\documentclass{amsart}

\newcommand{\invo}{\overline{\rule{0.8em}{0mm}\rule{0pt}{5pt}}}

\DeclareMathOperator{\disc}{disc}
\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\ad}{ad}
\DeclareMathOperator{\End}{End}
\DeclareMathOperator{\Trd}{Trd}
\DeclareMathOperator{\Nrd}{Nrd}
\DeclareMathOperator{\Int}{Int}
\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\tr}{tr}

\newtheorem{lem}{Lemma}[section]
\newtheorem{prop}[lem]{Proposition}
\newtheorem{cor}[lem]{Corollary}
\newtheorem{thm}[lem]{Theorem}

\theoremstyle{remark}
\newtheorem*{remark}{Remark}

\title{Decomposable Quadratic Forms and Involutions}
\author{D.W. Lewis\and M.G. Mahmoudi\and J.-P. Tignol\and O. Villa}
%
\address{Department of Mathematics, University College Dublin, Belfield, Dublin 4, Ireland}
\email{david.lewis@ucd.ie}
\email{mohammad.mahmoudi@ucd.ie}
%
\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}
\urladdr{http://www.math.ucl.ac.be/membres/tignol}
%
\address{Department of Mathematics, Eidgen\"ossische Technische Hochschule, CH-8092 Z\"urich, Switzerland}
\email{oliver.villa@math.ethz.ch}
%
\thanks{The second and third authors were partially funded by IRCSET Basic Research Grant SC/02/265. The third author was partially supported by the National Fund for Scientific Research (Belgium). He gratefully acknowledges the hospitality of University College Dublin while this paper was written. The fourth author acknowledges the financial support provided through the European Community's Human Potential Programme under contract HPRN-CT-2002-00287, KTAGS}
%
\date{21 February 2005}

\begin{document}

\begin{abstract}
In his book on compositions of quadratic forms, Shapiro asks whether a quadratic form decomposes as a tensor product of quadratic forms when its adjoint involution decomposes as a tensor product of involutions on central simple algebras. We give a positive answer for quadratic forms defined over local or global fields and produce  counterexamples over fields of rational fractions in two variables over any formally real field.
\end{abstract}

\maketitle

\section{Introduction}

Every nondegenerate symmetric or skew-symmetric bilinear form $b$ on a finite-dimensional vector space $V$ induces an involution (i.e.\ an anti-automorphism of period~$2$) on the endomorphism algebra $\End V$. This involution, known as the \emph{adjoint involution of $b$} and denoted by $\ad_{b}$, is characterized by the following property:
\[
b(x,f(y))=b(\ad_{b}(f)(x),y)
\qquad\text{for $x$, $y\in V$ and $f\in\End V$}.
\]
If the form $b$ is the tensor product of two nondegenerate symmetric or skew-symmetric bilinear forms, i.e.
\[
V=V_{1}\otimes V_{2}\qquad\text{and}\qquad
b=b_{1}\otimes b_{2},
\]
then it is easy to see that the adjoint involution also decomposes,
\[
\End V=(\End V_{1})\otimes(\End V_{2})
\qquad\text{and}\qquad
\ad_{b}=\ad_{b_{1}}\otimes\ad_{b_{2}},
\]
(see \cite[Corollary~6.10]{Shap}), which we denote
\[
(\End V,\ad_{b})=(\End V_{1}\otimes\End V_{2},\ad_{b_{1}}\otimes\ad_{b_{2}}).
\]
In \cite[p.~201]{Shap}, Shapiro raises the following question: if the adjoint involution has a nontrivial decomposition of the form
\[
(\End V,\ad_{b})\simeq
(A_{1},\sigma_{1})\otimes(A_{2},\sigma_{2})
\]
for certain involutions $\sigma_{1}$, $\sigma_{2}$ on central simple algebras $A_{1}$, $A_{2}$, does it follow that $b$ decomposes nontrivially into a tensor product of bilinear forms? He showed that the answer is positive if $A_{1}$ and $A_{2}$ are split, see \cite[Corollary~6.10]{Shap} or Proposition~\ref{decsplit.prop} below.

In this paper, we give an affirmative answer to Shapiro's question over certain fields of characteristic different from~$2$, including local and global fields, and give examples where the answer is negative. Our examples are involution trace forms of central simple algebras of degree~$2m$, where $m$ is an odd integer, see Subsection~\ref{Shapironeg.subsec}.

Throughout the paper, the characteristic of the base field $F$ is assumed to be different from~$2$. Therefore, symmetric bilinear forms correspond bijectively to quadratic forms, and this correspondence is used to define the tensor product of quadratic forms. For any (nondegenerate) quadratic form $q$, we write $\ad_{q}$ for the adjoint involution of the polar form of $q$.

In Section~\ref{forms.sec}, we collect results on factorizations of quadratic forms. In particular, we determine the decomposable forms of dimension at most~$8$. The decomposition of involutions is discussed in Section~\ref{Shapiro.sec}.

\section{Decomposable quadratic forms}
\label{forms.sec}

All the quadratic forms considered in this paper are nondegenerate. A quadratic form $q$ over a field $F$ is called \emph{decomposable} if there is a factorization $q\simeq q_{1}q_{2}$ where $q_{1}$ and $q_{2}$ are quadratic forms with $\dim q_{1}$, $\dim q_{2}\geq2$. In particular, the hyperbolic forms of dimension $n\geq4$ are decomposable since they are isometric to
\[
\langle1,-1\rangle\otimes\bigl({\textstyle\frac{n}{2}}\langle1\rangle\bigr).
\]
\relax From the outset, note that the notion of decomposable form is not compatible with Witt equivalence, since every quadratic form $q$ is Witt-equivalent to the decomposable form $q\otimes\langle1,1,-1\rangle$.

After reviewing below some general properties of factorizations of quadratic forms, we consider factorizations of forms of small dimension, and determine the indecomposable forms over some special fields.

\subsection{General properties}

\begin{lem}
\label{prodevendim.lem}
If $q_{1}$, $q_{2}$ are even-dimensional quadratic forms, the discriminant and
Clifford invariant of the product $q_{1}q_{2}$ satisfy
\[
\disc(q_{1}q_{2})=1\qquad\text{and}\qquad
c(q_{1}q_{2})=(\disc q_{1},\disc q_{2})_{F}.
\]
\end{lem}

\begin{proof}
For $i=1$, $2$, let $d_{i}\in F^\times$ be such that $\disc
q_{i}=d_{i}F^{\times2}$. Since $q_{i}$ is even-dimensional, we have
$q_{i}\equiv\langle1,-d_{i}\rangle\bmod I^2F$, hence
\[
q_{1}q_{2}\equiv\langle1-d_{1}\rangle\langle1,-d_{2}\rangle
\bmod I^{3}F.
\]
Therefore, 
\[ 
\disc(q_{1}q_{2})=\disc(\langle1,-d_{1}\rangle
\langle1,-d_{2}\rangle)=1
\] 
and 
\[
c(q_{1}q_{2})=c(\langle1,-d_{1}\rangle
\langle1,-d_{2}\rangle)=(d_{1},d_{2})_{F}.
\]
\end{proof}

For any quadratic form $q$ over $F$, let $q_{a}$ denote the anisotropic kernel
of $q$, and let
\[
\dim_{a}q=\dim q_{a}.
\]

\begin{lem}
\label{trivdec.lem} If a quadratic form $q$ satisfies $\dim q=n_{1}n_{2}$ for
some integers $n_{1}$, $n_{2}$ subject to
\begin{gather}
\label{1.eq} n_{1}\equiv\dim_{a}q\bmod2,\qquad n_{2}\equiv1\bmod2,
\\
\label{2.eq} n_{1}\geq\dim_{a}q,\qquad n_{1},n_{2}\geq2,
\end{gather} then $q$ is decomposable.
\end{lem}

\begin{proof} 
Letting $k_{1}=\frac12(n_{1}-\dim_{a}q)$ and
$k_{2}=\frac12(n_{2}-1)$, we have
\[
q=(q_{a}\perp k_{1}\langle1,-1\rangle)\otimes(\langle1\rangle\perp
k_{2}\langle1,-1\rangle).
\]
\end{proof}

