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\begin{document}

\title{Levels of Quaternion Algebras}


\author{Detlev W. Hoffmann}



\address{%
School of Mathematical Sciences\\
University of Nottingham\\
University Park\\ 
Nottingham, NG7 2RD\\
UK}
 
\email{detlev.hoffmann@nottingham.ac.uk}

\thanks{Supported in part by
the European research network HPRN-CT-2002-00287 ``Algebraic
$K$-Theory, Linear Algebraic Groups and Related Structures''}

\subjclass[2000]{Primary 11E04; Secondary 11E25, 12D15, 16K20}
\keywords{Quaternion algebra, level, sublevel, quadratic form, 
function field of a quadratic form}

\date{January 31, 2007}



\begin{abstract}
The level of a ring $R$ with $1\neq 0$ is the smallest positive integer $s$
such that $-1$ can be written as a sum of $s$ squares in $R$, provided
$-1$ is a sum of squares at all.  D.W.~Lewis showed that any value of type $2^n$ or 
$2^n+1$ can be realized as level of a quaternion algebra, and he
asked whether there exist quaternion algebras whose levels are not of 
that form. Using function fields of quadratic forms, we construct such examples.
\end{abstract}

\maketitle


\section{Introduction}
\label{intro}
A famous result by E.~Artin and O.~Schreier \cite{AS} says that if $F$
is a field (of characteristic $\neq 2$) then $F$ has at least one ordering
if and only if $-1$ cannot be written as a sum of squares in $F$. In this
situation, the field $F$ is said to be {\it formally real}, or {\it real}
for short. Naturally, one might then ask how many squares are actually 
needed to write $-1$ as a sum of squares in a nonreal field $F$.
This has lead to the definition of the level which we formulate for 
arbitrary (possibly nonassociative) rings with $1\neq 0$:

\begin{definition}  Let $R$ be a ring with $1\neq 0$. Let {\em level} $s(R)$
is defined as follows:  
\begin{enumerate}
\item If $-1$ is not a sum of squares in $R$, then $s(R)=\infty$.
\item If $-1$ is a sum of squares in $R$, then
$$s(R)=\min\{n\,|\,\exists x_1,\cdots ,x_n\in R\,:\ 
-1=x_1^2+\cdots +x_n^2\}$$
\end{enumerate}
\end{definition}

In the early 1930s, Van der Waerden asked which values can arise as 
level of a field.  At the time, all fields where the level was known and
finite had level $1$, $2$ or $4$.  H.~Kneser \cite{Kneser} proved in 1934 that
the only possible finite values were of the form $1$, $2$, $4$, $8$ or
certain multiples of $16$, though there were still no known fields of finite
level $>4$.  The complete solution to the level question
was finally given by Pfister \cite{P1} who showed that the level of a field,
if finite, must always be a $2$-power, and that all these values could
in fact be realized.

The level question for integral domains was solved in 1980 by Z.D.~Dai, T.Y.~Lam 
and  C.K.~Peng \cite{DLP} who proved that any positive integer can occur as level
of an integral domain, more precisely, they showed that the integral domain
$$R=\RR [X_1,\cdots ,X_n]/(1+X_1^2+\cdots +X_n^2)$$
has level $n$.  The proof is topological in nature and invokes the Borsuk-Ulam 
theorem.  Incidentally, the quotient field $F=\Quot(R)$ has level $2^k$
where $k$ is such that $2^k\leq n <2^{k+1}$, yielding Pfister's original 
examples of fields whose level is a prescribed $2$-power.

Levels of noncommutative rings have been studied by various authors, see
Lewis's survey \cite{Lew3} and the list of references there.  There are
variations of the notion of level in the noncommutative case such as 
the product level, where squares are replaced by products of squares, and for
rings $R$ with involution $\sigma$ such as the hermitian level, where squares
are replaced by ``hermitian squares'' $x\sigma(x)$.  But we will only 
consider the level as defined above.

Most of the study of levels in the noncommutative case concerns 
central simple algebras, and quaternion algebras in particular.
Obviously, we may assume that the centers of these algebras are
fields of characteristic different from $2$ in order for the level problem
to be of any interest.
Lewis \cite{Lew1} constructed examples of quaternion division algebras whose 
levels take any given prescribed value of the form $2^k$ or $2^k+1$, while indicating 
that it is not known whether other values can be realized. 

