernel $\Zm2$ from \eqref{LX}.
\end{rmk}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Examples}

\begin{eg}     \label{P.prop}
Let $B$ be a central simple algebra of degree divisible by $4$, endowed with an orthogonal involution $\rho$, and let $Q$ be a quaternion algebra.  We have:
     \[
     \D \left[ (Q, \binv) \ot (B, \rho) \right]  = [Q] \cdot (\disc\rho),
     \]
cf.~\cite{SerhirTig}.
This can be proved by induction on the symbol length $\ell$ of $B$.  If $\ell$ is 0 or 1, then $B$ is not division and the lemma is \cite[Example 3]{BMT}.  The case of larger $\ell$ is reduced to the case $\ell = 1$ by the Albert quadric technique as in the proof of Theorem \ref{inv}. 
\end{eg}

\begin{eg} \label{dec.eg}
Suppose that $\As$ is completely decomposable as defined in the introduction, i.e., isomorphic to $\ot_{i=1}^n (Q_i, \s_i)$, and $n$ is at least 3. By \cite[5.2]{KPS:16} we can write $\As$ as in Example \ref{P.prop} with $(B,\rho)$ completely decomposable. Then $\disc \rho = 1$ by \cite[7.3]{KMRT}, which gives: \emph{If $\As$ is completely decomposable and the degree of $A$ is at least $8$, then $\D\As$ is zero.}
\end{eg}

\begin{eg} \label{P.2}
Let $B$ be a central simple algebra of degree divisible by 4 endowed with a symplectic involution $\sigma$, and let $Q$
be a quaternion algebra. For every orthogonal involution $\rho$ on $Q$,
\[
\Delta[(Q,\rho)\otimes(B,\sigma)]=
%\begin{cases}
0.
%[B]\cdot(\disc\rho)&\text{if $m$ is odd}.
%\end{cases}
\]
To prove this, we 
use induction on the symbol length of $B$ and the Albert quadric
technique to reduce to the case where $B$ is split or has 
index $2$. When $B$ is split, $\s$ and $\rho \ot \s$ are hyperbolic and the formula is clear.
If $B$ is Brauer-equivalent to a quaternion algebra $H$,
then Proposition 3.4 of \cite{BST} yields a decomposition
\[
(B,\sigma)\simeq(H,\binv)\otimes(M_m(F),\tau)
\]
for some orthogonal involution $\tau$. By Example \ref{P.prop} we have
\[
\Delta[(Q,\rho)\otimes(B,\sigma)]=
[H]\cdot(\disc\rho\otimes\tau)=
%\begin{cases}
0,%&\text{if $m$ is even},\\
%[B]\cdot(\disc\rho)&\text{if $m$ is odd}.
%\end{cases}
\]
as desired.
\end{eg}


\begin{prop}
     \label{dirsum.prop}
 If $(A,\sigma)=(B,\tau_1)\boxplus(B,\tau_2)$ for some
central simple algebra $B$ of degree divisible by $4$ and with symplectic
involutions $\tau_1$, $\tau_2$, then
$\Delta(A,\sigma)=
%\begin{cases}
\Delta_{\tau_1}(\tau_2)$.%&\text{if $m$ is even}\\
%\Delta_{\tau_1}(\tau_2)+[A]\cdot(-1)&\text{if $m$ is odd}.
%\end{cases}
\end{prop}

\begin{proof}
     Let $\hat \tau_{1}=t\otimes\tau_{1}$ on $M_{2}(B)=M_{2}(F)\otimes
     B$. The hypothesis means that we may identify
     \[
     A=M_{2}(B),\qquad
     \sigma=\Int\bigl(
     \begin{smallmatrix}
	1&0\\0& v
     \end{smallmatrix}
     \bigr)\hat\tau_{1}
     \]
     for some $v\in B^\times$ such that
     $\tau_{2}=\Int(v)\tau_{1}$. Then
     \[
     \Delta\As - \D(A, \hat\tau_{1})=\Delta_{\sigma}(\hat\tau_{1})=
     \bigl(\Nrp_{\sigma}\bigl(
     \begin{smallmatrix}
	1&0\\0& v
     \end{smallmatrix}
     \bigr)\bigr)\cdot [A]=
     (\Nrp_{\tau_{1}}(v))\cdot [B]=\Delta_{\tau_{1}}(\tau_{2}).
     \]
On the other hand, $(A, \hat\tau_1)$ can be decomposed as $(M_2(F), t) \ot (B, \tau_1)$, so $\D(A, \hat\tau_1)$ is zero by Example \ref{P.2}.
\end{proof}

\setlength{\unitlength}{.5cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Degree 3 invariants of $\psp_{2n}$}

Consider the following invariants that map $H^1(*, \grp) \ra H^d(*, \mmu2)$ for various $d$:
\begin{enumerate}
\item The constant invariant $\uo$ that sends every $\As \in H^1(K, \grp)$ to the 1 in the ring $H^\bullet(K, \mmu2)$.
\item The invariant $\delta \!: H^1(*, \grp) \ra H^2(*, \mmu2)$ that is given by the connecting homomorphism arising from the surjection $\pi \!: \Sp_{2n} \ra \grp$, where $\Sp_{2n}$ denotes the simply connected cover of $\grp$.  The map $\delta_K$ sends $\As \in H^1(K, \grp)$ to the class $[A]$ in $H^2(K, \mmu2)$.
\item The invariant $\D$ defined in Th.~\ref{inv}; it is defined when $n$ is divisible by 4.
\end{enumerate}

The collection of invariants $H^1(*, \grp) \ra H^\bullet(*, \mmu2)$ is a module over the ring $H^\bullet(F, \mmu2)$, and the invariants $\uo, \delta, \D$ are linearly independent over that ring.  We prove the following spanning result:

\begin{prop} \label{deg3inv}
Suppose that $n$ is even.  For $d \le 3$, every invariant $H^1(*, \psp_{2n}) \ra H^d(*, \mmu2)$ is in the span of $\uo, \delta, \D$ if $n$ is divisible by $4$, and in the span of $\uo, \delta$ otherwise.
\end{prop}

\begin{proof}
\emph{(i):} For $d = 1, 2$, the proposition is contained in \cite[31.15, 31.20]{KMRT}, so we consider the case $d = 3$.
Let $a$ be an invariant $H^1(*, \psp_{2n}) \ra H^3(*, \mmu2)$.  Consider the functor $H^1(*, \PGL_2) \ra H^1(*, \grp)$ that sends a quaternion algebra $Q$ to the algebra $M_n(Q)$ endowed with a hyperbolic involution (which exists because $n$ is even); composing this with $a$, we obtain an invariant $H^1(*, \PGL_2) \ra H^3(*, \mmu2)$ that is necessarily of the form
\[
Q \mapsto \la_3 + \la_1 \cdot [Q]
\]
for uniquely determined $\la_3 \in H^3(F, \mmu2)$ and $\la_1 \in H^1(F, \mmu2)$ by \cite[p.~43]{GMS}.

