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\title{NEW VALUES FOR THE LEVEL AND SUBLEVEL OF COMPOSITION ALGEBRAS}
\author{James O'Shea}
\date{}
\begin{document}


\maketitle

\begin{abstract} \noindent Constructions of quaternion and octonion algebras,
suggested to have new level and sublevel values, are proposed and
justified. In particular, octonion algebras of level and sublevel
$6$ and $7$ are constructed. In addition, Hoffmann's proof of the
existence of infinitely many new values for the level of a
quaternion algebra is generalised and adapted.
\end{abstract}
\medskip

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Is the above ok?
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Should say ``explicit constructions'' in abstract re. Thm~\ref{n7}
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%Hoffmann's method of proving the existence of infinitely many..

%The conjecture that this family contains members of
%level and sublevel $n$ for all values of $n$ is proposed. Octonion algebras of level $6$ and $7$ are
%constructed. These algebras are of sublevel $6$ and $7$ also.

\section{Introduction}

In \cite{LEW}, Lewis constructed quaternion algebras of level
$2^k$ and $2^k+1$, both of sublevel $2^k$, for all integers $k\geq
0$. Laghribi and Mammone recovered these values as the level of a
quaternion algebra using function field techniques (see
\cite{LM}). Pumpl\"un employed their methodology in \cite{P} to
construct octonion algebras of level $2^k$ and $2^k+1$. By
constructing a quaternion algebra of sublevel $3$, Kr\"uskemper
and Wadsworth produced the first example of a quaternion algebra
whose sublevel was not a power of $2$ (see \cite{KW}). This
construction was subsequently employed to prove the existence of
an octonion algebra of sublevel $3$, and indeed one of sublevel
$5$ was also produced (see \cite{O'S}). Heretofore, these remained
the only known values for the level and sublevel of quaternion and
octonion algebras.

In Section $3.1$, we introduce certain quaternion and octonion
algebras, denoted $Q(n)$ and $\sO(n)$ respectively, which we
conjecture to be of level and sublevel $n$, for all $n$.

In Section $3.1.1$, we proceed to precisely determine the level
and sublevel of these algebras for certain values of $n$. In
particular, we produce the first example of a composition algebra
whose known level is not equal to $2^k$ or $2^k+1$ for some $k$.

Section $3.1.2$ sees us outline the argument employed by Hoffmann
\cite{H3} to bound the level of $Q(n)$, proving the existence of
infinitely many values for the level of $Q(n)$ which are not of
the form $2^k$ or $2^k+1$ for some $k$. We extend this argument to
$\sO(n)$, and refine these bounds for particular $n$-values.

We conclude with Section $3.2$, wherein we introduce alternative
constructions of quaternion and octonion algebras, which we
suggest to be of level and sublevel $l\cdot 2^k$. Some merits of
these constructions are outlined.
\medskip

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Need to make mention of Section 3.3 now!
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\section{Preliminaries}

Let $F$ denote a field of characteristic $\ne 2$, $F_0$ a formally
real field and $C$ a unital, not necessarily associative,
$F$-algebra.

A map $*$ is called an \emph{involution} on $C$ if it is an
anti-automorphism of period $2$. We have $C=\Sym(C,*)\oplus
\Skew(C,*)$, with $\Sym(C,*)=\{x\in C| x^* = x\}$ and
$\Skew(C,*)=\{x\in C| x^* = -x\}$. An involution $*$ is called
\emph{scalar} if $x^*x \in F$ and $x^* + x\in F$ for all $x\in C$.
For an algebra $C$ with scalar involution $*$, we call
$t_C(x)=x+x^*$ the \emph{trace} of $C$ and the quadratic form
$T_C:C \to F,$ $T_C(x)=t_C(x^2)$ the \emph{trace form}.

The \emph{Cayley-Dickson doubling process} is an algorithm for
constructing new algebras with scalar involution from old ones.
Applying the process to an algebra $C$ with scalar involution $*$,
together with a chosen scalar $\mu$ $\in$ $F^\x$, produces a new
algebra, $\Cay (C,\mu)$, the \emph{Cayley-Dickson double} of $C$,
whose scalar involution we will also denote by $*$. We note that
$\Cay (C,\mu)$ is the $F$-module $C \times C$, with multiplication
defined by $(u,v)(u',v')=(uu'+\mu v'^*v,v'u+vu'^*)$ and involution
given by $(u,v)^*=(u^*,-v)$, for $u,u',v,v' \in C$.

An algebra $C$ is a $\emph{composition algebra}$ if there exists a
nondegenerate quadratic form $q$ on $C$ which allows composition,
that is $q(x\hspace{.05cm}y) = q(x)\hspace{.05cm}q(y)$ for all
$x, y$ $\in$ $C$. %In fact, $q$ can be taken to be the norm form of
%$C$, denoted $N_C$, with the standard (scalar) involution on $C$
%thus linked to $N_C$ as follows: $N_C(x) =
%\bar{x}\hspace{.05cm}x,$ $\forall$ $x$ $\in$ $C$.
Composition algebras are of rank $1$, $2$, $4$ or $8$. The
composition algebras of rank $2$ are the quadratic \'etale
$F$-algebras; the composition algebras of rank $4$ are the
(non-commutative) quaternion algebras and those of rank $8$ are
the (non-commutative and non-associative) octonion algebras. For
$a,b\in F^\times$, the \emph{quaternion algebra} $Q=\quat a b$
over $F$ is a 4-dimensional $F$-vector space with basis
$\{1,i,j,k\}$, satisfying $i^2=a, j^2=b$ and $ij=-ji=k$. For
$a,b,c\in F^\times$, the \emph{octonion algebra} $\oct a b c$ over
$F$ is defined as $\oct a b c :=\Cay\left(\quat a b,c\right)$. We
note that $\Cay\left(\quat a b,c\right)=\quat a b \oplus \quat a b
e$ is an $8$-dimensional $F$-vector space with basis
$\{1,i,j,k,e,ie,je,ke\}$, satisfying $i^2=a, j^2=b$ and $e^2=c$.
Applying the Cayley-Dickson doubling process to a composition
algebra $C$ over $F$, yields another composition algebra (if the
dimension of the new algebra is at most $8$), or alternatively
what is known as a \emph{generalised Cayley-Dickson algebra}.

%The following was in the above, a la las. However, we have not defined the norm form here, so I cut it out:
%Since two composition algebras are isomorphic if and only if
%their norm forms are isometric, changing the order of the slots
%$a,b,c$ in $\oct a b c$ yields an isomorphic octonion algebra.

For $C=\quat a b$, we have $T_C\simeq 2\langle 1, a, b,
-ab\rangle$, whereas for $C=\oct a b c$, $T_C\simeq 2\langle 1, a,
b, -ab, c, -ac, -bc, abc \rangle$. Since we are concerned with the
isotropy of trace forms, we may disregard this scalar factor of
$2$. We define the \emph{pure trace form} of a composition algebra
$C$, denoted $T_P$, via the following relation:
$T_C=\langle1\rangle \perp T_P$.

For $C$ a division algebra, we define the \emph{level} of $C$,
denoted $s(C)$, as the least integer $n$ such that $-1$ is a sum
of $n$ squares in $C$. If no such integer exists, we say that
$s(C)=\infty$. The \emph{sublevel} of $C$, denoted
$\underline{s}(C)$, is the least integer $n$ for which $0$ is a
sum of $n+1$ squares of elements in $C$. If $0$ is not expressible
in this manner, we say that $\underline{s}(C)=\infty$. Note that
$\underline{s}(C)\leq s(C)$.

In \cite{Pfister}, Pfister showed that the level of a field, if
finite, is a power of $2$, and moreover that any prescribed power
of $2$ may be realised as the level of a field. This
classification extends to the case where $C$ is a quadratic
\'etale $F$-algebra. A composition algebra is $\emph{split}$ if it
contains a composition subalgebra which is isomorphic to $F \oplus
F$, which is the case if and only if $C$ contains zero divisors.
Split quaternion and octonion algebras have level $1$ (a
consequence of a split quaternion algebra being isomorphic to
$M_2F$). Hence, in order to complete the classification of the
level of composition algebras, we consider quaternion and octonion
algebras which are division.

For $\varphi$ a regular $n$-dimensional quadratic form over $F$,
we can consider $\varphi$ as a homogeneous polynomial of degree
$2$, that is $\varphi(X)=\varphi(X_1,\ldots ,X_n)=\sum_{i,j}
\alpha_{ij}X_iX_j$, where $\alpha_{ij}\in F^\times$.  For $n>1$
and $\varphi\not\simeq\< 1,-1\>$, the \emph{function field} of
$\varphi$, denoted $F(\varphi)$, is defined to be the quotient
field of the integral domain $$F[X]/(\varphi(X))=F[X_1,\ldots
,X_n]/(\varphi(X_1,\ldots ,X_n)).$$

$\varphi$ is said to be \emph{isotropic} if there exists a
non-zero vector $x$ such that $\varphi(x)=0$. By construction,
every quadratic form is isotropic over its function field. The
\emph{Witt index} of $\varphi$, denoted $i_W(\varphi)$, is the
dimension of a maximal totally isotropic subform of $\varphi$. The
\emph{first Witt index} of $\varphi$, denoted $i_1(\varphi)$, is
the Witt index of $\varphi$ over its function field. The
\emph{essential dimension} of $\varphi$ is given by the following
relation: $\dim_{es}(\varphi)=\dim(\varphi)-i_1(\varphi)+1$. The
\emph{value set} of $\varphi$, denoted $D_F(\varphi)$, is the set
of elements of $F^\x$ which are represented by $\varphi$. By
$D_F(\varphi)D_F(\psi)$ we will denote the usual set product of
$D_F(\varphi)$ and $D_F(\psi)$.
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%It makes sense to throw in i_W and i_1 in the above, after
%"By construction, every quadratic form is isotropic over its function field".
%loosely: isotropic => contains hypic plane. i_w = # of hypic planes; with i_1 = i_w over F(\phi).
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Need to define dim_es too
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%Let $\cp$ denote the $F$-vector space spanned by ``pure
%octonions'', that is octonions with zero field component.
%We define the \emph{pure trace form}, denoted by $T_P$, to be the
%quadratic form $T_P:\cp\rightarrow F, p\mapsto p^2$. Hence, for
%$\sO=\oct a b c$, we have $T_P\simeq \langle a, b, -ab, c, -ac,
%-bc, abc \rangle$.

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Should I include a brief survey of the function field results I make use of, both elementary and deep?
%Need to introduce Witt Indices, Places and possibly stable birational equivalence
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

We highlight the following recent progress towards classifying
when one quadratic form becomes isotropic over the function field
of another:

\begin{theorem}\label{izh} \cite[Lemma 5.3]{I} Let $\psi_0,
\psi_1, \varphi_0, \varphi_1$ be anisotropic quadratic forms over
$F$. For $\psi:=\psi_0\perp T\psi_1$ and $\varphi:=\varphi_0\perp
T\varphi_1$, the following are equivalent:
\begin{itemize} \item $\psi$ isotropic over $F(T)(\varphi)$. \item
$\psi$ isotropic over $F((T))(\varphi)$. \item
$D_{F'}(\varphi_0)D_{F'}(\varphi_1)\subseteq
D_{F'}(\psi_0)D_{F'}(\psi_1)$ for every extension $F'$ of $F$.
\end{itemize}
\end{theorem}

\begin{theorem}\label{roussey} \cite[Theorem 5.1.5]{R} $\psi$ is isotropic over
$F(\varphi)$ $\iff$ $D_{F'}(\varphi)D_{F'}(\varphi)\subseteq
D_{F'}(\psi)D_{F'}(\psi)$ for every extension $F'$ of
$F$.\end{theorem}

In addition, Karpenko and Merkurjev employed advanced
algebro-geometric techniques involving Chow groups to prove the
following powerful result:
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Should I make the elementary/deep distinction between the above results and km, stating that I make use of the elementary
%results where possible (noting that km makes for more succinct proofs in section 3.1)?
%Could mention that I will often invoke Theorem~\ref{km}. Might say that since its proof lies outside of the
%realm of ``classical'' quadratic form theory, I offer alternative proofs where possible.
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


\begin{theorem}\label{km} \cite[Theorem 4.1]{KM} Let $\varphi$ and $\psi$ be anisotropic over $F$ and suppose
that $\psi$  is isotropic over $F(\varphi)$. Then
\begin{enumerate}[(i)]\item $\dim_{es}(\varphi)\leq\dim_{es}(\psi)$; \item moreover,
the equality $\dim_{es}(\varphi)=\dim_{es}(\psi)$ holds if and
only if $\varphi$ is isotropic over $F(\psi)$.
\end{enumerate}\end{theorem}

%In addition, we recall some classical results regarding function
%fields of quadrics which we will have recourse to:

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Is Cassels-Pfister well established enough already, allowing me to forego listing it? Think so.
%What about H, Thm 1? Or indeed, do I need to list the results below? I could just invoke them instead, with refs
%upon their first usage
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%\begin{theorem}\label{cp} (Cassels-Pfister Subform Theorem) Let
%$\psi$, $\varphi\simeq\langle 1\rangle\perp\varphi'$ be two
%quadratic forms over $F$. Assume that $\psi$ is anisotropic over
%$F$ and $\psi$ is hyperbolic over $F(\varphi)$. Then
%$\alpha\varphi$ is a subform of $\psi$ for all $\alpha \in
%D_F(\psi)$.
%\end{theorem}

%\begin{theorem}\label{el}\cite[Theorem 1.4]{EL} Let $n\geq 1$ be an integer. Let $\rho$ and $\pi$
%be quadratic forms, with $\pi$ isometric to an $n$-fold Pfister
%form, such that $\pi\otimes\rho$ is isotropic. Then, there exists
%an isotropic quadratic form $\rho'$ such that
%$\pi\otimes\rho\simeq\pi\otimes\rho'$. In particular,
%$i_W(\pi\otimes\rho)\geq \dim \pi$.
%\end{theorem}

%\begin{theorem}\label{elw} \cite[Theorem 3.5 and Remark 3.6]{ELW}
%Let $P$ be an ordering of $F$ and let $\varphi_i$ be a family of
%quadratic forms over $F$ of dimension at least $2$. Then $P$ can
%be extended to the free compositum of the $F(\varphi_i)$ if and
%only if each $\varphi_i$ is indefinite at $P$.
%\end{theorem}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Should I cite lam's book for a survey of the theory of function fields?????????????????????
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
For an overview of function fields of quadrics or for further
definitions and notation regarding quadratic forms, we refer the
reader to \cite{LAM}.
\medskip

\section{Results}

%\subsection{New values via explicit construction}
\subsection{$Q(n)$ and $\sO(n)$}\label{QandO}

In this section, we will introduce certain families of quaternion
and octonion algebras, before proceeding to investigate what
values their level and sublevel may take.

