DERIVED WITT GROUPS OF A SCHEME

Paul Balmer

Institut de Mathématiques, Université de Lausanne

Abstract. The Witt group of a triangulated category with duality is the quotient of the monoid of symmetric spaces by the submonoid of neutral forms. Neutral forms are defined in a traditional way, using lagrangians. To any noetherian scheme X is associated a derived category with duality, denoted by K(X). The Witt group of K(X) will be called the (derived) Witt group of X.

There is an isomorphism between the usual Witt group of a ring in which 2 is a unit and its derived Witt group. This approach allows us to compute the kernel of W(A) → W(Q), where A is a domain and Q its field of fractions. This kernel turns out to be the Witt group of some suitable triangulated category with duality. The point of view of derived categories seems particularly useful for localization. Let U be an open subscheme of a regular scheme X. It is not hard to establish that K(U) is a localization of K(X) with respect to a suitable multiplicative system. Denote by J the full subcategory of K(X) on the objects vanishing in K(U). We construct a connecting homomorphism from the Witt group of K(U) to some Witt group of J, associating skew-symmetric forms to symmetric ones. We prove that the kernel of this homomorphism is precisely the part of W(K(U)) coming from X. Using these results we obtain a very simple proof of purity in dimension 3.

Introduction.

In the study of the usual Witt group (Knebusch’s classical definition) of a regular ring, projective resolutions are very useful. In several cases, it would be very convenient to have forms on complexes and not only on projective modules. These considerations are not new and attempts to define analogues of the Witt group on these more general objects have been made for a long time. We could mention for instance A. Ranicki in [5], who defines, for any integer n, the Witt group of complexes of length n. It is then very easy to relate these groups to classical ones when n is zero !

We proceed in the opposite way. Allowing the complexes to be as wide as they want (bounded anyway), we gain in flexibility. Unfortunately, it becomes rather tricky to prove that this new group has something in common with the classical one. The purpose of this note is to establish the definition of these a derived” Witt groups and to relate them to the usual notions of quadratic spaces over schemes.

To define these Witt groups, we consider the obvious duality on complexes of projective modules. Actually, it is convenient to deal with a more general concept : any triangulated category with a suitable duality. In a recent paper, B. Youssin gives a definition of Witt groups of derived categories [7]. His point of view is slightly different from ours. In fact, the basic question is : which symmetric forms have to be considered trivial ? Youssin uses for this the notion of cobordism. Our definition uses the good old lagrangians of our grand-mothers. This point of view has the advantage to be very conceptual : trivial forms are those with a big” isotropic part. In fact, Youssin’s approach has more topological motivations and he doesn’t study the case of schemes. His reduction to the heart of a triangulated category with t-structure ([7], thm 7.4), using the functor H^{0}, does not apply to complexes of projective objects (whose homology is not projective). Compare with our homomorphism Ω : W_{der} → W_{us} in section 4 to understand this difficulty.

Our approach yields the framework of a more flexible theory, as it could be observed with the problem of localizing the Witt group of a scheme X to an open subscheme U. Actually, we can construct an obstruction for a element of W(U) to come from W(X). This obstruction lives in the Witt group W_{1}^{−1}(J) of a suitable triangulated category with duality J; the superscript -1 indicates skew-symmetry; the subscript 1 refers to a shifted duality. In other words, to each symmetric form φ on U, we can associate

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 11E81, 18E30, 19G12.

a skew-symmetric form in J, which is neutral if and only if φ is Witt-equivalent to the restriction of a form on X.

We adopt the notations and sign conventions of [6], where the reader may find all the material about triangulated categories necessary to understand this paper.

1. Duality, skew-duality, Witt groups.

1.1. Definition. Let K_{1} and K_{2} be triangulated categories and let T_{i} be the translation automorphism of K_{i}, i = 1, 2. Let δ = ±1. An additive contravariant functor

F : K_{1} → K_{2}

is δ-exact if F ∘T_{1} = T_{2}^{−1} ∘F and if for any exact triangle \xymatrix{A \ar[r]^{u} & B \ar[r]^{v} & C \ar[r]^{w} & T (A)} the following triangle is exact :

[figure: F (A); δ · T_{2}(F (w)); F (C); F (v); F (B); F (u)] .

1.2. Remark. By exact (resp. skew-exact) we mean 1-exact (resp. (-1)-exact). Observe that F is skewexact if and only if T_{2} ∘ F (or F ∘ T_{1}) is exact.

1.3. Remark. We say that a covariant functor commuting with translation is exact if it maps exact triangles to exact triangles and skew-exact if it maps exact triangles to skew-exact triangles (i.e. triangles with one sign changed). In fact, definition 1.1 endows the opposite category with a triangulation.

1.4. Definition. Consider a triangulated category K and denote by T its translation automorphism. Let δ = ±1. A δ-duality is a δ-exact contravariant functor # : K ⟶ K such that there exists an isomorphism

can : Id \underrightarrow{∼} # ∘ #

satisfying the following conditions :

can_{T (X)} = T (can_{X})  and  (can_{X})^{#} ∘ canX^{#} = IdX^{#}

for all X ∈ K. As before, a (-1)-duality will be called a skew-duality.

1.5. Remark. It is not hard to see that if # is a duality then T ∘ # is a skew-duality, using the fact that # ∘ T = T^{−1} ∘ #. More generally, if # is a duality then T^{n} ∘ # is a (−1)^{n}-duality for all n ∈ ℤ. This property of ±1-duality should not be confused with symmetry and skew-symmetry. These concepts have nothing in common a priori ! ±1-duality indicates the behaviour of # with respect to exact triangles and we will see that a usual” duality (e.g. on a ring) gives rise to a (+1)-duality.

1.6. Definition. Let K be a triangulated category and # : K → K a ±1-duality. A symmetric space is a pair (X, φ) where X is an object of K and φ : X →^{∼} X^{#} is an isomorphism such that

φ^{#} ∘ can_{X} = φ.

The morphism φ is usually called the (bilinear symmetric) form on X by pure nostalgia.

1.7. Classical definitions. Let (X_{i}, φ_{i}) be a symmetric space for i = 1, 2 and let f : X_{1} →^{∼} X_{2} be an isomorphism. We say that f is an isometry if f^{#} φ_{2} f = φ_{1} and we denote it by (X_{1}, φ_{1}) ≃ (X_{2}, φ_{2}).

It is easy to check that (X_{1} ⊕ X_{2}, ( \begin{matrix}φ_{1} & 0 \\ 0 & φ_{2} \end{matrix} ) ) is a symmetric space called the orthogonal sum of the

former ones and denoted by (X_{1}, φ_{1}) ⊥ (X_{2}, φ_{2}). Orthogonal sum is compatible with isometries.

1.8. Definitions. We will call a triple (K, #, can) a triangulated category with duality if the class of isometry classes of symmetric spaces in K is a set. Denote this set by MW(K, #, can) or simply by MW(K). This set is given a structure of abelian monoid with ⊥. We call it the Witt monoid of K.

1.9. Definition. Let # be a δ-duality on K. Let (X, φ) be a symmetric space. A pair (L, α), where L is an object of K and α : L → X is a morphism, is called a sublagrangian of (X, φ) if α^{#}φ α = 0. A triple (L, α, w) is called a lagrangian if the following triangle is exact :

\xymatrix{T^{−1}(L^{#}) \ar[r]^{w} & L \ar[r]^{α} & X \ar[r]^{α^{#} φ} & L^{#}}

and if w is δ-symmetric”, i.e.

T^{−1}(w^{#}) = δ · w.

1.10. Explanations.

Since T^{−1}(w^{#}) : T^{−1}(L^{#}) → T^{−1}(T^{−1}(L^{#})^{#}) = L^{##}, the identity T^{−1}(w^{#}) = δ · w should be understood as T^{−1}(w^{#}) = can_{L} ∘ δ · w. In fact, this condition can be expressed by saying that there exists a commutative diagram in which the first line is an exact triangle containing α and the second line is the dual of the first one :

[figure: T^{−1}(L^{#}); w; L; α; _{can_{L}}; X; β; φ; L^{#}; T^{−1}(L^{#}); δ·T^{−1}(w^{#}); L^{##}; β^{#}; X^{#}; α^{#}; L^{#} .]

Heuristically, a lagrangian is first of all a sublagrangian, which means that the form induced on L via α is zero. But moreover, in K_{0}(K) (see for instance [1]), we have that [X] = [L] + [L^{#}]. The classical notion of lagrangian finds here its natural generalization. We want to stress the fact that α needs not to be split in this definition !

1.11. Definition. We say that a symmetric space possessing a lagrangian is neutral.

1.12. Proposition. Let (X, φ) be a symmetric space. Then (X, φ) ⊥ (X, −φ) is neutral.

1.13. Proof : Denote by Δ : X → X ⊕ X the diagonal. Check that (X, Δ, 0) is a lagrangian of (X, φ) ⊥ (X, −φ). ♯

1.14. Definition. Clearly the orthogonal sum of two neutral symmetric spaces is neutral again and neutrality is preserved by isometry. Denote by NW(K, #, can) the submonoid of MW(K, #, can) consisting of classes of neutral spaces. Denote by

W(K, #, can) = \frac{MW(K, #, can)}{NW(K, #, can)}

the quotient of these two monoids. This is the Witt group of K. We denote by [X, φ] the class of (X, φ).

