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\begin{document}

\title{On ``horizontal" invariants attached to quadratic forms}
\author{Bruno Kahn}
\address{Institut de Math\'ematiques de Jussieu\\175--179 rue du
Chevaleret\\75013 Paris\\France}
\email{kahn@math.jussieu.fr}
\date{Jan. 9, 2005}
\keywords{Quadratic forms, dimension, $u$-invariant}
\subjclass[2000]{Primary: 11E04, 11E81}

\begin{abstract} We introduce series of invariants related to the 
dimension for quadratic forms over a field, study relationships between 
them and prove a few results about them.
\end{abstract}
\maketitle

{\small This is the \TeX-ing of a manuscript from 1993 entitled 
\emph{Quadratic forms and simple algebras of exponent two}. The original 
manuscript
contained an appendix that has appeared in \cite{K3}: I removed it
and replaced references to it by references to \cite{K3}. I also extended 
Section 2 somewhat, improved Proposition \ref{p3.3} a bit and removed a 
section that did not look too useful. Finally I changed the title to a 
better-suited one. These are essentially the only changes to the original 
manuscript.

The main reasons I have to exhume it are that 1) the notion of dimension 
modulo $I^{n+1}$ has recently been used very conceptually by Vishik (\eg 
\cite{vishik}, to which the reader is referred for lots of highly 
nontrivial computations) and 2) Question \ref{q1.1} has been answered 
positively by Parimala and Suresh \cite{P-S}. I have included a proof 
that is different from theirs (see Corollary \ref{ps}).

Vishik has suggested that the invariant $\lambda$ which is studied here 
might be replaced by a finer one: the ``geometric length", where one 
takes transfers into account. Namely, if $x\in I^nF/I^{n+1}F$, its 
geometric length is the smallest integer $\ell$ such that there exists an 
\'etale $F$-algebra $E$ of degree $\ell$ and a Pfister form $y=\ll 
u_1,\dots,u_n\gg\in I^nE/I^{n+1}E$ such that $x=Tr_{E/F}(y)$. This 
invariant should definitely be investigated as well.

Everything here is anterior to Voevodsky's proof of the Milnor
conjecture, which is not used.}

\section*{Introduction}

Let $F$ be a field of characteristic $\ne 2$. The $u$-invariant $u(F)$ of
$F$ is the least integer $n$ such that any quadratic form in more than $n$
variables over $F$ is isotropic, or $+\infty$ if no such integer exists.
(A finer version exists for formally real fields, but for simplicity we
shall not consider it.) In \cite{Me2} (see also \cite{Ti}), Merkurjev
disproved a long-standing conjecture of Kaplansky, asserting that the
$u$-invariant should always be a power of $2$. In fact, Merkurjev
produced for any integer $m \ge  3$ examples of fields of $u$-invariant
$2m$.

A remarkable feature of Merkurjev's examples is that they can be
contrived to have $2$-cohomological dimension $2$. This destroys a
na\"{\i}ve belief that $u(F)$ would be $2^{\nu(F)}$, where $\nu(F)$, the
$\nu$-invariant of $F$, is the largest integer $n$ such that $I^\nu F
\ne  0$, which was the case in all previously known examples (see \eg
\cite[th. 1]{K1}). It hints that a good understanding of the $u$-invariant
involves not only the $\nu$-invariant, but also `horizontal' invariants
attached to the quotients $I^nF/I^{n+1}F$.

Introducing such invariants and starting their study of is the object of
this paper. Given a quadratic form, one may approximate its anisotropic
dimension, the dimension of its kernel forms, by its `dimension modulo
$I^{n+1}$' for every $n \ge  1$ (the fact that this actually is an
approximation is a consequence of the Arason-Pfister theorem). Given an
element of $I^nF/I^{n+1}F$, one may study its `length' or `linkage index',
the smallest number of classes of Pfister forms necessary to express it.
The suprema $u_n(F)$ of the former invariants (`$u$-invariant modulo
$I^{n+1}$') approximate the $u$-invariant; the suprema $\lambda^n(F)$ of
the latter help giving
upper bounds for the former. More precisely, one can bound $u_n(F)$ in
terms of $\lambda^n(F)$ and $u_{n-1}(F)$ (Proposition \ref{p1.2}).

The only case in which I can prove a converse to these bounds is $n = 2$,
where $u_2(F) = 2\lambda^2(F) + 2$. However, it is not impossible that all
the $\lambda^n(F)$, as well as $u(F)$ when $F$ is not formally real, can
actually be bounded in terms of $\lambda^2(F)$ (and $n$ for
$\lambda^n(F)$). At least this is the case when $\lambda^2(F) = 1$, by a
theorem of Elman-Lam \cite[th. XI.4.10]{Lam}. I partially generalize this
theorem in one direction (Propositions \ref{p3.2} and \ref{p3.3}), but the
general case seems quite open.

In all this paper, we use Lam's \cite{Lam} notations for Pfister forms,
\ie $\ll a_1,\dots,a_n\gg  = \langle  1,a_1\rangle  \otimes \dots\otimes \langle  1,a_n\rangle  $. We write
$\cong$  (resp. $\sim$) for isometry (resp. Witt-equivalence) of
quadratic forms.