\subsubsection{Introducing $Q(n)$ and $\sO(n)$}

%The principal method of approach to the classification of the
%level and sublevel of quaternion algebras, and accordingly
%octonion algebras, has been to consider the relationship between
%these quantities and the isotropy of associated quadratic forms.
As motivated above, we will alternate between considering
quaternion and octonion algebras which are division. We will let
$C$ denote a quaternion or octonion algebra which is division. For
those occasions in which we will need to discriminate between
these alternatives, we will use $Q$ to denote a quaternion
division algebra and $\sO$ an octonion division algebra. By $T_P$
we will denote the pure trace form of the algebra in question,
which should be clear from the context.

Since $D_F\left(n\times T_P\right)=\{\sum_{i=1}^n {p_i}^2|p_i\in
\Skew(C,*)\}$ for $*$ given by conjugation, if $\langle 1\rangle
\perp n\times T_P$ is isotropic then $-1$ is expressible as a sum
of $n$ squares of pure elements in $C$, yielding that $s(C)\leq
n$.

For $n=2^k-1$, Lewis stated that the converse holds for quaternion
division algebras (see \cite{LEW2}). This result extends to
composition algebras (see \cite[Lemma 3.9]{O'S}), allowing us to
conclude that $$s(C)\leq n\iff\langle 1\rangle\perp n\times
T_P\hspace{.1cm}\text{is isotropic for}\hspace{.1cm} n=2^k-1.$$
Furthermore, for $n=2^k$ Lewis showed that $s(Q)\leq n$ implies
that $\langle 1\rangle\perp n\times T_P$ or $(n+1)\times\langle
1\rangle\perp (n-1)\times T_P$ is isotropic (see \cite[Lemma
3]{LEW2}), with Leep proving the converse in \cite[Theorem
2.2]{L}. This equivalence also admits generalisation to the class
of composition algebras:
\begin{equation}\label{eq:level2k} s(C)\leq 2^k \iff \langle
1\rangle\perp 2^k\times
T_P\hspace{.1cm}\text{or}\hspace{.1cm}(2^k+1)\times\langle
1\rangle\perp (2^k-1)\times T_P\hspace{.1cm}\text{is isotropic}
\end{equation}

In light of the above results, a natural question which then arose
was whether $s(C)\leq n$ implies that $\langle 1\rangle\perp
n\times T_P$ is isotropic for all $n$. By producing a quaternion
algebra of level $2^k$ such that $\langle 1\rangle\perp 2^k\times
T_P$ is anisotropic for all $k$, Koprowski \cite{K} answered this
question in the negative. Pumpl\"un later showed that there also
exist octonion algebras of level $2^k$ such that $\langle
1\rangle\perp 2^k\times T_P$ is anisotropic for all $k$ (see
\cite{P}).

More specifically, Koprowski's example and its octonion algebra
analogue are directly relevant to (\ref{eq:level2k}), proving that
if $s(C)\leq 2^k$ it remains necessary, for all values of $k$, to
consider whether $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times
T_P$ becomes isotropic in consequence. As an interesting aside, we
note that if $s(C)\leq 2^k$, it is actually the consideration of
the isotropy of $\langle 1\rangle\perp 2^k\times T_P$ which
becomes redundant for $k$ sufficiently large, as the following
attests:

%Similarly, there exist octonion algebras of level $2^k$ such that
%$\langle 1\rangle\perp 2^k\times T_P$ is anisotropic:

%\begin{theorem}\label{octkop} For any positive integer $k$ there
%exists an octonion algebra $\sO$ such that $-1$ is a sum of $2^k$
%squares of octonions but $-1$ is not a sum of $2^k$ squares of
%pure octonions. Moreover, $s(\sO)$ is precisely $2^k$.
%\end{theorem}

%\begin{proof} Suppose $F_0$ is of Pythagoras
%number $2^k+1$. Let $c_0 \in F$ be of length $2^k+1$ (that is
%$c_0$ is a sum of $2^k+1$ squares in $F$ but no fewer), and
%consider $\sO=\openoct {-c_0} x y {F_0(x,y)}$. Since
%$(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ is
%isotropic over $F_0(x,y)$, $s(\sO)\leq 2^k$ (\cite[Proposition
%2.12]{P}).

%Now $\langle 1\rangle\perp 2^k\times\langle -c_0, x, c_0x, y,
%c_0y, -xy, -c_0xy \rangle$ is anisotropic over $F_0(x,y)$ if and
%only if $\langle 1\rangle\perp 2^k\times\langle -c_0, x,
%c_0x\rangle$ and $2^k\times\langle 1, c_0, -x, -c_0x\rangle$ are
%anisotropic over $F_0(x)$ by Springer's Theorem (see
%\cite[p.209]{S}), which in turn is equivalent to $\langle
%1\rangle\perp 2^k\times\langle -c_0\rangle$ and $2^k\times\langle
%1, c_0\rangle$ being anisotropic over $F_0$. Now $\langle
%1\rangle\perp 2^k\times\langle -c_0\rangle$ isotropic over $F_0$
%implies that $c_0$ is a sum of $2^k$ squares, which is a
%contradiction. $2^k\times\langle 1, c_0\rangle$ is also
%anisotropic over $F_0$, since it is a positive definite form over
%a formally real field. Thus $\langle 1\rangle\perp
%2^k\times\langle -c_0, x, c_0x, y, c_0y, -xy, -c_0xy\rangle$ is
%anisotropic over $F_0(x,y)$. Thus, $\langle 1\rangle\perp
%(2^k-1)\times\langle -c_0, x, c_0x, y, c_0y, -xy, -c_0xy\rangle$
%is also anisotropic, and hence $s(\sO)\nleq 2^k-1$ (see
%\cite{LEW2}).
%***************************************************************************
%Thus $s(\sO)=2^k$. For $k\geq 2$, $\underline{s}(\sO)=2^k$
%also, since $\underline{s}(\sO)\leq s(\sO)$ and
%$\underline{s}(\sO)\leq 2^k-1\iff \< 1\>\perp (2^k-1)\x T_P$ is
%isotropic.
%\end{proof}

%For $Q$ a quaternion division algebra, $s(Q)\leq 2^k \iff
%(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ or $\langle
%1\rangle\perp 2^k\times T_P$ is isotropic (Lewis proved the left
%to right implication in \cite{LEW2}, with Leep proving the
%converse in \cite{L}). Indeed, this equivalence admits the
%following generalisation: $s(C)\leq 2^k \iff (2^k+1)\times\langle
%1\rangle\perp (2^k-1)\times T_P$ or $\langle 1\rangle\perp
%2^k\times T_P$ is isotropic. Koprowski's example and its octonion
%algebra analogue prove that it is necessary to consider the
%isotropy of $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$
%for all values of $k$, that is $s(C)\leq 2^k$ is not equivalent to
%$\langle 1\rangle\perp 2^k\times T_P$ is isotropic. In fact, it is
%the consideration of the isotropy of $\langle 1\rangle\perp
%2^k\times T_P$ that becomes redundant for certain values of $k$,
%as the following result attests:

\begin{theorem}\label{sreduction} For $Q$ a quaternion division algebra, $s(Q)\leq 2^k\iff (2^k+1)\times\langle 1\rangle\perp
(2^k-1)\times T_P$ is isotropic, where $k \geq 2$. For $\sO$ an
octonion division algebra, $s(\sO)\leq 2^k\iff
(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ is
isotropic, where $k \geq 3$.
\end{theorem}

\begin{proof} Since $s(C)\leq 2^k \iff
(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ or $\langle
1\rangle\perp 2^k\times T_P$ is isotropic, it suffices to prove
that the isotropy of $\langle 1\rangle\perp 2^k\times T_P$ implies
that of $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$,
for $k$ as above. Suppose $\langle 1\rangle\perp 2^k\times T_P$ is
isotropic, implying that $2^k\times T_C\simeq (2^k\times\< 1\>)
\otimes T_C$ is isotropic. We know that $i_W(2^k\times T_C)\geq
2^k$ by \cite[Proposition 1.4]{EL}. For $k$ as above, $2^k\times\<
1\>\perp (2^k-1)\times T_P$ is a subform of $2^k\times T_C$ of
codimension $<2^k$, and thus is also isotropic.
\end{proof}

%\begin{cor}
%\end{cor}
%NAH!!!!

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%NOTE: Lewis may have posed the level n and 1+n*TP equivce as a question. Seem to remember it (word form)
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%$D_F(\langle 1\rangle \perp T_P)$ corresponds to the field
%component of the square of a quaternion, the following relations
%arise immediately from the definition of the level and sublevel:
%\begin{itemize}\item $s(\sO)\leq n\Rightarrow \langle 1\rangle \perp n\times(\langle
%1\rangle \perp T_P)$ is isotropic. \item $\underline{s}(\sO)\leq
%n\Rightarrow (n+1)\times(\langle 1\rangle \perp T_P)$ is
%isotropic. \item $\langle 1\rangle \perp n\times T_P$ isotropic
%$\Rightarrow s(\sO)\leq n$ for all $n$.\end{itemize}

%In general, we cannot make any stronger statement regarding the
%isotropy of $T_P$ or $\< 1\>\perp T_P$. For specific values of $n$
%however, we have the following results (see \cite{LEW},
%\cite{LEW2}, \cite{L}, \cite{P}, \cite{O'S}):
%\begin{equation}\label{eq:level2k} s(\sO)\leq 2^k \iff (2^k+1)\times\langle
%1\rangle\perp (2^k-1)\times
%T_P\hspace{.1cm}\text{or}\hspace{.1cm}\langle 1\rangle\perp
%2^k\times T_P\hspace{.1cm}\text{is isotropic.}\end{equation}
%\begin{equation}\label{eq:level2k-1} s(\sO)\leq 2^k-1 \iff \langle 1\rangle\perp
%2^k-1\times T_P\hspace{.1cm}\text{is isotropic.}\end{equation}
%\begin{equation}\label{eq:sub2k-1} \underline{s}(\sO)\leq 2^k-1 \iff \langle 1\rangle\perp
%2^k-1\times T_P\hspace{.1cm}\text{is isotropic, where $k\geq
%2$.}\end{equation}

%In \cite{K}, Koprowski produced an example of a quaternion algebra
%of level $2^k$ (and sublevel $2^k$ for $k\geq 2$) such that
%$\langle 1\rangle \perp 2^k\times T_P$ is anisotropic. A similar
%construction is possible for octonion algebras:%To prove this, we
%will invoke the following result:
%
%\begin{theorem}\label{spring} (Springer) Let $\varphi_1$, $\varphi_2$ be two
%quadratic forms over $F$, and $F(x)$ the rational function field
%in one variable over $F$. Then $\varphi_1\perp x\varphi_2$ is
%isotropic over $F(x)$ if and only if $\varphi_1$ or $\varphi_2$ is
%isotropic over $F$.
%\end{theorem}

%\begin{prop} For any positive integer $k$ there
%exists an octonion algebra $\sO$ such that $-1$ is a sum of $2^k$
%squares of octonions but $-1$ is not a sum of $2^k$ squares of
%pure octonions. Moreover, $s(\sO)=2^k$.
%\end{prop}

%\begin{proof} Suppose $F_0$ is of Pythagoras
%number $2^k+1$. Let $c_0 \in F$ be of length $2^k+1$ (that is
%$c_0$ is a sum of $2^k+1$ squares in $F$ but no fewer), and
%consider $$\sO=\openoct {-c_0} x y {F_0(x,y)}.$$

%Since $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ is
%isotropic over $F_0(x,y)$, $s(\sO)\leq 2^k$ (\cite[Proposition
%2.12]{P}).

%Now $\langle 1\rangle\perp 2^k\times\langle -c_0, x, c_0x, y,
%c_0y, -xy, -c_0xy \rangle$ is anisotropic over $F_0(x,y)$ if and
%only if $\langle 1\rangle\perp 2^k\times\langle -c_0, x,
%c_0x\rangle$ and $2^k\times\langle 1, c_0, -x, -c_0x\rangle$ are
%anisotropic over $F_0(x)$ by Springer's Theorem, which in turn is
%equivalent to $\langle 1\rangle\perp 2^k\times\langle -c_0\rangle$
%and $2^k\times\langle 1, c_0\rangle$ being anisotropic over $F_0$.
%Now $\langle 1\rangle\perp 2^k\times\langle -c_0\rangle$ isotropic
%over $F_0$ implies that $c_0$ is a sum of $2^k$ squares, which is
%a contradiction. $2^k\times\langle 1, c_0\rangle$ is also
%anisotropic over $F_0$, since it is a positive definite form over
%a formally real field. Thus $\langle 1\rangle\perp
%2^k\times\langle -c_0, x, c_0x, y, c_0y, -xy, -c_0xy\rangle$ is
%anisotropic over $F_0(x,y)$. Thus, $\langle 1\rangle\perp
%2^k-1\times\langle -c_0, x, c_0x, y, c_0y, -xy, -c_0xy\rangle$ is
%also anisotropic, and hence $s(\sO)\nleq 2^k-1$ by a previous
%remark. Thus $s(\sO)=2^k$. For $k\geq 2$, $\underline{s}(\sO)=2^k$
%also, since $\underline{s}(\sO)\leq s(\sO)$ and
%$\underline{s}(\sO)\leq 2^k-1\iff \< 1\>\perp (2^k-1)\x T_P$ is
%isotropic.
%\end{proof}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Want a sensible way to recap that neither
%$s(\sO)=n$ nor $\underline{s}(\sO)=n \Rightarrow \< 1\>\perp
%n\times T_P$ is isotropic for all $n$.
%Should I bother introducing equivce for 2^k-1 case here, as I'll need it later anyway????
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Actually, sublevel is a weaker statement!

Since $s(C)=2^k \Rightarrow \underline{s}(C)=2^k$ (\cite[Theorem
2.5]{L} for $C=Q$ and \cite[Theorem 3.11]{O'S} for $C=\sO$),
Koprowski's example and its octonion algebra analogue prove that
neither $s(C)=n$ nor $\underline{s}(C)=n$ implies that $\<
1\>\perp n\times T_P$ is isotropic for all $n$. Of course, this
does not preclude the existence of composition algebras of level
and sublevel $n$ such that $\< 1\>\perp n\times T_P$ is isotropic,
for all values of $n$.

As we have seen, the isotropy of $\nTP$ encodes $n$ as an upper
bound for the level and sublevel of a composition algebra. Indeed,
since $D_F\left(n\times T_P\right)=\{\sum_{i=1}^n {p_i}^2|p_i\in
\Skew(C,*)\}$, $\nTP$ is the most natural quadratic form whose
isotropy bounds the level.  Thus it seems reasonable to suggest
that if one were to construct composition algebras, subject to
$\nTP$ being isotropic, in a \emph{suitably general fashion}, then
their level and sublevel might actually equal $n$.

%Indeed for $n=6$ and $7$, our construction of an octonion algebra
%of level and sublevel $n$ explicitly employs the isotropy of
%$\nTP$.

