% 5-7-06         
   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%                                                %%%%
%%%%           Graded Hermitian forms and           %%%%      
%%%%               Springer's Theorem               %%%% 
%%%%                                                %%%%
%%%%                  by                            %%%%
%%%%                                                %%%%
%%%% J.-F. Renard, J.-P. Tignol,  and  A. Wadsworth %%%%
%%%%                                                %%%%  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
 
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%\DeclareMathOperator*{\minn}{\sl{min}}
\DeclareMathOperator*{\perpp}{\perp}
\DeclareMathOperator*{\opluss}{\oplus}

\def \div{\; \big{|} \; }
\def \an{\text{\sl an}}
 \def \average{\text{\sl av}}
\def \Aut{\text{\sl Aut}}
\def \Alt{\text{\sl Symd}}
\def \Br{\text{\sl Br}}
\def \charr{\text{\sl char}}
\def \cor{\text{\sl cor}}
\def \deg{\text{\sl deg}}
\def \dim{\text{\sl dim}}
\def \expp{\text{\sl exp}}
\def \gr{\text{\sl gr}} 
\def \gcd{\text{\sl gcd}}
\def \E{ \text{\it \v{E}}}
\def \F{ \text{\it \v{F}}}
\def \End{\text{\sl End}}
\def \Hom{\text{\sl Hom}}
\def \id{\text{\sl id}} 
\def \im{\text{\sl im}} 
\def \ind{\text{\sl ind}} 
\def \inn{\text{\sl int}} 
\def \inff{\text{\sl inf}} 
\def \kerr{\text{\sl ker}}

\def \maxx{\text{\sl max}}
\def \minn{\text{\sl min}}
\def \minlim#1{\underset{#1} \minn}
\def \res{\text{\sl res}}
\def \symd{\text{\sl Symd}}
\def \Trd{\text{\sl Trd}}
\def\ov{\overline}
\def\wi{\widetilde}
\def\wh{\widehat}
\def\vg#1{\Gamma_{#1}}
\def\ff{\mathbb F}
\def\qq{\mathbb Q}
\def\zz{\mathbb Z}
\def\GMF{\Gamma_F}
\def\gal{\mathcal G}
\def\inv{^{-1}}
\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\Ga{\Gamma}
\def\GE{\Gamma_E}
\def\GS{\Gamma_S}
\def\de{\delta}
\def\ep{\epsilon}
\def\la{\lambda}
\def\si{\sigma}
\def\br#1{[#1]}
\def\brga{{\br\ga}}
\def\brde{{\br\de}}
\def\kgam{k_{\brga}}
\def\Ssub#1{S_{\br{#1}}}
\def\Tsub#1{T_{\br{#1}}}
\def\Usub#1{U_{\br{#1}}}
% \def\Alt#1#2{\text{\sl Alt}({#1},{#2})}
\def\av#1#2{\average_{{#1},{#2}}}
\def\GHerm#1#2{\mathcal G\mathcal H(E, #1, \ep;\br#2)}
\def\GHermep#1#2#3{\mathcal G\mathcal H(E, #1, #2;\br#3)}
\def\GHermepp#1#2#3{\mathcal G\mathcal H^+(E, #1, #2;\br#3)}
\def\GHermp#1#2{\mathcal G\mathcal H^+(E, #1, \ep;\br#2)}
\def\GHerma{\mathcal G\mathcal H(E, \si, \ep)}
\def\GHermap{\mathcal G\mathcal H^+(E, \si, \ep)}
\def\Herm#1#2{\mathcal H(E_0, #1, #2)}
\def\Hermp#1#2{\mathcal H^+(E_0, #1, #2)}
\def\HermD#1#2{\mathcal H(D, #1, #2)}
\def\HermDp#1#2{\mathcal H^+(D, #1, #2)}
\def\bd{\prec}
\def\cmp{\,\,\overline\prec\,\,}
\def\halp{h'_\al}
\def\hbep{h'_\be}
\def\lbep{\ell'_\be}
\def\sh{^\sharp}
\def\dtl{(D,\tau,\la)}
\def\half{\frac12}
\def\bigperp{\text{\Large$\perp$}}

\def\proclaimfont{\sl}
 
\title{
%\large \bf 
Graded Hermitian Forms and Springer's Theorem}  
\date{}
%\author{J.-F.~Renard\and J.-P.~Tignol\and A.~R.~Wadsworth}
\author{\scshape{J.-F.~Renard, J.-P.~Tignol, and A.~R.~Wadsworth}}
%\normalsize J.-F.~Renard, J.-P.~Tignol, and A.~R.~Wadsworth}
\address{Institut de Math\'ematique Pure et Appliqu\'ee\\
Universit\'e catholique de Louvain\\
B-1348 Louvain-la-Neuve\\
Belgium}
\email{jfrenard@gmail.com, tignol@math.ucl.ac.be}
% \address{Institut de Math\'ematique Pure et Appliqu\'ee\\ 
% Universit\'e catholique de Louvain\\
% B-1348 Louvain-la-Neuve\\
% Belgium}
% \email{tignol@math.ucl.ac.be}
\address{Department of Mathematics\\
University of California, San Diego\\
La Jolla, CA-92093-0112\\
USA}
\email{arwadsworth@ucsd.edu}
\thanks{The second author is
partially supported by the National Fund for Scientific Research
(Belgium) and by the European Community under contract
HPRN-CT-2002-00287,
KTAGS.  The third author would like to thank the second 
author and UCL for their hospitality while the work for this
paper was carried out.}


% \ \ AMS subject classification: 16K20}  }
\begin{document}
\begin{abstract}
An analogue of Springer's theorem on the Witt group of quadratic forms over a
complete discretely valued field is proved for Hermitian forms over division
algebras over a Henselian field, including some cases where the residue
characteristic is 2. Residue forms are defined by means of vector space
valuations as Hermitian forms on the graded modules associated with the
induced filtrations.
\end{abstract}
\maketitle
 
\setcounter{theorem}{0}
\setcounter{equation}{0}

\setcounter{section}{0}

%\section*{\large %0\ \ 
%\hfil Introduction\hfil}
\section*{Introduction}
In the algebraic theory of quadratic forms, a fundamental result due to
Springer \cite{Sp1} (see also \cite[Ch.~6, \S2]{S}) yields an isomorphism from the
Witt group of any complete discretely
valued field $F$ onto the direct sum of two copies of the Witt group of the
residue field $\overline{F}$, provided the characteristic of $\overline{F}$ is
different from~$2$:
\begin{equation}
\label{Springer.eq}
W(F)\simeq W(\overline{F})\oplus W(\overline{F}).
\end{equation}
(Springer also considered the case where $\overline{F}$ is a perfect field of
characteristic~$2$ and the characteristic of $F$ is $0$, but his result in
this case has a different form.)
Springer's theorem has been generalized in various
ways, most recently by Larmour \cite{L}, who proved an analogue for the Witt
group of Hermitian or skew-Hermitian forms over division algebras with
involution over a field with Henselian valuation with residue characteristic
not $2$.  In this paper we give another approach to Larmour's generalization.
We work in terms of valuations on vector spaces and the  graded
structures arising from the filtrations determined by the valuations.
Valuations on vector spaces were used  in Springer's original papers
\cite{Sp1}, \cite{Sp2},  and also appear in the work of 
Goldman and Iwahori \cite{GI} and of Bruhat and Tits \cite{BT1}, \cite{BT2}.
But, the use of associated graded structures is new here, and it seems
to considerably illuminate the earlier approaches.  Besides reproving 
Larmour's theorem, we are able to prove the analogous result in many cases where
the residue characteristic is $2$.  See Def.~\ref{good} for a precise description 
of these cases---they appear to be all the cases where 
our approach yields
% one could hope to have  
a result like Springer's theorem. However, our results 
do \emph{not} cover the very complicated case of quadratic forms over 
valued fields of residue 
characteristic~$2$, as treated for instance by Jacob in \cite{J} and
Aravire and Jacob in \cite{AJ}.
 
% The aim
% of this paper is to give another proof of Larmour's generalization of
% Springer's theorem, using vector space valuations as in Springer's original
% approach. This generalization is cast in terms of graded structures associated
% with valuations, which 
We think our approach sheds an interesting light even on the
classical case. Indeed, a discrete valuation on a field $F$ defines a
$\mathbb{Z}$-filtration whose associated graded ring $\gr(F)$ is
$\overline{F}[t,t\inv]$, the ring of Laurent polynomials in one indeterminate
over the residue field; for a suitably defined Witt ring
$W_{g}\bigl(\gr(F)\bigr)$ of graded 
forms over $\gr(F)$, the isomorphism \eqref{Springer.eq} can be viewed
as an isomorphism
\[
W(F)\simeq W_{g}\bigl(\gr(F)\bigr).
\]
% Our method also applies in many cases where the residue characteristic
% is~$2$; see Section~4 for the precise description of those cases. However, it
% does \emph{not} cover the very complicated case of quadratic forms over 
% valued fields of residue
% characteristic~$2$, as treated for instance by Jacob in \cite{J} and
% Aravire and Jacob in \cite{AJ}.

The graded rings associated with the filtration induced by a valuation on a
division algebra have the property that every homogeneous element is
invertible; they are therefore called \emph{graded division rings} (although
they are not division rings). The first section develops the theory of graded
Hermitian forms over graded division rings with involution. It is well-known 
(cf.~\cite[Ch.~7]{S}, \cite[Ch.~1, \S6]{K}) that the fundamental properties of 
Hermitian forms over a division ring of  characteristic not 2, such as Witt 
cancellation, hold also for even  (also called trace-valued) forms over 
a division ring of characteristic $2$.  We show in Prop.~\ref{cancel} that 
analogues of these fundamental properties hold also for graded Hermitian
forms over a graded division ring; again, we must restrict to even forms when
the characteristic is $2$.  It is convenient to state our results in terms of 
even forms.  But, \lq\lq even" is only a restriction in characteristic $2$;
for, when the characteristic is different from $2$, all forms are even.
Given a graded
division ring $E$ with torsion-free abelian grade group $\Gamma_{E}$, an
involution $\sigma$ on $E$ preserving the gradation, and a central element
$\epsilon$ such that $\epsilon\sigma(\epsilon)=1$, we define the Witt group
$W_{g}^+(E,\sigma,\epsilon)$ of even 
% (or trace-valued) 
nondegenerate graded
$\epsilon$-Hermitian forms for $\sigma$ over $E$. 
% (When $\charr(E) = 2$,
% we must restrict to consideration only of even forms.  When $\charr(E)
% \ne 2$, this is no restriction at all, as all forms are then even.)
% (The condition that the form
% be even is not a restriction if the characteristic is different from~$2$.)
Graded $\epsilon$-Hermitian forms have a canonical orthogonal decomposition
determined by the grade group,
which yields a (non-canonical) decomposition of $W_{g}^+(E,\sigma,\epsilon)$
into a direct sum indexed by $\frac12\Gamma_{E}/\Gamma_{E}$ of Witt groups
of the homogeneous component of $E$ of degree~$0$, with respect
to various involutions, see Prop.~\ref{witt}.
% We also show that most of the familiar properties of even $\epsilon$-Hermitian
% forms on vector spaces
% hold in the graded case (Prop.~\ref{cancel}). 
The main difference in the graded setting is
that graded hyperbolic planes are not all isometric: They are isometric if and
only if they have the same grade set.

In Section~2 we discuss value functions, which are analogues of valuations for
vector spaces, and their associated graded vector spaces. For a vector space
$M$ over a division ring $D$, the most useful value functions
$\alpha\colon\,M\to\Gamma\cup\{\infty\}$ are those for which there exists a
base $\{\,m_{1},\ldots,m_{k}\,\}$ such that
\[
\al\big(\textstyle\sum\limits_{i=1}^k m_i d_i\big)\ =  \ 
\minlim{1\le i\le k}
\big(\al(m_id_{i})\big)\, .
\]
Such a base is called a \emph{splitting base} of $\alpha$, and value functions for
which there exists a splitting base are called \emph{norms}. Given any two norms
$\alpha$ and $\beta$ on a vector space $M$, we show the existence of a common
splitting base (Th.~\ref{common}) and use it to define a norm which we call
the average of $\alpha$ and $\beta$. Our principal results in this section are 
known in the complete
discrete case; they were observed by Goldman and Iwahori \cite{GI} and by
Bruhat and Tits \cite{BT1}. 
% Our approach using associated graded vector spaces
% seems to be new, however, and it puts the earlier results in a very natural
% perspective.

The main results of this paper are given in Sections~3 and 4.
In Section~3, we consider norms $\alpha$ on vector spaces over a valued
division algebra $D$ with involution $\tau$
which are compatible
with a given $\lambda$-Hermitian form $h$, in the sense that there is an
induced nondegenerate graded $\lambda'$-Hermitian form $h'_{\alpha}$ on the
associated graded vector space (for the induced involution $\tau'$ on $\gr(D)$).
However, there is a fundamental obstruction when the residue characteristic is $2$,
in that the form $h'_{\alpha}$ induced by an even form 
$h$ may not be even. In Prop.~\ref{preseven}, we spell out conditions
on the valuation and on the pair $(\tau,\lambda)$ which guarantee that the
form $h'_{\alpha}$ is even for every even form $h$ and every compatible norm
$\alpha$. Under these conditions, we show in Th.~\ref{springer} that the Witt
equivalence class of $h'_{\alpha}$ does not depend on the choice of compatible
norm $\alpha$, and that the correspondence $h\mapsto h'_{\alpha}$ yields a
well-defined and canonical group epimorphism
\begin{equation}\label{theta}
\Theta\colon\,W^+(D,\tau,\lambda)\to W^+_{g}(\gr(D),\tau',\lambda').
\end{equation} 
The graded form $h'_{\alpha}$ may be viewed as
a generalized residue form of $h$; it actually encapsulates \emph{all} the
residue forms of $h$, which appear as the components in  the canonical orthogonal
decomposition of $h'_{\alpha}$.

The notion of compatible norm is due to Springer \cite{Sp1}, \cite{Sp2}, 
though it was not expressed in terms of associated graded forms.  This notion
also appears in \cite{GI} and \cite{BT2}. Our definition in Def.~\ref{comp.def}
follows \cite{BT2} rather than \cite{Sp2} and \cite{GI} in that we require that
$
\alpha(m)+\alpha(n)\leq v\bigl(h(m,n)\bigr)$ for
all $m$, $n$ in the vector space,
instead of
$2\alpha(m)\leq v\bigl(h(m,m)\bigr) 
$.

The results in Section~3 do not require a Henselian hypothesis. In Section~4,
we obtain the analogue of Springer's theorem,
Th.~\ref{isothm}, which asserts that the map
$\Theta$ of \eqref{theta} 
is an isomorphism when the subfield of the center of $D$ fixed under
$\tau$ is Henselian and the residue characteristic is different from~$2$. 
Furthermore,
a form is anisotropic iff its associated graded form is anisotropic.
These
results also hold in the  good cases when the residue characteristic is $2$:
For these, we need a tameness assumption on~$D$ and 
that the isometry group of the forms  be
of unitary or symplectic type, see Def.~\ref{good}. 
When $D$ is tame, the good cases for residue characteristic $2$ are
exactly those cases where induced graded forms of even forms are always even.
Finally, under the same
hypotheses as in our generalization of Springer's theorem, we show in
Prop.~\ref{isometry} that the residues $h'_{\alpha}$, $\ell'_{\beta}$ of two
Hermitian forms $h$, $\ell$ with respect to compatible norms $\alpha$, $\beta$
are isometric if and only if there is an isometry between $h$ and $\ell$ which
preserves the norms.

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\vskip0.25truein

\section{Graded division rings, vector spaces, and Hermitian
forms}

%\section*{\large 1\ \ Graded division rings, vector spaces, and Hermitian
%forms}

\ \indent
Let $\Ga$ be a divisible torsion-free abelian group.  Let $E =
\bigoplus\limits_{\ga\in\Ga} E_\ga$ be a $\Ga$-graded ring, i.e., $E$ is an
associative ring with each $E_\ga$  an additive subgroup of $E$ and 
$E_\ga \cdot E_\de \subseteq E_{\ga + \de}$ for all
$\ga,\de \in \Ga$.  The set of homogeneous elements
of $E$ is $E^h = \bigcup \limits_{\ga \in \Ga} E_\ga$.  The grade set of $E$ is
$\GE = \{\, \ga \in \Ga\,|\ E_\ga \ne (0)\,\}$.

Assume now that the graded ring $E$ is a {\it graded division ring}, i.e.,
every nonzero homogeneous element of $E$ is a unit.  Then, $E_0$ is a
division ring,  and for each $\ga \in \GE$, $E_\ga$ is a $1$-dimensional left
and  right  $E_0$-vector space.  Also, $\GE$ is a subgroup of $\Ga$.  Note that
the center of $E$, denoted $Z(E)$, inherits a  grading from $E$, and $Z(E)$ is a
graded field, i.e, a commutative graded division ring.


Let $S = \bigoplus\limits_{\ga\in\Ga} S_\ga$ be a graded right $E$-module;
that is, $S$ is a right $E$-module with each $S_\ga$ an additive 
subgroup of $S$ and $S_\ga \cdot E_\de \subseteq S_{\ga +\de}$ for all
$\ga,\de \in \Ga$.  The homogeneous elements of $S$ are those in $
\bigcup\limits_{\ga\in\Ga} S_\ga$.
Since $E$ is a graded division ring, slight variations of
the usual ungraded arguments show:  $S$~is a free $E$-module with a base
consisting of homogeneous elements;  every two such bases have the same
cardinality; every homogeneous generating set of $S$ as an $E$-module 
contains a base;  every set of $E$-independent homogeneous elements of $S$ can be
enlarged to a homogeneous base.  All this is easy to prove, and is 
well-known (see, e.g. \cite[\S 1]{HW}). Because of
these analogues with the ungraded case, $S$ is called a {\it graded right
$E$-vector space}, and
$\dim_E(S)$ is defined to be the number of elements in any homogeneous base of $S$.

The grade set of $S$ is $\GS = \{\, \ga\in\Ga\,|\ S_\ga\ne (0)\}$.  Note that
$\GS$ need not be a subgroup of $\Ga$, but it is a union of cosets of $\GE$. 
Indeed, there is a canonical decomposition of $S$ according to
the cosets of
$\GE$  in $\GS$:  For $\ga\in\Ga$, let $\br\ga =\ga +\GE\subseteq \Ga$.
Let 
$$
S_{\br\ga}\ = \textstyle\bigoplus \limits_{\de \in \GE} S_{\ga +\de}\,.
$$
Then, $S_{\br\ga}$ is clearly a graded $E$-subspace of $S$, and if 
$\Ssub \ga \ne (0)$, then $\Ga_{\Ssub \ga} =\br\ga$.  We call $\Ssub \ga$
the {\it $\br\ga$-component} of $S$.  Observe that $\dim_{E}(\Ssub\ga) =
\dim_{E_0}(S_{\ga +\de})$ for each $\de \in \GE$.
We have
\begin{equation}\label{canonical}
S\ = \ \textstyle \bigoplus\limits_{\br\ga\in\GS/\GE}\Ssub\ga \,.
\end{equation}
It is easy to see that if $T$ is another graded right $E$-vector space, 
then $T \cong S$ (graded, i.e., grade-preserving, $E$-vector space isomorphism) iff
$\Ga_T =\GS$ and $\dim_E(T_{\br\ga}) = \dim_E(S_{\br\ga})$ for each $\ga \in
\Ga_T$.

  
Let $\si\colon E \to E$ be a {\it graded
involution} on $E$, i.e., $\si$ is an antiautomorphism of $E$ with
$\si\circ \si = \id$ and $\si(E_\ga) = E_\ga$ for each $\ga \in \GE$. As usual,
$\si$ is said to be of the {\it first kind} if $\si|_{Z(E)} = \id$, and of the 
{\it second kind} otherwise.  
(If $\si$ is of the first kind, then it is of either orthogonal type or symplectic
type.  This is discussed after Remark.~\ref{residue} below.)
 Take any $\ep \in Z(E)_0$ with $\ep \si(\ep) =
1$.  (Of course, if $\si$ is of the first kind, then necessarily $\si(\ep) =
\ep$, so $\ep = \pm 1$.)  A {\it graded $\ep$-Hermitian form} for
$\si$ on a  finite-dimensional graded right $E$-vector space $S$
is a bi-additive function
$k\colon S\times S \to E$ such that for all $s,t\in S$, $c,d\in E$, $\ga,
\de\in \GS$,
\begin{subequations}\label{gherm} 
\begin{align}
 k(sc,td) \ & = \  \si(c) k(s,t) d\,;\\
 k(t,s) \ & =\    \ep \si(k(s,t))\,;\\
 k(S_\ga,S_\de)\ & \subseteq \ E_{\ga+\de}\, .
\end{align}
\end{subequations}
%% \begin{equation*}
% \begin{enumerate}
% \item[{\rm(a)}]\quad 
% $k(sc,td) \, = \, \si(c) k(s,t) d$\,;
% \item[{\rm(b)}]
% \quad $k(t,s) \, =\,   \ep \si(k(s,t))\,$;
% \item[{\rm(c)}]
% \quad $k(S_\ga,S_\de) \subseteq E_{\ga+\de}\,$.
% \end{enumerate}
%\end{equation*}

Let $T$ be a graded subspace of $S$, i.e., $T$ is an $E$-submodule of 
$S$ with $T = \bigoplus\limits _{\ga\in \Ga} T_\ga$, with each $T_\ga$
~a subgroup of $S_\ga$. 
% For any graded subspace $T$ of $S$, s
Set $T^\perp = \{\, s\in S\,|\ k(s,t) = 0
\text{ for all } t\in T\,\}$.  Clearly $T^\perp$ is a graded subspace of
$S$.  As in the ungraded case, we have $\dim_E(T) + \dim_E(T^\perp) = \dim
_E(S) + \dim _E(S^\perp)$;  hence $T^{\perp\perp} = T$ for any graded subspace
$T$ of $S$.  We say that $k$ is {\it nondegenerate} if $S^\perp = (0)$. We say
that $k$ is {\it isotropic} if it has an isotropic vector, i.e., a
nonzero $s\in S$ with $k(s,s) = 0$. A significant fact that follows from the 
assumption that $\Gamma$ is torsion-free is that whenever $k$ is isotropic,
it has a {\it homogeneous} isotropic vector.  For, since $\Gamma$ is torsion-free,
it can be given a total ordering making it into an ordered abelian group.  Then,
with respect to this ordering, any nonzero 
$s = \sum s_\ga$ 
%is an isotropic vector 
(with each $s_\ga \in S_\ga$) has a leading term, which is
the 
nonzero $s_\de$ with the smallest $\de$.   Clearly,
if $s$ is isotropic, then its leading term is a homogeneous isotropic vector.

If $k$ is nondegenerate, we say that $k$ is {\it metabolic} if $S$ has a 
totally isotropic graded subspace $T$ (i.e., $k(T,T) = 0$) with $\dim_E(T) =
\frac 12 \dim(S)$.  We say that $k$ is {\it hyperbolic} if it is
nondegenerate and $S$ has two complementary totally isotropic graded
subspaces. Clearly, every hyperbolic space is an orthogonal sum of
two-dimensional hyperbolic graded subspaces.  If $\ell\colon U\times U\to E$ is
another graded
$\ep$-Hermitian form for $\si$, we write $k\cong \ell$ if $k$~and~$\ell$
are {\it graded isometric}, i.e., there is a  graded (i.e.,
grade-preserving) $E$-vector space isomorphism $f\colon S \to U$ 
 with $f$ an isometry between  $k$  and $\ell$.  We write $k \perp \ell$ for the 
orthogonal sum of $k$ an $\ell$ on $S\oplus U$:  $(k\perp \ell)\big((s,u),
(s',u')\big) = k(s,s') + \ell(u,u')$.