\section{A few quadratic invariants}

As in Arason \cite{A}, for a quadratic form $q$ we denote by $\diman(q)$ the
rank of the unique anisotropic quadratic form whose class in $W(F)$ equals
the class of $q$ (the kernel form of $q$). As will be seen in Proposition
\ref{p1.1}, Definition \ref{d1.1} generalises this definition.

\begin{defn}\label{d1.1} a) Let $q$ be a quadratic form over $F$ and $J$ an ideal of $W(F)$. The \emph{$J$-dimension} of $q$ is $\dim_J(q) = \inf \{\dim(q') \mid q' \equiv  q \pmod{J}\}$.\\
b) Let $n\ge 0$ be an integer. The \emph{$n$-dimension} of $q$ is $\dim_n(q) = \dim_{I^{n+1}F}(q)$.\footnote{One should be careful that Vishik's notation in \protect\cite[Def. 6.8]{vishik} is different: our $\dim_n(q)$ is his $\dim_{n+1}(q)$.}
\end{defn}

\begin{rk}\label{r1.1}
$\dim_n(q)$ only depends on the class of $q$ modulo $I^{n+1}F$. For all
$n$, one has $\dim_n(q) \le  \diman(q)$.
\end{rk}

\begin{defn}\label{d1.2} A quadratic form $q$ is \emph{anisotropic modulo
$I^{n+1}F$} (or \emph{$n$-anisotropic}) if $\dim_n(q) = \dim(q)$.
\end{defn}

It is clear that for two quadratic forms $q, q'$, one has
$\dim_n(q\perp q')\le  \dim_n(q) + \dim_n(q')$. The following lemma
strengthens this result when $q' \in   I^nF$. Despite its simplicity, it
is basic in much of this section.

\begin{lemma} \label{l1.1} Let $(q,q') \in   W(F)\times   I^nF$ with $q,
q'\ne  0$. Then $\dim_n(q\perp q')
\le  \dim_n   (q) + \dim_n   (q') - 2$.
\end{lemma}

\prf Without loss of generality, we may assume that $q$ and $q'$ are
anisotropic modulo $I^{n+1}F$. Since $q' \in   I^nF$, $q' \equiv  \langle  a\rangle  q'
\pmod{I^{n+1}F}$ for any $a\in   F^*$. For suitable $a$, the form $q\perp
\langle  a\rangle  q'$ is isotropic. Hence
\begin{multline*}
\dim_n   (q\perp  q') = \dim_n   (q\perp  \langle  a\rangle  q') \le 
\diman(q\perp  \langle  a\rangle  q')\\
\le  \dim(q) + \dim(q') - 2 = \dim_n   (q) +
\dim_n   (q') - 2.
\end{multline*}
\qed

\begin{lemma}\label{l1.2} Let $q \in   W(F)$ be anisotropic modulo
$I^{n+1}F$. Then its only subforms belonging to $I^nF$ are $0$ and
possibly $q$.
\end{lemma}

\prf This follows from Lemma \ref{l1.1}.\qed

\begin{defn}\label{d1.3} Let $n \ge  0$ and $x \in   I^nF/I^{n+1}F$. The
\emph{length} of $x$ is 
\[\lambda(x) = \inf \{r \mid x \text{ is a sum of  $r$
classes of $n$-fold Pfister forms}\}.\] 
If $q\in   I^nF$, we write $\lambda(q)$ for $\lambda(x)$, where $x$ is the
image of $q$ in $I^nF/I^{n+1}F$.
\end{defn}

\begin{prop}\label{p1.1} a) $\dim_0(q) =	
\begin{cases}
0& \text{if $\dim(q)$ is even}\\
1& \text{if $\dim(q)$ is odd.}
\end{cases}$\\
b) $\dim_1(q) =
\begin{cases}
0& \text{if $\dim(q)$ is even and $d_\pm   (q) = 1$}\\
1& \text{if $\dim(q)$ is odd}\\
2& \text{if $\dim(q)$ is even and $d_\pm   (q) \ne  1$.}
\end{cases}$\\
c) If $q \in   I^nF - I^{n+1}F$, then $2^n \le  \dim_n   (q) \le  (2^n -
2)\lambda(q) + 2$.\\
 d) If $q \not\in  I^nF$, then $\dim_n   (q) \le  (2^n-2)\lambda(q') +
\dim_{n-1}(q)$ for some $q' \in   I^nF$.\\ 
e) If $q \in   I^2F - I^3F$, then $\dim_2(q) = 2\lambda(q) + 2$.\\ 
f) If $2^n \ge  \diman(q)$, then $\dim_n(q)=\diman(q)$.
\end{prop}

\prf a) is obvious. Let $\dim(q)$ be odd. Then, for the right choice of
$\varepsilon = \pm   1$, $q\perp  \langle  \varepsilon d_\pm   (q)\rangle   \in   I^2F$, so
$\dim_1(q) = 1$. Let $\dim(q)$ be even. Then $q \in   I^2F$ if and only if
$d_\pm   (q) = 1$; if $q
\in  I^2F$, then $q\perp  \langle  -1,d_\pm   (q)\rangle   \in   I^2F$, so $\dim_1(q) = 2$.
This proves b).

In c), the lower bound for $\dim_n   (q)$ is a consequence of the
Arason-Pfister theorem \cite{AP}. The upper bounds in c) and d)
follow from Lemma \ref{l1.1} by induction on $\lambda(q)$. Let us prove
equality in the case $n = 2$ (\cf \cite[lemma]{Me1}). We may assume
$\dim(q) = \dim_2(q) = 2m$. We argue by induction on $m$. The case $m = 1$
is impossible. Assume $m
\ge  2$. We may write $q = \langle  a,b,c\rangle  \perp  q'$. Then $q$ is Witt-equivalent
to $\langle  a,b,c,abc\rangle  \perp  (\langle  -abc\rangle  \perp  q')$. The first summand is
$\langle  a\rangle  \ll ab,ac\gg 
\equiv  \text{$\ll ab,ac\gg$} \pmod{I^3F}$, while the second one $q'' = \langle  -abc\rangle  \perp 
q'$ has dimension $2m-2$, so that $\dim_2(q'') \le  2m-2$. By induction,
$2\lambda(q'')+2 \le  \dim_2(q'')$, so $2\lambda(q)+2 \le  2\lambda(q'')+4
\le  \dim_2(q'')+2 \le  \dim_2(q)$.