This has then been formulated explicitly as a question by Leep 
\cite[Question (2)]{Leep}, and again by Lewis \cite[Open Question 1]{Lew3}:
\begin{question*} What integers can occur as $s(D)$ for a quaternion
division algebra $D$~?  In particular, can $s(D)$ take values that are not 
of the form $2^k$ or $2^k+1$~?
\end{question*}

In this note, we will give a partial answer to that question by showing that
there are infinitely many values not of the form $2^k$ or $2^k+1$ 
that occur as level of quaternion division algebras.  More precisely,

\begin{theorem}\label{mainthm}  Let $m$ be a nonnegative integer.  Then there
exists a quaternion division algebra $D$ with $m+1\leq s(D)\leq m+1+[m/3]$
(where $[x]$ ($x\in\RR$) denotes the largest integer $\leq x$).
\end{theorem}
For example, for $m=5$, we deduce the existence of a 
quaternion division algebra $D$ with $s(D)\in\{ 6,7\}$.
For $m=17$ and $m=23$, we get such algebras $D$, $D'$ with 
$18\leq s(D)\leq 23$ and $24\leq s(D')\leq 31$.  More generally, for
any $k\geq 4$, we can in a similar way find disjoint intervals $I$ 
and $I'$ inside $[2^k+2, 2^{k+1}-1]$ and construct quaternion division
algebras $D$ and $D'$ with $s(D)\in I$ and $s(D')\in I'$, showing that 
in each interval bounded by large enough consecutive $2$-powers, there
exist at least two values not of the form $2^k$ or $2^k+1$ that can be 
realized as level of quaternion division algebras.

Our method of proof uses function fields of quadrics and facts about
isotropy of quadratic forms over such function fields.
The arguments as such in our proofs are rather elementary. However, at one
point we do have to invoke a very deep result by Karpenko and Merkurjev, 
Theorem \ref{karp-merk}.  The drawback of our elementary approach is that
we can only give bounds for the  level of the constructed
quaternion algebra and not its exact value.  It should also be noted that our 
methods are in spirit similar to those employed by
Laghribi and Mammone \cite{LM} who gave a construction of quaternion 
algebras of level $2^k$ resp. $2^k+1$ different from Lewis's original 
construction of such algebras.

We should note that James O'Shea \cite{O} has used similar methods to study the 
level of octonion algebras and to construct octonion algebras
of prescribed level.

The paper is structured as follows.  In the next section, we collect results from
quadratic form theory that will be needed in our construction.  In section 3, we
motivate and sketch the idea underpinning our construction and explain its limitations.
The construction itself will be given in section 4, where we also derive explicit bounds 
on the level of the produced quaternion algebras.  In section 5, we sketch a result by
O'Shea \cite{O} concerning quaternion algebra whose so-called sublevels are
bounded in a similar way.

\section{Some facts about quadratic forms}

Throughout this article, fields are assumed to be of characteristic $\neq 2$.
All what we need from the algebraic theory of quadratic forms over fields
can be 
found in Lam's book \cite{Lam}.  By `quadratic form' over a field $F$ 
(or `form' for short), we will always mean
a finite-dimensional nonsingular quadratic form defined over $F$.  
Recall that a form over $F$
is called anisotropic if it does not represent $0$ nontrivially over $F$, 
and hyperbolic if 
it is isometric to an orthogonal sum of hyperbolic planes $\HH = \qf{1,-1}$.

An orthogonal sum $\phi\perp\cdots\perp\phi$ of $n$ copies of a form $\phi$ over $F$
will be denoted by $n\times\phi$, whereas $a\phi$ with $a\in F^*$
denotes the form $\phi$ scaled by the factor $a$. 

By Witt decomposition,
any form $\phi$ decomposes up to isometry in a unique way as an 
orthogonal sum  $\phi\cong\phi_{an}\perp\phi_h$ with $\phi_{an}$ 
anisotropic and $\phi_h$ hyperbolic.  The Witt index of $\phi$ is defined to be 
$i_W(\phi)=\frac{1}{2}\dim\phi_h$, and it is in fact nothing else but the dimension
of a maximal totally isotropic subspace of $\phi$.

A form $\pi$ is called an $n$-fold Pfister form if $\pi$ is isometric to a 
tensor product of $n$ binary forms: 
$\pi\cong\qf{1,-a_1}\otimes\cdots\otimes\qf{1,-a_n}$, $a_i\in F^*$.
We write $\pff{a_1,\cdots ,a_n}$ for short.

If $K/F$ is a field extension and $\phi$ is a form over $F$, we denote by
$\phi_K$ the form obtained from $\phi$ by scalar extension to $K$.

The function field $F(\phi)$ of a quadratic form $\phi$ over $F$ is defined as follows.
If $\dim\phi\geq 3$, then $F(\phi)$ is the function field of the projective quadric
$\phi = 0$.  If $\phi\cong\qf{a,b}$ is anisotropic 
(i.e. $\phi\not\cong\HH$), we put $F(\phi)=F(\sqrt{-ab})$. If $\phi\cong\HH$ or
if $\dim\phi =1$, we put $F(\phi)=F$.

It is well known that for $\dim\phi\geq 2$, $F(\phi)$ can be realized as a 
purely transcendental extension of transcendence degree $\dim\phi -2$, followed
by a quadratic extension, and that $F(\phi)/F$ is purely transcendental
if and only if $\phi$ is isotropic.  

