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\topmatter
\title
Triangular Witt groups.\break
Part I\,: The 12-term localization exact sequence.\break
* * * Full Version * * *
\endtitle
\author
Paul Balmer
\endauthor
\address
Department of Mathematics,
Rutgers University,
PISCATAWAY,
NJ 08854,
USA
\endaddress
\address
e-mail address: balmer\@math.rutgers.edu
\endaddress
\date May 1999\enddate
\abstract
To a short exact sequence of triangulated categories with duality, we
associate a long exact sequence of Witt groups. For this, we introduce
{\it higher Witt groups} in a very algebraic and explicit way.
Since those Witt groups are 4-periodic, this long exact sequence reduces to
a cyclic 12-term one. Of course, in addition to higher Witt groups, we
need to construct connecting homomorphisms, hereafter called {\it
residue homomorphisms}.
\endabstract
\subjclass 11E81, 18E30, 19G12\endsubjclass
\endtopmatter


\document
\rightheadtext{TWG I\,:\ The 12-term localization exact sequence}% simplifies
                                                       % the top text
                                                       % on right pages

\espaceavanttruc
\centerline{\bigsmc Introduction.}
\espaceaprespar
\espaceavanttruc


The usual Witt group of a scheme is obtained by considering symmetric
vector bundles (up to isometry) modulo the bundles possessing a
maximal totally isotropic sub-bundle (called a {\it lagrangian}, see
Knebusch [\the\kneb, definition p.\,133]). This is an
invariant whose behavior with respect to localization is pretty
tough.

Several attempts have been made to define higher Witt groups. 
For instance, one might mention the famous contributions of Karoubi (see
[\the\kara]), Pardon (see [\the\pard]) and Ranicki
(see [\the\rani]).
However, the existence of a localization sequence for Witt groups over
arbitrary schemes and open subschemes is unknown, even in the affine
case. My opinion on the question is that there is no natural kernel of
the localization map and therefore no obvious result to
conjecture, beyond the naive idea that some role has to be
attributed to the closed complement of the open subscheme.

In part II of this series, I shall explain how the Witt group of a
scheme $X$ can be computed
in terms of the derived category of bounded complexes of locally free
$\Cal O_X$-modules of finite type. The affine case has already been treated
in [\the\bala, theorem 4.1] and [\the\balb, {\sl th\'eor\`eme} 3.29,
p.\,89]. These results lead us to rephrase the
localization problem in terms of triangulated categories.

It is a real pleasure to see how everything can be expressed in this
language and how the axioms of triangulated categories (including of
course the octahedron axiom) are sufficient to establish the 12-term
exact sequence. Nevertheless, the results are not always easier to
prove. On the contrary, some of the questions are harder to tackle in
the triangulated case than in the usual one (the better example being
the {\it sub-lagrangian construction} treated in \S\,3). On the other
hand, this new approach gains in flexibility. Observe for instance
that stably metabolic forms are metabolic (theorem 2.5), which is
known to be wrong in the classical framework, even over rings.

One great advantage of the triangulated point of view is the existence
of a {\it kernel category} for localization. Let me make this more precise.

First of all, recall that a triangulated category $K$ possesses a {\it
duality} if it is endowed with a contravariant exact functor\,:
$$\#:K\too K$$
such that $\#^2\cong\Id$. Exactness means that $\#$
sends exact triangles of $K$ to exact triangles. The Witt group of $K$
is defined to be the quotient of the monoid of isometry classes of
symmetric spaces by the sub-monoid formed by those spaces possessing a
{\it lagrangian}. The notion of lagrangian used here generalizes the
classical one in a natural way. For all of this, see \S\,1.

Now, consider a localization of triangulated categories\,:
$$q:\ K\too S\inv K$$
and suppose that both are equipped with a duality in a compatible way
($\#\circ q=q\circ\#$). Then this localization has a {\it kernel}\ :
$$\lab{1.5}{\text{LOC}}0\too J\too K\too S\inv K\too0$$
namely the
full subcategory $J$ of $K$ on the objects $M$ such that $q(M)$ is
isomorphic to zero in $S\inv K$. Note that such an exact sequence does
not apply to {\it exact} categories and is a typically triangular
strategy. Using the notion of {\it saturated} subcategory (\cf
[\the\verd]), one can start from $J$ and see $S\inv K$ as $K/J$. A
kernel category like above is always saturated. I prefer to stick to
the presentation of (LOC) because I have localization of schemes in
mind, but both point of vue are equivalent ({\sl loco citato} II.2.1).

If we have an exact sequence like (LOC) above and if the dualities on
$K$ and $S\inv K$ are compatible, it is obvious that the
duality $\#$ of  $K$ restricts to a duality on $J$. Now the conjecture
is clear\,:
\vskip2mm

{\leftskip=30pt\rightskip=30pt\it\noindent To such a localization exact sequence of
triangulated categories with duality, we can associate a long
exact sequence of Witt groups (introducing higher and lower Witt
groups).\par}

\vskip2mm
This is the result we are going to establish (see theorem 5.1). There
is no assumption to be made on the triangulated category except a very
light {\it noetherianity} hypothesis (see definition 3.3).

The {\it higher} Witt groups are defined as follows. They are
the Witt groups of the same category with {\it shifted} dualities
$T^n\circ\#$, for $n\in\Z$ (see definition 1.13). In other
words, these higher Witt groups can be defined without topology and
without metaphysical considerations on the category of small triangulated
categories (e.g., no use of {\it flasque} triangulated categories).
This means that the theory of Witt groups over triangulated
categories is in some sense {\it complete with respect to
localization}. Moreover the simplicity of their definition allows the
computation of some of these higher Witt groups. 

It will be our next duty to interpret this 12-term exact
sequence in the special case of the derived category of a scheme (or
the derived category of any exact category with duality of course). This
requires additional work and is not presented here but will be part of this series.

One word about the proofs. The main result depends upon two facts\,:
first, stably neutral spaces are neutral (\cf already mentioned
theorem 2.5) and, second, we can more or less generalize the classical
sub-lagrangian construction to the derived framework (\cf theorem
3.17).  The exposition is such that the reader can follow the
reasoning step by step. The only facts needed about triangulated
categories are those recalled in \S\,0 and the four basic axioms.
Especially the octahedron is used with no mercy.

Paragraph 1 recalls the definitions of $(\pm1)$-dualities and Witt groups
in the language of triangulated categories.  Paragraphs 2 and 3 contain results on
symmetric spaces to be used in paragraphs 4 and 5, where the Witt
groups return and finally triumph at the satisfaction of the reader.





\leavevmode\espaceavantpar
\centerline{\bigsmc Table of Contents\,:}
\espaceaprespar
\espaceavanttruc

{\parindent=0pt
0. Background on triangulated categories.
\espaceaprespar

1. The four Witt groups of a triangulated category with duality.
\espaceaprespar

2. Neutrality and stable neutrality.
\espaceaprespar

3. The sub-lagrangian construction.
\espaceaprespar

4. Localization context and residue homomorphism.
\espaceaprespar

5. Main result.
\espaceaprespar

\leavevmode\hphantom{6. }Bibliography.
}






\parno=-1
\paragraph{Background on triangulated categories}

\espaceavanttruc
Here is a little {\sl baise-en-ville} of the triangulated
mathematician. Details can be found in [\the\verd], for instance.
%
Let $K$ denote a triangulated category. Recall that $K$ is additive
and endowed with a {\it translation} additive functor $T:K\to K$ which is an
automorphism. We have exact triangles\,:
$$(\Delta)\kern2cm\xytriangle A;B;C;u;v;w;$$
which replace the exact sequences of the abelian (or exact) framework. We
do not want to write down the axioms but we would like to recall
some very elementary and useful techniques. First, if $\Delta$ is
exact, so is any isomorphic triangle (axiom) and in particular any
triangle obtained from $\Delta$ by changing the sign of two of the
three morphisms.

To explain how triangles replace
exact sequences, recall that necessarily the composition of two 
consecutive morphisms in $\Delta$ is trivial\,: $v\,u=0,\ w\,v=0$ but
also $T(u)\,w=0$, $u\,T\inv(w)=0$ and so on. Let us focus on the middle morphism
$v$. Then $(A,u)$ is a {\it weak kernel} of $v$ in the sense that
$v\,u=0$ and that any morphism $f:X\to B$ such that $v\,f=0$ factors (non
uniquely) through $A$\,:
$$\xymatrix{A\ar[r]^-{\D u}
&B\ar[r]^-{\D v}
&C\ar[r]^-{\D w}
&T(A).\cr
&X\ar[u]_(.55){\D f}\ar[ru]_{\D v\,f=0}\ar@{-->}[lu]^{\D\exists\,\tilde f}}$$
As well, $\big(T(A),w\big)$ is a {\it weak cokernel} of $v$\,: for
$g:C\to Y$, we have $g\,v=0$ if and only if $g$ factors (non
uniquely) through $T(A)$\,:
$$\xymatrix{A\ar[r]^-{\D u}
&B\ar[r]^-{\D v}\ar[rd]_{\D g\,v=0}
&C\ar[r]^-{\D w}\ar[d]_(.45){\D g}
&T(A)\ar@{-->}[ld]^{\D\exists\,\bar g}\cr
&&Y.}$$

Another very common trick is the following. Given a morphism of exact
triangles\,:
$$\xymatrix{A\ar[r]^-{\D u}\ar[d]_{\D f}
&B\ar[r]^-{\D v}\ar[d]_{\D g}
&C\ar[r]^-{\D w}\ar[d]_{\D h}
&T(A)\ar[d]_{\D T(f)}
\cr
A'\ar[r]_-{\D u'}
&B'\ar[r]_-{\D v'}
&C'\ar[r]_-{\D w'}
&T(A')}$$
if two among $f$, $g$ and $h$ are isomorphisms so is the third
(analogue of the five lemma). This
is an easy consequence of the following also useful remark\,: Given
any endomorphism of an exact triangle of the form\,:
$$\xymatrix{A\ar[r]^-{\D u}\ar[d]_{\D0}
&B\ar[r]^-{\D v}\ar[d]_{\D0}
&C\ar[r]^-{\D w}\ar[d]_{\D k}
&T(A)\ar[d]_{\D0}
\cr
A\ar[r]_-{\D u}
&B\ar[r]_-{\D v}
&C\ar[r]_-{\D w}
&T(A),}$$
we have necessarily $k^2=0$. This, in turn, is an easy consequence of the above
remarks about weak kernels and cokernels.

This implies that the triples $(C,v,w)$ such that
$\Delta$ is an exact triangle (for $u:A\to B$ fixed) are all
isomorphic. Any such triple (or sometimes only the object $C$) is
called {\it the cone} of $u$ and will be written as
$\cone(u)$.
%
The cone  plays weakly but simultaneously the role of
kernel and cokernel. It contains essentially all the
{\it homological} information. A very important question is then to know
how to compare the cone of a composition $u'\circ u$ with the cones of
$u'$ and $u$. The answer is the famous axiom of the octahedron, which
says\,:
there is a {\it good} exact triangle linking the cone of the composition of two
morphisms and the cones of these morphisms. Any more precise formulation
should be close to the classical one due to Verdier.





\paragraph{The four Witt groups of a triangulated category with duality}

\truc{Once and for all}
Let $K$ denote a triangulated category and $T$ be its {\it translation
automorphism}. Here, we will always suppose that $\half\in K$. This means
that the abelian group $\Hom_K(A,B)$ is uniquely 2-divisible for all
objects $A$ and $B$ in $K$.

\defi
Let $\delta=\pm1$. An additive contravariant functor $\#:K\to K$ is said to be
{\it $\delta$-exact} if $T\circ \#=\#\circ T^{-1}$ and if, for any exact
triangle
$$\spxytriangle A;B;C;T(A),;u;v;w;$$
the following triangle is exact:
$$\xymatrix@C=15pt{\di C\ar[rr]^-{\D\di v}
&&\di B\ar[rr]^-{\D\di u}
&&\di A\ar[rrr]^-{\D\delta\cdot T(\di w)}
&&&T(\di C)
}$$
where $\di{(-)}$ trickily denotes $\#(-)$. Suppose, moreover, that
there exists an isomorphism of functors $\varpi:\Id\isoto\#\circ\#$
such that $\varpi_{T(M)}=T(\varpi_M)$ and 
$\di{(\varpi_M)}\circ\varpi_{\di M}=\Id_{\di M}$
for any object $M$ of $K$. Then the triple $(K,\#,\varpi)$ is called a
{\it triangulated category with $\delta$-duality}. In case $\delta=1$,
we shall simply talk about a {\it duality} and in case $\delta=-1$, we
shall use the word {\it skew-duality}.

\truc{Easy and important exercise}
Let $(K,\#,\varpi)$ be a triangulated category with
$\delta$-duality ($\delta=\pm1$).
\roster
\item Let $n\in\Z$. Prove that $(K,T^n\circ\#,\varpi)$ is again a triangulated category with $\big((-1)^n\cdot\delta\big)$-duality.
\item Prove that $(K,\#,-\varpi)$ is again a triangulated category
with $\delta$-duality. Same $\delta$\,!
\endroster

\truc{Notions using only the additive structure}
Let $(K,\#,\varpi)$ be a triangulated category with $\delta$-duality
($\delta=\pm1$). In particular, $K$ is an additive category with
duality. A {\it symmetric space} is a pair $(P,\varphi)$ such that
$P$ is an object in $K$ and $\varphi:P\isoto\di P$ is a {\it symmetric
form}, meaning that $\di\varphi\circ\varpi_P=\varphi$. Orthogonal sum
and isometries are defined as usual. A {\it skew-symmetric form}
$\varphi:P\isoto\di P$ verifies $\di\varphi\circ\varpi_P=-\varphi$ or, in
other words, is a symmetric form in $(K,\#,-\varpi)$. Pay attention\,:
the sign $\delta=\pm1$ has nothing to do {\sl a priori} with symmetry
and skew-symmetry.




\rem
Our $K$ is more than an additive category\,: we have the cone data. If
$u$ is symmetric morphism (\ie $\di u=u$ {\it without} assuming $u$ to
be an isomorphism), then the cone of $u$ is more
than the homological information about $u$, it also carries a 
symmetric structure. Let us make this more precise. 

This is a central point of our study\,: do not to skip this
part\,! It will be useful  for the definitions of {\it metabolic}
spaces and  of the connecting homomorphisms in the
long exact sequence of localization.

\thm Let $\delta=\pm1$. Let $(K,\#,\varpi)$ be a triangulated category
with $\delta$-duality. Suppose that $\half\in K$.  Let $A$ be an
object of $K$ and $u: A\to\di A$ be a symmetric morphism, that is $\di
u\circ\varpi_A=u$.
\roster
\item Choose any exact triangle over $u$\,:
$$\spxytriangle A;\di A;C;T(A).;u;u_1;u_2;$$
Then there exists an isomorphism $\psi$ such that the diagram\,:
$$\lab1{\Gamma}\vcenter{\xymatrix@C=45pt{%
A\ar[r]^-{\D u}\ar@{=}_{\D\delta\cdot\varpi_A}[d]
&\di A\ar[r]^-{\D u_1}\ar@{=}[d]
&C\ar[r]^-{\D u_2}\ar[d]_{\D\simeq}^{\D \psi}
&T(A)\ar@{=}^{\D\delta\cdot T(\varpi_{A})}[d]
\cr
A^{\#\#}\ar[r]_-{\D\delta\cdot\di u}
&\di A\ar[r]_-{\D-T(\di u_2)}
&T(\di C)\ar[r]_-{\D T(\di u_1)}
&T(A)
}}
$$
commutes and such that
$$T\big(\di\psi\big)\circ\varpi_C=(-\delta)\cdot\psi.$$
\vskip2mm

\item Other choices of $(C,u_1,u_2)$ and $\psi$ satisfying (1) give an
  isometric space $(C,\psi)$.
\endroster
\endthm

\goodbreak

\proof
Choose an exact triangle like in (1). Then dualize it using definition 1.2 to get\,:
$$\magnxytriangle25;\di C;A^{\#\#};\di A;\di u_1;\di u;\delta\cdot T(\di u_2);$$
and rotate it to put $\di u$ in the first place. Make an even number of sign changes to get the second line of $(\Gamma)$. Now the symmetry hypothesis on $u$ can be expressed as the commutativity of the left square of $(\Gamma)$. By an axiom of triangulated categories, one can complete it with some morphism $\psi$ satisfying\,:
$$\left\{\eqalign{\psi\,u_1&=-T(\di u_2)\cr
T(\di u_1)\,\psi&=\delta\cdot T(\varpi_{A})\,u_2.
}\right.$$
Apply $T\circ\#$ to those equations to obtain\,:
$$\left\{\eqalign{T(\di u_1)\,T(\di\psi)&=-u_2^{\#\#}\cr
T(\di\psi)\,u_1^{\#\#}&=\delta\cdot T(\di u_2)\,\di\varpi_{A}.
}\right.$$
Compose on the right the first line with $\varpi_C$ and the second with $\varpi_{\di A}$. Use definition of $\varpi\,:\Id\simeq\#\circ\#$ to find\,:
$$\left\{\eqalign{T(\di
    u_1)\,\Big(T(\di\psi)\circ\varpi_C\Big)&=-T(\varpi_{A})\,u_2\cr
    \Big(T(\di\psi)\circ\varpi_C\Big)\,u_1&=\delta\cdot T(\di u_2).
    }\right.$$
In other words, $\big(-\delta\cdot
T(\di\psi)\circ\varpi_C\big)$ can replace $\psi$ in the diagram
$(\Gamma)$. Since $\half\in K$, we can replace $\psi$ by
$\half\big(\psi-\delta\cdot T(\di\psi)\circ\varpi_C\big)$. This
morphism is necessarily an isomorphism (\cf \S 0) and satisfies (1).
\bigbreak

Let us prove (2). Part (1) says that $\psi$ is symmetric in
$\big(K,T\circ\#,(\delta)\cdot\varpi\big)$ which is a triangulated
category with the $(-\delta)$-duality
$$*\,:=T\circ\#.$$
Pay attention that ``$(-\delta)$-duality'' does not
come from $(\delta)\cdot\varpi$ but from exercise 1.3 part 1 for
$n=1$\,!