Besides these (essentially trivial) factorizations, a main source of
decompositions is the following well-known result (due to Pfister):

\begin{lem}
\label{quadext.lem}
An anisotropic quadratic form $q$ splits over a quadratic extension
$F(\sqrt{d})$ (i.e., becomes hyperbolic over $F(\sqrt{d})$) if and only if
$q\simeq\langle1,-d\rangle\otimes q'$
for some quadratic form $q'$.
\end{lem}

\begin{proof}
See \cite[Chapter~2, Theorem~5.2]{Scharlau}.
\end{proof}

\begin{cor}
\label{dim2p.cor}
Let $q$ be a decomposable quadratic form of dimension $\dim q=2p$ for some odd
prime number $p$. If $\disc q=1$, then $q$ is hyperbolic. If $\disc q\neq 1$,
then $\dim_{a}q\equiv2\bmod4$ and $q$ splits over $F(\sqrt{\disc q})$.
\end{cor}

\begin{proof}
Let $q=q_{1}q_{2}$ be a nontrivial decomposition of $q$. We may assume $\dim
q_{1}=2$ and $\dim q_{2}=p$. Comparing discriminants, we obtain $\disc q=\disc
q_{1}$ since $p$ is odd. Therefore, if $\disc q=1$, then $q_{1}$ is hyperbolic,
hence $q$ also is hyperbolic. If $\disc q=dF^{\times2}\neq1$, then $F(\sqrt{d})$
splits $q_{1}$, hence also $q$ and $q_{a}$. By Lemma~\ref{quadext.lem}, it
follows that $q_{a}\simeq\langle1,-d\rangle\otimes q'$ for some quadratic form
$q'$. If $\dim q'$ is even, then $\disc q_{a}=1$, a contradiction since $\disc
q=\disc q_{a}$. Therefore, $\dim_{a}q\equiv2\bmod4$.
\end{proof}

The results above yield necessary and sufficient conditions for the decomposability
of forms $q$ with $\dim_{a}q\leq 4$, as we now show.

\begin{prop}
\label{odddim.prop}
A quadratic form $q$ of odd dimension is decomposable whenever its dimension
has a nontrivial factorization $\dim q=n_{1}n_{2}$ with $n_{1}\geq\dim_{a}q$.
In particular, if $\dim_{a}q\leq3$, then $q$ is decomposable if and only if
$\dim q$ is neither $1$ nor a prime number.
\end{prop}

\begin{proof}
If $\dim q$ is odd, then $\dim_{a}q$ is odd and the conditions \eqref{1.eq}
hold for any factorization $\dim q=n_{1}n_{2}$. The proposition follows from
Lemma~\ref{trivdec.lem}.
\end{proof}

\begin{prop}
\label{dima2.prop}
If $\dim_{a}q=2$, then $q$ is indecomposable if and only if $\dim q$ is a
power of $2$.
\end{prop}

\begin{proof}
Note that $\dim q\equiv\dim_{a}q\bmod 2$, so the hypothesis implies $\dim q$
is even. If it has an odd prime factor $p$, let $n_{1}=(\dim q)/p$ and
$n_{2}=p$. Then $n_{1}$ is even, so $q$ is decomposable by
Lemma~\ref{trivdec.lem}. Therefore, $q$ is decomposable if $\dim q$ is not a
power of $2$.

Conversely, if $\dim q$ is a power of $2$, then every nontrivial factorization
of $q$ has the form $q=q_{1}q_{2}$ with $\dim q_{1}$, $\dim q_{2}$ even. Then
$\disc q=1$ by Lemma~\ref{prodevendim.lem}, which is impossible since
$\dim_{a}q=2$.
\end{proof}

\begin{prop}
\label{dima4.prop}
If $\dim_{a}q=4$, then $q$ is indecomposable if and only if one of the
following conditions holds:
\begin{enumerate}
\item[(a)]
$\dim q=2p$ for some odd prime $p$,
\item[(b)]
$\dim q$ is a power of $2$ and $\disc q\neq1$.
\end{enumerate}
\end{prop}

\begin{proof}
Since $\dim_{a}q=4$, it follows that $\dim q$ is even. In case~(a), 
$q$ is indecomposable by Corollary~\ref{dim2p.cor}.
If $\dim q$ is a power of $2$, then every nontrivial factorization of $q$ is
into a product of forms of even dimension. The existence of such a
factorization implies $\disc q=1$ by Lemma~\ref{prodevendim.lem}, hence $q$ is
indecomposable in case~(b).
Conversely, Lemma~\ref{trivdec.lem} shows that $q$ is decomposable whenever
$\dim q$ has a factorization $\dim q=n_{1}n_{2}$ with $n_{1}$ even, $n_{2}$
odd and $n_{1}\geq4$, $n_{2}\geq3$. Since $\dim q$ is even, such a
factorization exists unless $\dim q$ is a power of $2$ or of the form $2p$ for
some odd prime $p$. To prove that either (a) or (b) holds when $q$ is
indecomposable, it only remains to show that $q$ is decomposable if $\dim q$
is a power of $2$ and $\disc q=1$. In this case, $q_{a}$ has a diagonalization
\[
q_{a}=\langle a,b,c,d\rangle
\]
with $abcd\in F^{\times2}$, so 
\[
q_{a}=\langle a,b\rangle\otimes\langle1,ac\rangle
\]
and, for $k=\frac14(\dim q)-1$,
\[
q=\langle a,b\rangle\otimes\bigl(\langle1,ac\rangle\perp 
k\langle1,-1\rangle\bigr).
\]
\end{proof}

\subsection{Forms of small dimension}

We may also use Propositions~\ref{dima2.prop} and \ref{dima4.prop}, together
with Corollary~\ref{dim2p.cor}, to discuss the decomposability of forms of
dimension at most~$8$. Of course, quadratic forms whose dimension is $1$ or a
prime number are indecomposable. Therefore, we need to consider only forms of
dimension $4$, $6$ or $8$.

\begin{prop}
\label{dim4.prop}
A quadratic form $q$ of dimension~$4$ is indecomposable if and only if one of
the following conditions holds:
\begin{enumerate}
\item[(a)]
$\dim_{a}q=2$,
\item[(b)]
$\dim_{a}q=4$ and $\disc q\neq1$.
\end{enumerate}
\end{prop}

\begin{proof}
This readily follows from Propositions~\ref{dima2.prop} and \ref{dima4.prop}.
\end{proof}

\begin{prop}
A quadratic form $q$ of dimension~$6$ is indecomposable if and only if one of
the following conditions holds:
\begin{enumerate}
\item[(a)]
$\dim_{a}q=4$,
\item[(b)]
$\dim_{a}q=6$ and $\disc q=1$,
\item[(c)]
$\dim_{a}q=6$, $\disc q\neq1$ and $q$ is not split by $F(\sqrt{\disc q})$.
\end{enumerate}
\end{prop}

\begin{proof}
The form $q$ is decomposable if $\dim_{a}q=2$, by
Proposition~\ref{dima2.prop}, and indecomposable if $\dim_{a}q=4$, by
Proposition~\ref{dima4.prop}. Therefore, it only remains to consider the case
where $\dim_{a}q=6$, i.e., $q$ is anisotropic. Corollary~\ref{dim2p.cor} shows
that $q$ is indecomposable in cases~(b) and (c), and Lemma~\ref{quadext.lem}
proves that $q$ is decomposable if $\disc q\neq1$ and $q$ is split by
$F(\sqrt{\disc q})$.
\end{proof}