1.15. Remark. Recall that if H ⊂ G is a submonoid of an abelian monoid G, we can define a relation on G by setting x ∼ x′ if there exist h, h′ ∈ H such that x + h = x′ + h′. The set G /∼ is again an abelian monoid. An immediate corollary of proposition 1.12 is that W(K, #, can) is really a group as claimed in the above definition; the opposite of [X, φ] is [X, −φ].

1.16. Definition. Two symmetric spaces (X, φ) and (Y, ψ) are said to be Witt-equivalent if [X, φ] = [Y, μ] in W(K).

1.17. Remark. It is convenient to observe that if can : Id → # ∘ # is an isomorphism of functors, then so is −can : Id → #∘#. A symmetric space in (K, #, −can) is simply a skew-symmetric space. Therefore we treat simultaneously both cases without particular distinction. Since we confess that we intend to often forget the mention of can : Id → # ∘ # in the following, this sign will be recalled as superscript in our notation. For instance W^{−1}(K) will denote the Witt group of skew-symmetric spaces.

1.18. Notation. Let ϵ = ±1 and n ∈ ℤ. As T^{n} ∘ # is a (−1)^{n}-duality, we introduce the notation :

W_{n}^{ϵ}(K) = W(K, T^{n} ∘ #, ϵ · can).

By the Witt group of K, we mean W_{0}^{1}(K) = W(K, #, can), which we simply abbreviate W(K).

1.19. Functoriality. Let K_{i} be a triangulated category with duality #_{i} for i = 1, 2. Suppose #_{1} and #_{2} are both exact or both skew-exact. A morphism of triangulated categories with duality is a covariant additive functor, exact or skew-exact (see remark 1.3) satisfying the following conditions :

F ∘ #_{1} = #_{2} ∘ F  and  F (can_{1}) = can_{2}.

In this case, if (X, φ) is a symmetric space for #_{1} then (F (X), F (φ)) is a symmetric space for #_{2}. It is very easy to prove that neutral forms are mapped to neutral forms : if (L, α, w) is a lagrangian of the starting space, then (F (L), F (α), δ · F (w)) is a lagrangian of its image, where δ = ±1 comes from the δ-exactness of F. Hence F induces a group homomorphism

W(F) : W(K_{1}, #_{1}, can_{1}) → W(K_{2}, #_{2}, can_{2}).

The Witt group is a functor from triangulated categories with duality (resp. with skew-duality) to abelian groups.

1.20. Proposition. For all n ∈ ℤ and ϵ = ±1 we have a canonical isomorphism :

W_{n}^{ϵ}(K) \underrightarrow{∼} W_{n+2}^{ϵ}(K).

1.21. Proof : According to the above considerations, the translation functor T : K → K is a morphism of triangulated categories from (K, #, ϵ can) to (K, T^{2}∘#, ϵ can) and more generally from (K, T^{n}∘#, ϵ can) to (K, T^{n+2} ∘ #, ϵ can). The isomorphism is W(T). ♯

2. Witt groups of a scheme.

2.1. Construction. Let X be a noetherian scheme. Denote by A(X) the abelian category of coherent O_{X}-modules and by L(X) the additive full subcategory of A(X) consisting of locally free O_{X}-modules. Denote by F : A(X) → A(X) the functor HomO_{X} (−, O_{X}) and observe that F (L(X)) ⊂ L(X). We still denote by F the natural extension of F to the triangulated category of bounded complexes with morphisms up to homotopy :

\begin{matrix}F : K^{b}(A(X)) & ⟶ K^{b}(A(X)) \\ (M_{i}, ∂_{i})_{i=−∞}^{∞} & ↦ (F (M_{−i}), F (∂_{−i+1}))_{i=−∞}^{∞} \end{matrix} (See 4.4).

Observe the following points :

(i) F (K^{b}(L(X))) ⊂ K^{b}(L(X)),

(ii) F is exact in the sense of definition 1.1,

(iii) if M ∈ K^{b}(L(X)) is acyclic (i.e. H_{i}(M) = 0 for all i ∈ ℤ) then so is F (M),

(iv) when M ∈ K^{b}(L(X)), the canonical morphism can_{M} : M → F (F (M)) is an isomorphism. Hence F induces an exact functor on the localization of K^{b}(L(X)) with respect to quasi-isomorphisms, which we will denote by D^{b}(L(X)).

2.2. Notation. Let X be a noetherian scheme. With the above notations, we put

K(X) = D^{b}(L(X))

and we denote by #_{X} : K(X) → K(X) the localization of F.

2.3. Proposition. Let X be a noetherian scheme. Then (K(X), #_{X}) is a triangulated category with duality.

2.4. Proof : On K^{b}(L(X)) there exists an isomorphism of functors can : Id →^{∼} F ∘ F. It is easy to see that (K^{b}(L(X)), F, can) is a triangulated category with duality. One checks as well the same identities in the derived category. More generally, we can define the localization of a triangulated category with duality, with respect to a multiplicative system of morphisms compatible with the duality. Here K(X) turns out to be the localization of the triangulated category K^{b}(L(X)) with respect to the multiplicative system of quasi-isomorphisms. ♯

2.5. Definition. Let X be a noetherian scheme. Let n ∈ ℤ and ϵ = ±1. We put

W_{n}^{ϵ}(X) = W_{n}^{ϵ}(K(X)).

These are the Witt groups of X.

2.6. Notation. The Witt group of X will be W(X) = W_{0}^{1}(X). To avoid confusion, we denote the usual Witt group of X by W_{us}(X). When X = Spec(A), we write W(A) for W(Spec(A)); and so on …

2.7. Remark. The reader could verify the functoriality of these groups W_{n}^{ϵ}(X) as a familiarizing exercise.

2.8. Remark. Suppose our scheme X is regular in the following sense : every coherent O_{X}-module has a finite resolution by locally free coherent O_{X}-modules. In that case, it is well known that our category K(X) is equivalent to the derived category D^{b}(A(X)) and #_{X} is the derived functor R^{b}F of the left exact functor F of 2.1 (see [6] Thm 10.5.9, p. 393).

2.9. Notation. Denote by i_{0} : L(X) → D^{b}(L(X)) the functor that associates to any O_{X}-module the complex concentrated in degree 0:

E ↦ i_{0}(E) := ⋯0 ⟶ 0 ⟶ E ⟶ 0 ⟶ 0⋯

2.10. Proposition. The map (E, φ) ↦ (i_{0}(E), i_{0}(φ)) induces a group homomorphism

ι_{X} : W_{us}(X) → W(X).

2.11. Proof : It is obvious that #_{X}(i_{0}(E)) = i_{0}(E^{∗}). Isometric forms map to isometric forms, orthogonal sum to orthogonal sum and so on. We only have to prove that if there exists an exact sequence of the form :

0 ⟶ F ⟶^{γ} E ⟶^{γ^{∗}∘φ} F ⟶ 0

with F ∈ L(X) then (M, ψ) := (i_{0}(E), i_{0}(φ)) is neutral. Set L = i_{0}(F) and α = i_{0}(γ). We are going to find w such that (L, α, w) is a lagrangian of (M, ψ).

Let C denote the mapping cone of α. We have an exact triangle :

\xymatrix{L \ar[r]^{α} & M \ar[r]^{j} & C \ar[r]^{k} & T (L)}

where j and k are the usual morphisms. Here k is simply :

[figure: C =; k; ⋯0; F; −γ; −1; E; 0; 0⋯; T (L) =; ⋯0; F; 0; 0; 0⋯]

Consider the quasi-isomorphism :

\xymatrix{C = \ar[d]_{s} & ⋯0 \ar[r] & F \ar[r]^{−γ} \ar[d] & E \ar[r] \ar[d]^{γ^{∗}∘φ} & 0 \ar[r] \ar[d] & 0⋯ \\ L^{#} = & ⋯0 \ar[r] & 0 \ar[r] & F^{∗} \ar[r] & 0 \ar[r] & 0⋯}

We now have an exact triangle in D^{b}(L(X)) :

\xymatrix{L \ar[r]^{α} & M \ar[r]^{s∘j} & L^{#} \ar[r]^{k∘s^{−1}} & T (L)} .

Observe that this triangle would not exist in K^{b}(L(X)).

A direct calculation gives s ∘ j = α^{#} ∘ ψ and if we set w = −T^{−1}(k ∘ s^{−1}) we have an exact triangle :

\xymatrix{T^{−1}(L^{#}) \ar[r]^{w} & L \ar[r]^{α} & M \ar[r]^{α^{#}∘ψ} & L^{#}} .

It suffices to show that T^{−1}(w^{#}) = w, which is equivalent to T (s^{#}) ∘ k = T (k^{#}) ∘ s in D^{b}(L(X)). An immediate verification shows that these two morphisms of complexes are homotopic. Hence (L, α, w) is a lagrangian of (M, ψ). ♯

2.12. Remark. As noticed in this proof, the natural map sending locally free O_{X}-modules with symmetric forms to complexes concentrated in degree 0 in K^{b}(L(X)) would not send neutral forms (in the usual sense) to neutral ones. This is the main reason for introducing D^{b}(L(X)) instead of K^{b}(L(X)).

3. Techniques in a triangulated category.

As astonishing as it may seem, it is possible to proceed to some calculations even in such an abstract setting as triangulated categories. We give here a few results that we will use in other sections.

From now on K denotes a triangulated category with a 1-duality #. The following results can easily be adapted to skew-dualities and we will use them in this case as well.