Suppose that you make another choice of $\psi$, call it $\chi$,
satusfying (1). Then $\chi\inv\psi$ fits in an endomorphism of the
triangle over u\,:
$$\xymatrix@C=45pt{%
A\ar[r]^-{\D u}\ar@{=}[d]
&\di A\ar[r]^-{\D u_1}\ar@{=}[d]
&C\ar[r]^-{\D u_2}\ar[d]_{\D\simeq}^{\D \chi\inv\psi}
&T(A)\ar@{=}[d]
\cr
A\ar[r]_-{\D u}
&\di A\ar[r]_-{\D u_1}
&C\ar[r]_-{\D u_2}
&T(A).
}
$$
By \S\,0, $\chi\inv\psi=1+k$ with $k^2=0$. It is easy to check that
$\chi\,k=\dis k\,\chi$. Therefore,
$$(1+\half k)^*\,\chi\,(1+\half k)=\chi\,(1+\half
k)^2=\chi\,(1+k)=\psi.$$
That is, $\chi$ and $\psi$ are isometric. It
is even easier to prove that the isometry class of $(C,\psi)$ does not
depend on the choice of the exact triangle.  \qed

\defi As suggested by the above result and by its proof, it is
convenient to make the following convention. Let $(K,\#,\varpi)$ be a
triangulated category with $\delta$-duality for $\delta=\pm1$. Then
the {\it translated} (or {\it shifted}) structure of triangulated
category with $(-\delta)$-duality is
$$T\big(K,\#,\varpi\big):=\big(K,\,T\circ\#\,,\,(-\delta)\cdot\varpi\,\big).$$

\truc{Remark and definition} 
In other words, if $\#$ is a duality, we consider $-\varpi$ as
the identification $\Id\isoto(T\circ\#)^2$. But, if $\#$ is a skew-duality,
then we use $\varpi$ again as the identification $\Id\isoto(T\circ\#)^2$.
This construction is invertible and we can define\,: for any
triangulated category with $\delta$-duality $(K,\#,\varpi)$,
$$T\inv\big(K,\#,\varpi\big):=\Big(K,\,T\inv\circ\#\,,\,(+\delta)\cdot\varpi\,\Big)$$
which is a structure of triangulated category with
$(-\delta)$-duality. Check that this is really the inverse
construction\,! If confusion with signs occurs, redo exercise 1.3.

For example, if $(K,\#,\varpi)$ is a triangulated category with
duality (\ie $\delta=+1$) then
$T^n(K,\#,\varpi)=\big(K,T^n\circ\#,(-1)^{n(n+1)\over2}\cdot\varpi\big)$
is a triangulated category with $(-1)^n$-duality, for all $n\in\Z$.

With this terminology, theorem 1.6 can be rephrased as\,: {\it The
  cone of a symmetric morphism for $(K,\#,\varpi)$ inherits a
  symmetric form for the translated duality $T(K,\#,\varpi)$. This
  symmetric form is well defined up to isometry}. This justifies the
following definition.

\defi
Let $(K,\#,\varpi)$ be a triangulated category with $\delta$-duality (for
$\delta=\pm1$). Let $A$ be
an object in $K$ and $u:A\to\di A$ be a symmetric
morphism ($\di u\circ\varpi_A=u$). The {\it cone} of $(A,u)$ is defined to be the
symmetric space
$$\cone(A,u):=(C,\psi)$$
fitting in the diagram $(\Gamma)$ of theorem 1.6. This is a space for the shifted
duality $T(K,\#,\varpi)$. It is well defined up to (non-unique) isometry.

\rem
Consider $\#$ as the translated duality of $T\inv\circ\#$. Theorem 1.6
allows us to construct a lot of trivial forms for $\#$
starting with symmetric morphisms $u$ for the {\it previous} skew-duality
$T\inv\circ\#$. These are the forms for the duality $\#$ which we
are going to ignore in the Witt group. In [\the\bala] and [\the\balb],
we chose to call these forms ``neutral'' because we also had to deal with classical
metabolic forms and we wanted to avoid the confusion.

\defi
Let $(K,\#,\varpi)$ be a triangulated category with $\delta$-duality
($\delta=\pm1$). A symmetric space $(P,\varphi)$ is {\it
neutral} (or {\it metabolic}) if there exist an object $A$ and a
morphism $u:A\to T\inv(\di A)$ such that
$$T\inv(\di u)\circ(\delta\cdot\varpi)=u\link1{and}(P,\varphi)=\cone(A,u).$$
It is easy to convince oneself that this is the natural
generalization of Knebusch's definition of {\it metabolic forms} (see
[\the\kneb]). Compare with [\the\balb, {\sl d\'efinition} 2.18, p.\,32] and the
subsequent remarks or with lemma 2.1 (2) below for the
``optical'' approach to the analogy.

\defi
Let $(K,\#,\varpi)$ be a triangulated category with $\delta$-duality
($\delta=\pm1$). We define the {\it Witt monoid} of $K$ to be the
monoid of isometry classes of symmetric spaces endowed with the
orthogonal sum. We write it as
$$\MW(K,\#,\varpi).$$

Obviously, since this cone-construction is additive and since
neutrality is preserved by isometry, the set of isometry classes of
neutral spaces form a  well defined sub-monoid of $\MW(K,\#,\varpi)$
that we shall call
$$\NW(K,\#,\varpi).$$

The quotient monoid is a group (for quotient of abelian monoids, see
[\the\balb, {\sl remarque} 2.27, p.\,35] if necessary), called the
{\it Witt group} of $(K,\#,\varpi)$ and written as
$$\W(K,\#,\varpi)={\MW(K,\#,\varpi)\over\NW(K,\#,\varpi)}.$$
If $(P,\varphi)$ is a symmetric space, we write $[P,\varphi]$ for
its class in the corresponding Witt group. We say that two symmetric
spaces are {\it Witt-equivalent} if their classes in the Witt groups
are the same.

Referring to exercise 1.3 and definitions 1.8 and 1.9, we define
$$\W^{\,n}(K,\#,\varpi):=\W\big(T^n(K,\#,\varpi)\big)$$
for all $n\in\Z$. These are the {\it shifted} Witt groups of $(K,\#,\varpi)$.


\prop
Let $(K,\#,\varpi)$ be a triangulated category with duality ($\delta=\pm1$).
The translation functor $T:K\to K$ induces isomorphisms\,:
$$\W(K,T^n\circ\#,\varpi)\isoto\W(K,T^{n+2},\varpi)$$
for all $n\in\Z$. In particular, we have the four-periodicity\,:
$$\W^{\,n}(K)\isoto\W^{\,n+4}(K).$$

\endthm

\proof
See [\the\bala, proposition 1.20] or [\the\balb, {\sl proposition}
2.38, p.\,37] or do it as an easy exercise. To see the periodicity, recall that definition 1.8 gives
$$T^2(K,\#,\varpi)=(K,T^2\circ\#,-\varpi).$$
Therefore the isomorphism of the proposition has to be used twice to reach
$(K,T^4\circ\#,\varpi)$ which is $T^4(K,\#,\varpi).$
\qed



\rem
Some people like to call symmetric forms for $(K,\#,-\varpi)$ {\it skew-symmetric}, considering that $(K,\#,\varpi)$ is somehow fixed. Using the previous proposition and our convention for the translation of a duality, the Witt group of skew-symmetric forms is nothing but $\Wm(K,\#,\varpi)$. 

The reader should keep in mind the following interpretation of this proposition:
on a given triangulated category with a $(+1)$-duality $\#$, there is
essentially one {\it associated} skew-duality, namely $T\circ\#$. In
terms of Witt groups, it means that we have two Witt groups for the
given duality and two Witt groups for {\it the} associated
skew-duality, since in both cases one can consider symmetric or
skew-symmetric forms. Then we keep four Witt groups. We have the
following collection\,: 
\vskip2mm

\leavevmode\noindent\rlap{$\W(K)$}\kern46pt$=\W(K,\#,\varpi)$ the {\it natural}\ \,Witt
group of symmetric forms.
\vskip2mm

\leavevmode\noindent\rlap{$\Wm(K)$}\kern46pt$\cong\W(K,\#,-\varpi)$ the {\it natural}\ \,Witt
group of skew-symmetric forms.
\vskip2mm

\hangindent=70pt\hangafter=1
\leavevmode\noindent\rlap{$\Wum(K)$}\kern46pt$=\W(K,T\circ\#,-\varpi)$ the {\it associated}\ \,Witt
group of skew-symmetric forms, using {\it the} skew-duality associated to $\#$.
\vskip2mm

\leavevmode\noindent\rlap{$\Wu(K)$}\kern46pt$\cong\W(K,T\circ\#,\varpi)$ the
{\it associated}\ \,Witt group of symmetric forms.
\vskip2mm

To make a long story short, considering all dualities $T^n\circ\#$, $n\in\Z$, and all identifications $\epsilon\cdot\varpi$, $\epsilon=\pm1$, you obtain four Witt groups. We choose to put them in the above order because of the long exact localization sequence we shall obtain.

\exa
Let $X$ be a scheme. Consider $\Dblf(X)$ the derived category of
bounded complexes of locally free $\Cal O_X$-modules of finite rank.
This is naturally endowed with a structure of triangulated category
with duality by localizing the usual duality on the exact category of
locally free $\Cal O_X$-modules of finite rank. Its Witt group is the
same as the usual one, as we shall establish in part II of this
series, under weak hypotheses. This generalises to any exact category with duality and we shall prove it in this generality.

\rem
The above example places the {\it usual} Witt group in the new
framework of triangulated categories with duality. It is very
interesting to notice that there exists some kind of a Witt group for
finitely generated (or coherent) modules also.

To understand this, note that under certain hypotheses $\Dbcoh(X)$ is
endowed an essentially unique duality as explained in [\the\hart, chapter V,
pp.\,252-301]. In other words, this is a $G_0$ (or $K_0'$) type of
Witt group, concept unknown to me until now. Even for rings, it is
quite unusual. The point is that any coherent module becomes
reflexive in the derived category $\Dbcoh(X)$, which is pretty wrong
in the abelian category of coherent modules.

This is another example of the generality of the triangular framework.
Of course, both types of Witt groups will coincide over regular
separated schemes.






\paragraph{Neutrality and stable neutrality}



\lem
Let $(K,\#)$ be a triangulated category with duality. Let
$(P,\varphi)$ be a symmetric space. The following conditions are
equivalent\,:
\roster
\item $(P,\varphi)$ is neutral;
\item there exist $L$, $\alpha$ and $w$ such that the triangle\,:
$$\magnspxytriangle 25;T\inv(\di L);L;P;\di L;w;\alpha;\di\alpha\varphi;$$
is exact and such that $T\inv(\di w)=w$;
\item there exists an exact triangle\,:
$$\magnspxytriangle 25;T\inv(\di M);L;P;\di M;\nu_0;\nu_1;\nu_2;$$
and an isomorphism $h:L\isoto M$ such that the following diagram
commutes\,:
$$\xymatrix@C=40pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di h)}^{\D\simeq}
&L\ar[r]^-{\D\nu_1}\ar[d]^{\D h}_{\D\simeq}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar[d]^{\D\di h}_{\D\simeq}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}
&M\ar[r]_-{\D\di\nu_2}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L.
}$$
\endroster

\proof
Using definition 1.12 for $\delta=+1$ and diagram $(\Gamma)$ of theorem 1.6 (recall that
$T\inv\#$ is a skew-duality), it is obvious that condition (1) and (2)
are equivalent. Clearly (2) implies (3) by taking $h=\Id$.

Suppose now that there exists a diagram like in (3). The triangle
$$\magnspxytriangle 25;T\inv(\di M);M;P;\di M;h\circ\nu_0;\nu_1\circ
h\inv;\nu_2;$$
is isomorphic to the exact triangle of (3) and is therefore exact. Let
$w=h\circ\nu_0$ and $\alpha=\nu_1\circ h\inv$. Then the triple
$(L,\alpha,w)$ satisfies (2).
\qed

\defi
Given a neutral symmetric space $(P,\varphi)$, a triple $(L,\alpha,w)$
satisfying condition (2) of the above lemma is called a {\it lagrangian} of
$(P,\varphi)$. In particular, $\di\alpha\varphi\,\alpha=0$.

Given any symmetric space $(P,\varphi)$, a pair $(L,\alpha)$ is called
a {\it sub-lagrangian} of $(P,\varphi)$ if
$\di\alpha\varphi\,\alpha=0$. We shall deal with sub-lagrangians in
paragraph 3.

\rem
A symmetric space $(P,\varphi)$ is said to be {\it stably neutral} if
there exists a neutral space $(R,\psi)$ such that
$$(P,\varphi)\bot (Q,\chi)\simeq(R,\psi)\bot (Q,\chi)$$
for some symmetric space $(Q,\chi)$. This is the same as saying that
$(P,\varphi)\ \bot$ some neutral space is neutral (by adding
$(Q,-\chi)$ on both side). This is also the same as saying that
$[P,\varphi]=0$ in $\W(K)$. Obviously neutral spaces are stably
neutral. The converse is wrong in the {\it usual} framework (there exist
non-metabolic stably metabolic spaces) but is true in the
{\it triangulated} one.

\thm
Let $K$ be a triangulated category with duality. Suppose that
$\half\in K$. Any stably neutral symmetric space is neutral.
\endthm


\proof
Let $(P,\varphi)$ be a stably neutral space. 
By the above remark, there exists a space $(Q,\chi)$
such that the space
$$(P,\varphi)\bot (Q,\chi)\bot (Q,-\chi)$$
is neutral. Since $(Q,\chi)\bot (Q,-\chi)$ is isometric to
$(Q\oplus\di Q,\pmatrix 0&1\\1&0\endpmatrix)$, this means that the space
$$(P\oplus Q\oplus\di Q,\pmatrix
\varphi&0&0\\0&0&1\\0&1&0\endpmatrix)$$
is neutral. By condition (2) of lemma 2.1 applied to the above
symmetric space, there exists an exact triangle\,:
$$\xymatrix{T\inv(\di L)\ar[r]^-{\D w}
&L\ar[r]^-{\pmatrix a\\ b\\ c\endpmatrix}
&P\oplus Q\oplus\di Q
\ar[rrr]^-{\pmatrix \di a\varphi &\di c &\di b\endpmatrix}
&&&\di L
}$$
with $w=T\inv(\di w)$ for some morphisms $a,\ b$ and $c$ as above. Chose an exact triangle containing $b$\,:
$$\xytriangle L;Q;T(M);b;b_1;b_2;$$
(we choose to call $T(M)$ the cone of $b$).
Now, apply the octahedron axiom to the relation
$$\pmatrix 0&1&0\endpmatrix\cdot%
\pmatrix a\\ b\\ c\endpmatrix=b.$$
This gives the following\,:
\vskip-5mm
$$\xymatrix@C=34pt @R=45pt{%
&&&Q\ar[ld]^-{\D 0}
\ar[rrdd]^-{\D b_1}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
P\oplus Q\oplus\di Q\ar[rrru]^-{\pmatrix 0&1&0\endpmatrix}
\ar[rrdd]_-{\pmatrix \di a\varphi & \di c & \di b\endpmatrix}
&&T(P)\oplus T(\di Q)\ar[ll]|(.4){\pmatrix1&0\\0&0\\0&1\endpmatrix}
\ar[dd]|(.6){\pmatrix T(\di a\varphi)&T(\di b)\endpmatrix}
\\
&&&L\ar[lllu]|!{[lu];[ld]}\hole|-{\pmatrix a\\ b\\ c\endpmatrix}
\ar[uu]|!{[rr];[lu]}\hole_(.55){\D b}
&&T(M)\ar[ll]|(.6){\D b_2}
\ar[lllu]_-{\D\exists\,g}
\\
&&\di L\ar[ru]|-{\D -T(w)}
\ar[rrru]_-{\D\exists\,f}
}$$
for some morphisms $f:\di L\to T(M)$ and $g:T(M)\to T(P)\oplus T(\di
Q)$ such that the above diagram is an octahedron (meaning that faces
are alternatively commutative or exact
triangles and that the two ways from the left to the right
(respectively the two ways from the right to the left)
coincide). Confer [\the\verd] for the octahedron axiom.

In particular, the condition $\pmatrix1&0\\0&0\\0&1\endpmatrix\cdot g=
T(\pmatrix a\\ b\\ c\endpmatrix)\cdot b_2$
forces $g$ to be $\pmatrix T(a)b_2\\ T(c)b_2\endpmatrix$. Thus the
above octahedron reduces to the existence of a morphism $f:\di L\to
T(M)$ such that\,:
\roster
\item the following triangle is exact\,:
$$\xymatrix{%
\di L\ar[r]^-{\D f}
&T(M)\ar[rr]^-{\pmatrix T(a)b_2\\ T(c)b_2\endpmatrix}
&&T(P)\oplus T(\di Q)\ar[rrr]^-{\pmatrix T(\di a\varphi)&T(\di
b)\endpmatrix}
&&&T(\di L)
}$$
\item $b_2\, f=-T(w)$;
\item $f\,\di c=b_1$.
\endroster
\goodbreak

It will appear that $(M,a\circ T\inv(b_2),m)$ is a lagrangian of
$(P,\varphi)$ for a suitable morphism $m:T\inv(\di M)\to M$ (more or
less). Therefore we baptize\,:
$$\alpha:=a\circ T\inv(b_2)\,:\,M\too P.$$
The reluctant reader can motivate himself by checking that
$\di\alpha\varphi\,\alpha=0$.
The octahedron axiom will help us to establish an exact triangle over
$\alpha$. Let us consider the following identity\,:
$$\pmatrix \di a\varphi &\di b\endpmatrix\cdot\pmatrix 0\\ 1\endpmatrix=\di
b$$
and apply to this the octahedron axiom to get\,:
$$\xymatrix@C=34pt @R=45pt{%
&&&\di L\ar[ld]^-{\D f}
\ar[rrdd]^-{\D -T(\di b_2)}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
P\oplus\di Q\ar[rrru]^-{\pmatrix \di a\varphi&\di b\endpmatrix}
\ar[rrdd]_-{\pmatrix 1&0\endpmatrix}
&&T(M)\ar[ll]|(.4){\pmatrix -T(a)b_2\\ -T(c)b_2\endpmatrix}
\ar[dd]|(.6){\D-T(\alpha)}
\\
&&&\di Q\ar[lllu]|!{[lu];[ld]}\hole|-{\pmatrix 0\\ 1\endpmatrix}
\ar[uu]|!{[rr];[lu]}\hole|(.55){\D\di b}
&&\di M\ar[ll]|(.55){\D T(\di b_1)}
\ar[lllu]_-{\D\exists\,g'}
\\
&&P\ar[ru]^(.57){\D 0}
\ar[rrru]_-{\D\exists\,f'}
}$$
The exact triangles over $\pmatrix \di a\varphi &\di b\endpmatrix$ and
$\di b$ are obtained respectively by rotating (1) and by dualizing the exact
triangle over $b$ from the beginning.

Here, the relation
$$f'\circ\pmatrix 1&0\endpmatrix=
-T(\di b_2)\circ\pmatrix \di a\varphi&\di b\endpmatrix$$
forces $f'=-T(\di b_2)\,\di a\varphi
=-\di{\big(a\,T\inv(b_2)\big)}\varphi=-\di\alpha\varphi$. 
Thus the information of the octahedron reduces to the existence of some
$g':\di M\to T(M)$ that we astutely re-baptize $g'=T(m)$ for $m:T\inv(\di M)\to
M$ such that\,:
\roster
\item[4] the following triangle is exact\,:
$$\spxytriangle T\inv(\di M);M;P;\di M;m;\alpha;\di\alpha\varphi;$$
\item $m\, \di b_2=-T\inv f$
\item $c\,T\inv(b_2)\,m=-\di b_1$.
\endroster

To make $(M,\alpha,m)$ a lagrangian of $(P,\varphi)$ we still need
$T\inv(\di m)=m$. So let us consider the triangles over $m$ and over
$T\inv(\di m)$.