If $\phi$ is an anisotropic form over $F$ of dimension $\geq 2$, then $\phi$ is isotropic 
over $F(\phi)$, and the first Witt index $i_1(\phi )$ of $\phi$ is defined
to be $i_W(\phi_{F(\phi)})$.  It is a well known fact due to the generic nature
of the function field that for any field extension $K/F$ with $\phi_K$ isotropic,
one has $i_W(\phi_K)\geq i_1(\phi)$.

Our construction uses in a crucial way the isotropy behaviour of a quadratic form
when passing to the function field of another form.

The next lemma is folklore.

\begin{lemma}\label{folk}  Let $\phi$ and $\psi$ be forms over $F$, and
let $F(x)$ be the rational function field in the variable $x$ over $F$.
Then $i_W(\phi\perp x\psi)=i_W(\phi)+i_W(\psi)$.  In particular, $\phi\perp x\psi$
is anisotropic over $F(x)$ if and only if $\phi$ and $\psi$ are anisotropic
over $F$.
\end{lemma}
\begin{proof} Using a standard degree argument, one readily checks
that $\phi\perp x\psi$ is anisotropic over $F(x)$ if and only if $\phi$ and $\psi$ are 
anisotropic over $F$ (this is in fact \cite[Exercise 3, p.~313]{Lam}).
The statement about the Witt indices for arbitrary $\phi$ and $\psi$ now follows by 
applying Witt decomposition to $\phi$ and $\psi$ over $F$.
\end{proof}

%%\begin{corollary}\label{folk-cor}  Let $\phi$, $\psi$, $\gamma$ be forms
%%over $F$.  Then $$i_W((\phi\perp x\psi)_{F(x)(\gamma)})=
%%i_W(\phi_{F(\gamma)})+i_W(\psi_{F(\gamma)})\ .$$
%%\end{corollary}
%%
%%\begin{proof}  Note that $F(x)(\gamma)\cong F(\gamma)(x)$ over $F$.  Now apply 
%%Lemma \ref{folk}.\end{proof}

The next result is a deep theorem due to Karpenko and Merkurjev \cite{KM}.
It is the crucial ingredient in our construction.

\begin{theorem}[Karpenko-Merkurjev]\label{karp-merk}  
Let $\phi$ and $\psi$ be anisotropic forms
over $F$.  Suppose that $\phi_{F(\psi)}$ is isotropic.  Then
$\dim\phi - i_1(\phi)\geq \dim\psi -i_1(\psi)$, and equality holds if and only if
$\psi_{F(\phi)}$ is isotropic as well.
\end{theorem}

The next lemma is also well-known, we include the straighforward proof for the
reader's convenience.

\begin{lemma}\label{karp-merk-lemma}
Let $\phi$ be a form over $F$ and let $\psi$ be a subform of $\phi$, i.e.
$\phi\cong\psi\perp\tau$ for some form $\tau$.
\begin{enumerate}
\item[(i)]  If $\phi$ is isotropic and $\dim\psi >\dim\phi -i_W(\phi )$, then
$\psi$ is isotropic as well.
\item[(ii)] Let $\phi$ be an anisotropic form over $F$ and let $\psi$ 
be a subform of $F$ with $\dim\psi >\dim\phi -i_1(\phi)$.  
Then $\dim\phi - i_1(\phi)=\dim\psi -i_1(\psi)$.
\end{enumerate}
\end{lemma}

\begin{proof} (i) Consider $\psi$ as the restriction of $\phi$ to a sub-vector
space $W$ of dimension $\dim\psi$ of the underlying vector space $V$ of $\phi$.
Now any maximal totally isotropic subspace of $\phi$ is of dimension $i_W(\phi)$
and intersects therefore nontrivially with $W$ by an easy dimension count.  Therefore,
$W$ contains a nonzero isotropic vector.

(ii) By assumption on $\dim\psi$ and by (i), using that
$i_1(\phi)=i_W(\phi_{F(\phi)})$, we have that $\psi$ becomes isotropic over
$F(\phi)$.  On the other hand, $\psi$ and therefore also $\phi$ becomes 
isotropic over $F(\psi)$.  The result now follows from Theorem~\ref{karp-merk}
\end{proof}

In certain situations, one can say a little more about $i_1$.  This will allow
us to refine the main theorem slightly.

\begin{lemma}\label{multipfister}  Let $\pi$ be an anisotropic $n$-fold
Pfister form over $F$ and let $\phi$ be another form over $F$.
\begin{enumerate}
\item[(i)] There exist forms $\phi_0$ and $\phi_h$ such that $\pi\otimes\phi_0$
is anisotropic, $\phi_h$ is hyperbolic, and $\pi\otimes\phi\cong
\pi\otimes\phi_0\perp\pi\otimes\phi_h$.  In particular, $i_W(\pi\otimes\phi)=
2^{n-1}\dim\phi_h=2^ni_W(\phi_h)$.
\item[(ii)]  Suppose $\pi\otimes\phi$ is anisotropic, and let $\psi$ be a subform
of $\pi\otimes\phi$ and $\dim\psi >2^n(\dim\phi -1)$.  Then
$\dim\psi-i_1(\psi)\leq 2^n(\dim\phi -1)$.
\end{enumerate}
\end{lemma}

\begin{proof} (i) This is well-known, see e.g. \cite[Theorem 2]{WS}.