For any $s\in S$, condition (\ref{gherm}b) shows that $k(s,s) = \ep \si(k(s,s))$.
We say that the form $k$ is {\it even}~if 
\begin{equation}\label{even}
\text {for every $s\in S$ there is $c\in E$ with } \ \  k(s,s) = c + \ep \si(c)\,.
\end{equation}
If $\charr(E) \ne 2$, then every form is even (take $c = \frac12k(s,s)$).
This is also true whenever $\si$ is of the second kind. (For, then there is 
$z\in Z(E)$ with $z+ \si(z) = 1$. Then take $c = zk(s,s)$.) 
Just as in the ungraded case, we will see that many results holding when
$\charr(E) \ne 2$ continue to be true for even forms when $\charr (E) = 2$. 

The compatibility of the graded Hermitian form $k$ with the
gradings on $S$ and
$E$ assures that $k$~is well-behaved with respect to the canonical
decomposition (\ref{canonical}) of $S$.  Note that because $E_\rho =
(0)$ for $\rho\notin \GE$, condition (\ref{gherm}c) shows  
\begin{equation}\label{sperp}
k(\Ssub \ga , \Ssub\de) = 0 \quad \text{ whenever } \ga +\de\notin \GE\,. 
\end{equation}
For $\ga \in \Ga$, we write $\kgam$ for $k|_{S_\brga}$.
% Formula \eqref{sperp} yields:   


\begin{proposition}\label{kdecomp} 
Assume $k$ is nondegenerate. Then,
\begin{enumerate}
\item[{\rm(i)}]
If $\gamma\in \frac12\Gamma_E$, then $\kgam$ is nondegenerate and
${S_{[\gamma]}}^\perp = \bigoplus\limits_{[\delta]\ne[\gamma]}S_{[\delta]}$.
  \item[{\rm(ii)}]
If $\gamma\notin \frac12\Gamma_E$, then $\kgam$ is totally
isotropic,
${S_{\br\ga}}^\perp = \bigoplus\limits_{[\delta]\ne[-\gamma]}S_{[\delta]}$,
$\dim_E (S_{[-\gamma]}) = \dim_E (S_{[\gamma]})$, and 
$k|_{S_{[\gamma]}+S_{\br{-\ga}}}$ is nondegenerate and hyperbolic.
  \item[{\rm(iii)}]
$S \ =\   
\underset{\br\ga \in \frac 12 \GE/\GE}\bigperp S_{\br\ga} \ 
\perp \
%\bigoplus\limits^\perp
\underset{\br\de}\bigperp (S_{\br\de} +S_{\br{-\de}})$, 
where the second orthogonal sum is
taken with one summand for each pair $\br\de,\br{-\de}$ with $\br
\de\notin
\frac 12\GE/\GE$.
  \item[{\rm(iv)}]
$k$ is anisotropic iff $\Ga_S\subseteq \frac 12\GE$ and 
$\kgam$ is anisotropic for each $\ga\in \frac12\GE$.
  \item[{\rm(v)}]  $k$ is metabolic (resp.~hyperbolic, resp.~even) iff
$\kgam$ is metabolic (resp.~hyperbolic, resp.~even) for each $\ga\in
\frac12\GE$.

\end{enumerate}
\end{proposition}

\begin{proof}
 (i) \ Suppose $\ga\in \frac12 \GE$.  Formula (\ref{sperp}) shows that 
$\bigoplus\limits_{[\delta]\ne[\gamma]}S_{[\delta]} \subseteq
\Ssub\ga^\perp$; this inclusion is an equality by dimension count.
Then, $\kgam$ is nondegenerate, as $\Ssub\ga \cap
\Ssub\ga^\perp = (0)$.

(ii) \ Suppose $\ga\notin \frac12 \GE$.  
In this case, formula (\ref{sperp}) shows that 
$\bigoplus\limits_{[\delta]\ne[-\gamma]}S_{[\delta]} \subseteq 
\Ssub\ga^\perp$.  Hence, 

\begin{equation}\label{sgammineq}
\dim_E(\Ssub\ga) \ = \ \dim_E(S) - \dim_E(\Ssub\ga^\perp)  \ \le \ 
\dim_E(S) - \textstyle\sum \limits_{\br\de \ne
\br{-\ga}}\dim_E(\Ssub\de) \ = \ \dim_E \Ssub{{-\ga}}\,.
\end{equation}
The same argument, using $-\ga$ in place of $\ga$, shows the reverse
inequality to (\ref{sgammineq}).  Hence, $\dim_E(\Ssub\ga) =
\dim_E(\Ssub{{-\ga}})$, and  equality holds in (\ref{sgammineq}).
Therefore, the inclusion for $\Ssub\ga^\perp$ is an equality.
Then, ${\big(\Ssub\ga + \Ssub{-\ga}\big)^\perp = \Ssub\ga^\perp \cap
\Ssub{-\ga}^\perp = \bigoplus\limits_{\br \de \ne \br{\pm
\ga}}\Ssub\de}$. Since this shows $\big(\Ssub\ga + \Ssub{-\ga}\big)
\cap \big(\Ssub\ga + \Ssub{-\ga}\big)^\perp = (0)$, $k |
_{\Ssub\ga + \Ssub{-\ga}}$ is nondegenerate. 
$\Ssub\ga+\Ssub{-\ga}$ is hyperbolic since it 
 contains the complementary totally isotropic graded subspaces 
$\Ssub\ga$ and $\Ssub{-\ga}$.

(iii) is clear from (\ref{sperp}).

(iv) is clear from (ii) and the fact that if $k$ is isotropic, then it
contains a homogeneous isotropic vector.

(v) \  Suppose $k$ is metabolic, say with totally isotropic graded 
subspace $W$ with ${\dim_E(W) =  \frac12 \dim_E(S)}$.  For each $\ga\in
\frac12\GE$, $W_{\br\ga}$ is a totally isotropic subspace of $\Ssub\ga$
with respect to the nondegenerate form $\kgam$, so 
${\dim_E(W_{\br\ga}) \le \frac12\dim_E(\Ssub\ga)}$. 
Likewise, $\dim_E(W_{\br\de} + W_{\br{-\de}}) \le 
\frac12\dim_E(S_{\br\de} + S_{\br{-\de}})$ for each $\de \in \Ga -
\frac12 \GE$.  Since $\sum\limits_{\rho\in \Ga} \dim_E(W_{\br\rho})
= \dim_E(W) = \frac12\sum\limits_{\rho\in \Ga} \dim_E(S_{\br\rho})$,
all these inequalities must be equalities. Hence,  $\kgam$
is metabolic for each $\ga \in \frac12\GE$. Likewise, if $k$ is
hyperbolic, with complementary totally isotropic graded subspaces 
$W$ and $U$, then for each $\ga\in \frac12\GE$, 
$W_{\br\ga}$ and $U_{\br\ga}$ are complementary totally isotropic
subspaces of $\Ssub\ga$, so $\kgam$ is hyperbolic. Also, any subform
of an even form is even. This proves
one direction of (v).  The converse is clear using (ii), since any 
orthogonal sum of metabolic (resp.~hyperbolic, resp.~even) forms is metabolic
(resp.~hyperbolic, resp.~even) and any hyperbolic form is even.  
\end{proof}



We now show how graded Hermitian forms for $E$ for $\si$  are related to 
Hermitian forms over the division ring $E_0$ with respect to various
involutions on $E_0$. 

% We assume that $\charr(E) \ne 2$.  This assures
% that for any $\ga \in \GE$ there is a nonzero $r\in E_\ga$ with $\si(r)
% = \pm r$.  For, take any nonzero  $s\in E_\ga$.  Then, we can take 
% $r = s+\si(s)$ or $r = s - \si(s)$.  Because $2s \ne 0$ at least one of
% these     choices for $r$ will be nonzero.
% Given our graded involution $\si$ on $E$, let $\si_0 = 
% \si|_{E_0}$, which is an involution on the division ring $E_0$. 
% For $\ga\in \frac 12\GE$, let
Let 
\begin{align*}
\GHerma\qquad&
\parbox[t]{.7\linewidth}{be the category of pairs $(S,k)$ where  
$S$ is a finite-dimensional graded right $E$-vector space  and 
$k\colon S\times S \to E$ is a nondegenerate graded 
$\ep$-Hermitian form on $S$ for $\si$;  the morphisms are
graded isometries.} 
\\
\intertext{For any $\ga \in \frac12\GE$, let}
\mathcal G\mathcal H(E,\si, \ep;\br\ga)\qquad&
\parbox[t]{.7\linewidth}{be the category of 
pairs $(S,k) \in\GHerma$ with $\GS = \ga +\GE$ or $S = (0)$;
the morphisms are graded isometries.}  
\\
\intertext{For any involution $\wh\si$ on $E_0$ and any $\wh\ep$ in $E_0$
with $\wh\ep\wh\si(\wh\ep) = 1$, let}
\mathcal H(E_0, \wh\si,\wh\ep)\qquad&
\parbox[t]{.7\linewidth}{be the category of pairs 
$(U,h)$ where $U$ is a finite-dimensional right $E_0$-vector
space and $h\colon U\times U\to E_0$ is a nondegenerate $\wh\ep$-Hermitian form on $U$
for $\wh\si$; the morphisms are isometries.} 
\\
\intertext{Let}
\GHermap\qquad&
\parbox[t]{.7\linewidth}{be the full subcategory of $\GHerma$
consisting of pairs $(S,k)$ with $k$~even.}
\end{align*}
Likewise, define $\GHermp\si\ga$ (resp.~$\Hermp{\wh\si}{\wh\ep}$) to be
the
subcategory of even forms in $\GHerm\si\ga$ (resp.~$\Herm{\wh\si}{\wh\ep}$).
Recall that if $\charr(E) \ne 2$, then $\GHermap = \GHerma$, and likewise for the
other two plus categories.  We will write $(S,k) \in \mathcal G\mathcal H(E,\si,
\ep)$ if $(S,k)$ is an object in this category.  We often abbreviate $(S,k)$ to
$k$.

If $S$ is any graded right $E$-vector space and $\de \in \Ga$, let $S(\de)$ 
denote the $\de$-shift of $S$, i.e., $S(\de) = S$ as a right $E$-vector space,
but with the grading shifted according to the rule 
$$
S(\de)_\ga\ = \ S_{\ga +\de}\,.
$$
Clearly, $\dim_E(S(\de) ) = \dim_E(S)$ and $\Ga_{S(\de)} = -\de +
\GS$. 

Note that for any $\ga \in \GE$ there is a nonzero $r\in E_\ga$ with 
$\si(r) = \pm r$. For, take any nonzero $s\in E_\ga$.  We can choose $r= s+\si(s)$
if  this is nonzero.  Otherwise, choose $r = s$. In  either case, there is a new 
graded involution $\wi \si$ on $E$, given by $\wi\si = \inn(r)\circ \si$, i.e., 
$\wi\si(c) = r\si(c) r\inv$.  If $\ga = -2\rho$ with $\rho \in \GE$, 
and $(S,k) \in \GHerma$, then we can define a form $\wi k$ on the shifted space
$S(\rho)$ by $\wi k(s,t) = r k(s,t)$.   It is easy to check that $\wi k$ is a
graded $\wi \ep$-Hermitian form  for $\wi \si$, where $\wi\ep = \ep$ if
$\si(r) = r$ and $\wi\ep = -\ep$ if $\si(r) = - r$; that is, $\wi\ep = \ep
\si(r) r\inv$. 


\begin{proposition}\label{category}
%Assume $\charr(E) \ne 2$.
{ \ }
\begin{enumerate}
\item[{\rm(i)}] 
Let $\si_0$ be the restriction of $\si$ to $E_0$, so $\si_0$ is an 
involution on the division ring $E_0$.
There is a canonical equivalence of categories $\Psi\colon\GHerm\si 0
\to 
\Herm{\si_0}\ep$ given by $(S,k) \mapsto (S_0, k|_{S_0})$.
  \item[{\rm(ii)}]
For any $\ga, \rho\in \frac 12\GE$, choose any nonzero $r\in E_{-2\rho}$
with $\si(r) = \pm r$. Let $\wi\si = \inn(r)\circ \si$ and 
$\wi\ep=  \ep \si(r)r\inv$.
There is an equivalence of categories $\GHerm\si\ga \to
\GHermep{\wi\si}{\wi\ep}{{\ga-\rho}}$ given by
$(S,k)
\mapsto (S(\rho), \wi k)$, where $\wi k(s,t) = rk(s,t)$. 
\end{enumerate}
\noindent These equivalences respect dimension and orthogonal sums and send
anisotropic (resp.~metabolic, resp.~hyperbolic, resp.~even) forms to anisotropic
(resp.~metabolic, resp.~hyperbolic, resp.~even) forms.
\end{proposition} 

\begin{proof}
(i)  \  If $(S,k) \in \GHerm\si 0$ then $(S_0, k|_{S_0})\in
\Herm{\si_0}\ep$. (To see that $k|_{S_0}$ is nondegenerate,
note that $S_0$ generates $S$ as a graded $E$-vector space; 
so for $s\in S_0$, $k(s, S_0) = 0$ implies $k(s, S) = 0$, hence $s=0$.)
The functor in the reverse direction $\Theta\colon
\Herm{\si_0}\ep \to \GHerm\si 0$ is given by scalar extension:
Map $(U,\ell) \in \Herm{\si_0}\ep$ to $(U\otimes_{E_0} E,k)\in 
\GHerm\si 0$, where $U\otimes_{E_0} E = \bigoplus\limits_{\ga \in \Ga}
U_\ga$ with $U_\ga\cong U \otimes _{E_0} E_\ga$; $k$ is the scalar
extension of $\ell$, defined by, for all $s_i, t_j\in U_0, d_i, c_j
\in E$,
$$
\textstyle k\big(\sum\limits _i s_i \otimes d_i, \, \sum\limits _j
t_j \otimes c_j \big) \ = \ \sum\limits_{i,j}\si(d_i) \ell (s_i, t_j)
c_j\,.
$$
It is routine to verify that $k$ is well-defined and satisfies axioms   
(\ref{gherm}a)--(\ref{gherm}c), and that ${k|_{(U\otimes_{E_0} E)_0}
\cong \ell}$ under the canonical isomorphism $(U\otimes_{E_0} E)_0\cong U$.
If $k$ were degenerate, then  the $0$-component of the graded vector
space $(U\otimes_{E_0} E)^\perp$ would correspond to a nonzero
$E_0$-vector space in $U^\perp$, contrary to the nondegeneracy
 of $\ell$.  Clearly, the compositions $\Theta \circ  \Psi$ and
$\Psi \circ \Theta$ are isomorphic to the identity functors on 
$\GHerm\si 0$ and $\Herm{\si_0}\ep$, so we have the desired equivalence of
categories.

(ii) \  This is clear.  The inverse morphism 
$\GHermep{\wi\si}{\wi\ep}{{\ga-\rho}}\to \GHerm\si\ga$ is given by 
$(S, \wi k) \mapsto (S(-\rho), \wh k)$, where $\wh k(s,t)
= r\inv \wi k (s,t)$.
\end{proof} 

If    $U$ is any right $E_0$-vector space, let $U^* =\Hom_{E_0}(
U, E_0)$, made into a right $E_0$-vector space via $\si_0$, i.e.,
for $u^*\in U^*$,
$y\in U$, and $c\in E_0$, $(u^*c)(y) = \si_0(c)u^*(y)$.     

\begin{remark}
Take any $\ga \in \Ga$ with $\ga \notin \frac 12 \GE$ and any 
finite-dimensional right graded $E$-vector space $S$ with $\GS = \br \ga\cup
\br{-\ga}$ and $\dim_{E_0}(S_\ga) = \dim_{E_0}(S_{-\ga})$.  If $k\colon
S\times S
\to E$ is any nondegenerate graded $\ep$-Hermitian form for $\si$ then
there is an
$E_0$-vector space isomorphism $\varphi\colon S_\ga \to(S_{-\ga})^*$ given
by 
\begin{equation}\label{eq1.4}
\varphi(s)(t) = k(s,t).
\end{equation}  
Then, $k$ is completely determined by $\varphi$, and every
$E_0$-isomorphism 
$\varphi\colon S_\ga \to(S_{-\ga})^*$ determines a unique $\ep$-Hermitian
form
$k$ on $S$ for $\si$ such that (\ref{eq1.4}) holds, given by
$k(s_1 + t_1, s_2 + t_2) = 
\varphi(s_1,t_2) + \ep \si(\varphi(s_2,t_1))$ for all $s_i\in \Ssub\ga$, $
t_i\in \Ssub{{-\ga}}$. 
Details are left to the reader. 
\end{remark}

Much of the theory of (even) Hermitian forms over division rings carries over to
graded Hermitian forms over graded division rings.  We collect here some of the
basic properties that we will need below.  The principal difference in the graded
setting is that hyperbolic forms of the same dimension need not be isometric.  
(One needs also that the underlying graded vector spaces be graded isomorphic.)

\begin{proposition}\label{cancel}   
%Assume $\charr(E) \ne 2$.
{\ }
\begin{enumerate}
\item[{\rm(i)}] 
$($isometry extension$)$ Let $(S,k)\in \GHermap$.
Suppose $T$ and $U$ are graded subspaces of $S$ such that $k|_T$ is 
nondegenerate, and that there is a graded isomorphism $f\colon T \to 
U$  which is a graded isometry between $k|_T$ and $k|_U$.  Then, 
there is a graded isomorphism $g\colon S \to S$ which is an isometry 
for $k$, such that $g|_T = f$.  
\item[{\rm(ii)}] 
$($Witt Cancellation$)$  For $k_1,k_2,k_3\in \GHermap$,  
 if ${k_1\perp k_3\cong k_2\perp k_3}$,
then $k_1\cong k_2$.
\item[{\rm(iii)}] 
% Every metabolic form in $\GHermap$  is
%  hyperbolic.
If $(S,k) \in \GHermap$ and $k$ is metabolic, with a totally isotropic 
graded subspace $W$ with $\dim_E(W) = \frac12 \dim_E(S)$, then $W$ has a
complementary totally isotropic subspace in $S$; hence, $S$~is hyperbolic.
  \item[{\rm(iv)}]
$($diagonalizability$)$ Every anisotropic form in $\GHermap$ is
isometric to an orthogonal sum of $1$-dimensional forms.
  \item[{\rm(v)}]
For any $k\in \GHermap$, $k\perp -k$ is hyperbolic.
  \item[{\rm(vi)}]
For any $k\in \GHermap$, we have  $k \cong k_{\text{\sl an}} \perp 
k_{\text{\sl hyp}}$, where
$k_{\text{\sl an}}$ is anisotropic and $k_{ \text{\sl hyp}}$  is
hyperbolic. The forms $k_{\text{\sl an}}$ and
$k_{ \text{\sl hyp}}$  are unique up to isometry.
  \item[{\rm(vii)}] 
If $k_1,k_2,  \ell_1,\ell_2\in \GHermap$ with $\ell_1$ and $\ell_2$
hyperbolic and ${k_1\perp\ell_1\cong 
k_2\perp \ell_2}$, then 
${k_{1\,\text{\sl an}} \cong k_{2\,\text{\sl an}}}$.  
  \item[{\rm(viii)}] 
For any $k_1,k_2\in \GHermap$, if  $k_1\perp -k_2$ is hyperbolic, then
$k_{1\,\text{\sl an}} \cong k_{2\,\text{\sl an}}$.
\end{enumerate}
\end{proposition}

\begin{proof}
These can presumably  be proved by mimicking the ungraded proofs as
in \cite{K} or  \cite{S}.  But, that is not necessary since we  
will instead use
the ungraded results and apply Prop.~\ref{category} to get the
corresponding graded ones.     

(i) \  In view of Prop.~\ref{kdecomp}, it suffices to prove that (a) 
for each $\ga \in \frac12 \GE$ the isometry $f|_{T_{\br\ga}}\colon
T_{\br\ga} \to U_{\br\ga}$ extends to an isometry $\Ssub\ga \to 
\Ssub\ga$; and (b) for each $\de\in \Ga$, $\de\notin \frac12 \GE$, 
the isometry $\Tsub\de+\Tsub{{-\de}} \to\Usub\de+\Usub{{-\de}}$ extends
to an isometry $\Ssub\de+\Ssub{{-\de}}\to \Ssub\de+\Ssub{{-\de}}$.
For case (a), we have from \cite[Cor.~(6.4.5)]{K}
 that the isometry extension result holds for forms in 
$\Hermp{\wh\si}{\wh \ep}$ for all involutions $\wh\si$ on $E_0$. Hence 
by Prop.~\ref{category}(i) we have isometry extension for forms in 
 $\GHermepp{\wi\si}{\wi\ep}0$   for all graded involutions $\wi\si$ on~$E$. Hence,
by Prop.~\ref{category}(ii) isometry extension holds for forms in 
$\GHermp\si\ga$ for any $\ga$ in~$\frac12\GE$. This settles case (a).
For case (b), take any $\de\in \Ga$ with $\de \notin \frac12\GE$,
and without loss of generality assume $S = \Ssub\de+\Ssub{{-\de}}$;
then, $\dim_E (\Ssub\de )= \dim_E( \Ssub{{-\de}})$, as $k$
is nondegenerate.  Because $f|_{\Tsub\de+\Tsub{{-\de}}}$ is
nondegenerate, we have $\Tsub\de\cap (\Tsub{{-\de}})^\perp = (0)$,
so by dimension count 
$\Ssub\de = \Tsub\de\oplus (\Tsub{{-\de}})^\perp$; likewise,
$\Ssub\de = \Usub\de\oplus (\Usub{{-\de}})^\perp$.  Let $f_1\colon
(\Tsub{{-\de}})^\perp \to (\Usub{{-\de}})^\perp$ be any 
graded $E$-vector space isomorphism, and let $g_0 = f|_{\Tsub\de}\oplus
f_1\colon
\Tsub\de\oplus (\Tsub{{-\de}})^\perp\to \Usub\de\oplus
(\Usub{{-\de}})^\perp$.  Let $\varphi\colon \Ssub\de \to
\Ssub{{-\de}}^*= \Hom_E(\Ssub{{-\de}}, E)$ be the map given by 
$\varphi(s)(t) = k(s,t)$; so $\varphi$ is an $E$-isomorphism when 
$\Ssub{{-\de}}^*$ is made a right $E$-module via~$\tau$.  Let
$g_1\colon \Ssub{{-\de}} \to\Ssub{{-\de}}$ be the unique graded $E$-
isomorphism satisfying the condition that for the dual map 
$g_1^*\colon \Ssub{{-\de}}^* \to \Ssub{{-\de}}^*$ we have 
$g_1^* = \varphi\circ g_0\inv \circ \varphi\inv$. Set $g = g_0\oplus
g_1\colon \Ssub\de + \Ssub{{-\de}} \to \Ssub\de +\Ssub{{-\de}}$.
The condition $g_1^*\circ \varphi\circ g_0 = \varphi$ says that 
$g$ is an isometry for $k$.   The definition of $g$ gives
 $g|_{\Tsub\de}
= f|_{\Tsub\de}$.  We need to verify the same equality on
$\Tsub{{-\de}}$.  Since $g$ is a $k$-isometry and 
$g((\Tsub{{-\de}})^\perp) = (\Usub{{-\de}})^\perp$, we have
$g(\Tsub{{-\de}}) = \Usub{{-\de}}$. Take any $t\in \Tsub{{-\de}}$
and any $u\in \Usub\de$, and let $t' = f\inv(u) \in \Tsub\de$.
Then, 
$$
k(u, g(t)) = k(f(t'),g(t)) = k(g(t'),g(t)) = k(t',t) 
= k(f(t'), f(t)) = k(u, f(t))\,.
$$
Thus, $g(t) - f(t) \in \Usub{{-\de}} \cap (\Usub\de)^\perp = (0)$.
Hence, $g|_{\Tsub{{-\de}}} = f|_{\Tsub{{-\de}}}$, completing the 
verification that $g|_T = f$.  This completes case (b).