To lighten a little bit the notations, we shall write
$$*=T\inv\circ\#.$$

Dualizing the exact triangle (4), we get the second line of the
following diagram, whose first line is simply exact triangle (4)\,:
$$\lab27\vcenter{\xymatrix@C=40pt{
\dis M\ar[r]^-{\D m}\ar@{=}[d]
&M\ar[r]^-{\D\alpha}\ar@{-->}[d]_{\D\exists\,1+x}^{\D\simeq}
&P\ar[r]^-{\D\di\alpha\varphi}\ar[d]_{\D\varphi}^{\D\simeq}
&\di M\ar@{=}[d]
\cr
\dis M\ar[r]_-{\D\dis m}
&M\ar[r]_-{\D\varphi\alpha}
&\di P\ar[r]_-{\D\di\alpha}
&\di M
}}$$
and it is an axiom of triangulated categories that the above morphism
of exact triangles can be completed. The third morphism is called $1+x$
because we are going to prove that $x$ is nilpotent. This is quite a
technical computation. Let us do it step by step, very carefully.
Directly from (7), we get\,:
\roster
\item[8] $x\,m=\dis m-m$
\item $\alpha\,x=0$.
\endroster
>From (8), we immediately have $\dis m\dis x=\dis{(x\,m)}=\dis{(\dis
m-m)}=m-\dis m=-x\,m$. Call this
$$\llap{(10)\kern2cm}\dis m\dis x=-x\,m.$$
%
Now, compose (5) on the right with $\dis c$ and use (3) to obtain
$$\llap{(11)\kern2cm}m\,\di b_2\dis c=-T\inv(b_1).$$
Applying $*$ to this last equality and using (6) gives
$$c\,T\inv(b_2)\,\dis m=-\di b_1=c\,T\inv(b_2)\,m.$$
Replacing in this equation $\dis m$ by $(1+x)\,m$, which is allowed
by (7), we obtain\,:
$$\llap{(12)\kern2cm}c\,T\inv(b_2)\,x\,m=0.$$
%
Dualizing this last relation and using (10) gives\,:
$$x\,m\,\di b_2\dis c=0.$$
But (11) allows us to replace $m\,\di b_2\dis c$ to get\,:
$$\llap{(13)\kern2cm}x\ T\inv(b_1)=0.$$

Now we are going to use the symmetry of the $w$ given at the beginning
to deduce some information about $x$. Compose relation (5) on the left
by $T\inv(b_2)$ and use (2) to obtain that
$$T\inv(b_2)\,m\,\di b_2=-T\inv(b_2)\,T\inv(f)=w.$$
Since $w=\dis w$, the left hand side of the above equation is also
$*$-symmetric. This gives
$$T\inv(b_2)\,m\,\di b_2=T\inv(b_2)\,\dis m\di b_2.$$
Again replacing $\dis m$ by $(1+x)\,m$, we obtain
$$T\inv(b_2)\,x\,m\,\di b_2=0.$$
But then (5) allows us to replace $m\,\di b_2$ and to have
$$\llap{(14)\kern2cm}T\inv(b_2)\,x\,T\inv(f)=0.$$

We are almost done. Recalling the definition of
$\alpha=a\,T\inv(b_2)$, composing (9) with $m$ on the right gives
$$a\,T\inv(b_2)\,x\,m=0.$$
Since, on the other hand (12) insures us that $c\,T\inv(b_2)\,x\,m=0$,
we have
$$\pmatrix a\,T\inv(b_2)\\ c\,T\inv(b_2)\endpmatrix\,x\,m=0.$$
Since triangle (1) is exact (rotate it a little bit, if you need),
there exists a morphism $y:\dis M\to\dis L$ such that
$$\llap{(15)\kern2cm}xm=T\inv(f)\,y.$$
\vskip2mm

>From relation (13) and the exact triangle over $b$ that we chose at
the very beginning of the proof, there exists a morphism $z:L\to M$
such that
$$\llap{(16)\kern2cm}x=z\,T\inv(b_2).$$

Now we are going to compute $x^3 m$ using the above relations. By (15)
and (16), we have
$$x^3m=x\cdot x\cdot (xm)=z\,T\inv(b_2)\cdot x\cdot T\inv(f)\,y.$$
But (14) insures that the composition of the 3 morphisms in the middle
is zero. Thus we have established
$$x^3m=0.$$
Of course, from (9), we have that $\alpha\,x^3=0$. In other words,
$x^3$ makes the following diagram commute\,:
$$\xymatrix@C=40pt{
\dis M\ar[r]^-{\D m}\ar[d]_{\D0}
&M\ar[r]^-{\D\alpha}\ar[d]^{\D x^3}
&P\ar[r]^-{\D\di\alpha\varphi}\ar[d]^{\D0}
&\di M\ar[d]^{\D0}
\cr
\dis M\ar[r]_-{\D m}
&M\ar[r]_-{\D\alpha}
&P\ar[r]_-{\D\di\alpha\varphi}
&\di M.
}$$
It is an easy exercise (\cf \S\,0.) to show that such endomorphisms have
trivial square\,:
$$x^6=0.$$

Therefore, $\D h:=1+\half\,x$ is an automorphism of $M$. Relation (10)
says that $\dis m\dis x=-x\,m$. This and the fact that $\dis
m=(1+x)\,m$ insures that 
$$\dis m\,\dis h=\dis m\dis{(1+\half x)}=\dis m+{\dis m\dis x\over 2}=(1+x)\,m-{x\,m\over 2}=
(1+\half\,x)\,m=h\,m.$$
In other words, the following diagram commutes (since, of course, we
still have $\alpha\,h=\alpha$ by (9))\,:
$$\xymatrix@C=40pt{
\dis M\ar[r]^-{\D m}\ar[d]_{\D\dis h}^{\D\simeq}
&M\ar[r]^-{\D\alpha}\ar[d]_{\D h}^{\D\simeq}
&P\ar[r]^-{\D\di\alpha\varphi}\ar[d]_{\D\varphi}^{\D\simeq}
&\di M\ar[d]_{\D\di h}^{\D\simeq}
\cr
\dis M\ar[r]_-{\D\dis m}
&M\ar[r]_-{\D\varphi\alpha}
&\di P\ar[r]_-{\D\di\alpha}
&\di M.
}$$
Lemma 2.1 insures that $(P,\varphi)$ is neutral.
\qed

\rem
The same proof goes through for a skew-duality.

\exo
Establish how a stably metabolic space in the usual sense
becomes neutral in the derived category. The above proof gives an
explicit way to find the lagrangian. The latter will not be (of
course) a complex concentrated in degree 0.






\vfill\eject

\paragraph{The sub-lagrangian construction}

\espaceavanttruc
In this paragraph, we fix the triangulated category $K$ and its
duality $\#$. The case of a skew-duality can be treated as well and
is left to the reader.

\truc{Presentation of the problem}

The classical theory of symmetric spaces as well as some
technical problems we shall encounter in the subsequent paragraphs
lead us to consider the following questions.
\vskip2mm

Consider a symmetric space $(P,\varphi)$. Suppose you have a
sub-lagrangian $(L,\nu_1)$, \ie $L$ is an object and $\nu_1:L\to P$ is
a morphism such that
$$\di\nu_1\varphi\,\nu_1=0.$$
Here are natural questions\,:
\roster
\item Can we {\it correctly} define $\orth L$, the orthogonal of $L$\,?

\item Can we map $L\to\orth L$?

\item Can we endow $\cone(L\to\orth L)$ with a structure of
symmetric space Witt-equivalent to $(P,\varphi)$?
\endroster

The first mental step is to renounce to the constraint for $L$ or for
$\orth L$ to be subobjects of $P$, as well as $L$ to be a subobject of
$\orth L$. In triangulated categories, it is
definitely a too strong hypothesis since any monomorphism (\ie any
morphism $\alpha$ such that $\alpha\,\beta=0$ forces $\beta=0$) is a
split inclusion as a direct summand. In other words, this would be the
same as forgetting the triangulated structure and focusing only on
the additive one.

In the classical case, $\orth L$ is defined to be the kernel of
$\di\nu_1\circ\varphi$. Since we know that $T\inv\big(\cone(u)\big)$
is a weak kernel of $u$, the natural analogue of $\orth L$ would be here 
$$T\inv\big(\cone(\di\nu_1\circ\varphi)\big)$$
which is the same as $\di{\big(\cone(\nu_1)\big)}$ as can be deduced
from exactness of $\#$ (\cf definition 1.2). In other words, if
we choose an exact triangle containing $\nu_1$\,:
$$\xytriangle L;P;C;\nu_1;\nu_2;\nu_3;$$
then $\orth L=\di C$. Therefore, we prefer to introduce directly the
following exact triangle (setting, if you prefer $\nu_0=-T\inv(\nu_3)$
and $M=\di C$)\,:
$$\lab2\Delta\spxytriangle T\inv(\di M);L;P;\di M;\nu_0;\nu_1;\nu_2;$$
in which $M$ is already the orthogonal of $L$.
%
Then, dualizing the above triangle, we get\,:
$$\lab{1.5}{\di\Delta}%
\xymatrix{%
T\inv(\di L)\ar[rr]^-{\D T\inv(\di\nu_0)}
&&M\ar[r]^-{\D\di\nu_2}
&\di P\ar[r]^-{\D\di\nu_1}
&\di L.}$$
This allows us to understand the more precise statement
$(L,\nu_1)^\bot=(M,\varphi\inv\di\nu_2)$.

We are going to construct a morphism of triangles between $\Delta$ and
$\di\Delta$. Its first part is given by $\varphi$\,:
$$\xymatrix@C=45pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}
&L\ar[r]^-{\D\nu_1}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}
&M\ar[r]_-{\D\di\nu_2}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L.
}$$
Since $\di\nu_1\circ(\varphi\,\nu_1)=0$ and since $\di\Delta$ is
exact, there exists a morphism 
$\alpha\,:L\to M$ (\ie we map $L$ in its orthogonal) such that
$\varphi\,\nu_1=\di\nu_2\,\alpha$ (see below). The third axiom of triangulated
categories insures us of the existence of a third morphism
$\beta\,:\di M\to \di L$ such that the following diagram commutes\,:
$$\xymatrix@C=45pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar@{-->}[d]_{\D T\inv(\beta)}
&L\ar[r]^-{\D\nu_1}\ar[d]_{\D\alpha}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar@{-->}[d]^{\D\exists\,\beta}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}
&M\ar[r]_-{\D\di\nu_2}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L.
}$$
Of course, we would prefer to have $\beta=\di\alpha$. This is easy to
obtain. It suffices to define
$$\eta_0:=\half(\alpha+\di\beta)$$
and to check, using the above diagram, that the following diagram commutes\,:
$$\xymatrix@C=45pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di\eta_0)}
&L\ar[r]^-{\D\nu_1}\ar[d]_{\D\eta_0}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar[d]^{\D\di\eta_0}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}
&M\ar[r]_-{\D\di\nu_2}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L.
}$$

In other words, we have mapped $L$ reasonably in its orthogonal, $M$. 
Now, we introduce the cone of this map. That is\,: we consider an
exact triangle over $\eta_0$\,:
$$\xytriangle L;M;Q;\eta_0;\eta_1;\eta_2;$$
and we use it (and its dual) to extend the last diagram to the following
key-diagram, which will be used several times in the subsequent
pages\,:
$$\lab1\Omega%
\vcenter{\xymatrix@C=45pt @R=40pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di\eta_0)}
&L\ar[r]^-{\D\nu_1}\ar[d]^{\D\eta_0}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar[d]^{\D\di\eta_0}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}\ar[d]_{\D\di\eta_2}
&M\ar[r]_-{\D\di\nu_2}\ar[d]^{\D\eta_1}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L
\cr
\di Q\ar[d]_{\D\di\eta_1}
&Q\ar[d]^{\D\eta_2}
\cr
\di M\ar[r]_-{\D T(\nu_0)}
&T(L).
}}$$

\truc{The question}

Of course, if you consider a diagram like $\Omega$, $(L,\nu_1)$ is
necessarily a sub-lagrangian of $(P,\varphi)$. Therefore, the ideal
sub-lagrangian question is now\,: 

{\leftskip=30pt\rightskip=30pt\it\noindent Given a commutative diagram $\Omega$ with exact lines
and columns, can we construct a symmetric form $\psi:Q\isoto\di Q$
such that $(P,\varphi)$ and $(Q,\psi)$ are Witt-equivalent\,?\par}
%
More precisely, we could also ask $\psi$ to fit in the above
diagram. As an exercise, the reader should try to solve the problem
for $L=0$. Even in this very case, it is immediate that we
cannot expect any form
making the above diagram commute to be Witt-equivalent to
$(P,\varphi)$. It appears also from this example that one cannot
simply complete the above diagram using the third axiom. The induced
morphism $\di Q\to Q$ might very well not be an isomorphism (can even be
zero). For these reasons and also for technical ones appearing in the
proof, it seems very important to consider the morphism\,:
$$s:=\nu_2\,\varphi\inv\di\nu_2\,:\,M\to\di M.$$
This is, in some
sense, the symmetric map induced by $\varphi$ on $M=\orth L$ (which is
sent in $P$ through $\varphi\inv\di\nu_2$). This morphism will play a
central role hereafter. For some reasons, the form $\psi$ we shall be
able to construct and that will fit in $\Omega$ will actually give the
opposite of $[P,\varphi]$ in the Witt group. Of course, this answers
also the question since you can replace $\psi$ by $-\psi$.

Observe that the choice of a diagram $\Omega$ is more than the
sub-lagrangian hypothesis, because there might be several such
morphisms $\eta_0$. The question is actually quite complicated because
of that precise point. In the application we shall make of this
sub-lagrangian construction to the 12-term sequence, we will have the
diagram $\Omega$ imposed by the opponent. The result goes as follows.

Given a sub-lagrangian $(L,\nu_1)$ of a symmetric space $(P,\varphi)$,
there exists always a way to construct a {\it good} $\eta_0$, that is
an $\eta_0$ such that the
corresponding diagram $\Omega$ can be completed and such that the form
on $Q$ is Witt-equivalent to the opposite of
$(P,\varphi)$. If, moreover, the morphism $\eta_0$ is imposed,
then there exists a way to modify it into a {\it good} one (in the above
sense) without modifying its cone too much. Since we proved in 3.1,
that there always exists an $\eta_0$  (not necessarily good), the
first assertion is a consequence of the second. This explains the
weird presentation of the results (theorem 3.17 and corollary 3.24).
\vskip2mm
As mentioned in the introduction, we were not able to establish the
sub-lagrangian construction without some finiteness hypothesis, called
here {\it noetherianity}. Let us introduce it.


\goodbreak

\defi
We say that a triangulated category $K$ is {\it \noeth} if it
satisfies the following two properties\,:

\roster
\item If $A$ and $B$ are objects in $K$ such that there exists an
isomorphism $A\oplus B\simeq A$ then $B\simeq 0$.
\item For any pair of fixed objects $A,\ B$ in
$K$ and for any sequence of exact triangles
$$\Big(\xytriangle A;B;C_i;u_i;v_i;w_i;
\Big)_{i\in\N}$$
over $A,\ B$ such that there exists split injections
$$\Big(f_i:C_i\hookrightarrow C_{i+1}\Big)_{i\in\N}$$
verifying $w_{i+1}\,f_i=w_i$ for all $i$,
this sequence of inclusions as direct summands
$$C_1\mathop{\hookrightarrow}_{f_1}
C_2\mathop{\hookrightarrow}_{f_2}
\cdots
C_i\mathop{\hookrightarrow}_{f_i}
C_{i+1}\mathop{\hookrightarrow}
\cdots$$
is stationary. In other words, there exists $n\in\N$ such that $f_i$
is an isomorphism for all $i\geq n$.
\endroster

\rem
It is an open question to give a good definition of
noetherian triangulated categories. The one given here has the
following property\,: Let $X$ be a noetherian scheme and consider
the derived category of bounded complexes of coherent $\Cal
O_{X}$-modules (resp.\ locally free $\Cal O_{X}$-modules of finite
type). Then a simple argument using long exact homology sequences
allows us to prove that this category is \noeth\ in the sense of
definition 3.3. The definition would be ``good'' if the converse was
true.
Anyway, for the future applications to schemes and open subschemes, it
will be sufficient to restrict ourselves to noetherian schemes to be
able to use the results of this paragraph. 

Under this noetherianity assumption, we can explain how the completion
of diagram $\Omega$ is sufficient to have a Witt-equivalent form on
the cone of $L\too\orth L$.

\thm
Let $(K,\#)$ be a \noeth\ triangulated category with duality.
Let $(P,\varphi)$ be a symmetric space and $(L,\nu_1)$ be a
sub-lagrangian of this space. Suppose that you have a commutative diagram with
exact rows and columns\,:
$$\lab1{\bar\Omega}%
\vcenter{\xymatrix@C=45pt @R=40pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di\eta_0)}
&L\ar[r]^-{\D\nu_1}\ar[d]^{\D\eta_0}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar[d]^{\D\di\eta_0}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}\ar[d]_{\D\di\eta_2}
&M\ar[r]_-{\D\di\nu_2}\ar[d]^{\D\eta_1}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L
\cr
\di Q\ar[d]_{\D\di\eta_1}
&Q\ar[d]^{\D\eta_2}\ar[l]_-{\D\psi=\di\psi}^-{\D\simeq}
\cr
\di M\ar[r]_-{\D T(\nu_0)}
&T(L)
}}$$
such that
$$\di\eta_1\psi\,\eta_1=-\nu_2\,\varphi\inv\di\nu_2\,:M\to\di M.$$
Then, $[P,\varphi]=[Q,-\psi]$ in $\W(K).$
\endthm

\proof
The proof is quite a long way and we shall establish partial results
first. As already asserted above, the symmetric morphism
$\nu_2\,\varphi\inv\di\nu_2$ plays a important role and we give it a
name\,:
$$s:=\nu_2\,\varphi\inv\di\nu_2\ :\ M\too\di M$$
once for all.
We shall also consider the morphism $\eta_0\,\nu_0:T\inv(\di
M)\to M$ which is symmetric for the skew-duality $T\inv\circ\#$. We
adopt the following notations\,:
$$*=T\inv\circ\#$$
and $w:=\eta_0\,\nu_0=\dis w:\dis M\to M$.
It is clear that
$\big(M,\pmatrix\varphi\inv\di\nu_2\\\eta_1\endpmatrix\big)$ is a
{\it sub}-lagrangian of $(P,\varphi)\bot(Q,\psi)$ but it is rather hard to
remove the prefix.

\vskip5mm

The first lemmas ignore the lower part of $\bar\Omega$ (\ie the
form $\psi$) and focus on $\Omega$.