Let $\As$ denote a versal $\psp_{2n}$-torsor defined over a field $K$.  Write $\psi_E$ for the composition 
\begin{equation} \label{n.odd}
\xymatrix{
H^1(E, \Sp\As)  \ar[r]^-{\pi} &  H^1(E, \psp\As) \ar[r]^-\sim_-{\mathrm{twist}}& H^1(E, \psp_{2n}) \ar[r]^-{a_E}& H^3(E, \mmu2)
}
\end{equation}
for extensions $E$ of $K$; this defines an invariant $\psi$ of torsors under $\Sp\As$.  By \cite[p.~127]{GMS} this invariant is of the form
\[
\eta \mapsto \la_0\,r_E(\eta) + a_E(A \ot E, \s).
\]
for fixed $\la_0$ equal to 0 or 1, where $r$ denotes the Rost invariant $H^1(*, \Sp\As) \ra H^3(*, \mmu2)$.  Recall from \cite[29.24]{KMRT} that the set $H^1(E, \Sp\As)$ is identified with a quotient of the set of $\s$-symmetric elements $s \in (A \ot E)^\times$.  For such an $s$, we have by \cite[31.46]{KMRT}:
\[
\psi_E(s) = \la_0 \cdot (\Nrp(s)) \cdot [A] + a_E(A \ot E, \s),
\]
which is $a_E(A \ot E, \Int(s) \tau)$ by the definition of $\psi$.  Whence, for every extension $E/K$ and every symplectic involution $\tau$ on $A \ot E$, we have 
\begin{equation} \label{psi}
a_E(A \ot E, \s) - a_E(A \ot E, \tau) = \la_0 \cdot \D_\tau(\s).
\end{equation}

\smallskip
\emph{(ii):}
Suppose first that $n$ is divisible by 4.
We claim that $a$ is the invariant $a' := \la_3 + \la_1 \cdot \delta + \la_0 \cdot \D$.  By \cite[p.~31]{GMS}, it suffices to check that $a$ and $a'$ take the same value on the versal torsor $\As \in H^1(K, \psp_{2n})$.  Arguing as in the proof of Th.~\ref{inv} above, we may find an extension $E/K$ such that $A \ot E$ has index 2 and the restriction $H^3(K, \mmu2) \ra H^3(E, \mmu2)$ is injective, so it suffices to prove that $a$ and $a'$ agree on $\res_{E/K} \As \in H^1(E, \grp)$.  As $A \ot E$ has index 2, it supports a hyperbolic involution $\tau$.  By construction (and property (1) of $\D$), the invariants $a$ and $a'$ both take the value $\la_3 + \la_1 \cdot [A]$ on $(A \ot E, \tau)$.  
On the other hand, by property (2) of $\D$ we have:
\[
a'_E(A \ot E, \s) - a'_E(A \ot E, \tau)= \la_0 \cdot \D_\tau(\s),
\]
which equals $a_E(A \ot E, \s) - a_E(A \ot E, \tau)$ by \eqref{psi}.  Combining the two preceding sentences, we conclude that $a$ and $a'$ agree on $(A \ot E, \s)$, hence are the same invariant.

\smallskip
\emph{(iii):} Suppose now that $n = 2m$ for some $m$ odd.  Replacing $a$ with $a - (\la_3 + \la_1 \cdot \delta)$, we may assume that $a$ vanishes on hyperbolic involutions.  (Note that an algebra of degree $4m$ with hyperbolic involution necessarily has index 1 or 2.)  We want to show that $a$ is the zero invariant.

We claim that the element $\la_0$ defined in \emph{(i)} is 0.  The algebra $A$ is Brauer-equivalent to a biquaternion division $K$-algebra $B$.  Fix a symplectic involution $\gamma$ on $B$ and define an involution $\theta$ on $A$ via
\[
(A, \theta) \cong (M_m(K), t) \otimes (B, \gamma).
\]
Let $X$ be the Albert quadric for $B$, and fix a hyperbolic symplectic involution $\tau$ on $A \ot K(X)$.  Because $a$ vanishes on hyperbolic involutions, we have:
%\[
\begin{align*}
\res_{K(X)/K} a(A, \theta) &= a(A \ot K(X), \theta) \\
&= a(A \ot K(X), \theta) - a(A \ot K(X), \s) \\
&\quad +\ a(A \ot K(X), \s) - a(A \ot K(X), \tau) \\
%\end{array}
\end{align*}
%\]
Applying \eqref{psi} twice, we find:
\begin{equation} \label{a.odd}
\res_{K(X)/K}(A, \theta) = \la_0 \cdot ( \D_\s(\theta) + \D_\tau(\s)) = \la_0 \cdot \D_\tau(\theta),
\end{equation}
where the second equality is by \cite[Prop.~1b]{BMT}.

The algebra $B \ot K(X)$ is Brauer-equivalent to a quaternion division algebra $H$ over $K(X)$ and $\theta$ is adjoint to a hermitian form
\[
\qform{1, 1, \ldots, 1} \ot \qform{1, -c}
\]
(with respect to the canonical involution on $H$) for some $c \in K(X)^\times$.  As in Remark \ref{biquat}, $\D_\tau(\theta) = (c) \cdot [H]$, which is an element of $H^3(K(X), \mmu2)$ that does not come from $H^3(K, \mmu2)$.  As $a(A, \theta)$ is an element of $H^3(K, \mmu2)$, we conclude via \eqref{a.odd} that $\la_0$ is zero.

By \eqref{psi}, we have:
\[
a(A \ot K(X), \s) = a(A \ot K(X), \tau) = 0.
\]
Hence $a\As$ is zero.  As $\As$ is a versal $\grp$-torsor, this proves the proposition.
\end{proof}

\begin{borel*} \label{par}
Suppose that $n$ is divisible by 4, and $a$ is a \emph{nonzero} invariant $H^1(*, \grp) \ra H^3(*, \mmu2)$ that satisfies condition (1) of Th.~\ref{inv}, i.e., such that $a\As$ is zero if $\s$ is hyperbolic.  It follows easily from Prop.~\ref{deg3inv} that $a$ equals $\D$.  Therefore, in the statement of Th.~\ref{inv}, condition (2) can be replaced with: \emph{$\D$ is nonzero}.
\end{borel*}


%\begin{rmk}
%Using the methods from the proof, it is easy to see that, in the statement of the theorem, property (1) could be replaced with:
%\[
%\parbox{4in}{($1'$)\ \ If $A$ is a central simple $K$-algebra of index 1 or 2 and $\s$ is a hyperbolic symplectic involution on $A$, then $\D\As = 0$.}
%\]
%\end{rmk}

\begin{rmk}
In the case $n = 2$,  the exceptional isomorphism $\mathsf{C}_2 = \mathsf{B}_2$ shows that $\psp_{2n}$ is the split special orthogonal group on a 5-dimensional space.  The invariants $H^1(*, \psp_4) \ra H^\bullet(*, \mmu2)$ were determined in \cite[p.~44]{GMS}; they make up a free $H^\bullet(F, \mmu2)$-module of rank 3.
\end{rmk}

The invariants of $\psp_{2n}$-torsors when $n$ is odd will be treated by Mark MacDonald in \cite{MacD}.  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Alternative definition of $\D$ for algebras of degree 8}

In the case $n = 4$ --- corresponding to algebras of degree 8 --- the invariant $\D$ given by Theorem \ref{inv} can be seen as a special case of the Rost invariant.  Most of what is written here is not used in the rest of this paper, so we omit many details.

\begin{borel}{The inclusion $\psp_8 \subset E_6$}
We view the split adjoint group $\psp_8$ as a subgroup of the split simply connected group $E_6$ of that type.  (This inclusion is well known, see e.g.\ \cite[p.~193]{Dynk:ssub}.)  We view $E_6$ as a Chevalley group given in terms of generators and relations as in \cite{St}.  Let $\alpha_1, \alpha_2, \ldots, \alpha_6$ be simple roots of $E_6$ numbered as in the diagram
\[
\begin{picture}(7,2.7)
    % put in the circles
    \multiput(1,0.7)(1,0){5}{\circle*{\darkrad}}
    \put(3,1.45){\circle*{\darkrad}}

    % put in the lines
    \put(1,0.7){\line(1,0){4}}
    \put(3,1.45){\line(0,-1){.75}}
    
    \put(1,0){\makebox(0,0.4)[b]{$1$}}
    \put(2,0){\makebox(0,0.4)[b]{$3$}}
    \put(3,0){\makebox(0,0.4)[b]{$4$}}
    \put(4,0){\makebox(0,0.4)[b]{$5$}}
    \put(5,0){\makebox(0,0.4)[b]{$6$}}
    \put(3.25,1.3){\makebox(0,0.4)[b]{$2$}}
\end{picture}
\]
Write $h_{\alpha_i} \!: \mathbb{G}_m \ra E_6$ for the homomorphism corresponding to $\alpha_i$ such that the images of the $h_{\alpha_i}$ generate a maximal torus in $E_6$.  Write $\phi$ for the outer automorphism of $E_6$ that permutes the generators in a manner corresponding to the non-identity automorphism of the Dynkin diagram.  The subgroup $H$ of $E_6$ consisting of elements fixed by the automorphism
\[
\Int(h_{\alpha_1}(-1) h_{\alpha_4}(-1) h_{\alpha_6}(-1)) \cdot \phi
\]
is connected and reductive \cite[8.1]{St:end}.  The root system of $H$ is explicitly described by the recipe from p.~275 of \cite{St:coll}; it is of type $C_4$.