(ii) \ This is immediate from (i).

(iii) \ Let $W$ be a graded subspace of $S$ which is totally isotropic with respect
to the form $k$, with $\dim_E(W) = \frac12\dim_E(S)$.  Then, for each $\ga\in 
\frac12\GE$,  the proof of Prop.~\ref{kdecomp}(v) shows that $W_\brga$ is a
totally isotropic subspace of $S_\brga$ with $\dim_E(W_\brga) =
\frac12\dim_E(S_\brga)$.  Because maximal totally isotropic subspaces of
metabolic forms in $\Hermp{\wh\si}{\wh\ep}$ have complementary totally isotropic
subspaces \cite[proof of Prop.~(3.7.1)]{K}, it follows by Prop.~\ref{category} that
this is also true in 
$\GHermepp {\si}{\ep}\ga$. Hence, there is a  totally isotropic graded subspace
$U_\brga$ of 
$S_\brga$ which is complementary to $W_\brga$.  Also,  for $\de \notin 
\frac12\GE$, the proof of Prop.~\ref{kdecomp}(v) shows that $\dim_E(W_{\br
 \de} +W_{\br{-\de}}) = \frac12\dim_E(S_{\br\de} +S_{\br{-\de}})$.  Since 
$W_{\br{-\de}}\subseteq W_\brde^\perp \cap S_{\br{-\de}}$, dimension count 
shows that this inclusion is an equality.  Choose any complementary graded
subspace~$U_\brde$ of $W_\brde$ in $S_\brde$, and set $U_{\br{-\de}} =
U_\brde^\perp \cap S_{\br{-\de}}$.  Then, $U_\brde + U_{\br{-\de}}$ is a maximal
totally isotropic graded subspace of $S_\brde + S_{\br{-\de}}$. Moreover,
$$  
U_{\br{-\de}}\cap W_{\br{-\de}} \ = \ \big(U_\brde^\perp \cap S_{\br{-\de}}\big)
\cap \big(W_\brde ^\perp \cap S_{\br{-\de}}\big) \ = \ \big(U_\brde +W_\brde\big)
^\perp\cap S_{\br{-\de}} \ = \ S_\brde^\perp \cap S_{\br{-\de}}\ = (0)\,.
$$
Therefore, $U_\brde + U_{\br{-\de}}$ is a totally isotropic graded subspace of 
$S_\brde + S_{\br{-\de}}$ which is complementary to $W_\brde + W_{\br{-\de}}$. Let
$U$ be the sum of the $U_\brga$ for $\ga \in \frac12\GE$ and the $U_\brde
+U_{\br{-\de}}$, one for each pair $\brde, \br{-\de}$ with 
$\de\notin \frac12\GE$.  Then $U$ is totally isotropic since the summands are
totally isotropic and pairwise orthogonal, and $U$ is complementary to $W$ in $S$.


% Let $k$ be a metabolic $\ep$-Hermitian form for $\si$ on
% a graded $E$-vector space $S$, say with totally isotropic graded 
% subspace $W$ with $\dim_E(W) =  \frac12 \dim_E(S)$.  
% Then, by Prop.~\ref{kdecomp}(v),   $k|_{\Ssub\ga}$
% is metabolic for each $\ga \in \frac12\GE$.  Because metabolic forms
% in $\Herm{\wh \si}$ are hyperbolic for all involutions $\wh \si$
% on $E_0$ by \cite[???]{S}, Prop.~\ref{category}(i) and (ii) show that  
% $k|_{\Ssub\ga}$ is  hyperbolic for each $\ga\in \frac12\GE$. For each
% $\de \notin \frac12\GE$ we know already by Prop.~\ref{kdecomp}(ii) that
% $k|_{\Ssub\ga +\Ssub{-\ga}}$ is hyperbolic.  Thus,
% Prop.~\ref{kdecomp}(iii) yields that $k$ is an orthogonal sum of
% hyperbolic subspaces, so $k$ is hyperbolic. 
 
(iv) \ It is known \cite[Prop.~(6.2.4)]{K} that every form in
$\Hermp{\wh\si}{\wh\ep}$ is diagonalizable, except when $\wh\si$~is trivial on 
$E_0$ and $\wh\ep = -1$.  But, in that case, all the forms in 
$\Hermp{\wh\si}{\wh\ep}$ are hyperbolic. Hence by Prop.~\ref{category}(i)
all the anisotropic forms in $\GHermepp{\wi\si}{\wi\ep}0$ are diagonalizable,
for every graded involution~$\wi \si$.  Hence, by
Prop.~\ref{category}(ii) the same is true for anisotropic forms 
in $\GHermp\si\ga$ for each $\ga \in \frac12\GE$. The desired result
then holds by Prop.~\ref{kdecomp}(iv) and (iii). 

(v) \ For any $(S, k) \in \GHermap$ and any $\ga \in \frac12 \GE$, 
% write
% $k_{\br\ga}$ for $k|_{\Ssub\ga}   $, the $\br\ga$-piece of $k$. Then, 
we have $(k\perp -k)_{\br\ga} = 
k_{\br\ga}\perp - k_{\br\ga}$, 
which is hyperbolic by virtue of
Prop.~\ref{category} and the corresponding result for $\Hermp{\wh\si}{\wh \ep}$
for  each involution $\wh\si$ on~$E_0$ \cite[Prop.~(3.5.3)]{K}. For each $\de
\notin
\frac12\GE$, the $\br\de \cup \br{-\de}$-piece of $k\perp -k$ is
automatically hyperbolic, see Prop.~\ref{kdecomp}(ii).  Hence,
$k\perp -k$ is an orthogonal sum of hyperbolic forms, so it is 
hyperbolic.

(vi) \ By Prop.~\ref{category} and the corresponding result for every
$\Hermp{\wh \si}{\wh\ep}$, see  \cite[Prop.~(6.3.2)]{K}, for each $\ga \in
\frac12\GE$ we have $k_\brga \cong k_{\brga\,\text{\sl an}} \perp
k_{\brga\,\text{\sl hyp}}$ with $k_{\brga\, \text{\sl an}}$ anisotropic and
$k_{\brga\,\text{\sl hyp}}$ hyperbolic, each of them in $\GHermp\si\ga$, and
$k_{\brga\,\text{\sl an}}$ and $k_{\brga\,\text{\sl hyp}}$ are unique up to
isometry.
  Set $k_{\text{\sl an}} = \ \underset {\brga\in \frac12\GE/\GE}\perp
k_{\brga\,\text{\sl an}}$ and $k_{\text{\sl hyp}} = \ \underset {\brga\in
\frac12\GE/\GE}\perp k_{\brga\,\text{\sl hyp}} \ \perp \ k_{\text{\sl
irr}}$, where $k_{\text{\sl irr}}$ is the sum of the $k_{\br\de}$
for $\br\de \in \Ga/\GE - \frac12 \GE / \GE$.  Then, $k\cong
k_{\text{\sl an}}\perp k_{\text{\sl hyp}}$, and by  Prop.~\ref{kdecomp} (iv)
and (v),  $k_{\text{\sl an}}$ is anisotropic and 
$k_{\text{\sl hyp}}$ is hyperbolic.  For the uniqueness, suppose 
$k \cong k_1 \perp k_2$ with $k_1$ anisotropic and $k_2$ hyperbolic.
Then, for $\ga\in\frac12\GE$, 
 $k_{\brga\,\text{\sl an}} \cong (k_1 \perp k_2)_{\brga\,\text{\sl an}}
\cong k_{1\,\brga}$, by the uniqueness in $\GHermp\si\ga$, since 
$k_{1\,\brga}$ is anisotropic and $k_{2\,\brga}$ is hyperbolic, by 
Prop.~\ref{kdecomp} (iv) and (v).  Hence, $k_1\cong k_{\text{\sl an}}$
by Prop.~\ref{kdecomp}(iv); then $k_2\cong k_{\text{\sl hyp}}$ by 
part (ii) above.

(vii) \ 
% $k \cong k_{\sl\text{an}} \perp k_{\sl\text{hyp}}$
 Since $k_{1\,\text{\sl hyp}}\perp\ell_1$ is hyperbolic, the 
uniqueness part of (vi) shows that 
$$
k_{1\,\text{\sl an}}
\cong (k_1 \perp \ell_1)_{\text{\sl an}} \cong
(k_2 \perp \ell_2)_{\text{\sl an}} \cong k_{2\,\text{\sl an}}\,.
$$
 

(viii) \ Apply (vii) to $k_1 \perp (k_2 \perp-k_2) \cong k_2 \perp 
(k_1\perp -k_2)$ using (v). 
\end{proof}


For each of the categories defined preceding Prop.~\ref{category} there is an
associated Witt group:  Let $\mathcal C$ be any of $\GHermap$,
$\GHermp\si\ga$, or $\Hermp{\wh\si}{\wh\ep}$.  For $(S,k) \in \mathcal C$, we
write 
$c\ell(k)$ for the isometry class of $(S,k)$ (meaning graded isometry class in
the graded case). The set $\mathcal Iso(\mathcal C)$ of isometry classes of
forms in $\mathcal C$ is a cancellative monoid with respect to the operation
induced by orthogonal sum.  The Witt group $W(\mathcal C)$ is the group 
$\mathcal Iso(\mathcal C)/ {\sim}$ of equivalence classes with respect to the
equivalence relation $c\ell(k_1) \sim c\ell(k_2)$ iff there are hyperbolic
forms $\ell_1$ and $\ell_2$ in $\mathcal C$ with $k_1\perp \ell_1
\cong k_2\perp \ell_2$.  Let $\br k$ denote the equivalence class of
$c\ell(k)$.  Prop.~\ref{cancel}(vii) in the graded case or \cite[Prop.~(6.3.2)]{K}
in the ungraded case show that $\br{k_1} = \br{k_2}$ in $W(\mathcal C)$
iff $k_{1\,\text{\sl an}}\cong k_{2\,\text{\sl an}}$.  The (well-defined,
associative) operation in $W(\mathcal C)$ is: $\br{k_1}+\br{k_2} = 
\br{k_1\perp k_2}$, and Prop.~\ref{cancel}(v) shows that $W(\mathcal C)$
is actually a group.   
Set 
\begin{align*}
W_g^+(E, \si, \ep) \ &= \ W(\GHermap)\,;\\
W_g^+(E, \si, \ep;\brga) \ &= \ W(\GHermp\si\ga)\,,\text{ for each }\brga\in
\textstyle\frac12\GE/\GE ;\\
W^+(E_0, \wh\si, \wh\ep)\ &= \ W(\Hermp{\wh\si}{\wh\ep})\,, \text{ the usual even
Witt group of }
\Hermp{\wh\si}{\wh\ep}\,.  
\end{align*}


\begin{proposition}\label{witt}
% Assume $\charr(E) \ne 2$.  Then, 
For any graded involution $\si$ on $E$ and any
$\ep \in Z(E_0)$ with $\ep \si(\ep) = 1$, 
\begin{enumerate}
  \item[{\rm(i)}]
$W_g^+(E, \si,\ep;\br 0) \ \cong  \ W^+(E_0,\si|_{E_0},\ep)$, canonically.
  \item[{\rm(ii)}] 
$W_g^+(E, \si, \ep) \ \cong \ \bigoplus\limits_{\brga \in \frac12\GE/\GE}
W_g^+(E, \si, \ep;\brga)$, canonically.
  \item[{\rm(iii)}]
For any $\ga, \rho\in \frac12\GE$, choose any nonzero $r\in E_{-2\rho}$ with 
$\si(r) = \pm r$, and let $\wi\si = \inn(r) \circ \si$
and $\wi\ep = \ep\si(r)r\inv$.  Then, 
$W_g^+(E, \si, \ep;\brga)  \ \cong \ W_g^+(E, \wi\si,\wi\ep;\br{\ga-\rho})$.   
  \item[{\rm(iv)}]
For each $\brga \in \frac12\GE/\GE$, choose any $\ga \in \brga$ and any
nonzero $r_\ga\in E_{-2\ga}$ with $\si(r) = \pm r$,  let 
$\si_\brga$~be the graded involution $\inn (r_\ga) \circ \si$, and let 
$\si_{\brga \,0}= \si_\brga|_{E_0}$ and $\ep_\brga = \ep\si(r_\brga)r_\brga\inv$. 
Then, 
$$
W_g^+(E, \si, \ep) \ \cong \ \textstyle\bigoplus\limits_{\brga \in
\frac12\GE/\GE} W^+(E_0, \si_{\brga\,0}, \ep_\brga)\,.
$$  
\end{enumerate}
\end{proposition}

\begin{proof}
(i) is immediate from  Prop.~\ref{category}(i).

(ii) \ For $[k] \in W^+_g(E, \si,\ep)$  the map $\br k\mapsto 
\big([k_\brga]\big)_{\brga \in \frac12\GE/\GE}$ is well-defined and 
gives the asserted isomorphism, by Prop.~\ref{kdecomp}.

(iii) is immediate from Prop.~\ref{category}(ii).

(iv)  \ We have by (iii) and (i), $W^+_g(E, \si,\ep;\brga)\cong W^+_g(E,
\si_\brga,\ep_\brga;\br 0) \cong W^+(E_0, \si_{\brga\,0},\ep_\brga)$.  With this,
(iv) follows from (ii).  
\end{proof}

%\setcounter{equation}{0}
%\setcounter{theorem}{0}
%\setcounter{section}{2}
%
\vskip0.25truein
%\section*{\large 2\ \ Norms on vector spaces over valued fields}
\section{Norms on vector spaces over valued fields}
\ \indent
Let $D$ be a division ring finite dimensional over its center $F$, and 
suppose $D$ has a valuation~$v$. That is, we have a divisible totally 
ordered
abelian group $\Ga$ and an element $\infty$ (with $\infty >\ga$ for all $\ga
\in \Ga$
and $\ga+\infty = \infty + \ga = \infty + \infty = \infty$) and $v\colon
D\to 
\Ga\cup \{\infty\}$ is a function satisfying, for all $c,d\in D$,
\begin{subequations}\label{val}
\begin{align}
& v(d) =\infty \text{ iff } d = 0;\\
&v(cd) = v(c) + v(d);\\
&v(c+d)\ge \minn\big(v(c),v(d)\big).
\end{align}
\end{subequations}
It is immediate that $v(1) = v(-1) = 0$, and if $v(c) \ne v(d)$ then 
$v(c+d) = v(c-d) = \minn\big (v(c),v(d)\big)$.  Let $D^\times = D-\{\,0\,\}$.  Let 
$\vg D = v(D^\times)$, the value group of $v$, which is a subgroup of
$\Ga$. 

For each $\ga \in \vg D$ define the abelian groups
$$
D^{\ge \ga} = \{ \, d\in D\,|\ v(d) \ge \ga\,\},\quad
D^{> \ga} = \{ \, d\in D\,|\ v(d) > \ga\,\},\quad
\text{ and } D_\ga = D^{\ge \ga}\big/ D^{>\ga}\,.
$$
The {\it associated graded ring} of ($v$ on) $D$ is
$$ 
\gr(D) \ =\ \textstyle\bigoplus\limits_{\ga\in \Ga} D_\ga \,,
$$
with the multiplication 
induced by the multiplication in $D$.  For any $d\in D^\times$, we
write
$d'$ for the  image $d+D^{>v(d)}$ of $d$ in $D_{v(d)}$, and
$(0_D)'= 0$ in $\gr(D)$.    Property \ref{val}b implies that
$$
(cd)'\ =\ c'd'\quad \text{ for all } c,d\in D\,.
$$
So, in particular, we have $d'(d\inv)'  = 1'$ for any $d\in D^\times$.  This
shows that $\gr(D)$ is a graded division ring, as described in \S 1.
Clearly the grade group $\vg{\gr(D)} = \vg D$.  Note also that
the valuation ring of $v$ is $V_D = D^{\ge 0}$, and the unique
maximal left (and right) ideal $M_D$ of $V_D$ is $M_D = D^{>0}$. 
Hence, $D_0 = V_D\big/M_D = 
\ov D$, the residue division ring of $D$.  

Now, let $M$ be any finite-dimensional right $D$-vector space.  
We call a function $\al\colon M \to \Ga \cup \{\,\infty\,\}$ a 
{\it value function} (with respect to $v$ on $D$) if for all
$m,n\in M$ and $d\in D$,
\begin{subequations} \label{valf}
\begin{align}
&\al(m) =\infty \text{ \ iff \  } m = 0;\\
&\al(md) = \al(m) + v(d);\\
&\al(m+n)\ge \minn \big(\al(m),\al(n)\big)\,. 
\end{align}
\end{subequations}
It is immediate that $\al(-m) = \al(m)$ and that if 
$\al(m) \ne \al(n)$ then $\al(m+ n) =\al(m-n) =\minn\big(\al(m),\al(n)\big)$.
The value set of $\al$ is $\vg M = \{\,\al(m)\,|\ m\in M,  \ m\ne0\,\}
\subseteq \Ga$.  This $\vg M$ need not be a group, but it is a
union of  cosets of $\vg D$.  For each $\ga \in \Ga$, define the
abelian groups
$M^{\ge \ga}$, $M^{>\ga}$, and $M_\ga$ just as for $D$ above.  
The {\it associated graded vector space} of ($\al$ on) $M$ is
$$
\gr(M)  \ =\  \textstyle \bigoplus\limits_{\ga\in \Ga} M_\ga \,.
$$
When we need to specify the value function, we write $\gr_\al(M)$.
The module action of $D$ on $M$ induces a well-defined module action
of $\gr(D)$ on $\gr(M)$, making $\gr(M)$ into a graded right vector space over
$\gr(D)$.  For  nonzero $m\in M$, we write $m'$ for the image 
$m+M^{>\al(m)}$ of $m$ in $M_{\al(m)}$.  We write $(0_M)' = 0$ in 
$\gr(M)$.  Clearly, for all $m\in M$, $d\in D$, we have  
$$
(md)' = m'd'\, .
$$ 
Also, for nonzero $m,n\in M$, we frequently use the obvious fact that
\begin{equation}
% &m',  \text {\quad\quad\ \   if \ } \al(m) < \al(n);\\
 (m+n)' \ = 
\begin{cases} \ m'& \text { if \,} \al(m) < \al(n);\\
\ n'& \text {  if \,} \al(m) > \al(n);\\
\  m'+n'& \text{  if \,} \al(m) = \al(n) \text{ and } m' + n' \ne 0\,.
% \cases m', \text { if } \al(m) < \al(n);\\
% n', \text { if } \al(m) > \al(n);\\
% m'+n', \text{ if } $\al(m) = \al(n) \text{ and } m' + n' \ne 0\,. \endcases
\end{cases}
\end{equation}

Here is a fundamental way of constructing a value function on
$M$:  Take any  base $\{\, m_1, \ldots, m_k\,\}$~of~$M$ as 
$D$-vector space, and take  any $\ga_1, \ldots, \ga_k \in \Ga$. 
Then, define $\al\colon M\to
\Ga\cup \{\infty\}$ by $\al(\sum m_id_i) = 
\minlim{1\le i\le k}  
\big(\ga_i+v(d_i)\big)$.  That is, $\al(m_i) = \ga_i$ and for all $d_1, \ldots,
d_k\in D$,
\begin{equation}\label{normcond}
\al\big(\textstyle\sum\limits_{i=1}^k m_i d_i\big)\ =  \ 
\minlim{1\le i\le k}
\big(\al(m_i) + v(d_i)\big)\, .
% \underset X   A.
\end{equation}

It is easy to check that $\al$ satisfies the axioms for a value function on
$M$.  In fact, we will be exclusively interested in the value functions
arising this way.  

\begin{definition}
Given a value function $\al$ on $M$, a base $\{\, m_1, \ldots , m_k\,\}$ of 
$M$ for which formula $($\ref{normcond}$)$ holds is called a {\it splitting
base} of $\al$.  We say that the value function $\al$ is a {\it norm} on $M$
$($with respect to the valuation $v$ on $D$$)$ if there is a splitting base
for 
$\al$.
\end{definition}

The associated graded vector space elucidates the notion of splitting bases:

\begin{proposition} \label{linind}
Let $\al$ be a value function on $M$, and let $m_1, \ldots, m_\ell\in M-\{0\}$.
Then,
\begin{enumerate}
  \item[{\rm(i)}]
$m_1',\ldots, m_\ell'$ are $\gr(D)$-linearly independent in $\gr(M)$ iff 
$\al\big(\sum\limits_{i=1}^\ell  m_i d_i\big)\ =  \ 
\minlim{1\le i\le \ell}
\big(\al(m_i) + v(d_i)\big)$ for all $d_1, \ldots, d_\ell\in D$.
  \item[{\rm(ii)}]
If $m_1',\ldots, m_\ell'$ are $\gr(D)$-linearly independent in
$\gr(M)$, then $m_1, \ldots, m_\ell$ are $D$-linearly independent
in $M$.
\end{enumerate}
\end{proposition}

\begin{proof}
Suppose $m_1', \ldots, m_\ell'$  are $gr(D)$-linearly dependent in
$\gr(M)$, say $\sum\limits_{i=1}^\ell m_i'c_i= 0$ in $\gr(M)$, with some
$c_i\ne 0$.  Then, each homogeneous component of the sum is $0$.  
Suppose  some $m_i'c_i$ has nonzero $\gamma$-component. Then, after
throwing out the summands  with trivial $\ga$-component and renumbering
the summands, we have $0 = \sum\limits_{i=1}^j m_i'd_i' = 
\sum\limits_{i=1}^j (m_id_i)'$ in $M_\ga$, for some nonzero $d_i\in D$ with
each $\al(m_id_i) = \ga$. This means that $\al \big(\sum\limits_{i=1}^j
m_id_i\big) > \ga = \minlim{1\le i\le j} \big(\al(m_i) + v(d_i)\big)$, so
the  condition on the $m_i$ in (i) fails to hold. 