\begin{prop}
A quadratic form $q$ of dimension~$8$ is indecomposable if and only if one of
the following conditions holds:
\begin{enumerate}
\item[(a)]
$\dim_{a}q=2$,
\item[(b)]
$\dim_{a}q=4$ and $\disc q\neq1$,
\item[(c)]
$\dim_{a}q=6$,
\item[(d)]
$\dim_{a}q=8$ and $q$ is not split by any quadratic extension of $F$.
\end{enumerate}
\end{prop}

\begin{proof}
If $\dim_{a}q=2$, then $q$ is indecomposable by Proposition~\ref{dima2.prop}.
If $\dim_{a}q=4$, then $q$ is indecomposable if and only if $\disc q\neq1$, by
Proposition~\ref{dima4.prop}. Suppose next $\dim_{a}q=6$. If $\disc q\neq1$,
then $q$ is not decomposable by Lemma~\ref{prodevendim.lem}. If
$\disc q=1$, then $q$ is Witt-equivalent to an anisotropic Albert form, hence
its Clifford invariant has index~$4$, by \cite{BoI}. On the other hand, forms
which decompose into a product of forms of even dimension have a Clifford
invariant of index at most~$2$ by Lemma~\ref{prodevendim.lem}; therefore $q$ is
indecomposable.
Finally, the case where $\dim_{a}q=8$ follows from Lemma~\ref{quadext.lem}.
\end{proof}

\subsection{Forms over special fields}

Recall from \cite[Chapter~2,\S16]{Scharlau} that the $u$-invariant of a field $F$ is the supremum of the dimensions of anisotropic quadratic forms over $F$.
\begin{cor}
\label{u4.cor}
If $u(F)\leq4$, a quadratic form $q$ over $F$ is indecomposable if and only if
one of the following conditions holds:
\begin{enumerate}
\item[(a)]
$\dim q$ is a prime number,
\item[(b)]
$\dim_{a}q=2$ and $\dim q$ is a power of $2$,
\item[(c)]
$\dim_{a}q=4$ and $\dim q=2p$ for some odd prime $p$,
\item[(d)]
$\dim_{a}q=4$, $\dim q$ is a power of $2$ and $\disc q\neq1$.
\end{enumerate}
In particular, every quadratic form $q$ with $\dim q\equiv0\bmod4$ and $\disc
q=1$ is decomposable.
\end{cor}

\begin{proof}
The hypothesis $u(F)\leq 4$ implies $\dim_{a}q\leq4$. The corollary therefore
readily follows from Propositions~\ref{odddim.prop},
\ref{dima2.prop} and \ref{dima4.prop}. For the last statement, observe that
$\dim_{a}q\neq2$ if $\disc q=1$.
\end{proof}

The corollary applies for instance to $p$-adic fields or to non-formally real
number fields. Note that case~(d) does not arise for $p$-adic fields since
every anisotropic quadratic form of dimension~$4$ then has trivial
discriminant.

Suppose now $F$ is a real-closed field. Quadratic forms over $F$ are
classified by their dimension and signature, which are related by
\[
\vert\sgn q\vert\leq\dim q
\qquad\text{and}\qquad
\sgn q\equiv\dim q\bmod 2.
\]

\begin{prop}
A quadratic form $q$ over a real-closed field $F$ is decomposable if and only
if $\dim q$ and $\sgn q$ have factorizations of the form
\[
\dim q=n_{1}n_{2},\qquad\sgn q=s_{1}s_{2}
\]
with
\[
n_{i}\geq2,\qquad
\vert s_{i}\vert\leq n_{i}\quad\text{and}\quad
s_{i}\equiv n_{i}\bmod2\quad\text{for $i=1$, $2$}.
\]
\end{prop}

\begin{proof}
If $q=q_{1}q_{2}$ is a nontrivial decomposition, then
\[
\dim q=\dim q_{1}\dim q_{2}
\qquad\text{and}\qquad
\sgn q=\sgn q_{1}\sgn q_{2}
\]
are factorizations of the required form. Conversely, given factorizations as
in the statement of the proposition, one may find quadratic forms $q_{1}$,
$q_{2}$ over $F$ with $\dim q_{i}=n_{i}$ and $\sgn q_{i}=s_{i}$. The forms $q$
and $q_{1}q_{2}$ have the same dimension and signature, hence $q=q_{1}q_{2}$.
\end{proof}

\begin{cor}
\label{Shapreal.cor}
If a quadratic form $q$ over a real-closed field $F$ satisfies
\[
\dim q\equiv 0\bmod4
\qquad\text{and}\qquad
\disc q=1,
\]
then $q=\langle1,1\rangle\otimes q'$ for some quadratic form $q'$. In
particular, $q$ is decomposable.
\end{cor}

\begin{proof}
Since $\dim q$ is even and $\disc q=1$, it follows that $\sgn q\equiv
0\bmod4$. Therefore, we may apply the proposition above with $n_{1}=s_{1}=2$.
\end{proof}

The case of quadratic forms of dimension divisible by~$4$ and trivial
discriminant is particularly relevant for Shapiro's question, see
Section~\ref{Shapiro.sec}. The following result (of which
Corollary~\ref{Shapreal.cor} is a special case) gives further examples where
these forms have a nontrivial decomposition.

\begin{prop}
\label{linked.prop}
Suppose the field $F$ satisfies the following properties:
\begin{itemize}
\item
$F$ is linked, i.e., every tensor product of quaternion $F$-algebras is
Brauer-equivalent to a quaternion algebra,
\item
every quaternion $F$-algebra is split by a quadratic extension $K/F$ such that
$I^3K=0$.
\end{itemize}
Then every quadratic form $q$ over $F$ such that $\dim q\equiv0\bmod4$ and
$\disc q=1$ decomposes as $q=\langle1,a\rangle\otimes q'$ for some $a\in F^\times$
and some quadratic form $q'$. If $F$ is formally real, the element $a$ may be
chosen to be totally positive, i.e., $a>_{P}0$ for every ordering $P$ of $F$.
\end{prop}

\begin{proof}
Since $F$ is linked, the Clifford invariant $c(q)$ is Brauer-equivalent to a
quaternion $F$-algebra. Let $K/F$ be a quadratic extension with $I^3K=0$ which
splits $c(q)$. As was shown by Elman and Lam \cite[Theorem~3.11]{EL}, quadratic forms over $K$ are classified by their dimension, discriminant and Clifford invariant, hence $q$ splits over $K$. If $K=F(\sqrt{d})$, Lemma~\ref{quadext.lem}
yields
\[
q_{a}=\langle1,-d\rangle \otimes q_{0}
\]
for some quadratic form $q_{0}$. Since $\disc q=\disc q_{a}=1$, it follows
that $\dim q_{0}$ is even, hence $\dim q_{a}\equiv0\bmod4$. Therefore, the
Witt index $i_{W}(q)$ is even. Letting $i_{W}(q)=2m$, we have
\[
q=q_{a}\perp2m\langle1,-1\rangle=
\langle1,-d\rangle\otimes (q_{0}\perp m\langle1,-1\rangle).
\]
We obtain the required factorization with $a=-d$ and $q'=q_{0}\perp
m\langle1,-1\rangle$. Note that the condition $I^3K=0$ implies $K$ is not
formally real, hence $d<_{P}0$ for every ordering $P$ of $F$ (if any).
\end{proof}

Recall that the virtual cohomological $2$-dimension of a field $F$
is the cohomological $2$-dimension of $F(\sqrt{-1})$. Fields with virtual
cohomological $2$-dimension at most~$1$ satisfy the conditions in
Proposition~\ref{linked.prop} since $I^2F(\sqrt{-1})=0$. 