Let $\Sp_8$ be the simply connected cover of $H$.  Since it is simply connected, the cocharacter group of a maximal torus is identified with the coroot lattice.  The same holds for $E_6$, and we can describe the arrow $\Sp_8 \ra E_6$ on the level of maximal tori by computing the corresponding map on coroot lattices.  Number the simple roots of $C_4$ as $\g_1, \g_2, \g_3, \g_4$ as in the diagram
\[
\begin{picture}(5,2)
    % put in the horizontal lines
    \put(1,1){\line(1,0){2}} %multiput(1,1)(1,0){2}{\line(1,0){1}}
%    \put(4,1){\line(1,0){1}}
    \put(3,1.04){\line(1,0){1}}
    \put(3,0.96){\line(1,0){1}}
 
    % label horizontal circles
    \put(1,0.3){\makebox(0,0.5)[b]{$1$}}
    \put(2,0.3){\makebox(0,0.5)[b]{$2$}}
    \put(3,0.3){\makebox(0,0.5)[b]{$3$}}
    \put(4,0.3){\makebox(0,0.5)[b]{$4$}}
%    \put(5,1.2){\makebox(0,0.4)[b]{$\ell-1$}}
%    \put(6, 0.6){\makebox(0,0.4)[b]{$\ell$}}
    
    \put(3.5,0.83){\makebox(0,0.4)[b]{$<$}}
 

    % put in the dots
%    \multiput(3.3,1)(0.2,0){3}{\circle*{0.01}}

%    % put in the diagonal lines
%    \put(5.07,0.93){\line(1,-1){0.56}}
%    \put(5.07,1.07){\line(1,1){0.56}}

    % put in the horizontal circles
    \multiput(1,1)(1,0){4}{\circle*{\darkrad}}
%   \put(4,1){\circle{\lrad}}

%    % put in the diagonal circles
%    \put(5.7,0.3){\circle{0.2}}
%    \put(5.7,1.7){\circle{0.2}}
%    \put(5.7,-0.1){\makebox(0,0.4)[b]{$\ell^{-}$}}
%    \put(5.7,1.9){\makebox(0,0.4)[b]{$\ell^{+}$}}
\end{picture}
\]
It follows form the explicit description of the root system of $H$ that the map of a maximal torus in $\Sp_8$ to one in $E_6$ is given by
\begin{equation} \label{coroots}
\begin{array}{c@{\ \mapsto\ }l@{\qquad}c@{\ \mapsto\ }l}
\gch_1 & 2 \ach_2 + \ach_3 + 2\ach_4 + \ach_5  &
\gch_2 &  \ach_1 + \ach_6 \\
\gch_3 &  \ach_3 + \ach_5 &
\gch_4 &  \ach_4
\end{array}
\end{equation}
Notice that the short coroot $\gch_4$ maps to the short coroot $\ach_4$.  By \cite[p.~122]{GMS}, this implies:
\begin{equation} \label{rinv.comp}
\parbox{4in}{\emph{The composition of $\Sp_8 \ra E_6$ with the Rost invariant of $E_6$ is the Rost invariant of $\Sp_8$.}}
\end{equation} 

Finally, we show that $H$ is adjoint, i.e., is $\psp_8$.  Write $\rho \!: E_6 \ra GL_{27}$ for the 27-dimensional Weyl module of $E_6$ with highest weight $\omega_1$ (the fundamental dominant weight dual to $\alpha_1$); it is minuscule, so the representation is irreducible in all characteristics.  By the explicit description of inclusion of coroot lattices from \eqref{coroots}, we see that the restriction of $\omega_1$ to $\Sp_8$ is zero on $\gch_i$ for $i \ne 2$ and is 1 on $\gch_2$.  That is, it restricts to the fundamental weight $\mu$ of the root system $C_4$ dual to $\g_2$.  (This is standard, see e.g.\ \cite[p.~298]{McKP}.) Note that this weight belongs to the root lattice and so vanishes on the center of $\Sp_8$.  Therefore, the center of $\Sp_8$ is in the kernel of the composition
\[
\Sp_8 \ra H \subset E_6 \xrightarrow{\rho} GL_{27}.
\]
However, $\rho$ is injective, so it follows that the center of $\Sp_8$ is sent to zero in $H$.  This proves that $H$ is adjoint.
\end{borel}

%Indeed, the split group $\psp_8$ is a subgroup of the split simply connected group $E_6$ of that type, and the composition of the map $H^1(F, \psp_8) \ra H^1(F, E_6)$ with the Rost invariant of $E_6$ is easily seen to take values in $H^3(F, \mmu2)$ and satisfy property (2) of the theorem.  To see that the composition also satisfies property (1), one uses a concrete description of the embedding $\psp_8 \injects E_6$ 
%(which can be found in \cite[p.~193] {Dynk:ssub} or reconstructed using methods from \S8 of \cite{St:end}) to see that the centralizer in $E_6$ of a maximal split torus in $\psp_8$ cannot contain a group of type $D_4$, hence every $\As \in H^1(F, \psp_8)$ with $\s$ hyperbolic is sent to zero in $H^1(F, E_6)$.

%Each $\eta \in H^1(F, \psp_8)$ is killed by an extension of degree a power of 2, and it follows that twisting $E_6$ by $\eta$ produces a group with Tits index
%\[
%\begin{picture}(5,2)
%    % put in the circles
%    \multiput(0.5,0.5)(1,0){5}{\circle*{\darkrad}}
%    \put(2.5,1.5){\circle*{\darkrad}}

%    % put in the lines
%    \put(0.5,0.5){\line(1,0){4}}
%    \put(2.5,1.5){\line(0,-1){1}}

%    % put in the open circles
%    \put(0.5,0.5){\circle{\lrad}}
%    \put(4.5,0.5){\circle{\lrad}}
%\end{picture}
%\]
%The semisimple anisotropic kernel of this group is the spin group of a 3-Pfister quadratic form $\phi$, and $\D(\eta)$ is the Arason invariant of $\phi$.  In particular, $\D(\eta)$ is a symbol in $H^3(F, \mmu2)$.

\begin{borel}{Alternative definition of the invariant}
We define an invariant $\D'$ to be the composition 
\[
\D' \!: H^1(*, \psp_8) \ra H^1(*, E_6) \ra H^3(*, \mathbb{Q}/\Z(2)),
\]
where the first map comes from the inclusion $\psp_8 \subset E_6$ and the second map is the Rost invariant of $E_6$.  The image of the second map is contained in $H^3(*, \Zm6)$ by \cite[p.~150]{GMS}.
\end{borel}

\begin{eg} \label{split.eg}
If $A$ is split, then $\As$ corresponds to the neutral class in $H^1(F, \psp_8)$.  Since the maps in the definition of $\D'$ are maps of pointed sets, $\D' \As$ is zero.
\end{eg}

\begin{prop}\label{E6}
For every field $F$, the image of $\D'$ lies in $H^3(F, \mmu2)$ and consists of symbols.
\end{prop}

\begin{proof}
For every $\As \in H^1(F, \psp_8)$, there is an extension $K/F$ of degree a power of 2 that kills $\As$, hence also kills $\D' \As$.  It follows that the order of $\D' \As$ belongs to the 2-primary torsion of $H^3(F, \Zm6)$, i.e., to $H^3(F, \mmu2)$.