On the other hand,
suppose $m_1', \ldots, m_\ell'$ are $\gr(D)$-linearly independent in   
$\gr(M)$.  Take any linear combination $m = \sum\limits_{i=1}^\ell
m_i d_i$ with the $d_i\in D$ and some $d_i\ne 0$.  Let $\ga =\minlim{1\le
i\le l}\big(\al(m_id_i)\big)$.  After reordering the $m_i$, there is a 
$j\ge 1$ with $\al(m_id_i) = \ga$ for $1\le i\le j$ and $\al(m_id_i) >
\ga$ for $i >j$.  Let $n = \sum\limits_{i=1}^j m_i d_i$. Since all the
$\al(m_id_i) = \ga$ for $1\le i \le j$ and $\sum\limits_{i=1}^j 
(m_id_i)' =\sum\limits_{i=1}^j m_i' d_i' \ne 0 $ in $M_\ga$ by the
independence of the~$m_i'$, it follows that  $\al(n) = \ga$.   
% and $n'= \sum\limits_{i=1}^j m_i' d_i' \ne 0$ in $M_\ga$.
We have $m - n = \sum\limits_{i=j+1}^\ell m_i d_i\in M^{>\ga}$.  
Since ${\al(m-n) >\ga = \al(n) = \al(-n)}$, we have $m-n\ne -n$, 
so $m\ne 0$, and $\al (m) = \al (n +(m-n)) = \minn\big(\al (m),
\al(m-n)\big) = \ga$. This shows that the $m_i$ satisfy the condition
in (i), completing the proof of~(i).  It also shows (as we saw $m \ne 0$)
that $m_1, \ldots, m_\ell$ are $D$-linearly independent in $M$, proving~
(ii). 
\end{proof}

\begin{corollary}\label{normdim}
Let $\al$ be a value function on $M$. Then,
\begin{enumerate}
  \item[{\rm(i)}]
$\dim_{\gr(D)}\gr(M) \le \dim_D(M)$, and equality holds iff $\al$ is a norm.
  \item[{\rm(ii)}]
Suppose $\al$ is a norm.  Then $\{\, m_1, \ldots, m_k\,\}$ is a splitting base
for $\al$ iff $\{\, m_1',\ldots, m_k'\,\}$ is a homogeneous base of 
$\gr(M)$ as a graded $\gr(D)$-vector space.
  \item[{\rm(iii)}]
Suppose $\al$ is a norm with splitting base $\{\, m_1, \ldots, m_k\,\}$.  Take any
nonzero $n = \sum \limits_{i=1}^km_id_i\in M$.  For any $j$ with $\al(m_jd_j) =
\al(n)$ the set $\{n\} \cup \{\,m_i\ |\ i\ne j\,\}$ is a splitting base for $\al$.
\end{enumerate} 
\end{corollary}

\begin{proof}
(i) and (ii) are  immediate from Prop.~\ref{linind}. 

(iii) \ Let $\ga = \al(n)$. Then, the image $n'$ of $n$ in $\gr(M)$ lies in 
$M_\ga$, and we have $n' = \sum\limits_{i\in I} m_i'd_i'$, where $I = \{\,i\,| \
\al(m_id_i) = \ga\,\}$, and each of these summands is nonzero.  By hypothesis, one
of the summands is~$m_j'd_j'$. We can use this equation to express $m_j'$ as a
linear combination of $n'$ and the $m_i'$ with $i\ne j$. Thus, the usual exchange
argument applies to show that $\{n'\}\cup \{\,m_i'\,|\ i\ne  j\,\}$ is a
homogeneous $\gr(D)$-base of $\gr(M)$.  Therefore, by part(ii), 
$\{n\}\cup \{\,m_i\,|\ i\ne  j\,\}$ is a splitting base for $\al$.
\end{proof}

\begin{remark}\label{valex} Let $F\subseteq L$ be fields with 
$[L\!: \! F]<\infty$, let $v$ be any valuation on $F$, and let 
$\al$ be any valuation on $L$ extending $v$. Let $\vg F$, $\vg L$ be the value 
groups of $v$ and $\al$, and let $\ov F$ and $\ov L$ be the residue fields of the
associated valuation rings.  Of course, $\al$ is a value function on $L$, 
viewed as an $F$-vector space, with respect to $v$.
It is easy to prove, and well-known (cf.~\cite{HW} or \cite{Bl}) 
that 
% \begin{equation}\label{greq} 
$[\gr(L)\!: \!\gr(F)] = [\ov L\!: \!\ov F]\, |\vg L\!: \!\vg F|$.
% \end{equation}  
The quantities on the
right are the residue degree and the ramification index of $\al/v$.  The
Fundamental Inequality in valuation theory (see, e.g., \cite[VI.8.1, Lemma 2]{B})
says that
\begin{equation}\label{fund}
[\ov L\!: \!\ov F]\, |\vg L\!: \!\vg F| \ \le \ [L\!: \!F]\,. 
\end{equation}
Thus,
\begin{equation}\label{greq}
[\gr(L)\!: \!\gr(F)]\ \le\ [L\!: \!F]\,, 
\end{equation}
which agrees with Cor.~\ref{normdim}(i).     
Now, it is standard in valuation theory that whenever 
$v$ has more than one extension to $L$, then the inequality \eqref{fund} is 
strict, so we have a strict inequality in \eqref{greq}; then the value 
function $\al$ on $L$ is not a norm, by Cor.~\ref{normdim}.  This provides an
abundant source of examples of value functions which are not norms. On the 
other hand, suppose $\al$ is the only extension of $v$ to $L$. One says
that  
$\al/v$ is {\it defectless} if equality holds in \eqref{fund}.  Ostrowski's
Theorem, deducible from  \cite[Cor.~5.3.8]{EP}, yields that $\al/v$ is defectless
whenever
$\charr(\ov F)\nmid [L\!:\!F]$.  If $v$ is maximally complete (i.e., $v$~has no
immediate extensions to any larger field), then it is known by \cite[Th.~5.2.5]
{EP}  and \cite[Th.~11, p.~55]{Sg}
that for any  field $L \supseteq F$ with $[L\!:\!F] < \infty$, $v$ has a
unique extension  to $L$ and equality holds in \eqref{fund}.  In fact, a variation
on the argument in \cite[proof of Th.~11, p.~55]{Sg}  shows that when $v$ is
maximally complete, then every value function for $v$ on any finite-dimensional
vector space over 
$F$ is a norm.  This is indicated in \cite[p.~299]{BT1}.  Examples of maximally
complete valuations are complete discrete valuations and the usual valuations on 
iterated Laurent series fields $K((x_1))\ldots ((x_n))$. 
 
\end{remark}

\begin{proposition}\label{subspace}
Suppose $\al$ is a norm on $M$. Let $N$ be any nonzero $D$-subspace of $M$.  
Then, $\al|_N$ is a norm on $N$.  Moreover, any splitting base of $\al|_N$
can be enlarged to a splitting base of $\al$.
\end{proposition}

\begin{proof}
Let $k = \dim_D(M) =\dim_{\gr(D)}(\gr(M))$.  For the first assertion 
we argue by induction on $\dim_D(N)$.  If $\dim_D(N) = 1$, then 
$1 \le \dim_{\gr(D)}(\gr(N)) \le \dim_D(N) = 1$ by 
Cor.~\ref{normdim}(ii); hence equality holds, so $\al|_N$ is a norm.
Now assume $\dim_D(N) >1$.  Take any nonzero $n\in N$.  Then, $n'\ne 0$
in $\gr(M)$, so there exist $m_2, \ldots , m_k \in M$ such that $\{
\,n', m_2', \ldots, m_k'\,\}$ is a homogeneous base for the  graded
$gr(D)$-vector space $\gr(M)$.  Let $P = D$-span of $\{\, m_2, \ldots,
m_k\,\}$.  By Prop.~\ref{linind}(ii) $\dim_D(P) = k-1$. Since 
$\dim_{\gr(D)}(\gr(P)) \le \dim_D(P) = k-1$,  
by Cor.~\ref{normdim}(i), the $gr(D)$-linearly
independent elements $\{\, m_2',
\ldots, m_k'\,\}$  span $\gr(P)$. Since $n' \notin \gr(P)$, we have
$N\not \subseteq P$.  Let $Q = N\cap P$; so $\dim_D(Q) = \dim_D(N)-1$.
By induction we have $\al|_Q$ is a norm, so 
$\dim_{\gr(D)} (\gr(Q)) = \dim_D(N) -1$.  But, $n' \notin \gr(Q)$,
as $\gr(Q)\subseteq \gr(P)$.  Hence, $\dim_{\gr(D)}(\gr(N)) \ge 1 
+ \dim_{\gr(D)}(\gr(Q)) = \dim_D(N)$.  By Cor.~\ref{normdim}(i), 
$\al|_N$ is a norm.  

The second assertion of the proposition follows easily from
Cor.~\ref{normdim}(ii).
\end{proof}

\begin{remark}\label{comprem}
If $\al$ is a norm on $M$ and $N$ is a $D$-subspace of $M$, then a
complementary subspace~$P$ of $N$ in $M$ (i.e., $P\cap N = (0)$
and 
$P+N= M$) is called a {\it splitting complement} of $N$ if 
$\al(n+p)=  \minn\big(\al(n),\al(p)\big)$ for all $n\in N$, $p\in P$. It is
easy to see that splitting complements always exist.  Indeed, a complement
$P$ of $N$ is a splitting complement iff $\gr(P)$ is a complement of 
$\gr(N)$ as graded subspaces of $\gr(M)$. 
\end{remark}

Let $M$ and $N$ be finite-dimensional right $D$-vector spaces with respective norms
$\al$ and $\be$, and let  $f\colon M\to N$ be any nonzero $D$-linear map.  Define
\begin{equation}\label{jdef}
j_{\al,\be}(f) \ =\ \minn\{\,\be(f(m)) - \al(m)\,|\ m\in M,\ m\ne0\,\}\,.
\end{equation}
 We  show  that this minimum exists, so $j_{\al,\be}$ is well-defined:
Let $\{\, m_1, \ldots, m_k\,\}$ be any splitting base of $M$ for $\al$.
For $m \in M$, $m \ne 0$, write $m = \sum\limits_{i = 1}^k m_i d_i$.  
Then, 
\begin{equation}\label{jcalc}
\begin{split}
\be(f(m))  \ &\ge  \ \minlim{1\le i\le k} \big(\be(f(m_i)) + v(d_i)\big)
 \ = \   \minlim{1\le i\le k} \big(\be(f(m_i)) - \al(m_i) + \al(m_i) +v(d_i) \big)\\
&\ge  \  \minlim{1\le i\le k} \big(\be(f(m_i)) - \al(m_i)\big) 
+  \minlim{1\le i\le k}\big(\al(m_i) +v(d_i)\big)\\
&= \ \minlim{1\le i\le k} \big(\be(f(m_i)) - \al(m_i)\big) + \al(m)\,.
\end{split}\end{equation}
Thus, $\be(f(m)) -\al(m) \ge \minlim{1\le i\le k} \big(\be(f(m_i)) - \al(m_i)\big)$
for all nonzero $m\in M$.  
This shows that $j_{\al,\be}$~exists, and that 
\begin{equation}\label{jformula}
j_{\al,\be}(f) \ = \ \minlim{1\le i\le k} \big(\be(f(m_i)) - \al(m_i)\big)\,.
\end{equation}
 
For short, let $j = j_{\al,\be}$. By definition of $j$, for each $\ga \in \Ga$, 
we have $f(M^{\ge \ga}) \subseteq N^{\ge\ga+j}$, so also 
$f(M^{> \ga}) \subseteq N^{>\ga +j}$.
% We say that a $D$-linear map $f\colon M\to N$ is {\it norm
% compatible} if $\be(f(m)) \ge \al(m)$ for every $m\in M$.  That is, for each
% $\ga\in \Ga$, we have $f(M^{\ge \ga}) \subseteq N^{\ge\ga}$, so also 
% $f(M^{> \ga}) \subseteq N^{>\ga +j}$. 
Therefore, $f$ induces a
well-defined map 
$M_\ga \to N_{\ga+j}$ for each $\ga\in \Ga$;  these combine to give the {\it
associated graded map} $f'\colon\gr_\al(M) \to \gr_\be(N)$.  This $f'$ is 
given on homogeneous elements by, for any $m\in M$,
\begin{equation}\label{assoc}
f'(m+M_{\al(m)}) \ = \ f(m) + N_{\al(m)+j} \,. 
\end{equation}
 It is easy to check that $f'$ is a  $\gr(D)$-module homomorphism
which shifts all grades by $j$.
Hence, $\kerr(f')$ (resp.~$\im(f')$) is a graded subspace of $\gr(M)$
(resp.~$\gr(N)$).
\begin{proposition}\label{grmap}
Let $f\colon M\to N$ be a nonzero $D$-linear map,  and let $f'\colon\gr_\al(M) \to
\gr_\be(N)$ be the associated graded map just described. Then,
\begin{enumerate}
  \item[{\rm(i)}] 
$\gr(\kerr(f)) \subseteq \kerr(f')$ \ and \ $\im(f') \subseteq \gr(\im(f))$.
  \item[{\rm(ii)}]
$\im(f') = \gr(\im(f))$ \ iff \ $\dim_{\gr(D)}(\im(f')) =
\dim_{\gr(D)}(\gr(\im(f)))$ 
\ iff \ $\gr(\kerr(f)) = \kerr(f')$  \ iff \ for every $n\in \im(f)$ there is $m\in
M$ with $f(m) = n$ and $\al(m) = \be(n)-j_{\al,\be}(f)$.
\end{enumerate}
\end{proposition}

\begin{proof} (i) is clear from the definitions.  Because the subspaces involved
are graded, one has only to check the inclusions for homogeneous elements.

(ii) \ The fact that each condition implies the next is obvious from dimension
considerations, except that the next to last implies the last.  We now prove that.
Suppose $\gr(\kerr(f)) = \kerr(f')$.  Let $P$ be a splitting complement of
$\kerr(f)$ for $\al$.  Then $f|_P\colon P \to \im(f)$ is an isomorphism.
Since $\gr(P) \cap \kerr(f') = \gr(P) \cap \gr(\kerr(f)) =
(0)$, we have $f'|_{\gr(P)}$ is injective.  Take any splitting base
$\{\, p_1, \ldots, p_k\,\}$ for $\al|_P$. Since $f'(p_i') \ne 0$, we have 
${\be(f(p_i)) = \al(p_i)+j}$, where $j = j_{\al,\be}(f)$, and $f(p_i)' = f'(p_i')$.
The injectivity of $f'|_{\gr(P)}$ shows that the set $\{\,f(p_1)', 
\ldots, f(p_k)'\,\}$ is $\gr(D)$-independent in $\gr(N)$.   
For any $n\in \im(f)$, 
take the $m \in P$ with $f(m) = n$.  Write ${m = \sum\limits_{i=1}^k p_id_i}$. 
Then 
$n =  \sum\limits_{i=1}^k f(p_i)d_i$.  From the $\gr(D)$-independence of the
$p_i'$
and of the $f(p_i)'$, Prop.~\ref{linind}(i) yields
$$
\be(n)  \ =  \ \minlim{1\le i \le k} \big( \be(f(p_i)) +v(d_i)\big)
 \ =  \ \minlim{1\le i \le k} \big( \al(p_i)+ j+v(d_i)\big) 
 \ = \  \al(m)+j\,.
$$
This proves the last condition in (ii).  Since the last condition in (ii) clearly
implies the first, the cycle of implications is now complete.
\end{proof}

\begin{theorem}\label{common}
Let $M$ be a finite-dimensional $D$-vector space with two norms $\al $ and $\be$. 
Then, there is a subset of $M$ which is a splitting base for $\al$ and also a
splitting base for $\be$.
\end{theorem}

\begin{proof}
We argue by induction on $\dim_D(M)$.  Since the $1$-dimensional case is clear,
assume ${\dim_D(M) >1}$.  Let $\{\,m_1,\ldots, m_k\,\}$ be a splitting base of
$M$ for $\al$ and $\{\,n_1, \ldots, n_k\,\}$ a splitting base for
$\be$.  Write each
$n_j = \sum \limits_{i=1}^km_i d_{ij}$.  Choose $s$, $1\le s\le k$, so that
$\minlim{1\le i\le k}\big(\al(n_i) - \be(n_i)\big) = \al(n_s) - \be(n_s)$. 
Choose $r$, $1\le r\le k$, so that $\al(n_s) = \al(m_r d_{rs})$; so $d_{rs}\ne
0$.  By Cor.~\ref{normdim}(iii), $\{\,n_s\,\}\cup \{\,m_i\,|\ i\ne r\,\}$ is a
splitting base for $\al$.  Let $P = \sum \limits_{i\ne r}m_iD$.  So, $P$ is a 
splitting complement to $n_sD$ for $\al$.  We show that this is also true for
$\be$.  For $j\ne s$, let
$$
p_j\ = \ n_j \, - \, n_sd_{rs}\inv d_{rj} \, \in P\,. 
$$
Now, for any $j$, we have
$$
\al(m_r) +v(d_{rj})-\be (n_j) \ \ge  \ \al(n_j) - \be(n_j) \ \ge \ \al(n_s) -
\be(n_s) \ = \ \al(m_r) + v(d_{rs}) - \be(n_s)\,.
$$
Hence,
$$
\be(n_j) \ \le \ \be(n_s) -v(d_{rs}) + v(d_{rj}) \ = \ \be (n_sd_{rs}\inv d_{rj})\,.
$$
This shows that for $j\ne s$ we have in $\gr_\be(M)$, $p_j' = n_j'$ or 
$p_j' = n_j' - n_s'(d_{rs}\inv d_{rj})'$. We know from Cor.~\ref{normdim}(ii)
that 
$\{ \,n_1', \ldots, n_k'\,\}$ is a homogeneous $\gr(D)$-base for $\gr_\be(M)$. 
Whichever values the $p_j'$ take
 it is clear that the set $\{\,n_s'\,\} \cup \{\,p_j'\,|\ j\ne s\,\}$ spans
$\gr_\be(M)$, so it is a homogeneous base.  Therefore, by Cor.~\ref{normdim}(ii) 
$\{\,n_s\,\} \cup \{\,p_j\,|\ j\ne s\,\}$ is a splitting base of $M$ for $\be$. 
The $D$-linearly independent set $\{ \,p_j\,|\ j\ne s\,\}$ must span $P$, 
since $\dim_D(P) = k-1$. Therefore, $P$ is a splitting complement to $n_sD$ for~
$\be$.  By induction, $P$ has a simultaneous splitting base for $\al$ and $\be$. 
This set combined with $n_s$ gives a simultaneous splitting base for $\al$ and
$\be $ on $M$. 
\end{proof}

The existence of common splitting bases for norms in the case of complete discrete
valuations with finite residue fields was proved by Goldman and Iwahori in
\cite[Prop.~1.3]{GI} by an argument attributed to Weil.  It was noted more generally
for arbitrary rank $1$ valuations by Bruhat and Tits in \cite[Prop.~1.26;
App.]{BT1}.


If $\al$ and $\be$ are two norms on the same $D$-vector space $M$, we
define
\begin{equation}\label{normle}
\al \le \be \text { \ \ if } \al (m)\le \be (m) \text { for all } m\in M.
\end{equation}


The existence of common splitting bases allows one to define convex combinations
of norms.  In the next section we will use the average:

\begin{definition}\label{average}
Let $\al$ and $\be$ be two norms on a $D$-vector space $M$, and choose some
subset  
$\{\,m_1, \ldots, m_k\,\}$ of $M$ which is a splitting base    for both $\al$
and
$\be$. Define the {\it average} of $\al$ and $\be$, ${\av\al\be \colon M\to 
\Gamma\cup\{\infty\}}$ by 
$$\av\al\be\big(\textstyle\sum\limits_{i=1}^km_id_i\big) \ =\ \minlim{1\le i\le
k}\big(\half
\al(m_i) + \half\be(m_i) + v(d_i) \big)\,.
$$ 
Thus, $\av\al\be$ is the norm on $M$ with splitting base $\{\,m_1, \ldots,
m_k\,\}$ such that ${\av\al\be( m_i) = 
\half \al(m_i) + \half\be(m_i)}$ for all $i$.   
\end{definition}

\begin{proposition}\label{avunique}
Let $\al$ and $\be$ be norms on  a $D$-vector space $M$.
\begin{enumerate}
  \item[{\rm(i)}]
For all $m\in M$, \ $\av\al\be(m) \ge \half \al(m) + \half \be(m)$.
  \item[{\rm(ii)}]
The definition of $\av\al\be$ is independent of the choice of common splitting
base of $M$ for $\al$~and~$\be$.  Any common splitting base for $\al$ and $\be $ is
also a splitting base for $\av\al\be$.
\end{enumerate}
\end{proposition}

\begin{proof} Assume $\av\al\be$ has been defined using the common 
splitting base $\{\,m_1, \ldots, m_k\,\}$.

(i) \ For any $m = \sum\limits_{i=1}^k m_id_i\in M$, we have
\begin{align*}
\av\al\be(m) \ &= \ \minlim{1\le i\le k}\big(\textstyle\half \al(m_i) + \half
\be(m_i)  + v(d_i) \big)\\ 
&\ge \ \textstyle\half\, \minlim{1\le i \le k}\big(\al(m_i)
+ v(d_i)\big) +\half\, \minlim{1\le j \le k}\big(\be(m_j) + v(d_j)\big) \ 
= \ \half \al(m) 
+ \half \be(m)\,.
\end{align*}

(ii)  \ Let $\{\,n_1,\ldots, n_k\,\}$ be any common splitting base for $\al$ and
$\be$, and define $\mu\colon M\to \Ga \cup \{\infty\}$ by 
 $\sum\limits_{j=1}^k n_jc_j\mapsto \minlim{1\le j\le k}\big(
\half \al (n_j) + \half\be(n_j) + v(c_j)\big)$.  For each $n_j$, we have 
${\mu(n_j) = \half\al(n_j) + \half \be(n_j) \le \av\al\be(n_j)}$ by (i).
 Hence, for any 
$m =  \sum\limits_{j=1}^k n_jc_j$ in $M$, 
\begin{align*}
\mu(m) \ & =\  \minlim{1\le j\le k}\big (\mu(n_j) + v(c_j)\big) \ 
\le \ \minlim{1\le j\le k} \big(\av\al\be(n_j) + v(c_j)\big)\ 
= \ \minlim{1\le j\le k} \big(\av\al\be(n_jc_j)\big)\\
&\le \av\al\be\big(\textstyle \sum\limits_{j=1}^k n_jc_j\big)
= \av\al\be(m)\,. 
\end{align*}
Thus, $\mu\le\av\al\be$.  Symmetrically, we have $\av\al\be \le \mu$, so equality
holds. This shows that the definition of $\av\al\be$ is independent of the choice
of common splitting base for $\al$ and $\be$.  If we take any common splitting
base for $\al$ and $\be$, we could  use that splitting base for defining
$\av\al\be$ and it is then clear that that base is also a splitting base for 
$\av\al\be$.
\end{proof}



%\setcounter{equation}{0}
%\setcounter{theorem}{0}
%\setcounter{section}{3}
%
\vskip0.25truein
%\section*{\large 3\ \ Graded Hermitian forms induced by norms}
\section{Graded Hermitian forms induced by norms}
\, \indent 
Let $D$ be a division ring finite-dimensional over its center $F$, 
and suppose $D$ has a valuation ${v\colon D \to \Ga \cup \{\infty\}}$.
% Assume also that the residue division ring $\ov D$ has characteristic
% not $2$.  
Let $\tau$ be an involution on $D$ such that $v(\tau(d)) =
v(d)$ for all $d\in D$.  Let $K = Z(D)$, and let $F = 
K^\tau= \{\, c\in K \,| \ \tau(c) = c\,\}$, which is a subfield of 
$K$ with $[K\!:\!F] < 2$ and $K$ Galois over $F$. We say $\tau$ is of the 
{first kind} if $F= K$, and of the second kind otherwise. 
Because $v\circ\tau$ is a valuation on~$D$
and every valuation on $F$ has at most one extension to $D$ by 
\cite{E} or \cite{W}, the hypothesis that $v\circ \tau = v$ reduces to $v\circ
\tau|_F =v|_F$.  This holds automatically if $\tau$ is of the first kind;
when $\tau$ is of the second kind, this condition is equivalent to:
$v|_F$ has a unique extension to a valuation on $K$.
Because $\tau$ is compatible with $v$,  $\tau$ induces a well-defined graded
involution $\tau'$ on 
$\gr(D)$. Fix some $\la\in D$ with $\la\tau(\la) = 1$ (so $v(\la) = 0$).
Let $\la'$ be the image of $\la$ in $\gr(D)_0 = \ov D$.