To obtain further examples, we use the Effective Diagonalization Property (ED), which was characterized in various ways by Prestel and Ware \cite{PW}. For instance, a field $F$ is ED if and only if for every valuation $v$ with formally real residue field $\overline{F}$, the group $2v(F^\times)$ has index at most $2$ in $v(F^\times)$, and $\overline{F}$ is euclidean when $2v(F^\times)\neq v(F^\times)$. This readily shows that number fields are ED.

\begin{lem}
\label{quatalg.lem}
Every quaternion division algebra over an ED field $F$ is split by a non-formally
real quadratic extension of $F$.
\end{lem}

\begin{proof}
Let $Q=(a,b)_{F}$ be a quaternion division $F$-algebra. Since $F$ is ED, Theorem~2 of \cite{PW} shows that the form $\langle1,a,b,-ab\rangle$ is almost isotropic, which means that for every diagonalization $\langle1,a,b,-ab\rangle\simeq\langle c_{1},c_{2},c_{3},c_{4}\rangle$, there is an integer $n$ such that $n\langle c_{1}\rangle\perp\langle c_{2},c_{3},c_{4}\rangle$ is isotropic. In particular, there is an integer $n$ such that $n\langle1\rangle\perp\langle a,b,-ab\rangle$ is isotropic, hence $\langle a,b,-ab\rangle$ represents an element $u\in F^\times$ such that $-u$ is a sum of squares. The algebra $Q$ is then split by $F(\sqrt{u})$, which is a non-formally real quadratic extension of $F$.
\end{proof}

\begin{cor}
\label{Shapnumber.cor}
If a quadratic form $q$ over a global field $F$ satisfies
\[
\dim q\equiv 0\bmod4
\qquad\text{and}\qquad
\disc q=1,
\]
then $q=\langle1,a\rangle\otimes q'$ for some totally positive $a\in F^\times$ and
some quadratic form $q'$. In particular, $q$ is decomposable.
\end{cor}

\begin{proof}
Global fields are linked and ED, and non-formally real global fields $K$ satisfy
$I^3K=0$. Therefore, the corollary follows from Proposition~\ref{linked.prop}
and Lemma~\ref{quatalg.lem}.
\end{proof}

\section{Decomposable adjoint involutions}
\label{Shapiro.sec}

Let $q$ be a quadratic form on an $F$-vector space $V$ and let $\ad_{q}$ be the
adjoint involution on $\End_{F}V$. In this section, we consider decompositions
of algebras with involution
\begin{equation}
\label{decompinvo.eq}
(\End_{F}V,\ad_{q})\simeq(A_{1},\sigma_{1})\otimes_{F}(A_{2},\sigma_{2})
\end{equation}
where $A_{1}$, $A_{2}$ are central simple $F$-algebras and $\sigma_{1}$,
$\sigma_{2}$ are involutions on $A_{1}$ and $A_{2}$ respectively. Since
$\End_{F}V$ represents the trivial element in the Brauer group of $F$, the
Brauer classes of $A_{1}$ and $A_{2}$ are opposite. However, $\sigma_{1}$
yields an isomorphism between $A_{1}$ and its opposite algebra, hence $A_{1}$ and
$A_{2}$ are also Brauer-equivalent. Since $\ad_{q}$ is an orthogonal involution,
the involutions $\sigma_{1}$ and $\sigma_{2}$ must be both orthogonal or both
symplectic, by \cite[Proposition~6.9]{Shap} or \cite[(2.23)]{BoI}.

In \cite[p.~201]{Shap}, Shapiro asks whether a nontrivial decomposition as in \eqref{decompinvo.eq} implies that $q$ is decomposable.

\subsection{Positive results on Shapiro's question}

The case where $A_{1}$ and $A_{2}$ are split and $\sigma_{1}$, $\sigma_{2}$
are orthogonal is discussed in Shapiro's book \cite{Shap}:

\begin{prop}
\label{decsplit.prop}
Let $q$, $q_{1}$, $q_{2}$ be quadratic forms on $F$-vector spaces $V$,
$V_{1}$, $V_{2}$. The following statements are equivalent:
\begin{enumerate}
\item[(a)]
$(\End_{F}V,\ad_{q})\simeq
(\End_{F}V_{1},\ad_{q_{1}})\otimes_{F}(\End_{F}V_{2},\ad_{q_{2}})$;
\item[(b)]
there exists $\lambda\in F^\times$ such that $q\simeq\langle\lambda\rangle
q_{1}q_{2}$.
\end{enumerate}
\end{prop}

\begin{proof}
See \cite[Corollary~6.10, p.~112]{Shap}.
\end{proof}

\begin{cor}
\label{Shapsplit.cor}
Suppose that one of the algebras $A_{1}$, $A_{2}$ in \eqref{decompinvo.eq} is
split. Then $A_{1}$ and $A_{2}$ are both split and the quadratic form $q$ is
decomposable if $\deg A_{1}$, $\deg A_{2}>1$. Moreover, if $\sigma_{1}$ and
$\sigma_{2}$ are symplectic, then $q$ is hyperbolic.
\end{cor}

\begin{proof}
The first part is clear, since $A_{1}$ and $A_{2}$ are Brauer-equivalent. Let
$A_{1}=\End_{F}V_{1}$ and $A_{2}=\End_{F}V_{2}$ for some $F$-vector spaces
$V_{1}$, $V_{2}$. If $\sigma_{1}$ and $\sigma_{2}$ are orthogonal, then there
exist quadratic forms $q_{1}$ on $V_{1}$ and $q_{2}$ on $V_{2}$ such that
$\sigma_{1}=\ad_{q_{1}}$ and $\sigma_{2}=\ad_{q_{2}}$. By
Proposition~\ref{decsplit.prop}, it follows from~\eqref{decompinvo.eq} that
$q=\langle\lambda\rangle q_{1}q_{2}$ for some $\lambda\in F^\times$, hence $q$
is decomposable if $\dim V_{1}$, $\dim V_{2}>1$, i.e.\ if $\deg A_{1}$, $\deg
A_{2}>1$.

If $\sigma_{1}$ and $\sigma_{2}$ are symplectic, then they are adjoint to some
skew-symmetric bilinear forms $b_{1}$ on $V_{1}$ and $b_{2}$ on $V_{2}$. By
\cite[Corollary~6.10]{Shap}, the polar form $b_{q}$ of $q$ satisfies
$b_{q}\simeq\lambda b_{1}\otimes b_{2}$ for some $\lambda\in F^\times$, hence
$q$ is hyperbolic. Since hyperbolic forms of dimension at least $4$ are
decomposable, the proof is complete.
\end{proof}

We now use results from Section~\ref{forms.sec} to derive
from~\eqref{decompinvo.eq} the decomposability of $q$ over special fields.