\lem
Given a diagram $\Omega$ like in 3.1, there exists a morphism $y\,:Q\to\di M$ such
that
\roster
\item $T(\nu_0)\,y=\eta_2$;
\item the following triangle is exact where $w=\eta_0\,\nu_0$\,:
$$\xymatrix{T\inv(\di M)\ar[r]^-{\D w}
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
&&P\oplus Q\ar[rr]^-{\pmatrix \nu_2 & y\endpmatrix}
&&\di M.
}$$
\endroster

\proof
>From the commutativity of $\Omega$, we have
$\varphi\,\nu_1\,T\inv(\eta_2)=\di\nu_2\,\eta_0\,T\inv(\eta_2)=0$ and
since $\varphi$ is an isomorphism, we have
$$\nu_1\circ\big(-T\inv(\eta_2)\big)=0.$$
Apply the forth axiom of triangulated categories to get the following
octahedron, in which we immediately used the commutativity of the faces
containing $\pmatrix 1\\0\endpmatrix$ and $\pmatrix
0&1\endpmatrix$\,:

$$\xymatrix@C=35pt @R=40pt{%
&&&P\ar[ld]|-{\D\nu_2}
\ar[rrdd]^-{\pmatrix 1\\0\endpmatrix}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
L\ar[rrru]^-{\D\nu_1}
\ar[rrdd]_-{\D\eta_0}
&&\di M\ar[ll]_(.35){\D-T(\nu_0)}
\ar[dd]|-{\D-T(\eta_0)\,T(\nu_0)}
\\
&&&T\inv(Q)\ar[lllu]|!{[lu];[ld]}\hole|(.65){\D-T\inv(\eta_2)}
\ar[uu]|!{[rr];[lu]}\hole_(.55){\D 0}
&&P\oplus Q\ar[ll]|(.55){\pmatrix 0&1\endpmatrix}
\ar[lllu]_-{\D\exists\pmatrix \nu_2 & y\endpmatrix}
\\
&&M\ar[ru]|-{\D\eta_1}
\ar[rrru]_-{\D\exists\pmatrix x\\ \eta_1\endpmatrix}
}$$
for some $x\,:M\to P$ and $y\,:Q\to \di M$ such that the triangle
$$\xymatrix{T\inv(\di M)\ar[r]^-{\D w}
&M\ar[r]^-{\pmatrix x\\ \eta_1\endpmatrix}
&P\oplus Q\ar[rr]^-{\pmatrix \nu_2 & y\endpmatrix}
&&\di M
}$$
is exact (recall that $w=\eta_0\,\nu_0$) and such that
$$\eqalign{x\,\eta_0&=\nu_1\cr
T(\nu_0)\,y&=\eta_2.}$$
Observe that $x\,\eta_0=\nu_1=\varphi\inv\di\nu_2\eta_0$ by $\Omega$. In
other words, $(x-\varphi\inv\di\nu_2)\,\eta_0=0$. Using the exact
triangle containing $\eta_0$, we find some $a\,:Q\to P$ such that
$$x-\varphi\inv\di\nu_2=a\,\eta_1.$$
This can also be expressed by saying that the following diagram commutes\,:
$$\xymatrix{T\inv(\di M)\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix x\\ \eta_1\endpmatrix}\ar@{=}[d]
&&P\oplus Q\ar[rrr]^-{\pmatrix \nu_2 & y\endpmatrix}
\ar[d]^-{\pmatrix 1&-a\\0&1\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
T\inv(\di M)\ar[r]_-{\D w}
&M\ar[rr]_-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
&&P\oplus Q\ar[rrr]_-{\pmatrix \nu_2 & y+\nu_2\,a\endpmatrix}
&&&\di M.
}$$
But then, since the first triangle is exact, so is the second one. It
suffices now to replace $y$ by $\bar y:=y+\nu_2\,a$ which still satisfies
$$T(\nu_0)\circ\bar y=
\underbrace{T(\nu_0)\,y}_{\D=\eta_2}+\underbrace{T(\nu_0)\,\nu_2}_{\D=0}\,a
=\eta_2.$$
This argument will be used again later with less details.
\qed

\lem
Consider the diagram $\Omega$ and suppose moreover that
$$\nu_2\,\varphi\inv\di\nu_2=0.$$
Then $\Omega$ can be completed in an $\bar\Omega$ like in theorem
3.5 in such a way that the following triangle is exact\,:
$$\xymatrix{T\inv(\di M)\ar[r]^-{\D w}
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
&&P\oplus Q\ar[rr]^-{\pmatrix \nu_2 & \di\eta_1\psi\endpmatrix}
&&\di M
}$$
where $w=\eta_0\,\nu_0$. In particular, $(P,\varphi)$ and $(Q,-\psi)$ are Witt-equivalent.

\proof
By the previous lemma, we have an exact triangle
$$\lab10\xymatrix{\dis M\ar[r]^-{\D w}
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
&&P\oplus Q\ar[rr]^-{\pmatrix \nu_2 & y\endpmatrix}
&&\di M
}$$
for some morphism $y\,:Q\to\di M$ such that
$T(\nu_0)\,y=\eta_2$. Since $\dis w=w$, we can consider
$\chi:=\cone(\dis M,w)$ as defined in 1.10, \ie a neutral symmetric form
$$\chi=\pmatrix a&b\\ \di b& c\endpmatrix=\di\chi$$
such that the following diagram commutes (in which the second line is
the dual of the first)\,:
$$\lab11\vcenter{\xymatrix{\dis M\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}\ar@{=}[d]
&&P\oplus Q\ar[rrr]^-{\pmatrix \nu_2 & y\endpmatrix}
\ar[d]^-{\pmatrix a&b\\ \di b&c\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
\dis M\ar[r]_-{\D \dis w}
&M\ar[rr]_-{\pmatrix \di\nu_2\\ \di y\endpmatrix}
&&\di P\oplus\di Q\ar[rrr]_-{\pmatrix \nu_2\,\varphi\inv & \di\eta_1\endpmatrix}
&&&\di M.
}}$$
We have $a=\di a$, $c=\di c$ and
$$\eqalign{a\,\varphi\inv\di\nu_2+b\,\eta_1&\lab42=\di\nu_2
\cr
\di b\varphi\inv\di\nu_2+c\,\eta_1&\lab43=\di y.
}$$
%
Composing (2) on the right with $\eta_0$ and using $\eta_1\,\eta_0=0$,
we have
$$a\,\underbrace{\varphi\inv\di\nu_2\eta_0}_{\D=\nu_1}=\di\nu_2\eta_0=
\varphi\nu_1$$
by $\Omega$. In other words, $(a-\varphi)\,\nu_1=0$ which means that
there exists a morphism $d\,:\di M\to \di P$ such that
$$a=\varphi+d\,\nu_2.$$
Therefore, using our extra hypothesis that $\nu_2\,\varphi\inv\di\nu_2=0$,
we have\,:
$$a\,\varphi\inv\di\nu_2=(\varphi+d\,\nu_2)\,\varphi\inv\di\nu_2=
\di\nu_2+d\,\underbrace{\nu_2\,\varphi\inv\di\nu_2}_{\D=0}=\di\nu_2.$$
This last equality compared with (2) forces
\vskip2mm
$$\lab14b\,\eta_1=0.$$
%
Putting (3) and (4) together, we obtain the commutativity of the
following diagram \big(compare with (1)\big)\,:
$$\lab15\vcenter{\xymatrix{\dis M\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}\ar@{=}[d]
&&P\oplus Q\ar[rrr]^-{\pmatrix \nu_2 & y\endpmatrix}
\ar[d]^-{\pmatrix \varphi&b\\ \di b&c\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
\dis M\ar[r]_-{\D \dis w}
&M\ar[rr]_-{\pmatrix \di\nu_2\\ \di y\endpmatrix}
&&\di P\oplus\di Q\ar[rrr]_-{\pmatrix \nu_2\,\varphi\inv & \di\eta_1\endpmatrix}
&&&\di M.
}}$$
In other words, we can replace $a$ by $\varphi$ in (1). By definition,
the following form over $P\oplus Q$ is neutral\,:
$$\pmatrix\varphi&b\\ \di b&c\endpmatrix.$$
But, of course, we can get rid of the $b$ through an isometry of the type\,:
$$h:=\pmatrix1&-\varphi\inv b\\0&1\endpmatrix\,:\ P\oplus Q\too^\sim P\oplus Q.$$
This gives the following. Define $\psi\,:Q\to\di Q$ to be
$$\psi:=c-\di b\varphi\inv b$$
and check that we have the following isometry\,:
$$\lab16\pmatrix1&0\\-\di b\varphi\inv&1\endpmatrix\cdot
\pmatrix\varphi&b\\ \di b&c\endpmatrix\cdot
\pmatrix1&-\varphi\inv b\\0&1\endpmatrix=
\pmatrix\varphi&0\\0&\psi\endpmatrix.$$
Use this automorphism $h$ of $P\oplus Q$ to modify the triangle (0)\,:
$$\xymatrix@R=30pt{\dis M\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}\ar@{=}[d]
&&P\oplus Q\ar[rrr]^-{\pmatrix \nu_2 & y\endpmatrix}
\ar[d]^-{\D h\inv=\pmatrix 1&\varphi\inv b\\0&1\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
\dis M\ar[r]_-{\D w}
&M\ar[rr]_-{\D w_1}
&&P\oplus Q\ar[rrr]_-{\D w_2}
&&&\di M.
}$$
The first line is exact triangle (0) and therefore the second is
exact, defining $w_1$ and $w_2$ such that the diagram commutes.
By (4), a direct computation gives\,:
$$w_1=\pmatrix1&\varphi\inv b\\0&1\endpmatrix\cdot\pmatrix
\varphi\inv\di\nu_2\\ \eta_1\endpmatrix=
\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix.$$
Now, the easiest way to compute $w_2$ is by using the fact that the
induced form is $\varphi\bot\psi$, \ie concretely\,:
$$\eqalign{w_2&=\pmatrix \nu_2 & y\endpmatrix\circ h
\equal{5}
\pmatrix \nu_2\,\varphi\inv & \di\eta_1\endpmatrix\cdot
\pmatrix\varphi&b\\ \di b&c\endpmatrix\circ h=
\cr
&\equal{4}
\pmatrix \nu_2\,\varphi\inv & \di\eta_1\endpmatrix\cdot
\pmatrix1&0\\-\di b\varphi\inv&1\endpmatrix\cdot
\pmatrix\varphi&b\\ \di b&c\endpmatrix\circ h=
\cr
&\equal{6}
\pmatrix \nu_2\,\varphi\inv & \di\eta_1\endpmatrix\cdot
\pmatrix \varphi&0\\0&\psi\endpmatrix=
\pmatrix \nu_2& \di\eta_1\psi\endpmatrix.
}$$
This gives the announced triangle.
\vskip2mm

Use this triangle and condition (2) of lemma 2.1 to understand that
$\varphi\bot\psi$ is neutral. Then
$[P,\varphi]=-[Q,\psi]=[Q,-\psi]$ which is the result.
\qed

\lem
Let $(P,\varphi)$ be a symmetric space in $K$ such that
there exists an exact triangle
$$\spxytriangle T\inv(\di L);L;P;\di L;z;\alpha;\beta;$$
satisfying all the following properties\,:
\roster
\item $\di \alpha\varphi\,\alpha=0$
\item $\beta\,\varphi\inv\,\di \beta=0$
\item $T\inv(\di z)=z$.
\endroster
Then, there exists a morphism $\xi:L\to L$, a morphism $\eta:L\to Q$,
where $Q=\cone(\xi)$ and a symmetric form $\psi$ over $Q$ such that
the following triangle is exact\,:
$$\xymatrix{T\inv(\di L)\ar[r]^-{\D w}
&L\ar[rr]^-{\pmatrix \varphi\inv\di\beta\\ \eta\endpmatrix}
&&P\oplus Q\ar[rr]^-{\pmatrix \beta & \di\eta\psi\endpmatrix}
&&\di L
}$$
where $w=\xi\,z$ is moreover symmetric\,: $T\inv(\di w)=w$.

In particular $(P,\varphi)$ and $(Q,-\psi)$ are Witt-equivalent.

\proof
This is an immediate corollary of the previous lemma and of the fact
that there exists a diagram $\Omega$ by discussion 3.1 applied to the case where $M=L$. 
This was only stated to make the proof of the next proposition easier.
\qed


\prop
Let $K$ be \noeth. Let
$(P,\varphi)$ be a symmetric space in $K$. Suppose that
there exists an exact triangle
$$\spxytriangle T\inv(\di L);L;P;\di L;z;\alpha;\beta;$$
such that all the following properties are satisfied\,:
\roster
\item $\di \alpha\varphi\,\alpha=0$
\item $\beta\,\varphi\inv\,\di \beta=0$
\item $T\inv(\di z)=z$.
\endroster
Then, $(P,\varphi)$ is neutral.

\proof
We are going to apply several times the previous lemma. It is not
necessary but more convenient to proceed {\sl ab absurdo}. So, let us
suppose that $(P,\varphi)$ is {\it not} neutral. By theorem 2.5, this is
the same as saying that $(P,\varphi)$ is not stably neutral.
\vskip2mm
Let us first of all introduce a neutral form on $P$ that will be
useful in the proof. Recall that, since $\dis z=z$, there exists a
neutral symmetric structure on $P$ defined by\,:
$$\lab24\vcenter{\xymatrix
{T\inv(\di L)\ar[r]^-{\D z}\ar@{=}[d]
&L\ar[r]^-{\D\alpha}\ar@{=}[d]
&P\ar[r]^-{\D\beta}\ar[d]_{\D\simeq}^{\D\chi=\di\chi}
&\di L\ar@{=}[d]
\cr
T\inv(\di L)\ar[r]_-{\D\dis z}
&L\ar[r]_-{\D\di\beta}
&\di P\ar[r]_-{\D\di\alpha}
&\di L.
}}$$
Keep this in mind.
\vskip2mm



By lemma 3.11, formally applied, there exists a morphism $\xi_1:L\to
L$, an object $Q_1$ (the cone of $\xi_1$), a morphism $\eta_1:L\to Q_1$ and
a symmetric form $\psi_1$ on $Q_1$ such that the following triangle
is exact\,:
$$\lab15
\xymatrix{T\inv(\di L)\ar[r]^-{\D z_1}
&L\ar[rr]^-{\pmatrix \varphi\inv\di\beta\\ \eta_1\endpmatrix}
&&P\oplus Q_1\ar[rr]^-{\pmatrix \beta & \di\eta_1\psi_1\endpmatrix}
&&\di L
}$$
where $z_1=\xi_1\,z=\dis z_1$. Moreover, $(Q_1,\psi_1)$ is Witt-equivalent to
$(P,-\varphi)$ and therefore is not neutral by hypothesis. Call this
$$\lab26(Q_1,\psi_1)\kern5mm\hbox{is\ not\ neutral.}$$
In particular, $Q_1$ is not zero.
%
Note that $0\equal{5}
\pmatrix \beta & \di\eta_1\psi_1\endpmatrix\cdot
\pmatrix \varphi\inv\di\beta\\ \eta_1\endpmatrix=
\beta\,\varphi\inv\di\beta+\di\eta_1\psi_1\,\eta_1
\equal{2}\di\eta_1\psi_1\,\eta_1.$ Call this 
$$\lab27\di\eta_1\psi_1\,\eta_1=0.$$
\vskip2mm

Define $P_1:=P\oplus Q_1$ and endow it with the form 
$$\varphi_1:=\chi\bot\psi_1$$
(pay attention\,: the neutral form $\chi$ replaces $\varphi$ on $P$\,!). Keep an eye on
triangle (5) while defining
$$\alpha_1:=\pmatrix \varphi\inv\di\beta\\ \eta_1\endpmatrix
\link1{and}
\beta_1:=\pmatrix \beta & \di\eta_1\psi_1\endpmatrix.$$
We want to check that $P_1,\varphi_1,L,z_1,\alpha_1,\beta_1$ satisfy
the hypothesis of the proposition, in place of
$P,\varphi,L,z,\alpha$ and $\beta$. First, the exact triangle is simply
(5). We already have property (3)\,: $\dis z_1=z_1$ and it suffices to check that 
$\di\alpha_1\varphi_1\,\alpha_1=0$ and that
$\beta_1\,\varphi_1\inv\,\di \beta_1=0$. This is easy. Recall that (2)
says that $\beta\circ(\varphi\inv\,\di \beta)=0$. Then, exactness of the
triangle of the proposition insures the existence of a morphism
$\mu:L\to L$ such that $\varphi\inv\,\di \beta=\alpha\,\mu$. Then
$$\lab18\di\alpha_1\varphi_1\,\alpha_1=
(\kern-2mm\underbrace{\beta\,\varphi\inv}_{\D=\di{(\alpha\,\mu)}}\kern-2mm)\,\chi\,
(\underbrace{\,\varphi\inv\di\beta}_{\D\vphantom{\di{a}}=\alpha\,\mu})+
\underbrace{\di\eta_1\psi_1\,\eta_1}_{\D\equal{7}0}
=\di\mu\underbrace{\di\alpha\chi\,\alpha}_{\D\equal{4}0}\mu=0$$
and
$$\lab19\beta_1\,\varphi_1\inv\,\di \beta_1=
\underbrace{\beta\,\chi\inv\di\beta}_{\D\equal{4}0}
+\underbrace{\di\eta_1\psi_1\eta_1}_{\D\equal{7}0}=0.
$$
Now we can apply the previous lemma to this new space and this new
triangle.

So there exists a morphism $\xi_2:L\to L$, an object $Q_2$ (the cone
of $\xi_2$), a morphism $\eta_2:L\to Q_2$ and a symmetric form
$\psi_2$ over $Q_2$ such that the following triangle is exact\,:

$$\lab1{10}
\xymatrix{T\inv(\di L)\ar[r]^-{\D z_2}
&L\ar[rr]^-{\pmatrix \varphi_1\inv\di\beta_1\\ \eta_2\endpmatrix}
&&P_1\oplus Q_2\ar[rr]^-{\pmatrix \beta_1 & \di\eta_2\psi_2\endpmatrix}
&&\di L
}$$
where $z_2=\xi_2\,z_1=\dis z_2$. Note that $(P_1,\varphi_1)$ cannot be
neutral, otherwise (since $\chi$ is) so  would be $(Q_1,\psi_1)$,
in contradiction with (6). Moreover, $(Q_2,\psi_2)$ is Witt-equivalent
to $(P_1,-\varphi_1)$ and therefore isn't neutral neither. Call this 
$$\lab2{11}(Q_2,\psi_2)\kern5mm\hbox{is\ not\ neutral.}$$
In particular, $Q_2$ is not zero. But we can deduce something else
from the fact that $(Q_2,\psi_2)$ is Witt-equivalent to
$(P_1,-\varphi_1)$, namely that
$$\underbrace{\chi\bot\psi_1}_{\D=\varphi_1}
\bot\psi_2\kern1cm\hbox{is neutral}$$
and therefore, since $\chi$ is neutral, we have that
$$\lab1{12}\psi_1\bot\psi_2\kern1cm\hbox{is neutral.}$$
%
A last property that we can get from (10) is that
$$\lab1{13}0=\pmatrix \beta_1 & \di\eta_2\psi_2\endpmatrix\cdot
\pmatrix\varphi_1\inv\di\beta_1\\ \eta_2\endpmatrix=
\underbrace{\beta_1\,\varphi_1\inv\di\beta_1}_{\D\equal{9}0}
+\di\eta_2\psi_2\,\eta_2=\di\eta_2\psi_2\,\eta_2.$$


Finally, a direct computation gives that
$\varphi_1\inv\di\beta_1=\pmatrix\chi\inv\di\beta \\
\eta_1\endpmatrix=\pmatrix\alpha \\
\eta_1\endpmatrix$, by (4).
%
In other words, replacing this in (10) gives the following exact triangle\,:
$$\lab1{14}
\xymatrix{T\inv(\di L)\ar[r]^-{\D z_2}
&L\ar[r]^-{\pmatrix \alpha \\ \eta_1\\ \eta_2\endpmatrix}
&P\oplus Q_1\oplus Q_2
\ar[rrr]^-{\pmatrix \beta& \di\eta_1\psi_1 & \di\eta_2\psi_2\endpmatrix}
&&&\di L.
}$$
Using (12), $\varphi\bot\psi_1\bot\psi_2$ is not neutral, otherwise so
would be $\varphi$. But this new form $\varphi\bot\psi_1\bot\psi_2$ 
satisfies again the hypothesis of
the proposition. In fact, the exact triangle is (14) and since
$z_2=\dis z_2$, we only have to check conditions (1) and (2). These
are all immediate consequences of (1), (2), (7) and (13). The good
reader checks this carefully.
\vskip5mm

Consider this as the first step of the proof. To a (non-neutral)
symmetric space 
like in the proposition, we associate these two spaces $(Q_1,\psi_1)$
and $(Q_2,\psi_2)$ and an exact triangle
$$\xymatrix{T\inv(\di L)\ar[r]^-{\D z_2}
&L\ar[rr]^-{\D\alpha_2:=\pmatrix \alpha \\ \eta_1\\ \eta_2\endpmatrix}
&&P\oplus Q_1\oplus Q_2
\ar[rrrr]^-{\D\beta_2:=\pmatrix \beta& \di\eta_1\psi_1 & \di\eta_2\psi_2\endpmatrix}
&&&&\di L
}$$
where the three morphisms $z_2$, $\alpha_2$, $\beta_2$ satisfy the
same hypotheses (1), (2) and (3) for the symmetric space
$$\Big(P\oplus Q_1\oplus Q_2,\varphi\bot\psi_1\bot\psi_2\Big).$$
This space is not neutral. Moreover, $Q_1$ and $Q_2$ are not zero.
\vskip2mm

Then, of course, do this again and again. You  will get an infinite
tower of spaces always fitting in triangles over $T\inv(\di L)$ and $L$ and
such that you have the compatibility condition of definition
3.3 ($w_{i+1}\,f_i=w_i$). To see this, observe that the first
component of $\beta_2$ is $\beta$ (this was wrong for $\beta_1$ and is
the reason why we used twice lemma 3.11\,!).