Let $M$ be a finite dimensional right $D$-vector space, and let 
$h\colon M\times M\to D$ be a nondegenerate $\la$-Hermitian form for
$\tau$.  Let $\al\colon M\to \Ga  \cup \{\infty\}$ be a norm on $M$.

\begin{definition}
\label{comp.def}
 $($a$)$ We say $\al$ is {\it bounded by} $h$, denoted
$\al \bd h$, if for all $m,n\in M$,
\begin{equation}
\al(m) +\al(n)\  \le \ v(h(m,n))\,.
\end{equation}
$($b$)$  If $\al \bd h$, we say that $\al$ is {\it compatible with} $h$, 
denoted $\al \cmp h$, if for each $m\in M$ there is $n\in M$
with $\al(m) +\al(n) = v(h(m,n))$.
\end{definition}

The condition that $\al \bd h$ can be restated: For all $\ga,\de \in
\vg D$,
\begin {equation} \label{hcond}
h(M^{\ge \ga}, M^{\ge \de})\ \subseteq \ D^{\ge \ga +\de}\,.
\end{equation}
\eqref{hcond} shows that $h$ also maps $M^{> \ga}\times   M^{\ge \de}$ and
$M^{\ge \ga}\times   M^{> \de}$ into $D^{> \ga +\de}$.  Therefore,
$h$  induces a 
well-defined bi-additive map $\halp\colon M_\ga\times M_\de \to D_{\ga
+\de}$.  This $\halp$ is given by: For any $m,n\in M$ with $\al(m) =
\ga$ and $\al(n) = \de$,
\begin{equation}\label{defh'}
\halp(m',n') \ = 
 \begin{cases}
\ h(m,n)'& \text{ if \  } v(h(m,n)) = \al(m) + \al(n)\,,\\
\ 0& \text { if \ } v(h(m,n)) > \al(m) + \al(n)\,.
\end{cases}
\end{equation}
Now extend $\halp$ biadditively to a map also denoted $\halp\colon
\gr(M)\times \gr(M) \to \gr(D)$. Easy calculations show that $\halp$
is a  graded $\la'$-Hermitian form on $\gr(M)$ for the graded involution
$\tau'$ on $\gr(D)$.  

\begin{remark}\label{nondeg}
The condition that $\al\bd h$ is exactly what is needed to assure that
the associated graded form $\halp$ defined by (\ref{defh'}) is
well-defined.  The stronger condition that $\al \cmp h$ can be restated:
For every $\ga\in \vg{\gr(D)}$ and every nonzero $m'\in M_\ga$ (where
$m\in M$ with $\al(m) = \ga$)  there is a nonzero $n \in N$ with 
$\halp(m',n')\ne 0$.  This is equivalent to:  $\halp$ is a
nondegenerate form. It is clear that if $\al\cmp h$ and $N$ is any subspace of
$M$, then $\al|_N\bd h|_N$.  We have $\al|_N\cmp h|_N$ iff $\halp|_{\gr(N)}$ is
nondegenerate.    
\end{remark}


\begin{lemma}\label{basetest}
Let $\al$ be a norm on $M$.   
If $\{m_1,\ldots,m_k\}$ is any splitting base of $M$ for $\al$, then 
$\al\bd h$ iff for all $i,j$,
\begin{equation}\label{sbbound}
\al(m_i)+\al(m_j)\  \le \ v(h(m_i,m_j))\,.
\end{equation} 

\end{lemma}

\begin{proof}
 Condition (\ref{sbbound}) holds by definition if $\al \bd h$.
Conversely, suppose we have (\ref{sbbound}).  For any $m,n\in M$,
write $m = \sum m_id_i$ and $n = \sum m_i c_i$ with all $d_i, c_i\in
D$. Then,
\begin{align*} 
v(h(m,n))\ &= \ v\big(\textstyle \sum
\limits_{i,j}\tau(d_i)h(m_i,n_i)c_j\big) \ \ge \ \minlim{i,j}
\big( v(\tau(d_i)) + v(h(m_i,m_j) )+ v(c_j)\big)\\
&\ge\ \minlim{i,j}\big( v(d_i) + \al (m_i) +\al(m_j) + v(c_j)\big) \\
&\ge\ \minlim i\big( v(d_i) + \al(m_i)\big) \ + \ \minlim
j\big(\al(m_j) + v(c_j)\big) \ = \ \al (m) + \al (n)\,. 
\end{align*}
So, $\al \bd h$. 
\end{proof}

For any $\la$-Hermitian form on $M$ and any norm $\al$ on $M$ there is the $h$-dual
norm $\al\sh$ defined by 
 \begin{equation} 
 \al\sh(n)\  = \  \minn\{ v(h(n,m)) - \al(m)\,|\ m\in M\}\,.  
\end{equation}
To see that $\al\sh$ is well-defined, note that $h$ induces an isomorphism
$\varphi\colon M \to M^* = \Hom_D(M,D)$ given by $\varphi(m)(n)
= h(m,n)$. This $\varphi$ is actually a right $D$-vector space isomophism 
when we turn the left $D$-vector space $M^*$ into a right $D$-vector space via $\tau$,
i.e., for $d\in D$, $f^*\in M^*$ define $ f^*\cdot d$ by 
$ (f^*\cdot d)(m) = \tau(d)f^*(m)$. 
Observe that when we view $v$ as a $D$-norm on $D$ with respect to $v$,
then $\al\sh(n) = j_{\al,v}(\varphi(n))$ for the function $j_{\al,v}$ defined 
in \eqref{jdef} (with $N = D$ and $\be = v$). The well-definition of $j_{\al,v}$,
proved in the calculation preceding \eqref{jformula}, yields that $\al\sh$ 
is well-defined.  Formula \eqref{jformula} shows that for any splitting base
$\{\,m_1, \ldots, m_k\,\}$ of $M$ and any $n\in M$, we have
\begin{equation}\label{sharpsp}
\al\sh(n)\ = \ \minlim{1\le i\le k}\big(v(h(n,m_i)) - \al(m_i)\big)\,.
\end{equation}
Lemma~\ref{sharp}(i) below shows that $\al\sh$ is  a norm on $M$.




\begin{lemma}\label{sharp}
Let $\al,\be$ be norms on $M$.  Then, 
\begin{enumerate}
  \item[{\rm(i)}]
If $\{\,m_1,\ldots,m_k\,\}$ is any splitting base for $\al$, let $\{\,m_1^\sharp,
\ldots,m_k^\sharp\,\}$ be the $h$-dual base of $M$, defined by 
$h(m_i^\sharp, m_j) = \de_{ij}$ (Kronecker delta).  Then, $\{\,m_1^\sharp,
\ldots,m_k^\sharp\,\}$ is a splitting base of $M$ for $\al^\sharp$, and
$\al^\sharp(m_i^\sharp) = -\al(m_i)$, for all $i$. Hence, $\al\sh$ is a 
norm on $M$.
  \item[{\rm(ii)}]
$\al^{\sharp\sharp} = \al$\,.
  \item[{\rm(iii)}]
If $\al \le \be$ \,then\, $\be\sh \le \al \sh$\,.
  \item[{\rm(iv)}] 
$(\av\al\be)^\sharp \ =\  \av{\al^\sharp}{\be^\sharp}\,.$
\end{enumerate}
\end{lemma}

\begin{proof} 
(i) \ This follows by an easy direct calculation, using \eqref{sharpsp}. 

(ii) follows from (i) since the $h$-dual of the $h$-dual base $\{\,m_1\sh, \ldots ,
m_k\sh\,\}$ of (i) is $\{\,m_1^{\sharp\sharp}, \ldots, m_k^{\sharp\sharp}\,\}$,
where each $m_i^{\sharp\sharp} = \la m_i$.

(iii) is clear from the definition.

(iv) \ Let $\{\,m_1,\ldots,m_k\,\}$ be a common splitting base for $\al$ and $\be$,
which exists by Th.~\ref{common}.  By Prop.~\ref{avunique}(ii)  this set is also a
splitting base for $\av\al\be$.  Then by (i) $\{\,m_1^\sharp,\ldots,m_k^\sharp\,\}$
is a splitting base for $\al\sh$ and $\be \sh$ (so also for
$\av{\al\sh}{\be\sh})$ and for $(\av\al\be)\sh$.  Part (i) shows that
$\av{\al\sh}{\be\sh}$ and $(\av\al\be)\sh$ agree on the $m_i\sh$, so they must
coincide.
\end{proof}

\begin{proposition}\label{sharpcomp}
Let $\al$ be a norm on $M$.
\begin{enumerate}
  \item[{\rm(i)}]
$\al\bd h$ iff $\al \le \al\sh$\,.
  \item[{\rm(ii)}]
$\al \cmp h$ iff $\al = \al\sh$\,.
  \item[{\rm(iii)}]
If $\al\cmp h$, then $\al$ is maximal in $\{\,\be\,| \ 
\be \text{ is a norm of $M$ and } \be \bd h\,\}$. 
\item[{\rm(iv)}] 
$\av\al{\al^\sharp} \cmp h$.
\item[{\rm(v)}]
If $\al\bd h$, then $\al \le \av\al{\al^\sharp}$. 
\end{enumerate}
\end{proposition}

\begin{proof}
(i) and (ii) are immediate from the definitions. 


(iii) \ Suppose $\al\cmp h$ and $\be$ is a norm with $\be\bd h$
and $\al \le \be$.  Then, $\al \le \be\le \be\sh\le \al \sh =\al$, by (i), 
Lemma~\ref{sharp}(iii), and (ii) of this proposition.  Hence, $\be = \al$.

(iv) \ By Lemma~\ref{sharp}(iv) and (ii), $(\av\al{\al\sh})\sh=
\av{\al\sh}{\al^{\sharp \sharp}} = \av{\al\sh}\al = \av\al{\al\sh}$.
So, (iv) follows from (ii) of this proposition.

(v)  Let $\{\, m_1, \ldots, m_k\,\}$ be a common splitting base for $\al$ and 
$\al\sh$, so it is also a splitting base for $\av\al{\al\sh}$ by 
Prop.~\ref{avunique}(ii).  Because $\al \bd h$ we have from (i) above that
$\al(m_i) \le \al \sh(m_i)$.  Hence, ${\al(m_i) \le \half \al(m_i) + \half
\al\sh(m_i) = \av\al{\al\sh}(m_i)}$.  Since this is true for all the $m_i$ in a
common splitting base for $\al$ and $\av\al{\al\sh}$, we must have $\al \le
\av\al{\al\sh}$.
\end{proof}

Prop.~\ref{sharpcomp} shows that the norms compatible with $h$ are precisely the 
ones that are maximal among the norms bounded by $h$.  Moreover, parts (iv) and
(v) show that every norm bounded by $h$ is less than or equal to a norm compatible
with $h$.  This was shown previously for discrete valuations  by Bruhat and Tits in
\cite[pp.~160--162]{BT2}, where the norms we have defined as compatible with $h$ 
are called ``maximinorantes" for $h$, i.e., maximal among norms bounded
by $h$.  Earlier still, norms maximal among those bounded by $h$ were considered
by Springer in \cite{Sp2} and by Goldman and Iwahori in \cite{GI} for certain
complete discrete valuations;   but their definition was somewhat different, since
for $\al\bd h$ they require only the weaker condition that $2\al(m) \le v(h(m,m))$
for all~$m \in M$.  

\begin{corollary}\label{existence}
If $h$ is any nondegenerate $\la$-Hermitian form for 
$\tau$ on a $D$-vector space $M$, then there is a norm $\al$
on $M$ with $\al\cmp h$.  
\end{corollary}

\begin{proof} 
This is immediate from Prop.~\ref{sharpcomp}(iv).
\end{proof}

\begin{examples}\label{excomp}
(i) \ Suppose $\{m_1, \ldots, m_k\,\}$ is an orthogonal base for 
$h$ on $M$. For any $\ga_1,\ldots,\ga_k\in \Ga$, let~$\al$ be the norm
on $M$ with splitting base $\{m_1, \ldots, m_k\,\}$ such that each
$\al(m_i) = \ga_i$.  Then, by Lemma~\ref{basetest},
$\al \bd h$ iff each $\al(m_i)+\al(m_i)\  \le \ v(h(m_i,m_i))$, 
iff each $\ga_i\le \frac12 v(h(m_i,m_i))$.  When this holds,
we have $\halp(m_i',m_j') = 0$ whenever $i\ne j$ and
\begin{equation*}
\halp(m_i',m_i') = \begin{cases} \ h(m_i,m_i)' & \text{ if \  } 
v(h(m_i,m_i)) = 2\ga_i\,,\\ 
\ 0& \text { if  \ } v(h(m_i,m_i)) > 2\ga_i\,.
\end{cases}
\end{equation*}
Since the diagonal form $\halp$ 
is nondegenerate iff  each $\halp(m_i',m_i') \ne 0$, we have
$\al\cmp h$ iff
each ${\ga_i=\frac12 v(h(m_i,m_i))}$.


(ii) \ Suppose $h$ is a hyperbolic $\la$-Hermitian form on $M$.  Then,
$M$ has complementary totally isotropic subspaces $N$ and $P$ of the same
dimension.  Let $\{n_1, \ldots, n_\ell\}$ be any $D$-vector space 
base of $N$, and let $\{p_1, \ldots, p_\ell\}$ be the corresponding base 
of $P$ such that $h(n_i, p_j) = \de_{ij}$ (Kronecker delta) for all
$i,j$. Take any $\ga\in \Ga$.  Let $\al$ be the norm on $M$ with
splitting base $\{\, n_1, \ldots, n_\ell, p_1, \ldots, p_\ell\,\}$ such
that $\al(n_i) = \ga$ and $\al(p_i) = - \ga$ for $1 \le i \le
\ell$.  Then, $\al \cmp h$, and $\gr(M) = \gr(N) \oplus \gr(P)$, 
with $\vg{\gr(N)} = [\ga]$ and $\vg{\gr(P)} = [-\ga]$. Also, $\gr(N)$ 
and $\gr(P)$ are complementary totally $\halp$-isotropic subspaces of 
$\gr(M)$, so $\halp$ is hyperbolic.  To see that $\al \bd h$, one can
check (\ref{sbbound}) for the given splitting base of $\al$.  Since 
$\halp(n'_i, p'_j) = \de_{ij}$ and $\halp(n'_i, n'_j) = 0  =\halp(p'_i,
p'_j) $ for all $i,j$ it is clear that $\halp$ is nondegenerate, which
verifies $\al \cmp h$.

 (iii) \ Suppose $M$ and $N$ are finite-dimensional right 
$D$-vector spaces with respective nondegenerate 
$\lambda$-Hermitian forms  (for the involution $\tau$
on $D$) $h$
and $k$, and respective norms $\al$~and~$\be$.  Then on $M\oplus N$
we have the  nondegenerate
$\la$-Hermitian form $h\perp k$ for $\tau$
given by 
%\break
$$
{(h\perp k) \big((m_1,n_1),(m_2,n_2)\big)  \ = \  h(m_1,m_2) 
+k(n_1,n_2)\, .}
$$  
There is also the  value function $\al\oplus \be$
on $M\oplus N$ 
given by $(\al \oplus \be) (m,n) = \minn(\al(m), \be(n))$.  Then, 
 $\al\oplus\be$ is a norm on $M\oplus N$, with 
$\gr(M\oplus N) \cong \gr(M) \oplus \gr(N)$.  Furthermore, if
$\al \bd h$ and $\be \bd k$, then $\al\oplus\be \bd h\perp k$.
When this occurs, we have $(h\perp k)'_{\al\oplus\be}
\cong \halp\perp k'_\be$.
Since an orthogonal sum of nondegenerate graded Hermitian forms is
nondegenerate,  it follows that if $\al \cmp h$ and $\be \cmp k$, then
$\al\oplus\be \cmp h\perp k$.   All this is easy to verify.

\end{examples}



These examples give another way of seeing the existence of norms compatible with 
any given $\la$-Hermitian form $h$.
For, by \cite[p.~259, Th.~6.3; p.~264, Th.~8.1]{S}, $h$ is  diagonalizable or 
 hyperbolic. 
%   furthermore, since $h_1$ is anisotropic, it is diagonalizable.  
The first case is covered by Ex.~\ref{excomp}(i) and the second by
Ex.~\ref{excomp}(ii).


\begin{proposition}\label{grperp}
Let $h$ be a nondegenerate $\la$-Hermitian form for $\tau$ on $M$
and let $\al$ be a norm on $M$ with $\al \cmp h$. Let $N$ be any
subspace of $M$.  
Let $\gr(N)^\perp$ be the orthogonal of $\gr(N)$ in $\gr(M)$ with
respect to $\halp$. Then,
\begin{enumerate}
\item[{\rm(i)}]
$\gr(N)^\perp = \gr(N^\perp)$ in $\gr(M)$.
\item[{\rm(ii)}]  Suppose $h|_{N}$ is nondegenerate. Then,
$\al|_{N} \cmp h|_{N}$ iff $N^\perp$ is a splitting complement
of $N$ with respect to $\al$, iff $\al|_{N^\perp}\cmp h|_{N^\perp}$\,.
\end{enumerate}
\end{proposition}

\begin{proof}
(i)  \ It is clear from (\ref{defh'}) that $\gr(N^\perp) \perp \gr(N)$
with  respect to $\halp$, i.e., $\gr(N^\perp) \subseteq \gr(N)^\perp$.
Since $\al\cmp h$, we have $\halp$ is a nondegenerate form on $\gr(M)$. 
Hence, as $\al|_{N}$ and $\al|_{N^\perp}$ are norms by
Prop.~\ref{subspace}, we have 
 \begin{align*}
 \dim_{\gr(D)}(\gr(N^\perp)) \ &= \ \dim_{\gr(D)}(\gr(M)) 
-\dim_{\gr(D)}(\gr(N)) \ =\  \dim_D(M) - \dim_D(N)\\
&= \ \dim_D(N^\perp) \ = \ \dim_{\gr(D)}(\gr(N^\perp))\,.
 \end{align*}
Hence, $\gr(N^\perp) = \gr(N)^\perp$.

(ii)  \ The nondegeneracy of $h|_N$ implies that $N\cap N^\perp = (0)$,
so $h|_{N^\perp}$ is also nondegenerate.
Since $\halp$~is nondegenerate, we have 
$\al|_{N} \cmp h|_{N}$ iff $\gr(N) \cap\gr(N)^\perp = (0)$, iff
(by (i)) $\gr(N) \cap \gr(N^\perp) =(0)$, iff (by Remark~\ref{comprem})
$N^\perp$ is a splitting complement of $N$. This condition is symmetric in
$N$ and $N^\perp$.  So it holds iff $\al|_{N^\perp}\cmp h|_{N^\perp}$.
\end{proof}

\begin{proposition}\label{hyp}
Suppose the $\la$-Hermitian form $h$ on $M$ is hyperbolic.  For any 
norm $\al$ on $M$ with $\al \cmp h$, the associated graded form 
$\halp$ is metabolic.  If $\halp$ is even, then it is hyperbolic.
\end{proposition}

\begin{proof}  Since $h$ is hyperbolic, $M$ has a totally isotropic 
subspace $N$ with $\dim_D(N) = \frac12\dim_D(M)$.  Then, $\gr(N)$~
is a graded subspace of $\gr(M)$ with $\dim_{\gr(D)}(\gr(N)) =
\frac12\dim_{\gr(D)}(\gr(M))$.  Furthermore, $\gr(N)$~is totally isotropic
for $\halp$, as $\halp(m',n')=0$ for all homogeneous elements 
$m',n'$ of $\gr(N)$.  Hence, $\halp$ is metabolic. If $\halp$ is even, it is also
 hyperbolic, by Prop.~\ref{cancel}(iii).
 \end{proof}

Prop.~\ref{hyp} indicates the importance of knowing that associated graded forms
are even.  

\begin{definition}\label{pef}
For a  division algebra $D$ with valuation $v$ and involution  $\tau$ 
compatible with $v$ and any $\la \in
D$ with
$\la\tau(\la) = 1$, we say that $(v,\tau,\la)$ {\it preserves even forms} if for
any $(S,h)\in \HermDp\tau\la$, and any norm $\al$ on $S$ with $\al \cmp h$, the
associated graded form $\halp$ is even. 
\end{definition}

It is clear that whenever $\charr(\ov D)\ne 2$, $(v,\tau,\la)$ preserves even forms
since  all forms are even  in characteristic different from $2$.  We will show in 
Prop.~\ref{preseven} below other significant cases where $(v,\tau,\la)$ preserves
even forms.

 
\begin{theorem}\label{springer} 
{\ }
\begin{enumerate}
  \item[{\rm(i)}]
Let $h$ be a nondegenerate $\la$-Hermitian form for $\tau$ on $M$.
If $\al$ and $\be$ are any two norms on~$M$ with $\al \cmp h$ and
$\be\cmp h$, and if $\halp$ and $\hbep$ are even,
then the anisotropic parts of $\halp$ and $\hbep$
are isometric.
  \item[{\rm(ii)}] Suppose $(v,\tau,\la)$ preserves even forms.  Then, the map $h
\mapsto h'_{\al\,
\an}$ (for any  norm $\al$ with $\al \cmp h$) gives a well-defined group 
epimorphism $\Theta\colon W^+(D,\tau,\la) \to W^+_g(\gr(D), \tau', \la')$.
Furthermore, there is a canonical \lq\lq first residue map" $W^+(D,\tau,\la)
\to W^+(\ov D, \ov \tau, \la')$, where $\ov \tau$ is the involution on $\ov D$
induced by $\tau$.
 %  \item[{\rm(iii)}] If $v$ is Henselian and $(D,\tau,\la)$ preserves even forms,
% then $W^+(D,\tau,\la)\cong W^+_g(\gr(D),\tau', \la')$.
 \end{enumerate}
\end{theorem} 

\begin{proof}
(i)  \ On the $D$-vector space $M\oplus M$ there is the hyperbolic
$\la$-Hermitian form $h\perp -h$ and the norm $\al\oplus \be$.
Since $\al\cmp h$ and $\be \cmp -h$, Ex.~\ref{excomp}(iii) shows that
$\al\oplus \be \cmp h\perp -h$ and $(h\perp -h)'_{\al\oplus \be}
\cong \halp \perp (-h)'_{\be}  \cong \halp \perp  -h'_{\be}$.  
Therefore, by Prop.~\ref{hyp} $\halp \perp  -h'_{\be}$ is hyperbolic.
Hence, $\halp$ and  $h'_{\be}$ have isometric anisotropic parts,
by Prop.~\ref{cancel}(viii).

(ii) \ Take any $(M,h), (N, \ell)\in \mathcal H^+\dtl)$ with $\ell$ hyperbolic, 
and norms $\al$ on $M$, $\be $ on $N$, and $\de$ on $M\oplus N$ with
$\al \cmp h$, $\be \cmp \ell$, and $\de\cmp h\perp \ell$. We have, by (i),
$$
(h\perp \ell)_{\de\, \an}' \ \cong\ (h\perp\ell)_{\al\oplus \be\,\an}' 
\ \cong \ (\halp\perp\ell_\be')_\an\ \cong h_{\al\,\an}'\,,
$$
as $\ell_\be'$ is hyperbolic by Prop.~\ref{hyp}. Therefore, the map $\Theta$ is
well-defined.  It is clearly a group homomorphism.  It is surjective since any
anisotropic form in $\mathcal G\mathcal H^+(\gr(D), \tau', \la')$ is an
orthogonal sum of $1$-dimensional forms (see Prop.~\ref{cancel}(iv)) whose Witt
group classes clearly lie in $\im (\Theta)$ (see Ex.~\ref{excomp}(i)). 