%aiming for just such a construction in our definition of q(n) and o(n)
In defining $Q(n)$ and $\sO(n)$, we are aiming for just such a
construction. In an attempt to make $Q(n)$ and $\sO(n)$ suitably
general, the better to eradicate factors which could serve to
reduce the level or sublevel below $n$, we choose the generators
of our algebra to be transcendental over a formally real ground
field, and extend to the function field of $\nTP$, the natural
method of generically encoding its isotropy. Hence, we let
$$Q(n):=\openquat x y {F}\otimes_{F}F\left(\nTP\right)$$ and
$$\sO(n):=\openoct x y z {F}\otimes_{F}F\left(\nTP\right),$$ where
$T_P$ is the pure trace form of $\openquat x y {F}$ and $\openoct
x y z {F}$ respectively,
and $F$ denotes $F_0(x,y)$ and $F_0(x,y,z)$ respectively. %We
%assume that $p(F)\geq n$, as otherwise $F(\nTP)\cong F(\< 1\>\perp
%p(F)\times T_P)$ implying that $s(Q(n))\leq p(F)<n$. We posit the
%following conjecture:

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%The above statement re $p(F)$ seems unnecessary in light of new proofs. Indeed, it may even be wrong.
%Consider $p(F)$ vs. $p(F(\phi))$
%Will comment out for now
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%\emph{Conjecture.}
We posit the following conjecture:

\begin{conj}\label{bigconj}
$s\left(Q(n)\right)=\underline{s}\left(Q(n)\right)=s\left(\sO(n)\right)=\underline{s}\left(\sO(n)\right)=n$
for all $n$.
\end{conj}

\subsubsection{Determined Values}

We begin our investigation of $Q(n)$ and $\sO(n)$ by noting that
if the sublevel of these algebras equals $n$, then their level and
sublevel must coincide.

%\begin{lemma}\label{obvious} Let $C$ be a composition algebra of sublevel $n$,
%$x_i \in C$. For $\sum_{i=1}^{n+1}{x_i}^2=0$, if $x_i \in F$ for
%some $i$, then $s(C)=\underline{s}(C)$.\end{lemma}

%\begin{proof} Recall that $\underline{s}(C)\leq s(C)$. If
%$\sum_{i=1}^{n+1}{x_i}^2=0$, where $x_1 \in F$ without loss of
%generality, then $\sum_{i=2}^{n+1}{x_i}^2=-x_1^2 \Rightarrow
%\sum_{i=2}^{n+1}{\left(\frac {x_i} {x_1}\right)}^2=-1 \Rightarrow
%s(C) \leq n = \underline{s}(C)$.\end{proof}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Actually, we can see the above by noting that $s(C)\leq n$. Hence, $\underline{s}(C)\nleq n-1 \Rightarrow s(C)\nleq n-1$.
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

We will now proceed to establish the first seven cases of the
octonionic component of Conjecture~\ref{bigconj}. In doing so, we
prove that $6$ and $7$ are realisable as both the level and
sublevel of octonion algebras. We note that these are the first
examples of composition algebras whose level is not of the form
$2^k$ or $2^k+1$ for some $k$.

\begin{theorem}\label{n7} $s\left(\sO(n)\right)=\underline{s}\left(\sO(n)\right)=n$, for $n\leq 7$.\end{theorem}

\begin{proof} For ease of notation we let $\varphi:=\nTP$.

By construction $\underline{s}\left(\sO(n)\right)\leq n$.

Suppose $\underline{s}\left(\sO(n)\right)\leq n-1$. Hence
\begin{equation}\label{eq:1sub6} \hspace{0cm}\sum_{i=1}^{n}{c_i}^2+
\sum_{i=1}^{n}{p_i}^2=0\end{equation} and
\begin{equation}\label{eq:2sub6}
\hspace{0cm}\sum_{i=1}^{n}{c_ip_i}=0,\end{equation} where $c_i \in
F\left(\varphi\right)$ and $p_i \in \cp$, the
$F\left(\varphi\right)$-vector space spanned by the pure
octonions.

If $c_i=0$ $\forall$ $i$, (\ref{eq:1sub6}) implies that $n\times
T_P$ is isotropic over $F\left(\varphi\right)$.

Alternatively, if there exists an $i$ such that $c_i\neq 0$,
(\ref{eq:2sub6}) implies the existence of $V$, an
$(n-1)$-dimensional $F\left(\varphi\right)$-subspace of $\cp$,
containing $p_1,\ldots, p_n$. Let $\beta:V\rightarrow
F\left(\varphi\right)$, $p\mapsto p^2$. Hence, $\beta$ is an
$(n-1)$-dimensional subform of $T_P$ and (\ref{eq:1sub6}) implies
that $\vartheta:=n\times (\langle 1\rangle\perp\beta )$ is
isotropic over $F\left(\varphi\right)$.

Note that $n\times T_P$ and $8\times(\langle -1\rangle\perp T_P)$
are anisotropic over $F$ by Springer's Theorem.

If $n\times T_P$ becomes isotropic over $F\left(\varphi\right)$,
then the Pfister form $8\times (\langle 1\rangle\perp -T_P)$
becomes hyperbolic over $F\left(\varphi\right)$. Invoking the
Cassels-Pfister Subform Theorem (see \cite[p.155]{S}),
$\alpha(\varphi)$ should be a subform of $8\times (\langle
1\rangle\perp -T_P)$ for any $\alpha\in D_F\left(8\times (\langle
1\rangle\perp -T_P)\right).$ In particular, since $-x\in
D_F\left(8\times (\langle 1\rangle\perp -T_P)\right),$
$-x(\varphi)\simeq\langle -x\rangle\perp n\times \langle
-1,-xy,y,-xz,z,xyz,-yz\rangle$ should be a subform of $8\times
(\langle 1\rangle\perp -T_P)$, implying that $n\times \langle
-1\rangle$ is also a subform, which is clearly false. Hence,
$n\times T_P$ also remains anisotropic over
$F\left(\varphi\right)$.

Thus, $\vartheta$ must be isotropic over $F\left(\varphi\right).$
By Theorem~\ref{roussey} we must have
$$D_{F'}\left(\varphi\right)D_{F'}\left(\varphi\right)\subseteq
D_{F'}\left(\vartheta\right)D_{F'}\left(\vartheta\right)\hspace{.1cm}\text{for
every extension}\hspace{.1cm}F'
\hspace{.1cm}\text{of}\hspace{.1cm}F.$$

Let $F'=F$.

Since $T_P\simeq\langle x,y,-xy,z,-xz,-yz,xyz\rangle$, we have
$x\in D_{F}\left(\varphi\right)D_{F}\left(\varphi\right)$. %for every extension
%$F'$ of $F$, and thus in particular for $F'=F$.
Suppose $\langle x\rangle$ is not a subform of $\beta$. Hence, all
multiples of $x$ in
$D_{F}\left(\vartheta\right)D_{F}\left(\vartheta\right)$ are of
the form $-x\left(\sum_{i=1}^m a_{i}^2\right)$, for $a_i \in F$
(note that $\langle (y)(-xy)\rangle \simeq \langle (z)(-xz)\rangle
\simeq \langle (-yz)(xyz)\rangle \simeq \langle -x\rangle$). If
$x$ is positive with respect to some ordering on $F$, then
$-x\left(\sum_{i=1}^m a_{i}^2\right)$ is necessarily negative, and
vice-versa. Thus $x \notin
D_{F}\left(\vartheta\right)D_{F}\left(\vartheta\right)$, which is
a contradiction.
%Then, by examining the
%products of two of the remaining possible $1$-dimensional subforms
%of $\beta$ (noting that $\langle (y)(-xy)\rangle \simeq \langle
%(z)(-xz)\rangle \simeq \langle (-yz)(xyz)\rangle \simeq \langle
%-x\rangle$, which is not $\simeq \langle x\rangle$ since $F_0$ is
%formally real), we conclude that $x \notin D_{F}(n\times (\langle
%1\rangle\perp\beta ))D_{F}(n\times (\langle 1\rangle\perp\beta
%))$, which is a contradiction.
Hence $\langle x\rangle$ is a subform of $\beta$.

Note that $y\in
D_{F}\left(\varphi\right)D_{F}\left(\varphi\right)$. Now suppose
that $\langle y\rangle$ is not a subform of $\beta$. Hence, all
multiples of $y$ in
$D_{F}\left(\vartheta\right)D_{F}\left(\vartheta\right)$ are of
the form $-y\left(\sum_{i=1}^m a_{i}^2\right)$, for $a_i \in F$
(note that $\langle (x)(-xy)\rangle \simeq \langle (z)(-yz)\rangle
\simeq \langle (-xz)(xyz)\rangle \simeq \langle -y\rangle$).
Again, if $y$ is positive with respect to some ordering on $F$,
then $-y\left(\sum_{i=1}^m a_{i}^2\right)$ is negative, and
vice-versa. Thus $y \notin
D_{F}\left(\vartheta\right)D_{F}\left(\vartheta\right)$, which is
a contradiction.
%which is not $\simeq \langle
%y\rangle$ since $F_0$ is formally real), we conclude that $y
%\notin D_{F}(n\times (\langle 1\rangle\perp\beta ))D_{F}(n\times
%(\langle 1\rangle\perp\beta ))$, which again is a contradiction.
Hence $\langle y\rangle$ is a subform of $\beta$.


Similarly, since $-xy, z, -xz, -yz$ and $xyz \in
D_{F}\left(\varphi\right)D_{F}\left(\varphi\right)$, $\langle
-xy\rangle, \langle z\rangle, \langle -xz\rangle, \langle
-yz\rangle$ and $\langle xyz\rangle$ must also be subforms of
$\beta$.

Thus, $T_P\simeq \langle x\rangle\perp\langle y\rangle\perp\langle
-xy\rangle\perp\langle z\rangle\perp\langle -xz\rangle\perp\langle
-yz\rangle\perp\langle xyz\rangle$ is a subform of $\beta$, which
contradicts the fact that $\beta$ is $(n-1)$-dimensional.

Thus $\underline{s}\left(\sO(n)\right)\nleq n-1$, implying that
$\underline{s}\left(\sO(n)\right)=n$.
\end{proof}

%$s(\sO)\leq n$ since $\langle 1\rangle \perp n \times T_P$ is
%isotropic, by construction. Moreover, since
%$\underline{s}(\sO)\leq s(\sO)$, we conclude that $s(\sO)=n$.

%Lemma~\ref{obvious} implies that $s\left(\sO(n)\right)=n$ also.
%
%NEED TO FIX EQUATION ALIGNMENT IN THE ABOVE!!!!!!!!!!!!

Showing that $\vartheta$ remains anisotropic over
$F\left(\varphi\right)$ is a key step in the above proof. The
machinery with which we successfully endeavoured to achieve this
end is Roussey's Theorem~\ref{roussey}. Hence, all the ingredients
in the above proof lie within the ``classical'' theory of
quadratic forms.

If we instead choose to invoke Theorem~\ref{km} of Karpenko and
Merkurjev, whose proof relies upon deep algebro-geometric
techniques, in order to prove anisotropy of forms over
$F\left(\langle 1\rangle\perp n\times T_P\right)$, we obtain a
more succinct, if less elementary, argument:

\begin{proof} As detailed above, to prove that $\underline{s}\left(\sO(n)\right)\nleq
n-1$ it suffices to show that $n\times T_P$ and $n\times (\langle
1\rangle\perp\beta )$ are anisotropic over $F\left(\langle
1\rangle \perp n \times T_P\right)$. Since
$i_1\left(\nTP\right)=1$ (see Lemma~\ref{Oi1}), the fact that
$\dim \left(\nTP\right)>\dim \left(n\times T_P\right)$ implies
that $n\times T_P$ is anisotropic over $F\left(\langle 1\rangle
\perp n \times T_P\right)$ by Theorem~\ref{km}. Similarly since
$\dim \left(\nTP\right)>\dim \left(n\times (\langle
1\rangle\perp\beta )\right)$, we can conclude that $n\times
(\langle 1\rangle\perp\beta )$ is also anisotropic over
$F\left(\langle 1\rangle \perp n \times T_P\right)$ by
Theorem~\ref{km}. Hence $\underline{s}\left(\sO(n)\right)\nleq
n-1$.
\end{proof}

%, whereby Lemma~\ref{obvious} implies that
%$s\left(\sO(n)\right)=n$ also.


Both of the above methods of proof apply to quaternion algebras as
well, yielding:

\begin{theorem}\label{n3} $s\left(Q(n)\right)=\underline{s}\left(Q(n)\right)=n$ for $n\leq 3$.
\end{theorem}

We note that the above results may be extended to their respective
next cases. Indeed, the following result shows that
Conjecture~\ref{bigconj} is true for all $n=2^k$:

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Do I want to mention the possible confusion arising from LM's paper here?
%I guess that if I'm going to go into that level of detail, then I'll need to highlight
%the fact that it's K&W's trick which makes Theorem~\ref{n7} possible. This smacks of overkill.
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\begin{theorem}\label{both2^k} $s\left(Q(n)\right)=\underline{s}\left(Q(n)\right)
=s\left(\sO(n)\right)=\underline{s}\left(\sO(n)\right)=n$, for $n=2^k$.
\end{theorem}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Note: I don't have to use Lemma~\ref{obvious} below, as I'm really proving both things at once.
%Worth changing?
%Could argue directly with Lewis' 2^k-1 equivce...wouldn't require $k\geq 2$ bit.
%Is it worth maintaining the current method
%solely in order to keep the reference to \cite[Theorem 3.5]{O'S} intact?
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\begin{proof} We may assume that $k\geq 2$ by Theorem~\ref{n7} and
Theorem~\ref{n3}.

Suppose $\underline{s}\left(Q(2^k)\right)\leq 2^k-1$. Thus, $\<
1\>\perp(2^k-1)\x T_P$ is isotropic over $F\left(\varphi\right)$
by \cite[Theorem 3.5]{O'S}, where $\varphi:=\langle 1\rangle \perp
2^k \times T_P$. Since $\psi_{\< x\>}:=\langle 1\rangle \perp 2^k
\times \< x\>$ is a subform of $\varphi$, there exists an
$F$-place from $F\left(\varphi\right)$ to $F\left(\psi_{\<
x\>}\right)$, by \cite[Theorem 3.3]{Kn}. Thus $\< 1\>\perp
(2^k-1)\x T_P$ is isotropic over $F\left(\psi_{\< x\>}\right)$, by
\cite[Theorem 3.3]{Kn}. Springer's Theorem implies that either $\<
1\>\perp (2^k-1)\x \< x\>$ or $(2^k-1)\times \< 1,-x\>$ is
isotropic over $F_0(x)\left(\psi_{\< x\>}\right)$. However, $\<
1\>\perp (2^k-1)\x \< x\> $ is anisotropic over
$F_0(x)\left(\psi_{\< x\>}\right)$ by \cite[Theorem 1]{H1}. Hence,
$(2^k-1)\times \< 1,-x\>$ must be isotropic over
$F_0(x)\left(\psi_{\< x\>}\right)$, implying that $2^k\times \<
1,-x\>$ is hyperbolic over $F_0(x)\left(\psi_{\< x\>}\right)$. The
Cassels-Pfister Subform Theorem implies that $\psi_{\< x\>}$ is a
subform of $2^k\times \< 1,-x\>$, giving a desired contradiction.
Hence $\underline{s}\left(Q(2^k)\right)=2^k$.