We wittingly omit the canonical isomorphism can : Id →^{∼} # ∘ # to lighten notations and therefore consider # ∘ # = Id. The masochist reader may divert himself in restoring everywhere the mention of the canonical isomorphism.

3.1. Hypothesis. From now on we suppose that \frac{1}{2} belongs to our category K. This simply means that we can divide by 2 in every group of morphisms Hom_{K}(A, B).

3.2. Lemma. Let \xymatrix{A \ar[r]^{u} & B \ar[r]^{v} & C \ar[r]^{w} & T (A)} be an exact triangle and let (f, g, h) be a morphism from this triangle to itself. Suppose that two of these endomorphisms are nilpotent. Then so is the third one.

3.3. Proof : Suppose that f and g are nilpotent. Composing (f, g, h) with itself a sufficient number of times, we may suppose f = 0 and g = 0. In that case, w ∘ h = 0 gives h = v ∘ h̃ for some h̃ : C → B and then h^{2} = h ∘ v ∘ h̃ = 0 since h ∘ v = 0. ♯ 3.4. Lemma. Let \xymatrix{T^{−1}(L^{#}) \ar[r]^{s} & L \ar[r]^{s_{1}} & X \ar[r]^{s_{2}} & L^{#}} be an exact triangle such that T^{−1}(s^{#}) = s. Suppose that φ, ψ : X → X^{#} are two forms on X, both making the following diagram commute :

[figure: T^{−1}(L^{#}); s; 1; L; s_{1}; 1; X; s_{2}; φ,ψ; ≃; L^{#}; T^{−1}(L^{#}); T^{−1}(s^{#}); L; s_{2}^{#}; X^{#}; s_{1}^{#}; L^{#}] .

Then φ and ψ are isometric.

3.5. Proof : Since (1, 1, φ) and (1, 1, ψ) are isomorphisms, (1, 1, ψ^{−1}φ) is an automorphism of the first triangle and (0, 0, −1 + ψ^{−1}φ) is an endomorphism. If we define h := −1 + ψ^{−1}φ, lemma 3.2 gives : h^{2} = 0. We have ψ(1 + h) = φ and therefore h^{#}ψ = ψ h. Hence (1 + \frac{h}{2})^{#}ψ(1 + \frac{h}{2}) = ψ(1 + h) = φ. ♯

3.6. Remark. It is well known that in a triangulated category, if (f, g, h) is a morphism of exact triangles and if two of these morphisms are isomorphisms so is the third. Nevertheless, given f and g, the h such that (f, g, h) is a morphism is not unique !

The very easy lemma 3.4 is then a key result, allowing us to pass through this classical problem in the theory of Witt groups. In particular, lemma 3.4 implies that a neutral form is characterized up to isometry by w : T^{−1}L^{#} → L such that T^{−1}w^{#} = w (see 1.9 and 1.10).

3.7. Sublagrangian construction.

It is natural to wonder whether we could divide out any symmetric space by an isotropic subspace. This means constructing a form on a reduced space. In the classical context, the orthogonal subspace of a given subspace is well defined. In triangulated categories, since “cones” are not unique, this notion has to be more flexible. In fact, it turns out that under a small assumption on the sublagrangian this construction can be carried out independently of the choice of an orthogonal. Let us be more precise.

3.8. Lemma. Let (X, φ) be a symmetric space and (L, α) a sublagrangian. Choose any triangle on α, say

\xymatrix{ & M^{#} \ar[dl]_{−T^{−1}(α_{0})} \\ L \ar[rr]_{α} & & X \ar[ul]_{α_{2}}} .

There exists η_{0} : L → M such that the following diagram commutes :

\xymatrix{T^{−1}(M^{#}) \ar[r]^{α_{0}} \ar[d]_{T^{−1}(η_{0}^{#})} & L \ar[r]^{α} \ar[d]_{η_{0}} & X \ar[r]^{α_{2}} \ar[d]^{φ}_{≃} & M^{#} \ar[d]^{_{η_{0}_{#}}} \\ T^{−1}L^{#} \ar[r]_{T^{−1}(α_{0}^{#})} & M \ar[r]_{α_{2}^{#}} & X^{#} \ar[r]_{α^{#}} & L^{#}} .

where the second line is the dual triangle of the first one. M could be understood as an orthogonal of L.

3.9. Proof : Since α^{#} ∘ (φ α) = 0, there exists η : L → M such that φ α = α_{2}^{#} ∘ η. Complete the diagram with a morphism (ν, η, φ). Observe that (T^{−1}(η^{#}), T^{−1}(ν^{#}), φ) is also a morphism between these two exact triangles. Putting

η_{0} = \frac{η + T^{−1}ν^{#}}{2} ,

we get the desired one. ♯ 3.10. Question. Choosing any exact triangle containing η_{0}, say

\xymatrix{L \ar[r]^{η_{0}} & M \ar[r]^{η_{1}} & Y \ar[r]^{η_{2}} & T (L)} ,

is there a symmetric form μ = μ^{#} : Y →^{∼} Y^{#} such that [X, φ] = [Y, μ] in W(K) ? Heuristically, Y plays the role of L^{⊥}/L.

3.11. Theorem. Let (X, φ) be a symmetric space and α : L → X such that α^{#} φ α = 0. Suppose Hom_{K}(L, T^{−1}L^{#}) = 0. Then, with the above notations, there exists a symmetric form μ : Y → Y^{#} Witt-equivalent to (X, φ).

3.12. Proof : Keep the notations of lemma 3.8. Define s := η_{0} α_{0} : T^{−1}(M^{#}) → M and notice that T^{−1}(s^{#}) = s.

Observe that α ∘ T^{−1}(η_{2}) = 0. Applying the axiom of the octahedron to this relation, we get an exact triangle :

[figure: T^{−1}(M^{#}); s; M; ^{( \begin{matrix}x \\ η_{1} \end{matrix} )}; X ⊕ Y; ^{(^{α_{2} y})}; M^{#}]

for some morphisms x : M → X and y : Y → M^{#} satisfying:

x η_{0} = α  and  η_{2} = T (α_{0}) y.

Since (x−φ^{−1} α_{2}^{#}) η_{0} = 0 there exists h : Y → X such that x+h η_{1} = φ^{−1} α_{2}^{#}. Using the automorphism

( \begin{matrix}1 & h \\ 0 & 1 \end{matrix} ) of X ⊕ Y, we may suppose x = φ^{−1} α_{2}^{#}, keeping η_{2} = T (α_{0}) y.

Now we construct another octahedron for the identity ( 1 0 ) · ( \begin{matrix}φ^{−1} α_{2}^{#} \\ η_{1} \end{matrix} ) = φ^{−1} α_{2}^{#} and we get an

exact triangle :

Y ⟶^{y} M^{#} ⟶^{y_{1}} L^{#} ⟶^{T (η_{1}) α_{0}^{#}} T (Y)

such that y_{1}α_{2} = α^{#}φ. From this equality, we get that (y_{1}−η_{0}^{#}) α_{2} = 0 and therefore y_{1}−η_{0}^{#} = k∘T (α_{0}) for a morphism k : T L → L^{#}. Our hypothesis on L insures that k = 0. That is y_{1} = η_{0}^{#}.

After this, compare the two triangles containing η_{0}^{#} to get a non necessarily symmetric isomorphism ψ : Y^{#} → Y

[figure: T^{−1}L^{#}; −η_{2}^{#}; Y^{#}; ^{η_{1}^{#}}; ∃ψ; ≃; M^{#}; ^{η_{0}^{#}}; L^{#}; T^{−1}L^{#}; −η_{1}∘T^{−1}(α_{0}^{#}); Y; y; M^{#}; _{η_{0}_{#}}; L^{#}]  (1)

There exists a morphism δ : M → M such that the following diagram commutes :

[figure: T^{−1}(M^{#}); s; M; ^{( \begin{matrix}φ^{−1} α_{2}^{#} \\ η_{1} \end{matrix} )}; X ⊕ Y; ^{(^{α_{2} y})}; M^{#}; T^{−1}(M^{#}); T^{−1}s^{#}=s; M; _{( \begin{matrix}α_{2} & ^{#} \\ y^{#} \end{matrix} )}; ∃ 1+δ; X^{#} ⊕ Y^{#}; _{( \begin{matrix}φ^{−1} & 0 \\ 0 & ψ \end{matrix} )}; _{(_{α_{2} φ^{−1} η_{1}^{#}} )}; M^{#}]  (2)

We are going to show that δ^{3} = 0. From diagram (2) we get that α_{2}^{#} δ = 0 and then δ = T^{−1}(α_{0}^{#}) δ̃ for some δ̃ : M → T^{−1}L^{#}. Since Hom_{K}(L, T^{−1}L^{#}) = 0, we have that δ̃η_{0} = 0 and then δ = δ̄η_{1} for some δ̄: Y → M. Hence δ^{3} = (δ̄η_{1}) ∘ δ ∘ (T^{−1}(α_{0}^{#}) δ̃) and it suffices to show that η_{1} ∘ δ ∘ T^{−1}(α_{0}^{#}) = 0. To prove this remember that T (α_{0}) y = η_{2} and use it in the following computation : η_{1} T^{−1}(α_{0}^{#}) =^{(1)} ψη_{2}^{#} =

ψ y^{#} T^{−1}(α_{0}^{#}) =^{(2)} η_{1} (1 + δ) T^{−1}(α_{0}^{#}) = η_{1} T^{−1}(α_{0}^{#}) + η_{1} ∘ δ ∘ T^{−1}(α_{0}^{#}). Hence δ^{3} = 0.