(ii) By (i) and the definition of $i_1$, we have 
$\dim (\pi\otimes\phi) - i_1(\pi\otimes\phi)\leq 2^n\dim\phi -2^n<\dim\psi$.
By Lemma \ref{karp-merk-lemma}(ii), we have
$$\dim\psi-i_1(\psi)=\dim (\pi\otimes\phi) - i_1(\pi\otimes\phi)
\leq 2^n(\dim\phi -1)\ .\eqno\qedhere$$
\end{proof}

We also need a few facts about orderings and signatures.  Recall that
an ordering $P$ on a field $F$ is a subset $P\subsetneq F$ such that
$P+P\subseteq P$, $P\cdot P\subseteq P$ and $P\cup (-P)=F$.
One can readily check that if $P$ is an ordering of $F$, then $-1\not\in P$,
$P\cap (-P)=\{ 0\}$, and every sum of squares is in $P$.  In fact, by
Artin-Schreier, $F$ is real if and only if $F$ has an ordering.

If $P$ is an ordering on $F$ and if $K/F$ is a field extension, then we say
that $P$ extends to $K$ if there is an ordering $Q$ on $K$ such that
$Q\cap F=P$.

Given an ordering $P$ on $F$, we obtain a total ordering ``$\leq_P$'' on $F$
by defining $a\geq_Pb$ if $a-b\in P$.

Let $\phi\cong\qf{a_1,\cdots ,a_n}$ be a quadratic form over a real field $F$, 
and let $P$ be an ordering on $F$.  The signature of $\phi$ at $P$ is defined
by 
$$\sgn_P(\phi)=|\{ i\,|\,a_i>_P0\}| - |\{ i\,|\,a_i<_P0\}|$$
It is an invariant for the isometry class of $\phi$ (this is nothing but
Sylvester's law of inertia).  We say that $\phi$ is indefinite at $P$ if
$\dim\phi >|\sgn_P(\phi)|$.
%%, definite otherwise (with the obvious definitions
%%for positive resp. negative definite).

Using Witt decomposition and the
fact that $\sgn_P(\HH)=0$, it is obvious that
$\dim\phi_{an}\geq |\sgn_P(\phi)|$.

\begin{lemma}\label{order}  Let $\phi$ be a quadratic 
form over a real field $F$, $\dim\phi\geq 2$, and let $P$ be 
an ordering on $F$.  Then $P$ extends to $F(\phi)$ if and only
if $\phi$ is indefinite at $P$.

In this situation and if $\psi$ is another form over $F$, then
$\dim (\psi_{F(\phi)})_{an}\geq |\sgn_P(\psi)|$.
\end{lemma}

\begin{proof}  The first part is due to Elman-Lam-Wadsworth 
\cite[Theorem 3.5]{ELW} and independently Knebusch \cite[Lemma 10]{GS}.

The second part follows readily from  the remark preceding
the proposition, using the fact that if $Q$ is an ordering on $F(\phi)$
extending $P$, then $\sgn_P(\psi)=\sgn_Q(\psi_{F(\phi)})$.
\end{proof}

\section{Some observations about the levels of quaternion algebras}
Recall that a quaternion algebra over $F$ is a $4$-dimensional central 
simple $F$-algebra.  If $Q$ is a quaternion algebra, one can find 
$a,b\in F^*$ and an $F$-basis $\{ 1,i,j,k\}$ of $Q$ such that $i^2=a$, $j^2=b$,
and $ij=-ji=k$, in which case we denote $Q$ by $(a,b)_F$.

$Q=(a,b)_F$ is a division algebra if and only if the Pfister form
$\pff{a,b}$ is anisotropic over $F$, and $(a,b)_F\cong (c,d)_F$
as $F$-algebras if and only if $\pff{a,b}\cong\pff{c,d}$
(see \cite[Ch.III, Theorems 2.5, 2.7]{Lam}).