Note that this argument fails for $n \ge  3$.

Let us prove f). We may assume that $q$ is anisotropic. Assume that
$\dim_n   (q) <   \dim(q)$. Then there exists $q'$ with $q' \equiv  q
\pmod{I^{n+1}F}$ and $\dim(q') <   \dim(q)$. Therefore, $q\perp  -q' \in  
I^{n+1}F$. But $\dim(q\perp  -q') <   2\dim(q) \le  2^{n+1}$: therefore, by the
Arason-Pfister theorem \cite{AP}, $q\perp  -q'$ is hyperbolic. This means
that $q$ is Witt-equivalent to $q'$: this is impossible, since $q$ is
anisotropic and $\dim(q') <   \dim(q)$.\qed

\begin{defn} Let $n \ge  1$ and $k \le  n$. The \emph{$k$-restricted
$u$-invariant modulo $I^{n+1}$ of $F$} is $u_n^k(F) = sup \{\dim_n(q) \mid
q\in   I^k(F)\}$. If $k = 0$, we write $u_n(F)$ for $u_n^k(F)$ and call it
the \emph{$u$-invariant of $F$ modulo $I^{n+1}$}.
\end{defn}

\begin{rk}\label{r1.9} When $k$ is fixed, the $u_n^k(F)$'s form a
non-decreasing sequence for increasing $n$. In particular the $u_n(F)$
form a non-decreasing sequence. Similarly, when $n$ is fixed, the
$u_n^k(F)$ form a non-increasing sequence for increasing $k$.
\end{rk}

\begin{defn} Let $n \ge  1$. The \emph{$n$-th $\lambda$-invariant} of $F$
is $\lambda^n(F) = \sup \{\lambda(x) \mid x
\in   I^nF/I^{n+1}F\}.$
\end{defn}

\begin{prop} \label{p1.2} a) For $k > 0$ and $n \ge  k$, $I^kF \ne  0$
$\iff$
$u_n^k(F)
\ne  0$ $\iff$ $u_n^k(F)
\ge  2^k$. If $I^nF \ne  0$, $u_n^k(F) \ge  2^n$. In particular, if $I^nF
\ne  0$, $u_n(F)
\ge  2^n$.\\
b) For $k \ge  0$, $\sup_{n\ge k} u_n^k(F) = u_k(F) := \sup
\{\diman(q) \mid q\in   I^k(F)\}$; in particular, $\sup_{n\ge 0} u_n(F) =
u(F)$.\\
c) $u_n^n(F) \le  (2^n - 2)\lambda^n(F) + 2$.\\
d) For any $k <   n$, $u_n^k(F)
\le  (2^n - 2)\lambda^n(F) + u^k_{n-1}(F)$.\\
e) $u_0(F) = 1$.\\
f) If $IF \ne  0$, $u_1(F) =
u_1^1(F) = 2$.\\
g) If $I^2F \ne  0$, then $u_2(F) = u_2^2(F) = 2\lambda^2(F) + 2$.
\end{prop}

(I am indebted to O. Gabber for pointing out d).)

\prf It is clear that $I^kF = 0$ $\Rightarrow$ $u_n^k(F) = 0$ for all $n
\ge  k$. By Remark \ref{r1.9}, $u_n^k(F) \ge  u_k^k(F)$ when $n \ge  k$.
Assume $I^kF = I^{k+1}F$. Then any $k$-fold Pfister form belongs to
$I^{k+1}F$. By \cite{AP}, such a form must be hyperbolic, hence $I^kF =
0$. This shows that if $I^kF \ne  0$, there exists a form $q \in   I^kF -
I^{k+1}F$. By Proposition \ref{p1.1} c), we have $\dim_k(q)\ge  2^k$, hence
$u_k^k(F) \ge  2^k$ and $u_n^k(F) \ge  2^k$. The last two claims of a)
follow by Remark \ref{r1.9} ($u_n(F)\ge  u_n^k(F) \ge  u_n^n(F)$ when $k
\le  n$). This proves a).

To prove b), first assume that $u_k(F)$ is finite. Let $n$ be such that
$2^n\ge  u_k(F)$. By Proposition \ref{p1.1} f), $\dim_n(q) = \diman(q)$ for
any $q\in   I^kF$. In particular, $u_n^k(F) = u^k(F)$ for all such $n$.
Assume now that the sequence $(u_n^k(F))_{n\ge k}$ is bounded, say by
$N$. Let $q\in   I^kF$ and choose $n$ such that $2^n \ge  \dim(q)$.
Applying Proposition \ref{p1.1} f) again, we have $\dim_n   (q) =
\diman(q)$. This shows that $\diman(q) \le  N$, hence that $u^k(F) \le 
N$.

Part c) is an immediate consequence of Proposition \ref{p1.1} c). To
prove d), we may assume that $IF \ne  0$, otherwise it is trivial. We may
further assume that $I^kF \ne  0$. Let $q \in   I^kF$. We distinguish two
cases:

\begin{thlist}
\item $q \in  I^nF$. By Proposition \ref{p1.1} d), $\dim_n(q)  \le 
(2^n-2)\lambda^n(F) +\dim_{n-1}(q) \le  (2^n-2)\lambda^n(F) +
u^k_{n-1}(F)$. 
\item $q \in   I^nF$.
By Proposition \ref{p1.1} c), $\dim_n(q) \le  (2^n - 2)\lambda^n(F) + 2$. This is $\le 
(2^n-2)\lambda^n(F) + u^k_{n-1}(F)$ provided $u^k_{n-1}(F) \ge  2$. If $k
\ge  1, u^k_{n-1}(F)\ge  2^k \ge  2$ by a). If $k = 0$, $u_{n-1}(F) \ge 
u_1(F)\ge  2$ since $IF \ne  0$ (see f)).
\end{thlist}

e) and f) follow from Proposition \ref{p1.1} a) and b). It remains to prove g). First
we prove that $u_2(F)$ cannot be odd, i.e. $u_2(F) = u_2^1(F)$. This is a
consequence of Proposition \ref{p1.1} e), Proposition 1.2 e) and the
following lemma.