In the end, taking the mean between the morphism of diagram (2) and its dual, we get the wanted symmetric isomorphism on Y :

[figure: T^{−1}(M^{#}); s; M; ^{( \begin{matrix}φ^{−1} α_{2}^{#} \\ η_{1} \end{matrix} )}; X ⊕ Y; ^{(^{α_{2} y})}; M^{#}; T^{−1}(M^{#}); T^{−1}s^{#}; 1+ \frac{1}{2} T^{−1}δ^{#}; ≃; M; _{( \begin{matrix}α_{2} & ^{#} \\ y^{#} \end{matrix} )}; 1+ \frac{1}{2} δ; ≃; X^{#} ⊕ Y^{#}; _{( \begin{matrix}φ^{−1} & 0 \\ 0 & \frac{1}{2} (ψ + ψ^{#}) \end{matrix} )}; _{(_{α_{2} φ^{−1} η_{1}^{#}} )}; M^{#}; 1+ \frac{1}{2} δ^{#}; ≃]

Since δ s = 0, we have also s ∘ T^{−1}δ^{#} = 0. The proof reduces then to next lemma. ♯

3.13. Lemma. Let (Z, ξ) be a symmetric space and let

\xymatrix{T^{−1}M^{#} \ar[r]^{s} & M \ar[r]^{s_{1}} & Z \ar[r]^{s_{2}} & M^{#}}

be an exact triangle with s = T^{−1}s^{#}. Let h : M →^{∼} M be such that the following diagram commutes :

\xymatrix{T^{−1}(M^{#}) \ar[r]^{s} \ar[d]^{T^{−1}h^{#}}_{≃} & M \ar[r]^{s_{1}} \ar[d]^{h}_{≃} & Z \ar[r]^{s_{2}} \ar[d]^{ξ=ξ^{#}}_{≃} & M^{#} \ar[d]^{h^{#}}_{≃} \\ T^{−1}(M^{#}) \ar[r]_{T^{−1}s^{#}} & M \ar[r]_{_{s_{2}_{#}}} & Z^{#} \ar[r]_{_{s_{1}_{#}}} & M^{#}}

and such that h ∘ s = s.

Then (Z, χ) is neutral.

3.14. Proof : From h ∘ s = s, we get an isomomorphism of triangles :

[figure: T^{−1}(M^{#}); s; M; s_{1}; h; ≃; Z; s_{2}; M^{#}; T^{−1}(M^{#}); s; M; s_{1}∘h^{−1}; Z^{#}; s_{2}; M^{#}]

Since the first triangle is exact, so is the second. We check that (M, s_{1} ∘ h^{−1}, s) is a lagrangian of (Z, ξ). It remains to see that (s_{1} ∘ h^{−1})^{#} ∘ ξ = s_{2} which is immediate :

(s_{1} ∘ h^{−1})^{#} ∘ ξ = (h^{−1})^{#} ∘ s_{1}^{#} ∘ ξ = (h^{−1})^{#} ∘ h^{#} ∘ s_{2} = s_{2}.
♯

4. The isomorphism.

4.1. Theorem. Let A be a noetherian ring in which 2 is a unit. Then the homomorphism

ι_{A} : W_{us}(A) → W(A)

is surjective.

4.2. Proof : In the construction of section 2, the scheme X is Spec(A), the locally free O_{X}-modules are the projective A-modules : L(X) = P and it is well known that D^{b}()P ≃ K^{b}()P.

It is obvious that any bounded complex endowed with a symmetric form φ:

[figure: X =; φ; ⋯0; P_{n}; ∂_{n}; φ_{n}; P_{n−1}; ∂_{n−1}; φ_{n−1}; ⋯; ∂_{−n+2}; P_{−n+1}; ∂_{−n+1}; φ_{−n+1}; P_{−n}; φ_{−n}; 0⋯; X^{#} =; ⋯0; P_{−n}^{∗}; ∂_{−n+1}^{∗}; P_{−n+1}^{∗}; ∂_{−n+2}^{∗}; ⋯; ∂_{n−1}^{∗}; P_{n−1}^{∗}; ∂_{n}^{∗}; P_{n}^{∗}; 0⋯]

admits the following sub-lagrangian when n ≥ 1 :

[figure: L =; α; ⋯0; 0; ⋯; 0; P_{−n}; 0⋯; X =; ⋯0; P_{n}; ∂_{n}; ⋯; ∂_{−n+2}; P_{n−1}^{∗}; ∂_{n}^{∗}; P_{−n}; 0⋯]

It is clear that Hom(L, T^{−1}L^{#}) = 0. So we may apply theorem 3.11. But with the notations of the sublagrangian construction, we may choose here for Y a complex shorter than X. To see this, choose your exact triangle containing α, observing that X = C(α_{0}) is the mapping cone of the morphism of complexes :

[figure: T^{−1}M^{#} :=; α_{0}; ⋯0; 0; P_{n}; −∂_{n}; ⋯; −∂_{−n+2}; P_{−n+2}; −∂_{−n+2}; P_{−n+1}; −∂_{−n+1}; 0⋯; L =; ⋯0; 0; 0; ⋯; 0; P_{−n}; 0⋯]

Then verify that Y is of the form :

⋯0 ⟶ 0 ⟶ Q_{n−1} ⟶^{∂′_{n−1}} ⋯ ⟶^{∂′_{−n+1}} Q_{−n} ⟶ 0· · ·.

As we have a homotopy equivalence ψ : Y → Y^{#}, Id_{Y} is homotopic to a map factorising as Y → Y^{#} → Y. This map is (by the structure of Y) zero in degree −n. This implies that ∂′_{−n+1} is a split epimorphism. Hence Y is homotopically equivalent to a shorter complex.

By induction we come down to a complex concentrated in degree zero. This gives the result. ♯

In order to prove that ι_{A} is an isomorphism we are going to construct an inverse Ω_{A} : W(A) → W_{us}(A). This is the goal of the end of this section. We will proceed step by step, the first step being definition 4.4.

4.3. Remark. Given a symmetric form φ on a bounded complex of projective A-modules, since \frac{1}{2} ∈ A, we may suppose that φ is strongly symmetric in the sense that φ_{−i}^{∗} = φ_{i} for all i ∈ ℤ.

4.4. Definition. Let (X, φ, φ̄, ϵ) be as follow : X is a bounded complex of projective A-modules, φ : X → X^{#}, φ̄ : X^{#} → X are such that φ^{#} = φ, φ̄^{#} = φ̄, and φ̄ ∘ φ ∼ Id with homotopy ϵ.

[figure: X =; φ; ⋯0; P_{n}; ∂_{n}; φ_{n}; P_{n−1}; ∂_{n−1}; φ_{n−1}; ⋯; ∂_{−n+2}; P_{−n+1}; ∂_{−n+1}; φ_{−n+1}; P_{−n}; φ_{−n}; 0⋯; X^{#} =; φ̄; ⋯0; P_{−n}^{∗}; ∂_{−n+1}^{∗}; φ̄_{n}; P_{−n+1}^{∗}; ∂_{−n+2}^{∗}; φ̄_{n−1}; ⋯; ∂_{n−1}^{∗}; P_{n−1}^{∗}; ∂_{n}^{∗}; φ̄_{−n+1}; P_{n}^{∗}; φ̄_{−n}; 0⋯; X =; ⋯0; P_{n}; ∂_{n}; P_{n−1}; ∂_{n−1}; ⋯; ∂_{−n+2}; P_{−n+1}; ∂_{−n+1}; P_{−n}; 0⋯]

[figure: X =; φ̄ ∘ φ; ⋯0; P_{n}; ∂_{n}; φ̄_{n} φ_{n}; P_{n−1}; ∂_{n−1}; φ̄_{n−1} φ_{n−1}; ϵ_{n−1}; ⋯; ∂_{−n+2}; P_{−n+1}; ∂_{−n+1}; φ̄_{−n+1} φ_{−n+1}; P_{−n}; φ̄_{−n} φ_{−n}; ϵ_{−n}; 0⋯; X =; ⋯0; P_{n}; ∂_{n}; P_{n−1}; ∂_{n−1}; ⋯; ∂_{−n+2}; P_{−n+1}; ∂_{−n+1}; P_{−n}; 0⋯]

Suppose φ and φ̄ strongly symmetrical : φ_{−i}^{∗} = φ_{i} and φ̄_{−i}^{∗} = φ̄_{i} for all i ∈ ℤ. We have Id_{X} = φ̄ φ + ∂_{X} ϵ + ϵ ∂_{X} and therefore Id_{X^{#}} = φ^{#} φ̄^{#} + ϵ^{#} ∂_{X^{#}} + ∂_{X^{#}} ϵ^{#}, where we have to pay attention to the indices of ϵ^{#} which are shifted by 1 (that is (ϵ^{#})_{i} := ϵ_{−i−1}^{∗}). We set