If $Q$ is not division, then $Q\cong M_2(F)$ and we have $s(Q)=1$.
Indeed,
$$\left(\begin{array}{rr} 0 & 1 \\ -1 & 0\end{array}\right)^2=
\left(\begin{array}{rr} -1 & 0 \\ 0 & -1\end{array}\right)\ .$$

\begin{remark}
This also readily implies that for any split even-dimensional central simple
algebra $A$, i.e. $A\cong M_{2n}(F)$ for some $n\geq 1$, we get
$s(A)=1$. In the odd-dimensional split case, the situation is more
complicated.  The full result for split algebras $M_m(F)$, $m\geq 2$, reads

$$s(M_m(F))=\left\{\begin{array}{ll} 1 & \mbox{if $m\geq 2$ is even}\\
\min\{ 3,s(F)\} & \mbox{if $m\geq 3$ is odd}\end{array}\right.$$
This follows readily from a more general result due to Richman \cite{R} that states
that every element $x\in M_m(F)$ is a sum of two squares provided $x$ is not of
type $cI_m$ with $m$ odd and $c\in F^*$ not a sum of two squares.
Such elements $cI_m$ can be written as sums of three but not two squares
(here, $I_m$ denotes the $m\times m$ identity matrix).
\end{remark}

Let us from now on assume that $Q=(a,b)_F$ is a division quaternion algebra
over $F$ (with $F$-basis $\{ 1,i,j,k\}$ as above). We start with some elementary
observations.

Let $\zeta\in Q$.  Then $\zeta =x+yi+zj+wk$ with $x,y,z,w\in F$, and we call
$x$ the {\em scalar part} of $\zeta$, and $\zeta'=yi+zj+wk$ its {\em pure part}.
If $x=0$, then $\zeta=\zeta'$ is called a {\em pure quaternion}.
One readily computes that
$$\zeta^2=x^2+ay^2+bz^2-abw^2+2x\zeta'\ ,$$
in particular, the scalar part of $\zeta^2$ is represented by the quadratic
form $\qf{1,a,b,-ab}$, whereas $\zeta'^2=ay^2+bz^2-abw^2$ is an element in $F$
represented by the form $\qf{a,b,-ab}$.  This immediately leads to the following
observation:

\begin{lemma}\label{observation}\begin{enumerate}
\item[(i)] If $-1$ is a sum of $n$ squares in $Q$, then $-1$ is represented by 
$n\times\qf{1,a,b,-ab}$ over $F$,
i.e., $\qf{1}\perp (n\times\qf{1,a,b,-ab})$ is isotropic over $F$.
\item[(ii)] $-1$ is a sum of $n$ squares of pure quaternions in $Q$ if and only if
$-1$ is represented by $n\times\qf{a,b,-ab}$ over $F$, i.e.,
$\qf{1}\perp (n\times\qf{a,b,-ab})$ is isotropic over $F$.
\end{enumerate}
\end{lemma}

For $n\geq 1$, let us put 
$$\begin{array}{rcl}\psi_n & := & \qf{1}\perp (n\times\qf{1,a,b,-ab})\ ,\\
\phi_n & := & \qf{1}\perp (n\times\qf{a,b,-ab})\ .\end{array}$$
Note that $\dim\psi_n=4n+1$ and $\dim\phi_n=3n+1$.

The previous lemma readily implies the following criterion.

\begin{corollary}\label{level-criterion}  Let $n>m\geq 0$ be integers.
Let $Q=(a,b)_F$ and $\phi_n$ and $\psi_m$ be as above.  If $\phi_n$ is isotropic and
$\psi_m$ is anisotropic, then $s(Q)\in [m+1,n]$.
\end{corollary}

\begin{remark}\label{construction-sketch}
With this criterion at hand, one might try the following `generic' construction
of quaternion algebras of a given level in $[m+1,n]$.  

Start with a quaternion division
algebra $(a,b)_F$ over a suitably chosen $F$ (in which $-1$ is not a square) 
such that $\phi_n$ and
$\psi_{m}$ are anisotropic ($m<n$).   Let now $K_n=F(\phi_n)$.  

The $4$-dimensional
form $\pff{a,b}$, which is anisotropic over $F$ since $(a,b)_F$ is division, will
stay anisotropic over $K_n$.  Indeed, otherwise $\pff{a,b}$ would become hyperbolic
over $K_n$, thus $\phi_n$ would be similar to a subform of $\pff{a,b}$ over $F$
(see, e.g., \cite[Ch.~X, Cor.~4.9]{Lam}).  This is impossible for dimension reasons
if $n\geq 2$, and it is also impossible for $n=1$ by comparing determinants because
$\det\phi_1=\det\qf{1,a,b,-ab}=-1=-\det\pff{a,b}\in F^*/F^{*2}$ with $-1$ not a square in $F$.
In particular $Q_{K_n}=(a,b)_{K_n}$ will be division.

Clearly, $(\phi_n)_{K_n}$ is isotropic, in particular, $s(Q_{K_n})\leq n$.  
If one can now show that $\psi_{m}$ stays
anisotropic over $K_n$, then by the previous corollary, $s(Q_{K_n})\in [m+1,n]$.
\end{remark}

If this reasoning worked for $m=n-1$, we could deduce that $s(Q_{K_n})=n$.
And although this construction might indeed yield a quaternion algebra of level $n$,
we cannot, unfortunately, employ the above reasoning for $m=n-1$ in general as the 
following result shows (which generalizes Proposition 2.5 in \cite{LM}).

