ON THE DIMENSION OF ANISOTROPIC FORMS IN I^{n}

A. VISHIK

1.

Let k be a field of characteristic not 2, W (k) be Witt ring of quadratic forms, and I ⊂ W (k) be an ideal of even-dimensional forms. Ideal I generates the filtration W (k) ⊃ I ⊃ I^{2} ⊃ ⋯ ⊃ I^{n} ⊃ …, significance of which is illustrated, for example, by the fact that the associated graded ring gr_{I^{·}}^{∗}(W (k)) is naturally isomorphic to the Milnor’s ring K_{∗}^{M}(k)/2 and to the Galois cohomology H_{et}^{∗}(k, ℤ/2)of the field k (by the result of V.Voevodsky). An important problem is to describe the possible dimensions of anisotropic forms from I^{n}.

Basic here is the following:

Hauptsatz (Arason-Pfister).

If q ∈ I^{n} is anisotropic quadratic form, then dim(q) is either 0, or ⩾ 2^{n}.

The natural question arises: do we still have some restrictions on dim(q), if dim(q) ⩾ 2^{n}.

The following Conjecture is well-known (was formulated, for example, by B.Kahn - see [6], Conjecture 9)

Conjecture 1.1 .

Let q ∈ I^{n}(k) be anisotropic quadratic form. Then dim(q) is either 0, or 2^{n}, or ⩾ 2^{n} + 2^{n−1}.

This conjecture is settled for n = 3 by a theorem of A.Pfister - see [12], Satz 14 and Zusatz, and for n = 4 by a result of D.Hoffmann - see [4], Main Theorem.

In the present paper we will prove this Conjecture for all n for fields of characteristic 0 - see the Main Theorem .

The main tool here is the following lemma describing the possible dimension of “binary” direct summands in the motives of quadrics.

Lemma 1.2 .

Let char(k) = 0. Let Q be anisotropic quadric. Suppose L is a direct summand in M(Q), and L|\overline{k} = ℤ ⊕ ℤ(i)[2i] for some i (in other words, L is “binary”). Then i = 2^{r} − 1 for some r.

The proof of Lemma 1.2 will be given in the next section. This proof is based on the technique of V.Voevodsky.

Conjecture 1.1 implies some other conjectures. For example, we get the description of the so-called twisted Pfister forms - see [3], Conjecture 3.9, Proposition 3.11(i).

The same methods permit us to find all possible splitting patterns of forms of height 2 - see Theorem 3.1 , and to prove some interesting result about the degree of the maximal pure divisor of two symbols in Milnor’s K-theory (mod 2) - see Corollary 4.8 . Finally we formulate the analogue of Conjecture 1.1 , describing all possible dimensions of anisotropic forms from I^{n} - see Conjecture 4.11 .

Some notations and conventions. Except for the proof of Lemma 1.2 , we will be working in the category of effective Chow motives Chow^{eff.}(k) (see the Appendix). For smooth projective variety R we will denote it’s motive as M(R). We will denote Tate-motives as ℤ(i)[2i], in particular, M(Spec(k)) will be denoted simply as ℤ. By this reason, we use notation Z for groups and rings ℤ.

In Section 2 we will be working in the triangulated category of mixed motives DM_{eff.}^{−}(k). We refer the reader to [18] and [19] for basic facts and notations.

As J^{n} ⊂ W (k) we will denote the ideal of forms of degree ⩾ n (see [9]). From Voevodsky’s methods it follows that I^{n} = J^{n}, but we will not use this fact here.

Everywhere in the text we will assume that fields have characteristic not 2.

Acknoledgements. I would want to thank Oleg Izhboldin for the very fruitfull discussions. Also, I want to thank D.Hoffmann and B.Kahn for suggesting me the name for this article. This work was done during my stay at Max-Planck-Institut für Mathematik, and I would like to express my gratitude to this institution for the support, excellent working conditions and the very stimulating atmosphere.

The Main result.

Main Theorem .

Let k be a field, s.t. char(k) = 0. Let q ∈ I^{n}(k) be anisotropic quadratic form. Then dim(q) is either 0, or 2^{n}, or ⩾ 2^{n} + 2^{n−1}.

Now, we can show, how to deduce Main Theorem from Lemma 1.2 . Lemma 1.3 . Let k be any field of characteristic not 2.

Suppose q be some quadratic form, s.t. M(Q) contains ℤ(l)[2l] as a direct summand. Suppose m = min(l, dim(Q) − l). Then q is m + 1times isotropic.

Proof

If M(Q) contains ℤ(l)[2l] as a direct summand, then it also contains ℤ(dim(Q)−l)[2 dim(Q)−2l] (if pr ∈ CH^{dim(Q)}(Q×Q) is the corresponding projector, then just consider the dual one: pr^{∨}, obtained by switching of factors in Q × Q). So, we can assume that m = l ⩽ dim(Q)/2.

Let M(Q) contains ℤ(l)[2l] as a direct summand.

So, we have maps: pr : M(Q) → ℤ(l)[2l] and s : ℤ(l)[2l] → M(Q), s.t. pr ∘ s = id_{ℤ(l)[2l]}. Via identification: Hom(M(Q), ℤ(l)[2l]) = CH^{l}(Q) and Hom(ℤ(l)[2l], M(Q)) = CH_{l}(Q), our maps pr and s correspond to cycles A ∈ CH^{l}(Q), and B ∈ CH_{l}(Q). Then the composition pr ∘ s ∈ Hom(ℤ(l)[2l], ℤ(l)[2l]) = CH_{0}(Spec(k)) = Z is given by the degree of intersection A ∩ B ∈ CH_{0}(Q). So, we have that degree(A ∩ B) = 1. This implies: if l < dim(Q)/2, then degree(B) is odd, and if l = dim(Q)/2 then at least one of degree(A), degree(B) is odd. Now, everything follows from:

Sublemma 1.3.1 .

Let 0 ⩽ l ⩽ dim(Q)/2, and Q has an l-dimensional cycle of odd degree. Then q is l + 1-times isotropic.

Proof

If l = 0 then, by Springer’s Theorem (see [10], VII, Theorem 2.3), we get a rational point on Q. So, q is isotropic.

Suppose we proved the statement for any quadratic form p, and for any 0 ⩽ a < l.

By taking intersection of A with plane section of codimension l, we get a zero-cycle of odd degree on Q. So, q is isotropic: q = ℍ ⊥ q′, for some form q′.

Let x be any rational point on q, then Q′ can be identified with projective quadric of lines on Q, passing through x. And the union of all lines on Q passing through x is the cone over Q′ with vertex x, and it is just the intersection R := Q ∩ T_{x}, where T_{x} is a tangent space to Q at x. We have natural projection φ : R\x → Q′.

We can always take x outside A (since the set of rational points on isotropic quadric is dense). Then φ_{∗}(A∩T_{x}) will be an l−1-cycle of odd degree on Q′. By induction, q′ is l-times isotropic. So, q is l + 1-times isotropic.

Lemma 1.4 .

Let k be a field of characteristic 0.

Let q, p ∈ J^{n}(k) be anisotropic quadrics, height(q) = height(p) = 2, and dim(q) = dim(p) < 2^{n+1}. Suppose q|_{k(P)} and p|_{k(Q)} are isotropic. Then for arbitrary field extension K/k, q|_{K} is hyperbolic if and only if p|_{K} is.

Proof By Theorem A.16 , M(Q), M(P) contain isomorphic indecomposable direct summands U and V, s.t. CH_{0}(U|\overline{k}) ≠ 0.

By Theorem A.17 , M(Q) also contains direct summands U(f)[2f] for any 0 ⩽ f < i_{1}(q). Since U, …, U(i_{1}(q) − 1)[2i_{1}(q) − 2] are not isomorphic, by Theorem A.12 , ⊕_{0⩽f<i_{1}(q)}U(f)[2f] is isomorphic to a direct summand of M(Q). Since i_{1}(q) < dim(Q)/2 (notice, that height(q) = 2 > 1), we have: CH_{f}(U|\overline{k}) = 0, for any 0 < f < i_{1}(q) (otherwise, rank(CH_{f}(Q|\overline{k})) would be ⩾ 2). Since U(i_{1}(q)−1)[2i_{1}(q)− 2] is a direct summand in M(Q), we have: CH_{l}(U) = 0, for any dim(Q) − i_{1}(q) + 1 < l ⩽ dim(Q).

If there is no i_{1}(q) ⩽ l ⩽ dim(Q) − i_{1}(q), s.t. CH_{l}(U|\overline{k}) ≠ 0, then CH_{s}(U|\overline{k}) can be nonzero only for s = 0 and dim(Q) − i_{1}(q) + 1, and rank(CH(U|\overline{k})) ⩽ 2.

By Lemma 1.2 , we have: dim(Q) − i_{1}(q) + 1 = 2^{r} − 1 for some r. This is impossible, since if dim(q) = 2^{n} + 2i, where 0 < i < 2^{n−1}, then i_{1}(q) = i (we remind that q has height = 2, and q ∈ J^{n}(k)), so, dim(Q) − i_{1}(q) + 1 = dim(q) − i_{1}(q) − 1 = 2^{n} + i − 1 ≠ 2^{r} − 1 for any r.

This shows, that there exists i_{1}(q) ⩽ l ⩽ dim(Q)−i_{1}(q), s.t. CH_{l}(U|\overline{k}) ≠ 0. This means, that for any field extension K/k the following conditions are equivalent:

1) U|_{K} is a direct sum of Tate-motives.

2) q|_{K} is hyperbolic.

Really, it is evident that 2) implies 1) (since the motive of a hyperbolic quadric is a sum of Tate-motives). In the other direction: if U|_{K} is a direct sum of Tate-motives, then in M(Q|_{K}) we have a direct summand, isomorphic to ℤ(l)[2l], where i_{1}(q) ⩽ l ⩽ dim(Q) − i_{1}(q). Let m = min(l, dim(Q) − l). Since in M(Q|_{K}) we have a direct summand, isomorphic to ℤ(l)[2l], by Lemma 1.3 , we get: q|_{K} is m + 1-times isotropic. Since m + 1 > i_{1}(q), and height(q) = 2, we get: q|_{K} is hyperbolic.

Applying similar considerations to p, we get that for any field extension K/k the following conditions are equivalent:

1) V |_{K} is a direct sum of Tate-motives.

2) p|_{K} is hyperbolic.

Since U is isomorphic to V we get that q|_{K} is hyperbolic if and only if p|_{K} is.

Theorem 1.5 .

Let k be a field of characteristic 0.

Suppose q ∈ J^{n}(k) be of dim(q) < 2^{n} + 2^{n−1}. Then dim(q_{an}) ⩽ 2^{n}. Proof Suppose, it is not the case. Let us choose counterexample of the

smallest possible dimension (among all forms over all fields of characteristic 0). Abusing notations, we will denote the corresponding field as k, and corresponding form as q.

Since q is counterexample of the smallest possible dimension, we have: height(q) = 2.

Let s ⊂ q be any subform of dimension 2^{n−1}. By [2], Main Lemma, there exists field extension F/k, and some n-fold Pfister form π, that s|_{F} is a subform of π, and for any anisotropic form φ over k, the restriction φ|_{F} is anisotropic. In particular, if p := (q|_{F} ⊥ −π)_{an}, then dim(p) ⩽ dim(q), and q|_{F} is anisotropic.

Notice, that p ∈ J^{n}(F), and dim(p) ⩽ d.

Let E/F be any field extension. Then if dim((p|_{E})_{an}) < dim(q) = d, then by definition of d, dim((p|_{E})_{an}) ⩽ 2^{n}, i.e.: (p|_{E})_{an} is proportional to an n-fold Pfister form over E. Then (q|_{E})_{an} is proportional to a generalized Albert form, and by the result of R.Elman and T.Y.Lam (see [1], Theorem 4.5), dim((q|_{E})_{an}) is either ⩾ 2^{n} + 2^{n−1} > d, or ⩽ 2^{n}. That means, that if dim((p|_{E})_{an}) < d, then q|_{E} is isotropic.

In particular, dim(p) = d, height(p) = 2, and q|_{F (p)} is isotropic. Changing q by p in the considerations above, we get that if E/F

be any field extension, and q|_{E} is isotropic, then p|_{K} is isotropic. In particular, p|_{F (q)} is isotropic.

By Lemma 1.4 , for any field extension E/F, q|_{E} is hyperbolic if and only if p|_{E} is.

Over the field E = F (q), the forms (q|_{E})_{an} and (p|_{E})_{an} are proportional to some n-fold Pfister forms β and γ. We have: c · γ = b · β − π|_{E} in W (E).

Since dim(q) > 2^{n} = dim(π), we have: π|_{E} is anisotropic.

Take K = E(β). We have: q|_{K} is hyperbolic, but p|_{K} = cγ|_{K}, and this form is nonhyperbolic, since γ ≠ β.

We get a contradiction. Theorem 1.5 is proven.

Proof of the Main Theorem

Theorem 1.5 , evidently, implies the Main Theorem in view of Hauptsatz , and the fact that I^{n} ⊂ J^{n} (see [9]).

2. Proof of Lemma 1.2

Lemma 1.2 was, actually, proven (but not formulated) in [15] - see the proof of Statement 6.1 there. We will reproduce it’s proof here for the reader’s convenience.