Ω(M, φ, φ̄, ϵ) = [Y, ψ] ∈ W_{us}(A)

where

[figure: Y =; ψ; ⋯ ⊕ P_{−3}^{∗} ⊕ P_{2} ⊕ P_{−1}^{∗} ⊕ P_{0} ⊕ P_{1}^{∗} ⊕ P_{−2} ⊕ P_{3}^{∗} ⊕ ⋯; _{( \begin{matrix}& & & & & & & & . .. & . .. & . .. \\ & 0 & 0 & 0 & 0 & 0 & 0 & ∂_{−2} & φ̄_{−3} & −ϵ_{−4} \\ & 0 & 0 & 0 & 0 & 0 & ∂_{2}^{∗} & φ_{2} & −ϵ_{2}^{∗} & 0 \\ & 0 & 0 & 0 & 0 & ∂_{0} & φ̄_{−1} & −ϵ_{−2} & 0 & 0 \\ ⋯ & 0 & 0 & 0 & ∂_{0}^{∗} & φ_{0} & −ϵ_{0}^{∗} & 0 & 0 & 0 & ⋯ \\ & 0 & 0 & ∂_{2} & φ̄_{1} & −ϵ_{0} & 0 & 0 & 0 & 0 \\ & 0 & ∂_{−2}^{∗} & φ_{−2} & −ϵ_{−2}^{∗} & 0 & 0 & 0 & 0 & 0 \\ & ∂_{4} & φ̄_{3} & −ϵ_{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ . .. & . .. & . .. \end{matrix} )}; Y^{∗} =; ⋯ ⊕ P_{−3} ⊕ P_{2}^{∗} ⊕ P_{−1} ⊕ P_{0}^{∗} ⊕ P_{1} ⊕ P_{−2}^{∗} ⊕ P_{3} ⊕ ⋯]

We also define the homomorphism ψ̄ : Y^{∗} → Y by

ψ̄ = ( \begin{matrix}& & & & & & & & . .. & . .. & . .. \\ & 0 & 0 & 0 & 0 & 0 & 0 & ϵ_{−3}^{∗} & φ_{3} & −∂_{−3}^{∗} \\ & 0 & 0 & 0 & 0 & 0 & ϵ_{1} & φ̄_{2} & −∂_{3} & 0 \\ & 0 & 0 & 0 & 0 & ϵ_{−1}^{∗} & φ_{1} & −∂_{−1}^{∗} & 0 & 0 \\ ⋯ & 0 & 0 & 0 & ϵ_{−1} & φ̄_{0} & −∂_{1} & 0 & 0 & 0 & ⋯ \\ & 0 & 0 & ϵ_{1}^{∗} & φ_{−1} & −∂_{1}^{∗} & 0 & 0 & 0 & 0 \\ & 0 & ϵ_{−3} & φ̄_{−2} & −∂_{−1} & 0 & 0 & 0 & 0 & 0 \\ & ϵ_{3}^{∗} & φ_{−3} & −∂_{3}^{∗} & 0 & 0 & 0 & 0 & 0 & 0 \\ . .. & . .. & . .. \end{matrix} )

4.5. Lemma. With the above notations, ψ^{∗} = ψ, ψ̄^{∗} = ψ̄ and ψ ψ̄ = ( \begin{matrix}1 & & & & 0 \\ & . & . & . \\ ⋆ & & & & 1 \end{matrix} ). Hence, (Y, ψ) is a

symmetric space.

4.6. Proof : Direct computation using ∂ ∘ ∂ = 0, ∂ φ = φ ∂, Id_{X} = φ̄ φ + ∂_{X} ϵ + ϵ ∂_{X}, and so on … ♯

4.7. Important remark. In fact the above construction could be carried out with any homotopy equivalence f : X → X′ with inverse g : X′ → X and homotopies Id_{X} = g f + dϵ + ϵd and Id_{X′} = f g + d′ϵ′ + ϵ′d′. We construct in a similar way an isomorphism (!) from an alternate sum of terms of X and terms of X′ to the alternate sum of the remaining terms. Replacing in the above definition φ by f, φ̄ by g, ϵ by ϵ and ϵ^{∗} by ϵ′, you obtain two homomorphisms ρ and ρ̄ with zeros outside the three diagonals. A direct computation gives that ρ · ρ̄ is lower triangular with 1 in the diagonal.

It is not the subject of this note, but this construction might induce a homomorphism from the “derived” K_{1} of A to the usual K_{1}(A). Apart from this, we will use this construction in the end of the proof.

Here comes now a very useful result, which allows us to define Ω independently of any choice. It shows that we can change ψ under the three diagonals without changing the isometry class of ψ.

4.8. Lemma. With the above notations, consider ψ̃: Y → Y^{∗} such that ψ̃^{∗} = ψ̃ and

ψ̃ − ψ = ( \begin{matrix}0 & & & 0 \\ & & . .. & 0 \\ & . .. & . .. \\ 0 & 0 & & ⋆ \end{matrix} ) .

Then ψ̃ ≃ ψ. In particular, ψ̃ is an isomorphism.

4.9. Proof : We have ψ̃ = ψ + a, where a is the matrix in the statement. That is ψ̃ = (1 + b)ψ, where b = a · ψ^{−1}. Since ψ ψ̄ = E, where E is the matrix of lemma 4.5, ψ^{−1} = ψ̄ E^{−1}. As E^{−1} is also lower triangular with 1 in the diagonal, using the hypothesis on a it is easy to check that b is lower triangular with 0 in the diagonal. Hence b is nilpotent. Since \frac{1}{2} ∈ A, there exists a polynomial P ∈ A[X] such that P (b)^{2} = 1 + b. Then one verifies that P (b) ψ P (b)^{∗} = ψ̃. ♯

For the moment, Ω(X, φ, φ̄, ϵ) is well defined for strongly symmetric φ and φ̄ and for a homotopy ϵ : φ̄ φ ∼ Id. First of all, we are going to show that Ω is independent of the choice of ϵ.

4.10. Lemma. With the above notations, suppose we are given μ : X → X of degree +1 such that ∂ μ + μ ∂ = 0. Then Ω(X, φ, φ̄, ϵ) = Ω(X, φ, φ̄, ϵ + μ).

4.11. Proof : Denote by ψ̌ the form induced on Y in the definition of Ω(X, φ, φ̄, ϵ + μ). Consider h : Y^{∗} → Y^{∗} defined by :

h = ( \begin{matrix}... & ... & ... \\ 0 & 1 & 0 & & & 0 \\ −μ_{−2} ϵ_{−3} & −μ_{−2} φ̄_{−2} & 1 + μ_{−2} ∂_{−1} & 0 \\ 0 & 0 & 0 & 1 & 0 \\ & 0 & −μ_{0} ϵ_{−1} & −μ_{0} φ̄_{0} & 1 + μ_{0} ∂_{1} & 0 \\ 0 & & 0 & 0 & 0 & 1 & & 0 \\ ... & ... & ... & ... \end{matrix} ) .

Direct computation gives that h ψ h^{∗} = ψ̌ + a, where a is as in lemma 4.8. Since ψ̌ is a form, lemma 4.8 implies that h ψ h^{∗} is isometric to ψ̌ and in particular is an isomorphism. Then h is onto and (as h is an endomorphism) it is an isomorphism. Thus, ψ ≃ ψ̌. ♯

4.12. Corollary. Ω(X, φ, φ̄, ϵ) does not depend on the choice of ϵ.

4.13. Proof : Two such ϵ would differ by a μ as in lemma 4.10. ♯

We now want to prove that Ω does not depend on the choice of the strongly symmetric φ representing the form.

4.14. Lemma. With the above notations, suppose we are given μ : X → X^{#} of degree +1 such that μ = μ^{#} (paying attention to indices, this equality means that μ_{−i+1}^{∗} = μ_{i} for all i ∈ ℤ). Then

Ω(X, φ, φ̄, ϵ) = Ω(X, φ + ∂^{∗}μ + μ∂, φ̄ , ϵ − φ̄μ).

4.15. Proof : Observe first of all that φ̌ := φ + ∂^{∗}μ + μ∂ is another strongly symmetric form homotopic to φ and that ϵ − φ̄μ gives a homotopy φ̌ φ̄ ∼ Id. Hence Ω(X, φ + ∂^{∗}μ + μ∂, φ̄ , ϵ − φ̄μ) is well-defined. Denote by ψ̌ the form on Y induced by φ̌. Define h : Y → Y by

h = ( \begin{matrix}... & μ_{2} \\ & 1 & 0 & & & 0 \\ & & 1 & μ_{0} \\ & & & 1 & 0 \\ 0 & & & & 1 & μ_{−2} \\ . & . & . \end{matrix} ) .

Lemma 4.8 applied to h^{#} ψ h and to ψ̌ gives the result. ♯

4.16. Corollary. Ω(X, φ, φ̄, ϵ) depends only on the space X and on the homotopy class of φ : X → X^{#}.

4.17. Proof : Two strongly symmetric φ differ by a null homotopic morphism with a homotopy that might be chosen symmetric because \frac{1}{2} ∈ A. Hence we apply lemma 4.14. We prove in the same way that Ω(X, φ, φ̄, ϵ) does not depend on the choice of the strongly symmetric inverse φ̄. ♯

4.18. Notation. Let (X, f) be a symmetric space in K^{b}(P). We put

Ω(X, f) = Ω(X, φ, φ̄, ϵ) ∈ W_{us}(A)

where f = [φ], f^{−1} = [φ̄], φ and φ̄ are strongly symmetric and where ϵ is any homotopy Id = φ̄ φ+∂ϵ+ϵ∂. This is well defined by corollaries 4.12 and 4.16.

4.19. Remark. At this point, we have a well defined map Ω on symmetric spaces. It is obvious that Ω((X, φ) ⊥ (X′, φ′)) = Ω(X, φ) ⊥ Ω(X′, φ′). We now have to prove that Ω is invariant by isometry (isomorphisms are homotopy equivalences !) and that Ω sends neutral forms to zero.