Further, let $G$ be the simply connected group of type $E_6$ over $F$ obtained by twisting $E_6$ by the image of $\As$.  As $G$ is split by $K$, it is split (hence $\D' \As$ is zero) or has Tits index
\[
\begin{picture}(5,2)
    % put in the circles
    \multiput(0.5,0.5)(1,0){5}{\circle*{\darkrad}}
    \put(2.5,1.5){\circle*{\darkrad}}

    % put in the lines
    \put(0.5,0.5){\line(1,0){4}}
    \put(2.5,1.5){\line(0,-1){1}}

    % put in the open circles
    \put(0.5,0.5){\circle{\lrad}}
    \put(4.5,0.5){\circle{\lrad}}
\end{picture}
\]
by \cite[p.~58]{Ti:Cl}.
In the second case, the semisimple anisotropic kernel is a simply connected group of type $^1\!D_4$ with trivial Tits algebras, i.e., $\Spin(q)$ for a 3-Pfister form $q$.  It follows that $\D' \As$ is the Arason invariant of $q$.
\end{proof}


\begin{eg}[Hyperbolic involutions] \label{hyp.E6}
Suppose that $\s$ is hyperbolic.  We will prove that $\D' \As$ is zero.  Indeed, by \cite[p.~56]{Ti:Cl} or the kind of reasoning in \cite{CG}, the Tits index of $\psp\As$ is
\[
\begin{picture}(5,2)
    % put in the horizontal lines
    \put(1,1){\line(1,0){2}} %multiput(1,1)(1,0){2}{\line(1,0){1}}
%    \put(4,1){\line(1,0){1}}
    \put(3,1.04){\line(1,0){1}}
    \put(3,0.96){\line(1,0){1}}
 
    % label horizontal circles
    \put(1,0.3){\makebox(0,0.5)[b]{$1$}}
    \put(2,0.3){\makebox(0,0.5)[b]{$2$}}
    \put(3,0.3){\makebox(0,0.5)[b]{$3$}}
    \put(4,0.3){\makebox(0,0.5)[b]{$4$}}
%    \put(5,1.2){\makebox(0,0.4)[b]{$\ell-1$}}
%    \put(6, 0.6){\makebox(0,0.4)[b]{$\ell$}}
    
    \put(3.5,0.83){\makebox(0,0.4)[b]{$<$}}
 

    % put in the dots
%    \multiput(3.3,1)(0.2,0){3}{\circle*{0.01}}

%    % put in the diagonal lines
%    \put(5.07,0.93){\line(1,-1){0.56}}
%    \put(5.07,1.07){\line(1,1){0.56}}

    % put in the horizontal circles
    \multiput(1,1)(1,0){4}{\circle*{\darkrad}}
   \put(4,1){\circle{\lrad}}

%    % put in the diagonal circles
%    \put(5.7,0.3){\circle{0.2}}
%    \put(5.7,1.7){\circle{0.2}}
%    \put(5.7,-0.1){\makebox(0,0.4)[b]{$\ell^{-}$}}
%    \put(5.7,1.9){\makebox(0,0.4)[b]{$\ell^{+}$}}
\end{picture}
\]
or has more vertices circled.  That is, if we fix a set of roots of $\psp \As$ relative to a maximal $F$-torus containing a maximal $F$-split torus, the intersection of the kernels of the simple roots $\gamma_1, \gamma_2, \gamma_3$ contains a rank 1 split torus $S$, namely one corresponding to 
\[
\gch_1 + 2 \gch_2 + 3 \gch_3 + 4 \gch_4
\]
in the coroot lattice.  

Put $G$ for the group obtained by twisting $E_6$ by $\As$.  The image of $S$ in $G$ corresponds to the coroot
\[
2\ach_1 + 2 \ach_2 + 4 \ach_3 + 6 \ach_4 + 4 \ach_5 + 2 \ach_6
\]
by \eqref{coroots}, i.e., to $2\omega_2 - 2\alpha_2$.  The roots of $G$ that are orthogonal to $\omega_2 - \alpha_2$ are those whose $\alpha_2$ and $\alpha_4$-coordinates agree, i.e., the linear combinations of the simple roots $\alpha_1, \alpha_2 + \alpha_4, \alpha_3, \alpha_5, \alpha_6$.
The centralizer $Z$ of $S$ in $G$ is reductive, and its semisimple part $Z^{\mathrm{ss}}$ has root system of type $A_5$ consisting of the roots orthogonal to $\omega_2 - \alpha_2$.  Further, $Z^{\mathrm{ss}}$ contains the semisimple anisotropic kernel of $G$.  

Suppose for the sake of contradiction that $G$ is not split, hence  the semisimple anisotropic kernel of $G$ is of type $D_4$.  Over an algebraic closure of $F$, this gives an inclusion of split simply connected groups $D_4 \subset A_5$, hence $D_4$ has a faithful representation of dimension 6.  But this is impossible.  Therefore $G$ is split and the claim is proved.
\end{eg}

We have:
\begin{equation}
\parbox{4in}{\emph{The invariant $\D' \!: H^1(*, \psp_8) \ra H^3(*, \mmu2)$ is the invariant $\D$ given by Theorem \ref{inv}.}}
\end{equation}
To see this, we note that $\D'$ is an invariant, it satisfies property (1) of Theorem \ref{inv} by Example \ref{hyp.E6}, and it is nonzero by \eqref{rinv.comp}.  Therefore, $\D'$ equals $\D$ by \ref{par}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Square-central elements} \label{sq.central}

Let now $\As$ be a central simple $F$-algebra of degree 8 with symplectic involution.  In this section, we find a $\s$-symmetric, trace zero element $u \in A$ such that $u^2$ belongs to $\Fx$. 

\begin{lem} \label{u}
Suppose $A$ is division.  Then: 
\begin{enumerate}
\item There exists an element $s \in A$ such that $\s(s) = -s$ and $s^2 \in \Fx$.
\item Any such element $s$ can be embedded in a triquadratic extension $F(s_1, s_2, s_3) \subset A$ such that $\s(s_i) = -s_i$ and $s_i^2 \in \Fx$ for $i = 1, 2, 3$ and $s = s_1 s_2 s_3$.
\end{enumerate}
\end{lem}

The lemma sharpens --- and is proved using --- a result of Rowen from \cite{Rowen:CSA} (or see \cite[\S5.6]{Jac:fd}) that says that every $A$ as in the lemma contains a triquadratic extension of $F$. If $A$ is not division, there exist elements $s\in A$ such that $\sigma(s)=-s$, $\Trd_A(s)=0$, and $s^2\in F^\times$; part~(2) holds for any such element $s$. The proof is omitted, since this result is not used in the sequel.