\begin{lemma} a) $u_2(F) = \infty$ iff $\lambda^2(F) = \infty$.\\
b) Assume that $u_2(F) <   \infty$ and $IF \ne  0$. Then $u_2(F) =
u_2^1(F)$.
\end{lemma}

\prf a) is a consequence of Proposition \ref{p1.1}, b) d) and e). For b), let $q$ be such that $\dim(q) = \dim_2(q) =
u_2(F)$. We show that
$\dim(q)$ cannot be odd, unless $IF = 0$. If it is, then as in the proof
of Proposition \ref{p1.1} b) we choose $\epsilon = \pm   1$ such that $q' = q\perp 
\langle  \epsilon d_\pm  (q)\rangle  \in   I^2F$. By assumption, $\dim_2(q') \le 
\dim(q)$, so that $q' \equiv  q'' \pmod{I^3F}$, where $q'' \in   I^2F$ is
such that $\dim(q'')\le  \dim(q)$. Then $q \equiv  q''\perp  \langle  -\epsilon
d_\pm(q)\rangle   \pmod{I^3F}$. If $q'' = 0$, then $u_2(F) = 1$, which implies $IF
= 0$ (Proposition 1.2, b) and remark 1.2). Otherwise, by Lemma \ref{l1.1}
we have
$\dim_2(q) \le  \dim(q'') - 1 <   \dim(q)$, a contradiction.\qed

We now prove that $u_2^1(F) = u_2^2(F)$. By lemma 1.3 we may assume that
$u_2(F) <   \infty$. Let $q \in   IF$ be such that $\dim_2(q) = u_2^1(F)$.
We may assume that $q$ is anisotropic modulo $I^3F$. Then $q' = q\perp 
\langle  -1,d_\pm(q)\rangle   \in   I^2F$, with $\dim(q') = \dim(q) + 2$. Since $\dim(q) =
u_2^1(F) = u_2(F)$, we have $q\perp  \langle  -1\rangle   \equiv  q'' \pmod{I^3F}$, where
$q''$ is such that $\dim(q'')\le  \dim(q)$. But $\dim(q'') \equiv 
\dim(q\perp  \langle  -1\rangle  ) \pmod{2}$, so that $\dim(q'') \langle   \dim(q)$. Therefore,
$q'\equiv  q''' \pmod{I^3F}$, with $\dim(q''') \le  \dim(q) =
\dim_2(q)$, and $q \equiv  q'''\perp  \langle  1,-d_\pm   (q)\rangle   \pmod{I^3F}$. If
$q''' = 0$, $u_2^1(F) = 2$, but then $I^2F = 0$. Otherwise, since
$q'''\in   I^2F$, $\dim_2(q) =
\dim_2(q'''\perp  \langle  1,-d_\pm   (q)\rangle  ) \le  \dim_2(q''')$ by Lemma
\ref{l1.1}. Hence $\dim_2(q''') = \dim_2(q)$ and $u_2^1(F) =
u_2^2(F)$.\qed

\begin{rk} The proof of g) is borrowed from \cite[ch XI, proof of lemma
4.9]{Lam}.
\end{rk}

\begin{rk} The statement $u_n(F) = u_n^1(F)$ is equivalent to ``$u_n(F)$
is even".
\end{rk}

\begin{rk} These proofs prompt the definition of quadratic forms
universal modulo $I^{n+1}$, round modulo $I^{n+1}$. This is left to the
reader.
\end{rk}

\begin{cor}[cf \protect{\cite[ch. XI, lemma 4.9]{Lam}}] \label{c1.1} If
$u_2(F) > 1$, it is even.
\end{cor}

\begin{ex} \label{e1.1} For all $n \ge  0$, $u_n(\R) = 2^n$.
\end{ex}

\begin{qn} \label{q1.1} For $n > 2$, can one bound $\lambda^n(F)$ in terms
of $n$ and $u_n(F)$?
\end{qn}

\begin{sloppypar}
For $n = 3$, one would like to prove this by using the following generic argument. Let $k$ be a base field, $m$ a fixed integer, $F_0 = k(T_1,\dots,T_{2m})$, $Q$ the quadratic form $\langle  T_1,\dots,T_{2m}\rangle  $ over
$F_0$,
$F_1 = F_0(\sqrt{(-1)^mT_1\dots T_{2m}})$ 
and $F_2$ the function field of
the Severi-Brauer variety of the Clifford algebra of $Q_{F_1}$. Then, by Merkurjev's theorem \cite{Me}
$Q_{F_2}
\in   I^3F_2$ and is a `generic element of rank $2m$
in $I^3$'. Show that, for any field $F$ containing $k$
and any $q \in   I^3F$ of rank $2m$, one has $\lambda(q)\le
\lambda(Q_{F_2})$. 

This has been achieved by Parimala and Suresh \cite{P-S} with a general position argument: we shall give a different argument avoiding general position in the next section (see Cor. \ref{ps}).
\end{sloppypar}

\begin{qn} \label{q1.2} By Elman-Lam \cite{EL2}, if $F$ is not formally
real and $\lambda^2(F) = 1$ then $u(F) = 1,2,4$ or $8$. Is there a
nonformally real field $F$ such that $\lambda^2(F) = 2$ and $u(F) =
\infty$?
\end{qn}

In the next section, we give some evidence that the answer to this
question might be `no'.