4.20. Lemma. Let L be a bounded complex of projective A-modules and let a morphism of complexes w : T^{−1}L^{#} → L be strongly symmetric in the sense that T^{−1}w^{#} = w as morphisms of complexes. Let X = C(w) be the mapping cone of w and let φ = C(Id, can) the map given by the naturality of the mapping cone construction (see 1.10). Then φ is a form on X and Ω(X, φ) = 0.

4.21. Proof : Let w : T^{−1}L^{#} → L be given by :

[figure: L :=; w; ⋯0; P_{n}; ∂_{n}; P_{n−1}; ∂_{n−1}; w_{n−1}; ⋯; ∂_{−n+2}; P_{−n+1}; ∂_{−n+1}; w_{−n+1}; P_{−n}; w_{−n}; 0; 0⋯; T^{−1}(L^{#}) =; ⋯0; 0; P_{−n}^{∗}; −∂_{−n+1}^{∗}; ⋯; −∂_{n−2}^{∗}; P_{n−2}^{∗}; −∂_{n−1}^{∗}; P_{n−1}^{∗}; −∂_{n}^{∗}; P_{n}^{∗}; 0⋯]

and satisfy w_{−i−1}^{∗} = w_{i} for all i ∈ ℤ. We may give φ explicitly by :

φ_{i} : \underbrace{P_{−i−1}^{∗} ⊕ P_{i−1}}_{C(w)_{i}} \overrightarrow{( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )} \underbrace{P_{i−1} ⊕ P_{−i−1}^{∗}}_{(C(w)^{#})_{i}}

for all i ∈ ℤ. It is then a quite easy exercise to compute Ω(X, φ) = 0 and to show that it is metabolic. This is left to the reader. ♯

4.22. Remark. This does not insure that neutral forms map to zero since the form of the above lemma is very particular. A neutral form is isometric to such a form by lemma 3.4. The very point is to prove that Ω is well defined up to homotopy equivalence.

4.23. Lemma. Let (X, φ) be a form and h : Y → X be an isomorphism of complexes (that is h_{i} = isomorphism for all i ∈ ℤ). Then, Ω(X, φ) = Ω(Y, h^{#} φ h).

We have Ω(X, −φ) = −Ω(X, φ) as well.

4.24. Proof : These are easy exercises with the following indications.

For the first assertion, use the obvious diagonal isometry (with h_{i}, h_{i}^{−1}, h_{i}^{∗} or (h_{i}^{−1})^{∗} in the diagonal) and corollary 4.12.

For the second one, use the isometry ( \begin{matrix}... & & & 0 \\ & −1 \\ & & 1 \\ & & & −1 \\ & 0 & & & ... \end{matrix} ). ♯

4.25. Lemma. Let X and Y be bounded complexes and let α : X →^{∼} Y^{#} be a homotopy equivalence (an isomorphism in our category K^{b}()P). Then

Ω(X ⊕ Y, ( \begin{matrix}0 & α^{#} \\ α & 0 \end{matrix} )) = 0.

4.26. Proof : Expliciting the homomorphism of definition 4.4 and regrouping the parts coming from X and the parts coming from Y, it is not too hard to see that

Ω(X ⊕ Y, ( \begin{matrix}0 & α^{#} \\ α & 0 \end{matrix} )) ≃ ( \begin{matrix}0 & ρ^{∗} \\ ρ & 0 \end{matrix} ) ,

where ρ is the isomorphism associated to α through the construction of remark 4.7. ♯

4.27. Corollary. Let (X, φ) be a form and let h : Y →^{∼} X be now any isomorphism (h is a homotopy equivalence, compare with lemma 4.23). Then Ω(X, φ) = Ω(Y, h^{#} φ h).

4.28. Proof : Observe that :

Ω(h^{#} φ h) − Ω(φ) =^{lemma 4.23} Ω(h^{#} φ h) + Ω(−φ) =^{rem 4.19} Ω(( \begin{matrix}h^{#} φ h & 0 \\ 0 & −φ \end{matrix} )).

Using elementary transformations and the fact that \frac{1}{2} ∈ A, we can find an isomorphism of complexes g such that

g^{#} ( \begin{matrix}h^{#} φ h & 0 \\ 0 & −φ \end{matrix} ) g = ( \begin{matrix}0 & −h^{#} φ \\ −φ h & 0 \end{matrix} ) .

By lemma 4.23, it suffices to show that Ω(( \begin{matrix}0 & −h^{#} φ \\ −φ h & 0 \end{matrix} )) = 0. In lemma 4.25, take the homotopy

equivalence α to be −φ h to get the result. ♯

4.29. Theorem. The group homomorphism

ι_{A} : W_{us}(A) ⟶ W(A)

is an isomorphism whose inverse is induced by Ω.

4.30. Proof : From corollary 4.27 we get a well defined additive map

Ω : MW(K^{b}(P)) → W_{us}(A).

Corollary 4.27 again and lemma 4.20 implies that Ω(NW(K^{b}(P))) = 0 (see remark 4.22). It is clear that Ω ∘ ι_{A} = Id and since ι_{A} is already known to be an epimorphism (theorem 4.1), we have the conclusion. ♯ 4.31. Corollary. If ϵ = ±1

W_{0}^{ϵ}(A) = W_{us}^{ϵ}(A).
♯

4.32. Remark. Using proposition 1.20, we only have four distinct Witt groups, namely : W_{0}^{1}(X), W_{0}^{−1}(X), W_{1}^{1}(X) and W_{1}^{−1}(X). The first two seem to be strongly related to W_{us}(X) and W_{us}^{−1}(X). However, we have no idea of a classical interpretation of the other two, except over a field, where it is possible to show that they are trivial.

4.33. Example. Let A be a noetherian domain in which 2 is a unit. Denote by Q its field of fractions. Let J denote the full subcategory of K^{b}()P whose objects are those which become zero in K^{b}(Q). These are the complexes X_{•} ∈ K^{b}()P such that Q ⊗_{A} H_{i}(X) = 0 for all i ∈ ℤ.

Let x ∈ ker(W(A) → W(Q)). By theorem 4.29, we may suppose that

x ∈ ker(W_{us}(A) → W_{us}(Q)).

Then it is easy to see that there exists a classical form (P, φ), a free module L and a homomorphism α : L → P with x = [P, φ] and such that the following complex, say Y :

⋯ ⟶ 0 ⟶ L ⟶_{α} P ⟶_{α^{∗} φ} L^{∗} ⟶ 0 ⟶ ⋯

is exact when tensorized by Q over A. Apply theorem 3.11 to the sublagrangian (L, α) to find a form ψ on Y such that [Y, ψ] = [X, φ] = x. Note that (L, α) is a sublagrangian in our sense but not in the usual one, since α is not split. This shows an interesting aspect of the general setting of derived categories.

We have proved the following refreshing result :

4.34. Theorem. The sequence

W(J) ⟶ W(A) ⟶ W(Q)

is exact. ♯

4.35. Remark. Moreover, for other reasons, we trust that W(J) → W(A) is injective. This is not proved yet, although it is possible to show that forms on J becoming neutral in A are already trivial in W(J). The fact that this kernel is the Witt group of a suitable triangulated (sub-) category shows that this extension of the definition of Witt groups leads us to a more complete theory.

5. Localization.

5.1. Definition. Let (K, #) be a triangulated category with duality. A multiplicative system of morphisms S is called compatible with the duality if #(s) ∈ S for all s ∈ S.

5.2. Notation. Let S^{−1}K be the localization of K with respect to this system of morphisms. Denote by q : K → S^{−1}K the localization functor. Denote by J(S) the full subcategory of K on the objects X ∈ K such that q(X) = 0.

5.3. Hypothesis. We suppose S to be saturated in the following sense : if q(t) is an isomorphism then t ∈ S.

5.4. Proposition. Let (K, #) be a triangulated category with duality and S saturated and compatible with #. Then J(S) endowed with the restriction of # and S^{−1}K endowed with the localization of # are triangulated categories with duality. Moreover, the natural morphisms

ι : J(S) → K  et  q : K → S^{−1}K

are covariant morphisms of triangulated categories with duality.

5.5. Proof : Left to the reader. Everything can be made explicit, using the construction of S^{−1}K with fractions (see [6] § 10.3). ♯

5.6. Remark. By the above proposition, we have a group homomorphism, still denoted by

q : W(K) ⟶ W(S^{−1}K).

The goal of this section is to identify the part of W(S^{−1}K) coming from W(K). In order to do this, we are going to define a homomorphism ∂ : W(S^{−1}K) → W_{1}^{−1}(J).

Choose a form in S^{−1}K, say (X, φ). Since the objects of S^{−1}K are those of K, X is in K. A priori φ is a fraction but, up to isometry, it might be taken of the form q(f) with f in K. This f might not be symmetric in K, but (since it is symmetric in S^{−1}K) there exists some s ∈ S such that f^{#} s = f s. Therefore, replacing f by s^{#} f s, we may suppose any form of S^{−1}K to be the localization q(f) of some symmetric f in K. Since q(f) is an isomorphism, f belongs to S.

Putting all together, we have proved that any isometry class x ∈ MW(S^{−1}K) contains a space of the form (X, q(s)) where s : X → X^{#}, is such that s = s^{#} and s ∈ S. We are going to construct the connecting homomorphism ∂ on these pairs (X, s) (definition 5.9). Then we will have to prove that this does not depend on the choice of (X, s) in x (theorem 5.15).