\begin{prop}
\label{Shapos.prop}
Suppose the adjoint involution $\ad_{q}$ decomposes as
in~\eqref{decompinvo.eq} with $\deg A_{1}$, $\deg A_{2}>1$. Then the quadratic
form $q$ is decomposable if the field $F$ satisfies any of the following
conditions:
\begin{itemize}
\item
the Brauer group of $F$ is trivial;
\item
$u(F)\leq4$;
\item
$F$ is real-closed;
\item
$F$ is a global field.
\end{itemize}
\end{prop}

\begin{proof}
Whenever $A_{1}$ and $A_{2}$ are split---which is certainly the case if the
Brauer group of $F$ is trivial---the decomposability of $q$ follows from
Corollary~\ref{Shapsplit.cor}. We may therefore restrict our discussion to the
case where $A_{1}$ and $A_{2}$ are not split. The Brauer class of $A_{1}$  and
$A_{2}$ then has order~$2$, hence $\deg A_{1}$ and $\deg A_{2}$ are even, and
therefore $\dim q\equiv0\bmod4$. Moreover, by \cite[Lemma~10.25]{Shap} or
\cite[(7.3)]{BoI}, $\disc q=1$. The decomposability of $q$ follows from
Corollary~\ref{u4.cor} if $u(F)\leq4$, from Corollary~\ref{Shapreal.cor} if
$F$ is real-closed, and from Corollary~\ref{Shapnumber.cor} if $F$ is global.
\end{proof}

The decomposability of $q$ can also be derived from certain types of
decompositions of $\ad_{q}$:

\begin{prop}
Suppose the adjoint involution $\ad_{q}$ decomposes as
in~\eqref{decompinvo.eq} with $\deg A_{1}$, $\deg A_{2}>1$. Then the quadratic
form $q$ is decomposable in each of the following cases:
\begin{enumerate}
\item[(a)]
$A_{1}$ and $A_{2}$ are Brauer-equivalent to a quaternion $F$-algebra $Q$ and
$\sigma_{1}$, $\sigma_{2}$ are symplectic;
\item[(b)]
$\deg A_{1}=2$ and $\sigma_{1}$, $\sigma_{2}$ are orthogonal;
\item[(c)]
$\deg A_{1}=4$ and $\sigma_{1}$, $\sigma_{2}$ are symplectic;
\item[(d)]
$\deg A_{1}=4$, $\sigma_{1}$, $\sigma_{2}$ are orthogonal and
$\disc\sigma_{1}=1$;
\item[(e)]
$\deg A_{1}=8$, $\sigma_{1}$, $\sigma_{2}$ are symplectic and $A_{1}$ is not a
division algebra;
\item[(f)]
$\deg A_{1}=8$, $\sigma_{1}$, $\sigma_{2}$ are orthogonal, $\disc\sigma_{1}=1$
and one of the factors of the Clifford algebra $C(A_{1},\sigma_{1})$ is split.
\end{enumerate}
Moreover, in case~\emph{(a)} the form $q$ is divisible by the norm form $n_{Q}$
of $Q$, and in case~\emph{(b)} it is divisible by the form
$\langle1,-\disc\sigma_{1}\rangle$.
\end{prop}

\begin{proof}
In case~(a), we may assume $Q$ is a division algebra, since the proposition
follows from Corollary~\ref{Shapsplit.cor} if $A_{1}$ and $A_{2}$ are split.
Let $\gamma$ be the conjugation involution on $Q$. Since $\sigma_{1}$ and
$\sigma_{2}$ are symplectic, it follows from \cite[Proposition, p.~202]{Shap}
or \cite[Proposition~3.4]{BST} that there exist quadratic forms $q'_{1}$,
$q'_{2}$ on $F$-vector spaces $V'_{1}$, $V'_{2}$ such that
\[
(A_{1},\sigma_{1})\simeq(Q,\gamma)\otimes_{F}(\End_{F}V'_{1},\ad_{q'_{1}}),
\qquad
(A_{2},\sigma_{2})\simeq(Q,\gamma)\otimes_{F}(\End_{F}V'_{2},\ad_{q'_{2}}).
\]
On the other hand, by \cite[(11.1)]{BoI}, we have
\[
(Q,\gamma)\otimes_{F}(Q,\gamma)\simeq(\End_{F}Q,\ad_{n_{Q}}).
\]
Therefore, \eqref{decompinvo.eq} implies
\[
(\End_{F}V,\ad_{q})\simeq
(\End_{F}Q,\ad_{n_{Q}})\otimes_{F}
(\End_{F}V'_{1},\ad_{q'_{1}})\otimes_{F}
(\End_{F}V'_{2},\ad_{q'_{2}}).
\]
It follows by \cite[Corollary~6.10]{Shap} that $q\simeq\langle\lambda\rangle
n_{Q}q'_{1}q'_{2}$ for some $\lambda\in F^\times$, hence $q$ is decomposable
if $\dim V'_{1}$ or $\dim V'_{2}$ is at least~$2$. If $\dim V'_{1}=\dim
V'_{2}=1$, then $q$ is a multiple of $n_{Q}$ hence it is decomposable by
Proposition~\ref{dim4.prop}.

In case~(b), the quaternion algebra with involution $(A_{1},\sigma_{1})$
becomes hyperbolic over $F(\sqrt{\disc\sigma_{1}})$, hence $q$ also becomes
hyperbolic over this extension. By Lemma~\ref{quadext.lem}, there exists a
quadratic form $q'$ such that
\[
q_{a}\simeq\langle1,-\disc\sigma_{1}\rangle\otimes q'.
\]
Since $\disc q=1$ by \cite[Lemma~10.25]{Shap} or \cite[(7.3)]{BoI}, the dimension of $q'$ is even. Therefore,
the Witt index $i_{W}(q)$ is even. Letting $i_{W}(q)=2m$, we have
\[
q=q_{a}\perp2m\langle1,-1\rangle\simeq
\langle1,-\disc\sigma_{1}\rangle\otimes(q'\perp m\langle1,-1\rangle).
\]
Cases (c)---(f) reduce to (a) or (b) since in each case there is a
decomposition of the form
\[
(A_{1},\sigma_{1})\simeq(Q,\sigma)\otimes_{F}(A',\sigma')
\]
for some quaternion $F$-algebra $Q$: see \cite[Proposition~10.21]{Shap} for
case~(c), \cite[Proposition~10.26]{Shap} for case~(d), \cite[Theorem~7]{BMT}
for case~(e) and \cite[(42.11)]{BoI} for case~(f).
\end{proof}

\subsection{Negative results on Shapiro's question}
\label{Shapironeg.subsec}

For any central simple algebra with orthogonal or symplectic involution
$(A,\sigma)$ over a field $F$, the involution trace form
$T_{\sigma}\colon\,A\to F$ is defined by
\[
T_{\sigma}=\Trd_{A}(\sigma(x)x),
\]
where $\Trd_{A}$ is the reduced trace. By \cite[(11.1)]{BoI}, the map
$\sigma_{*}\colon\,A\otimes_{F}A\to\End_{F}A$ defined by 
$\sigma_{*}(a\otimes b)(x)=ax\sigma(b)$ for $a$, $b$, $x\in A$ carries the involution
$\sigma\otimes\sigma$ on $A\otimes_{F}A$ to the adjoint involution of
$T_{\sigma}$, so
\[
(\End_{F}A,\ad_{T_{\sigma}})\simeq(A,\sigma)\otimes(A,\sigma).
\]
Therefore, $\ad_{T_{\sigma}}$ is decomposable if $\deg A>1$. However, we show
in this subsection that the quadratic form $T_{\sigma}$ is indecomposable for
certain orthogonal involutions $\sigma$ on central simple algebras $A$ of
degree $\deg A\equiv2\bmod4$, providing a negative solution to Shapiro's
question.

\begin{remark}
If the base field $F$ is a global field (or a real-closed field, or satisfies $u(F)\leq4$), Proposition~\ref{Shapos.prop} shows that the form $T_{\sigma}$ is decomposable when $\deg A>1$, since its adjoint involution is decomposable.
\end{remark}

We start our construction of examples where $T_{\sigma}$ is indecomposable with a lemma from elementary number theory.