\section{Discrete valuations; iterated power series}

Let $A$ be a complete discrete (rank 1) valuation ring, $E$ its quotient
field anf $F$ its residue field. We assume that $\car F \ne  2$.

\begin{prop} \label{p2.1} a) For all $n \ge  1$, $\lambda^n(E) \le 
\lambda^n(F) + \lambda^{n-1}(F)$.\\
b) For all $n \ge  1$ and $0 \le  k \le 
n$, $u_n^k(E)\le  u_n^{k-1}(F) + u_{n-1}^{k-1}(F)$.
\end{prop}

Note. In b), one should interpret $u_n^{-1}(F)$ as $u_n(F)$.

\prf Let $\pi$ be a prime element of $E$. Every quadratic form $q$ over
$E$ can be written $q \cong  q_1\perp  \pi q_2$, where $q_1$ and $q_2$
are classes of unimodular forms over $A$. Alternatively, $q$ can be
written up to Witt equivalence $q'_1\perp  \ll \pi\gg q_2$, still with $q'_1$
integral. If $q\in   I^nE$, then $q'_1
\in   I^nA$ and $q_2 \in   I^{n-1}A$ \cite{S}. Since $W(A)\iso W(F)$ is a
filtered isomorphism, this proves a).

To see b), let $q \in   I^nE$ and $q'_1$, $q_2$ as above. Let $\bar q_2$
be the residue image of $q_2$ over $F$. Take $\bar q'_2 \equiv  \bar q_2
\pmod{I^nF}$ with $\bar q'_2\in I^{k-1}F$ and $\dim(\bar q'_2) \le 
u_{n-1}^{k-1}(F)$. Let $q'_2$ be a lift of $\bar q'_2$ to $A$. Then $q'_2
\equiv  q_2 \pmod{I^nA}$ and $q\equiv  q'_1\perp  \ll \pi\gg q'_2 = q'_1\perp 
-q'_2\perp  \langle  \pi\rangle  q'_2 \pmod{ I^{n+1}F}$. Now let $q''_1 = q'_1\perp  -q'_2
\in   I^{k-1}A$; choose $q'''_1 \in   I^{k-1}A$ such that $\bar q'''_1
\equiv  \bar q''_1
\pmod{ I^{n+1}F}$ and $\dim(\bar q'''_1) \le  u_n^{k-1}(F)$. Then $q'''_1
\equiv  q''_1\pmod{ I^{n+1}A}$, $q \equiv  q'''_1\perp  \langle  \pi\rangle  q'_2 \pmod{
I^{n+1}F}$ and $\dim(q'''_1\perp  \langle  \pi\rangle  q'_2) \le  u_n^{k-1}(F) +
u_{n-1}^{k-1}(F)$.\qed

\noindent{\bf	An example.} Start from a field $K$ and set $K_d =
K((t_1))\dots((t_d))$, a field of iterated power series. Then $K_d$ is
complete for a discrete valuation, with residue field $K_{d-1}$.
Proposition \ref{p2.1} allows one to get upper bounds for the invariants
of $K_d$ in terms of $d$ and those of $K$, by induction on $d$.

However, these inductive bounds are by no means sharp in general.
Computing, or at least estimating $\lambda^n(K_d)$ and $u_n(K_d)$
turns out to be ``global" in $d$. To illustrate this, we now consider the case where $K$ is algebraically closed.

\begin{defn} \label{d2.1} Let $k$ be a field, $V$ a $d$-dimensional
$k$-vector space and $n$ an integer $\le  d$. Let $ \Lambda^n(V)$ be the $n$-th exterior power of $V$ and let $x \in   \Lambda^n(V)$
be an $n$-vector. The \emph{length} of $x$ is the smallest integer
$\ell(x)$ such that $x$ is the sum of $\ell(x)$ pure $n$-vectors. We
denote by $N(k,d,n)$ the supremum of $\ell(x)$ when $x$ runs through $V$
(this is independent of $V$).
\end{defn}

The following proposition is clear.

\begin{prop} \label{p2.2} a) Let $V = K_d^*/K_d^{*2}$, viewed as an
$\F_2$-vector space with basis $t_1,\dots,t_d$. Then there are canonical
isomorphisms 
\[I^nK_d/I^{n+1}K_d\simeq  \Lambda^n(V)\] 
mapping Pfister forms to pure
$n$-vectors.\\  
b) For $x\in   I^nK_d/I^{n+1}K_d$, with image $x'$ in
$\Lambda^n(V)$,
$\lambda(x) =\ell(x')$, where
$\lambda(x)$ is as in Definition \ref{d1.3} and $\ell(x')$ is as in
Definition \ref{d2.1}.\\
c) $\lambda^n(K_d) = N(\F_2,d,n)$.\qed
\end{prop}

The following information on $N(k,d,n)$ is collected from \cite{K2}.