The noetherianity condition insures that this tower should stop, that
is some $Q_i=0$, which leads to a contradiction.
\qed



\truc{Proof of theorem 3.5}
Hereafter we shall label by (1) the first hypothesis of theorem
3.5, namely the fact that we have the commutative diagram $\bar\Omega$
with exact lines and columns.

The second hypothesis of theorem 3.5 will be labeled as
$$\lab22\di\eta_1\psi\,\eta_1=-\nu_2\,\varphi\inv\di\nu_2.$$

Forgetting for a while the form over $Q$, we have a diagram $\Omega$
for $(P,\varphi)$. Therefore, lemma 3.7 insures the existence of an
exact triangle\,:
$$\lab13\xymatrix{T\inv(\di M)\ar[r]^-{\D w}
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
&&P\oplus Q\ar[rr]^-{\pmatrix \nu_2 & y\endpmatrix}
&&\di M
}$$
for some morphism $y:Q\to\di M$ such that
$$\lab14T(\nu_0)\,y=\eta_2.$$
Recall that $w=\eta_0\,\nu_0=\dis w$.

By the above proposition, it suffices to prove that\,:
$$\lab1i\di\alpha(\varphi\bot\psi)\,\alpha=0
\link1{where}\alpha:=\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix$$
and
$$\lab1{ii}\beta\,(\varphi\bot\psi)\inv\,\di\beta=0
\link1{where}\beta:=\pmatrix \nu_2 & y\endpmatrix.$$
That is what we are going to verify.
\vskip2mm

Condition (i) is immediate\,:
$$\di\alpha(\varphi\bot\psi)\,\alpha=
\pmatrix \nu_2\,\varphi\inv& \di\eta_1\endpmatrix\cdot
\pmatrix\varphi&0\\0&\psi\endpmatrix\cdot
\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix=
\nu_2\,\varphi\inv\di\nu_2+\di\eta_1\psi\,\eta_1
\equal{2}0.$$
\vskip2mm

Dualizing triangle (3) gives the first line of the following diagram,
which is an isomorphism of triangles\,:
$$\xymatrix{T\inv(\di M)\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix \di\nu_2\\ \di y\endpmatrix}\ar@{=}[d]
&&\di P\oplus \di Q\ar[rrr]^-{\pmatrix \nu_2\,\varphi\inv & \di\eta_1\endpmatrix}
\ar[d]^-{\pmatrix \varphi\inv&0\\0&\psi\inv\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
T\inv(\di M)\ar[r]_-{\D w}
&M\ar[rr]_-{\pmatrix \varphi\inv\di\nu_2\\ \psi\inv\di y\endpmatrix}
&&P\oplus Q\ar[rrr]_-{\pmatrix \nu_2 & \di\eta_1\psi\endpmatrix}
&&&\di M.
}$$
Then the second line is also exact. Call this exact triangle (5).
\vskip2mm

Now, from (4) and the lower commutative square of $\bar\Omega$, we
have that\,:
$$T(\nu_0)\,y=\eta_2=T(\nu_0)\,\di\eta_1\psi.$$
Therefore $T(\nu_0)\,(y-\di\eta_1\psi)=0$. This gives, using the exact
triangle over $T(\nu_0)$, the existence of a morphism
$$\lab26r\,:Q\to P\link1{such that}y=\di\eta_1\psi+\nu_2\,r.$$
This is the same as saying that the following diagram commutes\,:
$$\xymatrix{T\inv(\di M)\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
&&P\oplus Q\ar[rrr]^-{\pmatrix \nu_2 & y\endpmatrix}
\ar[d]^-{\pmatrix 1&r\\0&1\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
T\inv(\di M)\ar[r]_-{\D w}
&M\ar[rr]_-{\pmatrix \varphi\inv\di\nu_2\\ \psi\inv\di y\endpmatrix}
&&P\oplus Q\ar[rrr]_-{\pmatrix \nu_2 & \di\eta_1\psi\endpmatrix}
&&&\di M.
}$$
The first line of this diagram is exact triangle (3) and the second
one is exact triangle (5). It is then an axiom of triangulated
categories that you can complete this into a commutative diagram\,:
$$\lab17\vcenter{\xymatrix{T\inv(\di M)\ar[r]^-{\D w}\ar@{=}[d]
&M\ar[rr]^-{\pmatrix \varphi\inv\di\nu_2\\ \eta_1\endpmatrix}
\ar@{-->}[d]_{\D\exists\,}^{\D g}
&&P\oplus Q\ar[rrr]^-{\pmatrix \nu_2 & y\endpmatrix}
\ar[d]^-{\pmatrix 1&r\\0&1\endpmatrix}_\simeq
&&&\di M\ar@{=}[d]
\\
T\inv(\di M)\ar[r]_-{\D w}
&M\ar[rr]_-{\pmatrix \varphi\inv\di\nu_2\\ \psi\inv\di y\endpmatrix}
&&P\oplus Q\ar[rrr]_-{\pmatrix \nu_2 & \di\eta_1\psi\endpmatrix}
&&&\di M.
}}$$
and the morphism $g$ is necessarily an isomorphism. In particular, we
have from the middle commutative square\,:
$$\eqalign{\varphi\inv\di\nu_2+r\,\eta_1&\lab58=\varphi\inv\di\nu_2g
\cr
\eta_1&\lab59=\psi\inv\di y g.
}$$
%
Let $s:=\nu_2\,\varphi\inv\di\nu_2
\equal{2}-\di\eta_1\psi\,\eta_1$. Note that $\di
s=s$. Directly from (3), we have that
$$0=\pmatrix\nu_2& y\endpmatrix\cdot
\pmatrix\varphi\inv\di\nu_2\\ \eta_1\endpmatrix=
\nu_2\,\varphi\inv\di\nu_2+y\,\eta_1$$
and therefore
$$\lab2{10}y\,\eta_1=-s.$$
%
Then composing the equation (9) by $\di\eta_1\psi$ on the left and
using (10), we obtain\,:
$$\underbrace{\di\eta_1\psi\,\eta_1}_{\D=-s}=
\di\eta_1\psi\,\psi\inv\di y g=
\di{\big(\underbrace{y\,\eta_1}_{\D=-s}\big)}g=-s\,g$$
which means that
$$\lab2{11}s\,g=s.$$
%
Now, apply $\nu_2$ on the left to (8) to obtain\,:
$$\underbrace{\nu_2\,\varphi\inv\di\nu_2}_{\D=s}+\nu_2\,r\,\eta_1=
\underbrace{\nu_2\,\varphi\inv\di\nu_2}_{\D=s}\,g.$$
This, compared with (11), forces
$$\lab2{12}\nu_2\,r\,\eta_1=0.$$
%
We are now able to compute
$$\eqalign{y\,\psi\inv\di y&
\equal{6}(\di\eta_1\psi+\nu_2\,r)\,\psi\inv\di y
\equal{9}(\di\eta_1\psi+\nu_2\,r)\,(\eta_1\,g\inv)=\cr
&=\di\eta_1\psi\,\eta_1\,g\inv+\nu_2\,r\,\eta_1\,g\inv
\equal{12}\underbrace{\di\eta_1\psi\,\eta_1}_{\D=-s}\,g\inv
\equal{11}-s=\cr
&=-\nu_2\,\varphi\inv\di\nu_2.}$$
%
Using this last equation, we can now check condition (ii)\,:
$$\beta\circ(\varphi\bot\psi)\inv\circ\di\beta\ 
\equal{\text{def}}\ 
\pmatrix \nu_2 & y\endpmatrix\cdot
\pmatrix \varphi&0\\0&\psi\endpmatrix\inv\cdot
\pmatrix \di\nu_2\\\di y\endpmatrix
=\nu_2\,\varphi\inv\di\nu_2+y\,\psi\inv\di y=0.$$

This gives the result by proposition 3.13.\qed





\truc{Comment}
%
So far, we have established that given a diagram $\Omega$ which has
been completed in a diagram $\bar\Omega$ (meaning that we have
constructed a form on $Q$), then the new form is Witt-equivalent to
(the opposite of) the form we started with.


Unfortunately, this is not enough for the application we want to do of
this in the localization exact sequence (see lemma 5.3). We need to be able to
construct such a form on the cone of
\underbar{an}y\kern-3pt\underbar{\,} morphism
$\eta_0$ such that $\Omega$ commutes. We do not know if the latter is
true or not\,! Nevertheless, the following result will be sufficient.

\thm
Suppose that $K$ is a \noeth\ triangulated category.
Let $(P,\varphi)$ be a symmetric space and $(L,\nu_1)$ be a
sub-lagrangian of it. Choose an exact triangle over $\nu_1$\,:
$$\spxytriangle T\inv(\di M);L;P;\di M.;\nu_0;\nu_1;\nu_2;$$
Let $\eta_0:L\to M$ be a morphism such that the diagram $\Omega$ commutes
(\cf 3.1). Then there exists another morphism $\mu_0:L\to M$ such
that\,:
\roster
\item there exists a form on the cone of $\mu_0$ Witt-equivalent to
$(P,\varphi)$;
\item $\cone(\mu_0)$ and $\cone(\eta_0)$ are stably isomorphic, namely
$$P\oplus\cone(\mu_0)\simeq P\oplus\cone(\eta_0).$$
\endroster


\proof
First we shall consider two intermediate results. The proof is in 3.23.
Notation\,: once again $*=T\inv\circ\#$.


\lem
Under the hypothesis of theorem 3.17, there exists a morphism
$\lambda:L\to T\inv(\di L)$ such that
\roster
\item[3] $T\inv(\di\nu_0)\,\lambda\,\nu_0=0$
\item setting $\mu_0:=\eta_0+T\inv(\di\nu_0)\lambda$, the following triangle
is exact\,:
$$\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\ -\di\mu_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
&&&T(M)}$$
where $s=\nu_2\,\varphi\inv\di\nu_2$.
\endroster

\lem
Under the hypothesis of theorem 3.17, let $s=\nu_2\,\varphi\inv\di\nu_2$.
Given any morphism $\mu_0:L\to M$ such that the triangle
$$\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\ -\di\mu_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
&&&T(M)}$$
is exact and given any exact triangle over $\mu_0$\,:
$$\spxytriangle L;M;R;T(L),;\mu_0;\mu_1;\mu_2;$$
there exists a symmetric form $\psi$ on the cone $R$ of $\mu_0$ such
that the following diagram commutes\,: 
$$\lab1{\bar{\bar\Omega}}\vcenter{\xymatrix@C=45pt @R=40pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di\mu_0)}
&L\ar[r]^-{\D\nu_1}\ar[d]^{\D\mu_0}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar[d]^{\D\di\mu_0}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}\ar[d]_{\D\di\mu_2}
&M\ar[r]_-{\D\di\nu_2}\ar[d]^{\D\mu_1}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L
\cr
\di R\ar[d]_{\D\di\mu_1}
&R\ar[d]^{\D\mu_2}\ar[l]_-{\D\psi=\di\psi}^-{\D\simeq}
\cr
\di M\ar[r]_-{\D T(\nu_0)}
&T(L)
}}$$
and such that
$$\di\mu_1\psi\,\mu_1=-\nu_2\,\varphi\inv\di\nu_2.$$
In other words, such a $\mu_0$ is a good way to send $L$ in its
orthogonal, $M$ (see 3.2).
\endthm


\truc{Proof of lemma 3.19}
Let us give a number to the upper part of $\Omega$ for $\eta_0$\,:
$$\lab15%
\vcenter{\xymatrix@C=45pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di\eta_0)}
&L\ar[r]^-{\D\nu_1}\ar[d]^{\D\eta_0}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D \varphi}_{\D \simeq}
&\di M\ar[d]^{\D\di\eta_0}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}
&M\ar[r]_-{\D\di\nu_2}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L
}}$$
and another number, say the next available one, to the exact triangle\,:
$$\lab16\spxytriangle L;M;Q;T(L).;\eta_0;\eta_1;\eta_2;$$
%
We are going to study exact triangles over $s=\nu_2\,\varphi\inv\di\nu_2$.
\vskip2mm

Consider the composition axiom for the relation\,:
$$\di\nu_1\circ(\varphi\,\nu_1)=0$$
to obtain the following octahedron (the exact triangles over $\di\nu_1$
and over $\varphi\,\nu_1$ are obtained from the one over $\nu_1$)\,:
$$\xymatrix@C=35pt @R=40pt{%
&&&\di L\ar[ld]|-{\D\di\nu_0}
\ar[rrdd]^-{\pmatrix0\\1\endpmatrix}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
\di P\ar[rrru]^-{\D\di\nu_1}
\ar[rrdd]_-{\D-\nu_2\,\varphi\inv}
&&T(M)\ar[ll]|{\D T(\di\nu_2)}
\ar[dd]|-{\D-T(s)}
\\
&&&L\ar[lllu]|!{[lu];[ld]}\hole_-{\D\varphi\,\nu_1}
\ar[uu]|!{[rr];[lu]}\hole_(.55){\D0}
&&T(L)\oplus\di L\ar[ll]|(.55){\pmatrix1&0\endpmatrix}
\ar[lllu]_-{\D\exists \pmatrix c&d\endpmatrix}
\\
&&\di M\ar[ru]|-{\D T(\nu_0)}
\ar[rrru]_-{\D\exists \pmatrix a\\ b\endpmatrix}
}$$
for some morphisms $a,b,c,d$. The relations involving the
injection $\pmatrix0\\1\endpmatrix$ and the projection
$\pmatrix1&0\endpmatrix$ do force $a=T(\nu_0)$ and $d=\di\nu_0$. As
well, the relation
$$b\circ(-\nu_2)=\di\nu_1\varphi
\equal{1}\di\eta_0\nu_2$$
insures the existence of a morphism $e:T(L)\to\di L$ such that\,:
$$b=-\di\eta_0+e\,T(\nu_0).$$
As in the proof 3.8, one can use the automorphism $\pmatrix1&0\\-e&1\endpmatrix$
of $T(L)\oplus\di L$ to replace $b$ by $-\di\eta_0$. Replacing $c$ by
$c+\di\nu_0e$, and calling this $T(\beta)$, everything reduces to the
existence of an exact triangle\,:
$$\lab{.5}7\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\beta)&\di\nu_0\endpmatrix}
&&&T(M)}$$
for some morphism $\beta:L\to M$ such that\,:
$$T(\di\nu_2)\,T(\beta)=T(\varphi)\,T(\nu_1)$$
because this was true for $c$ and because $T(\di\nu_2)\,\di\nu_0=0$.
This is equivalent to\,:
$$\lab28\di\nu_2\beta=\varphi\,\nu_1.$$
Of course, the aesthete would prefer to have $\beta=\eta_0$. This is
not true in general but it appears that one can replace $\eta_0$ by
some $\mu_0$ such that the following triangle is exact\,:
$$\lab{.5}9\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\mu_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
&&&T(M).}$$
That is what we are going to do now.
\vskip2mm
>From property (8) and from (5), we deduce that
$$\di\nu_2\beta=\varphi\,\nu_1=\di\nu_2\eta_0.$$
In other words, $\di\nu_2(\beta-\eta_0)=0$. Then there
exists a morphism $l:L\to T\inv(\di L)$ such that
$$\lab1{10}\beta=\eta_0+T\inv(\di\nu_0)\,l.$$
%
Dualizing exact triangle (7), recalling that $\di s=s$, gives the following exact triangle\,:
$$\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix \di\beta\\ T(\nu_0)\endpmatrix}
&&\di L\oplus T(L)\ar[rrr]^-{\pmatrix -\di\nu_0&+T(\eta_0)\endpmatrix}
&&&T(M).}$$
Then using the isomorphism $\pmatrix 0&1\\-1&0\endpmatrix$ from $\di
L\oplus T(L)$ to $T(L)\oplus\di L$, we get the following exact
triangle\,:
$$\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\beta\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\eta_0)&\di\nu_0\endpmatrix}
&&&T(M).}$$
Put this last triangle as the second row of the following diagram, in
which the first is simply (7)\,:
$$\lab{.5}{11}\vcenter{\xymatrix{M\ar[r]^-{\D s}\ar@{=}[d]
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
\ar@{-->}[d]_{\D\exists\,}^{\D1+h}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\beta)&\di\nu_0\endpmatrix}
\ar[d]^{\pmatrix1&0\\T(l)&1\endpmatrix}
&&&T(M)\ar@{=}[d]
\cr
M\ar[r]_-{\D s}
&\di M\ar[rr]_-{\pmatrix T(\nu_0)\\-\di\beta\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]_-{\pmatrix T(\eta_0)&\di\nu_0\endpmatrix}
&&&T(M)}}$$
and whose right square commutes because of (10). Since it has exact rows, it is an
axiom that there exists a morphism such that the entire diagram
commutes and nobody can keep us from
calling it $1+h$ for some morphism $h:\di M\to\di M$. We have the following
relations\,:
$$\eqalign{h\,s\lab3{12}&=0\cr
T(\nu_0)\,h\lab3{13}&=0\cr
-\di\beta(1+h)\lab3{14}&=T(l)\,T(\nu_0)-\di\eta_0.}$$
Compose (14) on the right with $\nu_2$ gives\,:
$$-\kern-4mm\underbrace{\di\beta\nu_2}_{\D\kern5mm\equal{8}\di\nu_1\varphi}
-\di\beta h\,\nu_2=
T(l)\underbrace{T(\nu_0)\,\nu_2}_{\D=0}-
\kern-4mm\underbrace{\di\eta_0\nu_2}_{\D\kern5mm\equal{5}\di\nu_1\varphi}$$
which forces obviously
$$\lab1{15}\di\beta h\,\nu_2=0.$$
>From (12) and the exact triangle presented in the second line of
(11), we deduce that
$$\lab1{16}h=\bar h\,\pmatrix T(\nu_0)\\-\di\beta\endpmatrix$$
for some morphism $\bar h:T(L)\oplus\di L\to\di M$.
Similarly, from (13) and the exact triangle in the first line of (5),
we deduce that\,:
$$\lab1{17}h=\nu_2\,\tilde h$$
for some morphism $\tilde h:\di M\to P$.