\begin{lem}
\label{el.lem}
Suppose $m$, $k$, $l$, $r$, $s\geq1$ are integers such that
\begin{gather}
\label{el1.eq}
m^2=kl,\qquad (m-1)^2=rs,\\
\label{el2.eq}
k\geq r\geq1,\qquad l\geq s\geq1.
\end{gather}
If $m\equiv s\equiv1\bmod2$, then $l=1$.
\end{lem}

\begin{proof}
The hypotheses show that $l$ is odd. Let $k-r=a$ and $l-s=2b$. By \eqref{el2.eq}, we have $k>a\geq0$ and $l>2b\geq0$, hence, by the first equation in \eqref{el1.eq},
\begin{equation}
\label{el3.eq}
m^2>2ab\geq0.
\end{equation}
By definition of $a$ and $b$,
\[
2ab=(k-r)(l-s)=(k-r)l+k(l-s)+rs-kl,
\]
hence, by \eqref{el1.eq},
\[
2ab=al+2bk+(m-1)^2-m^2.
\]
It follows that
\[
al+2bk=2m-1+2ab,
\]
hence, using the first equation in \eqref{el1.eq},
\[
(al-2bk)^2=(al+2bk)^2-8abkl=
(2m-1+2ab)^2-8abm^2.
\]
This equation shows that
\[
(2m-1+2ab)^2\geq8abm^2,
\]
hence
\begin{equation}
\label{el4.eq}
(2ab-1)(2ab-1+4m-4m^2)\geq0.
\end{equation}

By \eqref{el3.eq} we have
\[
2ab-1+4m-4m^2<4m-1-3m^2,
\]
and the right side is easily seen to be nonpositive for all $m\geq1$. Therefore, \eqref{el4.eq} yields $2ab\leq1$, hence $a=0$ or $b=0$ since $a$ and $b$ are nonnegative integers. If $a=0$, then \eqref{el1.eq} yields
\[
m^2=kl\qquad\text{and}\qquad(m-1)^2=ks.
\]
Since $m^2$ and $(m-1)^2$ are relatively prime, it follows that $k=1$, hence $s=(m-1)^2\equiv0\bmod2$, a contradiction. Therefore, $b=0$ and \eqref{el1.eq} implies $l=1$ since $m^2$ and $(m-1)^2$ are relatively prime.
\end{proof}

\begin{thm}
\label{indecom.thm}
Let $Q$ be a quaternion algebra over a field $F$ and let $m$ be an odd integer. If an orthogonal involution $\sigma$ on the matrix algebra $M_{m}(Q)$ satisfies the following conditions:
\begin{enumerate}
\item[(a)]
the biquaternion algebra $Q\otimes_{F}(\disc\sigma,-1)_{F}$ is division,
\item[(b)]
there is an ordering $P$ on $F$ such that $\sgn_{P}T_{\sigma}=4(m-1)^2$,
\end{enumerate}
then $T_{\sigma}$ is an indecomposable form.
\end{thm}

\begin{proof}
Suppose $T_{\sigma}=q_{1}q_{2}$ for some quadratic forms $q_{1}$, $q_{2}$. We use condition~(a) to show that $\dim q_{1}$ and $\dim q_{2}$ cannot be both even, and condition~(b) to show that the decomposition is trivial if one of $\dim q_{1}$, $\dim q_{2}$ is odd.

Suppose first $\dim q_{1}\equiv\dim q_{2}\equiv 0\bmod2$. By Lemma~\ref{prodevendim.lem}, the Clifford invariant $c(q_{1}q_{2})$ has index at most $2$. On the other hand, the Hasse invariant of $T_{\sigma}$ was computed by Lewis in \cite{L} and Qu\'eguiner in \cite{Q}. From their computation, and from the relations between the Hasse and the Clifford invariant (see \cite[p.~81]{Scharlau}), it follows that $c(T_{\sigma})$ is represented by the biquaternion algebra $Q\otimes_{F}(\disc\sigma,-1)_{F}$, which has index~$4$ if condition~(a) holds. Therefore, this case leads to a contradiction.

Suppose next that $\dim q_{2}$ is odd. Since $\dim T_{\sigma}=4m^2$, the dimension of $q_{1}$ is a multiple of $4$. Let $\dim q_{1}=4k$ and $\dim q_{2}=l$, so
\[
m^2=kl,\qquad
k,l,m\geq1\text{ and }m\equiv1\bmod2.
\]
Comparing signatures at $P$, we also have
\[
\sgn_{P}T_{\sigma}=4(m-1)^2=\sgn_{P}q_{1}\sgn_{P}q_{2},
\]
with $\sgn_{P}q_{1}\equiv\dim q_{1}\equiv0\bmod2$ and $\sgn_{P}q_{2}\equiv\dim q_{2}\equiv 1\bmod2$. Therefore, $\sgn_{P}q_{1}$ is a multiple of $4$. Letting $\vert\sgn_{P}q_{1}\vert=4r$ and $\vert\sgn_{P}q_{2}\vert=s$, we have
\[
(m-1)^2=rs,\qquad r,s\geq1\text{ and }s\equiv1\bmod2.
\]
Lemma~\ref{el.lem} shows that $l=1$, so the decomposition $T_{\sigma}=q_{1}q_{2}$ is trivial.
\end{proof}

Let $\invo$ be the conjugation involution on a quaternion division $F$-algebra $Q$. For $x\in M_{m}(Q)$, we define $\overline{x}$ by letting $\invo$ act on each entry of $x$, so the map $x\mapsto\overline{x}^t$ is a symplectic involution on $M_{m}(Q)$. To obtain orthogonal involutions, consider diagonal matrices of pure quaternions
\[
\Delta=\diag(\alpha_{1},\ldots,\alpha_{m})
\]
with $\overline{\alpha_{i}}=-\alpha_{i}\neq0$ for $i=1$, \ldots, $m$, and define $\sigma_{\Delta}\colon\,M_{m}(Q)\to M_{m}(Q)$ by
\begin{equation}
\label{sigma.eq}
\sigma_{\Delta}(x)=\Delta^{-1}\cdot\overline{x}^t\cdot\Delta
\qquad\text{for $x\in M_{m}(Q)$}.
\end{equation}
Since $\overline{\Delta}^t=-\Delta$, the map $\sigma_{\Delta}$ is an orthogonal involution, see \cite[(2.7)]{BoI}. Moreover, since every orthogonal involution on the endomorphism algebra of a finite-dimensional $Q$-vector space is adjoint to some skew-hermitian form with respect to $\invo$ (see \cite[(4.2)]{BoI}), every orthogonal involution on $M_{m}(Q)$ is conjugate to an involution $\sigma_{\Delta}$ for some $\Delta$ as above.

\begin{prop}
\label{discsgn.prop}
The discriminant of $\sigma_{\Delta}$ and the signature of $T_{\sigma_{\Delta}}$ at an ordering $P$ of $F$ are determined as follows:
\begin{enumerate}
\item[(a)]
$\disc\sigma_{\Delta}=\alpha_{1}^2\ldots \alpha_{m}^2F^{\times2}\in F^\times/F^{\times2}$.
\item[(b)]
Let $F_{P}$ be a real closure of $F$ at $P$. If $Q$ is not split by $F_{P}$, then 
\[
\sgn_{P}T_{\sigma_{\Delta}}=0.
\] 
If $Q$ is split by $F_{P}$,
choose a pure quaternion $\beta\in Q^\times$ such that $\beta^2<_{P}0$, let $m_{1}$ be the number of indices $i$ such that $\alpha_{i}^2<_{P}0$ and $\Trd_{Q}(\alpha_{i}\beta)>_{P}0$, and let $m_{2}$ be the number of indices $i$ such that $\alpha_{i}^2<_{P}0$ and $\Trd_{Q}(\alpha_{i}\beta)<_{P}0$; then
\[
\sgn_{P}T_{\sigma_{\Delta}}=4(m_{1}-m_{2})^2.
\]
\end{enumerate}
\end{prop}

\begin{proof}
The arguments\footnote{%
  There is a misprint in the statement of \cite[Proposition~(7.3)(2)]{BoI}: the formula given there yields $\det(\Int(u)\circ\sigma)$ instead of $\disc(\Int(u)\circ\sigma)$.}
in the proof of \cite[(7.3)(2)]{BoI} show that
\[
\disc\sigma_{\Delta}=(-1)^m\Nrd_{M_{m}(Q)}(\Delta)F^{\times2}=
(-1)^m\Nrd_{Q}(\alpha_{1})\ldots\Nrd_{Q}(\alpha_{m})F^{\times2}.
\]
Part~(a) follows, since $\alpha_{i}^2=-\Nrd_{Q}(\alpha_{i})$.