\begin{sloppypar}
\begin{prop} \label{p2.3} a) $N(k,d,n) = N(k,d,d-n)$.\\
b) $N(k,d,0) = N(k,d,1) = 1$.\\
c) $N(k,d,2) = [d/2]$.\\
d) $N(k,6,3) = 3$ for any field $k$.\\
e) If $k$ is algebraically closed, $N(k,7,3) \le  4$ (probably $= 4$);
$N(\R,7,3) = 5$; for any field $k$, $N(k,7,3) \le  6$ (probably $\le 
5$).\\ 
f) If $k$ is algebraically closed of characteristic $0$, $N(k,8,3) = 5$;
$N(\R,8,3)\le  8$; for any $k$, $N(k,8,3) \le  10$.\\
g) If $k$ is algebraically closed of characteristic $0$, $N(k,9,3) \le 
9$.\\ 
h) There exists a polynomial $f_n$ of degree $\le  n-2$ such that, for
any field $k$, $N(k,d,n) \le  \displaystyle\frac{d^{n-1}}{2(n-1)!} +
f_n(d)$. In particular, for any $k$, $\limsup_{d\to
\infty}\displaystyle\frac{N(k,d,n)}{d^{n-1}}
\le \frac{1}{2(n-1)!}$.\\ 
i) If $k$ is infinite, $N(k,d,n)
\ge  \displaystyle\frac{\binom{d}{n}}{n(d-n)+1}$. If $k$ is finite with
$q$ elements,
$N(k,d,n)\ge \displaystyle \frac{\binom{d}{n}}{n(d-n)+1+\epsilon(q)}$,
where
\[\epsilon(q) =\displaystyle \log_q(\prod_{i=2}^\infty(1-q^{-i})-1).\] 
(So $\epsilon(2)\approx  0.75$.) In particular, for any field
$k$, $\liminf_{d\to\infty}\displaystyle\frac{N(k,d,n)}{d^{n-1}} \ge 
\frac{1}{n.n!}$.
\end{prop}
\end{sloppypar}

This proposition enables us to list values of $\lambda^n(K_d)$ for $d \le 
6$:
\bigskip

\begin{center}
\begin{tabular}{| c | c | c | c | c  | c | c | c |}\hline$d\backslash n$ &0  &1   &2 &3&4&5&6\\\hline1 &1 &1\\\cline{1-4}2 &1 &1 &1\\\cline{1-5}3 & 1 &1&1&1\\\cline{1-6}4 &1 &1 &2&1&1\\\cline{1-7}5 &1 &1 &2 &2&1 &1\\\cline{1-8}6 &1 &1 &3 &3&3&1&1\\ \hline\end{tabular}
\end{center}
\bigskip

From this table, we see that the inequality 
\[u(K_d)\le \sum_{n=2}^d (2^n-2)\lambda^n(K_d)+2\] 
from Proposition \ref{p1.2} d) quickly becomes completely inaccurate.

More generally, let us consider a field $F$ provided with a discrete valuation $v$ of rank $d$. Let $K$ be the residue field of $v$, $L$ the value group of $v$ and $\hat F$ the completion of $F$ at $v$. Assume that $\car K\ne 2$. Then we have
\[\hat{F}\simeq K((t_1))\dots((t_d))\]
where $t_1,\dots,t_d\in \hat{F}$ are elements such that $v(t_1),\dots, v(t_d)$ form a basis of $L$.

Let $\bar K$ be an algebraic closure of $K$ and $\tilde F=\bar K((t_1))\dots((t_d))$. By weak approximation, the map $F^*/F^{*2}\to \hat{F}^*/\hat{F}^{*2}$ is surjective, hence so is the composition
\[I^nF/I^{n+1}F\to I^n\hat{F}/I^{n+1}\hat{F}\to I^n\tilde{F}/I^{n+1}\tilde{F}\]
for any $n\ge 1$. It follows that $N(\F_2,d,n)$ is a \emph{lower bound} to $\lambda^n(F)$. I don't know if it is even true that
\[\lambda^n(F)\ge \lambda^n(F_0)+N(\F_2,d,n)?\]

We also have:

\begin{prop} \label{lred} Let  $\mu$ denote the place from $F$ to $F_0$ associated to $v$. Let $Q\in I^nF$ have good reduction at $\mu$. Then $\lambda(\mu_*Q)\le \lambda(Q)$.
\end{prop}

\begin{proof} Since $v$ is a composition of discrete valuations of rank $1$, we may reduce to $v$ of rank $1$. Pick a prime element $\pi$. Then we have a ring homomorphism (\eg see \cite[Ch. IV, \S 1]{MH})
\begin{align*}
\Delta_{v,\pi}:W(F)&\to W(F_0)\\
\langle \pi^ru\rangle&\mapsto \langle\bar u\rangle
\end{align*}
where $u$ is a unit and $\bar u$ is its image in $F_0^*$. Clearly $\Delta_{v,\pi}$ preserves $n$-fold Pfister forms, hence $\lambda(\Delta_{v,\pi}(Q))\le \lambda(Q)$ for any $Q\in I^nF$, and the claim is a special case.
\end{proof}

\begin{cor}\label{ps} The answer to Question \ref{q1.1} is positive for $n=3$.
\end{cor}

Indeed, the strategy outlined just after the question works as follows: given $q=\langle a_1,\dots,a_{2m}\rangle\in I^3F$, consider the place $\mu_0$ from $F_0$ to $F$ sending $T_i$ to $a_i$. It is well-known that $F_2$ is a generic splitting field for the Clifford algebra of $Q_{F_1}$; hence $\mu_0$ extends to a place $\mu$ from $F_2$ to $F$ sending $Q_{F_2}$ to $q$. Since the valuation associated to $\mu_0$ is discrete, so is the one associated to $\mu$, and we may apply Proposition \ref{lred}.