Now, compute $h^3$ using (16) and (17)\,:
$$h^3=\bar h\,\pmatrix T(\nu_0)\\-\di\beta\endpmatrix\circ h\circ(\nu_2\,\tilde h)
=\bar h\,\pmatrix
T(\nu_0)\,h\,\nu_2\\-\di\beta\,h\,\nu_2\endpmatrix\,\tilde h.$$
It follows then immediately from (13) and (15) that $h^3=0.$ Therefore
$1+\half h$ is an isomorphism\,:
$$1+\half h\,:\di M\isoto\di M.$$

But, dualizing (11) and using as before the isomorphism $\pmatrix
0&1\\-1&0\endpmatrix$, we get the following morphism of exact
triangles\,:
$$\xymatrix{M\ar[r]^-{\D s}\ar[d]_{\D1+\di h}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
\ar@{=}[d]
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\beta)&\di\nu_0\endpmatrix}
\ar[d]^{\pmatrix1&0\\-\di l&1\endpmatrix}
&&&T(M)\ar[d]^{\D1+T(\di h)}
\cr
M\ar[r]_-{\D s}
&\di M\ar[rr]_-{\pmatrix T(\nu_0)\\-\di\beta\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]_-{\pmatrix T(\eta_0)&\di\nu_0\endpmatrix}
&&&T(M).}$$
If the reader prefers, this last diagram can be obtained directly from
the relations (10), (12), (13) and (14) by applying the duality.
Taking the mean of this last morphism with (11), we get the
following\,:
\vskip5mm
\noindent(18)
%\vskip-15mm
$$\xymatrix@C=22pt @R=35pt{%
M\ar[r]^-{\D s}\ar[d]_{\D1+\half\di h}^\simeq
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
\ar[d]^{\D1+\half h}_\simeq
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\beta)&\di\nu_0\endpmatrix}
\ar[d]^{\pmatrix1&0\\\half(T(l)-\di l)&1\endpmatrix}
&&&T(M)\ar[d]^{\D1+\half T(\di h)}_\simeq
\cr
M\ar[r]_-{\D s}
&\di M\ar[rr]_-{\pmatrix T(\nu_0)\\-\di\beta\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]_-{\pmatrix T(\eta_0)&\di\nu_0\endpmatrix}
&&&T(M).}$$
%
Let us give names\,:
$$f:=1+\half \di h\,:M\isoto M
\link{1.5}{and}
\bar l:={l-\dis l\over 2}=-\dis{\bar l}:L\to T\inv(\di L).$$
%
Pushing $f$ up and $\di f$ down in (18), we obtain the following
morphism of exact triangles\,:
$$\xymatrix@R=30pt{M\ar[rr]^-{\D s\,f\inv}\ar@{=}[d]
&&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
\ar@{=}[d]
&&T(L)\oplus\di L\ar[rrr]^-{\D T(f)\circ\pmatrix T(\beta)&\di\nu_0\endpmatrix}
\ar[d]^{\pmatrix1&0\\T(\bar l)&1\endpmatrix}
&&&T(M)\ar@{=}[d]
\cr
M\ar[rr]_-{\D (\di f)\inv s}
&&\di M\ar[rr]_-{\D\pmatrix T(\nu_0)\\-\di\beta\endpmatrix\circ\di f}
&&T(L)\oplus\di L\ar[rrr]_-{\pmatrix T(\eta_0)&\di\nu_0\endpmatrix}
&&&T(M).}$$
>From (12) and the definition of $f$, we have $s\,f=s$ and from (13) we
have $T(\nu_0)\,\di f=T(\nu_0)$. Use this in the above diagram to
obtain the following one\,:
$$\lab{.5}{19}\vcenter{\xymatrix@R=30pt{M\ar[r]^-{\D s}\ar@{=}[d]
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
\ar@{=}[d]
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\gamma)&\di\nu_0\endpmatrix}
\ar[d]^{\pmatrix1&0\\T(\bar l)&1\endpmatrix}
&&&T(M)\ar@{=}[d]
\cr
M\ar[r]_-{\D s}
&\di M\ar[rr]_-{\pmatrix T(\nu_0)\\-\di\gamma\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]_-{\pmatrix T(\eta_0)&\di\nu_0\endpmatrix}
&&&T(M)}}$$
where $\gamma:=f\,\beta$. Note that the exact triangle of the first
line gives\,:
$$0=\pmatrix T(\gamma)&\di\nu_0\endpmatrix
\cdot\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix
=T(\gamma\,\nu_0)-\di\nu_0\di\eta_0$$
which implies
$$\lab2{20}\gamma\,\nu_0=\dis\nu_0\dis\eta_0
\equal{5}\eta_0\,\nu_0.$$
Now, since the third commutative square of (19) gives
$$\lab2{21}\gamma=\eta_0+\dis\nu_0\bar l$$
we have, composing this last equation with $\nu_0$ on the right and
comparing with (20), that\,:
$$\lab2{22}\dis\nu_0\bar l\,\nu_0=0.$$
Consider now the following commutative diagram\,:
$$\lab{.5}{23}\vcenter{\xymatrix@R=30pt{M\ar[r]^-{\D s}\ar@{=}[d]
&\di M\ar[rrr]^-{\pmatrix T(\nu_0)\\-\di\eta_0\endpmatrix}
\ar@{=}[d]
&&&T(L)\oplus\di L\ar[rrrr]^-{\pmatrix T(\gamma)&\di\nu_0\endpmatrix}
\ar[d]^{\pmatrix1&0\\\half T(\bar l)&1\endpmatrix}_\simeq
&&&&T(M)\ar@{=}[d]
\cr
M\ar[r]_-{\D s}
&\di M\ar[rrr]_-{\pmatrix T(\nu_0)\\\half T(\bar l\,\nu_0)-\di\eta_0\endpmatrix}
&&&T(L)\oplus\di L\ar[rrrr]_-{\pmatrix T(\gamma-\half\dis\nu_0\bar l)&\di\nu_0\endpmatrix}
&&&&T(M)}}$$
in which the first row (and therefore also the second) is an exact
triangle (see 19).
Define
$$\mu_0:=\eta_0+\half\dis\nu_0\bar l.$$
Observe that (21) implies
$$\gamma-\half\dis\nu_0\bar l=\eta_0+\half\dis\nu_0\bar l=\mu_0$$
and that (here we use $\bar l=-\dis{\bar l}$ which is immediate from the
definition of $\bar l$)\,:
$$\di\mu_0=\di\eta_0+\half\di{\bar l}T(\nu_0)=
\di\eta_0+\half T(\dis{\bar l}\nu_0)
=\di\eta_0-\half T(\bar l\nu_0).$$
Replace these two last facts in the second line of (23) to obtain the
announced exact triangle (4)\,:
$$\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\ -\di\mu_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
&&&T(M).}$$
Define $\lambda=\half\bar l$, so that $\mu_0=\eta_0+\dis\nu_0\lambda$
as required. Condition (3) asks $\dis\nu_0\lambda\,\nu_0=0$, which is
a direct consequence of (22).
\qed


\truc{Proof of lemma 3.20}
%
Recall that $*=T\inv\circ\#$ and that $s$ denotes
$s=\nu_2\,\varphi\inv\di\nu_2:M\to\di M$. Note that $s=\di s$. Call
(1) the exact triangle of the hypothesis\,:
$$\lab2{1}\vcenter{\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\ -\di\mu_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
&&&T(M).}}$$

Apply the composition axiom to the relation\,:
$$\mu_0=\pmatrix\mu_0&\dis\nu_0\endpmatrix\cdot\pmatrix1\\0\endpmatrix$$
to construct the following octahedron \big(where the exact triangle over
$\pmatrix\mu_0&\dis\nu_0\endpmatrix$ comes from (1)\big)\,:
$$\xymatrix@C=35pt @R=45pt{%
&&&M\ar[ld]^-{\D s}
\ar[rrdd]^-{\D\mu_1}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
L\oplus T\inv(\di L)\ar[rrru]^-{\pmatrix\mu_0&\dis\nu_0\endpmatrix}
\ar[rrdd]_-{\pmatrix0&1\endpmatrix}
&&\di M\ar[ll]|(.4){\pmatrix-T(\nu_0)\\+\di\mu_0\endpmatrix}
\ar[dd]|-{\D\di\mu_0}
\\
&&&L\ar[lllu]|!{[lu];[ld]}\hole^(.52){\pmatrix1\\0\endpmatrix}
\ar[uu]|!{[rr];[lu]}\hole_(.55){\D\mu_0}
&&R\ar[ll]|(.55){\D\mu_2}
\ar[lllu]_-{\D\exists\,z}
\\
&&T\inv(\di L)\ar[ru]|-{\D0}
\ar[rrru]_-{\D\exists\,y}
}$$
The relation
$y\circ\pmatrix0&1\endpmatrix=\mu_1\circ\pmatrix\mu_0&\dis\nu_0\endpmatrix$
forces $y=\mu_1\,\dis\nu_0$. Then, the octahedron reduces to the
existence of a morphism $z:R\to\di M$ such that
\roster
\item[2] the following triangle is exact\,:
$$\spxytriangle T\inv(\di L);R;\di M;\di
L;\mu_1\,\dis\nu_0;z;\di\mu_0;$$
\item $z\,\mu_1=s$
\item $-T(\nu_0)\,z=\mu_2$.
\endroster
Dualizing (2) we obtain the exact triangle\,:
$$\lab2{5}\xymatrix{L\ar[r]^-{\D\mu_0}
&M\ar[r]^-{\D\di z}
&\di R\ar[rr]^-{\D-T(\nu_0)\,\di\mu_1}
&&T(L).}
$$
But we already have an exact triangle over $\mu_0$, so let us compare
them\,:
$$\lab2{6}\vcenter{\xymatrix{L\ar[r]^-{\D\mu_0}\ar@{=}[d]
&M\ar[r]^-{\D\mu_1}\ar@{=}[d]
&R\ar[rr]^-{\D\mu_2}\ar@{-->}[d]_{\D\exists\,\chi}^{\D\simeq}
&&T(L)\ar@{=}[d]
\cr
L\ar[r]_-{\D\mu_0}
&M\ar[r]_-{\D\di z}
&\di R\ar[rr]_-{\D-T(\nu_0)\,\di\mu_1}
&&T(L).}}$$
There exists an isomorphism, non necessarily symmetric,
$\chi\,:R\isoto\di R$ making diagram (6) commute. We cannot use the
classical argument because $\di\chi$ does not make diagram (6)
commute\,!

Since
$\chi\,\mu_1=\di z$, we obtain by composing on the right with
$-\dis\nu_0$ that
$$\chi\circ(-\mu_1\,\dis\nu_0)=-\di z\,\dis\nu_0.$$
Using the dual of (4), the right hand side of the above equation may
be replaced by $\di\mu_2$. Apply duality to this new equation to
obtain\,: $-T(\nu_0)\,\di\mu_1\big)\circ\di\chi=\mu_2$
which should be compared with (6). We can complete it into\,:
$$\lab2{7}\vcenter{\xymatrix{L\ar[r]^-{\D\mu_0}\ar@{=}[d]
&M\ar[rr]^-{\D\mu_1}\ar@{-->}[d]_{\D\exists\,}^{\D1+h}
&&R\ar[rr]^-{\D\mu_2}\ar[d]^{\D\di\chi}
&&T(L)\ar@{=}[d]
\cr
L\ar[r]_-{\D\mu_0}
&M\ar[rr]_-{\D\di z}
&&\di R\ar[rr]_-{\D-T(\nu_0)\,\di\mu_1}
&&T(L)}}$$
for some morphism $h:M\to M$. We immediately have\,:
$$\lab2{8}h\,\mu_0=0.$$
>From (6), we took $\chi\,\mu_1=\di z$. Dualize it\,:
$\di\mu_1\di\chi=z$. Compose this last equality by $\mu_1$ on the
right and use (3) to obtain\,:
$$\lab1{9}\di\mu_1\di\chi\,\mu_1=z\,\mu_1\equal{3}s.$$
But (7) tells us that $\di\chi\,\mu_1=\di z\,(1+h)$. Replacing this
in (9) gives\,:
$$s=\di\mu_1\,\di z\,(1+h)=
\di{(\underbrace{z\,\mu_1}_{\D\equal{3}s})}\,(1+h)=
\di s\,(1+h)=s\,(1+h)$$
and therefore\,:
$$\lab2{10}s\,h=0.$$

We still haven't used the third commutative square of (6) which says
that $\mu_2=-T(\nu_0)\,\di\mu_1\,\chi$. Dualizing gives\,:
$$\di\mu_2=-\di\chi\mu_1\,\dis\nu_0.$$
We can use (7) to replace $\di\chi\mu_1$ by $\di z\,(1+h)$\,:
$$\di\mu_2=-\di z\,(1+h)\,\dis\nu_0=
-\kern-13mm\underbrace{\di z\,\dis\nu_0}_{\kern15mm\D=\di{\big(T(\nu_0)\,z\big)}}
-\di z\,h\,\dis\nu_0
\equal{4}\,\di\mu_2-\di z\,h\,\dis\nu_0$$
from which one finds\,:
$$\lab2{11}\di z\,h\,\dis\nu_0=0.$$

It is now easy to get $h^3=0$. Use (8) and exact triangle (5) to write
$$h=\bar h\,\di z$$
for some morphism $\bar h:\di R\to M$. Similarly, use (10) and exact
triangle (1) to write
$$h=\pmatrix\mu_0&\dis\nu_0\endpmatrix\circ\tilde h$$
for some morphism $\tilde h:M\to L\oplus T\inv(\di L)$. Now simply compute
$$h^3=\bar h\,\di z\circ
h\circ\pmatrix\mu_0&\dis\nu_0\endpmatrix\,\tilde h=
\bar h\circ\pmatrix\di z\,h\,\mu_0&\di z\,h\,\dis\nu_0\endpmatrix\circ\tilde h
$$
Equations (8) and (11) insure the central matrix to be
zero. Therefore $h$ is nilpotent and then $1+\half h$ is an
isomorphism. Taking the mean between (6) and (7), we obtain the
following commutative diagram with exact rows\,:
$$\lab2{12}\vcenter{\xymatrix{L\ar[r]^-{\D\mu_0}\ar@{=}[d]
&M\ar[rr]^-{\D\mu_1}\ar[d]_{\D\simeq}^{\D1+\half h}
&&R\ar[rr]^-{\D\mu_2}\ar[d]^{\D\half(\chi+\di\chi)}
&&T(L)\ar@{=}[d]
\cr
L\ar[r]_-{\D\mu_0}
&M\ar[rr]_-{\D\di z}
&&\di R\ar[rr]_-{\D-T(\nu_0)\,\di\mu_1}
&&T(L).}}$$
In other words, the symmetric morphism
$$\psi:=\ \ \text{\bf- \kern-5pt- \kern-5pt-}\ \ \half(\chi+\di\chi):R\to\di R$$
is an isomorphism. Diagram (12) gives\,:
$$T(\nu_0)\,\di\mu_1\psi=\mu_2.$$
This means that the form $\psi$ makes the diagram $\bar{\bar\Omega}$
commute. We still have to check
$$\di\mu_1\psi\mu_1=-\nu_2\,\varphi\inv\di\nu_2 \kern5mm(=-s)$$
but this is immediate from the mean of relation (9) and its dual.
\qed


\truc{Proof of theorem 3.17}
%
Choose $\lambda$ as in lemma 3.19 and define of course
$$\mu_0=\eta_0+\dis\nu_0\lambda.$$
Lemma 3.20 insures the existence of a completion $\bar{\bar\Omega}$
and theorem 3.5 implies that
$(R,-\psi)$ is Witt-equivalent to $(P,\varphi)$ which is expected
property (1) of theorem 3.17. To see how to obtain property
(2), observe that\,:
$$\mu_0\,\nu_0\equal{\text{def}}
(\eta_0+\dis\nu_0\lambda)\,\nu_0=
\eta_0\,\nu_0+\underbrace{\dis\nu_0\lambda\,\nu_0}_{\D\equal{3}0}
=\eta_0\,\nu_0.$$
Then it suffices to apply twice lemma 3.7, once to $\eta_0$ and once 
to $\mu_0$. But in both case the morphism
$w=\eta_0\,\nu_0=\mu_0\,\nu_0$ is the same. Therefore $\cone(w)$ is $P\oplus
Q$ or $P\oplus R$ as well. This gives (2) and then theorem 3.17.
\qed

\espaceavanttruc
By discussion 3.1, there always exists a morphism $\eta_0$ such that $\Omega$
commutes. Therefore the above proof gives in particular the following
corollary.


\cor For any symmetric space
$(P,\varphi)$ and any sub-lagrangian $(L,\nu_1)$ of $(P,\varphi)$, for
any exact triangle over $\nu_1$\,:
$$\spxytriangle T\inv(\di M);L;P;\di M;\nu_0;\nu_1;\nu_2;$$
there exists a morphism $\mu_0\,:L\to M$ and a form $\psi$ on
$R:=\cone(\mu_0)$ such that diagram $\bar{\bar\Omega}$ of lemma 3.20
commutes, has exact lines and columns and such that
$\di\mu_1\psi\,\mu_1=-\nu_2\,\varphi\inv\di\nu_2$. In particular
$(R,\psi)$ is Witt-equivalent to $(P,-\varphi)$.\qed
\endthm

\rem
This result, although it sounds very nice, is probably useless in
general. In the classical case, it was the ideal formulation because
there was only one way to send a sub-lagrangian into its
orthogonal. Here, there are several ($\eta_0$, $\mu_0,\,\ldots$), some
better ($\mu_0$) than others ($\eta_0$). Theorem 3.17 will be more useful.

The results of this paragraph might be used to establish some link
between the Witt group I propose here and the quotient of it defined
by Youssin in [\the\yous].