%, with Lemma~\ref{obvious} implying that
%$s\left(Q(2^k)\right)=2^k$ also.

Now suppose $\underline{s}\left(\sO(2^k)\right)\leq 2^k-1$. Thus,
$\< 1\>\perp(2^k-1)\x T_P$ is isotropic over $F\left(\langle
1\rangle \perp 2^k \times T_P\right)$ by \cite[Theorem 3.5]{O'S}.
Since $\psi_{\< x,y,-xy\>}:=\langle 1\rangle \perp 2^k \times \<
x,y,-xy\>$ is a subform of $\langle 1\rangle \perp 2^k \times
T_P$, there exists an $F$-place from $F\left(\langle 1\rangle
\perp 2^k \times T_P\right)$ to $F\left(\psi_{\<
x,y,-xy\>}\right)$, by \cite[Theorem 3.3]{Kn}. Thus $\< 1\>\perp
(2^k-1)\x T_P$ is isotropic over $F\left(\psi_{\<
x,y,-xy\>}\right)$, by \cite[Theorem 3.3]{Kn}. Springer's Theorem
implies that either $\< 1\>\perp (2^k-1)\x \< x,y,-xy\>$ or
$(2^k-1)\times \< 1,-x,-y,xy\>$ is isotropic over
$F_0(x,y)\left(\psi_{\< x,y,-xy\>}\right)$. In proving the
quaternion case above, we showed that $\< 1\>\perp (2^k-1)\x \<
x,y,-xy\> $ is anisotropic over $F_0(x,y)\left(\psi_{\<
x,y,-xy\>}\right)$. Hence, $(2^k-1)\times \< 1,-x,-y,xy\>$ must be
isotropic over $F_0(x,y)\left(\psi_{\< x,y,-xy\>}\right)$,
implying that $2^k\times \< 1,-x,-y,xy\>$ is hyperbolic over
$F_0(x,y)\left(\psi_{\< x,y,-xy\>}\right)$. The Cassels-Pfister
Subform Theorem implies that $\psi_{\< x,y,-xy\>}$ is a subform of
$2^k\times \< 1,-x,-y,xy\>$, furnishing a contradiction. Hence
$\underline{s}\left(\sO(2^k)\right)=2^k$.
\end{proof}

%, with Lemma~\ref{obvious} implying that
%$s\left(\sO(2^k)\right)=2^k$ also.


%\begin{proof} By construction $\underline{s}(\sO(2^k))\leq
%2^k$.

%Suppose $\underline{s}(\sO(2^k))\leq 2^k-1$. Thus, $\< 1\>\perp
%(2^k-1)\x T_P$ is isotropic over $F(\langle 1\rangle \perp 2^k
%\times T_P)$ by \cite[Theorem 3.5]{O'S} (we may assume that $k\geq
%2$, since the case where $k=1$ is covered by Theorem~\ref{n7}
%above). Since $\< 1\>\perp (2^k-1)\x\<x,y,-xy\>,
%(2^k-1)\x\<1,-x,-y,xy\>, \< 1\>\perp 2^k\x\<x,y,-xy\>$ and
%$2^k\x\<1,-x,-y,xy\>$ are all anisotropic over $F_0(x,y)$ by
%Springer's Theorem, we may invoke Theorem~\ref{izh}. Thus, we have
%$$D_{K}(\< 1\>\perp 2^k\x\<x,y,-xy\>)D_{K}(2^k\x\<1,-x,-y,xy\>)$$
%$$\subseteq D_{K}(\< 1\>\perp
%(2^k-1)\x\<x,y,-xy\>)D_{K}((2^k-1)\x\<1,-x,-y,xy\>),$$ for all
%extensions $K$ of $F_0(x,y)$. Consider the case where $K =
%F_0(x,y)$. Let $\sum_{i=1}^{2^k}{t_i}^2$ be of length $2^k$, for
%$t_i\in K$. Now $\sum_{i=1}^{2^k}{t_i}^2\in D_{K}(\< 1\>\perp
%2^k\x\<x,y,-xy\>)D_{K}(2^k\x\<1,-x,-y,xy\>),$ and hence belongs to
%$D_{K}(\< 1\>\perp
%(2^k-1)\x\<x,y,-xy\>)D_{K}((2^k-1)\x\<1,-x,-y,xy\>)$. Thus,
%$\exists$ $p',\ldots ,{w_{2^k-1}}'\in K$ such that
%$\sum_{i=1}^{2^k}{t_i}^2$ equals
%$$\left({p'}^2+x\sum_{i=1}^{2^k-1}{{q_i}'}^2
%+y\sum_{i=1}^{2^k-1}{{r_i}'}^2-xy\sum_{i=1}^{2^k-1}{{s_i}'}^2\right)\left(\sum_{i=1}^{2^k-1}{{t_i}'}^2
%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2\right).$$

%Multiplication by $\sum_{i=1}^{2^k-1}{{t_i}'}^2
%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2$
%yields that $$\sum_{i=1}^{2^k}{{t_i}''}^2
%-x\sum_{i=1}^{2^k}{{u_i}''}^2-y\sum_{i=1}^{2^k}{{v_i}''}^2+xy\sum_{i=1}^{2^k}{{w_i}''}^2
%=\left({p'}^2+x\sum_{i=1}^{2^k-1}{{q_i}'}^2
%+y\sum_{i=1}^{2^k-1}{{r_i}'}^2-xy\sum_{i=1}^{2^k-1}{{s_i}'}^2\right)a^2$$
%where ${t_1}'',\ldots ,{w_{2^k}}''\in K$, since $D_K(2^k\x\< 1\>)$
%is a multiplicative group. %$a=\sum_{i=1}^{2^k-1}{{t_i}'}^2
%%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2$ and

%Hence, $\sum_{i=1}^{2^k}{{t_i}''}^2={p'}^2a^2=(p'a)^2$. This is a
%contradiction, since
%$$\sum_{i=1}^{2^k}{{t_i}''}^2=\left(\sum_{i=1}^{2^k}{{t_i}}^2\right)\left(\sum_{i=1}^{2^k-1}{{t_i}'}^2\right)\notin
%{K^{\times}}^2,$$ as a necessary condition for a product of two
%numbers to equal a square is that they are of the same length
%(recall that multiplication by a square does not alter length).

%Thus, $\underline{s}(\sO(2^k))\nleq 2^k-1$, implying that
%$\underline{s}(\sO(2^k))=2^k$.

%Lemma~\ref{obvious} implies that $s(\sO(2^k))=2^k$ also.

%\end{proof}

Karpenko and Merkurjev's result, Theorem~\ref{km}, provides
another method of proof:

\begin{proof} Since $i_1\left(\< 1\>\perp
2^k\x T_P\right)=1$ (see Lemma~\ref{i1} and Lemma~\ref{Oi1}), the
fact that $\dim \left(\< 1\>\perp 2^k\x T_P\right)>\dim \left(\<
1\>\perp (2^k-1)\x T_P\right)$ implies that $\< 1\>\perp (2^k-1)\x
T_P$ remains anisotropic over $F\left(\< 1\>\perp 2^k\x
T_P\right)$ by Theorem~\ref{km}. Hence
$\underline{s}\left(Q(2^k)\right)\nleq 2^k-1$ and
$\underline{s}\left(\sO(2^k)\right)\nleq 2^k-1$.
\end{proof}

%Lemma~\ref{obvious} implies that
%$s\left(Q(2^k)\right)=s\left(\sO(2^k)\right)=2^k$ also.

%\emph{Remark.} Again the analogous result holds for quaternion
%algebras, that is $s(Q(2^k))=\underline{s}(Q(2^k))=2^k$.

In their construction of a quaternion algebra of level $2^k+1$ for
$k\geq 1$, Laghribi and Mammone considered $Q(2^k+1)$, proving
that $s\left(Q(2^k+1)\right)=2^k+1$ \cite[Theorem 2.1]{LM}.
Furthermore, Pumpl\"un adapted their argument to show that
$s\left(\sO(2^k+1)\right)=2^k+1$ for $k\geq 1$ \cite[Theorem
3.1]{P}. We note that these results extend to the case where
$k=0$, as this is merely the level component of
Theorem~\ref{both2^k} when $k=1$. We now present a simplified
method of proving these results:

\begin{theorem}\label{level2^k+1} $s\left(Q(n)\right)=s\left(\sO(n)\right)=n$, for $n=2^k+1$.
\end{theorem}

\begin{proof} Suppose $s\left(Q(2^k+1)\right)\leq 2^k$. Hence $(2^k+1)\times\langle 1\rangle\perp
(2^k-1)\times T_P$ is isotropic over $F\left(\< 1\>\perp
(2^k+1)\times T_P\right)$ by Theorem~\ref{sreduction} (we can
assume that $k\geq 2$ by Theorem~\ref{n3}). Since $y(2^k+1)\times
\< 1,-x\>$ is a subform of $\< 1\>\perp (2^k+1)\times T_P$, there
exists an $F$-place from $F\left(\< 1\>\perp (2^k+1)\times
T_P\right)$ to $F\left((2^k+1)\times \< 1,-x\>\right)$, by
\cite[Theorem 3.3]{Kn}. Thus $(2^k+1)\times\langle 1\rangle\perp
(2^k-1)\times T_P$ is isotropic over $F\left((2^k+1)\times \<
1,-x\>\right)$, by \cite[Theorem 3.3]{Kn}. Springer's Theorem
implies that either $(2^k+1)\times\langle 1\rangle\perp
(2^k-1)\times \< x\>$ or $(2^k-1)\times \< 1,-x\>$ is isotropic
over $F_0(x)\left((2^k+1)\times \< 1,-x\>\right)$. However, both
forms are anisotropic by \cite[Theorem 1]{H1}. Hence
$s\left(Q(2^k+1)\right)\nleq 2^k$.

The same method applies in the octonion case.
%Suppose $s(\sO(2^k+1))\leq 2^k$. Hence $(2^k+1)\times\langle
%1\rangle\perp (2^k-1)\times T_P$ is isotropic over $F(\< 1\>\perp
%(2^k+1)\times T_P)$ (we can assume that $k\geq 3$ by
%Theorem~\ref{n7}). Since $z(2^k+1)\times \< 1,-x,-y,xy\>$ is a
%subform of $\< 1\>\perp (2^k+1)\times T_P$, there exists an
%$F$-place from $F(\< 1\>\perp (2^k+1)\times T_P)$ to
%$F((2^k+1)\times \< 1,-x,-y,xy\>)$, by \cite[Theorem 3.3]{Kn}.
%Thus $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ is
%isotropic over $F((2^k+1)\times \< 1,-x,-y,xy\>)$, by
%\cite[Theorem 3.3]{Kn}. Springer's Theorem implies that either
%$(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times \<x,y,-xy\>$ or
%$(2^k-1)\times \< 1,-x,-y,xy\>$ is isotropic over
%$F_0(x,y)((2^k+1)\times \< 1,-x,-y,xy\>)$. However, both forms are
%anisotropic by \cite[Theorem 1]{H1}. Hence $s(\sO(2^k+1))\nleq
%2^k$.
\end{proof}

%\emph{Remark.} The same method shows that $s(Q(n))=n$, for
%$n=2^k+1$.

%\begin{proof} Since $\langle 1\rangle \perp (2^k+1) \times
%T_P$ is isotropic over its function field, $s(\sO)\leq 2^k+1$.

%Suppose $s(\sO)\leq 2^k$. Thus, either $(2^k+1)\times\langle
%1\rangle\perp (2^k-1)\times T_P$ or $\langle 1\rangle\perp
%2^k\times T_P$ is isotropic over $F(\langle 1\rangle \perp (2^k+1)
%\times T_P)$ by (\ref{eq:level2k}). Since the associated subforms
%are anisotropic over $F_0(x,y)$ by Springer's Theorem, we may
%invoke Theorem~\ref{izh} as a consequence of these statements.

%Suppose first that $\langle 1\rangle\perp 2^k\times T_P$ is
%isotropic over $F(\langle 1\rangle \perp (2^k+1) \times T_P)$. By
%Theorem~\ref{izh} we have $$D_{K}(\< 1\>\perp
%(2^k+1)\x\<x,y,-xy\>)D_{K}((2^k+1)\x\<1,-x,-y,xy\>)$$ $$\subseteq
%D_{K}(\< 1\>\perp
%(2^k)\x\<x,y,-xy\>)D_{K}((2^k)\x\<1,-x,-y,xy\>),$$ for all
%extensions $K$ of $F_0(x,y)$. Consider the case where $K =
%F_0(x,y)$. Let $\sum_{i=1}^{2^k+1}{t_i}^2$ be of length $2^k+1$,
%for $t_i\in K$. Now $\sum_{i=1}^{2^k+1}{t_i}^2\in D_{K}(\<
%1\>\perp (2^k+1)\x\<x,y,-xy\>)D_{K}((2^k+1)\x\<1,-x,-y,xy\>),$ and
%hence belongs to $D_{K}(\< 1\>\perp
%2^k\x\<x,y,-xy\>)D_{K}(2^k\x\<1,-x,-y,xy\>)$. Thus, $\exists$
%$p',\ldots ,{w_{2^k}}'\in K$ such that $\sum_{i=1}^{2^k+1}{t_i}^2$
%equals $$\left({p'}^2+x\sum_{i=1}^{2^k}{{q_i}'}^2
%+y\sum_{i=1}^{2^k}{{r_i}'}^2-xy\sum_{i=1}^{2^k}{{s_i}'}^2\right)\left(\sum_{i=1}^{2^k}{{t_i}'}^2
%-x\sum_{i=1}^{2^k}{{u_i}'}^2-y\sum_{i=1}^{2^k}{{v_i}'}^2+xy\sum_{i=1}^{2^k}{{w_i}'}^2\right).$$

%Multiplication by $\sum_{i=1}^{2^k}{{t_i}'}^2
%-x\sum_{i=1}^{2^k}{{u_i}'}^2-y\sum_{i=1}^{2^k}{{v_i}'}^2+xy\sum_{i=1}^{2^k}{{w_i}'}^2$
%yields that $$\sum_{i=1}^{2^{k+1}}{{t_i}''}^2
%-x\sum_{i=1}^{2^{k+1}}{{u_i}''}^2-y\sum_{i=1}^{2^{k+1}}{{v_i}''}^2+xy\sum_{i=1}^{2^{k+1}}{{w_i}''}^2
%=\left({p'}^2+x\sum_{i=1}^{2^k}{{q_i}'}^2
%+y\sum_{i=1}^{2^k}{{r_i}'}^2-xy\sum_{i=1}^{2^k}{{s_i}'}^2\right)a^2$$
%where ${t_1}'',\ldots ,{w_{2^{k+1}}}''\in K$, since
%$D_K(2^{k+1}\x\< 1\>)$
%is a multiplicative group. %$a=\sum_{i=1}^{2^k-1}{{t_i}'}^2
%%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2$ and

%Hence, $\sum_{i=1}^{2^{k+1}}{{t_i}''}^2={p'}^2a^2=(p'a)^2$, which
%is a contradiction, since
%$$\sum_{i=1}^{2^{k+1}}{{t_i}''}^2=\left(\sum_{i=1}^{2^k+1}{{t_i}}^2\right)\left(\sum_{i=1}^{2^k}{{t_i}'}^2\right)\notin
%{K^{\times}}^2.$$

%Thus, $\langle 1\rangle\perp 2^k\times T_P$ is anisotropic over
%$F(\langle 1\rangle \perp (2^k+1) \times T_P)$.