We have canonical maps, 
\begin{align*}
W_g^+(\gr(D), \tau',\la') \ &\overset{\cong}\longrightarrow \ 
\textstyle\bigoplus\limits_{\brga \in \frac12\vg{\gr(D)}/\vg{\gr(D)}}
W_g^+(\gr(D), \tau',\la';\brga) \ 
\longrightarrow\ W_g^+(\gr(D), \tau',\la';[0]) \\
&\overset{\cong}\longrightarrow W^+(\gr(D)_0, \tau'|_{\gr(D)_0}, \la')\ 
\overset{=}\longrightarrow \ W^+(\ov D, \ov \tau, \la')\,,
\end{align*} 
where the first map is the isomorphism of Prop.~\ref{witt}(ii), 
the second is projection onto the $[0]$-component, the third is the isomorphism 
of Prop.~\ref{witt}(i), and the fourth expresses the equality $\gr(D)_0 =
\ov D$, in which $\tau'|_{\gr(D)_0} =\ov \tau$.   
The first
residue map is the composition of these maps with~$\Theta$.
\end{proof}

\begin{remark}\label{residue}
For a diagonal $\la$-Hermitian form $h = \langle d_1, \ldots , d_k\rangle$ for
$\tau$, with each $d_i\in \Alt\dtl$, the image of $h$ under the first residue map
of Th.~\ref{springer}(ii) is computable as follows:  We can reorder the~$d_i$ so
that
$v(d_1), \ldots, v(d_j)\in 2\vg D$ and $v(d_{j+1}), \ldots, v(d_k) \notin 2\vg D$.
For each $i\le j$ choose any $s_i\in D$ with $v(s_i) = -\frac12v(d_i) \in \vg D$.
Let $c_i = \tau(s_i)d_is_i$; so $v(c_i) = 0$.  Then the first residue of $h$ is the
class of $\langle \ov{c_1}, \ldots, \ov {c_j}\rangle$ in $W^+(\ov D, \ov \tau, \ov
\la)$.  The theorem shows that the Witt class of this form is well-defined,
independent of the choice of diagonalization of $h$ and independent of the choices
of the $s_i$.  There are also second residue maps obtainable by projection onto
the other components in the direct sum of Prop.~\ref{witt}(ii).  But these are not
canonical because of the choices of the $r_\ga$.  Notice also that these second
residues live in $W^+(\ov D, \ov{\tau_\brga},\ov{\la_\brga})$ where the
involutions  on $\ov D$ and the $\ov\la$'s in~$\ov D$ can vary for the different 
$\brga \in \frac 12\vg D/\vg D$.

\end{remark}

 


When the involution $\tau$ on $D$ is of the first kind, it is of either 
symplectic type or orthogonal type (see the definitions in
\cite[Def.~(2.5)]{KMRT}),  and $\la = \pm 1$.  We say that $(\tau,\la)$ is a {\it
symplectic pair} if $\tau$  is of symplectic type and $\la = 1$ or $\tau$ is of
orthogonal type and $\la \ne 1$.  (So, when $\charr(D) = 2$, $(\tau,\la)$ is  a
symplectic pair iff $\tau$ is of symplectic type.)  This terminology is used because
$(\tau,\la)$ is a symplectic pair iff
the isometry groups of all $\la$-Hermitian forms for $\tau$ are symplectic groups.
% iff $(\tau,\la)$ is a symplectic pair.  


For any kind or type of involution, in investigating preservation of even forms,
we need to work with the set of $\la$-symmetrized elements of $D$: Set 
\begin{equation}\label{altdef}
\Alt(D, \tau,\la) \ = \ \{\,d+\la\tau(d)\,|\ d\in D\,\}\,.
\end{equation}
Then, $\Alt(D, \tau,\la)$ is  a vector space over   
 $F = K^\tau$, where $K = Z(D)$.
It is known (see \cite[Prop.~(2.6), Prop.~(2.17), Prop.~(2.7)]{KMRT}) that if
$\dim_K(D) = n^2$, 
 
\begin{equation}\label{altdim} 
%\!
\dim_F(\Alt(D, \tau,\la))  \ = \  
\begin{cases}
\ \ \ \ \ \ n^2 &\text{if $\tau$ is of the second kind};  \\
n(n-1)/2 &\text{if $\tau$ is of the first kind with $(\tau,\la)$   a
symplectic pair}; \\ 
  n(n-1)/2&\text {if $\tau$ is of the first kind and
$\charr(D) = 2$};\\ n(n+1)/2 &\text{otherwise}.
\end{cases}
\end{equation} 

The analogous result holds in the graded situation: As in \S  1, let $E$ be a graded
division ring finite dimensional over its center $Z(E)$, let $\si$ be a graded
involution on $E$, and let $\ep\in Z(E)_0$ with $\ep \si(\ep) = 1$.  Let $R =
Z(E)$, and let $S = Z(E)^\si$, which is a graded subfield of $R$ with 
$[R\!:\!S] =1$ or $2$, depending on whether $\si$ is of the first or the second
kind.  The graded division ring $E$ has no zero divisors, as $\vg E$ is totally
ordered.  (If we had $ab = 0$ for nonzero
elements $a,b$ of $E$, then the product of their least degree homogeneous
components would be $0$; but $E$ has no homogeneous zero divisors.) Thus, the
integral domain $S$ has a quotient field, call it $Q$.  We have  $E\otimes _SQ$ has
no zero divisors and is finite dimensional over its center $R\otimes _S Q$, so it
is a division ring, with $\dim_{R\otimes _SQ}(E\otimes _SQ) = \dim_R(E)$.  The
involution $\si$ on~$E$ extends to an involution $\wi\si=\si\otimes \id$ on
$D\otimes _SQ$, and clearly $\wi\si$ is of the same kind (first or second)
as $\si$.  When $\si$ is of the first kind, we define the type of $\si$ (orthogonal
or symplectic)  to be that of $\wi\si$.  We say that $(\si, \ep)$ is a
symplectic pair if $(\wi\si,\ep)$ is a symplectic pair for $E\otimes _SQ$. 
Analogously to \eqref{altdef}, define 
\begin{equation}\label{graltdef}
\Alt(E, \si, \ep)\ = \ \{\, c+\ep\si(c) \ | \ c\in E\,\} \,.
\end{equation}
Clearly $\Alt(E, \si, \ep)$ is a graded 
$S$-vector subspace of $E$ and ${\Alt(E, \si, \ep) \otimes_S Q \cong
\Alt(E\otimes _S Q, \wi\si, \ep)}$. By applying \eqref{altdim}
to $\Alt(E\otimes _S Q, \wi\si, \ep)$, it follows that if $n^2
=\dim_R(E) = \dim_{R\otimes _SQ}(E\otimes _S Q)$ then $\dim_S(\Alt(E, \si, \ep))$
satisfies the formulas analogous to those in \eqref{altdim} in all four cases.  
Notice also that if $\charr(E)\ne 2$ and $\si$ is of the first kind, then
$\dim_S(\Alt(E, \si, \ep))$ distinguishes the type of $\si$ directly within
$E$ without reference to $Q$.  When $\charr(E) = 2$ and $\si$ is of the first kind,
then one has (see \cite[Prop.~(2.6)(2)]{KMRT}) that $\wi\si$ is symplectic
iff $\Trd(c) = 0$ for all
$c\in E\otimes _SQ$ such that $\wi\si(c) = c$, where $\Trd$ is the reduced trace.
Furthermore, this holds iff $1\in \Alt(E\otimes _S Q, \wi\si, \ep)$.  The
corresponding criteria apply within $E$  for $\si$ to be symplectic, since 
for the reduced trace $\Trd$ on $E\otimes _S
Q$, we have  $\Trd(E_\ga) \subseteq R_\ga$ for each $\ga\in \vg E$.  (This follows
because the minimal polynomial over $R\otimes _S Q$ of a homogeneous element of~$R$
has homogeneous coefficients in $R$, as shown by the proof of
\cite[Prop.~2.2]{HW1}.)
 


The following proposition will be needed in determining preservation of even forms 
when ${\charr(\ov D) = 2}$.  Since $\Alt\dtl$ is an $F$-vector subspace of 
$D$, the valuation $v$ restricts to a value function on $\Alt\dtl$ with respect to
the valuation $v|_F$ on $F$.  So, it has an associated graded $\gr(F)$-vector space
$\gr(\Alt\dtl))$, which is a graded subspace of $\gr(D)$. We say that $D$ is
{\it defectless} over $F$ if $\dim_{\gr(F)}(\gr(D)) =  \dim_F(D)$.  Equivalently,
$D$ is defectless over $F$ iff $v$ is a norm on $D$ with respect to $v|_F$ when 
$D$ is viewed as a vector space over $F$.  When this occurs, $v$ restricts to a norm
on $\Alt\dtl$, by Prop.~\ref{subspace}.

%\bigskip 


\begin{proposition}\label{grequal}
{\ }
\begin{enumerate} 
  \item[{\rm(i)}]
$\Alt(\gr(D), \tau',\la') \ \subseteq \ \gr(\Alt\dtl) \,.$
\item[{\rm(ii)}]  
Suppose $D$ is defectless over $F$.  Then, the following 
conditions are equivalent:
\begin{enumerate}
   \item[{\rm(a)}]
$(v,\tau,\la)$ preserves even forms.
  \item[{\rm(b)}]
$\Alt(\gr(D), \tau',\la') \ = \ \gr(\Alt\dtl) \,.$
  \item[{\rm(c)}]
$\dim_{\gr(D)}\big(\Alt(\gr(D), \tau',\la')\big)  \ = \  \dim_D(\Alt(\dtl)\big)$\,.
  \item[{\rm(d)}]
For every $d\in \Alt\dtl$ there is an
$a\in D$ with $a+\la\tau(a) = d$ and $v(a) = v(d)$.
\end{enumerate}
\end{enumerate}
\end{proposition}

\begin{proof}
(i) \ For $d\in D^\times$ with 
image $d'$ in $\gr(D)$, we have $d'+ \la'\tau'(d') = \big(d+\la\tau(d)\big)'$
if ${v(d+\la\tau(d)) = v(d)}$, and  $d'+ \la'\tau'(d') = 0$ otherwise.
This proves the desired inclusion for homogeneous elements; the inclusion then
holds throughout these graded vector spaces.

(ii) (The defectless assumption  is not needed for (a) $\Leftrightarrow$ (b).) 
 (b) $\Rightarrow$ (a)
Suppose condition (b) holds.  Take any $(M,h)\in \HermDp\tau\la$ and any norm
$\al$ on~$M$ with $\al \cmp h$, and form $\gr(M)$ with respect to~$\al$.  For any
nonzero homogeneous element $\wi m$ of $\gr(M)$ there is a nonzero $m\in M$ with
$m' = \wi m$.  We have $\halp(\wi m, \wi m) = h(m,m)'$ or $= 0$, by \eqref{defh'}.
In either case, $\halp(\wi m, \wi m) \in \gr(\Alt\dtl)$. 
% When equality holds in
% \eqref{inclusion}we have 
Condition~(b) yields $\halp(\wi m,\wi m) \in \Alt(\gr(D),\tau',\la')$.
Because $\gr(M)$
is generated as an abelian group by its homogeneous elements, it follows that
$\halp(s,s) \in \Alt(\gr(D),\tau',\la')$ for all $s\in \gr(M)$; so $\halp$ is an
even form,
proving (a).  

(a) $\Rightarrow$ (b) 
Suppose (b) does not hold.  Then, there is a homogeneous $\wi
a\in \gr(\Alt\dtl)$ with $\wi a\notin\Alt(\gr(D),\tau',\la')$. We have $\wi a =
a'$ for some 
$a\in \Alt\dtl$.  On the $1$-dimensional $D$-vector space $D$, define an even
$\la$-Hermitian form $h$ for $\tau$ by $h(d,e) = \tau(d) a e$.  
% Because $D$ is tame over $F$, we have $\dim_F(D) =\dim_{\gr(F)}(\gr(D))$ (see
% \eqref{tame}; thus, $v$ is a $v|_F$-norm for $D$ when $
Any norm $\al$ on
$D$ is defined by choosing $\ga \in \Ga$ and setting $\al(d) = v(d) +\ga$ for
all $d\in D$. Then, $\{\,1\,\}$ is a splitting base of $\al$ and an orthogonal
base for $h$, so Ex.~\ref{excomp}(i) shows that $\al\cmp h$ iff $\ga = \frac12
v(h(1,1)) = \frac12v(a)$.  When this holds, we have $\halp(1',1') = h(1,1)' = 
\wi a \notin \Alt(\gr(D),\tau',\la')$.  So, $\halp$ is not even for the unique
$v$-norm on $D$ which is compatible with $h$; thus, (a) does not hold.  

(b) $\Leftrightarrow$ (c) As noted above, since $D$ is defectless over $F$,
$v$ is a $v|_F$-norm for the $F$-vector space~$D$ and for its subspace $\Alt\dtl$.
Hence, $\dim_{\gr(F)}\big(\gr(\Alt\dtl)\big) = \dim_F(\Alt\dtl)$.  Therefore, 
(c) is equivalent to:  $\dim_{\gr(F)}\big(\gr(\Alt\dtl) \big) =  
\dim_{\gr(F)}\big(\Alt(\gr(D), \tau',\la') \big)$. In view of~(i), this is 
clearly equivalent to (b). 

(b) $\Leftrightarrow$ (d)
We again use the fact that $v$ is a $v|_F$-norm for $D$.
% Since $D$ is defectless over $F$, we have that $v$ is a norm for $D$ viewed as a vector
% space over~$F$ with respect to the valuation $v|_F$ on $F$, see \eqref{tame}.  
Let $f\colon D \to D$ be the $F$-linear map given by 
$c\mapsto c+\la\tau(c)$.  In the notation of \eqref{jdef}, let 
$$
j \ = \ j_{v,v}(f) = \minn\{\, v(c+\la\tau(c))  - v(c)\,|\ c\in D^\times\,\} \,.
$$
Clearly, $j \ge 0$.  Suppose first that $j>0$.  
% Then (taking $c= 1$), $v(1+\la) \ge j
% >0$, i.e. $\la' = -1$.    
So, for any $d\in D^\times$, we have $v(d+\la\tau(d)) \ge v(d) + j > v(d)$.
Therefore, condition (c) holds only for $d = 0$. Also, taking $d = 1$, we have 
$v(1 + \la) >v(1) =0$,
 i.e. $\la' = -1$. Furthermore,  for any $d\in D^\times$,
% and  $v((1+\la)\tau(c))> v(\tau(c))$, 
we have $v(d-\tau(d)) = v(d+\la\tau(d)-(1+
\la)\tau(d)) > v(d)$.  This means that $\tau'(d') = d'$ in $\gr(D)$, and hence
$\Alt(\gr(D), \tau',\la') = 0$.    Thus, condition (b) holds iff $\Alt\dtl = 0$
iff condition (c) holds.

Now, suppose $j = 0$.  Then, the graded map $f'$ on $\gr(D)$ induced by 
$f$ (see \eqref{assoc}) maps $D_\ga$ into $D_\ga$ for each $\ga \in \vg D$, and
for $c\in D^\times$,  
\begin{equation*}
f'(c') \ = \ 
\begin{cases} 
\ (c+\la\tau(c))'\!\!& = \ c' + \la'\tau'(c'), \quad \text{if $\,v(c+\la\tau(c)) =
v(c)$};\\
\ \ \ \ \ \ \ 0 & = \ c' + \la'\tau'(c'), \quad \text{if $\,v(c+\la\tau(c)) >
v(c)$}.
\end{cases}
\end{equation*}
So, $\im(f') = \Alt(\gr(D), \tau',\la')$, while clearly $\gr(\im(f)) =
\gr(\Alt\dtl)$.  Thus, condition (b) holds iff $\im(f') =
\gr(\im(f))$, iff, by Prop.~\ref{grmap}(ii), 
% Hence, in the good cases for $\charr(\ov D) = 2$,
% Lemma~\ref{grequal}(ii) shows that $\im(f') =
% \gr(\im(f))$.  Therefore, Prop.~\ref{grmap} yields that 
for every $d\in \Alt\dtl = 
\im(f)$ there is an $a\in D$ with $d = f(a) = a+ \la\tau(a)$ and $v(a) = v(d) + j =
v(d)$, which is condition (d). 
\end{proof}

In some cases, preservation of even forms requires an assumption of tameness 
of the valuation. Let $K^h$ be the Henselization of $K$ with respect
to $v|_K$.  (If $v|_K$ is a discrete valuation, we could replace $K^h$~
by the completion of $K$ with respect to $v$.)  Let $D^h = D\otimes_KK^h$.  
By Morandi's theorem \cite[Th.~2]{M}, because $v|_K$ extends to a valuation on 
$D$, $D^h$ is a division ring; furthermore,  the Henselian valuation $v^h$ on $K^h$
extends uniquely to a valuation on $D^h$ with $\ov{D^h} \cong \ov D$ and $\vg{D^h}
=\vg D$.  Tameness of division algebras over Henselian fields is described in
\cite[\S 6]{JW} and in \cite[\S 3]{W2}.  We say that $D$~is {\it tame} with
respect to $v$ if $D^h$ is tame, i.e., if $D^h$ is split by the maximal tamely
ramified extension field of $K^h$.  By 
\cite[Cor.~4.4]{Bl} or  \cite[Prop.~4.3]{HW} (applied to $D^h$), we
have
\begin{equation}\label{tame} 
D \text{ is tame \  iff \ $\dim_{\gr(K)}(\gr(D)) \ = \  \dim_K(D)$ and $Z(\gr(D))
= \gr(K)$\,.}
\end{equation} 
In many  cases arising here, we have  $\dim_K(D)$  a power of $2$ and $\charr(\ov D)
=2$; the condition of tameness is then equivalent to: $D$ is split by the maximal
unramified extension of $K^h$; equivalently, $D^h$ has a maximal subfield which 
is unramified over $K^h$.   When the involution $\tau$ is of the second kind, 
we sometimes require that $D$ be tame over $F = K^\tau$.  This means that $D$
is tame and  $K$ is tame over $F$ i.e., as $[K\!:\! F] = 2$, 
either $[\ov K\!:\!\ov F] = 2$ and $\ov K$ is separable over $\ov F$, or 
$|\vg K\!:\!\vg F| = 2$, with the latter case not allowed if $\charr(\ov F) = 2$.

\begin{remark}\label{samekind}
Whenever $D$ is tame over $F$ 
the involutions $\tau$ and $\tau'$ are of the 
same kind. For, if $\tau$ is of the first kind, then $\tau'|_{Z(\gr(D))} = \id$, as
$Z(\gr(D)) = \gr(Z(D))$ by  \eqref{tame}.  On the other hand, if $\tau$ is of the 
second kind and 
$\charr(\ov D) = 2$, then the tameness implies that $K$~is unramified over $F$.  Since
$\tau$ induces the nontrivial automorphism of $K/F$, the residue involution
$\ov \tau$ induces the 
nontrivial automorphism of $\ov K/\ov F$, so $\tau'$ is not the identity 
on $\gr(Z(D))$.  If $\tau$ is of the second kind and $\charr(\ov D)\ne 2$, then 
there is $c \in K$ with $\tau(c) = - c$.  So, in $\gr(K) = Z(\gr(D))$, we have
$\tau'(c') = -c' \ne c'$, showing  that $\tau'$ is of the second kind.
\end{remark}


\begin{proposition}\label{preseven}
$(v,\tau,\la)$ preserves even forms in each of the following cases:
\begin{enumerate}
  \item[{\rm(i)}]
$\charr(\ov D) \ne 2$.
  \item[{\rm(ii)}] 
$\tau'$ is of the second kind.
  \item[{\rm(iii)}]
$\charr(D) = 2$ and $D$ is tame over $F$. 
% and $\tau$ and $\tau'$ are of the same kind.
% $\tau$ and $\tau'$ are each of the second kind.
\item[{\rm(iv)}] 
$\charr(D)= 0$, $\charr(\ov D) = 2$, $D$ is tame, 
% $\tau$ and $\tau'$ are each of the first kind,  
and $(\tau,\la)$ is a symplectic pair.
% either $($a$)$~$\tau$~is of orthogonal type and
% $\la = -1$ or $($b$)$ $\tau$ is of symplectic type and $\la = 1$.
\end{enumerate}
\end{proposition}

\begin{proof} (i) and (ii) \ 
When $\charr(\ov D) \ne 2$ or $\tau'$ is of the second kind, then all forms in 
$\mathcal H(\gr(D),\tau',\la')$ are even.

(iii) and (iv) \ In case (iii), we may assume that $\tau$ (hence also $\tau'$)
is of the first kind, since 
the second kind case is covered by (ii). 
% Since $D$ is tame over $F$, it is 
% defectless over $F$ (see \eqref{tame}).
% We show that
% We claim that $(v,\tau,\la)$ preserves even forms iff
% equality holds in \eqref{inclusion}.
% condition (c) of Prop.~\ref{grequal}(ii) holds.  Then, condition~(a) also holds,
% which is what we need here.
% \begin{equation}\label{inclusion}
%  \dim_{\gr(F)}(\Alt(\gr(D),\tau',\la')) \ = \  \dim_{\gr(F)}(\gr(\Alt\dtl))\,.
%  \end{equation}
% In view of the inclusion of Prop.~\ref{grequal}(i) it follows that 
% $\Alt(\gr(D),\tau',\la') = \gr(\Alt\dtl)$.  Then, Prop.~\ref{grequal}(ii)
% shows $(v,\tau,\la)$ preserves even forms. 
Let $n^2 = \dim_K(D)$ and $n^{\prime 2} = \dim_{Z(\gr(D))}(\gr(D))$.
Because $D$ is tame over~$F$, we have $Z(\gr(D)) = \gr(K)$, $n = n'$, and 
$\dim_F(D) = \dim_{\gr(F)}(\gr(D))$ (see \eqref{tame}).  
 The last equality shows that $D$ is defectless over $F$.
% that $v$ is a $v|_F$-norm for $D$ when $D$ is viewed as an $F$-vector space. So,
% the restriction of $v$ to $\Alt\dtl$ is again a $v|_F$-norm, by
% Prop.~\ref{subspace}; hence, $\dim_{\gr(F)}(\gr(\Alt\dtl)) = \dim_F(\Alt\dtl)$, by
% Cor.~\ref{normdim}(i). Thus, to establish \eqref{inclusion} 
% since equality holds in (i) iff the two quantities
% have the same $\gr(F)$-vector space dimensions, 
% it suffices to verify that \break
% $  \dim_{\gr(F)}(\Alt(\gr(D), \tau',\la')) = \dim_F(\Alt\dtl)$.   
Since $n = n'$, the
dimension formula~\eqref{altdim} and the analogous graded formula yield 
in each case that 
$  \dim_{\gr(F)}(\Alt(\gr(D), \tau',\la')) = \dim_F(\Alt\dtl)$.
Hence, $(v,\tau, \la)$ preserves even forms, by Prop.~\ref{grequal}
(ii)(c)$\Rightarrow$(a).
\end{proof} 

Note that the proof of Prop.~\ref{preseven} shows that when $D$ is tame over 
$F$, $(v,\tau,\la)$ does {\it not} preserve even forms except in the cases
listed in the proposition. 