If $Q$ is not split by $F_{P}$, then $\sgn_{P}T_{\sigma_{\Delta}}=0$ by \cite[Theorem~1]{LT} (or \cite[(11.11)]{BoI}). For the rest of the proof, suppose $Q$ is split by $F_{P}$.
For $x=(x_{ij})_{1\leq i,j\leq m}\in M_{m}(Q)$ we have $\sigma_{\Delta}(x)=(x'_{ij})_{1\leq i,j\leq m}$ where
\[
x'_{ij}=\alpha_{i}^{-1}\overline{x_{ji}}\alpha_{j}.
\]
Therefore, 
\[
T_{\sigma_{\Delta}}(x)=\sum_{i,j=1}^m\Trd_{Q}(\alpha_{i}^{-1}
\overline{x_{ji}}\alpha_{j}x_{ji}).
\]
Letting $T_{ij}\colon\,Q\to F$ denote the quadratic form
\[
T_{ij}(x)=\Trd_{Q}(\alpha_{i}^{-1}\overline{x}\alpha_{j}x),
\]
we thus have
\[
T_{\sigma_{\Delta}}=\bigoplus^\perp_{1\leq i,j\leq m}T_{ij},
\]
hence
\[
\sgn_{P}T_{\sigma_{\Delta}}=\sum_{i,j=1}^m\sgn_{P}T_{ij}.
\]
If $\alpha_{i}$ and $\alpha_{j}$ anticommute, then $1$ and $\alpha_{i}$ span a totally isotropic subspace of $Q$ for $T_{ij}$, so $T_{ij}$ is hyperbolic and $\sgn_{P}T_{ij}=0$. Note that in this case we have either $\alpha_{i}^2>_{P}0$ or $\alpha_{j}^2>_{P}0$ since $Q$ splits over $F_{P}$. If $\alpha_{i}$ and $\alpha_{j}$ do not anticommute, pick a nonzero quaternion $\alpha'$ which anticommutes with $\alpha_{i}$. Computation shows that $(1,\alpha_{i},\alpha_{j}\alpha',\alpha_{j}\alpha_{i}\alpha')$ is an orthogonal basis of $Q$ for $T_{ij}$, and that $T_{ij}$ has the following diagonalization in this basis:
\[
T_{ij}=\langle T(\alpha_{i}^{-1}\alpha_{j})\rangle
\langle1,-\alpha_{i}^2,-\alpha_{j}^2{\alpha'}^2,\alpha_{i}^2\alpha_{j}^2{\alpha'}^2\rangle.
\]
Note that ${\alpha'}^2>_{P}0$ if $\alpha_{i}^2<_{P}0$ since $Q$ splits over $F_{P}$. It follows that $\sgn_{P}T_{ij}=0$ unless $\alpha_{i}^2<_{P}0$ and $\alpha_{j}^2<_{P}0$; in this case, $\sgn_{P}T_{ij}=\pm4$. Therefore, letting
\[
t_{ij}=\begin{cases}
0&\text{if $\alpha_{i}^2>_{P}0$ or $\alpha_{j}^2>_{P}0$,}\\
1&\text{if $\alpha_{i}^2<_{P}0$, $\alpha_{j}^2<_{P}0$ and $\Trd_{Q}(\alpha_{i}^{-1}\alpha_{j})>_{P}0$,}\\
-1&\text{if $\alpha_{i}^2<_{P}0$, $\alpha_{j}^2<_{P}0$ and $\Trd_{Q}(\alpha_{i}^{-1}\alpha_{j})<_{P}0$,}
\end{cases}
\]
we have
\[
\sgn_{P}T_{\sigma_{\Delta}}=4\sum_{i,j=1}^m t_{ij}.
\]

Now, let $\beta\in Q^\times$ be a pure quaternion such that $\beta^2<_{P}0$. If a pure quaternion $\alpha\in Q$ satisfies $\Trd_{Q}(\alpha\beta)=0$, then $\beta$ anticommutes with $\alpha$, hence $\alpha^2>_{P}0$ since $Q$ splits over $F_{P}$. For $i=1$, \ldots, $m$, let
\[
s_{i}=\begin{cases}
0&\text{if $\alpha_{i}^2>_{P}0$,}\\
1&\text{if $\alpha_{i}^2<_{P}0$ and $\Trd_{Q}(\alpha_{i}\beta)>_{P}0$,}\\
-1&\text{if $\alpha_{i}^2<_{P}0$ and $\Trd_{Q}(\alpha_{i}\beta)<_{P}0$,}
\end{cases}
\]
so that
\[
\sum_{i=1}^ms_{i}=m_{1}-m_{2}.
\]
To complete the proof, we have to show
\[
\sum_{i,j=1}^mt_{ij}=\Bigl(\sum_{i=1}^ms_{i}\Bigr)^2.
\]
It clearly suffices to prove $t_{ij}=s_{i}s_{j}$ for $i$, $j=1$, \ldots, $m$, which amounts to showing that if $\alpha_{i}^2<_{P}0$ and $\alpha_{j}^2<_{P}0$, then $\Trd_{Q}(\alpha_{i}^{-1}\alpha_{j})>_{P}0$ if and only if $\Trd_{Q}(\alpha_{i}\beta)$ and $\Trd_{Q}(\alpha_{j}\beta)$ have the same sign. Since $\alpha_{i}^2=-\Nrd_{Q}(\alpha_{i})$ and $\Trd_{Q}(\alpha_{i}\alpha_{j})=\alpha_{i}^2\Trd_{Q}(\alpha_{i}^{-1}
\alpha_{j})$, it is equivalent to show that if $\Nrd_{Q}(\alpha_{i})>_{P}0$ and $\Nrd_{Q}(\alpha_{j})>_{P}0$, then 
\[
\Trd_{Q}(\alpha_{i}\beta)\Trd_{Q}(\alpha_{j}\beta)
\Trd_{Q}(\alpha_{i}\alpha_{j})<_{P}0.
\]
To check this statement, we may extend scalars to a real closure of $F$ at $P$. We may therefore substitute for $Q$ a matrix algebra over a real-closed field.
The following lemma completes the proof:
\renewcommand{\qed}{}
\end{proof}

\begin{lem}
Let $K$ be an ordered field and let  $\alpha$, $\alpha'$, $\beta\in M_{2}(K)$ be matrices with trace zero. If  $\det\alpha$, $\det\alpha'$, $\det\beta>0$, then $\tr(\alpha\alpha')\tr(\alpha\beta)\tr(\alpha'\beta)<0$.
\end{lem}