%Hence, $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ is
%isotropic over $F(\langle 1\rangle \perp (2^k+1) \times T_P)$.
%Theorem~\ref{izh} implies that $$D_{K}(\< 1\>\perp
%(2^k+1)\x\<x,y,-xy\>)D_{K}((2^k+1)\x\<1,-x,-y,xy\>)$$ $$\subseteq
%D_{K}((2^k+1)\x\< 1\>\perp
%(2^k-1)\x\<x,y,-xy\>)D_{K}((2^k-1)\x\<1,-x,-y,xy\>),$$ for all
%extensions $K$ of $F_0(x,y)$. Consider the case where $K =
%F_0(x,y)$. Let $\sum_{i=1}^{2^k+1}{t_i}^2$ and
%$\sum_{i=1}^{2^k+1}{u_i}^2$ be of length $2^k+1$, for $t_i, u_i\in
%K$. Now, $\sum_{i=1}^{2^k+1}{t_i}^2-x\sum_{i=1}^{2^k+1}{u_i}^2\in
%D_{K}(\< 1\>\perp
%(2^k+1)\x\<x,y,-xy\>)D_{K}((2^k+1)\x\<1,-x,-y,xy\>),$ and hence
%belongs to $D_{K}((2^k+1)\x\< 1\>\perp
%(2^k-1)\x\<x,y,-xy\>)D_{K}((2^k-1)\x\<1,-x,-y,xy\>)$. Thus,
%$\exists$ ${p_1}',\ldots ,{w_{2^k-1}}'\in K$ such that
%$\sum_{i=1}^{2^k+1}{t_i}^2-x\sum_{i=1}^{2^k+1}{u_i}^2$ equals
%$$\left(\sum_{i=1}^{2^k+1}{{p_i}'}^2+x\sum_{i=1}^{2^k-1}{{q_i}'}^2
%+y\sum_{i=1}^{2^k-1}{{r_i}'}^2-xy\sum_{i=1}^{2^k-1}{{s_i}'}^2\right)\left(\sum_{i=1}^{2^k-1}{{t_i}'}^2
%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2\right).$$
%Hence ${t_i}'\neq 0$ and ${u_j}'\neq 0$ for some $i,j$ where
%$1\leq i,j\leq 2^k-1$, as otherwise
%$\sum_{i=1}^{2^k+1}{t_i}^2-x\sum_{i=1}^{2^k+1}{u_i}^2$ would not
%be attainable. Multiplication by $\sum_{i=1}^{2^k-1}{{t_i}'}^2
%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2$
%yields that
%$$\left(\sum_{i=1}^{2^k+1}{t_i}^2-x\sum_{i=1}^{2^k+1}{u_i}^2\right)\left(\sum_{i=1}^{2^k-1}{{t_i}'}^2
%-x\sum_{i=1}^{2^k-1}{{u_i}'}^2-y\sum_{i=1}^{2^k-1}{{v_i}'}^2+xy\sum_{i=1}^{2^k-1}{{w_i}'}^2\right)$$
%$$=\left(\sum_{i=1}^{2^k+1}{{p_i}'}^2+x\sum_{i=1}^{2^k-1}{{q_i}'}^2
%+y\sum_{i=1}^{2^k-1}{{r_i}'}^2-xy\sum_{i=1}^{2^k-1}{{s_i}'}^2\right)a^2.$$
%This is a contradiction, since
%$$-x\left[\left(\sum_{i=1}^{2^k+1}{u_i}^2\right)\left(\sum_{i=1}^{2^k-1}{{t_i}'}^2\right)
%+\left(\sum_{i=1}^{2^k+1}{t_i}^2\right)\left(\sum_{i=1}^{2^k-1}{{u_i}'}^2\right)\right]\neq
%x\left(\sum_{i=1}^{2^k-1}{{q_i}'}^2\right)a^2.$$

%Thus, $(2^k+1)\times\langle 1\rangle\perp (2^k-1)\times T_P$ is
%anisotropic over $F(\langle 1\rangle \perp (2^k+1) \times T_P)$,
%implying that $s(\sO)\nleq 2^k$.
%\end{proof}

%\begin{conj} For $\sO=\oct x y z \otimes_F F(\langle 1\rangle \perp n \times
%T_P)$, where $F=F_0(x,y,z)$, $s(\sO)=\underline{s}(\sO)=n$.
%\end{conj}

%Clearly for $\sO=\oct x y z \otimes_F F(\langle 1\rangle \perp n
%\times T_P)$, $s(\sO)\leq n$. It might be hoped that the
%construction of $\sO$ is sufficiently general to imply that no
%reduction of this bound takes place for any value of $n$.
%Theorem~\ref{n7}, Proposition~\ref{both2^k} and
%Proposition~\ref{level2^k+1} show that the conjecture holds in all
%cases where we can make stronger statements regarding the isotropy
%of $T_P$ than those which arise immediately from the definitions
%of level and sublevel.

%Given the similarities between sums of squares in quaternion and
%octonion algebras, it seems reasonable to propose the analogous
%conjecture for quaternion algebras also:

%\begin{conj} For $D=\quat x y \otimes_F F(\langle 1\rangle \perp n \times
%\< x,y,-xy\>)$, where $F=F_0(x,y)$, $s(\sO)=\underline{s}(\sO)=n$.
%\end{conj}

The values of $n$ specified in the above results remain the only
ones for which we can precisely determine $s\left(Q(n)\right),
\underline{s}\left(Q(n)\right), s\left(\sO(n)\right)$ or
$\underline{s}\left(\sO(n)\right)$. For other values of $n$,
bounds may be placed on these quantities via Theorem~\ref{km}, as
we will now proceed to outline.

\subsubsection{Bounding Undetermined Values}

Given the similarities between sums of squares in quaternion and
octonion algebras, that there exist octonion algebras whose level
is not of the form $2^k$ or $2^k+1$, as proved in
Theorem~\ref{n7}, suggests that the same is true for quaternion
algebras. Indeed, at a recent seminar in University College Dublin
\cite{H2}, Detlev Hoffmann kindly communicated his method of
showing the existence of infinitely many quaternion algebras whose
level is neither $2^k$ nor $2^k+1$ for some $k$, answering a
question posed by David Leep in \cite{L}. The family of quaternion
algebras that Hoffmann considers is $\left\{Q(n)\right\}$.

A key step in Hoffmann's argument is the following:

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Should I throw in "...answering a question of Leep (cite[Question blah]{LEEP})" above?
%Don't think Lewis asked the question
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\begin{lemma}\label{i1} \cite[Lemma 4.1]{H3} $i_1\left(\nTP\right)=1$ for all $n$, where $T_P$ represents the pure trace form of $Q(n)$.
\end{lemma}

Employing the same method of proof, one can obtain the
corresponding result for $\sO(n)$, as we will now proceed to
outline:

\begin{lemma}\label{Oi1} $i_1\left(\nTP\right)=1$ for all $n$, where $T_P$ represents the pure trace form of $\sO(n)$.
\end{lemma}

\begin{proof} For $P$ an ordering of $F_0$, let $P'$ represent an extension to $F$ such that $x,y$ and $z$ are negative.
Hence, $\nTP$ is indefinite with respect to $P'$, implying that
$P'$ extends to $F\left(\nTP\right)$ by \cite[Theorem 3.5 and
Remark 3.6]{ELW}. Now, since $\nTP$ has only one positive
coefficient with respect to $P'$, we can conclude that
$i_1\left(\nTP\right)=1$.
\end{proof}

Hoffmann invokes Lemma~\ref{i1} to prove the following result:

\begin{theorem}\label{HQ}\cite[Corollary 4.3]{H3} For $n=m+1+\lfloor\frac{m}{3}\rfloor$, $s\left(Q(n)\right)\in \left[m+1,
n\right]$.
\end{theorem}

The choice of $m+1+\lfloor\frac{m}{3}\rfloor$ for $n$ is
justifiable since it is the least value which ensures that the
essential dimension of $\nTP$ is greater than or equal to the
dimension of $\< 1\>\perp m\x T_Q$, thereby ensuring that
$s(Q)\nleq m$ via Theorem~\ref{km}. For the sake of consistency,
we reformulate the above result in the following manner:
$$s\left(Q(n)\right)\in \left[n-\left\lfloor\frac n
4\right\rfloor,n\right]\hspace{.1cm}\text{for
all}\hspace{.1cm}n.$$

Of course, we may similarly invoke Lemma~\ref{Oi1} to the same
effect, allowing us to obtain the analogous bounds for
$s\left(\sO(n)\right)$:

\begin{theorem}\label{sO} $s\left(\sO(n)\right)\in \left[n-\left\lfloor\frac n
8\right\rfloor,n\right]$ for all $n$.
\end{theorem}

\begin{proof} Since $i_1\left(\nTP\right)=1$ for all $n$, if $\dim
\left(\nTP\right)>\dim \left(\< 1\>\perp m\times
T_{\sO(n)}\right)$ then Theorem~\ref{km} implies that $\< 1\>\perp
m\times T_{\sO(n)}$ remains anisotropic over $F\left(\nTP\right)$,
bounding $s\left(\sO(n)\right)$ in $[m+1, n]$. Comparing
dimensions, we see that $n-\left\lfloor\frac n 8\right\rfloor -1$
is the greatest value of $m$ such that $\dim\left(\nTP\right)>\dim
\left(\< 1\>\perp m\times T_{\sO(n)}\right)$. Hence,
$s\left(\sO(n)\right)\in \left[n-\left\lfloor\frac n
8\right\rfloor,n\right]$.
\end{proof}

We note that the above result recovers the level component of
Theorem~\ref{n7}.

In a similar fashion, we may further apply Hoffmann's methodology
to considerations of the sublevel of $Q(n)$:

\begin{theorem}\label{subQ} $\underline{s}\left(Q(n)\right)\in \left[n-\left\lfloor\frac
{n+3} 4\right\rfloor,n\right]$ for all $n$.
\end{theorem}

\begin{proof} Since $i_1\left(\nTP\right)=1$ for all $n$, if $\dim
\left(\nTP\right)>\dim \left((m+1)\times T_{Q(n)}\right)$ then
Theorem~\ref{km} implies that $(m+1)\times T_{Q(n)}$ remains
anisotropic over $F\left(\nTP\right)$, bounding
$\underline{s}\left(Q(n)\right)$ in $[m+1, n]$. Comparing
dimensions, we see that $n-\left\lfloor\frac {n+3} 4\right\rfloor
-1$ is the greatest value of $m$ such that $\dim
\left(\nTP\right)>\dim \left((m+1)\times T_{Q(n)}\right)$. Hence,
$\underline{s}\left(Q(n)\right)\in \left[n-\left\lfloor\frac {n+3}
4\right\rfloor,n\right]$.
\end{proof}

The corresponding argument yields the following bounds for the
sublevel of $\sO(n)$:

\begin{theorem}\label{subO} $\underline{s}\left(\sO(n)\right)\in \left[n-\left\lfloor\frac
{n+7} 8\right\rfloor,n\right]$ for all $n$.
\end{theorem}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Might be nice to put in an example here
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Without placing any restrictions on the value that $n$ may take,
the above results represent the sharpest bounds on the level and
sublevel of $Q(n)$ and $\sO(n)$ currently available. For a large
class of values however, we may further reduce the intervals in
which these quantities are known to lie.

We achieve this reduction by exploiting Theorem~\ref{km} more
fully:

For $m=2^kd$, $m\x T_C\simeq 2^k\x\< 1\>\otimes(d\x T_C).$ In
consequence, $i_1\left(m\x T_{C}\right)\geq 2^k$ by \cite[Theorem
1.4]{EL}. Thus, invoking Theorem~\ref{km} in conjunction with this
observation, we may increase the lower bounds on
$\sub\left(Q(n)\right)$ and $\sub\left(\sO(n)\right)$ for certain
$n$, and consequently those on $s\left(Q(n)\right)$ and
$s\left(\sO(n)\right)$.

%For ease of comparison, we reformulate Theorem~\ref{subQ} and
%Theorem~\ref{subO} as follows:
%$$\sub\left(Q\left(m+\left\lfloor\frac {m-1}
%3\right\rfloor+1\right)\right)\in\left[m,m+\left\lfloor\frac {m-1}
%3\right\rfloor+1\right],$$ while
%$$\sub\left(\sO\left(m+\left\lfloor\frac {m-1}
%7\right\rfloor+1\right)\right)\in\left[m,m+\left\lfloor\frac {m-1}
%7\right\rfloor+1\right].$$

%We offer the following reductions:

\begin{theorem}\label{Qkmfull}$s\left(Q(n)\right)$ and $\sub\left(Q(n)\right)\in [m,n]$ for $n\geq m+\left\lfloor\frac
{m-2^k} 3\right\rfloor+1$, where $k$ is the $2$-adic order of $m$.
\end{theorem}

\begin{proof} \cite[Theorem 1.4]{EL} implies that $i_1\left(m\x T_{Q(n)}\right)\geq
2^k$. Comparing the essential dimensions of $\nTP$ and $m\x
T_{Q(n)}$, we see that $m+\left\lfloor\frac {m-2^k}
3\right\rfloor+1$ is the least value of $n$ such that
$\dim\left(\nTP\right)-1>\dim\left(m\x T_{Q(n)}\right)-2^k$.
Hence, for $n\geq m+\left\lfloor\frac {m-2^k} 3\right\rfloor+1$,
$m\x T_{Q(n)}$ is anisotropic over $F\left(\nTP\right)$ by
Theorem~\ref{km}, implying that $\sub\left(Q(n)\right)\in
\left[m,n\right]$.