Lemma 2.1 .

Let k be arbitrary field of characteristic not 2, and q be quadratic form over k.

Suppose L be a direct summand in M(Q), s.t. L|\overline{k} ≃ ℤ. Then L ≃ ℤ.

Proof

If Q is anisotropic, then by Sublemma A.17.8 , CH_{dim(Q)−i_{1}(q)+1}(L|\overline{k}) ≠ 0. Since CH_{r}(ℤ) = 0, for r > 0, we get: Q is isotropic, i.e.: q = ℍ ⊥ q′. Then, by [13], Proposition 1 (Theorem A.0 ), M(Q) = ℤ ⊕ M(Q′)(1)[2] ⊕ ℤ(dim(Q))[2 dim(Q)], and L = L_{0} ⊕ L′ ⊕ L_{dim(Q)}, where L_{i} is a direct summand of ℤ(i)[2i], and L′ is a direct summand of M(Q′)(1)[2]. Clearly, L′|\overline{k} = L_{dim(Q)}|\overline{k} = 0, and L_{0}|\overline{k} = ℤ. Hence, L_{0} = ℤ, on the other hand, by Rost Nilpotence Theorem ([14], Corollary 10)(see Theorem A.1 ), L′ = L_{dim(Q)} = 0. So, L = ℤ.

Lemma 2.1 is proven.

Proof of Lemma 1.2

The proof is based on the construction very close to that used by V.Voevodsky in [19].

Suppose, we have some direct summand N in M(Q), s.t. N|\overline{k} = ℤ ⊕ ℤ(n)[2n].

By Sublemma A.17.8 , n = dim(Q) − i_{1}(q) + 1.

Consider plane section P ⊂ Q of codimension i_{1}(q) − 1. Since from the existense of i_{1}(q) − 1-dimensional projective subspace on Q|_{E} follows the existense of rational point on P |_{E}, from the later follows the existence of rational point on Q|_{E}, and first and third are equivalent (since i_{1}(q) − 1 < i_{1}(q)), we have that q|_{k(P)} and p|_{k(Q)} are isotropic. Then by Theorem A.16 , M(P) contains a direct summand isomorphic to N. But now n = dim(P).

Chow^{eff.}(k) is a full additive subcategory (closed under taking of direct summands) in the triangulated category DM_{eff.}^{−}(k) - see [17]. The category DM_{eff.}^{−}(k) contains the “motives” of all smooth simplicial schemes over k. If P is smooth projective variety over k, we denote as X_{P}^{·} the standard simplicial scheme, corresponding to the pair P → Spec(k) - see [18], or [15], Definition 2.3.1. X_{P}^{m} := P × ⋯ × P (m + 1 times), with faces and degenerations maps given by partial projection and diagonals.

From the natural projection: X_{P}^{·} → Spec(k), we get a map: M(X_{P}^{·}) →^{pr} ℤ. One of the main properties of M(X_{P}^{·}) is: pr is an isomorphism if and only if P has a 0-cycle of degree 1 - see [15], Theorem 2.3.4. In particular, M(X_{P}^{·} |\overline{k}) = ℤ. From this point, we will denote M(X_{P}^{·}) simply as X_{P} (since we will not use simplicial schemes themselves any more).

Repeating the proof of [19], Theorem 4.4, we get that in DM_{eff.}^{−}(k), N = Cone[−1](X_{P} → X_{P} (n)[2n + 1]) (actually, this statement is the very particular case of [15], Lemma 3.23).

N := Cone[−1](X_{P} ⟶^{μ′} X_{P} (n)[2n + 1]), where μ′ is some (actually, the only) nontrivial (since, otherwise, X_{P} would be a direct summand in M(P), and by Lemma 2.1 , and Lemma 1.3 , p would be isotropic) element from Hom(X_{P} , X_{P} (n)[2n + 1]).

Let pr : X_{P} → ℤ be natural projection. Then, by [19], Lemma 4.7, pr_{∗} : Hom(X_{P} , X_{P} (a)[b]) → Hom(X_{P} , ℤ(a)[b]) is an isomorphism for any a, b.

Denote: μ := pr_{∗}(μ′) ∈ Hom(X_{P} , ℤ(n)[2n + 1]).

Sublemma 2.2 .

The map: (μ′)^{∗} : Hom(X_{P} , ℤ(c)[d]) → Hom(X_{P} , ℤ(c + n)[d + 2n + 1]) coinsides with multiplication by μ ∈ Hom(X_{P} , ℤ(n)[2n + 1]).

Proof It follows from the fact that Δ_{X_{P}} : X_{P} → X_{P} ⊗ X_{P} and π_{1} : X_{P} ⊗X_{P} → X_{P} are mutually inverse isomorphisms, μ·u = Δ_{X_{P}} (μ⊗u),

and \begin{matrix}X_{P} & ⟶^{μ} & ℤ(n)[2n + 1] \\ ⊗ & & ⊗ \\ X_{P} & ⟶^{u} & ℤ(c)[d] \end{matrix} coincides with the composition:

X_{P} ⟶^{μ′} X_{P} (n)[2n + 1] ⟶^{pr(n)[2n+1]} ℤ(n)[2n + 1]
⊗  ⊗  ⊗
X_{P} ⟶^{id}  X_{P}  ⟶^{u}  ℤ(c)[d]

which can be identified with the composition: X_{P} →^{μ′} X_{P} (n)[2n + 1] →^{u} ℤ(n + c)[2n + 1 + d], which is equal to (μ′)^{∗}(u).

Sublemma 2.3 .

Multiplication by μ performs an isomorphism: Hom(X_{P} , ℤ(c)[d]) → Hom(X_{P} , ℤ(c + n)[d + 2n + 1]) for any d − c > 0, and is surjective for d = c. The same is true about cohomology with ℤ/2-coefficients.

Proof Since N is a direct summand in M(P), and Hom(P, ℤ(i)[j]) = 0 for j − i > n = dim(P), we have: Hom(N, ℤ(a)[b]) = 0, for b > 2a, or b − a > n. Then the same is true about cohomology with ℤ/2coefficients.

Consider Hom’s from the exact triangle N → X_{P} →^{μ} X_{P} (n)[2n+1] → N[1] to ℤ(n+c)[2n+d+1]. We have: Hom(N, ℤ(n+c)[2n+d+1]) = 0, if d − c ⩾ 0, and Hom(N, ℤ(n + c)[2n + d]) = 0, if d − c > 0. This, combined with Sublemma 2.2 , implies the statement for ℤ-coefficients. Considering Hom’s to ℤ/2(n + c)[2n + d + 1], we get the ℤ/2-case.

We can also consider X̃_{P} := Cone[−1](X_{P} →^{pr} ℤ).

Sublemma 2.4 .

Hom(X_{P} , ℤ(a)[b]) is a 2-torsion group for b > a;

Hom(X_{P} , ℤ(a)[b]) embeds into Hom(X_{P} , ℤ/2(a)[b]) for b > a.

Also, the natural map: X̃_{P} →^{δ} X_{P} induces an isomorphism Hom(X_{P} , ℤ/2(a)[b]) →^{=} Hom(X̃_{P} , ℤ/2(a)[b]), for all b > a.

Proof For finite field extension E/k we have action of transfers on motivic cohomology. Transfer is a map

Tr : Hom(X|_{E}, ℤ(a)[b]) = Hom(X, M(E)(a)[b]) → Hom(X, ℤ(a)[b]),

where the equality is duality, and the second map is induced by the projection π : M(E) → ℤ (E/k-finite field extension); the main property of transfer is: Tr ∘j = ·[E : k], where

j : Hom(X, ℤ(a)[b]) → Hom(X, M(E)(a)[b]) = Hom(X|_{E}, ℤ(a)[b])

is induced by the map j : ℤ → M(E) dual to π via duality.

Quadric P has a point E of degree 2, and over E, X_{P} becomes ℤ (see [19], Lemma 3.8, or [15], Theorem 2.3.4), so we have that Hom(X_{P} |_{E}, ℤ(a)[b]) = 0 for b − a > 0.

Now, considering the composition Tr ∘j = ·[E : k] we get that Hom(X_{P} , ℤ(a)[b]) is a 2-torsion group for b > a.

In particular, this implies that the natural map Hom(X_{P} , ℤ(a)[b]) → Hom(X_{P} , ℤ/2(a)[b]) is an embedding, for b > a.

Since Hom(ℤ, ℤ/2(a)[b]) = 0 for any b > a (see [19], Corollary 2.3(1)), we also have that for b > a, δ^{∗} : Hom(X_{P} , ℤ/2(a)[b]) → Hom(X̃_{P} , ℤ/2(a)[b]) is an isomorphism.

We have the action of motivic cohomological operations Q_{i} on Hom(X_{P} , ℤ(∗)[∗′]) and Hom(X̃_{P} , ℤ(∗)[∗′]) - see [19], Theorems 3.16 and 3.17. The differential Q_{i} acts without cohomology on Hom(X̃_{P} , ℤ/2(∗)[∗′]) for any i ⩽ [log_{2}(n + 1)] - see [19], Theorem 3.25+Lemma 4.11 (notice, that in the proof of the later no specific of Pfister case is used).

Denote η := μ(mod2), i.e. the image of μ in cohomology with ℤ/2 coefficients. From Sublemma 2.4 it follows that η ≠ 0.

Denote r = [log_{2}(n)].

Sublemma 2.5 .

Q_{i}(η) = 0, for all i ⩽ r.

Proof Really, Q_{i}(η) ∈ Hom(X_{P} , ℤ/2(n + 2^{i} − 1)[2n + 2^{i+1}]), and the later group is an extension of 2-cotorsion in Hom(X_{P} , ℤ(n+2^{i}−1)[2n+ 2^{i+1}]), and 2-torsion in Hom(X_{P} , ℤ(n + 2^{i} − 1)[2n + 2^{i+1} + 1]).

But, by Sublemma 2.3 , μ performs a surjections Hom(X_{P} , ℤ(2^{i} − 1)[2^{i+1} −1]) → Hom(X_{P} , ℤ(n+2^{i} −1)[2n+2^{i+1}]), and Hom(X_{P} , ℤ(2^{i} − 1)[2^{i+1}]) → Hom(X_{P} , ℤ(n + 2^{i} − 1)[2n + 2^{i+1} + 1]).

And the groups: Hom(X_{P} , ℤ(2^{i} − 1)[2^{i+1} − 1]), Hom(X_{P} , ℤ(2^{i} − 1)[2^{i+1}]) are zero.

Really, from the exact triangle N → X_{P} → X_{P} (n)[2n + 1] → N[1], we have exact sequence: Hom(N, ℤ(2^{i}−1)[2^{i+1}−1]) ← Hom(X_{P} , ℤ(2^{i}− 1)[2^{i+1} −1]) ← Hom(X_{P} , ℤ(2^{i} −1−n)[2^{i+1} −2n]) The first group is zero since N is a pure motive (i.e.: a direct summand in the motive of some smooth projective variety), and (consequently) Hom(N, ℤ(a)[b]) = 0 for b > 2a (see [19], Corollary 2.3(2)). The third group is zero, since n > 2^{i} − 1 (see [19], Corollary 2.2(1)). So, the second is zero as well. Analogously, in the case of Hom(X_{P} , ℤ(2^{i} − 1)[2^{i+1}]).

So, Q_{i}(η) = 0.

Sublemma 2.6 .

Let 0 ⩽ j ⩽ r.

Then Q_{j} is injective on Hom(X̃_{P} , ℤ/2(c)[d]), if d − c = n + 1 + 2^{j}. Proof Let ṽ ∈ Hom(X̃_{P} , ℤ/2(c)[d]), where d − c = n + 1 + 2^{j}. If Q_{j}(ṽ) = 0,

then ṽ = Q_{j}(w̃), for some w̃ ∈ Hom(X̃_{P} , ℤ/2(c − 2^{j} + 1)[d − 2^{j+1} + 1]) (since Q_{j} acts without cohomology on Hom(X̃_{P} , ℤ/2(∗)[∗′])). Since (d−2^{j+1}+1)−(c−2^{j}+1) = n+1 > 0, we have that δ^{∗} : Hom(X_{P} , ℤ/2(c− 2^{j} + 1)[d − 2^{j+1} + 1]) → Hom(X̃_{P} , ℤ/2(c − 2^{j} + 1)[d − 2^{j+1} + 1]) is an isomorphism, and there exists w ∈ Hom(X_{P} , ℤ/2(c − 2^{j} + 1)[d − 2^{j+1} + 1]), s.t. w̃ = δ^{∗}(w).

By Sublemma 2.3 , w = η · u, for some u ∈ Hom(X_{P} , ℤ/2(c − 2^{j} + 1 − n)[d − 2^{j+1} − 2n]). By [19], Theorem 3.17(2), Q_{j}(η · u) = Q_{j}(η) · u + η · Q_{j}(u) + ∑{−1}^{x_{i}}φ_{i}(η) · ψ_{i}(u), where x_{i} > 0, and ϕ_{i}, ψ_{i} are cohomological operations of some bidegree (∗)[∗′], where ∗′ > 2∗ ⩾ 0.