%\setcounter{equation}{0}
%\setcounter{theorem}{0}
%\setcounter{section}{4}
%
\vskip0.25truein
%\section*{\large 4\ \ Henselian valuations} 
\section{Henselian valuations}
\ \indent 
Classically, Springer's Theorem \cite{Sp1} for  quadratic forms over a field with
complete discrete valuation (with residue characteristic not $2$) says that a form
is anisotropic iff its two residue forms are anisotropic, and that its class in the
Witt group is determined by the Witt classes of the residue forms.  This
corresponds to having not just a map $\Theta$ as in Th.~\ref{springer} but having 
$\Theta$ an isomorphism.  It is well known (see, e.g., \cite[p.~174]{AK}, 
\cite[Th.~12.1.5, sentence after (12.2.1)]{Kne})
that Springer's theorem is valid for
Henselian valuations (with  any value group) as well as for complete discrete
valuations.  We will in this section prove that when $v$ on $F$ is Henselian, then 
$\Theta$ is actually an isomorphism whenever $\charr(\ov F) \ne 2$, and 
sometimes even when ${\charr(\ov F) = 2}$. 

Recall that a valuation $v$ on a field $F$ is {\it Henselian} if Hensel's Lemma
holds for $v$.  Equivalently, (cf.~\cite[Th.~4.1.3]{EP}), $v$ is Henselian iff 
$v$ has a unique extension to a valuation on each field $L$ algebraic over $F$.  It
follows immediately that the extension of $v$ to any such $L$ is also Henselian. 
Furthermore, the uniqueness of the extension allows one to see that $v$ extends
uniquely to each division algebra finite-dimensional over $F$ (cf.~\cite{W}). 

Throughout this section, $D, v,\tau, \la, K, F, \tau', \la'$ will have the same
meaning as in \S 3.    When we say that $v$ is Henselian, we mean that $v|_F$
is Henselian.  This assures that the valuation on $K$ is also Henselian, and that
$v$ extends, uniquely, to a valuation on $D$.  The uniqueness of these valuations
guarantees that for any involution $\tau$ on $D$ with $K^\tau = F$, we have 
$v\circ \tau = v$.    

Our principal results in this section which hold whenever $\charr(\ov D)\ne
2$ also hold in the following cases when $\charr(\ov D) = 2$:   

\begin{definition}\label{good}
{\it The good cases when $\charr(\ov D) = 2$} are
 when $D$ is tame over $F$, and, if $\tau$ is of the first kind and $\charr(D) = 0$,
$(\tau,\la)$ is a symplectic pair.
\end{definition}

Note that when $D$ is tame over $F$ and $\charr(\ov D) = 2$, these good cases are
exactly those where $(v,\tau,\la)$ preserves even forms---see Prop.~\ref{preseven}
and the remark after its proof. Also recall Remark~\ref{samekind} that the tameness 
assumption guarantees that $\tau$ and $\tau'$ are of the same kind. 

\medskip

The case of anisotropic associated graded Hermitian forms is quite special, and
requires no Henselian assumption:

\begin{proposition}\label{graniso}
Let $(M,h)\in \mathcal H^+(D, \tau,\la)$.  Let $\be(m) = \frac12v(h(m,m))\in
\frac12\vg D\cup\{\infty\}$, for all
$m \in M$.  Then, the following conditions are equivalent:
\begin{enumerate}
  \item[{\rm(i)}] 
There is a norm $\al$ on $M$ with $\al\cmp h$ and $\halp$ anisotropic.
  \item[{\rm(ii)}]
$\be$ is a norm on $M$ with $\be \cmp h$.
  \item[{\rm(iii)}]
For all $m,n\in M$, \quad $2v(h(m,n)) \ \ge \ v(h(m,m)) \ + \
v(h(n,n))$.
\end{enumerate}
When these conditions hold, $\be$ is the only norm on $M$ which is compatible with
$h$, and $\hbep$ is anisotropic.
\end{proposition}

\begin{proof}
Note first that for any norm $\al$ on $M$ with $\al\cmp h$ and any nonzero $m\in
M$, \eqref{defh'} shows 
\begin{equation}\label{anisocond}
\halp(m',m') \ne 0 \quad \text{iff} \quad\al(m) = \be (m) \,. 
\end{equation}

(i) $\Rightarrow$ (ii)  \  If $\al\cmp h$ and $\halp$ is anisotropic, then
\eqref{anisocond} shows $\be = \al$.  Hence, $\be$ is a norm on $M$ with~
$\be\cmp h$.  

(ii) $\Rightarrow$ (i) and (iii)  \ Suppose (ii) holds.  Then, $\hbep$ has no
isotropic homogeneous elements, by \eqref{anisocond}, so 
$\hbep$ is anisotropic. Then (i) holds  with $\al = \be$.  Since $\be \bd h$,
we have, by  the definition, ${\be(m) + \be (n) \le v(h(m,n))}$ for all $m,n\in M$.
This is the inequality in (iii).

(iii) $\Rightarrow$ (ii)     
 \ Suppose (iii) holds.  By Cor.~\ref{existence} there is a norm $\al$ on $M$
with $\al \cmp h$.  Since $\al \bd h$, we have $\al(m)+\al(m) \le v(h(m,m))$
for all $m\in M$, i.e., $\al(m)\le \be (m)$.  Suppose there were a nonzero 
$n\in M$ with $\al (n) < \be(n)$.  Then, for every $m\in M$, the inequality in
(iii) says that ${v(h(n,m)) \ge \be(n) + \be(m) > \al(n) + \al(m)}$.  This
contradicts the definition of $\al\cmp h$.  Thus, we must have $\be = \al$,
proving (ii).

When the conditions (i)--(iii) hold, the proof of (iii) $\Rightarrow$ (ii)
shows that $\be$ is the only norm on $M$ compatible with $h$, and the proof of
(ii) $\Rightarrow$ (i) shows that $\hbep$ is anisotropic. 
\end{proof}

% For any $a_1, \ldots, a_k\in \Alt(D,\tau,\la)$, we write $\langle a_1,\ldots, a_k
% \rangle$ for the even $\la$-Hermitian form $h$ for $\tau$ on a $D$-vector space
% $M$ with an orthogonal base $\{\,m_1, \ldots, m_k\,\}$ such that each $h(m_i,m_i) =
% a_i$.
 
\begin{proposition}\label{presaniso} Suppose $(v,\tau,\la)$ preserves even forms.
Then, the following conditions are equivalent:\begin{enumerate}
 \item[{\rm(i)}]
For  every form $(M,h)\in \HermDp{\tau}{\la}$ and every norm $\al$ on $M$ with 
$\al\cmp h$, if $h$ is anisotropic, then $\halp$ is anisotropic.
 \item[{\rm(ii)}]
For  every form $(M,h)\in \HermDp{\tau}{\la}$ and every norm $\al$ on $M$ with 
$\al\cmp h$, if $\halp$ is hyperbolic, then $h$ is hyperbolic.
  \item[{\rm(iii)}]
The canonical map $\Theta\colon W^+(D,\tau,\la) \to W^+_g(\gr(D), \tau', \la')$ of
Th.~\ref{springer}(ii) is an isomorphism.
  \item[{\rm(iv)}] For each nonzero $a,c\in \Alt(D,\tau,\la)$ with 
$v(c) >v(a)$, the diagonal form ${\langle a, -(a+c)\rangle \in \HermDp{\tau}{\la}}$ 
is isotropic. 
  \item[{\rm(v)}] 
For each nonzero $a\in \Alt(D,\tau,\la)$ let $\tau_a = \inn (a\inv) \circ \tau$;
then for each $m\in \Alt(D,\tau_a, 1)$ with $v(m) >0$ there is $d\in D$
with $\tau_a(d)d = 1+m$. 
\end{enumerate}
\end{proposition}

\begin{proof} 
(i)$\Rightarrow$(ii) \ Suppose $(M,h) \in \mathcal H^+\dtl$ and $\al$ is a norm on
$M$ with $\al\cmp h$ and $\halp$  hyperbolic.  We argue by induction on
$\dim_D(M)$.  There is a two-dimensional graded subspace $N'$ of $\gr(M)$
with $\halp|_{N'}$ hyperbolic. Let $N$ be a subspace of $M$ with $\gr(N) = N'$.
Then $\al|_N \cmp h|_N$ by Remark~\ref{nondeg} since $\halp|_{\gr(N)}$ is
nondegenerate, and $\big(h|_N\big)_{\al|_N}'$ is hyperbolic, so isotropic.  By (i),
$h|_N$ is isotropic, hence metabolic, hence hyperbolic by Prop.~\ref{cancel}(iii),
since it is even.  By Prop.~\ref{grperp}, $\al|_{N^\perp} \cmp h|_{N^\perp}$
and $\gr(N^\perp) = N^{\prime\perp}$ in~$\gr(M)$.  Therefore, 
$\big(h|_{\gr(N^\perp)}\big)_{\al|_{\gr(N^\perp)}}'$
is hyperbolic, since $\halp$ and $\halp|_{N'}$ are hyperbolic. By induction, 
$h|_{N^\perp}$ is hyperbolic.  So, $h = h|_N\perp h|_{N^\perp}$ is hyperbolic.

(ii)$\Leftrightarrow$(iii) \ This is clear since the kernel of $W^+\dtl\to
W^+_g(\gr(D), \tau',\la')$ consists of Witt group equivalence classes of 
forms $h$ with compatible norm $\al$ with $\halp$ hyperbolic.

(ii)$\Rightarrow$(iv) \ Given $a$ and $c$ as in (iv), let $M$ be a two-dimensional
right $D$-vector space with base $\{\,m_1,m_2\,\}$ and let $h$ be the even
$\la$-Hermitian form for $\tau$ with 
$h(m_1,m_1) = a$, $h(m_1, m_2) = 0$, and $h(m_2,m_2) = -(a+c)$.  So, $h = 
\langle a, -(a+c)\rangle$.  Let $\al$ be the norm on $M$ with splitting base
$\{\,m_1,m_2\,\}$ such that $\al(m_1) = \al(m_2) =\frac12v(a) =
\frac12v(-(a+c))$. Then (see Ex.~\ref{excomp}(i)) $\al\cmp h$ and we have 
$\halp(m_1', m_1') = a' = -\halp(m_2,m_2)$ and $\halp(m_1',m_2') = 0$.  The
$2$-dimensional even form $\halp$ is clearly hyperbolic, so by~(ii) $h$ is
hyperbolic, so isotropic.     

(iv)$\Rightarrow$(i) \ Assume (iv) holds but not (i).  Then, there is 
$(M,h) \in \mathcal H^+\dtl$ and a norm $\al$ on~$M$ with $h \cmp\al$
with $h$ anisotropic but $\halp$ isotropic.  There is a $2$-dimensional
subspace $N'$ of $\gr(M)$ with $\halp|_{N'}$ hyperbolic.  Let $N$ be any
$2$-dimensional $D$-subspace of $M$ with $\gr(N) = N'$.  We have $\al|_N \cmp h|_N$
as $\halp|_{\gr(N)}$ is nondegenerate.  Take any orthogonal base $\{\,n_1,n_2\,\}$
of $N$ for $h|_N$, which exists as $h|_N$ is anisotropic (see
Prop.~\ref{cancel}(iv)). Say $h(n_1,n_1) =a$ and $h(n_2,n_2) = b$, with $a,b\in
\Alt\dtl$.  Let $\be$ be the norm on $N$ with splitting base $\{\,n_1,n_2\,\}$
such that $\be(n_1)= \frac12v(a)$ and $\be(n_2) = \frac12v(b)$.  Then $\be\cmp
h|_N$ by Ex.~\ref{excomp}(i), and for the form $\hbep$ on $\gr_\be(N)$ we have 
$\hbep(n_1',n_1') = a'$, $\hbep(n_2',n_2') = b'$, and $\hbep(n_1',n_2') = 0$.  
Because $\big(h|_N\big)_{\al|_N}'$ is isotropic, Prop.~\ref{graniso} shows that
$\hbep$ must also be isotropic;  so, it has a homogeneous isotropic vector of
the form  $n_1'd'+n_2'$ for some $d\in D^\times$.  Then, ${0 = \hbep(n_1'd'+n_2',
n_1'd'+n_2') = \tau'(d')a'd' +  b' = (\tau(d) a d)' + b'}$. Lifting back to~$D$,
this yields $b = -\tau(d)ad -e$ with $v(e)>v(\tau(d)ad) = v(a)+2v(d)$.  Then, by
replacing the base vector $n_2$ by $n_2d\inv$, we have 
$$
h|_N\ = \ \langle a,b\rangle\ \cong \  \langle a,\tau(d\inv)b d\inv \rangle
\ = \ \langle a, -a-\tau(d\inv) ed\inv\rangle   \ = \ \langle a, -(a+c)\rangle\,,
$$ 
where $c = \tau(d\inv) e d\inv$.  We have $v(c) = v(e) - 2v(d) >v(a)$, and 
$c\in \Alt\dtl$ since $a$ and $b$ and hence $e$ lie in $\Alt\dtl$. By (iv),
$h|_N$ is isotropic, contradicting the choice of $h$.

(iv)$\Leftrightarrow$(v) \ Let $h = \langle  a,-(a+c)\rangle  \in \mathcal
H^+\dtl$ with $a,c \in \Alt\dtl$ and $v(c) >v(a)$.  Then, the scaled form 
$a\inv h = \langle a\inv a, -a\inv(a+c)\rangle = \langle 1, -(1+a\inv
c)\rangle$ lies in
$\mathcal H^+(D,\tau_a, 1)$, where ${\tau_a = \inn(a\inv)\circ \tau}$.  We have
$1, (1+ac\inv)
\in \Alt(D,\tau_a, 1)$, so $a\inv c\in \Alt(D,\tau_a, 1)$, with $v(a\inv c)>0$.
If (v) holds, then $a^{-1}h$ is isotropic, so $h$ is isotropic, proving (iv). 
Conversely, for $a\in \Alt\dtl$ and ${m \in \Alt(D,\tau_a, 1)}$ with $v(m) >0$, 
we have $am\in \Alt\dtl$ with $v(am) >v(a)$.  If (iv) holds, then the form 
${\langle a, -(a+am)\rangle \in \mathcal H^+\dtl}$ is isotropic, so the scaled form 
$a\inv \langle a, -(a+am)\rangle = \langle 1, -(1+m)\rangle \in \mathcal H^+(D,
\tau_a, 1)$ is also isotropic.  Then, $1+m = \tau(d) d$ for some $d\in D^\times$,
proving (v). 
\end{proof}
 
\begin{lemma}\label{imnorm} 
Let $F$ be a field with a Henselian valuation $v$, and let $L$ be a finite
degree 
separable  extension field of $F$ with  $L$ unramified over $F$ with
respect to $v$. Let 
$\mathfrak m_F$ (resp.~$\mathfrak m_L$) be the maximal ideal of the valuation ring
of
$v$ on $F$ (resp.~$L$). Then, $1+\mathfrak m_F \subseteq N_{L/F}(1+\mathfrak m_L)$,
where  $N_{L/F}$ is the norm from $L$ to $F$.
\end{lemma}

\begin{proof}
This is known.  See \cite[Prop.~2]{E2}, where it is pointed out that there is a
proof of this contained in \cite[Lemma~4.1]{Y} which is valid for all Henselian
 valuations not just for discrete Henselian valuations.  For the
convenience of the reader we give the short proof.  That $L$ is unramified over $F$
means that the residue field
$\ov L$ is separable over $\ov F$  and $[\ov L \!:\! \ov F] = [L \!:\! F]$.  Take
any nonzero
$a\in \ov L$ with ${\ov L =
\ov F(a)}$.  Let ${\wi f = x^k + \wi c_{k-1}x^{k-1}+\cdots + \wi c_0 \in \ov F[x]}$
be the minimal polynomial of $a$ over $\ov F$.  Choose any ${c_0, \ldots ,c_{k-1}
\in F}$ with $v(c_i) \ge 0$ and $\ov {c_i} = \wi c_i$ in~$\ov F$, and let 
$f = x^k+c_{k-1}x^{k-1} + \cdots + c_0 \in F[x]$.  Since the image $\wi f$ of $f$
in $\ov L[x]$ has the simple root $a$ and $v$ on $L$ is Henselian,   by Hensel's
Lemma $f$ has a root $b$ in~$L$ with $v(b) = 0$ and $\ov b = a$ in $\ov L$.  
Then, $\ov{F(b)} = \ov L$, as $a\in \ov{F(b)}$, so $[F(b) \!:\! F]
\ge[\ov{F(b)}\!:\! \ov F] = $ $[\ov L \!:\! \ov F] = [L \!:\! F] = k$.  Therefore, 
$F(b) = L$ and $f$ must be the minimal polynomial of $b$ over $F$; hence, 
$N_{L/F}(b) = (-1)^{k-1} c_0$.   For any $m \in \mathfrak m_F$, let 
$g = x^k+c_{k-1}x^{k-1} + \cdots + c_1 x + c_0(1+m) \in F[x]$.  The same
reasoning
for $g$ as just given for $f$ shows that $g$ has a root $d$ in $L$ with $v(d) = 0$
and $\ov d = a$ in~$\ov L$, and $N_{L/F}(d) = (-1)^{k-1} c_0(1+m)$.  Then, 
$db\inv \in 1+\mathfrak m_L$ and $N_{L/F}(db\inv) = 1+m$.  
\end{proof}

In fact, the inclusion in Lemma~\ref{imnorm} is an equality.  The reverse inclusion
is not hard to prove, but not included here because we do not need it. 


\begin{proposition}\label{cent}
Let $A$ be a central simple algebra over a field $F$ and let $\si$ be a
symplectic involution on $A$.  Take any $a\in \Alt(A, \si, 1)$ such that $F(a)$
is a field.  Then, $\si$ restricts to a symplectic involution on 
the centralizer $C_A(F(a))$. 
\end{proposition} 

\begin{proof} This is known by \cite[Prop.~(4.12)]{KMRT} if $\charr(F) \ne 2$
(or if $a$ is separable over $F$;  it suffices in these cases that $\si(a)= a$).  
Thus, we may assume that $\charr(F) = 2$.  Let $L = F(a)$.  Assume the result is
false, i.e., that $\si|_{C_A(L)}$ is of orthogonal type.  There is a splitting
field $S$ of~$A$ with $S$ linearly disjoint to $L$ over $F$.  For example, we
could take $S$ to be the function field over $F$ of the Severi-Brauer variety
$SB(A)$, which is a regular extension of $F$ by \cite[Th.~3.2.11 and
Th.~3.7.12]{Ja},  so linearly disjoint to
every algebraic extension of $F$.  By replacing $A$ by $S\otimes_F A$, 
$L$ by $S\otimes _F L$, and $\si$ by 
$\id \otimes \si$ (which does not change the type of the involution), we may assume
that $A$ is split, say $A = \End_F(V)$ for some $F$-vector space $V$.  Because $L
\subset
\End_F(V)$, we may view
$V$ as a vector space over $L$.  Let $s\colon L\to F$ be any nonzero
$F$-linear map.  By \cite[Ex.~(4.11)]{KMRT}, there is a nondegenerate symmetric
$L$-bilinear form $b\colon V\times V \to L$  such that $\si$ is the adjoint
involution to the nondegenerate $F$-bilinear transfer form $s_*b = s\circ b\colon
V\times V \to F$. Then, $\si|_{C_A(L)}$ is the adjoint involution to $b$, by
\cite[Prop.~(4.7)]{KMRT}.  Because $\si|_{C_A(L)}$ is of orthogonal type, the form
$b$ is not alternating; so $V$ has an orthogonal $L$-vector space base $\{\,y_1,
\ldots, y_k\,\}$.   On the other hand, $s_*b$ is alternating, as $\si$ is
symplectic.  We claim:  for any $y\in V$, 
\begin{equation}\label{claim}
s_*b(y, ay) \ = \ 0\,.
\end{equation}
For, as $a\in \Alt(A,\si,1)$, we may write $a = c+\si(c)$ for some $c\in A$.  
Then,
\begin{align*}
s_*b(y,ay) \ &= \ s_*b(y, cy) +s_*b(y, \si(c)y) \ = \ 
s_*b(y, cy) + s_*b(cy,y)  \\ &= \  s_*b(y,cy) + s_*b(y, cy) \ = \ 0\,,
\end{align*} 
as claimed.  Then, for any integer $i\ge 0$, as $\si(a) = a$,
\begin{equation}\label{powers}
s_*b(y, a^iy) \ =\  \begin{cases}
\ s_*b(a^{i/2}y, a^{i/2}y) \  = \ 0, &\text{for $i$ even, as $s_*b$ is
alternating};\\
\ s_*b(a^{(i-1)/2}y, a^{(i-1)/2}ay) \ =\  0, &\text{for $i$ odd, by \eqref{claim}}.
\end{cases}
\end{equation}
Hence, $s_*b(y, Ly) = 0$ for every $y\in V$.  Thus, for any of the $y_i$, we have
$$
s_*b(y_i, V) \ = \ s_*b(y_i, Ly_i) \ +\ 
\textstyle\sum\limits_{j\ne i}s_*b(y_i,Ly_j) \ =\  0 + 0,
$$
by \eqref{powers} and the orthogonality of the $y_j$.  This contradicts
the nondegeneracy of $s_*b$. So $\si|_{C_A(L)}$ must be symplectic.
\end{proof}

\begin{theorem}\label{isothm} 
Let $D$ be a division algebra with valuation $v$, and involution $\tau$, with
$\la$, $K$, $F$, $\tau'$, and $\la'$  as defined at the beginning of \S 3.
Assume $v$ is Henselian.  Then, whenever $\charr(\ov D) \ne 2$ and also in the good
cases when $\charr(\ov D) = 2$   $($see Def.~\ref{good}$)$ 
the canonical
map  $\Theta\colon W^+(D,\tau,\la) \to W^+_g(\gr(D), \tau', \la')$ of
Th.~\ref{springer}$($ii$)$ is an isomorphism; the other conditions in 
Prop.~\ref{presaniso} also hold. 
% \begin{enumerate}
%    \item[{\rm(i)}]
% $\charr(\ov D)\ne 2$.
%   \item[{\rm(ii)}]
% $\tau$ and $\tau'$ are of the second kind, and $K$ is unramified over $F$.
%   \item[{\rm(iii)}]
% $\tau$ and $\tau'$ are of the first kind, $D$ is tame, and 
% $(\tau, \la)$ is a symplectic pair.
% \end{enumerate}
\end{theorem} 

\begin{proof}
When $\charr(\ov D)\ne 2$, and also in the good cases when $\charr(\ov D) = 2$,
 Prop.~\ref{preseven} shows that 
  $(v,\tau,\la)$ preserves even forms. 
We prove that condition (v) of Prop.~\ref{presaniso} holds.  Then, the 
theorem follows by  Prop.~\ref{presaniso}. The proof is divided into three cases.

Case I. \ Suppose $\charr(\ov D)\ne 2$.  For any $a \in \Alt\dtl$ and $m \in \Alt(D,
\tau_a,1)$ with $v(m)>0$, we have $\tau_a(m) = m$, so $\tau$ is the identity 
on $F(m)$.  By Hensel's Lemma (applied to $x^2-(1+m)\in F(m)[x]$), 
$1+m$ has a square root, say $b$, in $F(m)$. Then, $\tau_a(b) b = b^2 = 1+m$, as
desired.