\paragraph{Localization context and residue homomorphism}



\truc{Localization of triangulated categories with duality}

Let us consider what we can call an {\it exact sequence of
triangulated categories}\,:
$$0\too J\too K\too L\too0.$$
By that, we understand that $L$ is a localization of $K$, that is
$L=S\inv K$ for a class $S$ of morphisms. We can always suppose that
$S$ is saturated (\ie $S$ is formed by all the morphisms of $K$ which
become isomorphisms in $L$) and that $L$ is obtained by calculus of
fraction (nice exercise). Then $J$ is supposed to be the {\it kernel}
of this localization, \ie the full subcategory of $K$ over the objects
which become isomorphic to zero in $L$.

Suppose moreover that $K$ is endowed with a duality (or a
skew-duality) $\#$ such that $\#(S)=S$. Then $S\inv K$ inherits a
unique duality that can be called the localization of $\#$ and that we
shall lazily also denote by $\#$. It is then clear that $\#$ restricts
to a duality on $J$. In other words, we have an {\it exact sequence of
triangulated categories with duality}\,:
$$\text{(LOC)}\kern1cm
\vcenter{\xymatrix@C=30pt{
0\ar[r]
&J\ \ar[d]_\#\ar[r]^-j
&K\ar[d]_\#\ar[r]^-q
&S\inv K\ar[d]_\#\ar[r]
&0
\cr
0\ar[r]
&J\ \ar[r]_-j
&K\ar[r]_-q
&S\inv K\ar[r]
&0.
}}$$

\exa
Let $X$ be a noetherian regular separated scheme and $U$ an open
subscheme. Consider $\Dblf(X)$ the derived category of bounded complexes
of locally free $\Cal O_X$-modules of finite rank. Then $\Dblf(U)$ is a
localization of $\Dblf(X)$. It is moreover a localization of triangulated
categories with duality (\cf [\the\bala, {\sl th\'eor\`eme} 4.17, p.\,28]).

The kernel category $J$ is here the (derived) category of complexes whose
homology is concentrated on $Y$, the closed complement of $U$ in
$X$. It is a good question to wonder how the Witt groups of $J$ might
be related to the Witt groups of $Y$.

This very application of the present paper is my main motivation and
it will hopefully appear in forthcoming works. However, this and the
12-term localization sequence seem to me independent questions.
In other words, first, there is a localization sequence which is easy
to formulate in the triangulated framework and second, we shall try to
interpret $J$ in terms of $Y$. This enlights another point, namely the
difficulty usually encountered to define a good residue of a form on
$U$ with values in some Witt groups of $Y$. When existing, this
residue will actually be decomposable as the general residue presented
hereafter followed by some homomorphism between the Witt groups of $J$
and those of $Y$.







\rem
If we have an exact sequence (LOC), functoriality of Witt groups
naturally induces short complexes\,:
$$\xymatrix@C=25pt{\Wn(J)\ar[r]^-{\Wn(j)} &\Wn(K)\ar[r]^-{\Wn(q)}
  &\Wn(S\inv K) }$$
for all $n\in\Z$. To get a long (periodic) exact sequence, we are
going to construct connecting homomorphisms between $\Wn(S\inv K)$ and
$\Wnp(J)$. This will use the cone construction
of 1.6\,-\,1.10. First of all we shall recall from [\the\balb] the
construction of $\W(S\inv K)$ using $S$-spaces.

\defi
Let $(K,\#,\varpi)$ be a triangulated category with $\delta$-duality
($\delta=\pm1$). Let $S$ be a system of morphisms in $K$. A {\it
$S$-space} is a pair $(A,s)$ where $A$ is an object in $K$ and $s$ is
a morphism $s:A\to\di A$ such that\,: 
\roster
\item $s=\di s\,\varpi_A$
\item $s\in S$.
\endroster
We say that two $S$-spaces $(A,s)$ and $(B,t)$ are {\it $S$-isometric} if
there exists an object $C$ and morphisms $u,v$ in $S$ such that
$u:C\to A$, $v:C\to B$ and
$$\di us\,u=\di vt\,v.$$
Instead of $S$-spaces, the term {\it $S$-lattice} or {\it symmetric 
$S$-lattice} might be more conceptually acurate.

\prop
Let $\D J\mathop{\hookrightarrow}^j K\onto^qS\inv K$ be a localization of
triangulated categories with $\delta$-duality in the sense of 4.1.
The application $(A,s)\mapsto\big(q(A),q(s)\big)$ induces a well
defined monoid homomorphism between the set of $S$-isometry classes
of $S$-spaces and the Witt monoid $\MW(S\inv K)$ of $S\inv K$ (see
definition 1.13). This homomorphism is an isomorphism.

Through this application, $\NW(K)$ maps surjectively onto $\NW(S\inv
K)$. 

\proof
See [\the\balb, {\sl lemme} 2.11, p.\,98].\qed

\rem
Of course, a symmetric space is an $S$-space. The above proposition
describes the Witt monoid of $S\inv K$ in terms of $S$-spaces in
$K$. Neutral spaces in $S\inv K$ are obtained by the neutral spaces of
$K$. In other words, the Witt group of $S\inv K$ can be computed as
follows. Take $S$-isometry classes of $S$-spaces and divide out by
neutral spaces of $K$. As a consequence of that, to define an group
homomorphism 
$$W(S\inv K)\too G$$
to a group $G$, it suffices to define it additively on $S$-spaces, to check that
it is invariant under $S$-isometries and that it sends neutral spaces
of $K$ to zero in $G$.

We are going to check that such an homomorphism is induced by the cone
construction of 1.6\,-\,1.10.

\thm
Let $\D J\mathop{\hookrightarrow}^j K\onto^qS\inv K$ be a localization of
triangulated categories with $\delta$-duality in the sense of 4.1.
Let $(A,s)$ and $(B,t)$ be $S$-isometric $S$-spaces in $K$. Then
$$[\cone(A,s)]=[\cone(B,t)]\in\Wum(J).$$
\endthm

\proof
The proof goes in several steps. First of all, observe that the cone
construction is additive\,:
$$\cone\big((A,s)\bot(B,t))\simeq\cone(A,s)\bot\cone(B,t)$$
and is well defined up to strong isometry\,:
$$\cone(B,\di h\,s\,h)\simeq\cone(A,s)$$
if $h:B\isoto A$ is an isomorphism in $K$.

We shall only prove the case where $\#$ is a duality, that is
$\delta=+1$. The skew-duality goes as well.
It is convenient to abreviate $*=T\circ\#$ the shifted skew-duality. If
$(A,s)$ is an $S$-space for $\#$, then $\cone(A,s)$ is a symmetric
space for $(K,*,-\varpi)$. See, if necessary, $(\Gamma)$ in theorem 1.6 and
definition 1.10.

\lem 
Let $(P,\varphi)$ be a symmetric space in $K$ and let $(L,\nu_1)$
be a sublagrangian of this space such that $L$ belongs to $J$, the
kernel category. Choose an exact triangle
$$\spxytriangle T\inv(\di M);L;P;\di M.;\nu_0;\nu_1;\nu_2;$$
Then
$\nu_2\in S$ and $\orth L=M$ inherits from $\varphi$ a structure of
$S$-space, namely $(M,\nu_2\,\varphi\inv\di\nu_2)$, such that
$$\cone(M,\nu_2\,\varphi\inv\di\nu_2)=\big(T(L)\oplus\di L,%
\pmatrix0&1\\-1&0\endpmatrix\big).$$
In particular, $[\cone(M,\di\nu_2\varphi\inv\nu_2)]=0$ in $\Wum(J)$.

\proof
Define $s:=\nu_2\,\varphi\inv\di\nu_2:M\to \di M$.
By discussion 3.1, there exists a morphism $\eta_0:L\to M$ such that
diagram $\Omega$ commutes (\cf 3.1). By lemma 3.19, there exists an
exact triangle
$$\xymatrix{M\ar[r]^-{\D s}
&\di M\ar[rr]^-{\pmatrix T(\nu_0)\\ -\di\mu_0\endpmatrix}
&&T(L)\oplus\di L\ar[rrr]^-{\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
&&&T(M)}$$
where $\mu_0:L\to M$ is obtained by some modification of $\eta_0$ of
no importance here. Since one can use any exact triangle over $s$ to
construct $\cone(M,s)$, let us use the above one.

By definition, $\cone(M,s)=\big(T(L)\oplus\di L,\psi\big)$ for some
form $\psi:T(L)\oplus\di L\too\dis{\big(T(L)\oplus\di L\big)}=\di
L\oplus T(L)$ such that $\dis\psi=-\psi$ and such that the following
diagram commutes\,:
$$\xymatrix@C=35pt{M\ar[r]^-{\D s}\ar@{=}[d]
&\di M\ar[rr]^-{\D s_1:=\pmatrix T(\nu_0)\\ -\di\mu_0\endpmatrix}\ar@{=}[d]
&&T(L)\oplus\di L\ar[rrr]^-{\D s_2:=\pmatrix T(\mu_0)&\di\nu_0\endpmatrix}
\ar[d]^{\D\psi}
&&&T(M)\ar@{=}[d]
\\
M\ar[r]_-{\D s}
&\di M\ar[rr]_-{\D-\dis s_2=\pmatrix -\di\mu_0\\-T(\nu_0)\endpmatrix}
&&\di L\oplus T(L)\ar[rrr]_-{\D\dis s_1=\pmatrix \di\nu_0&-T(\mu_0)\endpmatrix}
&&&T(M).}$$
Clearly, it suffices to take $\psi=\pmatrix0&1\\-1&0\endpmatrix$.
\qed




\lem
Let $(P,\varphi)$ be a symmetric space in $K$ and let $s:C\to P$ be a
morphism in $S$. Then
$$[\cone(C,\di s\varphi\,s)]=0\link{.5}{in}\Wum(J).$$


\proof
Choose an exact triangle over $s$\,:
$$\lab21\spxytriangle C;P;D;T(C).;s;s_1;s_2;$$
Observe that $s\in S$ forces
$$\lab22D\in J.$$

Define 
$$t:=\di s\varphi\,s:C\to\di C.$$
 We want to prove that $\cone(C,t)$ is stably neutral in $(J,*)$. That is, we have to find some
object $N$ in $J$ and a morphism $w:N\to\di N$ (observe that
$T\inv\circ*=\#$) such that $w=\di w$ and $\cone(C,t)\bot\cone(N,w)$ is
neutral in $J$. Choose $N:=\di D$ and $w:=-s_1\varphi\inv s:\di D\to
D$. Observe immediately that $\di w=w$.

Choose an exact triangle over $t$ of the form\,:
$$\xytriangle C;\di C;T(L);t;t_1;t_2;$$
and observe that $t\in S$ implies
$$\lab{1.5}3 L\in J.$$
%
Write down the composition axiom for the relation
$$t=(\di s\varphi)\,s$$
where the exact triangle over $\di s$ (and then over $\di s\varphi$)
is constructed from exact triangle (1). This gives the following
octahedron\,: 
$$\xymatrix@C=35pt @R=45pt{%caca
&&&\di C\ar[ld]|-{\D-\dis s_2}
\ar[rrdd]^-{\D t_1}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
P\ar[rrru]^-{\D\di s\varphi}
\ar[rrdd]_-{\D s_1}
&&T(\di D)\ar[ll]|(.45){\D T(\varphi\inv)\,\dis s_1}
\ar[dd]|-{\D-T(w)}
\\
&&&C\ar[lllu]|!{[lu];[ld]}\hole^(.52){\D s}
\ar[uu]|!{[rr];[lu]}\hole_(.55){\D t}
&&T(L)\ar[ll]|(.55){\D t_2}
\ar[lllu]_-{\D\exists\,w_2}
\\
&&D\ar[ru]|-{\D s_2}
\ar[rrru]_-{\D\exists\,w_1}
}$$
That is, we have an exact triangle
$$\xytriangle \di D;D;T(L);w;w_1;w_2;$$
and relations among which\,:
$$\lab14T(\varphi\inv)\,\dis s_1 w_2=T(s)\,t_2.$$
Let us baptize $\nu_1:L\to P$ to be
$$\lab25\nu_1:=s\,T\inv(t_2)\equal{4}\varphi\inv\di
s_1T\inv(w_2).$$
We are going to use the previous lemma. To do that, observe that
$(L,\nu_1)$ is a sub-lagrangian of $(P,\varphi)$\,:
$$\di\nu_1\varphi\,\nu_1\equal{5}
\big(T(\di w_2)\,
s_1\,\varphi\inv\big)\,\varphi\,\big(s\,T\inv(t_2)\big)=0.$$
To apply the previous lemma, we have to construct an exact triangle
over $\nu_1$. To do that, observe that
$$t\circ(-T\inv s_2)=-\di s\varphi\,s\,T\inv(s_2)=0.$$
The octahedron gives\,:
$$\xymatrix@C=35pt @R=45pt{%
&&&\di C\ar[ld]|-{\D t_1}
\ar[rrdd]^-{\pmatrix1\\0\endpmatrix}
\ar@{-}[dd]|!{[rrdd];[ld]}\hole%useless line, but BUG in Nice
\\
C\ar[rrru]^-{\D t}
\ar[rrdd]_-{\D s}
&&T(L)\ar[ll]|(.45){\D t_2}
\ar[dd]|-{\D T(\nu_1)}
\\
&&&T\inv(D)\ar[lllu]|!{[lu];[ld]}\hole^(.52){\D -T\inv(s_2)}
\ar[uu]|!{[rr];[lu]}\hole_(.55){\D 0}
&&\di C\oplus D\ar[ll]|(.55){\pmatrix0&1\endpmatrix}
\ar[lllu]_-{\D\exists\,T(\nu_0)}
\\
&&P\ar[ru]|-{\D s_1}
\ar[rrru]_-{\D\exists\,\nu_2}
}$$
In particular, we have an exact triangle
$$\spxytriangle T\inv(\di C)\oplus T\inv(D);L;P;\di C\oplus
D;\nu_0;\nu_1;\nu_2;$$
where $\nu_2=\pmatrix x\\s_1\endpmatrix$ for some morphism $x:P\to \di
C$ such that $x\,s=t$. But $t=\di s\varphi\,s$. This means that
$$(\di s\varphi-x)\,s=0.$$
Using exact triangle (1), we have the existence of some morphism 
$a:D\to\di C$ such that $\di s\varphi-x=a\,s_1$, which, in turn,
can be seen as
$$\pmatrix1&a\\0&1\endpmatrix\pmatrix x\\s_1\endpmatrix=
\pmatrix \di s\varphi\\s_1\endpmatrix.$$
Using this endomorphism of $\di C\oplus D$, one can then suppose that 
$$\nu_2=\pmatrix \di s\varphi\\s_1\endpmatrix,$$
changing $\nu_0$ as well, but keeping the $\nu_1:L\to P$ we liked.
Now, having lemma 4.10 in mind, compute
$$\nu_2\,\varphi\inv\di\nu_2=\pmatrix \di
s\varphi\\s_1\endpmatrix\,\varphi\inv\,
\pmatrix \varphi\,s&\di s_1\endpmatrix
=\pmatrix t&0\\0&-w\endpmatrix.$$
Since $L\in J$ by (3), lemma 4.10 insures us that the cone of this form is neutral is
$J$. But it is obviously the orthogonal sum of the cone of $t$ and of
the cone of $-w:\di D\to D$, the latter being neutral in $J$ because $\di D\in J$ by (2). Therefore, $[\cone(t)]=0$ in $\Wum(J)$ as claimed. 
\qed


\truc{Proof of theorem 4.8}
Let $(A,s)$ and $(B,t)$ be $S$-isometric $S$-spaces. It suffices to do
the case when there exists a morphism $u:B\to A$, $u\in S$, such that
$$\lab21t=\di u s\, u.$$
Recall that the cone construction is additive and
invariant under strong isometries in $K$ (see 4.9). Now consider
$$\pmatrix1&0\\\di u&1\endpmatrix\cdot\pmatrix s&0\\0&-t\endpmatrix\cdot
\pmatrix1&u\\0&1\endpmatrix\equal{1}
\pmatrix s&s\,u\\\di u\,s&0\endpmatrix.$$
It suffices to prove that
$[\cone\big(A\oplus B,\pmatrix s&s\,u\\\di u\,s&0\endpmatrix\big)]=0$
in $\Wum(J)$. Now, consider the morphism
$$\pmatrix 1&0\\0&s\,u\endpmatrix:A\oplus B\too A\oplus \di A$$
which obviously belongs to $S$ and consider the symmetric form
$$\varphi=\pmatrix s&1\\1&0\endpmatrix$$
over $A\oplus\di A$. The result is now a direct consequence of lemma
4.18 since
$$\di{\pmatrix 1&0\\0&s\,u\endpmatrix}\cdot\varphi\cdot\pmatrix
1&0\\0&s\,u\endpmatrix=
\pmatrix s&s\,u\\\di u\,s&0\endpmatrix.$$
\qed


\truc{Mea culpa}
The carefull reader of [\the\bala] will object that theorem 5.15 there
already states that this connecting homomorphism is well
defined. The proof of lemma 5.11 {\sl loc.\ cit.}\ on which this theorem
relays, is incomplete and contains several typographic mistakes, as I
noticed too late. The present version should fill this gap.


\truc{Corollary and definition}
Let $\D J\mathop{\hookrightarrow}^j K\onto^qS\inv K$ be a localization of
triangulated categories with $\delta$-duality in the sense of 4.1.
The cone construction of 1.6 and 1.10 induce well defined
homomorphisms
$$\eqalign{\partial^{n}:
\Wn(S\inv K)\too&\Wnp(J)\cr
x\mapsto &[\cone(A,s)]\cr
&\text{where\ }x=[q(A),q(s)]
}$$
for all $n\in\Z$. They will be called the {\it residue homomorphisms}.


\proof
This is immediate from theorem 4.8 and remark 4.7. In
fact, the cone of any (neutral) form in $K$ is of course trivial since
the cone of an isomorphism is zero.

The theorem was stated for $n=0$ but this is of course enough: apply
it to $T^n(K,\#,\varpi)$.

\qed




\exo
Check that the $\partial^n:\Wn(S\inv K)\too\Wnp(J)$ commute with
the periodicity isomorphisms $\Wn\isoto\W^{\,n+4}$ of proposition
1.14.

\exo
Check that the sequence obtained by using the natural homomorphisms of
4.3 and the residue homomorphisms of 4.16 is a complex (very easy).

\exo
Understand the analogy between the residue homomorphisms presented
here and the classical (second) residue in the case of a valuation
ring, for instance. Understand first why an $S$-space is a good 
generalization of a lattice.








\paragraph{Main result}

\thm
Let $\D J\mathop{\hookrightarrow}^j K\onto^qS\inv K$ be a localization of
triangulated categories with duality in the sense of 4.1. Suppose that
$K$ and $S\inv K$ are \noeth\ and that $\half\in K$. Then, using the
natural homomorphisms $\Wn(j)$ and 
$\Wn(q)$ and the residue homomorphisms $\partial^n$ of
definition 4.16, we obtain the following long exact exact sequence\,:
$$\xymatrix@C=25pt{{\cdots}\ar[r]
&\W^{n-1}(S\inv K)\ar[r]^-{\partial^{n-1}}
&\Wn(J)\ar[r]^-{\Wn(j)}
&\Wn(K)\ar[r]^-{\Wn(q)}
&\Wn(S\inv K)\ar[r]^-{\partial^n}
&\Wnp(J)\ar[r]
&{\cdots}
}$$
of localization.