5.7. Construction. Let X be an object of K and s : X → X^{#} be a morphism in S such that s^{#} = s. Choose an exact triangle on s:

\xymatrix{X \ar[r]^{s} & X^{#} \ar[r]^{s_{1}} & E \ar[r]^{s_{2}} & T (X)}

and complete the following diagram (in which the second triangle is the dual of the first one) :

[figure: X; s; X^{#}; s_{1}; E; s_{2}; ∃ψ; ≃; T (X); X; s^{#}; X^{#}; −T s_{2}^{#}; T (E^{#}); T s_{1}^{#}; T (X)]

Observe that −T ψ^{#} also makes the diagram commute. Replacing ψ by \frac{1}{2}(ψ − T ψ^{#}) we may suppose that

ψ = −T ψ^{#} : E \underrightarrow{∼} T (E^{#}).

In the end notice that since s ∈ S, E ∈ J(S). Therefore (E, ψ) is a skew-symmetric space in J(S) with respect to the translated duality T ∘ # (see remark 1.5).

5.8. Notation. We set J = J(S) and abbreviate ∗ = T ∘ #.

5.9. Definition. Let s : X → X^{#} belong to S and satisfy s = s^{#}. Define ∂(X, s) to be the class of (E, ψ) in W_{1}^{−1}(J). An adapted version of lemma 3.4 insures that this is well defined.

5.10. Remark. This form (E, ψ) is not neutral, even if the diagram of construction 5.7 is precisely the one which insures that (E, ψ) is neutral when X ∈ J. To avoid confusion, recall what are the neutral forms for the skew-duality ∗. From definition 1.9, a neutral form in MW_{1}^{−1}(J) is in fact a pair (E, ψ) such that there exists an exact triangle in J (!) :

\xymatrix{D^{#} \ar[r]^{w} & D \ar[r]^{α} & E \ar[r]^{β} & T (D^{#})}

and a commutative diagram (in which the second triangle is the dual of the first one) :

[figure: D^{#}; w; D; α; −can_{D}; E; β; ψ; ≃; D^{∗}; D^{#}; −w^{#}; D^{∗∗}; β^{∗}; E^{∗}; α^{∗}; D^{∗}] .

This is equivalent to say that the following diagram commutes :

[figure: D^{#}; w; D; α; E; β; ψ; ≃; D^{∗}; D^{#}; w^{#}; D; −β^{∗}; E^{∗}; α^{∗}; D^{∗}] .

5.11. Lemma. Let f : X → X^{#} be such that f = f^{#} and let s : X → Y^{#} be any morphism in S. Then

∂(X ⊕ Y, ( \begin{matrix}f & s^{#} \\ s & 0 \end{matrix} )) = 0.

5.12. Proof : The proof is left as a non-trivial exercise, the solution of which will be given in a forthcoming article of mine (namely, Triangular Witt groups. Part I: The 12-term localization exact sequence Theorem 4.8).

5.13. Corollary. Suppose given (X, s), (Y, t) and u : Y → X such that s, t, u ∈ S and t = u^{#} s u. Then

∂(X, s) = ∂(Y, t).

5.14. Proof : It is easy to show that ∂ is additive with respect to orthogonal sums. Then it suffices to

prove that ∂(X ⊕ Y, ( \begin{matrix}−s & 0 \\ 0 & t \end{matrix} )) = 0. The result is clear if u is an isomorphism (use again lemma 3.4).

Consider h = ( \begin{matrix}1 & u \\ 0 & 1 \end{matrix} ) : X ⊕ Y \underrightarrow{∼} X ⊕ Y. It is clear that h^{#} · ( \begin{matrix}−s & 0 \\ 0 & t \end{matrix} ) · h is a matrix of the form of

the above lemma. This gives the result. ♯

5.15. Theorem. Let K be a triangulated category with duality and S a saturated system of morphisms compatible with #. For all x ∈ W(S^{−1}K) there exists an object X in K and a morphism s : X → X^{#} such that x = [q(X), q(s)]. The map sending x to ∂(X, s) induces a well-defined group homomorphism :

∂_{K,S} : W(S^{−1}K) ⟶ W_{1}^{−1}(J(S)).

5.16. Proof : The first assertion is proved in remark 5.6.

Define an equivalence relation on the pairs (X, s), with s = s^{#} : X → X^{#} in S, by :

(X, s) ∼ (Y, t)  iff  (X, s) and (Y, t) are isometric in S^{−1}K.

Check that this relation is generated by :

(X, s) ∼ (Y, t)  iff  there exists u : Y → X such that t = u^{#} s u.

Then use corollary 5.13 to induce a monoid homomorphism :

∂ : MW(S^{−1}K) ⟶ W_{1}^{−1}(J(S)).

To end the proof, we only have to show that ∂(NW(S^{−1}K)) = 0. This is very easy once we have observed that the natural homomorphism :

q : NW(K) ⟶ NW(S^{−1}K)

is surjective. This comes from the fact that a neutral form is characterized, up to isometry, by the w of definition 1.9 (see remark 3.6). Note that we may change w by T^{−1}(s^{#}) w s for any s ∈ S (use lemma 3.4 again !). Thus we may assume w is a morphism in K (and not a fraction). ♯ 5.17. Theorem. The sequence

W(K) ⟶^{W(q)} W(S^{−1}K) ⟶^{∂_{K,S}} W_{1}^{−1}(J(S))

is exact.

5.18. Proof : It is obvious that the composition is zero since the “cone” of an isomorphism is trivial.

Take an x ∈ ker(∂_{K,S}) and choose (X, s) such that x = [q(X), q(s)]. By construction, ∂(X, s) = 0. Choose an exact triangle :

\xymatrix{X \ar[r]^{s} & X^{#} \ar[r]^{s_{1}} & E \ar[r]^{s_{2}} & T (X)}

and a skew-symmetric form ψ on E as in the construction 5.7. In MW_{1}^{−1}(J) we have that (E, ψ) ⊥ some neutral is neutral. Adding to the second space its own opposite, we may suppose that (E, ψ) ⊥ some

(other) neutral is hyperbolic, which means isometric to some (L ⊕ T L^{#}, ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )). Call the (other)

neutral (D, μ). It appears in a triangle of the form

\xymatrix{N \ar[r]^{w=w^{#}} & N^{#} \ar[r]^{w_{1}} & D \ar[r]^{w_{2}} & T (N)}

with N ∈ J! (and w_{2} = w_{1}^{∗}μ). By definition of J, q(N) = 0, which means that our original x is equal to

[X ⊕ N, ( \begin{matrix}s & 0 \\ 0 & w \end{matrix} )] as well. We may now suppose that (E, ψ) is hyperbolic.

The conclusion follows from the next lemma : ♯

5.19. Lemma. Let s : X → X^{#} be in S and such that s = s^{#}. Suppose further that there exists an exact triangle of the form :

\xymatrix{X \ar[r]^{s} & X^{#} \ar[r]^{s_{1}=( \begin{matrix}t \\ −w^{∗} \end{matrix} )} & L ⊕ T L^{#} \ar[r]^{s_{2}=( w t^{∗})} & T (X)}

for some morphisms t : X^{#} → L and w : L → T (X) and for L ∈ J.

Then there exists a symmetric form (Z, φ) in W(K) such that q(Z, φ) and (q(X), q(s)) are isometric.

5.20. Proof : Choose an exact triangle containing t, say :

\xymatrix{Y \ar[r]^{t_{0}} & X^{#} \ar[r]^{t} & L \ar[r]^{t_{2}} & T (Y)}

Apply the octahedron axiom to the identity t = (1 0) · s_{1} to get an exact triangle

\xymatrix{X \ar[r]^{f_{1}} & Y \ar[r]^{w^{∗} t_{0}} & T L^{#} \ar[r]^{−T t^{#}} & T (X)}

for some morphism f_{1} : X → Y such that t_{2} = T (f_{1}) w and s = t_{0} f_{1}. The two above exact triangles contain t. Comparing them, we find a non necessarily symmetric isomorphism :

[figure: T^{−1}L; T^{−1}t_{2}; Y; t_{0}; ∃α; ≃; X^{#}; −t; L; T^{−1}L; t_{0}^{#} T^{−1}w; Y^{#}; f_{1}^{#}; X^{#}; −t; L]

Observe that q(α) is already symmetric. In fact, L ∈ J implies t_{0} ∈ S. Since t_{0} α^{−1} = f_{1}^{#}, we have

t_{0} α^{−1} t_{0}^{#} = f_{1}^{#} t_{0}^{#} = (t_{0} f_{1})^{#} = s^{#} = s.

As q(t_{0}) is an isomorphism, q(α^{−1}) is isometric to q(s), hence symmetric. Since for all symmetric form φ, φ ≃ φ^{−1}, q(α) is isometric to q(s).

We still have to improve our α (in fact, we will replace it by \frac{1}{2}(α + α^{#})) to be sure that it will be symmetric (and remain an isomorphism !).