If any element of $I^nF/I^{n+1}F$ has a generic splitting field also for $n>3$, then the above argument applies verbatim to answer Question \ref{q1.1} positively.

\begin{qn} Does Proposition \ref{lred} remain true when the valuation $v$ is not discrete?
\end{qn}


\section{Relationships between the $u$-invariant, the $\nu$-invariant and
the $\lambda$-invariants}

\begin{lemma}[\protect{\cite[th. 4.5]{EL1}}]\label{l3.1} Let $\phi$ be an
$m$-fold Pfister form and $\psi$ be an $n$-fold Pfister form. Assume that
$\phi$ and $\psi$ are $r$-linked but not $(r+1)$-linked. Then, for any
$a, b\in   F^*$, $\ind(\langle  a\rangle  \phi\perp  \langle  b\rangle  \psi) = 2^r$.\qed
\end{lemma}

\begin{prop}\label{p3.1} Assume that $u(F) \le  2^n$. Then $\lambda^n(F)
\le  1$.
\end{prop}

\prf Let $\phi$ and $\psi$ be two $n$-fold Pfister forms. Then $\psi$ is
universal, so $\psi\equiv  -\psi$, and $\ind(\phi\perp  \psi) =
\ind(\phi\perp  -\psi)\ge  2^{n-1}$. By Lemma \ref{l3.1}, $\phi$ and
$\psi$ are $(n-1)$-linked, so $\phi\perp  \psi$ is isometric to an
$n$-fold Pfister form.\qed

Using a result of Bloch and independently Kato, we can deduce a nice corollary to this proposition, generalising a well-kown result for global fields:

\begin{cor} \label{c2.1} Let $F$ be a function field in $n$ variables over
an algebraically closed field, or in $n-1$ variables over a finite field.
Then every element of  $H^n(F,\Z/2)$ is a symbol.
\end{cor}

\prf A theorem of Kato \cite[p. 609, prop. 3]{kato} (see also Bloch's argument in \cite[Lecture 5]{B}) shows that, for a field as in the statement, $H^n(F,\Z/2)$ is generated by
symbols. It is then sufficient to prove that every element of
$K_n^M(F)/2$ is a symbol. Notice that $F$ is $C_n$ in the sense of Lang
\cite{G}, hence $u(F)\le  2^n$. In view of Proposition \ref{p3.1}, it is
then sufficient to have:

\begin{lemma}[\protect{\cite[th. 6.1]{EL1}}] \label{l2.2} Let $F$ be a
field, $n\ge  1$ and $x, y, z
\in   K_n^M(F)/2$ be three symbols. Assume that $\nu_n(x)+\nu_n(y) =
\nu_n(z)$. Then $x+y = z$.\qed
\end{lemma}

In this lemma, $\nu_n:K_n^M(F)/2\to I^nF/I^{n+1}F$ is the homomorphism
defined in \cite{Mi}.

The following is no more than \cite[lemma 2.3 and cor. 2.5]{EL2}.

\begin{prop} \label{p3.2} Assume that $\lambda^n(F) \le  1$. Then
$\lambda^{n+1}(F)\le  1$. If furthermore $F$ is not formally real, then
$I^{n+2}F = 0$.\qed
\end{prop}

The next proposition is the main result of this section.

\begin{prop} \label{p3.3} Assume that $\lambda^n(F) = m < \infty$ and
that $F$ is not formally real. Then $I^{n(m+1)+1}F = 0$. If $-1$ is not a
square in $F$, then $I^{n(m+1)}F = 0$.
\end{prop}

In other words, if $F$ is not formally real then $\nu(F) \le  n(\lambda^n(F)+1)$ for all $n$.
For $n=2$, the right hand side is $u_2(F)$ by Proposition \ref{p1.2} d). When $-1$ is a square in $F$, this bound is improved to $\nu(F) \le  n(\lambda^n(F)+1)-1$.
\bigskip

\prf By Milnor \cite{Mi}, there are surjective homomorphisms
$\nu_{n}:K_{n}^M(F)/2\to I^{n}F/I^{n+1}F$, and $\nu_2$ is an
isomorphism. Let $K^{'M}_*(F)=K_*^M(F)/\{-1\}K_*(F)$: as explained in
\cite[Appendix]{K3}, the commutative ring $K^{'M}_*(F)/2$ enjoys graded
divided power operations $x\mapsto x^{[i]}$ which vanish on symbols: by
[the argument of the proof of] \cite[Prop. 1 (8)]{K3},
$K^{'M}_{n(m+1)}(F)/2=0$, hence every element of $I^{n(m+1)}F/I^{n(m+1)+1}F$ is a
multiple of (the $1$-fold Pfister form) $\ll1\gg $.

\begin{sloppypar}
If $-1$ is a square in $F$, then $\ll1\gg $ is hyperbolic and $I^{n(m+1)}F/I^{n(m+1)+1}F
= 0$; using the Arason-Pfister theorem, we deduce that $I^{n(m+1)}F = 0$.
Assume now that $-1$ is not a square in $F$; let $E = F(\sqrt{-1})$. By
\cite[Cor. 5.3]{BT}, $K_{n(m+1)+1}^M(E)$ is generated by symbols
$\{a_1,\dots,a_{n(m+1)+1}\}$, with $a_1,\dots,a_{n(m+1)}$ in $F^*$. Since every
element of $K_{n(m+1)}^M(F)/2$ is a multiple of $\{-1\}$, every element in
$K_{n(m+1)+1}^M(E)/2$ is a multiple of $\{-1\} = 0$, \ie $K_{n(m+1)+1}^M(E)/2 =
0$, hence $I^{n(m+1)+1}E/I^{n(m+1)+2}E = 0$ and $I^{n(m+1)+1}E = 0$. If now $F$ is not
formally real, \cite[Satz 3.6 (ii)]{A} implies that $I^{n(m+1)+1}F = 0$.\qed
\end{sloppypar}

\begin{rk} \label{r3.1} For $n=2$ Prop. \ref{p3.3} is optimal, at least when $-1$ is
a square in $F$. For example, let $F = \C((t_1))\dots((t_d))$ be the field
of iterated formal power series in $d$ variables over $\C$. Then $\nu(F)=d$ and, by Section 2, $\lambda^2(F) = [d/2]$ which is also the least integer greater than
$\displaystyle\frac{d-1}{2}$.
\end{rk}

\begin{qn} \label{q3.1} Is there a universal
bound for $\lambda^n(F)$ in terms of $n$ and $\lambda^2(F)$? In view of
Propositions \ref{p1.2} and \ref{p3.3}, this would provide a negative
answer to question \ref{q1.2}. For example, it seems plausible that if
$\lambda^n(F) = m$, then $\lambda^{nm}(F) \le  1$. This is true in the test example of Remark \ref{r3.1}. If one could prove it in general, then the estimate $\nu(F) \le  n(\lambda^n(F)+1)$ or $\nu(F) \le  n(\lambda^n(F)+1)-1$ of Proposition \ref{p3.3} would be improved to $\nu(F) \le  n\lambda^n(F)+1$ thanks to Proposition \ref{p3.2} (note that this is no improvement if $n=2$ and $-1$ is a square in $F$).