Since $\sub\left(Q(n)\right)\leq s\left(Q(n)\right)$,
$s\left(Q(n)\right)\in [m,n]$ also.
\end{proof}

Arguing in a likewise manner, we obtain the following bounds for
the level and sublevel of $\sO(n)$:

\begin{theorem}\label{Okmfull}$s\left(\sO(n)\right)$ and $\sub\left(\sO(n)\right)\in [m,n]$ for $n\geq m+\left\lfloor\frac
{m-2^k} 7\right\rfloor+1$, where $k$ is the $2$-adic order of $m$.
\end{theorem}



%\begin{theorem}\label{subreduction} $\sub\left(Q\left(m+\left\lfloor\frac {m-2^k}
%3\right\rfloor+1\right)\right)\in\left[m,m+\left\lfloor\frac
%{m-2^k} 3\right\rfloor+1\right]$ and \newline
%$\sub\left(\sO\left(m+\left\lfloor\frac {m-2^k}
%7\right\rfloor+1\right)\right)\in\left[m,m+\left\lfloor\frac
%{m-2^k} 7\right\rfloor+1\right],$ where $k$ is the $2$-adic order
%of $m$.
%\end{theorem}

%\begin{proof} \cite[Theorem 1.4]{EL} implies that $i_1\left(m\x T_{Q(n)}\right)\geq
%2^k$. Comparing the essential dimensions of $\nTP$ and $m\x
%T_{Q(n)}$, we see that $m+\left\lfloor\frac {m-2^k}
%3\right\rfloor+1$ is the least value of $n$ such that
%$\dim\left(\nTP\right)-1>\dim\left(m\x T_{Q(n)}\right)-2^k$.
%Hence, $m\x T_{Q(n)}$ is anisotropic over $F\left(\< 1\>\perp
%\left(m+\left\lfloor\frac {m-2^k} 3\right\rfloor+1\right)\x
%T_P\right)$ by Theorem~\ref{km}, implying that
%$\sub\left(Q(n)\right)\in \left[m,m+\left\lfloor\frac {m-2^k}
%3\right\rfloor+1\right]$.

%Similarly, $m+\left\lfloor\frac {m-2^k} 7\right\rfloor+1$ is the
%least value of $n$ such that
%$\dim\left(\nTP\right)-1>\dim\left(m\x T_{\sO(n)}\right)-2^k$.
%\end{proof}

%A bit more perhaps:

%Furthermore, these reductions have the following natural
%extensions:

%\begin{theorem}\label{Qi1reduction} $s\left(Q(n)\right)$ and $\sub\left(Q(n)\right)\in [m,n]$ for $n\geq m+\left\lfloor\frac
%{m-2^k} 3\right\rfloor+1$, where $k$ is the $2$-adic order of $m$.
%\end{theorem}

%\begin{proof} In the proof of Theorem~\ref{subreduction}, we showed
%that $m\x T_{Q(n)}$ is anisotropic over $F\left(\< 1\>\perp
%\left(m+\left\lfloor\frac {m-2^k} 3\right\rfloor+1\right)\x
%T_P\right)$. Since there exists an $F$-place from $F\left(\<
%1\>\perp n\x T_P\right)$ to $F\left(\< 1\>\perp
%\left(m+\left\lfloor\frac {m-2^k} 3\right\rfloor+1\right)\x
%T_P\right)$ for $n\geq m+\left\lfloor\frac {m-2^k}
%3\right\rfloor+1$, $m\x T_{Q(n)}$ is anisotropic over
%$F\left(\nTP\right)$, implying that $\sub\left(Q(n)\right)\in
%[m,n]$.

%Since $\sub\left(Q(n)\right)\leq s\left(Q(n)\right)$,
%$s\left(Q(n)\right)\in [m,n]$ also.
%\end{proof}

%\begin{theorem} $s\left(\sO(n)\right)$ and $\sub\left(\sO(n)\right)\in [m,n]$ for $n\geq m+\left\lfloor\frac
%{m-2^k} 7\right\rfloor+1$, where $k$ is the $2$-adic order of $m$.
%\end{theorem}

%\begin{proof} Argue as in the above proof.
%\end{proof}

Of course, if $n=2^k+h$ for $h\geq 0$ sufficiently small, we may
yet again increase the lower bounds on the levels and sublevels of
$Q(n)$ and $\sO(n)$ relative to those derived from the above
results:

\begin{theorem} $s\left(Q(n)\right),\underline{s}\left(Q(n)\right),s\left(\sO(n)\right)$
and $\underline{s}\left(\sO(n)\right)\in \left[2^k,n\right]$,
where $2^k$ is the largest $2$-power $\leq n$.
\end{theorem}

\begin{proof} In the proof of Theorem~\ref{both2^k}, we showed
that $\< 1\>\perp (2^k-1)\x T_P$ is anisotropic over $F\left(\<
1\>\perp 2^k\x T_P\right)$. For $n\geq 2^k$, $\< 1\>\perp 2^k\x
T_P$ is a subform of $\< 1\>\perp n\x T_P$, implying the existence
of an $F$-place from $F\left(\< 1\>\perp n\x T_P\right)$ to
$F\left(\< 1\>\perp 2^k\x T_P\right)$ by \cite[Theorem 3.3]{Kn}.
Hence, $\< 1\>\perp (2^k-1)\x T_P$ is anisotropic over $F\left(\<
1\>\perp n\x T_P\right)$ by \cite[Proposition 3.1]{Kn}, implying
the result.
\end{proof}

Similarly, if $n=2^k+h$ for $h\geq 1$ sufficiently small, we may
interpret Theorem~\ref{level2^k+1} as a further increase to
earlier lower bounds on the levels of $Q(n)$ and $\sO(n)$:

\begin{theorem} $s\left(Q(n)\right)$ and $s\left(\sO(n)\right)\in
\left[2^k+1,n\right]$, where $2^k$ is the largest $2$-power $< n$.
\end{theorem}

\begin{proof} As in the proof of Theorem~\ref{level2^k+1}, $(2^k+1)\x\< 1\>\perp (2^k-1)\x T_P$ is anisotropic over
$F\left(\< 1\>\perp (2^k+1)\x T_P\right)$. Hence, as in the
previous proof, it is also anisotropic over $F\left(\< 1\>\perp
n\x T_P\right)$ for $n\geq 2^k+1$.
\end{proof}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Might be nice to put in an improvement of the example here
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


%In seeking a further reduction of the intervals in which $s(Q(n))$
%and $s(\sO(n))$ lie, we can look to fully exploit
%Theorem~\ref{km}, asking whether there exist values of $m$ for
%which $i_1(\< 1\>\perp m\x T_{Q(n)})$ or $i_1(\< 1\>\perp m\x
%T_{\sO(n)})>1$, allowing for an increase in those values of $m$
%for which $\< 1\>\perp m\x T_{Q(n)}$ or $\< 1\>\perp m\x
%T_{\sO(n)}$ are anisotropic over $F(\nTP)$. Unfortunately, it
%appears that such values correspond to the cases where $(m+k)\x
%T_{Q(n)}$ or $(m+k)\x T_{\sO(n)}$ are divisible by a Pfister form,
%whereby the isotropy of $\nTP$ implies that of $\< 1\>\perp m\x
%T_{Q(n)}$ or $\< 1\>\perp m\x T_{\sO(n)}$.

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%The above is a bit confused. It needs a better exposition.
%In particular, I need to clarify set-up; why the isotropy of the superform implies that of $\< 1\>\perp m\x T_C$, via i_w;
%Hmmm, might be some margin for progress via $\< 1\>\perp m\x T_C < (m+k)\x T_C < \nTP$; where m+k is a xple of 2^k
%Worth investigating!!!!!!!!!!!!!!!!!
%$dim_es(\< 1\>\perp m\x T_C)=dim_es((m+k)\x T_C)$, so I could just compare $\nTP$ with $(m+k)\x T_C$.
%Should work...not really, actually: there is a reason why not.
%Is the above worth mentioning? Not sure.
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%check old sub bound first!!!!!!!!!!!!!!
%Rewrite Theorem~\ref{subQ} in terms of $m$. Make EL observation - reduction. new sub prop
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Might be nice to put in a further improvement of the example here
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Analogous statements in terms of 2-adic valuations are probably possible for all of the others too
%these incorporate earlier bounds when stated in terms of $m$
%might need to restate s bounds in terms of m too! not m+1.
%might recover level component of Theorem~\ref{both2^k} - i.e. incorporate it in this formulation
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


%We can further reduce the above sublevel intervals by detailing
%cases where $i_1(m\x T_{Q(n)})$ and $i_1(m\x T_{\sO(n)})>1$. In
%particular, if $m=2^kp$ then $i_1(m\x T_{Q(n)})$ and $i_1(m\x
%T_{\sO(n)})>2^k$ by \cite[Theorem 1.4]{EL}.

%It appears that $i_1=1$ in general, with $i_1>1$ coming from the
%Pfister form divisibility of a sufficiently ``close'' superform.
%However, further reduction is possible in sublevel intervals via
%$i_1>1$.

For certain values of $n$, the lower bounds obtained for
$s\left(Q(n)\right), \sub\left(Q(n)\right),s\left(\sO(n)\right)$
and $\sub\left(\sO(n)\right)$ in the above results are actually
optimal with respect to the standard isotropy tests.

For example, if we consider the case where $n=15$, we note that
$s\left(Q(15)\right)$ and $\sub\left(Q(15)\right)\in [12,15]$ as a
consequence of Theorem~\ref{Qkmfull}. We cannot hope to show that
$s\left(Q(15)\right)$ or $\sub\left(Q(15)\right)\neq 12$ via the
standard methodology, since both $\< 1\>\perp 12\x T_{Q(15)}$ and
$13\x T_{Q(15)}$ are isotropic over $F\left(\< 1\>\perp 15\x
T_P\right)$, as the following argument demonstrates:

$16\x T_{Q(15)}$ is clearly isotropic over $F\left(\< 1\>\perp
15\x T_P\right)$. \cite[Theorem 1.4]{EL} implies that
$i_W\left(16\x T_{Q(15)}\right)\geq 16$. Since $\< 1\>\perp 12\x
T_{Q(15)}$ and $13\x T_{Q(15)}$ are both subforms of $16\x
T_{Q(15)}$ of codimension $<16$, they too are isotropic over
$F\left(\< 1\>\perp 15\x T_P\right)$.

Hence, more refined quadratic form theoretic consequences of the
sublevels of $Q(n)$ and $\sO(n)$ being bound above by $n-1$ are
required, if we hope to prove Conjecture~\ref{bigconj}.

\subsection{Constructions for the case where $n=l\cdot 2^k$}

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Need to think about what I want to say about the $l=2^m$ case. Might be overkill, and indeed nonsensical, as is!!!!!
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%More philosophical. Frame in terms of our earlier
%constructions...begin by highlighting a ``bounding form'' which
%has been hitherto unexploited.

As outlined in Section $3.1.1$, if we seek to construct quaternion
and octonion algebras of level and sublevel $n$ for all values of
$n$, the most natural candidates to consider are $Q(n)$ and
$\sO(n)$. However, if we restrict our attention to certain values
of $n$, other constructions appear to be equally, if not more,
appropriate.

%In Conjecture~\ref{bigconj}, we proposed constructions Q(n) and
%$\sO(n)$ level and sublevel n. Clearly, in proposing such a
%conjecture, we are implicitly positing the following more general
%conjecture.

%\begin{conj} There exist quaternion and octonion algebras of level
%$n$, and of sublevel $n$, for all $n$.
%\end{conj}

%sections 1 and 2 support Conjecture~\ref{bigconj} implies support
%for this new guy. in addition, we may consider a new, more ad-hoc,
%construction

As a first step in arriving at our constructions of $Q(n)$ and
$\sO(n)$, we observed that the isotropy of the quadratic form
$\nTP$ placed an upper bound on the level and sublevel of these
algebras. Of course, it is possible to find other, though
admittedly less natural, quadratic forms whose isotropy encodes
such an upper bound on the level and sublevel of associated
quaternion or octonion algebras.

As highlighted earlier, one such quadratic form is
$(2^k+1)\times\<1\>\perp(2^k-1)\times T_P$, whose isotropy implies
that $s(C)\leq 2^k$ (\cite[Theorem 2.2]{L} for the $Q$ case and
\cite[Proposition 2.12]{P} for the $\sO$ case). This result, and
its method of proof, admit the following generalisation:
%We will consider the form
%$(2^k+1)\times\<1\>\perp(l\cdot 2^k-1)\times T_P$.

\begin{theorem} If $(2^h+1)\times\<1\>\perp(l\cdot 2^k-1)\times
T_P$ is isotropic, where $h$ denotes the $2$-adic order of $l\cdot
2^k$, then $s(C)\leq l\cdot 2^k$.
\end{theorem}

\begin{proof} Consider the case where $C$ is a quaternion algebra. If
$(2^h+1)\times\<1\>\perp(l\cdot 2^k-1)\times T_P$ is isotropic,
then there exists $-A\in D_F\left(\< 1\>\perp(l\cdot 2^k-1)\times
T_P\right)$ for some nonzero $A\in D_F\left(2^h\x\< 1\>\right).$
Hence for some $\alpha\in F$ and $B,C,D\in D_F\left((l\cdot
2^k-1)\x \< 1\>\right)\cup \{ 0\}$ we have
$-A=\alpha^2+aB+bC-abD$, that is $-1=\frac {1}
{A^2}\left(\alpha^2A+aAB+bAC-abAD\right)$. Let
$A=\sum_{\lambda=1}^{2^h} {x_{\lambda}}^2.$ We show that there
exist $y_{\lambda}\in F$ such that $\sum_{\lambda=1}^{l\cdot 2^k}
{y_{\lambda}}^2=AB$ and $\vec{x}\cdot\vec{y}=0$, where
$\vec{x}\cdot\vec{y}$ represents the scalar product of the two
vectors, and we take $x_{2^h+1},\ldots,x_{l\cdot 2^k}=0$ if
$2^h<l\cdot 2^k$. If $B=0$, let each $y_{\lambda}=0$. If $B\neq
0$, then $\< A,AB\>\cong A\cdot\< 1,B\>$ is a subform of
$A\cdot(l\cdot 2^k)\x \< 1\>\cong (l\cdot 2^k)\x \< 1\>$, since
$B\in D_F\left((l\cdot 2^k-1)\x \< 1\>\right)$ and $A\cdot2^h\x\<
1\>\cong 2^h\x \< 1\>$. Therefore such a $\vec{y}$ exists.
Similarly $\vec{z},\vec{w}$ exist such that
$\sum_{\lambda=1}^{l\cdot 2^k} {z_{\lambda}}^2=AC$,
$\sum_{\lambda=1}^{l\cdot 2^k} {w_{\lambda}}^2=AD$ and
$\vec{x}\cdot\vec{z}=\vec{x}\cdot\vec{w}=0.$ It follows that
$\sum_{\lambda=1}^{l\cdot 2^k}\left(\frac {\alpha x_\lambda} A +
\frac {y_\lambda} A i +\frac {z_\lambda} A j +\frac {w_\lambda} A
k\right)^2=\frac 1 {A^2}\left(\alpha^2A+aAB+bAC-abAD\right)=-1$.
Therefore $s(C)\leq l\cdot 2^k$.