We deduce from the above diagram that α^{#} f_{1} = t_{0}^{#}. Composing on the right with T^{−1}w, we get α^{#} T^{−1}t_{2} = t_{0}^{#} T^{−1}w. Then construct γ : X^{#} → X^{#} such that the following diagram commutes :

[figure: T^{−1}L; T^{−1}t_{2}; Y; t_{0}; α^{#}; ≃; X^{#}; −t; ≃; 1+γ; L; T^{−1}L; t_{0}^{#} T^{−1}w; Y^{#}; f_{1}^{#}; X^{#}; −t; L]

It remains to show that γ is nilpotent ! From t γ = 0, we have γ = t_{0} γ̃. From the first diagram we find also that w^{∗} γ t_{0} = 0. In the end, from the original property of f_{1} to satisfy s = t_{0} f_{1} and from s = s^{#},

we get γ s = 0. This last equality insures that γ = γ̌ ∘ ( \begin{matrix}t \\ −w^{∗} \end{matrix} ).

Now compute

γ^{3} = γ̌ ( \begin{matrix}t \\ −w^{∗} \end{matrix} ) γ t_{0} γ̃ = γ̌ ( \begin{matrix}t γ t_{0} \\ −w^{∗}γ t_{0} \end{matrix} ) γ̃ = 0.

Taking the mean of the morphisms (1, α, 1) and (1, α^{#}, 1 + γ), we get a morphism of exact triangles (1, \frac{1}{2}(α + α^{#}), 1 + \frac{1}{2}γ). Since 1 + \frac{1}{2}γ is an isomorphism (as well as 1), \frac{1}{2}(α + α^{#}) is a form as claimed above. Clearly, since q(α) was already symmetric, q(\frac{1}{2}(α + α^{#})) = q(α) which is isometric to s. ♯

5.21. Application to an open subscheme.

Let X be a regular scheme and U ⊂ X be an open subscheme. With the notations of section 2, consider the category A(U) of coherent O_{U}-modules and the category A(X) of coherent O_{X}-modules. Denote by i : U → X the inclusion. It induces a morphism Λ := i^{∗} : A(X) → A(U) which is simply the restriction. Since restriction is exact, this morphism induces an exact morphism on the corresponding derived categories for X and U.

5.22. Theorem. With notations of section 2, K(U) is a localization of K(X) with respect to a saturated multiplicative system of morphisms compatible with duality. Moreover, the localization of #_{X} to K(U) is #_{U}.

5.23. Proof : Since X is regular, K(X) =^{def}K^{b}(L(X)) ≃ D^{b}(A(X)). Since U is regular too, K(U) ≃ D^{b}(A(U)). Therefore, K(U) is the localization of Ch^{b}(A(U)) with respect to quasi-isomorphisms and it suffices to prove that A(U) is a localization of A(X) and that every quasi-isomorphism in Ch^{b}(A(X)) yields a quasi-isomorphism in Ch^{b}(A(U)). The second assertion is trivial since Λ is exact and the second is well known even for non-regular schemes.

Thus K(U) is a localization of K(X) with respect to all morphisms in K(X) whose restriction to K(U) is an isomorphism. Such morphisms are precisely those whose restriction to U is a quasi-isomorphism. Then this system arises from a cohomological functor in the sense of [6]. Namely, this cohomological functor is here H_{0} ∘ Λ. Therefore S is multiplicative.

The other assertions are left to the reader. ♯

5.24. Corollary. Let X be a regular scheme in which 2 is a unit and let U ⊂ X be an open subscheme. Let J denote the full subcategory of D^{b}(L(X)) on those complexes which are acyclic on U. Then there exists a group homomorphism :

∂_{X,U} : W(U) ⟶ W_{1}^{−1}(J)

such that the sequence

W(X) ⟶^{restr.} W(U) ⟶^{∂_{X,U}} W_{1}^{−1}(J)

is exact.

5.25. Proof : This is juxtaposition of theorems 5.17 and 5.22. ♯

6. Purity in dimension 3.

6.1. Theorem. Let (A, 𝔪) be a regular local ring of dimension 3 in which 2 is a unit, Q its field of fractions and U = Spec(A) \ {𝔪}. Consider the commutative diagram :

[figure: W_{us}(A); W_{us}(U); ι_{U}; W_{us}(Q); W(A); W(q); W(U); W(Q)]

Then, for all a ∈ W_{us}(U) there exists b ∈ W(A) such that a = b in W(Q).

6.2. Proof : Using the notations of section 5, we denote by J the full subcategory of K^{b}()P on the complexes whose restriction to U is acyclic. Abbreviate ∂ = ∂_{X,U}. By corollary 5.24, it suffices to prove that for all a ∈ W_{us}(U) we have ∂∘ι_{U}(a) = 0 in W_{1}^{−1}(J). We apply construction 5.7 to compute ∂(ι_{U}(a)).

Let a ∈ W_{us}(U). Let (E, φ) be a symmetric form on U with a = [E, φ]. Then E is a locally free O_{U}-module and φ = φ^{∗} : E →^{∼} E^{∗}. Set E = Γ(E) and f = Γ(φ), where Γ denotes global sections on U. We have (see [3]) that E is reflexive of projective dimension ≤ 1 and f : E →^{∼} E^{∗} is symmetric. Choose a projective resolution of E:

0 → P_{1} ⟶^{d_{1}} P_{0} ⟶^{d_{0}} E → 0

Check that ι_{U}(a) = [q(X), q(s)]) where :

[figure: X =; s; ⋯0; P_{1}; d_{1}; P_{0}; d_{0}^{∗} f d_{0}; 0; 0⋯; X^{#} =; ⋯0; 0; P_{0}^{∗}; d_{1}^{∗}; P_{1}^{∗}; 0⋯]

It is obvious that s = s^{#} and that Λ(s) = quasi-isomorphism. Therefore (see construction 5.7) ∂ ∘ι_{U}(a) = [Y, ψ] where :

[figure: Y =; ψ; ⋯0; P_{1}; −d_{1}; −1; P_{0}; −d_{0}^{∗} f d_{0}; −1; P_{0}^{∗}; d_{1}^{∗}; 1; P_{1}^{∗}; 1; 0⋯; T (Y^{#}) =; ⋯0; P_{1}; −d_{1}; P_{0}; d_{0}^{∗} f d_{0}; P_{0}^{∗}; d_{1}^{∗}; P_{1}^{∗}; 0⋯]

Using twice the periodicity of the Witt groups, we may suppose that Y “starts” in degree 0 and that ψ : Y → T^{−3}(Y^{#}). Verify that H_{0}(Y) = H_{1}(Y) = H_{2}(Y) = 0. Put M = H_{3}(Y). We may identify Ext^{3}(M, A) with H_{3}(T^{−3}(Y^{#})) and it is easy to show that ρ = H_{3}(ψ) is a skew-symmetric form on M with respect to the duality M̌ = Ext_{A}^{3}(M, A). See [4] for precisions.

Since the Witt group of skew-symmetric forms over finite length A-modules is zero : W_{lf}^{−1}(A) ≃ W_{lf}^{−1}(A/𝔪) = 0, this form (M, ρ) is metabolic. Use an exact sequence

0 ⟶ N ⟶^{i} M ⟶^{ǐρ} Ň ⟶ 0

and a resolution of N to prove that (Y, ψ) is neutral in J with duality ∗ = T ∘ #_{A}. ♯ 6.3. Remark. Details about the Witt group of skew-symmetric forms over finite length A-modules can be found in [4], where an ad hoc construction is used in place of our general theorem 5.17. In fact, our proof 6.2 is a “dramatic simplification” (M. Ojanguren dixit) of the one given there.

6.4. Corollary. The natural map

W(A) ⟶ W_{us}(U)

is surjective.

6.5. Proof : It suffices to know that W_{us}(U) ⟶ W(Q) is injective (see [2]) and to apply theorem 6.1. ♯

6.6. Remark. This could be applied to show that, under the hypotheses of theorem 6.1, the sequence :

W(A) ⟶ W(Q) ⟶_{residue}^{second} ⨁ ; 𝔭 ∈ Spec(A) ; of height 1 W(κ(𝔭))

is exact. In fact, it is not too hard to show that any a ∈ W(Q) which is in the kernel of the residue is in the image of W_{us}(U) → W(Q). Theorem 6.1 gives the result.

6.7. Post-scriptum. At the moment this article was going to press, the editors kindly allowed me some final modifications. I would like to use this opportunity to mention that in the forthcoming series “Triangular Witt groups”, the notations will be slightly changed into : W^{1} for the present W_{1}^{−} et cætera. This is justified by the long exact sequence and by compatibility with existing notations.

7. Bibliography.

[1] I. Bucur, Triangulated categories and algebraic K-theory, in: R. Moss and C. Thomas, eds, Algebraic K-theory and its geometric applications, Lecture Notes in Mathematics, Vol. 108 (Springer, Berlin, 1969) 28-54.

[2] F. Fernandez-Carmena, On the injectivity of the map of the Witt group of a scheme into the Witt group of its quotient field, Math.Ann. 277 (1987) 453-468.

[3] G. Horrocks, Vector bundles on the punctered spectrum of a local ring, Proc. London Math. Soc. 14 (1964) 689-713.

[4] M. Ojanguren, R. Parimala, R. Sridharan and V. Suresh, Witt groups of the punctured spectrum of a 3-dimensional regular local ring and a purity theorem, Proc. London Math. Soc., to appear.

[5] A. Ranicki, Algebraic L-Theory I. Foundations, Proc. Lond. Math. Soc. 27 (1973) 101-125.

[6] C. Weibel, An introduction to homological algebra (Cambridge University Press, 1994).

[7] B. Youssin, Witt Groups of Derived Categories, K-theory, vol 11 (1997) 373-395.