\begin{proposition}
Keep the notations as above.
Let $n=2^rs$ be a positive integer with $s$ odd, and let $t\geq 1$ be maximal
such that $s\equiv -1\bmod 2^t$.  Assume that $t\geq 3$ or 
$r\geq 2$, and that $\phi_n$ is isotropic.  Then so is $\psi_{n-1}$.  
\end{proposition}

\begin{proof}
Assume first that $t\geq 3$ and let $\tau = (n+2^r)\times\qf{1,a,b,-ab}$. 
Then $2^{r+t}$ divides $n+2^r$ 
(but not $2^{r+t+1}$ by the maximality of $t$),
and we can write $\tau\cong 2^{r+t}\times\sigma$ for some form
$\sigma$.  Note that $\phi_n\subset \tau$, so $\tau$ is isotropic.
If the $(r+t)$-fold Pfister $\pff{-1,\cdots ,-1}$ is isotropic and thus hyperbolic, 
then so is $\tau$
and we clearly have $i_W(\tau)\geq 2^{r+t}$.  This also holds if 
the $(r+t)$-fold Pfister $\pff{-1,\cdots ,-1}$
is anisotropic by  Lemma \ref{multipfister}.  But we also have that
$\psi_{n-1}\subset \tau$, and 
$$\dim\tau - \dim\psi_{n-1}=4(n+2^r)-(4(n-1)+1)=2^{r+2}+3<2^{r+3}\leq 
2^{r+t}\leq i_W(\tau)\ ,$$
hence $\psi_{n-1}$ is isotropic by Lemma \ref{karp-merk-lemma}.

Suppose now that $r\geq 2$.  Then we put $\tau =n\times\qf{1,a,b,-ab}$.
This time, $\tau$ is divisible by the $r$-fold Pfister $\pff{-1,\cdots ,-1}$,
$\phi_n$ and $\psi_{n-1}$ are subforms of $\tau$, and similarly as before
we get
$\dim\tau - \dim\psi_{n-1}=4n-(4(n-1)+1)=3< 2^r\leq i_W(\tau)$, and again,
$\psi_{n-1}$ is isotropic by Lemma \ref{karp-merk-lemma}.
\end{proof}
So, for example,
if $\phi_7$ is isotropic ($r=0$, $s=7$, $t=3$), then so is $\psi_6$, and
for any $n$ divisible by $4$, if $\phi_n$ is isotropic, then so is $\psi_{n-1}$.
%%This latter result in the case $n=2^r$ (with a similar argument as above) 
%%was shown in \cite{LM}.


We do not know in general whether the construction described in 
Remark \ref{construction-sketch} will actually yield 
$s(Q_{K_n})=n$ (except for small $n$ or
values of type $n=2^k+1$ for certain $F$, $a$, $b$, 
see Remark \ref{lag-mam} below).
However, the above reasoning to obtain $s(Q_{K_n})\in [m+1,n]$ will work (for certain
$a,b\in F^*$ over a suitably chosen $F$) if we make $n$ sufficiently large
compared to $m$, using Theorem \ref{karp-merk}.  We explain this construction in the
following section.



\section{The construction}

Throughout this section, let $F_0$ be any real field, 
and let $F=F_0(a,b)$ be the rational function field
in the two variables $a,b$ over $F_0$.  Let $Q=(a,b)_F$ and
let $\phi_n$, $\psi_m$ be as above, with integers $n>m\geq 0$.
\begin{lemma}\label{construction-lemma}  Let $K_n=F(\phi_n)$.
\begin{enumerate}
\item[(i)]  $Q$ is a division algebra over $F$ and stays a division algebra over $K_n$.
\item[(ii)] $\phi_n$ and $\psi_m$ are anisotropic.
\item[(iii)] $i_1(\phi_n)=1$.
\end{enumerate}
\end{lemma}

\begin{proof} It follows easily from $F_0$ being real together with
Lemma \ref{folk} that $\pff{a,b}$, $\phi_n$ and $\psi_m$ are anisotropic.
In particular, $Q=(a,b)_F$ is a division algebra and will stay so over $K_n$ as 
explained in Remark \ref{construction-sketch}.

To show (iii), let $P$ be any ordering on $F$ with $a<_P0$ and $b<_P0$
(such orderings always exist).
Then $-ab<_P0$ and we get
$$|\sgn_P(\phi_n)|=|\sgn_P((n\times\qf{a,b,-ab})\perp\qf{1})|=3n-1<3n+1=\dim\phi_n\ ,$$
so $\phi_n$ is indefinite at $P$ and it follows from Lemma \ref{order} that
$3n-1\geq \dim ((\phi_n)_{K_n})_{an}\geq |\sgn_P(\phi_n)|=3n-1$, therefore
$\dim ((\phi_n)_{K_n})_{an}=3n-1=\dim\phi_n-2$ and thus $i_1(\phi_n)=1$.
\end{proof}

\begin{corollary}\label{construction-cor1}  Let $n>m\geq 0$ be integers.
If $4m+1-i_1(\psi_m)<3n$, then $s(Q_K)\in [m+1,n]$.
\end{corollary}

\begin{proof}  We have $\dim\phi_n-i_1(\phi_n)=3n+1-i_1(\phi_n)=3n$ by 
Lemma \ref{construction-lemma}(iii), and $\dim\psi_m-i_1(\psi_m)=4m+1-i_1(\psi_m)$.
By Theorem \ref{karp-merk}, if $4m+1-i_1(\psi_m)<3n$, then $\phi_m$ stays
anisotropic over $K_n=F(\phi_n)$.  It follows from Remark \ref{construction-sketch}
that $s(Q_{K_n})\in [m+1,n]$.
\end{proof}