\proof
We are now well prepared to prove this theorem. We shall
only establish exactness of the sequence at $\W(J)$, $\W(K)$ 
and $\W(S\inv K)$ (the last one is already done in [\the\bala] and in
[\the\balb]). The same results for the shifted groups, using the
translated dualities go as well. The reader can establish the
analogue of the results of paragraphs 2 to 4 for 
skew-dualities. Details are left as an exercise.

\espaceavanttruc
\centerline{\smc Exactness at $\W(J)$.}
\espaceaprestruc

Let $x\in\Ker\big(\W(J)\to\W(K)\big)$. Choose a symmetric space such
that $x=[P,\varphi]$. Then $(P,\varphi)$ is stably neutral in $K$. By
theorem 2.5, $(P,\varphi)$ is neutral in $K$. This means by
definition 1.12 that
$$(P,\varphi)=\cone(A,u)$$
for some object $A$ in $K$ and some morphism $u:A\to T\inv(\di A)$
such that
$$T\inv(\di u)=u.$$
The reader puzzled down by the sign should re-read remark 1.9 and its neighbors.

Then $\cone(u)=P\in J$ and therefore $u$ becomes an isomorphism in
$S\inv K$ (where $P$ would be zero). Then, since $S$ is saturated,
$u\in S$. In other words, $(A,u)$ is an $S$-space for the skew-duality
$T\inv\circ\#$ and
$$(P,\varphi)=\cone(A,u).$$
By definition 4.16,
$$(P,\varphi)=\partial^{-1}\big(q(A),q(u)\big).$$
This is the announced result. Observe that we did not use that $K$ is \noeth. The key point is of course the very strong theorem 2.5.


\espaceavanttruc
\centerline{\smc Exactness at $\W(K)$.}
\espaceaprestruc

The first step is to give a criterion for symmetric spaces
over $K$ to become neutral in $S\inv K$. This should be compared with
lemma 2.1 condition (3). In $S\inv K$ the new isomorphisms are
essentially the morphisms of $S$ and therefore we obtain new neutral
forms using the criterion of the lemma. We prove hereafter that they
are the only one.


\lem
Let $\D J\mathop{\hookrightarrow}^j K\onto^qS\inv K$ be a localization
of triangulated categories with duality in the sense of 4.1. Suppose
that $\half\in K$. Then a symmetric space $(P,\varphi)$ in $K$ is
neutral in $S\inv K$ if and only if there exists a morphism of exact
triangles\,: 
$$\xymatrix@C=40pt{
T\inv(\di M)\ar[r]^-{\D\nu_0}\ar[d]_{\D T\inv(\di h)}
&L\ar[r]^-{\D\nu_1}\ar[d]^{\D h}
&P\ar[r]^-{\D\nu_2}\ar[d]^{\D\varphi}_{\D\simeq}
&\di M\ar[d]^{\D\di h}
\cr
T\inv(\di L)\ar[r]_-{\D T\inv(\di\nu_0)}
&M\ar[r]_-{\D\di\nu_2}
&\di P\ar[r]_-{\D\di\nu_1}
&\di L
}$$
with $h\in S$.
\proof
Suppose that there exists such a diagram. Then localize it (\ie see
it in $S\inv K$) and apply lemma 2.1 in $S\inv K$. You will obtain
neutrality of $\big(q(P),q(\varphi)\big)$.

\vskip2mm

The converse is the interesting part. The reader should pay attention
to distinguish diagrams in $S\inv K$ from those in $K$. Hereafter, $*$
shall denote $T\inv\circ\#$ from time to time.

Let $(P,\varphi)$ be a symmetric space in $K$ neutral in $S\inv K$. By
definition, there exists an exact triangle in $S\inv K$\,:
$$\lab21\spxytriangle T\inv(\di A);A;P;\di A;a_0;a_1;a_2;$$
such that
\roster
\item[2] $a_2=\di a_1\varphi$ in $S\inv K$;
\item $a_0=\dis a_0$ in $S\inv K$.
\endroster

\espaceavanttruc
\noindent{\bf Trivial Remark.}
{\it Consider an exact triangle in $S\inv K$ like in (1) satisfying (2) and
(3). Choose any morphism $s:B\to A$ in $K$ such that $s\in S$. 
Set $b_1=a_1\circ s$ and $w=s\inv a_0\,\dis{(s\inv)}$.
Then we have an exact triangle in $S\inv K$\,:
$$\spxytriangle T\inv(\di B);B;P;\di B;w;b_1;\di b_1\varphi;$$
with $w=\dis w$ in $S\inv K$.
}

To prove this, it suffices to observe the commutative diagram in
$S\inv K$\,:
$$\xymatrix{T\inv(\di A)\ar[r]^-{\D a_0}\ar[d]_{\D \dis s}^\simeq
&A\ar[r]^-{\D a_1}
&P\ar[r]^-{\D a_2}\ar@{=}[d]
&\di A\ar[d]_{\D \di s}^\simeq
\cr
T\inv(\di B)\ar[r]_-{\D w}
&B\ar[r]_-{\D b_1}\ar[u]^{\D s}_\simeq
&P\ar[r]_-{\D \di b_1\varphi}
&\di B
}$$
which commutes because, in $S\inv K$, we have\,:
$$\di b_1\varphi=\di{(a_1\,s)}\varphi=\di s\di a_1\varphi
\equal{2}\di sa_2.$$
Of course (3) implies that $\dis w=w$. This finishes the proof of the
trivial remark.
\vskip 2mm

Consider the exact triangle (1). Write $a_1$ as a fraction\,:
$$a_1=b_1\,s\inv\,:\,A\mathop{\longleftarrow}^{s}B\too^{b_1} P$$
for some object $B$ and some morphism $s:B\to A$ in $S$. We have
$b_1=a_1\,s$. Then the remark tells us that we can replace (1) by an
exact triangle in $S\inv K$\,:
$$\spxytriangle T\inv(\di B);B;P;\di B;w;b_1;\di b_1\varphi;$$
with $w=\dis w$ in $S\inv K$.
Then it is immediate that
$$\di b_1\varphi\,b_1=0\kern1cm\text{in\ }S\inv K.$$
This implies that there exists a morphism $s':B'\to B$ such that
$$\di b_1\varphi\,b_1\circ s'=0\kern1cm\text{in\ }K.$$
Use again the trivial remark to replace $b_1$ by
$b_1\,s'$. Summarizing up, we have the following situation\,:
\roster
\item[4] we have an exact triangle in $S\inv K$\,:
$$\spxytriangle T\inv(\di B);B;P;\di B,;w;b_1;\di b_1\varphi;$$
\item $w=\dis w$ in $S\inv K$,
\item $b_1$ is a morphism in $K$ \big(meaning $=q(b_1)$\big)
\item $\di b_1\varphi\,b_1=0$ $\underline{\text{in\ }K}$.
\endroster
\vskip2mm

Since we have (6), we can choose an exact triangle in $K$ containing
$b_1$\,:
$$\lab28\spxytriangle T\inv(C);B;P;C.;b_0;b_1;b_2;$$
In $S\inv K$, we can compare (4) and (8), or if you prefer, (4) and the
localization of~(8)\,:
$$\lab19\vcenter{
\xymatrix@C=35pt @R=15pt{T\inv(\di B)\ar[r]^-{\D w}\ar@{-->}[dd]_{\D \exists}^{\D\simeq}
&B\ar[r]^-{\D b_1}\ar@{=}[dd]
&P\ar[r]^-{\D \di b_1\varphi}\ar@{=}[dd]
&\di B\ar@{-->}[d]^{\D\exists\,\di t}
\cr
&&&\di D
\cr
T\inv(C)\ar[r]_-{\D b_0}
&B\ar[r]_-{\D b_1}
&P\ar[r]_-{\D b_2}
&C\ar@{-->}[u]_{\D\exists\,u}
}}$$
There exists an isomorphism $\di B\isoto C$, that can be expressed as
a fraction like displayed above. Moreover, the relation $\di t\di
b_1\varphi=u\,b_2$ is valid in $S\inv K$ and therefore in $K$ up to
some composition, say on the left, by a morphism in $S$. Up to a
modification of $D$, $t$ and $u$, one can suppose (9) and moreover\,:
$$\di t\di b_1\varphi=u\,b_2\kern1cm\underline{\text{in\ }K}.$$
Since $S$ is saturated, we have $t,u\in S$.
%
Define $c_1:=b_1\,t:D\to P$ so that the above relation becomes\,:

$$\lab2{10}\di c_1\varphi=u\,b_2\kern1cm\underline{\text{in\ }K}.$$
Choose an exact triangle over $c_1$ in $K$ and use the definition of
$c_1$, to construct the following morphism of exact triangles
$\underline{\text{in\ }K}$\,:
$$\lab1{11}\vcenter{
\xymatrix{T\inv(E)\ar[r]^-{\D c_0}\ar@{-->}[d]_{\D T\inv(v)}
&D\ar[r]^-{\D c_1}\ar[d]^{\D t}
&P\ar[r]^-{\D c_2}\ar@{=}[d]
&E\ar@{-->}[d]^{\D\exists\,v\in S}
\cr
T\inv(C)\ar[r]_-{\D b_0}
&B\ar[r]_-{\D b_1}
&P\ar[r]_-{\D b_2}
&C.
}}$$
There exists a morphism $v:E\to C$ making (11) commute and it is
necessarily in $S$, which is saturated. Now define
$$l:=u\,v:E\to\di D.$$
Note that $l\in S$. Compose on the left the third square of (11) with
$u$ to obtain $l\,c_2=u\,b_2$ and therefore, because of (10), we have\,:
$$\lab{1.5}{12}l\,c_2=\di c_1\varphi\kern1cm\underline{\text{in\ }K}.$$
Define also in $S\inv K$\,:
$$z:=t\inv\circ w\circ\dis{(t\inv)}:T\inv(\di D)\to D.$$
Relation (5) forces $\dis z=z$ in $S\inv K$. Now compute\,:
$$\eqalign{\kern2cmz\,T\inv(l)&=z\,T\inv(u\,v)=t\inv
\underbrace{w\,\dis{(t\inv)}\,T\inv(u)}_{\D\kern2mm\equal{9}b_0}\,T\inv(v)\cr
&=t\inv\,b_0\,T\inv(v)
\equal{11}c_0\kern3cm\text{in\ }S\inv K.}$$
Regrouping this last relation with (12), we have the following
commutative diagram in $S\inv K$\,:
$$\lab2{13}\vcenter{
\xymatrix{T\inv(E)\ar[r]^-{\D c_0}\ar[d]_{\D T\inv(l)}
&D\ar[r]^-{\D c_1}\ar@{=}[d]
&P\ar[r]^-{\D c_2}\ar@{=}[d]
&E\ar[d]^{\D l}
\cr
T\inv(\di D)\ar[r]_-{\D z}
&D\ar[r]_-{\D c_1}
&P\ar[r]_-{\D \di c_1\varphi}
&\di D.
}}$$

But relation (12) was true in $K$ and therefore can be expressed as
the commutativity of the following right hand square
$\underline{\text{in\ }K}$\,:
$$\xymatrix{T\inv(E)\ar[r]^-{\D c_0}\ar[d]_{\D T\inv(l)}
&D\ar[r]^-{\D c_1}
&P\ar[r]^-{\D c_2}\ar[d]^{\D\varphi}
&E\ar[d]^{\D l}
\cr
T\inv(\di D)\ar[r]_-{\D \dis c_0}
&\di E\ar[r]_-{\D \di c_2}
&\di P\ar[r]_-{\D \di c_1}
&\di D
}$$
where the first line is the exact triangle over $c_1$ we chose in (11)
and where the second line is simply the dual of the first. Now, choose
in $K$ a morphism $m:D\to \di E$ such that the following diagram
commutes\,:
$$\lab2{14}\vcenter{
\xymatrix{T\inv(E)\ar[r]^-{\D c_0}\ar[d]_{\D T\inv(l)}
&D\ar[r]^-{\D c_1}\ar@{-->}[d]^{\D\exists\,m}
&P\ar[r]^-{\D c_2}\ar[d]^{\D\varphi}
&E\ar[d]^{\D l}
\cr
T\inv(\di D)\ar[r]_-{\D \dis c_0}
&\di E\ar[r]_-{\D \di c_2}
&\di P\ar[r]_-{\D \di c_1}
&\di D.
}}$$
Observe that $m\in S$.
%
Define
$$h:=\half(m+\di l):D\to \di E.$$
It is obvious from (14) that the following diagram commutes
$\underline{\text{in\ }K}$ (easy exercise)\,:
$$\lab2{15}\vcenter{
\xymatrix{T\inv(E)\ar[r]^-{\D c_0}\ar[d]_{\D T\inv(h)}
&D\ar[r]^-{\D c_1}\ar[d]^{\D h}
&P\ar[r]^-{\D c_2}\ar[d]^{\D\varphi}
&E\ar[d]^{\D h}
\cr
T\inv(\di D)\ar[r]_-{\D \dis c_0}
&\di E\ar[r]_-{\D \di c_2}
&\di P\ar[r]_-{\D \di c_1}
&\di D.
}}$$
% 
The only non-trivial point is to check that $h\in S$, which is of
course not true simply because $l,m\in S$. To prove this last point,
we are going to establish that $\di h$ also makes diagram (13)
commute (in place of $l$). This is enough because $S$ is saturated. In
view of the definition of $h$, it is of course sufficient to establish
that $\di m$ makes diagram (13) commute.

In other words, we have to establish\,:
\roster
\item[16] $z\,T\inv(\di m)=c_0$ in $S\inv K$
\item $\di m\, c_2=\di c_1\varphi$ in $S\inv K$.
\endroster
But (17) is already true in view of the middle square of (14), by
applying $\#$. Now simply compute in $S\inv K$\,:
$$z\,\dis m=\dis z\,\dis m=
\dis{(m\kern-15mm\underbrace{z}_{\kern17mm\D\equal{13}c_0\,T\inv(l\inv)}\kern-15mm)}=
\dis{\big(m\,c_0\,T\inv(l\inv)\big)}\equal{14}\dis{(\dis c_0)}=c_0.$$
\vskip2mm
Putting everything together, we have the commutative diagram (15) with exact
lines in $K$, for a morphism $h:D\to\di E$ such that $h\in S$. This
is exactly the claim of the lemma.
The proof of this lemma is octahedron-free\,!
\qed

\espaceavanttruc
It is now easy to establish exactness of the sequence at $\W(K)$. It
suffices to prove that a form satisfying the condition of the lemma is
Witt-equivalent in $K$ to a form $(R,\psi)$ with $R\in J$. Theorem
3.17 applied to $\eta_0=h$, precisely asserts that there exists a form
on some $R=\cone(\mu_0)$, Witt-equivalent to $(P,\varphi)$ and such that
$$P\oplus R\simeq P\oplus\cone(h).$$
Localize the above isomorphism in $S\inv K$\,:
$$q(P)\oplus q(R)\simeq q(P)\oplus 
\underbrace{q\big(\cone(h)\big)}_{=0}\simeq q(P)$$
because $h\in S$. Therefore, since $S\inv K$ is \noeth (and we only
use condition (1) of definition 3.3),
$$q(R)\simeq0\kern1cm\text{in\ }S\inv K.$$
This means by definition that $R\in J$. Then, $[P,\varphi]\in\Im\big(\W(j)\big)$.




\espaceavanttruc
\centerline{\smc Exactness at $\W(S\inv K)$.}
\espaceaprestruc

This was established in [\the\bala, theorem 5.17] and in [\the\balb,
{\sl th\'eor\`eme} 2.21, p.\,103]. At that pre-walterian time, $\Wum$ was written $\W_1^{\,-}$ to recall the 1 from ``$T^1\circ\#$'' and the minus from ``$-\varpi$''.
\qed


\espaceavanttruc
Theorem 5.1, proposition 1.14 and exercise 4.18 immediately give the following corollary.


\cor
Let $\D J\mathop{\hookrightarrow}^j K\onto^qS\inv K$ be a localization of
triangulated categories with duality in the sense of 4.1. Suppose that
$K$ and $S\inv K$ are \noeth\ and that $\half\in K$. Then, we have the following\,:
$$\xymatrix@C=4pt{
&\W(J)\ar[rrrr]
\ar@{}[rrrrrrrrdddd]|{\D\text{12-term cyclic exact sequence.}}
&&&&\W(K)\ar[rrrr]
&&&&\W(S\inv K)\ar@/^1pc/[rd]
\\
\Wu(S\inv K)\ar@/^1pc/[ru]
&&&&&&&&&&\Wum(J)\dto
\\
\Wu(K)\ar[u]
&&&&&&&&&&\Wum(K)\dto
\\
\Wu(J)\ar[u]
&&&&&&&&&&\Wum(S\inv K)\ar@/^1pc/[ld]
\\
&\Wm(S\inv K)\ar@/^1pc/[lu]
&&&&\Wm(K)\ar[llll]
&&&&\Wm(J)\ar[llll]
}$$
\endthm
\vskip-3mm\qed





\goodbreak
\leavevmode
\espaceavantpar
\centerline{\bigsmc Bibliography}
\espaceaprespar


\bible Paul BALMER; Derived Witt groups of a scheme; Journal of pure
and applied algebra (to appear);

\bible Paul BALMER; Groupes de Witt d\'eriv\'es d'un sch\'ema; Th\`ese
de doctorat, Lausanne (1998);

\bible Robin HARTSHORNE;Residues and duality; Springer Lecture Notes in Mathematics 20 (1966);

\bible Max KAROUBI; Localisation de formes quadratiques I \& II; Annales
scientifiques de l'\'Ecole Normale Sup\'erieure de Paris, 4${}^e$
s\'erie, 7, pp.~359-404, (1974) \& 4${}^e$ s\'erie, 8, pp.~99-155, (1975);

\bible Manfred KNEBUSCH; Symmetric bilinear forms over
algebraic varieties; In\,: Conf. on Quadratic Forms, Kingston 1976,
Queen's Papers on Pure and Applied Mathematics 46, pp.~103-283, (1977);

\bible William PARDON; The exact sequence of a localization for Witt
groups; In\,: Algebraic K-theory, Evanston 1976, Springer Lecture
Notes in Mathematics 551, pp.~336-379, (1977);

\bible Andrew RANICKI; Exact Sequences in the Algebraic Theory of Surgery;
Mathematical Notes 26, Princeton University Press (1981);

\bible Jean-Louis VERDIER; Cat\'egories d\'eriv\'ees des cat\'egories ab\'eliennes; Ast\'erisque 239, Soci\'et\'e math\'ematique de France (1996);

\bible Boris YOUSSIN; Witt groups of derived categories; Journal of
K-theory, volume 11, No 4, pp.~373-395, (1997);

\espaceavantpar \centerline{\bf Acknowledgements} \espaceaprespar It
is a real pleasure to express here my profund gratitude to Bernhard
Keller, to Marco Schlichting to Charles Walter and to Chuck Weibel for
listening to expositions of some proofs, making valuable comments,
reading the first versions or for suggesting indispensable
modifications.  Last but not least, none of this would have been possible without the sitting Buddha, who will recognize himself.



\enddocument