We remark that part (1) of the lemma implies that $\s$ becomes hyperbolic over the quadratic extension $F(s)$ \cite[3.3]{BST}.

\begin{proof} We first prove (1), which is the crux.  Let $M$ be a triquadratic extension of $F$ contained in $A$; its existence is guaranteed by Rowen.  Note that because $\s$ is symplectic and $M$ is a maximal subfield, $M$ cannot consist of $\s$-symmetric elements \cite[4.13]{KMRT}.  Therefore, there is some $a \in M$ such that $a^2 \in \Fx$ and $a$ is not fixed by $\s$.  If $\s(a) = -a$, then we are done, so suppose not.

The elements $a$ and $\s(a)$ do not commute.  Indeed, otherwise we would have:
\[
0 = a^2 - \s(a)^2 = (a - \s(a)) (a + \s(a)).
\]
but neither of the terms in the product on the right are zero, contradicting the hypothesis that $A$ is division.

Let $Q$ be the $F$-subalgebra of $A$ generated by $a$ and $\s(a)$.  (Note that $a$ is not in $F$ because $\s$ does not fix $a$.) The center of $Q$ contains the field $Z$ generated over $F$ by the element $z := a \s(a) + \s(a) a$.  As a $Z$-algebra, $Q$ is generated by $a$ and 
\[
b := \s(a) - a\s(a) a^{-1} = 2\s(a) - za^{-1}.
\]
The element $b$ has square in $Z$ and anti-commutes with $a$, hence $Q$ is a quaternion algebra with center $Z$.  By the original definition of $Q$, we see that $Q$ is stable under $\s$, hence $\s$ restricts to an involution of the first kind on $Q$.  Further, $a$ has trace zero in $Q$, but $\s(a) \ne -a$, so $\s$ is orthogonal on $Q$.

We now consider the various cases for the dimension of $Z$ over $F$.  We remark that $Z$ cannot be 4-dimensional over $F$, because in that case the centralizer of $Z$ in $A$ is $Q$ itself, hence the restriction of $\s$ to $Q$ is symplectic, which is a contradiction. 

\smallskip
\emph{Case $Z = F$:} In case $Z = F$, the algebra $A$ is the tensor product of $Q$ with the centralizer $C_A(Q)$ of $Q$ in $A$, and $C_A(Q)$ is a central simple $F$-algebra of degree 4.  As $\s$ restricts to an orthogonal involution on $Q$, it restricts to a symplectic involution on $C_A(Q)$.  But any biquaternion division algebra with symplectic involution contains a quaternion subalgebra such that the restriction of the symplectic involution is the canonical involution on the quaternion subalgebra \cite[16.16]{KMRT}.  In particular, $C_A(Q)$ has a square-central and skew-symmetric element, so we are done in this case.

\smallskip
\emph{Case $[Z:F] = 2$:} By the Double Centralizer Theorem, the centralizer of $Z$ is a tensor product $Q \ot_Z Q'$ of $Q$ with another quaternion algebra $Q'$.
Further, the algebras $A \ot_F Z$ and $C_A(Z)$ are Brauer-equivalent (over $Z$).  Taking corestrictions of these two algebras, we find that
\[
\cores_{Z/F} [A \ot Z] = 2[A] = 0
\]
which equals
\[
\cores_{Z/F} [C_A(Z)] = \cores_{Z/F}[Q] + \cores_{Z/F}[Q'].
\]
But we can write $Q$ as a quaternion algebra over $Z$ with one of the slots equal to $a^2 \in \Fx$, so the corestriction of $Q$ has index at most 2 over $F$.  Necessarily the same holds for the index of the corestriction of $Q'$, so $Q'$ contains a trace zero element $s$ whose square lies in $F$ \cite[16.28]{KMRT}.

But the restriction of $\s$ to $C_A(Z)$ is symplectic, so the restriction of $\s$ to $Q'$ is also symplectic, hence $\s(s) = -s$, and we are done in this case.

\smallskip

The proof of (1) is now complete, and we use (1) to prove (2); let $s$ be the element provided by (1).  The centralizer $C_A(s)$ of $s$ in $A$ is a central simple $F[s]$-algebra of degree 4 on which $\s$ restricts to be a unitary involution.  

The discriminant algebra of $(C_A(s), \s)$ is $M_3(D)$ for some (possibly split) quaternion algebra $D$.  The canonical involution on it is orthogonal, i.e., is adjoint to a skew-hermitian form $h$ on a rank 3 $D$-module $V$.  We decompose $V$ as an orthogonal sum of rank 1 subspaces $V_1 \perp V_2 \perp V_3$ and observe that the tensor product $\ot_i C_0(V_i, h\vert_{V_i})$ of even Clifford algebras is an $F$-subalgebra of $C_0(h)$, which is in turn isomorphic to $(C_A(s), \s)$ by the exceptional isomorphism $\mathsf{A}_3 = \mathsf{D}_3$, cf.~\cite[15.24]{KMRT}.  Note that $C_0(V_i, h\vert_{V_i})$ is a quadratic extension $F[s_i]$ of $F$ where $s_i^2$ belongs to $\Fx$ and the canonical involution on the even Clifford algebra maps $s_i \mapsto -s_i$, which gives (2) except for the claim that $s = s_1 s_2 s_3$.  But the center 
of $C_A(s)$ is the quadratic extension $F[s]$, hence $s$ equals $\alpha s_1 s_2 s_3$ for some $\alpha \in \Fx$.  Replacing $s_1$ by $\alpha s_1$ completes the proof of (2).
\end{proof}

\begin{cor}\label{symm}
Let $\As$ be a central simple $F$-algebra of degree 8 with symplectic involution.  Then $A$ contains an element $u$ such that $\s(u) = u$, $u^2 \in \Fx$, and $\Trd_A(u) = 0$.
\end{cor}

\begin{proof}
If $A$ is division, one takes $u$ to be the product $s_1 s_2$, for elements $s_1 s_2$ as in Lemma \ref{u}(2).  The condition that $u$ has reduced trace zero follows from the fact that $A$ is division.

If $A$ is not division, we write $A$ as $M_2(B)$, endow $B$ with a symplectic involution $\tau$, and view $\s$ as adjoint to a $\tau$-hermitian form on a rank 2 $B$-vector space $V$.  We decompose $V$ as an orthogonal sum of two rank 1 spaces $V_1 \perp V_2$, and take $u \in A$ to be the element that fixes $V_1$ elementwise and acts as $-1$ on $V_2$.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The 10-dimensional quadratic form $q_u$} \label{tend}

We continue the notation of \S\ref{sq.central}: $\As$ is a central simple $F$-algebra of degree 8 with symplectic involution.

\begin{borel}{Definition of $q_u$} \label{qu.def}
Fix an element $u \in A$ as in Cor.~\ref{symm}.  We use it to construct a quadratic form $q_u$.

The centralizer $C_A(u)$ is semisimple of dimension $2^5$ and center $F[u]$, and we may consider the reduced
trace $\Trd_{C_{A}(u)}\colon C_{A}(u)\to F[u]$. Moreover, $\sigma$
restricts to a symplectic involution on $C_{A}(u)$. Consider the
$F$-vector space
\[
V_{u}=\{x\in C_{A}(u)\mid \sigma(x)=x\text{ and }\Trd_{C_{A}(u)}(x)=0\}.
\]
If $u^2\notin F^{\times2}$ (i.e., $F[u]$ is a field), then $C_{A}(u)$
is a central simple $F[u]$-algebra of degree~$4$, and it is proved
in \cite[\S15.C]{KMRT} that $x^2\in F[u]$ for $x\in V_{u}$. This
property also holds when $u^2\in F^{\times2}$ (see the proof of
Prop.~\ref{notfield.lem} below), so we may define a quadratic form
\[
q_{u}\colon V_{u}\to F
\]
as follows: let $t\colon F(u)\to F$ denote the $F$-linear map
such that $t(1)=0$ and $t(u)=1$, and set
\[
q_{u}(x) :=t(x^2)\qquad\text{for $x\in V_{u}$}.
\]
\end{borel}