It is clear that divided power operations have not been used up to
their full potential.
\end{qn}

\begin{comment}
\section{The invariant $i(q)$}

\begin{sloppypar}
\begin{defn} \label{d4.1} Let $q \in   I^2F$. The \emph{$i$-invariant} of
$q$ is $i(q) = \log_2(\ind(C(q)))$, where $C(q)$ is the Clifford algebra
of $q$.
\end{defn}
\end{sloppypar}

\noindent {\bf Warning.} The index of a central simple algebra $A$ is the
degree of the division algebra similar to $A$. The index of a quadratic
form $q$ is half the dimension of a maximal totally isotropic subspace of
$q$. The two notions have rather opposite meanings!

The following lemma is clear.

\begin{lemma} \label{l4.1} $i(q) \le  \lambda(q)$.\qed
\end{lemma}

\begin{defn} \label{l4.2} A form $q \in   I^2F$ such that $i(q) =
\lambda(q)$ is called a \emph{Merkurjev form}.
\end{defn}

\begin{rk} \label{r4.1} If $q$ is a Merkurjev form, then $\dim_2(q) =
dim(q)$.
\end{rk}

The following two theorems are, up to the language, due to Merkurjev.

\begin{thm} \label{t4.1} Let $q \in   I^2F$ and let $F(q)$ be the function
field of the projective quadric of equation $q = 0$.\\
a) If $q$ is not a
Merkurjev form, then for any division algebra $D$, central over $F$,
$D_{F(q)} := D\otimes_FF(q)$ is a division algebra. In particular, for
any $\phi\in   I^2F$, $i(\phi_{F(q)}) = i(\phi)$.\\
b) If $q$ is a Merkurjev form, write $C(q) = M_2(D(q))$: $D(q)$ is a
division algebra of degree $2^\lambda(q)$. If $D$ is a division algebra
central over $F$, then $D_{F(q)}$ is a division algebra, unless there
exists a division algebra $E$ such that $D\equiv  D(q)\otimes E$.  In
particular, $i(q_{F(q)}) < i(q)$, and for $\phi \in   I^2F$,
$i(\phi_{F(q)}) = i(\phi)$ if
$\dim_2(f) \le  dim(q)$ and $\phi \ne  q \pmod{ I^3F}$.
\end{thm}

\prf This is an immediate consequence of the Merkurjev-Tignol theorem
(\cite{Me2}, \cite[th.1]{Ti}).\qed

\begin{thm} \label{t4.2} If $cd_2(F) = 2$, then any anisotropic form $q
\in   I^2F$ is a Merkurjev form.
\end{thm}

\prf This is a translation of Merkurjev's theorem (see \cite[th.
3]{K1}).\qed
\end{comment}

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vecteurs d\'ecomposables}, preprint, Univ. Paris 7, 1991, not to be
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\bibitem[Me1]{Me1} A. S. Merkurjev Untitled manuscript, Leningrad,
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%\newpage

\begin{comment}
Dear Bruno,

Thank you very much for the very interesting text.

It seems that your `generic' argument after the Question 1.1
should work and really give generic element of rank $2m$ from $I^3$
(smooth places ...), and then inequality $\lambda(q)\leq\lambda(q_{F_2})$
would be automatic.

If one believes in the existence of the norm-varieties for arbitrary
element $h\in K^M_n/2$ (and one does ...), then the same works for arbitrary
$n$ (giving interesting invariant tower of fields associated to $Q$).
But it is not clear at all, how to estimate the length of the generic
element of $K^M_n$ corresponding to the form of rank $2m$.
For $q\in I^n$, $\dim(q)<2^{n+1}$ (that is, $=2^{n+1}-2^{i+1}$) it would be natural to expect that $\lambda^n(q)\leq n-i$. Notice, that there is
also ``geometric" length for elements of $K^M_n$ (taking transfers into
account), and abovementioned quadrics should have geometric length $\leq 2$
(proved by M. Rost for $n=3$). What happens for $\dim(q)\geq 2^{n+1}$ is less
clear.

Concerning ``geometric length". It seems that Proposition 3.3 should
be generalizable for $\lambda_{geom}$ instead of $\lambda$.
This would follow from the existence of ``multiplicative transfer"
operations. That is, let $\Lambda(y)=1+y^{[1]}+y^{[2]}+y^{[3]}+\dots$ be the full divided power operation, and $L(k)$ denote the set of series
$(1,x_1,x_2,\dots)$, where $x_i\in K^M_i(k)/2$. Then we are interested in
such operations $N: L(E) \to L(k)$ for the finite field extension $E/k$
that  $\Lambda(Tr(u))=N(\Lambda(u))$. In particular,
$N(u)_E$ should be just $\prod_{g\in G}g(u)$. Frankly speaking, (in a short time I thought about it) I did not prove its existence, but Markus
Rost was telling me something about some ``multiplicative transfer" a year ago (in some other connection). But if it would exist, your result could
be strengthened (for $\dim(q)<2^{n+1}$ quite visibly).

What do you think about it?

Sincerely yours,

Alexander
\end{comment}

\end{document}