Notice, that c−2^{j}+1−n = d−2^{j+1}−2n =: s. But by [19], Proposition 2.7+Corollary 2.13(2)+Theorem 4.1, we have that Hom(ℤ, ℤ/2(a)[b]) →^{pr^{∗}} Hom(X_{P} , ℤ/2(a)[b]) is an isomorphism for a ⩽ b. So, u = pr^{∗}(u_{0}), where u_{0} ∈ Hom(ℤ, ℤ/2(s)[s]) = K_{s}^{M}(k)/2. We have: Q_{j}(u_{0}) = 0 and ψ_{i}(u_{0}) = 0 (since Hom(ℤ, ℤ/2(a)[b]) = 0, for b > a).

But pr^{∗} commutes with Q_{j} and ψ_{i}. So, Q_{j}(u) = 0 and ψ_{i}(u) = 0. That means: Q_{j}(w) = Q_{j}(η · u) = Q_{j}(η) · u = 0, by Sublemma 2.5 .

We get: ṽ = Q_{j}(w̃) = Q_{j} ∘ δ^{∗}(w) = δ^{∗} ∘ Q_{j}(w) = 0. I.e., Q_{j} is injective on Hom(X̃_{P} , ℤ/2(c)[d]).

Denote: η̃ := δ^{∗}(η) ∈ Hom(X̃_{P} , ℤ/2(n)[2n + 1]). Since η ≠ 0, we have: η̃ ≠ 0.

Sublemma 2.7 .

Let 0 ⩽ m < r, and η̃ = Q_{m} ∘ ⋯ ∘ Q_{1} ∘ Q_{0}(η̃_{m}), for some η̃_{m} ∈ Hom(X̃_{P} , ℤ/2(n−2^{m+1} +m+2)[2n−2^{m+2} +m+4]), Then there exists η̃_{m+1}, s.t. η̃_{m} = Q_{m+1}(η̃_{m+1}).

Proof Since Q_{m+1} acts without cohomology on Hom(X̃_{P} , ℤ/2(∗)[∗′]), it is enough to show that Q_{m+1}(η̃_{m}) = 0.

Denote: ṽ := Q_{m+1}(η̃_{m}). We have: ṽ ∈ Hom(X̃_{P} , ℤ/2(n+m+1)[2n+ m + 3]). Since Q_{i} commutes with Q_{j} (see [19], Theorem 3.17(1)), we have: Q_{m} ∘ Q_{m−1} ∘ ⋯ ∘ Q_{0}(ṽ) = Q_{m} ∘ Q_{m−1} ∘ ⋯ ∘ Q_{0} ∘ Q_{m+1}(η̃_{m}) = Q_{m+1} ∘ Q_{m} ∘ Q_{m−1} ∘ ⋯ ∘ Q_{0}(η̃_{m}) = Q_{m+1}(η̃) = 0, by Sublemma 2.5 .

But, for any 0 ⩽ t ⩽ m, Q_{t−1} ∘ ⋯ ∘ Q_{0}(ṽ) ∈ Hom(X̃_{P} , ℤ(c)[d]), where d − c = n + 1 + 2^{t}, and Q_{t} is injective on Hom(X̃_{P} , ℤ(c)[d]), by Sublemma 2.6 . So, from the equality: Q_{m} ∘ Q_{m−1} ∘ ⋯ ∘ Q_{0}(ṽ) = 0, we get: ṽ = 0.

From Sublemma 2.7 it follows that η̃ = Q_{r} ∘ … Q_{1} ∘ Q_{0}(η̃_{r}). Denote: γ̃ := η̃_{r}. We have: γ̃ ∈ Hom(X̃_{P} , ℤ/2(n−2^{r+1}+2+r)[2n−2^{r+2}+4+r]).

But, (2n − 2^{r+2} + 4 + r) − (n − 2^{r+1} + 2 + r) = n − 2^{r+1} + 2, and r = [log_{2}(n)], so: 2^{r} ⩽ n < 2^{r+1}. Since we know, that Hom(X̃_{P} , ℤ/2(a)[b]) = 0, for a ⩾ b ([19], Proposition 2.7+Corollary 2.13(2)+Theorem 4.1), and η̃ ≠ 0, the only possible choice for n is n = 2^{r+1} − 1.

Lemma 1.2 is proven.

From the proof of Lemma 1.2 one can deduce more. Namely, since the group Hom(X̃_{P} , ℤ/2(n − 2 − 2^{r+1} + 2 + r)[2n − 2^{r+2} + 4 + r]) = Hom(X̃_{P} , ℤ/2(r + 1)[r + 2]) = Hom(X_{P} , ℤ/2(r + 1)[r + 2]) can be identified with Ker(K_{r+2}^{M}(k)/2 → K_{r+2}^{M}(k(P))/2) (by [18], Lemma 6.4 + [19], Theorem 4.1; or by [7], Theorem A.1), the nontrivial element γ̃ ∈ Hom(X̃_{P} , ℤ/2(r + 1)[r + 2]) gives us nontrivial element α ∈ Ker(K_{r+2}^{M}(k)/2 → K_{r+2}^{M}(k(P))/2) (compare with [5], Theorem 3.1). It follows from the proof above, that this element α has the property that for arbitrary field extension E/k, α|_{E} is zero if and only if γ̃|_{E} is zero, if and only if μ|_{E} is zero, if and only if P is isotropic. So, P is a norm-variety for α ∈ K_{r+2}^{M}(k)/2.

If α is a pure symbol {a_{1}, …, a_{r+2}}, then it is easy to see that p must be proportional to a neighbor of the Pfister form ⟨⟨a_{1}, …, a_{r+2}⟩⟩ (simply because ⟨⟨a_{1}, …, a_{r+2}⟩⟩|_{k(P)} is isotropic). Then our motive N must be isomorphic to a Rost motive (by [13], Proposition 4, and our Theorem A.9 (or [15], Lemma 3.21)).

Since it is quite difficult to imagine, that quadric can be a normvariety for the element from K_{r+2}^{M}(k)/2, which is not a pure symbol, the following conjecture is natural.

Conjecture 2.8 (compare with [5], Conjecture 3.2).

Let Q be anisotropic quadric, and N be a direct summand in M(Q), s.t. N|\overline{k} = ℤ(a)[2a] ⊕ ℤ(b)[2b]. Then N is isomorphic to a (shifted) Rost motive M_{α}(c)[2c] for some pure symbol α ∈ K_{t}^{M}(k)/2, and c = min(a, b).

Moreover, if a or b is < i_{1}(q) or > dim(Q)−i_{1}(q), then q is a neighbor of a Pfister form.

3. Splitting patterns of forms of height 2

Lemma 1.2 also permits to describe all possible dimensions of forms of height 2.

Theorem 3.1 .

Suppose k is a field of characteristic 0. Let q be anisotropic quadric over k, s.t. height(q) = 2.

Then the splitting pattern of q is:

1) for q ∈ I^{n}\I^{n+1}(k) for n ⩾ 1: either (2^{m}−2^{n}, 2^{n−1}), for m > n ⩾ 1, or (2^{n−1}, 2^{n−1}), or (2^{n−2}, 2^{n−1}) (for n ⩾ 2),

2) for q ∉ I(k): (2^{m} − 2^{n} + 1, 2^{n−1} − 1) for some m ⩾ n > 1.

Proof

Lemma 3.2 (compare with [15], Statement 6.2).

Let k be a field of characteristic not 2. Let q be anisotropic form over k, s.t. height(q) = 2 and i_{1}(q) does not divide i_{2}(q) (for example, i_{1}(q) > i_{2}(q)), then there exists an indecomposable direct summand N in M(Q), s.t. N|\overline{k} = ℤ ⊕ ℤ(dim(Q) − i_{1}(q) + 1)[2 dim(Q) − 2i_{1}(q) + 2].

Proof

Let N be indecomposable direct summand in M(Q), s.t. a(N) = 0 (see the definition of a(N) before Theorem A.17 ). By Theorem A.17 , in M(Q) we will have also direct summands N(i)[2i] for any 0 ⩽ i < i_{1}(q). Since N(i)[2i] is not isomorphic to N(j)[2j], if i ≠ j, by Theorem A.12 , we get a direct summand isomorphic to ⊕_{0⩽i<i_{1}(q)}N(i)[2i].

If L is a complementary direct summand to ⊕_{0⩽i<i_{1}(q)}N(i)[2i] in M(Q), then since i_{2}(q) is not divisible by i_{1}(q), we have that i_{1}(q) ⩽ a(L) ⩽ dim(Q)/2. By Theorem A.17 , for any i_{1}(q) ⩽ j ⩽ dim(Q)/2, L(j − a(L))[2j − 2a(L)] will be a direct summand in M(Q). Since a(N(i)[2i]) = i ≠ j = a(L(j − a(L))[2j − 2a(L)]), for any 0 < i_{1}(q) ⩽ j ⩽ dim(Q)/2, we have: L(j − a(L))[2j − 2a(L)] and N(i)[2i] are not isomorphic. By Theorem A.12 , (⊕_{i_{1}(q)⩽j⩽dim(Q)/2}L(j − a(L))[2j − 2a(L)]) ⊕ (⊕_{0⩽i<i_{1}(q)}N(i)[2i]) is isomorphic to a direct summand of M(Q). This implies (using Corollary A.17.12 ): CH_{r}(N|\overline{k}) = 0, for any r ≠ 0, or dim(Q) − i_{1}(q) + 1, i.e.: N|\overline{k} = ℤ ⊕ ℤ(dim(Q) − i_{1}(q) + 1)[2 dim(Q) − 2i_{1}(q) + 2].

From Lemma 3.2 and Lemma 1.2 it follows, that if q is of height = 2, then either dim(Q) − i_{1}(q) + 1 = 2^{r} − 1, or i_{1}(q) divides i_{2}(q).

Let us consider 2 cases: 1) degree(q) = n for n ⩾ 1; 2) q ∉ I(k).

1) If degree(q) = n ⩾ 1, then i_{2}(q) = 2^{n−1}. So, either i_{1}(q) = 2^{l} for 0 ⩽ l ⩽ n − 1, or dim(q) − i_{1}(q) − 1 = 2^{r} − 1.

In the former case, by Theorem 1.5 , we have: l = n − 1 or n − 2 (if n ⩾ 2), and we get splitting patterns (2^{n−1}, 2^{n−1}) and (2^{n−2}, 2^{n−1}).

In the later case, dim(q) − i_{1}(q) = 2^{r}, and since dim(q) = 2(i_{1}(q) + i_{2}(q)) = 2i_{1}(q) + 2^{n}, we have: i_{1}(q) = 2^{r} − 2^{n} for some r > n. We get the splitting pattern (2^{r} − 2^{n}, 2^{n−1}).

2) In this case, i_{2}(q) = 2^{n−1} − 1 for some n ⩾ 1 (since if odddimensional quadratic form is “hyperbolic” over generic point of the corresponding quadric, then it is proportional to a hyperplane section in some Pfister form).

Let p := q ⊥ − det(q) (here under det(⟨a_{1}, …, a_{s}⟩) I mean (−1)^{[dim(q)/2]}· ∏_{i=1}^{s} a_{i}). Then det(p) = 1, so: p ∈ I^{2}(k).

Let E/k be some field extension. Consider the following conditions:

a) dim((p|_{E})_{an}) < 2^{n}−2; b) dim((q|_{E})_{an}) < 2^{n}−1; c) dim((q|_{E})_{an}) = 1;

d) dim((p|_{E})_{an}) = 0.

Evidently, a) ⇒ b), and d) ⇒ a). Since i_{2}(q) = 2^{n−1} − 1, we get:

b) ⇒ c). Finally, since p ∈ I^{2}(k) (and, consequently, the last higher Witt index of p is at least 2), we have: c) ⇒ d). So, all the above conditions are equivalent. That means: the last higher Witt index of p is at least 2^{n−1} − 1. Since such index is always a power of 2, we have p ∈ J^{n}(k) (in the case n = 2 use the fact that p ∈ I^{2}(q)).

On the other hand, p ∉ J^{n+1}(k), since then over any field extension E/k we would have: dim((q|_{E})_{an}) is either ⩾ 2^{n+1} − 1, or = 1, which is not the case.

Then, by Theorem 1.5 , dim(p_{an}) is either A) ⩾ 2^{n+1}, or B) 2^{n+1} > dim(p_{an}) ⩾ 2^{n} + 2^{n−1}, or C) = 2^{n}.

In the case A): dim(q) ⩾ 2^{n+1} − 1, and i_{1}(q) ⩾ 2^{n−1} > 2^{n−1} − 1 = i_{2}(q). So, by Lemma 3.2 and Lemma 1.2 , dim(q)−i_{1}(q)−1 = 2^{m}−1, for some m > n. Since dim(q) = 2(i_{1}(q)+i_{2}(q))+1 = 2i_{1}(q)+2^{n}−1, we get: i_{1}(q) = 2^{m}−2^{n}+1, and the splitting pattern of q is (2^{m}−2^{n}+1, 2^{n−1}−1).