Note that the dimension of $V_u$ --- the vector space underlying $q_u$ --- is 10.

\begin{prop}  \label{notfield.lem}
$q_u$ is in $I^3F$.
\end{prop}

As a consequence of the proposition, $q_u$ is isotropic (because its dimension is 10 \cite[XII.2.8]{Lam}), and so $q_u$ is Witt-equivalent to  a multiple of a 3-Pfister form.

\begin{proof}[Proof of Prop.~\ref{notfield.lem}]
\emph{(i):} Suppose first that  $u^2=\lambda^2$ for some $\lambda\in F^\times$, and let
     \[
     e_{1}=\frac12(1-u\lambda^{-1}),\qquad
     e_{2}=\frac12(1+u\lambda^{-1}).
     \]
     The elements $e_{1}$, $e_{2}$ are $\sigma$-symmetric idempotents
     such that $e_{1}+e_{2}=1$ and
     $\Trd_{A}(e_{1})=\Trd_{A}(e_{2})=4$. Therefore, letting
     $\tau_{1}$ (resp.\ $\tau_{2}$) denote the restriction of $\sigma$
     to $e_{1}Ae_{1}$ (resp.\ $e_{2}Ae_{2}$), we have
     \[
     (A,\sigma)=(e_{1}Ae_{1},\tau_{1})\boxplus(e_{2}Ae_{2},\tau_{2}),
     \]
     and $e_{1}Ae_{1}\simeq e_{2}Ae_{2}$ is of degree~$4$. Moreover,
     \[
     C_{A}(u)=C_{A}(e_{1})=C_{A}(e_{2})=(e_{1}Ae_{1})\times
     (e_{2}Ae_{2}).
     \]
     Letting $\Sym(\tau_{i})^0=\{x\in\Sym(e_{i}Ae_{i},\tau_{i})
     \mid\Trd_{e_{i}Ae_{i}}(x)=0\}$ ($i=1$, $2$), and denoting by
     $s_{i}\colon\Sym(\tau_{i})^0\to F$ the quadratic form
     $s_{i}(x)=x^2$ (see \cite[\S15.C]{KMRT}), we have
     \[
     V_{u}=\Sym(\tau_{1})^0\times\Sym(\tau_{2})^0
     \quad\text{and}\quad
     q_{u}=s_{1}-s_{2}.
     \]
     %It is then clear that $\dim V_{u}=10$. 
     By \cite[16.18]{KMRT}, we
     have $q_{u}\in I^3F$.
 
 \medskip
 \emph{(ii):} By \emph{(i)}, we may assume $u^2\notin
     F^{\times2}$. Then $C_{A}(u)$ is a central simple $F[u]$-algebra
     of degree~$4$ on which $\sigma$ restricts to a symplectic
     involution. By \cite[\S15.C]{KMRT}, $V_{u}$ is an $F[u]$-vector
     space of dimension~$5$, endowed with the quadratic form
     \[
     s_{\sigma}(x)=x^2\in F[u].
     \]
     By definition, $q_{u}$ is the Scharlau transfer:
     \[
     q_{u}=t_{*}(s_{\sigma}).
     \]
     The quadratic form $\langle1\rangle\perp-s_{\sigma}$ is an Albert
     form for $C_{A}(u)$ (see \cite[(16.8)]{KMRT}). In particular, it lies in
     $I^2F(u)$. By commutativity of the diagram
     \[
     \begin{CD}
         I^2F(u) @>{t_{*}}>> I^2F\\
         @V{e_{2}}VV @VV{e_{2}}V\\
         \Br(F(u)) @>{\cores}>> \Br(F)
     \end{CD}
     \]
     we get
     \[
     e_{2}(t_{*}(\langle1\rangle\perp-s_{\sigma}))=\cores_{F(u)/F}(C_{A}(u)).
     \]
     Since $C_{A}(u)$ is Brauer-equivalent to $A\otimes_{F}F(u)$, and
     the Brauer class of $A$ is $2$-torsion, it follows that
     \[
     e_{2}(t_{*}(\langle1\rangle\perp-s_{\sigma}))=0,
     \]
     hence $t_{*}(\langle1\rangle\perp-s_{\sigma})\in I^3F$ by a
     theorem of Pfister (note that this form has dimension $12$), or
     of course by Merkurjev's theorem. Since
     $t_{*}(\langle1\rangle)=0$, it follows that $q_{u}$ is in $I^3F$.
\end{proof}

\begin{eg} \label{hyp.eg}
Consider the special case where $\As$ is a tensor product 
\[
(Q, \binv) \ot (M_2(F), t) \ot \left(M_2(F), \Int \left( \begin{smallmatrix}
0&1 \\
1&0
\end{smallmatrix} \right)
  t \right)
\]
where $Q$ is a quaternion algebra and $t$ denotes the transpose.
For every $x \in F^\times$, the element $u := 1 \ot 1 \ot \left( \begin{smallmatrix}
0&x\\
1&0 \end{smallmatrix}\right)$ is $\s$-symmetric, has reduced trace zero, and has square $x$.  We claim that $q_u$ is hyperbolic.

To see this, we put:
\[
v_1 := 1 \ot  
\left( \begin{smallmatrix}
0&1 \\
1&0
\end{smallmatrix} \right) 
\ot 1
\quad \text{and} \quad 
v_2  := 1 \ot \left( \begin{smallmatrix}
1&0 \\
0&-1
\end{smallmatrix} \right)
\ot 1.
\]
The elements $v_1$ and $v_2$ belong to $V_u$ and are linearly independent over $F[u]$.  
For $c = c_1 v_1 + c_2 v_2$ with $c_1, c_2 \in F[u]$, we have $c^2 = c_1^2 + c_2^2$.
The transfer of this 2-dimensional quadratic form $\qform{1,1}$ is a direct sum of two hyperbolic planes, and we deduce that the anisotropic part of $q_u$ has dimension at most 6.  Since $q_u$ belongs to $I^3F$, we conclude that $q_u$ is hyperbolic.
\end{eg}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Arason invariant of $q_u$ equals $\D\As$}

We continue the notation of \S\ref{sq.central} and \S\ref{tend}: $\As$ is a central simple $F$-algebra of degree 8 with symplectic involution, and  $u$ is a $\s$-symmetric element of reduced trace zero such that $u^2 \in \Fx$.  The purpose of this section is to prove the following result, where $e_3$ denotes the Arason invariant $I^3F \ra H^3(F, \mmu2)$.

\begin{prop} \label{e3d}
$e_3(q_u) = \D\As$.
\end{prop}

Combining Propositions \ref{e3d} and \ref{notfield.lem}, we see that $\D\As$ is a symbol in $H^3(F, \mmu2)$, which recovers Prop.~\ref{E6}.

\newcommand{\ts}{\tilde{\s}}
\newcommand{\ttau}{{\tilde{\tau}}}
\newcommand{\tA}{\tilde{A}}

\begin{proof}[Proof of Prop.~\ref{e3d}]
\emph{(i):} 
Suppose first that $F(u)$ is not a field. Using the notation of the proof of Prop.~\ref{notfield.lem}\emph{(i)}, we have
\[
(A,\sigma)=(e_1Ae_1,\tau_1)\boxplus(e_2Ae_2,\tau_2)\quad\text{and}\quad q_u=s_1-s_2.
\]
By \cite[16.18]{KMRT}, we have $e_3(s_1-s_2)=\Delta_{\tau_1}(\tau_2)$. On the other hand, Prop.~\ref{dirsum.prop} shows that $\Delta(A,\sigma)=\Delta_{\tau_1}(\tau_2)$. The proposition follows.

For the rest of the proof, assume $F(u)$ is a field.