Case II.  \  Suppose $\tau$ and $\tau'$ are of the second kind and $K$ is unramified
over
$F$.  For $a\in \Alt \dtl$, $\tau_a$ is also of the second kind with 
$K^{\tau_a} = F$, as $\tau_a|_K =\tau|_K$.  Take any $m \in \Alt(D, \tau_a,1)$
with $v(m) >0$; so 
$\tau_a(m) = m$. Let $L = F(m)$.  Because $\tau_a$ acts nontrivially on $K$ but
trivially on $L$, the field $K\cdot L$ is not $L$.  Hence, $[K\cdot L \!:\! L] =
2$ and
$\tau_a|_{K\cdot L}$ is the nonidentity $L$-automorphism of $K\cdot L$.  Also,
$K\cdot L$ is unramified over $L$ as $K$ is unramified over $F$ by
\cite[Th.~5.2.7, Rem.~5.2.8]{EP}. By Lemma~\ref{imnorm} applied to $K\cdot L/L$,
there is
$d\in K\cdot L$ with 
$1+m = N_{K\cdot L /L}(d) = \tau_a(d) d$, as desired.  

Case III. \ Now suppose $\charr(\ov D) = 2$, $\tau$ and $\tau'$ are of the first
kind, $D$ is tame.  If ${\charr(D) = 0}$, assume also  
that $(\tau,\la)$ is a symplectic pair.  
  For any nonzero $a\in \Alt\dtl$,
the pair
$(\tau_a,1)$ is  a symplectic pair by \cite[Prop.~(2.7)]{KMRT}, i.e., $\tau_a$
is a symplectic involution. (When $\charr(D) = 2$, this holds even if 
$(\tau,\la)$ is not a symplectic pair.)
For the case $\charr(D) = 2$, so $\la = 1$, to invoke \cite[Prop.~(2.7)]{KMRT} 
we need 
that $a\inv \in \Alt\dtl$.  But, if $a = e+\tau(e)$, then $\tau(a) = a$, so 
${a\inv = a\inv ea\inv +\tau(a\inv ea\inv) \in \Alt\dtl}$.)
Take any $m \in \Alt(D,\tau_a,1)$ with $v(m)>0$.  
By Prop.~\ref{cent}, $\tau_a|_{C_D(K(m))}$ is of symplectic type.
Let $L = K(m)$ and let $C= C_D(L)$.  Then, $L = Z(C)$ and $C\ne L$, since
$\tau_a|_C$ is a symplectic involution.  
 Moreover, we claim that $C$ is tame.
 We verify this using the criteria in \eqref{tame}:  Since $[\gr(D) \!:\! \gr(K)]=
 [D \!:\! K]$, $v$ is a $v|_K$ norm for $D$ viewed as a $K$-vector space; so
 $[\gr(L) \!:\! \gr(F)] = [L \!:\! F]$ by Prop.~\ref{subspace} and
 Cor.~\ref{normdim}(i).  Hence, $[\gr(D) \!:\! \gr(L)] = [\gr(D) \!:\! \gr(K)]\big/
 [\gr(L) \!:\! \gr(K)] = [D \!:\! K]\big/[L \!:\! K] = [D \!:\! L]$.
 Likewise, $[\gr(D) \!:\! \gr(C)] = [D \!:\! C]$; hence, $[\gr(C) \!:\! \gr(L)] = 
 [C \!:\! L]$.  Clearly, $\gr(C) \subseteq C_{\gr(D)}(\gr(L))$; in fact, equality
 holds here by dimension count, by the Double Centralizer Theorem and its graded
 analogue
 \cite[Prop.~1.5]{HW}.  Another application of the graded Double Centralizer
 Theorem then shows that $\gr(L) = Z(\gr(C))$.  Thus, $C$ is tame by \eqref{tame},
 as claimed.  (This follows also from \cite[p.~166, last line]{JW}.) 
We now have that $(C, \tau_a|_C, 1)$ is a good case for $\charr(\ov C) = 2$
as in Def.~\ref{good}.
Now, $1\in \Alt(C,\tau_a|_C, 1)$ as $\tau_a$ is a symplectic involution, 
by \cite[Prop.~2.6]{KMRT} if $\charr(C) = 2$, and trivially if $\charr(C) \ne 
2$ (then $1 = \half +\tau_a(\half)$). So, by applying 
Prop.~\ref{preseven}(iii) and (iv) and Prop.~\ref{grequal}(ii) to $C$,
we obtain  a $c\in C$ with $v(c) =
v(1) = 0$ and $c+\tau_a(c)= 1$.  
Let $M = L(c)$; so $\tau_a(M) = M$. Let $N =
M^{\tau_a}$.  Then, $[M \!:\! N] = 2$, as $c\notin N$, 
and $\tau_a$ is the nonidentity
automorphism of $M$.  The automorphism induced by $\tau_a$ on the residue field $\ov
M$ is  nontrivial, as it sends $\ov c$ to $\ov c+1$.  Therefore, $M$ is unramified
over
$N$.  Hence, by Lemma~\ref{imnorm}, there is $d\in M$ with $1+m = N_{M/N}(d) =
\tau_a(d)d$, as desired.

Cases I, II, and III cover all the cases stated in the theorem, since when $D$ is 
tame over $F$, $\tau$ and $\tau'$ are of the same kind, by Remark~\ref{samekind}.
%  Choose a  field
% $L$ maximal with respect to the properties that $K(m) \subseteq L \subseteq D$,
% $\tau_a|_L = \id$ and $\tau|_{C_D(L)}$ is  symplectic.  Let $Q = C_D(L)$.  Then,
% $Q \ne L$  and $Q$ must be a quaternion algebra over $L$. For, otherwise there
% would be a noncentral element $d$ in $\Alt(Q,\tau_a|_Q,1)$ (see formula
% \eqref{altdim}), and we could replace $L$ by $L(d)$ by using Lemma~\ref{cent}.
% Our $\tau_a$ on $Q$ must be the unique symplectic involution on this
% quaternion algebra, which is the conjugation map.  Because $\charr(\ov D) = 2$ and
% $\dim_K(D)$ is a $2$-power, the assumption that $D$ is tame implies that $D$ is
% split by the maximal unramified extension $K_{nr}$ of $K$, by \cite[???]{JW}.  
% Therefore $Q$ (which is Brauer equivalent to $D\otimes_F L$) is split by 
% $F_{nr}\cdot L$, which is the maximal unramified extension of $L$.  Hence, by 
% cite[???]{JW}, $Q$ has a maximal subfield $M$ which is unramified over $L$. 
% Because $M$ is separable over $L$, the conjugation involution on $Q$ (which is
% $\tau_a|_Q$) is the nonidentity $L$-automorphism of $M$.  By Lemma~\ref{imnorm}
% there is $d\in M$ with $1+m = N_{M/L}(d) = \tau_a(d) d$, as desired.
\end{proof}


\begin{example} 
Let $D = \left (\frac{-1,-1}{\qq_2}\right)$, the Hamilton quaternion division algebra 
over the
dyadic local field $\qq_2$. Let $\{\, 1, i,j,k\,\}$ be the standard base of 
$D$.  Let $u = (-1 +i+j+k)/2$ and $s = i-j$.  Then, 
$u^2 + u + 1= 0$, $\ sus\inv = -u-1$, and $s^2 = -2$.  From this it is clear that
$D$ is the cyclic algebra $(\qq_2(u)/\qq_2, \rho, -2)$,
where $\rho = \inn(s)|_{F(u)}$. The complete discrete (so 
Henselian) $2$-adic valuation on $\qq_2$ has value group $\vg{\qq_2} = \zz$ and 
residue field $\ov {\qq_2} = \ff_2$.  In the extension of $v$ to $D$, we must have
$v(u) = 0$ and $v(s) = \half$;  so $\ov D = \ff_4$ and $\vg D = \half \zz$.  Even though
$D$ is ramified over $\qq_2$, with ramification index equal to the residue characteristic,
$D$ is tame over~$\qq_2$ since it is inertially split, i.e., it has a maximal subfield
$\qq_2(u)$ which is unramified over $\qq_2$.  
The graded field  $\gr(\qq_2)$  of $\qq_2$ with respect to $v$ is the 
Laurent polynomial ring $\ff_2[t,t\inv]$, where $t = 2'$, which has grade $1$. We have
$\gr(D)_0 = \ov D = \ff_4$ and $\gr(D) = \ff_4\{s', s^{\prime-1}\}$, a twisted 
Laurent polynomial ring, where conjugation 
by $s'$ induces the Frobenius automorphism $\varphi$ on~$\ff_4$,  $s'^2 = t$, and 
$s'$ has grade equal to $v(s) = \half$.
Clearly, $Z(\gr(D)) = \ff_2[t,t\inv] = \gr(\qq_2)$.
Let $\tau$ be the unique symplectic involution on $D$ and let $\la = 1$, so $(\tau,\la)$~
 is a symplectic pair for $D$.   We have $\tau (u) = -u-1$ and $\tau(s) = -s$. The 
induced graded involution $\tau'$ on $\gr(D)$  is of the first kind, 
and is given by $\tau'(u') = u'+1$ (which 
shows that $\tau'$ is symplectic) and $\tau'(s') = s'$. 
We have $\tau'|_{\gr(D)_0} = \varphi$, which is an involution of the second
kind on $\gr(D)_0$, even though $\tau'$ itself is of the first kind. Since 
$s'\in \gr(D)_{\half}$ and $\tau'(s') = s'$, we can use $\wi \tau = 
\inn(s')\circ \tau'$ in computing the $[\frac 14]$-component of the Witt group.
Note that $\wi\tau|_{\gr(D)_0} = \id$.  Thus, 
for the Witt group of even (i.e., all) $\tau$-Hermitian forms on $D$, we have
by Th.~\ref{isothm} and Prop.~\ref{witt},
\begin{align*}
W^+(D, \tau, 1) \ \cong \ W^+_g(\gr(D), \tau', 1) \ &\cong \ 
W^+_g(\gr(D), \tau', 1;[0]) \ \oplus \  W^+_g(\gr(D), \tau', 1;[\textstyle\frac14])\\
&\cong \ W^+(\gr(D)_0, \tau',1) \ \oplus \  W^+(\gr(D)_0, \wi\tau,1)
\\ &= \ W^+(\ff_4,\varphi, 1) \ \oplus\ W^+(\ff_4,\id, 1) \ \cong \ 
\zz/2\zz \ \oplus \ (0)\,.
\end{align*}
For the last isomorphism, we use that $\Alt(\ff_4,\varphi, 1) = \ff_2$ and
$\Alt(\ff_4,\id, 1) = (0)$.   Of course, this Witt group could also have been
calculated using Jacobson's theorem \cite[Th.~1.7, p.~352]{S}, which gives an injection 
of $W^+(D, \tau, 1)$ into the Witt group of quadratic forms over 
$\qq_2$ via the transfer map.  

There are other involutions
on this $D$ as well, all of orthogonal type.  For example, let $d = i+j +k$ and let
$\wh \tau = \inn(d)\circ \tau$. So, $\wh \tau(u) = -u-1$ and $\wh \tau (s) = s$.
Note that even though $\wh \tau$ is orthogonal,
its associated graded involution $\wh \tau'$ coincides with $\tau'$, which is 
symplectic.  But, $(\wh\tau, 1)$ is not a symplectic pair, and we cannot hope 
to use $\wh \tau'$ to compute $W^+(D,\wh\tau, 1)$ since $(D,\wh\tau, 1)$ does 
not preserve even forms.  Indeed, $\Alt(D,\wh\tau, 1) = 
\{\, a + bs\,|\ a\in \qq_2,\ b\in \qq_2(u)\,\}$, which has dimension $3$ over
$\qq_2$, while $\Alt(\gr(D), \wh\tau',1) = \gr(\qq_2)$.  With respect to any
compatible norm $\al$ for  the even Hermitian form $h = \langle s \rangle$ for $\wh\tau$,
the associated graded form $\halp$ is not even.   
% as remarked after the proof of Prop.~\ref{preseven}.
On the other hand, $(\wh \tau, -1)$ is a symplectic pair, and Th.~\ref{isothm}
shows that $W^+(D, \wh\tau, -1) \ \cong \ W^+_g(\gr(D),\wh \tau', 1)$, which we 
just computed, as $\wh\tau' = \tau'$.
\end{example} 


\medskip

The isometry group of an even form $h$ acts on the family of norms compatible
with $h$.  We will show in Cor.~\ref{orbits} below that under the 
hypotheses of Th.~\ref{isothm} the action is as transitive as possible, in 
that two norms are in the same orbit iff they have isometric associated
graded forms.  The next lemma, giving a canonical 
form for norms compatible with hyperbolic
planes, is a building block for the group action result.


\begin{lemma}\label{plane}
Suppose we have  $\charr(\ov D) \ne 2$ or one of the good cases when $\charr(\ov D)
=2$.
Suppose $(M,h) \in \mathcal H^+\dtl$ with $\dim_D(M) = 2$  and  $h$~hyperbolic, and
let $\al$ be a norm on $M$ with $\al \cmp h$.  Then, there is a splitting base
$\{\,m,n\,\}$ for $\al$ with $m$ and $n$ isotropic, $h(m,n) = 1$, and 
$\al(n) = - \al(m)$. Furthermore, for any $\de \in \vg {\gr(M)}$, $m$ can be
chosen so that $\al(m) = \de$.
\end{lemma}

\begin{proof}
Take any isotropic vector $m\in M$.  Then, $m'$ is isotropic in $\gr(M)$.  By
Prop.~\ref{cancel}(iii) the maximal totally isotropic graded subspace $m'\gr(M)$
of
$\gr(M)$ has complementary totally isotropic graded subspace, call it $P$.  Take
any nonzero homogeneous element $\wh p \in P$.  Then, $\halp(m', \wh p) \ne 0$
since $(m'\gr(D))^\perp = m'\gr(D)$; also, $\halp(m',\wh p)$ is homogeneous
(hence a unit) in $\gr(D)$, since $m'$~and~$\wh p$ are each homogeneous in
$\gr(M)$.  Let $\wi p =\wh p\,\halp(m',\wh p)\inv$. Then $\wi p$ is homogeneous and
isotropic,
 and $\halp(m',\wi p) = 1$.  Because $\wi p$ is homogeneous and nonzero,
there is  a nonzero $p\in M$ with $p' = \wi p$.  Since $\halp(m',p') = 1\in 
\gr(D)_0$, by \eqref{defh'} $\al(m) + \al(p) = v(h(m,p)) = 0$.  If $p$ is
isotropic, then set $n = p\,h(m,p)\inv$; then, $n$ is isotropic, $h(m,n) = 1$,
and $\al(n) = \al(p) - v(h(m,p)) = -\al(m) - 0$, as desired.  

Now suppose $p$ is anisotropic.  By replacing $p$ by $p \,h(m,p)\inv$, we may
assume that $h(m,p) = 1$, while not changing $p'$.  
Because $h$ is even, there is $d\in D$ with $d+ \la\tau(d) = -h(p,p)$.  Moreover,
$d$ can be chosen with $v(d) = v(h(p,p))$.  This is clear if $\charr(\ov D) \ne 2$
(take $d = -\half h(p,p)$), and holds by 
%Lemma~\ref{image} 
Prop.~\ref{preseven}(iii) and (iv) and Prop.~\ref{grequal}(ii) 
 in the good cases when
$\charr(\ov D) = 2$.  
% Let $d = -\frac12h(p,p)$, so $\la\tau(d) = d$.  
Let $n = md+p$.  Then, ${h(n,n) = d+\la\tau(d) + h(p,p) =0}$.
 Because $\halp(p',p') = 0$, we have 
$2\al(p) <v(h(p,p) ) = v(d)$.
%, as $\charr(\ov D) \ne 2$.  
Hence, 
\begin{equation}\label{v(d)}
\al(m) \ + \ v(d) \ = \ -\al(p) + v(d) \ >\  \al(p) = -\al(m) \,.
\end{equation}
Because $\{\,m',p'\,\}$ is a  homogeneous base of $\gr(M)$, by
Cor.~\ref{normdim} $\{\,m,p\,\}$ is a splitting base for $\al$ on $M$. 
Therefore, by \eqref{v(d)},
$$
\al(n) \ = \ \minn\big(\al(md), \al(p)\big) \ = \ \minn\big(\al(m) + v(d),
-\al(m)\big) \ = \ -\al(m)\,.
$$
Since $h(m,n) = h(m,p) = 1$, we have $v(h(m,n)) = 0 = \al(m) + \al(n)$, so 
$\halp(m',n') = h(m,n)' \ne 0$.  Therefore, as $m'$ is isotropic, $m'$ and $n'$
must be $\gr(D)$-linearly independent in $\gr(M)$, so they form a homogeneous base
of $\gr(M)$.  Therefore, by Cor.~\ref{normdim}(ii) $\{\,m, n\,\}$ is a splitting
base for $\al$  on $M$.  Let $\ga = \al(m)$.  Then, $\vg{\gr(M)} = \brga \cup
\br{-\ga}$, where $\brga = \ga +\vg {\gr(D)}$.  Note that for any nonzero $c\in
D$, we have $\{\, mc, n\tau(c)\inv\,\}$ is a splitting base for $\al$ with 
$mc$ and $n\tau(c) \inv$ isotropic and $h(mc, n\tau(c)\inv) = 1$, and $\al(mc)=
\ga +v(c) = -\al(n\tau(c)\inv)$.  Therefore,  by interchanging $m$ and $n$ if
necessary and multiplying by a constant in $D$, we can arrange to find an $m$
with $\al(m)$ having any desired value in $\vg{\gr(M)}$.
\end{proof} 

\begin{proposition}\label{isometry}
Suppose $v$ is Henselian and we have $\charr(\ov D)\ne 2$ or one of the good 
cases when $\charr(\ov D) = 2$.  Let $(M,h)$, $(N,\ell) \in
\mathcal H^+\dtl$, and let 
$\al$ be a norm on~$M$ with $\al \cmp h$ and $\be$ a norm on $N$ with $\be \cmp
\ell$.  Then, there is an isometry $f\colon M\to N$ between $h$ and $\ell$ 
with $\al = \be \circ f$ iff $\halp \cong \lbep$ (graded isometry). 
\end{proposition}

\begin{proof}
$\Rightarrow$ \  If there is an isometry $f$ as described, then $f$ induces a map
$f'\colon \gr(M) \to \gr(N)$ which is clearly a graded isometry between $\halp$
and $\lbep$. 

$\Leftarrow$ \ Suppose there is a graded isometry $g\colon \gr(M) \to \gr(N)$.  
Consider first the special case where $\halp$~is anisotropic.  Then, $h$ is 
also anisotropic.  We have $\halp \perp - \lbep$ is hyperbolic, so $h \perp -\ell$
is hyperbolic, by Th.~\ref{isothm}(i) and Prop.~\ref{presaniso}.  Therefore, there
is an isometry
$f\colon M\to N$ between $h$ and $\ell$.  Because $\al$ and $\be\circ f$ are each
norms on $M$ compatible with   $h$, and $\halp$ is anisotropic, we have $\be\circ
f =\al$, by Prop.~\ref{graniso}.

Now consider another special case: Assume $\dim_D(M) = 2$ and $\halp$ is
hyperbolic.  Then, $h$ is hyperbolic by Th.~\ref{isothm}(i) and
Prop.~\ref{presaniso}.  For any $\ga \in
\vg{\gr(M)}$,  Lemma~\ref{plane} says there is a splitting base
$\{\,m_1,m_2\,\}$ for $\al$ on $M$ with $h(m_1,m_1) = h(m_2,m_2) = 0$,
$h(m_1,m_2) = 1$, and $\al(m_1) = \ga = -\al(m_2)$.  Since $\gr(N) = \gr(M)$
and $\lbep$ is also hyperbolic, there is a splitting base $\{\,n_1,n_2\,\}$
for $\be$ satisfying the same conditions relative to $\ell$ and $\be$. 
Then, the $D$-linear map $f\colon M\to N$ given by $f(m_i) = n_i$, $i= 1,2$
is an isometry between $f$ and $\ell$ with $\al = \be \circ f$.

We can now prove the general case.  We have by Prop.~\ref{cancel}(vi) and (iv), 
$\gr(M) =
\perpp\limits_{i=1}^k  M_i'$, where $\halp|_{M_0'}$ is anisotropic and for each 
$i \ge 1$, $\halp|_{M_i'}$ is hyperbolic with $\dim_{\gr(D)}(\gr(M_i')) = 2$.
We construct a \lq\lq good lift" of the $M_i'$ to $D$-subspaces $M_i$ of $M$.
Let $M_0$ be any $D$-subspace of $M$ with $\gr(M_0) = M_0'$ in $\gr(M)$. Then,
since $\halp|_{\gr(M_0)}$ is nondegenerate, Prop.~\ref{grperp} shows that 
$\gr(M_0^\perp) = M_0^{\prime \perp} = \bigoplus\limits _{i=1}^k M_i'$. Let 
$M_1$ be any subspace of $M_0^\perp$ with $\gr(M_1) = M_1'$.  Iterating this, we
see we can choose a subspace $M_j$ of $\big(\bigoplus\limits_{i = 0}^{j-1}
M_i\big)^\perp$ with $\gr(M_j) = M_j$.  Then, $M = \perpp\limits_{i=0}^k M_i$
with respect to $h$ and because $\gr(M)$ is a direct sum of the~$\gr(M_i)$, 
$\al = \opluss\limits _{i=0}^k\al|_{M_i}$. Likewise, let $N_i' = g(M_i')\subseteq
\gr(N)$.  Construct a good lift of the $N_i'$ to subspaces $N_i$ of $N$ with 
 $N = \perpp\limits_{i=0}^kN_i$ with respect to $\ell$ and $\gr(N_i) = N_i'$, so
$\be = \opluss\limits_{i=0}^k\be|_{N_i}$. By the special cases considered above,
for each $i$ there is $f_i\colon M_i \to N_i$ which is an isometry between
$h|_{M_i}$ and $\ell|_{N_i}$ with $\al|_{M_i} = \be|_{N_i} \circ f_i$. Then, the 
map $f = \oplus f_i$ has the desired properties.
\end{proof} 


\begin{corollary}\label{orbits}
Suppose $v$ is Henselian and we have $\charr(\ov D)\ne 2$ or one of the good 
cases when $\charr(\ov D) = 2$.  Let $(M,h)\in
\mathcal H^+\dtl$, and let $\al$, $\be$ be norms on $M$ with $\al\cmp h$ and
$\be \cmp h$.  Then, there is a $D$-linear map $f\colon M\to M$ which is an
isometry for $h$ with $\al = \be\circ f$ iff $\halp \cong \hbep$ (graded isometry).
\end{corollary}

\begin{proof}
This is immediate from Prop.~\ref{isometry}.
\end{proof} 

Cor.~\ref{orbits} shows that when $h$ is isotropic, the action of the isometry 
group of $h$ on the set of norms compatible with $h$ is not transitive.  This is
because hyperbolic forms of the same dimension in $\mathcal G \mathcal H^+(\gr(D),
\tau', \la')$ are not isometric unless they satisfy the added condition of being
isomorphic as graded vector spaces.  Ex.~\ref{excomp}(ii) shows that there are many
nonisomorphic possibilities for $\halp$ when $h$~is hyperbolic.  Note by contrast
that Springer in \cite{Sp2} and Goldman and Iwahori in \cite[Th.~4.16]{GI} did prove
transitivity of the isometry group action for their different notion of norms
compatible with a quadratic form.

 \vskip0.25truein 

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\smallskip

\bibitem[AJ]{AJ} R. Aravire and B. Jacob, Versions of Springer's theorem for
quadratic forms in characteristic~$2$, {\it Amer. J. Math.}, {\bf 118}  (1996), 
 235--261.
 
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\end{thebibliography} 


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