The same method of proof applies in the octonionic case.
\end{proof}

As the above form bounds the level of composition algebras, we may
adapt our constructions of $Q(n)$ and $\sO(n)$ by hardwiring in
the isotropy of this form as opposed to that of $\nTP$. We
introduce the following notation to signify this alteration: let
$$Q':=\openquat x y
{F}\otimes_{F}F\left((2^h+1)\times\<1\>\perp(l\cdot 2^k-1)\times
T_P\right)$$ and $$\sO':=\openoct x y z
{F}\otimes_{F}F\left((2^h+1)\times\<1\>\perp(l\cdot 2^k-1)\times
T_P\right),$$ where $h$ is the $2$-adic order of $l\cdot 2^k$,
$T_P$ is respectively the pure trace form of $\openquat x y {F}$
and $\openoct x y z {F}$ and $F$ denotes $F_0(x,y)$ and
$F_0(x,y,z)$ respectively.

By design, the level of these constructions is bounded above by
$l\cdot 2^k$. There is no a priori reason why the level or
sublevel of either $Q'$ or $\sO'$ might be strictly less than
$l\cdot 2^k$ for any values of $l$ or $k$. Moreover, given the
general fashion of these constructions, we propose that $Q'$ and
$\sO'$ are actually of level and sublevel $l\cdot 2^k$:

\begin{conj}\label{primeconj} $s\left(Q'\right)=\underline{s}\left(Q'\right)=s\left(\sO'\right)=\underline{s}\left(\sO'\right)=l\cdot 2^k$
for all $l$ and $k$.
\end{conj}

In the case where $l$ is a $2$-power, $Q'$ represents a
construction of Laghribi and Mammone (see \cite{LM}), with $\sO'$
coinciding with one of Pumpl\"un (see \cite{P}), allowing us to
conclude that $s(Q')=s(\sO')=2^k$. Moreover, we yield that
$\underline{s}(Q')=\underline{s}(\sO')=2^k$ immediately for $k\geq
2$, by \cite[Theorem 2.5]{L} for the $Q$ case and \cite[Theorem
3.11]{O'S} for the $\sO$ case, with an ad hoc argument proving the
case where $k=1$.

%Moreover, if $l$ is a $2$-power, $2^r$ say, then $l\cdot
%2^k=2^{r+k}$, whereby we will consider the form $(2^{r+k}+1)\x\<
%1\>\perp (2^{r+k}-1)\x T_P$ in preference to $(2^k+1)\x\< 1\>\perp
%(2^{r+k}-1)\x T_P$. The fact that $\dim_{es}\left((2^{r+k}+1)\x\<
%1\>\perp (2^{r+k}-1)\x T_P\right)>\dim_{es}\left((2^k+1)\x\<
%1\>\perp (2^{r+k}-1)\x T_P\right)$ justifies this preference. To
%see this we note that $\dim\left((2^{r+k}+1)\x\< 1\>\perp
%(2^{r+k}-1)\x T_P\right)>\dim\left((2^k+1)\x\< 1\>\perp
%(2^{r+k}-1)\x T_P\right)$, with the following result completing
%the argument:

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Note that the bones of the following proof are handled, perhaps more succinctly, in \cite{LM} and \cite{P}
%The proof is probably still alright for the thesis, I guess
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%\begin{theorem} $i_1\left((2^k+1)\times\<1\>\perp(2^k-1)\times
%T_P\right)=1$ for $T_P$ the pure trace form of $\quat x y$ or
%$\oct x y z$, where $F$ denotes $F_0(x,y)$ and $F_0(x,y,z)$
%respectively.
%\end{theorem}

%\begin{proof} For ease of notation, we let $\varphi_{\< x\>}$ and $\varphi_{T_P}$ denote $(2^k+1)\times\<1\>\perp
%(2^k-1)\times \< x\>$ and $(2^k+1)\times\<1\>\perp(2^k-1)\times
%T_P$ respectively.

%Let us first consider the quaternion algebra case. Suppose there
%exists $k$ such that $i_1\left(\varphi_{T_P}\right)>1$. Hence
%$2^k\times\<1\>\perp(2^k-1)\times T_P$ is isotropic over
%$F\left(\varphi_{T_P}\right)$, since it is a $1$-codimensional
%subform of $\varphi_{T_P}$. Since $\varphi_{T_P}$ is isotropic
%over $F\left(\varphi_{\< x\>}\right)$, there exists an $F$-place
%from $F\left(\varphi_{T_P}\right)$ to $F\left(\varphi_{\<
%x\>}\right)$ by \cite[Theorem 3.3]{Kn}. Thus
%$2^k\times\<1\>\perp(2^k-1)\times T_P$ is isotropic over
%$F\left(\varphi_{\< x\>}\right)$ by \cite[Proposition 3.1]{Kn}.
%Springer's Theorem implies that either
%$2^k\times\<1\>\perp(2^k-1)\times \< x\>$ or $(2^k-1)\times \<
%1,-x\>$ is isotropic over $F_0(x)\left(\varphi_{\< x\>}\right)$.

%Suppose $(2^k-1)\times \< 1,-x\>$ becomes isotropic over
%$F_0(x)\left(\varphi_{\< x\>}\right)$. Hence $2^k\x \< 1,-x\>$
%becomes hyperbolic over $F_0(x)\left(\varphi_{\< x\>}\right)$. The
%Cassels-Pfister Subform Theorem implies that $\varphi_{\< x\>}$ is
%a subform of $2^k\x \< 1,-x\>$, which is clearly false.

%Thus $2^k\times\<1\>\perp(2^k-1)\times \< x\>$ must be isotropic
%over $F_0(x)\left(\varphi_x\right)$. Since $\varphi_x$ is
%isotropic over $F_0(x)\left((2^k+1)\x\< 1\>\right)$, there exists
%an $F_0(x)$-place from $F_0(x)\left(\varphi_x\right)$ to
%$F_0(x)\left((2^k+1)\x\< 1\>\right)$ by \cite[Theorem 3.3]{Kn}.
%Thus $2^k\times\<1\>\perp(2^k-1)\times \< x\>$ is isotropic over
%$F_0(x)\left((2^k+1)\x\< 1\>\right)$ by \cite[Proposition
%3.1]{Kn}. Springer's Theorem implies that either $2^k\times\<1\>$
%or $(2^k-1)\times \< 1\>$ is isotropic over $F_0\left((2^k+1)\x\<
%1\>\right)$. However, both $2^k\times\<1\>$ and $(2^k-1)\times \<
%1\>$ remain anisotropic over $F_0\left((2^k+1)\x\< 1\>\right)$ by
%\cite[Theorem 1]{H1}.

%Hence, $2^k\times\<1\>\perp(2^k-1)\times \< x\>$ is anisotropic
%over $F_0(x)\left(\varphi_x\right)$, whereby
%$2^k\times\<1\>\perp(2^k-1)\times T_P$ is anisotropic over
%$F\left(\varphi_{T_P}\right)$, allowing us to conclude that
%$i_1\left(\varphi_{T_P}\right)=1$ for all $k$.

%Next we consider the case where $T_P$ represents the pure trace
%form of $\oct x y z$. We let $\varphi_{\< x,y,-xy\>}$ denote
%$(2^k+1)\times\<1\>\perp(2^k-1)\times \< x,y,-xy\>$, with
%$\varphi_{T_P}$ defined as above.

%Suppose there exists $k$ such that
%$i_1\left(\varphi_{T_P}\right)>1$. Hence
%$2^k\times\<1\>\perp(2^k-1)\times T_P$ is isotropic over
%$F\left(\varphi_{T_P}\right)$, since it is a $1$-codimensional
%subform of $\varphi_{T_P}$. Since $\varphi_{T_P}$ is isotropic
%over $F\left(\varphi_{\< x,y,-xy\>}\right)$, there exists an
%$F$-place from $F\left(\varphi_{T_P}\right)$ to
%$F\left(\varphi_{\< x,y,-xy\>}\right)$ by \cite[Theorem 3.3]{Kn}.
%Thus $2^k\times\<1\>\perp(2^k-1)\times T_P$ is isotropic over
%$F\left(\varphi_{\< x,y,-xy\>}\right)$ by \cite[Proposition
%3.1]{Kn}. Springer's Theorem implies that either
%$2^k\times\<1\>\perp(2^k-1)\times \< x,y,-xy\>$ or $(2^k-1)\times
%\< 1,x,y,-xy\>$ is isotropic over $F_0(x,y)\left(\varphi_{\<
%x,y,-xy\>}\right)$.

%As was proved above, $2^k\times\<1\>\perp(2^k-1)\times \<
%x,y,-xy\>$ is anisotropic over $F_0(x,y)\left(\varphi_{\<
%x,y,-xy\>}\right)$, implying that $(2^k-1)\times \< 1,x,y,-xy\>$
%must be isotropic over $F_0(x,y)\left(\varphi_{\<
%x,y,-xy\>}\right)$. However, if $(2^k-1)\times \< 1,x,y,-xy\>$ is
%isotropic over $F_0(x,y)\left(\varphi_{\< x,y,-xy\>}\right)$,
%$2^k\x \< 1,-x,-y,xy\>$ is hyperbolic over
%$F_0(x,y)\left(\varphi_{\< x,y,-xy\>}\right)$, whereby
%$\varphi_{\< x,y,-xy\>}$ is a subform of $2^k\x \< 1,-x,-y,xy\>$
%by the Cassels-Pfister Subform Theorem, which is clearly false.

%Hence, $2^k\times\<1\>\perp(2^k-1)\times T_P$ is anisotropic over
%$F\left(\varphi_{T_P}\right)$, allowing us to conclude that
%$i_1\left(\varphi_{T_P}\right)=1$ for all $k$.
%\end{proof}

%mention succinct recovery of 2^k construction with Theorem~\ref{km}

Since the conjecture is true in the above case, we will restrict
our attention to the situation where $l$ is not a $2$-power.
Moreover, we will henceforth assume without loss of generality
that $l$ is odd, whereby $k$ is the $2$-adic order of $l\cdot
2^k$.

%Tie the above together with the preceding comment...too disjointed
%at present. In fact, we should probably move this observation to
%where we use it, ie before i1 theorem.

Having introduced $Q'$ and $\sO'$, it is reasonable to wonder why
we might consider these algebras as opposed to $Q(l\cdot 2^k)$ and
$\sO(l\cdot 2^k)$ in seeking quaternion and octonion algebras of
level and sublevel $l\cdot 2^k$.

%In seeking to construct quaternion and octonion algebras of level
%$l\cdot 2^k$, one might more profitably consider $Q'$ and $\sO'$,
%as opposed to $Q(l\cdot 2^k)$ and $\sO(l\cdot 2^k)$, since they
%are generally of greater essential dimension.

%i'll have to look at the k values in light of the above...done
We begin our justification of this preference by noting that the
dimension of $(2^k+1)\x\< 1\>\perp (l\cdot 2^k-1)\x T_P$ is
greater than that of $\< 1\>\perp (l\cdot 2^k)\x T_P$ for $k\geq
1$ in the quaternionic case, and for $k\geq 2$ in the octonionic
case. Clearly, if we can show that
$i_1\left((2^k+1)\times\<1\>\perp(l\cdot 2^k-1)\times
T_P\right)=1$, then $(2^k+1)\x\< 1\>\perp (l\cdot 2^k-1)\x T_P$
will be of greater essential dimension than $\< 1\>\perp (l\cdot
2^k)\x T_P$ for $k$ as above. Theorem~\ref{km} would then allow
for the bounding of $s(Q')$ and $s(\sO')$ in a smaller interval
than that in which we can currently place $s(Q(n))$ and
$s(\sO(n))$.

The suggestion that $i_1\left((2^k+1)\times\<1\>\perp(l\cdot
2^k-1)\times T_P\right)=1$ is an entirely reasonable one. Consider
the form $\varphi\simeq\pi\otimes\rho$, with $\pi$ isometric to an
$n$-fold Pfister form where $n\geq 1$. Invoking \cite[Theorem
1.4]{EL}, we yield $i_1\left(\varphi\right)\geq \dim \pi$.
Theorem~\ref{km} implies that for $\varphi'$ a subform of
$\varphi$ whose codimension is strictly less than $\dim \pi$, we
have $i_1\left(\varphi'\right)>1$. Hence, we know that the class
of forms represented by $\varphi'$ have first Witt index strictly
greater than $1$. We suspect that such forms might actually be the
\emph{only} forms whose first Witt index is strictly greater than
$1$.

The following theorem, which constitutes the main result of this
section, shows that the first Witt index of
$(2^k+1)\times\<1\>\perp(l\cdot 2^k-1)\times T_P$ does indeed
equal $1$, as suspected:

%Highlight the following as being this section's main result. A
%discussion: unless el or similar, most forms probably have i1=1.

\begin{theorem}\label{longi1} For $Q'$ and $\sO'$ as above, $i_1\left((2^k+1)\times\<1\>\perp(l\cdot 2^k-1)\times
T_P\right)=1$. %where $2^k$ is the highest $2$-power dividing
%$l\cdot 2^k$. %need I qualify what values k and l may take? Ok now in light of l>1 hardwiring into construction
\end{theorem}

%Have I covered the octonion case in this proof??????????????????????????
\begin{proof} For ease of notation, we let $\psi_{\< x\>}$ and $\psi_{T_P}$ denote $(2^k+1)\times\<1\>\perp(l\cdot 2^k-1)\times \<
x\>$ and $(2^k+1)\times\<1\>\perp(l\cdot 2^k-1)\times T_P$
respectively.

Let us first consider the quaternion algebra case. Suppose there
exist $k$ and $l$ such that $i_1\left(\psi_{T_P}\right)>1$. Hence
$2^k\times\<1\>\perp(l\cdot 2^k-1)\times T_P$ is isotropic over
$F\left(\psi_{T_P}\right)$, since it is a $1$-codimensional
subform of $\psi_{T_P}$. Since $\psi_{T_P}$ is isotropic over
$F\left(\psi_{\< x\>}\right)$, there exists an $F$-place from
$F\left(\psi_{T_P}\right)$ to $F\left(\psi_{\< x\>}\right)$ by
\cite[Theorem 3.3]{Kn}. Thus $2^k\times\<1\>\perp(l\cdot
2^k-1)\times T_P$ is isotropic over $F\left(\psi_{\< x\>}\right)$
by \cite[Proposition 3.1]{Kn}. Springer's Theorem implies that
either $2^k\times\<1\>\perp(l\cdot 2^k-1)\times \< x\>$ or
$(l\cdot 2^k-1)\times \< 1,-x\>$ is isotropic over
$F_0(x)\left(\psi_{\< x\>}\right)$.

Suppose $(l\cdot 2^k-1)\times \< 1,-x\>$ becomes isotropic over
$F_0(x)\left(\psi_{\< x\>}\right)$. Hence $2^n\x \< 1,-x\>$
becomes hyperbolic over $F_0(x)\left(\psi_{\< x\>}\right)$, where
$2^n>l\cdot 2^k-1$. The Cassels-Pfister Subform Theorem implies
that $\psi_{\< x\>}$ is a subform of $2^n\x \< 1,-x\>$, which is
clearly false.