\emph{(ii):}
Suppose that $\tau$ is also a symplectic involution on $A$ that fixes $u$.  We write $q_u^\s, V_u^\s$ and $q_u^\tau, V_u^\tau$ for the 10-dimensional quadratic forms and their underlying vector spaces over $F$ deduced from $A, \s, u$ and $A, \tau, u$ respectively as in \ref{qu.def}; we will compute $e_3(q_u^\s - q_u^\tau)$.  

Write $s_u^\s, s_u^\tau$ for the quadratic forms $x \mapsto x^2$ on $V_u^\s, V_u^\tau$ respectively, and also $\ts, \ttau$ for the restrictions of $\s, \tau$ to $\tA := C_A(u)$.  By the exceptional isomorphism $\mathsf{B}_2 = \mathsf{C}_2$, the Arason invariant $e_3(s_u^\s - s_u^\tau)$ equals the relative invariant of symplectic involutions $\D_\ttau(\ts)$, cf.~\cite[16.18]{KMRT}.  Further, $\s = \Int(a) \tau$ for some $a \in A^\times$ that centralizes $u$, i.e., $a$ belongs to $\tA$.  We have:
\[
e_3(s^\s_u - s^\tau_u) = (\Nrp_{\tA}(a)) \cdot [\tA] \quad \in H^3(F(u), \mmu2).
\]
Taking the corestriction, we find:
\[
e_3(q^\s_u - q^\tau_u) = e_3(t_*(s^\s_u - s^\tau_u)) = \cores_{F(u)/F} \left( (\Nrp_{\tA}(a)) \cdot [\tA] \right).
\]
But $\tA$ is Brauer-equivalent to $A \ot_F F(u)$, so by the projection formula we have:
\[
e_3(q^\s_u - q^\tau_u) = \left( N_{F(u)/F} \Nrp_{\tA} (a) \right) \cdot [A] = (\Nrp_A(a)) \cdot [A] = \D_\tau(\s),
\]
where the middle equality is because $\Nrp$ is a square root of the reduced norm \cite[\S2]{KMRT}.

\smallskip
\emph{(iii):} We now prove the proposition.  If $A$ has index 1 or 2, then by Example \ref{hyp.eg} and Skolem-Noether, there is a hyperbolic symplectic involution $\tau$ on $A$ that fixes $u$, and the corresponding 10-dimensional quadratic form $q_u^\tau$ is hyperbolic.  By \emph{(ii)}, we find:
\[
e_3(q_u^\s) = \D_\tau(\s) = \D\As.
\]
If $A$ has index 4 or 8, we can apply the Albert quadric method to conclude that $e_3(q_u^\s)$ equals $\D\As$.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem \ref{decomp}: detecting decomposability}

We now prove Theorem \ref{decomp}, which asserts that $\D\As$ is zero if and only if $\As$ is completely decomposable.  The ``if" direction is  Example \ref{dec.eg}.  The ``only if" direction holds if $A$ is not division by
     \cite[Th.~8]{BMT}. Thus, we only need to consider the case
     where $A$ is division. In that case, the theorem follows from
     Prop.~\ref{e3d} and the following more precise result:

\begin{prop}
     \label{decomp.prop}
     Let $(A,\sigma)$ be a central division algebra of degree~$8$ with
     symplectic involution, and let $u\in A$ be such that $u^2\in
     F^{\times}$, $u\notin F$,  and $\sigma(u)=u$.
     \begin{enumerate}
     \item     There is a biquadratic extension $K/F$ such that $u\in
     K\subset\Sym(A,\sigma)$. 
     
     \item Let $K/F$ be an arbitrary extension of
     degree~$4$ such that $u\in K\subset\Sym(A,\sigma)$. If $q_{u}$ is
     hyperbolic, there is a decomposition of $(A,\sigma)$ into a
     tensor product of quaternion algebras with canonical involutions
     \[
     (A,\sigma)=(Q_{1},\gamma_{1})\otimes_{F}(Q_{2},\gamma_{2})\otimes_{F}
     (Q_{3},\gamma_{3})
     \]
     such that $K\subset Q_{1}\otimes Q_{2}$.
     \end{enumerate}
\end{prop}

\begin{proof}
     For (1), we may take $K=F(u,v)$ where $v\in V_{u}$ is
     any isotropic vector of $q_{u}$.

     For (2), suppose $K$ is an arbitrary quadratic extension of $F(u)$ in
     $\Sym(A,\sigma)$, and let $K=F(u)(x)$ where $x^2\in F(u)^\times$.
     
     Suppose $q_{u}$ is hyperbolic, and let $U\subset V_{u}$ be a
     maximal totally isotropic $F$-subspace. For $y\in U\cap uU$, we
     have $uy\in U$, hence also $(1+u)y\in U$. Since $y$ and $(1+u)y$
     are isotropic for $q_{u}$, we have
     \[
     y^2\in F\qquad\text{and}\qquad(1+2u+u^2)y^2\in F.
     \]
     This is impossible if $y^2\neq0$, hence $U\cap uU=\{0\}$ and
     therefore $V_{u}=U\oplus uU$. 
     
     Let $v$, $v'\in U$ be such that
     \[
     x=v+uv'.
     \]
     Substituting $xu$ for $x$ if necessary, we may assume $v\neq0$.
     If $v$ and $v'$ are linearly dependent over $F$, pick any nonzero
     $w\in v^\perp\cap U$. Otherwise, let
     \[
     w :=v'-vv'v^{-1}=2v'-v\frac{vv'+v'v}{v^2}.
     \]
     (Note that $w$ is nonzero by the second expression for $w$, because $v$ and $v'$ are linearly independent.)
     Since $v$, $v'$, and $v+v'$ are isotropic for $q_{u}$, we have
     $v^2$, ${v'}^2$, $vv'+v'v\in F$, hence $w\in U$ and $w^2\in
     F^\times$. Moreover, $w$ anticommutes with $v$, and the
     $F(u)$-span of $v$ and $w$ contains $x$. Thus, in both cases $v$
     and $w$ generate a quaternion $F$-algebra $H$ stable under
     $\sigma$, and $K\subset H\otimes_{F}F(u)$. The restriction of
     $\sigma$ to $H$ is an orthogonal involution since $\sigma(v)=v$
     and $\sigma(w)=w$. Therefore, the restriction of $\sigma$ to the
     centralizer $C_{A}(H)$ is symplectic. By \cite[(16.16)]{KMRT}, there is a
     decomposition into quaternion $F$-algebras
     \[
     (C_{A}(H),\sigma)=(H',\sigma')\otimes_{F}(Q_{3},\gamma_{3})
     \]
     where $u\in H'$, $\sigma'$ is an orthogonal involution and
     $\gamma_{3}$ is the canonical involution, hence
     \[
     (A,\sigma)=
     (H,\sigma\vert_{H})\otimes(H',\sigma')\otimes(Q_{3},\gamma_{3}).
     \]
     Now, we can find quaternion algebras $Q_{1}$, $Q_{2}$ with
     canonical involutions $\gamma_{1}$, $\gamma_{2}$ such that
     \[
     (H,\sigma\vert_{H})\otimes(H',\sigma')=
     (Q_{1},\gamma_{1})\otimes(Q_{2},\gamma_{2}).
     \]
     Since $K=F(u,x)\subset H\otimes H'=Q_{1}\otimes Q_{2}$, the proof
     is complete.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Application to the Pfister Factor Conjecture}

We now use our invariant to give an alternate proof
of the Pfister Factor Conjecture for tensor products of 6 quaternion algebras.