In the case B): we have n ⩾ 2, and, over some field extension E/k, height(p|_{E}) = 2, then, by 1), dim((p|_{E})_{an}) = 2^{n} + 2^{n−1}. Since dim((q|_{E})_{an}) > 2^{n} + 2^{n−1} − 1 > 2^{n} − 1, we have that height(q|_{E}) = 2, i.e.: q|_{E} is anisotropic. So, dim(q) is either 2^{n}+2^{n−1}−1, or 2^{n}+2^{n−1}+1 , and i_{1}(q) is either 2^{n−2}, or 2^{n−2}+1. We have two possibilities: I) n > 2; II) n = 2. I) In this case, i_{1}(q) does not divide i_{2}(q) = 2^{n−1} − 1 So, by Lemma 3.2 and Lemma 1.2 , dim(q) − i_{1}(q) − 1 = 2^{m} − 1 for some m, which leads to a contradiction. II) In this case, dim(p_{an}) = 6, i_{2}(q) = 1. So, clearly, we must have: i_{1}(q) = 1 and dim(q) = 5, since there are no 7−dimensional forms with i_{1} = 2 (one can use Lemma 3.2 and Lemma 1.2 here). We get the splitting pattern (1, 1) = (2^{2} − 2^{2} + 1, 2^{1} − 1).

In the case C): since i_{2}(q) = 2^{n−1} − 1 and height(q) = 2, we have dim(q) ⩾ 2^{n−1} + 1. On the other hand, since dim(p_{an}) = 2^{n}, we have: dim(q) ⩽ 2^{n−1} + 1. So, q has splitting pattern: (1, 2^{n−1} − 1) = (2^{n} − 2^{n} + 1, 2^{n−1} − 1).

Theorem 3.1 is proven.

4. Some other results

In this section we will give some other applications of Lemma 1.2 . Theorem 4.1 . Let k be a field of char(k) = 0. Let q ∈ J^{n}(k) be anisotropic, and

dim(q) < 2^{n+1}.

Let N be indecomposable direct summand in M(Q), s.t. a(N) = 0. Then M(Q) = ⊕_{0⩽i<i_{1}(q)}N(i)[2i].

Proof

We use induction on the height(q). If height(q) = 1, then q is proportional to a Pfister form ⟨⟨α⟩⟩, and, by [13], Proposition 4, M(Q) = ⊕_{0⩽i<i_{1}(q)}M_{α}(i)[2i], where M_{α} is a Rost motive (which is indecomposable)(instead, one can use Theorem A.17 , Theorem A.12 , and Sublemma A.17.8 ). Suppose now, r ⩾ 2, and the statement is true for all forms of height ⩽ r − 1 (over all fields). Let q ∈ J^{n}(k) be anisotropic form of height(q) = r, and dim(q) < 2^{n+1}. Then p := (q|_{k(Q)})_{an} is of height = r − 1, so, M(P) = ⊕_{0⩽j<i_{1}(p)}L(j)[2j], where L is indecomoposable direct summand in M(P), s.t. a(L) = 0.

Let N be indecomposable direct summand in M(Q), s.t. a(N) = 0 (such summand is defined up to isomorphism, by Theorem A.9 ). Then, by Theorem A.17 , N(i)[2i] will be a direct summand in M(Q) for all 0 ⩽ i < i_{1}(q). By Theorem A.12 , ⊕_{0⩽i<i_{1}(q)}N(i)[2i] will be a direct summand in M(Q). Let M be a complimentary summand (i.e.: M(Q) = M ⊕ (⊕_{0⩽i<i_{1}(q)}N(i)[2i])). Then a(M) ⩾ i_{1}(q). Then, by Corollary A.17.12 , b(M) ⩽ dim(Q) − i_{1}(q).

Suppose, M ≠ 0.

By [13], Proposition 1 (Theorem A.0 ), M(Q|_{k(Q)}) = ⊕_{0⩽i<i_{1}(q)}(ℤ(i)[2i]⊕ ℤ(dim(Q)−i)[2 dim(Q)−2i])⊕M(P)(i_{1}(q))[2i_{1}(q)]. Since a(L) ⩾ i_{1}(q), and b(L) ⩽ dim(Q) − i_{1}(q), we have that M|_{k(Q)} is isomorphic a direct summand in M(P)(i_{1}(q))[2i_{1}(q)]. So, M|_{k(Q)} = ⊕_{j∈Λ}L(i_{1}(q) + j)[2i_{1}(q) + 2j], where Λ ⊂ [0, …, i_{1}(p) − 1] is some subset.

Let U be indecomposable direct summand in M, s.t. a(U) = a(M). Clearly, i_{1}(q) ⩽ a(U) < i_{1}(q) + i_{1}(p) (since a(U) = i_{1}(q) + min(j ∈ Λ)). By Theorem A.17 , U(i_{1}(q) + j − a(U))[2i_{1}(q) + 2j − 2a(U)] is a direct summand of M(Q), for any 0 ⩽ j < i_{1}(p). Since L(i_{1}(q))[2i_{1}(q)] is indecomposable, and a(L(i_{1}(q))[2i_{1}(q)]) = i_{1}(q) = a(U(i_{1}(q)−a(L))[2i_{1}(q)− 2a(L)]|_{k(Q)}), we have: L(i_{1}(q))[2i_{1}(q)] is a direct summand in U(i_{1}(q)− a(L))[2i_{1}(q)−2a(L)]|_{k(Q)} (by Theorem A.9 ). Then M(P)(i_{1}(q))[2i_{1}(q)] = ⊕_{0⩽j<i_{1}(p)}L(i_{1}(q) + j)[2i_{1}(q) + 2j] is isomorphic to a direct summand of ⊕_{0⩽j<i_{1}(p)}U(i_{1}(q) + j − a(U))[2i_{1}(q) + 2j − 2a(U)]|_{k(Q)}.

Since a(N(i)[2i]) = i ≠ l = a(U(l − a(U))[l − 2a(U)]), for any 0 ⩽ i < i_{1}(q) ⩽ l < i_{1}(q) + i_{1}(p), by Theorem A.12 and above computations, (⊕_{0⩽i<i_{1}(q)}N(i)[2i])⊕M(P)(i_{1}(q))[2i_{1}(q)] is isomorphic to a direct summand of M(Q) (actually, to M(Q) itself). Then CH_{r}(N|\overline{k}) ≠ 0 only for r = 0 and r = dim(Q)−i_{1}(q)+1. Hence, N|\overline{k} = ℤ⊕ℤ(dim(Q)− i_{1}(q) + 1)[2 dim(Q) − 2i_{1}(q) + 2].

By Lemma 1.2 , dim(Q) − i_{1}(q) + 1 = 2^{t} − 1, for some t. This is impossible, since q ∈ J^{n}, and dim(q) < 2^{n+1}. Contradiction. So, M = 0, and Theorem 4.1 is proven.

Definition 4.2 .

Let N be direct summand of M(Q). We define Witt index i_{W} (N) of N as the number of Tate motives in the decomposition of N into indecomposables, divided by 2.

Let j_{0} < j_{1} < ⋯ < j_{h(N)} be all possible values of i_{W} (N|_{F}), for all field extensions F/k. We define the height of N as h(N), and the splitting pattern of N as the sequence (j_{0}, j_{1} − j_{0}, …, j_{h(N)} − j_{h(N)−1}).

Theorem 4.3 .

Let Q be anisotropic quadric, and N a direct summand of M(Q). Then the splitting pattern of N consists of integers.

Proof

Let F be some field, and Λ(F, q) := {r| s.t. either 0 ⩽ r < i_{W} (q|_{F}), or dim(Q) − i_{W} (q|_{F}) < r ⩽ dim(Q)}. Lemma 1.3 and [13], Proposition 1 (Theorem A.0 ), implies that i_{W} (N|_{F}) = rank(⊕_{r∈Λ(F,q)} CH_{r}(N|\overline{k})). This shows that to define splitting pattern of N, it is enough to consider F from the generalized splitting tower of M.Knebusch for q.

By induction, it is enough to show that i_{W} (N|_{k(Q)}) is integer. We can assume that N is indecomposable. In such a case, by Sublemma A.17.9 and Sublemma A.17.7 , either i_{W} (N|_{k(Q)}) = 0, or there exists subquadric P ⊂ Q and direct summand L of M(P), s.t. 1) i_{1}(p) = 1;

2) p|_{k(Q)} and q|_{k(P)} are isotropic; 3) a(L) = 0; 4) N ≃ L(a(N))[2a(N)]. Clearly, i_{W} (N|_{k(Q)}) = i_{W} (L|_{k(Q)}) = i_{W} (L|_{k(P)}), and the later is equal

to 1, by Sublemma A.17.5 (one should treat the evident case dim(P) = 0 separately).

Lemma 1.3 and [13], Proposition 1 (Theorem A.0 ) imply the following:

Observation 4.4 .

If N = M(Q), then i_{W} (M(Q)) = i_{W} (q), height(M(Q)) = height(q), and splitting pattern of M(Q) coinsides with that for q.

Theorem 4.5 .

Let char(k) = 0, and anisotropic q ∈ J^{n}(k) is of dimension < 2^{n+1}. Then splitting pattern of q is (0, 2^{s_{1}}, 2^{s_{2}}, …, 2^{s_{h}}), where 0 ⩽ s_{1} ⩽ s_{2} ⩽ … ⩽ s_{h} = n − 1.

Proof

Sublemma 4.5.1 .

Suppose char(k) = 0, and anisotropic q ∈ J^{n}(k) is of dimension < 2^{n+1}. Then i_{1}(q) = 2^{s_{1}} divides i_{r}(q) for 1 ⩽ r ⩽ h.

Proof

By Theorem 4.1 , M(Q) = ⊕_{0⩽i<i_{1}(q)}N(i)[2i], where N is indecomposable direct summand of M(Q), s.t. a(N) = 0.

Clearly, if (0, i_{1}(N), i_{2}(N), …, i_{h}(N)) is the splitting pattern of N, then the splitting pattern of M(Q) will be (0, i_{1}(q)·i_{1}(N), i_{1}(q)·i_{2}(N), …, i_{1}(q) · i_{h}(N)). Since i_{h}(q) = 2^{n−1}, we have: i_{1}(q) = 2^{s_{1}}, and i_{r}(q) is divisible by i_{1}(q), for 1 ⩽ r ⩽ h (we use Theorem 4.3 here).

Since i_{r+t}(q) = i_{1+t}((q|_{F_{r−1}})_{an}), where F_{r−1} is (r − 1)-st field in the generalized splitting tower of M.Knebusch for q, and (q|_{F_{r−1}})_{an} satisfy the conditions of our theorem, we have by Sublemma 4.5.1 : i_{r}(q) = 2^{s_{r}}, and i_{r}(q) divides i_{r′}(q), for r′ ⩾ r.

Main Theorem shows that in the sequence 0 ⩽ s_{1} ⩽ s_{2} ⩽ … ⩽ s_{h−1} ⩽ s_{h} = 2^{n−1} we nesessarily have: s_{h−1} = s_{h} − 1. This leads to the following natural:

Conjecture 4.6 .

Let q ∈ J^{n} be anisotropic form of dimension < 2^{n+1}. Then the splitting pattern of q is 2^{n−height(q)} · (0, 1, 2, 4, …, 2^{height(q)−1})

Theorem 4.5 permits to describe splitting patterns for (generalized) Albert forms:

Theorem 4.7 .

Let char(k) = 0. Let α, β ∈ K_{n}^{M}(k)/2 be pure symbols, and A_{α,β} := (⟨⟨α⟩⟩ − ⟨⟨β⟩⟩)_{an} be generalized Albert form, corresponding to α + β. Then the splitting pattern of A_{α,β} is 2^{r}(0, 1, 2, 4, …, 2^{n−r−1}), where r is the degree of common maximal pure divisor of α and β.

Proof

By the result of R.Elman and T.Y.Lam (see [1], Theorem 4.5), we have: dim((A_{α,β}|_{F} |_{an}) = 2^{n+1} − 2^{r(F)+1}, where r(F) is the degree of common maximal pure divisor of α|_{F} and β|_{F}. By Theorem 4.5 , for all 1 ⩽ l ⩽ h, there exists field F_{l}/k, s.t. dim((A_{α,β}|_{F_{l}})_{an}) = 2 · (2^{s_{l}} + ⋯ + 2^{s_{h}}). So, 2 · (2^{s_{l}} + ⋯ + 2^{s_{h}}) = 2^{n+1} − 2^{r(l)+1}, for all 1 ⩽ l ⩽ h, and some r(l). It is not difficult to see that this is possible only if (2^{s_{1}}, …, 2^{s_{h}}) = 2^{r}(1, 2, 4, …, 2^{n−r−1}).

Corollary 4.8 .

Let char(k) = 0. Let α, β ∈ K_{n}^{M}(k)/2 be pure symbols, and r is the degree of common maximal pure divisor of α and β. Then for any r ⩽ t ⩽ n, there exists field extension E_{t}/k, s.t. the degree of common maximal pure divisor of α|_{E_{t}} and β|_{E_{t}} is t.

Proof

Let k = F_{0} ⊂ F_{1} ⊂ ⋯ ⊂ F_{h} be generalized splitting tower of M.Knebusch for A_{α,β}. Put E_{t} := F_{t−r}. Then, by Theorem 4.7 , dim((A_{α,β}|_{E_{t}})_{an}) = 2^{n+1} − 2^{t+1}. By [1], Theorem 4.5, the degree of common maximal pure divisor of α|_{E_{t}} and β|_{E_{t}} is t.