\begin{proof}
Since $\tr(\alpha)=\tr(\beta)=0$, 
\[
\tr(\alpha\beta)=\alpha\beta+\beta\alpha.
\]
Therefore, letting $\alpha_{0}=\alpha-\beta\alpha \beta^{-1}$, we have
\begin{equation}
\label{t1.eq}
2\alpha=\tr(\alpha\beta)\beta^{-1}+\alpha_{0}
\qquad\text{and}\qquad
\tr(\alpha_{0}\beta^{-1})=0.
\end{equation}
The condition $\det \alpha>0$ then yields
\begin{equation}
\label{t2.eq}
\tr(\alpha\beta)^2\det \beta^{-1}+\det \alpha_{0}>0.
\end{equation}
Similarly, letting $\alpha'_{0}=\alpha'-\beta\alpha'\beta^{-1}$,
\begin{equation}
\label{t3.eq}
2\alpha'=\tr(\alpha'\beta)\beta^{-1}+\alpha'_{0},
\qquad
\tr(\alpha'_{0}\beta^{-1})=0,
\end{equation}
and
\begin{equation}
\label{t4.eq}
\tr(\alpha'\beta)^2\det \beta^{-1}+\det \alpha'_{0}>0.
\end{equation}
By \eqref{t1.eq} and \eqref{t3.eq},
\[
4\tr(\alpha\alpha')=\tr(\alpha_{0}\alpha'_{0})+
\tr(\alpha\beta)\tr(\alpha'\beta)\tr(\beta^{-2}).
\]
Since $\tr(\beta)=0$, we have $\beta^2=-\det\beta$, hence the preceding equation, together with \eqref{t2.eq} and \eqref{t4.eq}, yields
\begin{align*}
4\tr(\alpha\alpha')\tr(\alpha\beta)\tr(\alpha'\beta) & =
\tr(\alpha_{0}\alpha'_{0})\tr(\alpha\beta)\tr(\alpha'\beta)-2\tr(\alpha\beta)^2\tr(\alpha'\beta)^2\det \beta^{-1}\\
& < \tr(\alpha_{0}\alpha'_{0})\tr(\alpha\beta)\tr(\alpha'\beta)
+\tr(\alpha\beta)^2\det\alpha'_{0}\\
&\quad+\tr(\alpha'\beta)^2\det\alpha_{0}.
\end{align*}
Since $\tr(\alpha_{0})=\tr(\alpha'_{0})=0$, the right side of the last inequality is
\[
\det(\tr(\alpha\beta)\alpha'_{0}-\tr(\alpha'\beta)\alpha_{0}).
\]
To conclude the proof, it suffices to observe that the determinant of any matrix which anticommutes with $\beta$ is negative, otherwise $M_{2}(K)$ would be generated by two anticommuting matrices with negative square, i.e. $(-1,-1)_{K}\simeq M_{2}(K)$.
\end{proof}

\begin{remark}
It follows from \cite[Theorem~1]{LT} that the signature of the involution trace form $T_{\sigma}$ is a square for any orthogonal or symplectic involution $\sigma$ on a central simple algebra $A$. In the terminology of \cite{LT} (see also \cite[(11.10)]{BoI}), Proposition~\ref{discsgn.prop}(b) states that if $Q$ is split by $F_{P}$, then
\[
\sgn_{P}\sigma_{\Delta}=\vert m_{1}-m_{2}\vert.
\]
\end{remark}

We proceed to give examples satisfying the hypotheses of Theorem~\ref{indecom.thm}, for any odd integer $m\geq3$.

Let $F_{0}$ be an arbitrary ordered field and let $F=F_{0}(x,y)$, the field of rational fractions in two indeterminates over $F_{0}$. The following classical construction (see \cite[p.~75]{Prestel}) extends the ordering on $F_{0}$ to an ordering $P$ on $F$ such that $x<_{P}0$ and $0<_{P}y<_{P}1$: consider the $(x,y)$-adic valuation on $F$ defined on $F_{0}[x,y]\setminus\{0\}$ by
\[
v\Bigl(\sum_{i,j}a_{ij}x^{i}y^{j}\Bigr)=
\min\{(i,j)\mid a_{ij}\neq0\},
\]
where the minimum is for the lexicographic order on $\mathbb{Z}^2$. If $f\in F^\times$ satisfies $v(f)=(m,n)$, then $v((-x)^{-m}y^{-n}f)=0$ and we may consider the residue $\overline{(-x)^{-m}y^{-n}f}\in F_{0}$. Set
\begin{equation}
\label{ordering.eq}
f>_{P}0
\qquad\text{if and only if}\qquad
\overline{(-x)^{-m}y^{-n}f}>0\text{ in $F_{0}$}.
\end{equation}
Then $-x>_{P}0$, $y>_{P}0$ and $1-y>_{P}0$, so $x<_{P}0$ and $0<_{P}y<_{P}1$.

Consider the quaternion $F$-algebra $Q=(x,y)_{F}$. For any odd integer $m\geq3$, consider the following pure quaternions in $Q$:
\[
\alpha_{1}=i+k,\quad
\alpha_{2}=j,\quad
\alpha_{3}=\cdots=\alpha_{m}=i,
\]
and let $\sigma_{\Delta}$ be the involution on $M_{m}(Q)$ defined as in \eqref{sigma.eq}. Since $y>_{P}0$, the algebra $Q$ is split by any real closure of $F$ at $P$. To compute $\sgn_{P}T_{\sigma_{\Delta}}$, we use Proposition~\ref{discsgn.prop} with $\beta=i$. Since 
\[
\alpha_{1}^2=x(1-y)<_{P}0,\quad
\alpha_{2}^2=y>_{P}0\quad\text{and}\quad
\Trd_{Q}(i(i+k))=\Trd_{Q}(i^2)=2x<_{P}0,
\]
we have, in the notation of Proposition~\ref{discsgn.prop}, $m_{1}=0$ and $m_{2}=m-1$, so
\[
\sgn_{P}T_{\sigma_{\Delta}}=4(m-1)^2.
\]

By Proposition~\ref{discsgn.prop}, we also have
\[
\disc\sigma_{\Delta}=\alpha_{1}^2\ldots \alpha_{m}^2F^{\times2}=
(x-xy)yx^{m-2}F^{\times2}=(1-y)yF^{\times2}.
\]
To see that the hypotheses of Theorem~\ref{indecom.thm} hold, it remains to prove that the tensor product
\[
(x,y)_{F}\otimes_{F}(y(1-y),-1)_{F}
\]
is a division algebra, or, equivalently by a theorem of Albert (see \cite[(16.5)]{BoI}), that an associated Albert form such as
\[
q=\langle x,y,-xy,y(y-1),y(y-1),1\rangle
\]
is anisotropic. Since
\[
q=\langle1,y,y(y-1),y(y-1)\rangle\perp\langle x\rangle
\langle1,-y\rangle,
\]
a degree argument (see for instance \cite[Lemma~1.4(i)]{Hof}) shows that $q$ is isotropic over $F$ if and only if one of the forms $\langle 1,y,y(y-1),y(y-1)\rangle$, $\langle1,-y\rangle$ is isotropic over $F_{0}(y)$. The latter is clearly anisotropic since $y\notin F_{0}(y)^{\times2}$. Using the $(y-1)$-adic valuation on $F_{0}(y)$, one may construct as above (see \eqref{ordering.eq}) an ordering $P_{0}$ on $F_{0}(y)$ such that $y>_{P_{0}}1$. Then
\[
\sgn_{P_{0}}\langle1,y,y(y-1),y(y-1)\rangle=4,
\]
so $\langle1,y,y(y-1),y(y-1)\rangle$ is anisotropic over $F_{0}(y)$.

Thus, Theorem~\ref{indecom.thm} shows that the quadratic form $T_{\sigma_{\Delta}}$ is indecomposable, even though its adjoint involution is decomposable, as it is isomorphic to $\sigma_{\Delta}\otimes\sigma_{\Delta}$.

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