Note that $i_1(\psi_m)\geq 1$ for $m\geq 1$, and hence, we can always construct
$Q_{K_n}$ as above with $s(Q_{K_n})\in [m+1,n]$ whenever $4m<3n$ 
(this also clearly holds for $m=0$).  Thus, we obtain Theorem
\ref{mainthm} as corollary:

\begin{corollary}\label{mainthm-cor}  Let $n=m+1+[m/3]$.  
Then $s(Q_{K_n})\in [m+1,m+1+[m/3]]$.
\end{corollary}

\begin{remark}\label{lag-mam}  This corollary obviously allows us to 
get levels $1$, $2$, $3$.  It should furthermore be noted that it is
possible to show that in the above situation, $\psi_{2^k}$ stays anisotropic
over $K_n=F(\phi_n)$ with $n=2^k+1$, so that we then get $s(Q_{K_n})=2^k+1$.
This was shown in \cite{LM} and it yields a construction of quaternion algebras
of levels of the form $2^k+1$ that is different from Lewis's original one.
\end{remark}

The preceding corollary can be refined somewhat by getting more information on
$i_1(\psi_m)$ for certain values of $m$.

\begin{corollary}\label{refinement}  Let $m=2^km'$ with $m'$ odd, and let
$\ell\geq 1$ be maximal such that $m'\equiv -1\bmod 2^{\ell}$.  Let
$n=m+1+[\frac{m-2^k(2^{\ell}-4)}{3}]$.
Then $s(Q_{K_n})\in [m+1,n]$.
\end{corollary}

Before we prove this, let us remark that if $\ell =1$, then this yields an interval
with a larger upper bound than Corollary \ref{mainthm-cor}. 
%%(and it is therefore a weaker result in that case).  
If $\ell =2$, it yields the same interval, and if 
$\ell > 2$ it yields an interval with strictly smaller upper bound.
%% and thus gives us a stronger result in that last case.

\begin{proof} In view of the preceding remark, we may assume that $\ell\geq 2$.
Write $m=2^k(r2^\ell -1)$ with $r$ odd.
Consider the form $\tau\cong (m+2^k)\times\qf{1,a,b,-ab}$ (which is clearly anisotropic).
Since $m+2^k=r2^{k+\ell}$, it follows that $\tau$ is divisible
by the $(k+\ell)$-fold Pfister form $\pff{-1,\cdots,-1}$.  Hence, 
by Lemma \ref{multipfister}, we have $i_1(\tau)\geq 2^{k+\ell}$.

Now clearly $\psi_m\subset\tau$.  Furthermore,  $2^{k+2}\leq 2^{k+\ell}$ since
we assumed $\ell\geq 2$.  Hence
$$\dim\psi_m=4m+1 >4(m+2^k)-2^{k+\ell}=\dim\tau -2^{k+\ell}\geq \dim\tau -i_1(\tau)$$
and thus, by Lemma \ref{multipfister}, we get
$\dim\psi_m-i_1(\psi_m)=\dim\tau -i_1(\tau)\leq 4m-2^k(2^{\ell}-4)$.
Thus, if $n=m+t$ is such that $3(m+t)>4m-2^k(2^{\ell}-4)$, i.e. $3t>m-2^k(2^{\ell}-4)$,
we have that $s(Q_{K_n})\in [m+1,n]$.  The smallest integer $t$ for which this holds
is $t=1+[\frac{m-2^k(2^{\ell}-4)}{3}]$, and with this $t$ and $n=m+t$ we get 
$s(Q_{K_n})\in [m+1,n]=[m+1,m+1+[\frac{m-2^k(2^{\ell}-4)}{3}]]$.
\end{proof}

\begin{example}  For $m=94=2(3.2^4-1)$, we have $k=1$ and $\ell=4$.
Corollary \ref{mainthm-cor} yields a quaternion algebra whose level is in 
the interval $[95,95+[95/3]]=[95,126]$,
whereas Corollary \ref{refinement} yields a level in 
$[95,95+[(95-2(16-4))/3]]=[95,118]$.
\end{example}

%\begin{corollary}[O'Shea \cite{O}]\label{oshea} Let $v=2^uv'\in\NN$ with $v'$ odd.
%Then there exists a quaternion algebra $Q$ with $s(Q)\in [v,v+1+[(v-2^u)/3]]$.
%\end{corollary}