Definition 4.9 .

Let h ∈ K_{n}^{M}(k)/2. We define (algebraic) lenght of h as length(h) := min(s| h = α_{1} + ⋯ + α_{s}, α_{i} − pure symbol ).

Corollary 4.8 can be considered as a “derivative” of the following: Question 4.10 . Let h ∈ K_{n}^{M}(k)/2, and length(h) ⩾ s. Is it true that there exists

such field extension E/k, s.t. length(h|_{E}) = s ?

By the result of V.Voevodsky, Question 4.10 has positive answer for s = 1, which implies Milnor’s Conjecture on quadratic forms.

For the theory of quadratic forms it would be very useful to have positive answer for s = 2.

Remark

Corollary 4.8 can be considered as a “derivative” of Question 4.10 in the following sense. length gives increasing filtration on K_{n}^{M}(k)/2. It is natural to call h ∈ K_{n}^{M}(k)/2 infinitesimally close to 0, if length(h) ⩽ 2. In such neighborhood, we have more subtle filtration - by the degree of maximal pure divisor of h = α + β (which coinsides with the degree of maximal common pure divisor for α and β).

Finally, I will state the corrected version of Conjecture 1.1 .

Conjecture 4.11 .

Let q ∈ I^{n}(k) be anisotropic. Then dim(q) is either 2^{n+1} − 2^{i+1}, for some 0 ⩽ i ⩽ n, or is even ⩾ 2^{n+1}.

Clearly, Conjecture 4.11 follows from Conjecture 4.6 . Also, it is not difficult to show:

Theorem 4.12 .

In the situation of Conjecture 4.11 , all the prescribed values of dim(q) are possible for suitable k and q.

Appendix

In this Appendix we will list some results on the motives of quadrics, we use in the main part of the paper. All results are from (or are slight modifications of results from) [15] and [16]. We reproduce proofs here for the reader’s convenience.

All results in this section are valid under assumption: char(k) ≠ 2, which we will assume everywhere.

We start by reminding some facts about Chow^{eff.}(k) (see also [11]), and the structure of the motive of hyperbolic quadric.

An object of Chow^{eff.}(k) is a pair (X, p_{X}), where X = ⊔_{i}X_{i}, X_{i} is smooth connected projective variety over k, and p_{X} ∈ ×_{i,j} CH^{dim(X_{j})}(X_{i}× X_{j}) is a proejctor: p_{X}∘p_{X} = p_{X}, where the composition ∘ : CH^{dim(Z)}(Y × Z)⊗CH^{dim(Y)}(X ×Y) → CH^{dim(Z)}(X ×Z) is given by: π_{X,Z}_{∗}(π_{X,Y}^{∗} (−)∩ π_{Y,Z}^{∗}(−)). One defines Hom_{Chow^{eff.}(k)}((X, p_{X}), (Y, P_{Y})) as p_{Y} ∘CH^{dim(Y)}(X× Y)∘p_{X} ⊂ CH^{dim(Y)}(X×Y) with the composition given by ∘. Chow^{eff.}(k) is tensor additive category, where (X, p_{X})⊕(Y, p_{Y}) := (X⊔Y, (p_{X}, p_{Y})), and (X, p_{X}) ⊗ (Y, p_{Y}) := (X × Y, p_{X} × p_{Y}).

If X is smooth projective variety over k, then one defines the motive M(X) ∈ Ob(Chow^{eff.}(k)) as the pair (X, Δ_{X}). We denote M(Spec(k)) as ℤ.

One observes that M(ℙ^{1}) = (ℙ^{1}, ℙ^{1} × Spec(k)) ⊕ (ℙ^{1}, Spec(k) × ℙ^{1}). It is easy to see, that (ℙ^{1}, ℙ^{1} × Spec(k)) ≃ ℤ. One defines ℤ(1)[2] as (ℙ^{1}, Spec(k) × ℙ^{1}). And ℤ(l)[2l] := ℤ(1)[2]^{⊗l}. If U ∈ Ob(Chow^{eff.}(k)), we denote: U(l)[2l] := U ⊗ ℤ(l)[2l]. Also, we define: CH_{r}(U) := Hom_{Chow^{eff.}(k)}(ℤ(r)[2r], U). It is not difficult to show that CH_{r}(M(X)(l)[2l]) can be naturally identified with CH_{r−l}(X).

More generally, let P, Q, and R be some smooth projective varieties over some field k of dimension l, m, and n, respectively. Then Hom_{Chow^{eff.}(k)}(M(P)(a)[2a], M(Q)(b)[2b]) can be identified with CH^{m+b−a}(P × Q). Under this identification, the composition: Hom(M(P)(a)[2a], M(Q)(b)[2b]) × Hom(M(Q)(b)[2b], M(R)(c)[2c]) → Hom(M(P)(a)[2a], M(R)(c)[2c]) corresponds to the map: (−, −) : CH^{m+b−a}(P × Q) × CH^{n+c−b}(Q × R) → CH^{n+c−a}(P × R), (Φ, Ψ) := π_{1,3}_{∗}(π_{1,2}^{∗}(Φ) ∩ π_{2,3}^{∗}(Ψ)), where π_{1,2} : P × Q × R → P × Q, π_{2,3} : P × Q × R → Q × R, and π_{1,3} : P × Q × R → P × R are natural projection.

In particular, taking P = Spec(k), we get pairing: (−, −) : CH_{a}(Q)× CH^{n+c}(Q×R) → CH_{a−c}(R). In this case, we will denote (v, Φ) as Φ(v).

Let now char(k) ≠ 2, and Q be some quadric over k.

Choose some algebraic closure \overline{k} ⊃ k. Let h^{i} ⊂ Q|\overline{k} and l_{i} ⊂ Q|\overline{k}, for 0 ⩽ i < m/2 be cycles, given by plane section of codimension i, and projective subspace of dimension i on Q|\overline{k}, respectively. Let l_{m/2}^{1}, l_{m/2}^{2} ⊂ Q|\overline{k} be projective subspaces of middle dimension from two different families. For i = 1, 2, let us define: \overline{i} := i if m = 4t, and \overline{i} := 3 − i, if m = 4t + 2.

Then in CH^{m}(Q|\overline{k} × Q|\overline{k}) we have an identity:

Δ_{Q|\overline{k}} = ∑_{0⩽i<m/2}(h^{i} × l_{i} + l_{i} × h^{i}) + (l\overline{1} ; m/2 × l_{m/2}^{1} + l\overline{2} ; m/2 × l_{m/2}^{2}).

Moreover, h^{i} × l_{i}, and l_{i} × h^{i}, for 0 ⩽ i < m/2, as well as l\overline{1} ; m/2 × l_{m/2}^{1} and l\overline{2} ; m/2 × l_{m/2}^{2} are mutually orthogonal idempotents.

This defines a decomposition: M(Q|\overline{k}) = (⊕_{i=0,...,m}ℤ(i)[2i])⊕ℤ(m/2)[m]. In particular, for i ≠ m/2, CH_{i}(Q|\overline{k}) = Z, with generator l_{i}, if i < m/2, and h^{m−i}, if i > m/2; CH_{m/2}(Q|\overline{k}) = Z ⊕ Z, with generators l_{m/2}^{1} and l_{m/2}^{2}.

We have an isomorphism: End(M(Q|\overline{k})) ≃ ×_{r} End(CH_{r}(Q|\overline{k})). The above choice of generators for CH_{r}(Q|\overline{k}) gives isomorphism End(CH_{r}(Q|\overline{k})) ≃ Z, for r ≠ m/2, and End(CH_{m/2}(Q|\overline{k})) ≃ Mat_{2×2}(Z). For ψ ∈ End(M(Q)), we denote as ψ_{(r)} the coordinate, corresponding to the action of ψ|\overline{k} on CH_{r}(Q|\overline{k}).

Under the isomorphism above, the element γ ∈ CH^{m}(Q|\overline{k} × Q|\overline{k}), where γ = ∑_{0⩽i<m/2}(γ_{i} · h^{i} × l_{i} + γ_{m−i} · l_{i} × h^{i}) + (∑_{i=1,2; j=1,2} γ_{m/2}^{i,j} ·

l_{m/2}^{i} × l_{m/2}^{j}) has coordinates: γ_{(r)} = γ_{r}, for r ≠ m/2, and γ_{(m/2)}^{i,j} = γ_{m/2}^{\overline{i},j}.

Also, γ_{i}, resp. γ_{m−i}, is the degree of γ ∩ l_{i} × h^{i} ∈ CH_{0}(Q|\overline{k} × Q|\overline{k}), resp. of γ ∩ h^{i} × l_{i} ∈ CH_{0}(Q|\overline{k} × Q|\overline{k}), and γ_{(m/2)}^{i,j} is the degree of

γ ∩ l_{m/2}^{i} × l\overline{j} ; m/2 ∈ CH_{0}(Q|\overline{k} × Q|\overline{k}).

If q = l · ℍ ⊥ q′, then the motive of M(Q) can be decomposed in the following way:

Theorem A.0 (M.Rost, [13], Proposition 1).

M(Q) ≃ ⊕_{0⩽i<l}(ℤ(i)[2i]⊕ℤ(dim(Q)−i)[2 dim(Q)−2i])⊕M(Q′)(l)[2l]. End(M(Q)) = ×_{0⩽i<l}(End(ℤ(i)[2i])×End(ℤ(dim(Q)−i)[2 dim(Q)−

2i])) × End(M(Q′)(l)[2l]).

Remark Actually, [13], Proposition 1 does not contain the second statement, but it follows from the fact that there are no Hom-s between different direct summands above.

If N is a direct summand of M(Q), we will denote as p_{N} ∈ End(M(Q)) the corresponding projector, and as j_{N} : N → M(Q), φ_{N} : M(Q) → N the natural maps, s.t. φ_{N} ∘ j_{N} = id_{N}, and j_{N} ∘ φ_{N} = p_{N}. We define: CH_{r}(N) := Hom(ℤ(r)[2r], N). The composition with j_{N} identifies this group with p_{N} ·CH_{r}(Q) ⊂ CH_{r}(Q) = Hom(ℤ(r)[2r], M(Q)). We choose generators e_{w} of CH_{r}(N|\overline{k}) in the following way: if rank(CH_{r}(N|\overline{k})) = 1, then e_{w} is arbitrary generator of this group, and if rank(CH_{r}(N|\overline{k})) = 2 (i.e.: r = m/2, and CH_{r}(N|\overline{k}) = CH_{r}(Q|\overline{k})), then we take φ_{N}(l_{m/2}^{1}) and φ_{N}(l_{m/2}^{2}) as generators (usually, we will denote φ_{N}(l_{i}) and φ_{N}(h^{i}) simply as l_{i} and h^{i}). We denote: CH(N) := ⊕_{r} CH_{r}(N). We have: CH(N|\overline{k}) = ⊕_{w∈Ω(N)}e_{w} · Z.

We define map \overline{deg} : CH(N|\overline{k}) → Z/2 by the rule: \overline{deg}(e_{w}) := 1.

We have a natural isomorphism End(N|\overline{k}) ≃ ×_{r} End(CH_{r}(N|\overline{k})). The above choice of generators for CH(N) identifies End(N|\overline{k}) with the product of matrix rings. For ψ ∈ End(N), we will denote the corresponding coordinates as ψ_{(r)}.

All the results we prove in this Appendix are, in some sense, corollaries of the following result (Rost Nilpotence Theorem):

Theorem A.1 (M.Rost, [14], Corollary 10).

1) Let φ ∈ Hom(M(Q), M(Q)) be such that φ|\overline{k} = 0. Then φ^{n} = 0 for some n.

2) Let φ ∈ Hom(M(Q), M(Q)) be such that φ|\overline{k} is an isomorphism. Then φ is an isomorphism.

Remark In the case char(k) = 0, the nilpotence exponent n can be bounded by dim(Q) + 1 - see [15], Lemma 3.10.

Corollary A.2 ([15], Lemma 3.12). Let ξ ∈ Hom(M(Q), M(Q)) be some map, s.t. ξ|\overline{k} is a projector.

Then, for some d, ξ^{2^{d}} is a projector.

Proof Let x := ξ^{2} − ξ ∈ Hom(M(Q), M(Q)) = CH^{m}(Q × Q). Since

ξ|\overline{k} is idempotent, x|\overline{k} = 0. In particular, 2^{s} · x = 0, for some s, since Q is hyperbolic over some galois extension F/k of degree 2^{s}, and ∑_{σ∈Gal(F/k)}(σ(x|\overline{k})) = [F : k] · x.

By Theorem A.1 , we have: x^{t} = 0 for some t. That means, that for some large d: \binom{2^{d}}{j} · x^{j} = 0 for any j > 0. From the equality: ξ^{2} = (ξ+x) (and the fact that ξ and x commute),

we get: ξ^{2^{d+1}} = ∑_{0⩽j⩽2^{d}} \binom{2^{d}}{j}ξ^{2^{d}−j} · x^{j} = ξ^{2^{d}}.

So, ξ^{2^{d}} is a projector.