    Let $Q_{1}$, \ldots, $Q_{6}$ be quaternion $F$-algebras with
     canonical involutions $\gamma_{1}$, \ldots, $\gamma_{6}$. Assume
     $\bigotimes_{i=1}^6Q_{i}$ is split and let $\varphi$ be a $2^6$
     dimensional quadratic form representing~$1$ such that
     \[
     \gamma_{1}\otimes\cdots\otimes\gamma_{6}=\ad_{\varphi}.
     \]
     We have to show that $\varphi$ is a Pfister form.
     
     Put
           $A :=Q_{1}\otimes Q_{2}\otimes Q_{3}$ and $
           s = \g_1 \ot \g_2 \ot \g_3$.  The case where $A$ is not division is handled by \cite[Lemma 2]{SerhirTig}, so we assume below that $A$ is division.
           
                     Since
     $\bigotimes_{i=1}^6Q_{i}$ is split, we may identify
     $A$ also with $Q_{4}\otimes Q_{5}\otimes Q_{6}$. Let
     $\sigma :=\gamma_{1}\otimes\gamma_{2}\otimes\gamma_{3}$ and
     $\tau :=\gamma_{4}\otimes\gamma_{5}\otimes\gamma_{6}$, and let
     $s\in\Sym(A,\sigma)$ be such that
     \[
     \tau=\Int(s)\sigma,
     \]
     so that
     \[
     \ad_{\varphi}=\sigma\otimes\tau=\sigma\otimes(\Int(s)\sigma).
     \]
     
     Note that we may assume that $F(s)$ is a quadratic or biquadratic extension of $F$.
     Indeed, we may find an odd-degree extension $E/F$ such that
     $F(s)\otimes_{F}E$ is a $2$-extension of $E$. (Take for $E$ the
     cubic resolvent of $F(s)$ if $s$ has degree~$4$.) If we show that
     $\varphi_{E}$ is a Pfister form, then an easy argument (see
     \cite{Rost:desc}) shows that $\varphi$ also is a Pfister form. Therefore,
     substituting $E$ for $F$, we may henceforth assume $F(s)/F$ is quadratic or biquadratic. 
     
     Further, we may assume that $s$ belongs to $Q_1 \ot Q_2 \ot 1$.  Let $u\in F(s)\setminus F$ be a square-central
     element. Since $\sigma$ is decomposable, Example \ref{dec.eg} and
     Prop.~\ref{e3d}
     show that $q_{u}$ is hyperbolic, hence, by
     Proposition~\ref{decomp.prop}, we may find a decomposition
     \[
     (A,\sigma)=(Q'_{1},\gamma'_{1})\otimes(Q'_{2},\gamma'_{2})
     \otimes(Q'_{3},\gamma'_{3})
     \]
     such that $s\in Q'_{1}\otimes Q'_{2}$.  Replacing the $Q_i$ with $Q'_i$ for $i = 1, 2, 3$, we may assume that $s$ belongs to $Q_1 \ot Q_2$ so that
     \begin{equation}
	\label{4.eq}
	\tau=(\Int(s)(\gamma_{1}\otimes\gamma_{2}))\otimes\gamma_{3}.
     \end{equation}
     
     Writing $n_i$ for the norm form of $Q_i$, we have:
     \begin{equation} \label{big.eq}
     \ot_{i=1}^6 (Q_i, \g_i) =  (Q_1, \g_1) \ot \left[ (Q_2, \g_2) \ot (Q_1 \ot Q_2, \Int(s) (\g_1 \ot \g_2)) \right] \ot (Q_3, \g_3) \ot (Q_3, \g_3).
     \end{equation}
     The involution in brackets 
    is a symplectic involution on $Q_{2}\otimes Q_{1}\otimes
     Q_{2}$, so it is adjoint to a hermitian form over
     $(Q_{1},\gamma_{1})$. Using a diagonalization of this form, we
     get a $4$-dimensional quadratic form $\psi$ over $F$
     representing~$1$ such that
     \begin{equation}
	\label{2.eq}
	(Q_2, \g_2) \ot (Q_1 \ot Q_2, \Int(s)  (\g_1 \ot \g_2)) \cong (M_4(F), \ad_\psi) \ot (Q_1, \g_1).
	\end{equation}
%	\gamma_{2}\otimes(\Int(s)\circ(\gamma_{1}\otimes\gamma_{2}))=
%	\gamma_{1}\otimes\ad_{\psi}.
%     \end{equation}
     Comparing the value of $\Delta$ on each side, we obtain
     \[
     {[Q_{2}]}\cdot(\Nrd_{Q_{1}\otimes
     Q_{2}}(s))=[Q_{1}]\cdot(\disc\psi).
     \]
     Since
     \[
     ([Q_{1}]+[Q_{2}])\cdot(\Nrd_{Q_{1}\otimes
     Q_{2}}(s))=[Q_{1}\otimes Q_{2}]\cdot(\Nrd_{Q_{1}\otimes
     Q_{2}}(s))=0,
     \]
     it follows that
     \begin{equation}
	\label{3.eq}
	[Q_{1}]\cdot(\disc\psi) = [Q_{1}]\cdot(\Nrd_{Q_{1}\otimes
         Q_{2}}(s)).
     \end{equation}
  
  Plugging \eqref{2.eq} into \eqref{big.eq}, we find that $\varphi$ is isomorphic to $n_1 \ot n_3 \ot \psi$, and in particular belongs to $I^5F$.  We evaluate the invariant $e_5 \!: I^5F/I^6F \ra H^5(F, \mmu2)$ on $\varphi$ and find:
  \begin{equation} \label{e.eq}
  e_5(\varphi) = [Q_1] \cdot [Q_3] \cdot (\disc \psi).
  \end{equation}
  But $\tau$ is decomposable, so \eqref{4.eq} yields:
  \begin{equation} \label{e2.eq}
  [Q_3] \cdot (\Nrd_{Q_1 \ot Q_2}(s)) = \D(A, \tau) = 0.
  \end{equation}
  Combining equations \eqref{3.eq} through \eqref{e2.eq} gives that $e_5(\varphi)$ is zero.  That is, $\varphi$ belongs to $I^6F$. 
Therefore, $\varphi$ must be a
     $6$-fold Pfister form since $\dim\varphi=2^6$.  This completes the proof. $\hfill\qed$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{A question}

Given a central simple algebra $A$ of degree $4n$ (with $n \ge 2$) and exponent 2, one can define a corresponding element of the torsion of the Chow group $\CH^2(X_A)$ as follows, where $X_A$ denotes the Severi-Brauer variety of $A$.

Choose a symplectic involution $\s$ on $A$.  The discriminant $\D\As$ is in the kernel of the map
\[
\res \!: H^3(F, \mmu2) \ra H^3(F(X_A), \mmu2).
\]
Moreover, the class of $\D\As$ in 
\begin{equation} \label{CHt}
\ker(\res) / [A] \cdot H^1(F, \mmu2)
\end{equation}
depends only on $A$ and not on the choice of $\s$; we denote this class by $\D(A)$.

The quotient group \eqref{CHt} is identified with the torsion $\CH^2(X_A)_\tor$ in the Chow group $\CH^2(X_A)$, see \cite{Peyre:deg3}, and we may view $\D(A)$ as belonging to this group.  By a theorem of Karpenko \cite[5.3]{Karp:cod2}, this group is trivial if $A$ is decomposable of degree 8.  However, for the generic algebra $A^\gen$ of degree 8 and exponent 2, the group $\CH^2(X_{A^\gen})_\tor$ is $\Zm2$ by \cite[5.1]{Karp:cod2}.

\begin{ques} 
Is $\D(A^\gen)$ nonzero in $\CH^2(X_{A^\gen})_\tor$?  More generally, is $\D(A)$ nonzero for $A$ indecomposable of degree 8?
\end{ques}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
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\end{document}