%\begin{proof}  Put $m=v-1$.  If $u=0$, we have $v+1+[(v-2^u)/3]=m+2+[m/3]$
%and the result follows from Corollary \ref{mainthm-cor}.  If $u>0$ then
%$m=v'2^u-1$ and we can apply Corollary \ref{refinement} with $r=v'$, $\ell = u$ and
%$k=0$.
%\end{proof}

%So if $u>0$, Corollaries \ref{refinement} and \ref{oshea} yield the same interval, whereas
%Corollary \ref{refinement} yields smaller intervals in general when $u=0$.
%To prove Corollary \ref{oshea}, O'Shea used similar function field arguments but worked
%with the sublevel $\underline{s}(Q)$, see the following section.

\section{Some remarks on the sublevel}
The sublevel of a unitary ring $R$ is defined as follows:
\begin{enumerate}
\item If $0$ is not a sum of nonzero squares in $R$, then $\underline{s}(R)=\infty$.
\item If $0$ is a sum of nonzero squares in $R$, then
$$\underline{s}(R)=\min\{n\,|\,\exists x_1,\cdots ,x_n,x_{n+1}\in R, x_i^2\neq 0\,:\ 
0=x_1^2+\cdots +x_{n+1}^2\}$$
\end{enumerate}
Generally, $\underline{s}(R)\leq s(R)$, and one clearly has equality if $R$ is a field.
For a quaternion division algebra $Q=(a,b)_F$ over a field $F$ (of characteristic $\neq 2$),
it is still an open question how $s(Q)$ and $\underline{s}(Q)$ are related to
each other in general.  In all cases where the precise values are known, one has 
$s(Q)\in\{ \underline{s}(Q), \underline{s}(Q)+1\}$. 

Using a similar argument as before, one readily sees that if
$\widehat{\psi}_m\cong m\times\qf{1,a,b,-ab}$ is anisotropic, then 
$\underline{s}(Q)\geq m$.  On the other hand, if $\phi_n$ (as before) is isotropic,
then one readily sees that $0$ can be written in a nontrivial way as a sum of
$n+1$ squares and so $\underline{s}(Q)\leq n$.  It is now not surprising that
similar techniques as in our construction of certain new levels can be used to
generate new sublevels, and it was O'Shea \cite{O} who adapted our method accordingly
and gave explicit details.
We quickly sketch one of his results since all the tools are in place.

If, as in the previous section, $F=F_0(a,b)$ with real $F_0$ and variables $a,b$,
we see that $\widehat{\psi}_m$ and $\phi_n$ are anisotropic.  Again, let $K_n=F(\phi_n)$.
If we write $m=2^km'$ with $m'$ odd, then it follows from Lemma \ref{multipfister}
that $i_1(\widehat{\psi}_m)\geq 2^k$.  Recall that $i_1(\phi_n)=1$.  Thus,
by Theorem \ref{karp-merk}, we have that $\widehat{\psi}_m$ stays anisotropic over
$K_n$ if $\dim\phi_n -i_1(\phi_n)=3n > 4m-2^k\geq\dim\widehat{\psi}_m-i_1(\widehat{\psi}_m)$.
This is satisfied whenever $n\geq m+1+[(m-2^k)/3]$, so we obtain

\begin{proposition}[O'Shea \cite{O}]\label{o-prop} 
Let $Q$ and $m=2^km'$ ($m'$ odd) be as above. 
Put $n=m+1+[(m-2^k)/3]$ and let $K_n$ be as above. Then 
$m\leq \underline{s}(Q_{K_n})\leq s(Q_{K_n})\leq m+1+[(m-2^k)/3]$.
\end{proposition}

For example, $m=5$ yields a quaternion algebra of sublevel in $[5,7]$
This shows that the sublevel of a quaternion division algebra can take values other
than $3$ or powers of $2$, and it is clear that the above construction yields
infinitely many new values that are not of that type, 
thus partially answering Open Question 2 in \cite{Lew3}.

\begin{remark} If we are only interested in the level, then Proposition \ref{o-prop}
does not yield smaller intervals with given lower bound
than those obtained previously.  For if $k=0$, then this proposition yields the interval
$[m',m'+1+[(m'-1)/3]]$, whereas Corollary \ref{mainthm-cor} (with $m+1=m'$ there)
gives $[m',m'+[(m'-1)/3]]$ (and Corollary \ref{refinement} might still yield smaller upper
bounds depending on the $2$-adic value of $m'-1$).  If $k>0$, then one obtains the same
interval as in Corollay \ref{refinement} by replacing the tuple $(m,k,r,\ell)$ in the 
statement there by the tuple $(m-1,0,m',k)$, using that $m=(m-1)+1=(2^km'-1)+1$.
\end{remark}


For further results on levels and sublevels of quaternion and
octonion algebras obtained using variations of our methods, we refer to 
\cite{O}.



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