Corollary A.3 . Let N be a direct summand in M(Q), and ψ ∈ Hom(N, N) be such

that:

1) ψ|\overline{k} = 0. Then ψ^{n} = 0 for some n.

2) ψ|\overline{k} is a projector. Then ψ^{n} is a projector for some n.

3) ψ|\overline{k} is an isomorphism. Then ψ is an isomorphism. Proof

Let M(Q) = N ⊕ M. It is enough to consider φ = (\begin{matrix}ψ & 0 \\ 0 & ρ \end{matrix}), where

ρ = 0 in cases 1) and 2), and = id_{M} in case 3). In case 1) apply Theorem A.1 (1), in case 2) - Corollary A.2 , in case 3) - Theorem A.1 (2).

Lemma A.4 . Let L and N be direct summands in M(Q), s.t. p_{L}|\overline{k} ∘ p_{N}|\overline{k} = p_{N}|\overline{k} ∘

p_{L}|\overline{k} = p_{L}|\overline{k}.

Then there exists direct summand L̃ in N, s.t. L is isomorphic to L̃, and p_{L}|\overline{k} = pL̃|\overline{k}.

Proof

Let j_{L} : L → M(Q), j_{N} : N → M(Q), φ_{L} : M(Q) → L, φ_{N} : M(Q) → N be s.t.: φ_{L} ∘ j_{L} = id_{L}, φ_{N} ∘ j_{N} = id_{N}, and j_{L} ∘ φ_{L} = p_{L}, j_{N} ∘ φ_{N} = p_{N}.

Take: α := φ_{L} ∘ j_{N} : N → L, and β := φ_{N} ∘ j_{L} : L → N. If γ := α ∘ β : L → L, then γ|\overline{k} = id_{L}. By Corollary A.3 (2) and (1), γ^{s} = id_{L}, for some s. Consider ψ := φ_{N} ∘ p_{L} ∘ j_{N} : N → N. Then ψ^{s} is a projector, ψ^{s} = β ∘ α̃, where α̃ = α ∘ (β ∘ α)^{s−1}, and α̃ ∘ β = id_{L}. Then ψ^{s} defines direct summand L̃ in N, and for the corresponding projector in M(Q), pL̃ := j_{N} ∘ ψ^{s} ∘ φ_{N}, we have: pL̃|\overline{k} = p_{L}|\overline{k}.

Lemma A.5 (cf. [15], Lemma 3.13).

Let N be a direct summand in M(Q), dim(Q) = m.

1) Then there exists: κ_{r,N} ∈ End(N), s.t. (κ_{r,N})_{(s)} = 0, for any s ≠ r, and (κ_{r,N})_{(r)} = 2 · id_{CH_{r}(N|\overline{k})}.

2) If rank(CH_{m/2}(N|\overline{k})) = 2, then there exists: θ_{m/2,N} ∈ End(N),

s.t. (θ_{m/2,N})_{(m/2)} = (\begin{matrix}1 & 1 \\ 1 & 1 \end{matrix}); (θ_{m/2,N})_{(r)} = 0, for any r ≠ m/2.

Proof 1) Let dim(Q) = m. If r ≠ m/2, then we can take κ_{r,N} := φ_{N} ∘ (h^{r} × h^{m−r}) ∘ j_{N}.

If r = m/2, we can take κ_{r,N} := φ_{N} ∘ (2 · id_{M(Q)} − ∑_{0⩽i<m/2}(h^{i} × h^{m−i} + h^{m−i} × h^{i})) ∘ j_{N}.

2) Take θ_{m/2,N} := φ_{N} ∘ (h^{m/2} × h^{m/2}) ∘ j_{N}.

Lemma A.6 .

Let N_{i} is a direct summand of M(Q_{i}). Let for some odd number η, and some ψ ∈ Hom(N_{1}|\overline{k}, N_{2}|\overline{k}), we have: η·ψ ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|\overline{k}, N_{2}|\overline{k})). Then ψ ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|\overline{k}, N_{2}|\overline{k})).

Proof

Let F/k be Galois extension of degree 2^{n}, s.t. N_{i}|_{F} is a sum of Tate-motives. We have: Hom(N_{1}|_{F} , N_{2}|_{F}) → Hom(N_{1}|\overline{k}, N_{2}|\overline{k}) is an isomorphism. Let ψ_{F} be corresponding element of Hom(N_{1}|_{F} , N_{2}|_{F}).

We have: σ(ψ_{F}) = ψ_{F}, for any σ ∈ Gal(F/k). Really, if (σ(ψ_{F}) − ψ_{F}) ≠ 0, then η · (σ(ψ_{F}) − ψ_{F}) ≠ 0 too (since Hom(N_{1}|_{F} , N_{2}|_{F}) has no torsion), which contradicts to the fact that η·ψ ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|\overline{k}, N_{2}|\overline{k})). Then

2^{n} ·ψ_{F} = ∑_{σ∈Gal(F/k)} σ(ψ_{F}) ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|_{F} , N_{2}|_{F})).

Since η · ψ_{F}, 2^{n} · ψ_{F} ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|_{F} , N_{2}|_{F})), we have: ψ_{F} ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|_{F} , N_{2}|_{F})), which implies: ψ ∈ image(Hom(N_{1}, N_{2}) → Hom(N_{1}|\overline{k}, N_{2}|\overline{k})).

Definition A.7 .

Let N and N′ be indecomposable direct summands in M(Q). We say that N′ is a normal form of N, if N′ is isomorphic to N, and either m = dim(Q) is odd, or m is even and (p_{N′})_{(m/2)} is of one of the following types:

1) 0; 2) (\begin{matrix}1 & 1 \\ 0 & 0 \end{matrix}); 3) (\begin{matrix}0 & −1 \\ 0 & 1 \end{matrix}); 4) id.

Theorem A.8 (cf. [15], the proof of Lemma 3.21).

Each direct summand in M(Q) has a normal form.

Proof

The case of odd-dimensional quadric is trivial. So, we can assume that m := dim(Q) is even.

(p_{N})_{(m/2)} is an idempotent. So, if rank((p_{N})_{(m/2)}) is 0, or 2, we get cases 1) and 4), respectively.

So, we can assume that (p_{N})_{(m/2)} is a projector in Mat_{2×2}(Z) of rank = 1 (equivalently, det((p_{N})_{(m/2)}) = 0 and tr((p_{N})_{(m/2)}) = 1).

Sublemma A.8.1 .

Let N be indecomposable direct summand in M(Q), s.t. (p_{N})_{(m/2)} ≠ 0. Let ψ ∈ End(M(Q)) be such that ψ|\overline{k} ∘ p_{N}|\overline{k} = p_{N}|\overline{k} ∘ ψ|\overline{k} = ψ|\overline{k}, and tr(ψ_{(m/2)}) is odd. Then 2 · End(M(Q|\overline{k})) ⊂ image(End(M(Q)) → End(M(Q|\overline{k}))).

Proof If q is hyperbolic, then the map End(M(Q)) → End(M(Q|\overline{k}) is an isomorphism. So, we can assume that q is not hyperbolic.

Let τ ∈ Hom(M(Q), M(Q)) = CH^{m}(Q × Q) be the morphism given by the graph of “reflection” τ_{x} (with any (rational) center x ∈ ℙ^{m+1} ⊃ Q). Then τ^{2} = id_{M(Q)}. Also, τ_{(i)} = 1, for any i ≠ m/2, and τ_{(m/2)} =

(\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}).

Let ψ_{(m/2)} = (\begin{matrix}a & b \\ c & d \end{matrix}). Then degree(ψ(h^{m/2})) = a + b + c + d. Since

h^{m/2} is defined over k, we have that if a + b + c + d is odd, then on Q there is m/2-dimensional cycle of odd degree, which, by Sublemma 1.3.1 , implies that q is hyperbolic.

So, we can assume that a+b+c+d is even. Then in each pair (a, d), (b, c), one element is odd and another is even.

Changing ψ to ψ − ∑_{i≠m/2}[ψ_{(i)}/2] · κ_{i,Q}, we can assume that ψ_{(i)} are either 0’s or 1’s, for all i ≠ m/2. Notice, that this new ψ still satisfy the conditions of the sublemma.

We have two cases: A) a and b, or c and d are odd; B) a and c, or b and d are odd.

Let ψ^{∨} be the dual morphism. Denote: ψ̃ := ψ^{∨}, if m = 4t + 2, and

:= τ ∘ ψ^{∨} ∘ τ, if m = 4t. Then ψ̃_{(m/2)} = (\begin{matrix}d & b \\ c & a \end{matrix})

A) Put ε := ψ ∘ ψ̃ − κ_{m/2,Q} ∘ (ad · id + ab · τ)).

B) Put ε := ψ̃ ∘ ψ − κ_{m/2,Q} ∘ (ad · id + ac · τ)).

It is easy to see, that: ε_{(m/2)} = (\begin{matrix}0 & 0 \\ 2(cd − ab) & 0 \end{matrix}), in case A), and

= (\begin{matrix}0 & 2(bd − ac) \\ 0 & 0 \end{matrix}), in case B).

Clearly, ε_{(i)} = ψ_{(i)} · ψ^{∨}_{(i)}, for any i ≠ m/2. At the same time, ε^{2}_{(m/2)} = 0.

Since ε_{(i)} ∈ \{0, 1\}, ε^{2}|\overline{k} is a projector. By Corollary A.3 (2), ε^{2r} is a projector. Since ε^{2r}|\overline{k} = ε^{2}|\overline{k}, and ε^{2}|\overline{k} ∘ p_{N}|\overline{k} = p_{N}|\overline{k} ∘ ε^{2}|\overline{k} = ε^{2}|\overline{k}, by Lemma A.4 , we get a direct summand L̃ in N, s.t. pL̃|\overline{k} = ε^{2}|\overline{k}. Since ε^{2}_{(m/2)} = 0, and (p_{N})_{(m/2)} ≠ 0, we have: N ≠ L̃. Since N is indecomposable, we have: L̃ = 0. In particular, ε^{2}|\overline{k} = pL̃|\overline{k} = 0. This implies: ε_{(i)} = 0, for any i ≠ m/2.

Since (cd−ab) and (bd−ac) are odd in respective cases, we have, using ε, τ ∘ ε, ε ∘ τ, τ ∘ ε ∘ τ, and Lemma A.6 , that for any u ∈ 2 · Mat_{2×2}(Z), there exists φ ∈ Hom(M(Q), M(Q)), s.t. φ_{(i)} = 0, for any i ≠ m/2, and φ_{(m/2)} = u.

Since we also have κ_{i,Q} := h^{i} × h^{m−i} for i ≠ m/2, we get the statement.

Changing p_{N} by τ ∘ p_{N} ∘ τ, if nesessary (which does not change the isomorphism class of N), we can assume that in case A): a and b are odd, and in case B): b and d are odd. Since (p_{N})_{(m/2)} is an idempotent of rank = 1, we have: (p_{N})_{(m/2)} = γαγ^{−1}, where α, γ ∈ Mat_{2×2}(Z), and α is of type (2) in case (A), and of type (3) in case (B). Then it is easy to see that u := (γ − id) is in 2 · Mat_{2×2}(Z).

That means that p_{N}|\overline{k} = \overline{φ}∘π∘\overline{φ}^{−1}, where π_{(m/2)} = α is of type (2) in case (A), and of type (3) in case (B), and: (\overline{φ} − id) ∈ 2 · End(M(Q|\overline{k})).

Take ψ := p_{N}, then tr(ψ_{(m/2)}) = 1, and we can apply Sublemma A.8.1 . From Sublemma A.8.1 it follows that \overline{φ} is defined over k by some morphism φ. Then φ is an isomorphism, by Theorem A.1 (2), since it is over \overline{k}. ρ := φ^{−1} ∘ p_{N} ∘ φ is a projector (since p_{N} is), so, ρ = p_{N′} for some N′. Clearly, φ defines an isomorphism between N and N′, and (p_{N′})_{(m/2)} = α is of type (2), or (3).

Theorem A.9 (cf. [15], Lemma 3.21). Let N_{1} and N_{2} be nonisomorphic indecomposable direct summands

in M(Q). Let N′_{1} and N′_{2} be corresponding normal forms.

Then: p_{N′_{1}}|\overline{k} ∘ p_{N′_{2}}|\overline{k} = p_{N′_{2}}|\overline{k} ∘ p_{N′_{1}}|\overline{k} = 0. Proof Let φ := p_{N′_{1}} ∘ p_{N′_{2}}. Since N′_{a} is a normal form, we have: that φ_{(m/2)} is a projector in

Mat_{2×2}(Z), and φ_{(m/2)} · (p_{N′_{a}})_{(m/2)} = (p_{N′_{a}})_{(m/2)} · φ_{(m/2)} = φ_{(m/2)}. That means, φ|\overline{k} is a projector, and φ|\overline{k} ∘ p_{N′_{a}}|\overline{k} = p_{N′_{a}}|\overline{k} ∘ φ|\overline{k} = φ|\overline{k}. By Corollary A.3 , φ^{s} is a projector for some s, and if φ^{s} = p_{L}, for

some L, then by Lemma A.4 , L is isomorphic to a direct summand in N′_{a}. Since N_{a} is indecomposable, we have that either L is isomorphic to N′_{a}, or L = 0. Since N′_{1} is not isomorphic to N′_{2}, we have: L = 0. This implies: p_{N′_{1}}|\overline{k} ∘ p_{N′_{2}}|\overline{k} = φ|\overline{k} = φ^{s}|\overline{k} = 0

In the same way, considering ψ := p_{N′_{2}} ∘p_{N′_{1}}, we get: p_{N′_{2}}|\overline{k} ∘p_{N′_{1}}|\overline{k} = 0.