Thus $2^k\times\<1\>\perp(l\cdot 2^k-1)\times \< x\>$ must be
isotropic over $F_0(x)\left(\psi_{\< x\>}\right)$. Hence,
$\dim_{es}\left(2^k\times\<1\>\perp(l\cdot 2^k-1)\times \<
x\>\right)=\dim_{es}\left(\psi_{\< x\>}\right)$ by
Theorem~\ref{km}.

Consider $2^k\x \< 1\>\perp l\cdot 2^k\x \< x\>\simeq 2^k\x \<
1\>\otimes(\< 1\>\perp l\x \< x\>)$. Now $i_1\left(2^k\x \<
1\>\perp l\cdot 2^k\x \< x\>\right)\geq 2^k$ by \cite[Theorem
1.4]{EL}. In addition, choosing an ordering $P$ of $F_0(x)$ such
that $x$ is negative implies that $i_1\left(2^k\x \< 1\>\perp
l\cdot 2^k\x \< x\>\right)\leq 2^k$ by \cite[Theorem 3.5 and
Remark 3.6]{ELW}. Thus $i_1\left(2^k\x \< 1\>\perp l\cdot 2^k\x \<
x\>\right)=2^k$, implying that $2^k\times\<1\>\perp(l\cdot
2^k-1)\times \< x\>$ is isotropic over $F_0\left(2^k\x \< 1\>\perp
l\cdot 2^k\x \< x\>\right)$ and hence that $\dim_{es}\left(2^k\x
\< 1\>\perp l\cdot 2^k\x \<
x\>\right)=\dim_{es}\left(2^k\times\<1\>\perp(l\cdot 2^k-1)\times
\< x\>\right)$ by Theorem~\ref{km}.

We may therefore conclude that $i_1\left(\psi_{\<
x\>}\right)=2^k$. Consequently $(2^k+1)\times\<1\>\perp((l-1)\cdot
2^k)\times \< x\>$ is isotropic over $F_0\left(\psi_{\<
x\>}\right)$, since it is a subform of $\psi_{\< x\>}$ of
codimension $2^k-1$. Invoking Theorem~\ref{km} yields that
$i_1\left((2^k+1)\times\<1\>\perp((l-1)\cdot 2^k)\times \<
x\>\right)=1$.

However $l$ is odd, since $2^k$ is the highest $2$-power dividing
$l\cdot 2^k$, whereby $2^{k+1}$ divides $(l-1)\cdot 2^k$. Thus
$2^{k+1}\times\<1\>\perp((l-1)\cdot 2^k)\times \< x\>\simeq
2^{k+1}\x \< 1\>\otimes\left(\< 1\>\perp \frac {l-1} 2 \x \<
x\>\right)$. Now $i_1\left(2^{k+1}\times\<1\>\perp((l-1)\cdot
2^k)\times \< x\>\right)=2^{k+1}$ by \cite[Theorem 1.4]{EL} and
\cite[Theorem 3.5 and Remark 3.6]{ELW}. Hence
$(2^k+1)\times\<1\>\perp((l-1)\cdot 2^k)\times \< x\>$ becomes
isotropic over $F_0\left(2^{k+1}\times\<1\>\perp((l-1)\cdot
2^k)\times \< x\>\right)$, whereby
$\dim_{es}\left(2^{k+1}\times\<1\>\perp((l-1)\cdot 2^k)\times \<
x\>\right)=\dim_{es}\left((2^k+1)\times\<1\>\perp((l-1)\cdot
2^k)\times \< x\>\right)$ by Theorem~\ref{km}. Therefore we
conclude that $i_1\left((2^k+1)\times\<1\>\perp((l-1)\cdot
2^k)\times \< x\>\right)=2^k+1$, contradicting the supposition
that $2^k\times\<1\>\perp(l\cdot 2^k-1)\times \< x\>$ is isotropic
over $F_0(x)\left(\psi_{\< x\>}\right)$.
%our conclusion that $i_1((2^k+1)\times\<1\>\perp((l\cdot
%2^k)-1)\times \< x\>)=2^k$, which in turn contradicts our earlier
%conclusion that $\dim_{es}(2^k\times\<1\>\perp((l\cdot
%2^k)-1)\times \< x\>)=\dim_{es}((2^k+1)\times\<1\>\perp((l\cdot
%2^k)-1)\times \< x\>)$,

Hence $2^k\times\<1\>\perp(l\cdot 2^k-1)\times T_P$ is in fact
anisotropic over $F\left(\psi_{T_P}\right)$, contradicting the
supposition that there exist $k$ and $l$ such that
$i_1\left(\psi_{T_P}\right)>1$.

Next, we prove the result for $\sO'$. In this case, we let
$\psi_{\< x,y,-xy\>}$ denote $(2^k+1)\times\<1\>\perp(l\cdot
2^k-1)\times \< x,y,-xy\>$ and $\psi_{T_P}$ be defined as above,
with $T_P$ denoting the pure trace form of $\sO'$ here.

Again we suppose that there exist $k$ and $l$ such that
$i_1\left(\psi_{T_P}\right)>1$. Hence $2^k\times\<1\>\perp(l\cdot
2^k-1)\times T_P$ is isotropic over $F\left(\psi_{T_P}\right)$,
since it is a $1$-codimensional subform of $\psi_{T_P}$. Since
$\psi_{T_P}$ is isotropic over $F\left(\psi_{\<
x,y,-xy\>}\right)$, there exists an $F$-place from
$F\left(\psi_{T_P}\right)$ to $F\left(\psi_{\< x,y,-xy\>}\right)$
by \cite[Theorem 3.3]{Kn}. Thus $2^k\times\<1\>\perp(l\cdot
2^k-1)\times T_P$ is isotropic over $F\left(\psi_{\<
x,y,-xy\>}\right)$ by \cite[Proposition 3.1]{Kn}. Springer's
Theorem implies that either $2^k\times\<1\>\perp(l\cdot
2^k-1)\times \< x,y,-xy\>$ or $(l\cdot 2^k-1)\times \<
1,-x,-y,xy\>$ is isotropic over $F_0(x,y)\left(\psi_{\<
x,y,-xy\>}\right)$.

Suppose $(l\cdot 2^k-1)\times \< 1,-x,-y,xy\>$ becomes isotropic
over $F_0(x,y)\left(\psi_{\< x,y,-xy\>}\right)$. Hence $2^n\x \<
1,-x,-y,xy\>$ becomes hyperbolic over $F_0(x,y)\left(\psi_{\<
x,y,-xy\>}\right)$, where $2^n>l\cdot 2^k-1$. The Cassels-Pfister
Subform Theorem implies that $\psi_{\< x,y,-xy\>}$ is a subform of
$2^n\x \< 1,-x,-y,xy\>$, which is clearly false.

Thus $2^k\times\<1\>\perp(l\cdot 2^k-1)\times \< x,y,-xy\>$ must
be isotropic over $F_0(x,y)\left(\psi_{\< x,y,-xy\>}\right)$.
However, we know that $i_1\left(\psi_{\< x,y,-xy\>}\right)=1$
(this is what we proved in the first half of the proof), whereby
$\dim_{es}\left(\psi_{\<
x,y,-xy\>}\right)>\dim_{es}\left(2^k\times\<1\>\perp(l\cdot
2^k-1)\times \< x,y,-xy\>\right)$. Hence,
$2^k\times\<1\>\perp(l\cdot 2^k-1)\times \< x,y,-xy\>$ is
anisotropic over $F_0(x,y)\left(\psi_{\< x,y,-xy\>}\right)$ by
Theorem~\ref{km}, contradicting the supposition that there exist
$k$ and $l$ such that $i_1\left(\psi_{T_P}\right)>1$.
\end{proof}

Since the above argument relies upon the fact that $l$ is an odd
number $>1$, we cannot hope to apply it to determine the first
Witt index of $(2^k+1)\x\< 1\>\perp (2^k-1)\x T_P$. However, we
recall that \cite[Proposition 3.4]{LM} and \cite[Proposition
3.5]{P} state that $2^k\x\< 1\>\perp (2^k-1)\x T_P$ remains
anisotropic over $F\left((2^k+1)\x\< 1\>\perp (2^k-1)\x
T_P\right)$, where $T_P$ respectively represents the pure trace
form of $\openquat x y {F}$ and $\openoct x y z {F}$ and $F$
respectively denotes $F_0(x,y)$ and $F_0(x,y,z)$. Since $2^k\x\<
1\>\perp (2^k-1)\x T_P$ is a $1$-codimensional subform of
$(2^k+1)\x\< 1\>\perp (2^k-1)\x T_P$, Theorem~\ref{km} allows us
to conclude that $i_1\left((2^k+1)\x\< 1\>\perp (2^k-1)\x
T_P\right)=1$, for all $k$. Hence, we can incorporate into
Theorem~\ref{longi1} the case where $l$ is a $2$-power, whereby we
obtain: $$i_1\left((2^h+1)\times\<1\>\perp(l\cdot 2^k-1)\times
T_P\right)=1\hspace{.1cm}\text{for all values of}\hspace{.1cm}l.$$

%Clumbsy formulation above: all $k$ and all $l$ looks like a
%mistake.

As alluded to earlier, as an immediate consequence of
Theorem~\ref{longi1} we may increase the lower bounds on the level
of our prospective composition algebras of level $l\cdot 2^k$. As
$\dim_{es}\left((2^k+1)\x\< 1\>\perp (l\cdot 2^k-1)\x T_P\right)>
\dim_{es}\left(\< 1\>\perp (l\cdot 2^k)\x T_P\right)$ for $k\geq
1$ in the quaternionic case, and for $k\geq 2$ in the octonionic
case, employing Theorem~\ref{km} allows us to obtain such a
reduction.

For example, we know that $\underline{s}(Q(24))$ and
$s\left(Q(24)\right)\in [18,24]$ by Theorem~\ref{HQ} and
Theorem~\ref{subQ}, whereas for $l=k=3$ we can say that $\sub(Q')$
and $s(Q')\in [20,24]$:

Since $20\times T_{Q'}\simeq 4\x\< 1\>\otimes\left(5\x
T_{Q'}\right)$, $i_1\left(20\times T_{Q'}\right)\geq 4$ by
\cite[Theorem 1.4]{EL}. Hence, $\dim_{es}\left(20\times
T_{Q'}\right)\leq 76$. Theorem~\ref{longi1} implies that
$\dim_{es}\left(9\times \< 1\> \perp 23\x T_P\right)=77$. Hence,
$20\times T_{Q'}$ remains anisotropic over $F\left(9\times \< 1\>
\perp 23\x T_P\right)$ by Theorem~\ref{km}, whereby $\sub(Q')\in
[20,24]$.

Similarly, while we currently cannot make any stronger statement
regarding $Q(48)$ than $\sub\left(Q(48)\right)$ and
$s\left(Q(48)\right)\in [36,48]$, we have that
$\sub\left(Q'\right)$ and $s(Q')\in [40,48]$ for $l=3$ and $k=4$.

%what follows needs fixing:

As previously noted, Conjecture~\ref{primeconj} is true in the
case where $l$ is a power of $2$. As an additional application of
Theorem~\ref{longi1}, we conclude by resolving one further case of
this conjecture:

%Can probably do more cases in the following, i.e. for smaller values!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%Next, we consider the case where $l$ is any other
%$2$-power:

%\begin{theorem} $s(Q')=\underline{s}(Q')=s(\sO')=\underline{s}(\sO')=l\cdot 2^k$ for $l=2^m$.
%\end{theorem}

%\begin{proof} $s(Q')\leq l\cdot 2^k=2^{k+m}$ by construction. Now,
%$\underline{s}(Q')\leq 2^{k+m}\iff \< 1\>\perp (2^{k+m}-1)\x T_P$
%is isotropic, by \cite[Theorem 3.5]{O'S}. However,
%$dim_{es}\left(\< 1\>\perp (2^{k+m}-1)\x T_P\right)$ is clearly
%less than $dim_{es}\left((2^k+1)\x\< 1\>\perp (2^{k+m}-1)\x
%T_P\right)$ by Theorem~\ref{longi1}, whereby $\< 1\>\perp
%(2^{k+m}-1)\x T_P$ is anisotropic over $F\left((2^k+1)\x\<
%1\>\perp (2^{k+m}-1)\x T_P\right)$ by Theorem~\ref{km}.

%\end{proof}

%Finally, we can determine $s(\sO')$ and $\underline{s}(\sO')$ in
%one additional case:

\begin{theorem} For $l=3$ and $k=1$, $s(\sO')=\underline{s}(\sO')=l\cdot 2^k=6$.
\end{theorem}

\begin{proof} $s(\sO')\leq l\cdot 2^k=6$ by
construction. Suppose $\underline{s}(\sO')\leq 5$. Then, as in the
proof of Theorem~\ref{n7}, either $6\times T_P$ or $6\x (\<
1\>\perp \beta)$ is isotropic over $F(3\x \< 1\>\perp 5\x T_P)$,
where $\beta$ is some $5$-dimensional subform of $T_P$. Hence,
either $\dim_{es}(6\times T_P)$ or $\dim_{es}(6\x (\< 1\>\perp
\beta))\geq \dim_{es}(3\x \< 1\>\perp 5\x T_P)$ by
Theorem~\ref{km}. However, $\dim_{es}(6\times T_P)=32$ and
$\dim_{es}(6\x (\< 1\>\perp \beta))\leq 35$, whereas
$\dim_{es}(3\x \< 1\>\perp 5\x T_P)=37$ by Theorem~\ref{longi1},
yielding a contradiction.
\end{proof}
\medskip


%\stackrel{s.b.}{\sim}

%won't need to rely \emph{solely} on examples if i can prove the
%above in general. might be hard though. should still throw in an
%example (48 case say) for the sake of illustration.


%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Combine Leep forms with i1>1 for good reductions: try 20 and 24;
%48 and 56; etc.
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

%smaller intervals in contrast with earlier construction for these
%values of n (xples of 2-powers)

%example with 48, I guess: there exists a Q whose level lies
%between 40 and 48.

%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
%Works for sublevel too obviously
%!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\textbf{\Large{Acknowledgements}}

I would like to thank Detlev Hoffmann for communicating his recent
work, which made me aware of some of the latest techniques and
opened up new avenues of approach. I am hugely indebted to the
hospitality of Jean-Pierre Tignol and Jan Van Geel for stimulating
and enjoyable stays at Universit\'{e} catholique de Louvain and
Universiteit Gent, during which most of the above work took place.
In addition, I am very grateful to my supervisor, Thomas Unger,
for all the valuable suggestions he made regarding this article.
Useful comments from Susanne Pumpl\"un were also appreciated.
Finally, I gratefully acknowledge the support of my research by
the Irish Research Council for Science, Engineering and Technology
(scholarship RS Aug/2002/174-7).

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\medskip
\author{James O'Shea,\\ School of Mathematical Sciences, Science Centre, \\ University College Dublin,
Belfield, Dublin 4, Ireland.\\ james.oshea@ucd.ie}

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