Lemma A.10 . Let L and N be direct summands in M(Q), s.t. p_{L}|\overline{k} = p_{N}|\overline{k}. Then

L is isomorphic to N.

Proof Consider α := φ_{N} ∘ j_{L}, and β := φ_{L} ∘ j_{N}. Then (β ∘ α)|\overline{k} = id_{N|\overline{k}} and

(α ∘ β)|\overline{k} = id_{L|\overline{k}}. By Corollary A.3 (3), L is isomorphic to N.

Lemma A.11 . Let L_{1}, L_{2} be direct summands in M(Q), s.t. p_{L_{1}}|\overline{k} ∘ p_{L_{2}}|\overline{k} = p_{L_{2}}|\overline{k} ∘

p_{L_{1}}|\overline{k} = 0. Then there exists direct summand M in M(Q), s.t. M is isomorphic to L_{1} ⊕ L_{2}, and p_{M}|\overline{k} = p_{L_{1}}|\overline{k} + p_{L_{2}}|\overline{k}.

Proof Consider π := p_{L_{1}} + p_{L_{2}}. Then π|\overline{k} is a projector, and by Corollary

A.3 , π^{r} is a projector for some r, i.e. there exists direct summand M of M(Q), s.t. π^{r} = p_{M}.

By Lemma A.4 , there exists direct summand L̃_{1} in M, s.t. L̃_{1} is isomorphic to L_{1}, and pL̃_{1}|\overline{k} = p_{L_{1}}|\overline{k}. Then, for complimentary projector pL̃_{2} := p_{M} − p_{L_{1}}, we have: pL̃_{2}|\overline{k} = p_{L_{2}}|\overline{k}. By Lemma A.10 , L̃_{2} is isomorphic to L_{2}, and so, M ≃ L_{1} ⊕ L_{2}.

Theorem A.12 .

Let N_{1}, …, N_{s} be pairwise nonisomorphic indecomposable direct summands in M(Q). Then ⊕_{i=1}^{s}N_{i} is isomorphic to a direct summand in M(Q).

Proof

Let N′_{1}, …, N′_{s} be normal forms of N_{1}, …, N_{s}. The statement follows from Theorem A.9 and the inductive application of Lemma A.11 .

We remind, that if N is a direct summand in M(Q), then N|\overline{k} = ⊕_{w∈Ω}ℤ(w)[2w], and CH(N|\overline{k}) := ⊕_{r} CH_{r}(N|\overline{k}) = ⊕_{w∈Ω}e_{w} · Z, where e_{w} is the generator of CH_{r}(N|\overline{k}), if rank(CH_{r}(N|\overline{k})) = 1, and φ_{N}(l_{m/2}^{1}), φ_{N}(l_{m/2}^{2}), if rank(CH_{r}(N|\overline{k})) = 2. We have the natural map \overline{deg} : CH(N|\overline{k}) → Z/2, defined by the rule: \overline{deg}(e_{w}) = 1.

Theorem A.13 .

Let N be an indecomposable direct summand in M(Q), and ψ ∈ End(N) be some morphism. Then either \overline{deg} ∘ ψ = \overline{deg}, or \overline{deg} ∘ ψ = 0.

Proof

Sublemma A.13.1 .

Let N be a direct summand in M(Q). Let ψ ∈ End(N). Then either there exists idempotent ε ∈ End(N), s.t. (ε − ψ)|\overline{k} ∈ \{2 · End(N|\overline{k}) + θ_{m/2,N} · Z}, or Q is (even-dimensional) hyperbolic.

Proof

Let ψ_{(r,Z/2)} ∈ End(CH_{r}(N|\overline{k})/2) be the map induced by ψ.

If rank(CH_{r}(N|\overline{k})) = 1, then ψ_{(r,Z/2)} is always a projector.

If rank(CH_{r}(N|\overline{k})) = 2, then r = m/2, and either \overline{deg}(ψ(h^{m/2})) = 1, or one of: ψ_{(m/2,Z/2)}, (ψ + θ_{m/2,N})_{(m/2,Z/2)} is a projector. In the former case, Q is hyperbolic, by Sublemma 1.3.1 , since ψ(h^{m/2}) ∈ CH_{m/2}(Q) has odd degree.

Taking into account, that the map idempotents(End(N|\overline{k})) → idempotents(End(N|\overline{k})⊗Z/2) is surjective, we have: if Q is not (even-dimensional) hyperbolic, then there exists idempotent \overline{ε} ∈ End(N|\overline{k}), s.t. (ψ|\overline{k} − \overline{ε}) ∈ \{2 · End(N|\overline{k}) + θ_{m/2,N} · Z\}.

25

Let there exists r (equal to m/2, certainly), s.t. rank(CH_{r}(N|\overline{k})) = 2. Changing ψ by ψ + θ_{m/2,N}, if nesessary, we can assume that ψ|\overline{k} ≡ \overline{ε} ( mod 2). Then either 1) tr(ψ_{(m/2)}) is odd, or 2) ψ_{(m/2,Z/2)} = id.

1) In this case, by Sublemma A.8.1 , there exists ε ∈ End(N), s.t. ε|\overline{k} = \overline{ε}.

2) In this case, tr(ψ_{(m/2)}) is even, and det(ψ_{(m/2)}) is odd. Take: ε′ := −ψ ∘ (ψ − tr(ψ_{(m/2)}) · id_{N}) − (det(ψ_{(m/2)}) − 1) · id_{N} ∈ End(N), then (ε′)_{(m/2)} = id, and (ψ − ε′)_{(r)} ∈ 2 · End(CH_{r}(N|\overline{k})), for r ≠ m/2.

If rank(CH_{r}(N|\overline{k}) ⩽ 1, for all r, take: ε′ := ψ. Put ε := ε′ − ∑_{r≠m/2}([ε′_{(r)}/2] · κ_{r,N}), then ε|\overline{k} = \overline{ε}.

Let ε ∈ End(N) be an idempotent from Sublemma A.13.1 . Since N is indecomposable, ε is either 0, or id_{N}, which implies: that

either \overline{deg} ∘ ε = \overline{deg}, or \overline{deg} ∘ ε = 0. By Sublemma A.13.1 , the same is true for ψ. The hyperbolic case is evident.

Theorem A.14 (cf. [15], Lemma 3.25). Let N_{1} be indecomposable direct summand in M(Q_{1}), and N_{2} be in-

decomposable direct summand in M(Q_{2}). Suppose α ∈ Hom(N_{1}, N_{2}), and β ∈ Hom(N_{2}, N_{1}) be such maps, that the composition: \overline{deg} ∘β ∘α : CH(N_{1}|\overline{k}) → Z/2 is nonzero. Then N_{1} ⋍ N_{2}.

Proof Sublemma A.14.1 . In the situation of Theorem A.14 , for γ := β ∘ α ∈ End(N_{1}), we

have: γ|\overline{k} ∈ \{id_{N_{1}|\overline{k}} + 2 · End(N_{1}|\overline{k}) + θ_{m_{1}/2,N_{1}}|\overline{k} · Z\}.

Proof Let ε ∈ End(N_{1}) be projector from Sublemma A.13.1 , s.t. (ε − γ)|\overline{k} ∈ \{2 · End(N_{1}|\overline{k}) + θ_{m_{1}/2,N_{1}} · Z\}.

Clearly, \overline{deg} ∘ ε = \overline{deg} ∘ γ ≠ 0. Hence, ε ≠ 0. Since N_{1} is indecomposable, we get: ε = id_{N_{1}}.

As an evident consequence of Sublemma A.14.1 , we get: Sublemma A.14.2 . In the situation of Theorem A.14 , (β∘α)|\overline{k} : CH(N_{1}|\overline{k})/2 → CH(N_{1}|\overline{k})/2

is an isomorphism.

Sublemma A.14.3 . In the situation of Theorem A.14 , \overline{deg}(β ∘ α(y)) = \overline{deg}(y), for any

y ∈ CH(N_{1}|\overline{k}), and \overline{deg}(α ∘ β(z)) = \overline{deg}(z), for any z ∈ CH(N_{2}|\overline{k}). In particular, the conditions of Theorem A.14 are symmetric with respect to N_{1} and N_{2}.

Proof

From Sublemma A.14.1 it follows that \overline{deg}(β ∘ α(y)) = \overline{deg}(y).

Since (β ∘ α)|\overline{k} : CH(N_{1}|\overline{k})/2 → CH(N_{1}|\overline{k})/2 is an isomorphism (by Sublemma A.14.2 ), we have: α∘β(CH(N_{2}|\overline{k})/2) = (α∘β)^{2}(CH(N_{2}|\overline{k})/2) is isomorphic to CH(N_{1}|\overline{k})/2. Let ε ∈ End(N_{2}) be projector from Sublemma A.13.1 , s.t. (ε − α ∘ β)|\overline{k} ∈ \{2 · End(N_{2}|\overline{k}) + θ_{m_{2}/2,N_{2}} · Z\}. Then, since N_{2} is indecomposable, ε is either 0, or id_{N_{2}}.

If ε = 0, then (α ∘ β)^{2}|\overline{k} ∈ 2 · End(N_{2}|\overline{k}), and CH(N_{1}|\overline{k})/2 = 0, which clearly contradicts to the assumptions of Theorem A.14 . So, ε = 1. Then \overline{deg}(α ∘ β(z)) = \overline{deg}(z), for any z ∈ CH(N_{2}|\overline{k}).

Sublemma A.14.4 .

In the situation of Theorem A.14 , for any r, rank(CH_{r}(N_{1}|\overline{k})) = rank(CH_{r}(N_{2}|\overline{k})).

Proof

It follows from Sublemma A.14.2 , and the fact that the conditions of Theorem A.14 are symmetric with respect to N_{1} and N_{2} (Sublemma A.14.3 ).

Sublemma A.14.5 .

In the situation of Theorem A.14 , \overline{deg}(α(y)) = \overline{deg}(y), for any y ∈ CH(N_{1}|\overline{k}), and \overline{deg}(β(z)) = \overline{deg}(z), for any z ∈ CH(N_{2}|\overline{k}).

Proof

By Sublemma A.14.3 , it is enough to show that if \overline{deg}(y) = 0, then \overline{deg}(α(y)) = 0, and if \overline{deg}(z) = 0, then \overline{deg}(β(z)) = 0.

Ker(\overline{deg}) ⊂ CH(N_{1}|\overline{k}) is generated by: \{2 · e_{w}\}_{w∈Ω_{1}} and l_{m_{1}/2}^{1} + l_{m_{1}/2}^{2} = h^{m_{1}/2}. So, if for some y ∈ CH(N_{1}|\overline{k}), we have: \overline{deg}(y) = 0, and \overline{deg}(α(y)) = 1, then rank(CH_{m_{1}/2}(N_{1}|\overline{k})) = 2 and \overline{deg}(α(h^{m_{1}/2})) = 1. By Sublemma A.14.4 , rank(CH_{m_{1}/2}(N_{2}|\overline{k})) = 2, which implies: m_{1} = m_{2}, and \overline{deg} : CH_{m_{2}/2}(N_{2}|\overline{k}) → Z/2 coinsides with the usual degree (mod2). Then on Q_{2} there is m_{2}/2-dimensional cycle of odd degree. By Sublemma 1.3.1 , Q_{2} is hyperbolic. Then any indecomposable direct summand in M(Q_{2}) is a Tate-motive, which contradicts to the fact that rank(CH_{m_{2}/2}(N_{2}|\overline{k})) = 2.

Sublemma A.14.6 .

In the situation of Theorem A.14 , for any u ∈ Hom(N_{1}, N_{2}), either \overline{deg}∘u = \overline{deg}, or \overline{deg}∘u = 0. The same holds for any v ∈ Hom(N_{2}, N_{1}).

Proof

If \overline{deg} ∘ u ≠ 0, then the pair (u, β) satisfy the conditions of Theorem A.14 , by Sublemma A.14.5 . The statement follows now again from Sublemma A.14.5 .

Sublemma A.14.7 .

Let rank(CH_{r}(N_{1}|\overline{k})) = 2. Suppose α, and β are as in Theorem A.14 . Then there exists α′, β′, s.t. β′_{(r)} = λ · β_{(r)}, where λ is odd, det(α′_{(r)}) = det(β′_{(r)}), and (β′ ∘ α′)_{(r)} = l · id_{CH_{r}(N_{1}|\overline{k})}, where l is odd.

Proof

Let rank(CH_{r}(N_{1}|\overline{k})) = 2. Then, by Sublemma A.14.1 , γ_{(r)} ∈

\{id + 2 · Mat_{2×2}(Z) + (\begin{matrix}1 & 1 \\ 1 & 1 \end{matrix}) · Z\}. This implies that det(γ_{(r)}) = μ is

odd, and tr(γ_{(r)}) is even.

In End(CH_{r}(N_{1}|\overline{k})) we have an equality: −γ_{(r)} ∘ (γ_{(r)} − tr(γ_{(r)})) = μ · id_{CH_{r}(N_{1}|\overline{k})}. Take α′′ := −α ∘ (β ∘ α − tr(γ_{(r)}) · id_{N_{1}}). It is easy to see that (β ∘ α′′)_{(r)} = μ · id_{CH_{r}(N_{1}|\overline{k})}.

Let α′ := l · α′′, and β′ := det(α′′_{(r)}) · β. Then det(α′_{(r)}) = det(β′_{(r)}), and (β′ ∘ α′)_{(r)} = (μ^{2} · det(α′′_{(r)})) · id_{CH_{r}(N_{1}|\overline{k})}.

Sublemma A.14.8 .

In the situation of Theorem A.14 , for any r, s.t. rank(CH_{r}(N_{1}|\overline{k})) = 1, there exists κ_{r,1→2} ∈ Hom(N_{1}, N_{2}), s.t. (κ_{r,1→2})_{(r)} = 2, and (κ_{r,1→2})_{(s)} = 0, for any s ≠ r.

If for some r, rank(CH_{r}(N_{i}|\overline{k})) = 2, then there exist θ_{r,1→2} and

κ_{r,1→2} ∈ Hom(N_{1}, N_{2}), s.t. (θ_{r,1→2})_{(r)} = (\begin{matrix}1 & 1 \\ 1 & 1 \end{matrix}), (κ_{r,1→2})_{(r)} = (\begin{matrix}2 & 0 \\ 0 & 2 \end{matrix}),

and (θ_{r,1→2})_{(s)} = 0 = (κ_{r,1→2})_{(s)}, for any s ≠ r.

Proof

Let e_{i} be the generator of CH_{r}(N_{i}|\overline{k}). We can assume that N_{i} is in the normal form (see Definition A.7 ).

Using h^{m_{i}−r} (and Sublemma A.8.1 (with ψ = p_{N_{i}}), if r = m_{i}/2), we have: 2 · CH_{r}(M(Q_{i})|\overline{k}) ⊂ image(CH_{r}(M(Q_{i})) → CH_{r}(M(Q_{i})|\overline{k})). Since φ_{N_{i}} ∘ j_{N_{i}} = id_{N_{i}}, we have: 2 · CH_{r}(N_{i}|\overline{k}) ⊂ image(CH_{r}(N_{i}) → CH_{r}(N_{i}|\overline{k})). In particular, 2 · e_{i} ∈ image(CH_{r}(N_{i}) → CH_{r}(N_{i}|\overline{k})). Let 2 · e_{i} = \overline{g_{i}}.

By Sublemma A.14.4 , \overline{deg}(α(e_{1})) = \overline{deg}(e_{1}) = 1, and \overline{deg}(β(e_{2})) = \overline{deg}(e_{2}) = 1.

That means: e_{1} ∈ image(CH_{r}(N_{1}) → CH_{r}(N_{1}|\overline{k})) if and only if e_{2} ∈ image(CH_{r}(N_{2}) → CH_{r}(N_{2}|\overline{k})).

28

We have two cases: A) e_{i} ∈ image(CH_{r}(N_{i}) → CH_{r}(N_{i}|\overline{k})), and B) e_{i} ∉ image(CH_{r}(N_{i}) → CH_{r}(N_{i}|\overline{k})).

Define u ∈ Hom(N_{1}, ℤ(r)[2r]) as a composition: N_{1} →^{j_{N_{1}}} M(Q_{1}) →^{h^{m_{1}−r}} ℤ(r)[2r]. Define v ∈ Hom(ℤ(r)[2r], N_{2}) as: f_{2}, where \overline{f_{2}} = e_{2}, in case A), and as g_{2} in case B).

In case A): either r > m_{i}/2, or r = m_{i}/2 and N_{i} is of type (3) (see Definition A.7 ). In any case, we have: j_{N_{1}}(e_{1}) = ±h^{m_{1}−r}.

In case B): either r < m_{i}/2, or r = m_{i}/2 and N_{i} is of type (2) (see Definition A.7 ). In any case, we have: j_{N_{1}}(e_{1}) = ±l_{r}.

Clearly, u(e_{1}) (as a map from ℤ(r)[2r] to ℤ(r)[2r]) is equal to id times the degree of intersection of j_{N_{1}}(e_{1}) and h^{m_{1}−r}. So, it is ±2 in case A), and ±1 in case B).

Define κ_{r,1→2} := ±v ∘ u. Then κ_{r,1→2}(e_{1}) = ±u(e_{1}) · \overline{v} and it is equal to 2 · \overline{f_{2}} in case A), and to 1 · \overline{g_{2}} in case B). So, in any case, κ_{r,1→2}(e_{1}) = 2 · e_{2}, and so, (κ_{r,1→2})_{(r)} = 2. Since u(CH_{s}(N_{1}|\overline{k})) = 0, for any s ≠ r (since Hom(ℤ(s)[2s], ℤ(r)[2r]) = 0, for s ≠ r), we have: (κ_{r,1→2})_{(s)} = 0, for any s ≠ r.

Let now, for some r, rank(CH_{r}(N_{i}|\overline{k})) = 2. Then m_{1} = m_{2}, and we can define: θ_{r,1→2} := φ_{N_{2}} ∘ (h^{m_{1}/2} × h^{m_{1}/2}) ∘ j_{N_{1}}. It is easy to see that θ_{r,1→2} has needed properties.

To define κ_{r,1→2}, observe that if α′ and β′ are as in Sublemma A.14.7

, then β′_{(r)} = λ · (\begin{matrix}a & b \\ c & d \end{matrix}), and α′_{(r)} = λ · (\begin{matrix}d & −b \\ −c & a \end{matrix}).

If the pair (α, β) satisfy the conditions of Theorem A.14 , then the pair (α, β + θ_{r,2→1}) satisfy the conditions of Theorem A.14 too (by Sublemma A.14.6 ).

Then there exists odd μ, and α′′ ∈ Hom(N_{1}, N_{2}), β′′ ∈ Hom(N_{2}, N_{1}), s.t. β′′_{(r)} = μ·(β + θ_{r,2→1})_{(r)}, and det(α′′_{(r)}) = det(β′′_{(r)}), and (β′′ ∘ α′′)_{(r)} = l′′ · id_{CH_{r}(N_{1}|\overline{k})}.

Take: ε := λ · α′′ − μ · α′ + λμ · θ_{r,1→2}, then: ε_{(r)} = λμ · (\begin{matrix}2 & 0 \\ 0 & 2 \end{matrix}). The

composition \overline{deg} ∘ ε : CH_{r}(N_{1}|\overline{k}) → Z/2 is zero. By Sublemma A.14.6 , for any s, \overline{deg} ∘ ε : CH_{s}(N_{1}|\overline{k}) → Z/2 is zero. I.e. ε_{(s)} ∈ Z is even. Define: ε′ := ε − ∑_{s≠r}(ε_{(s)}/2) · κ_{s,1→2}. Then ε′_{(r)} = ε_{(r)}, and ε_{(s)} = 0, for s ≠ r. By Lemma A.6 , there exists κ_{r,1→2} ∈ Hom(N_{1}, N_{2}), s.t. λμ · κ_{r,1→2}|\overline{k} = ε′. Clearly, κ_{r,1→2} has desired properties.

Sublemma A.14.9 .

Suppose rank(CH_{r}(N_{i})) = 2, and α, β satisfy the conditions of Theorem A.14 . Then there exists α_{2} ∈ Hom(N_{1}, N_{2}), s.t. (α − α_{2})_{(r)} ∈ \{2·Hom(CH_{r}(N_{1}|\overline{k}), CH_{r}(N_{2}|\overline{k}))+θ_{r,1→2} ·Z\}, and (α_{2})_{(r)} = η ·A, where η is odd and A : CH_{r}(N_{1}|\overline{k}) → CH_{r}(N_{2}|\overline{k}) is an isomorphism.

Proof Since rank(CH_{r}(N_{1}|\overline{k})) = rank(CH_{r}(N_{2}|\overline{k})) = 2, we clearly have:

r = m_{1}/2 = m_{2}/2.

For any morphism u ∈ Hom(N_{i}, N_{j}), we denote as u^{∨} the dual morphism φ_{N_{i}} ∘ (j_{N_{j}} ∘ u ∘ φ_{N_{i}})^{∨} ∘ j_{N_{j}} ∈ Hom(N_{j}, N_{i}).

Let us denote as ũ ∈ Hom(N_{j}, N_{i}) the following morphism: ũ := u^{∨}, if m is not divisible by 4, and := τ_{i} ∘ u^{∨} ∘ τ_{j}, if m is divisible by 4.

If u_{(r)} = (\begin{matrix}x & y \\ z & t \end{matrix}), then ũ_{(r)} = (\begin{matrix}t & y \\ z & x \end{matrix}).

Suppose α′ and β′ be as in Sublemma A.14.7 . Let f := g.c.d.(c − b, d − a), and f = u · (d − a) + v · (c − b). Then for ψ := (ν(u − vτ_{2})) ∘

α′ + (ν(u + vτ_{2})) ∘ β̃′ − λ(ua + vb)κ_{r,1→2}, we have: ψ_{(r)} = λν · (\begin{matrix}2f & 0 \\ 0 & 0 \end{matrix}).

Let α_{1} := λν · α − ψ ∘ ([(a − d)/2f] + [(b − c)/2f] · τ_{1}). Then α_{1} =

λν · (\begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}), where (a_{1} − d_{1}), (b_{1} − c_{1}) ∈ \{0, f\}.

I.e.: either 1) a_{1} = d_{1}, or 2) b_{1} = c_{1}, or 3) a_{1} − d_{1} = b_{1} − c_{1}. Considering α_{1} ∘ τ_{1} (and α_{2} ∘ τ_{1}), we can reduce case 2) to case 1).

So, it is enough to consider cases 1) and 3).

1) Take α_{2} := α_{1}−λν((a_{1}−1)·θ_{r,1→2}−(b_{1}−a_{1}+1)/2·κ_{r,1→2}∘τ_{1}). Then

α_{2}_{(r)} = λν ·(\begin{matrix}1 & 0 \\ ∗ & 1 \end{matrix}) (notice that (b_{1} −a_{1} +1) ≡ (b−a+1) ≡ 0 ( mod 2)).

3) take α_{2} := α_{1}−λν(κ_{r,1→2}∘([(a_{1}+d_{1})/4]+[(b_{1}+c_{1})/4]·τ_{1})−[{(a_{1}+

d_{1})/4\} + \{(b_{1} + c_{1})/4\}] · θ_{r,1→2}). Then (α_{2})_{(r)} = λν (\begin{matrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix}), where

a_{2} − d_{2} = b_{2} − c_{2}; (a_{2} + d_{2}), (b_{2} + c_{2}) ∈ {0, 2}, and a_{2} + b_{2} + c_{2} + d_{2} = 2. Hence, in any case, det((α_{2})_{(r)}/λν) = ±1.

Sublemma A.14.10 . In the situation of Theorem A.14 , there exist α_{3} ∈ Hom(N_{1}, N_{2}),

and some odd number η, that for all s, (α_{3})_{(s)} = η · A_{s}, where A_{s} is invertible.

Proof If rank(CH_{s}(N_{1}|\overline{k})) ⩽ 1, for all s, take: α′ := α, and η := 1. If

rank(CH_{r}(N_{1}|\overline{k})) = 2, for some r, take: α′ := α_{2}, and take η from the Sublemma A.14.9 (here α_{2}, A are also from Sublemma A.14.9 ).

By Sublemma A.14.9 , \overline{deg} ∘ α′ : CH_{r}(N_{1}|\overline{k}) → Z/2 is nonzero. By Sublemma A.14.6 , for all s, s.t. rank(CH_{s}(N_{1}|\overline{k})) = 1, (α′)_{(s)} = λ_{s} is odd.

Let r be such that rank(CH_{s}(N_{1}|\overline{k})) = 2. Define α_{3} := α′−∑_{s≠r}(λ_{s}− η)/2 · κ_{s,1→2}, where κ_{s,1→2} are elements from Sublemma A.14.8 . Then (α_{3})_{(s)} = η, for any s ≠ r, and (α_{3})_{(r)} = η · A′, where A′ is invertible.

From Sublemma A.14.10 and Lemma A.6 it follows that, in the situation of Theorem A.14 , there exists α_{4} ∈ Hom(N_{1}, N_{2}), s.t. α_{4}|\overline{k} is an isomorphism. Since conditions of Theorem A.14 are symmetric with respect to N_{1} and N_{2} (by Sublemma A.14.3 ), we also have some β_{4} ∈ Hom(N_{2}, N_{1}), s.t. β_{4}|\overline{k} is an isomorphism. Then (β_{4} ∘ α_{4})|\overline{k} and (α_{4} ∘ β_{4})|\overline{k} are isomorphisms, and then β_{4} ∘ α_{4} and α_{4} ∘ β_{4} are, by Corollary A.3 (3).

Theorem A.14 is proven.

Theorem A.15 .

Let N_{1} is indecomposable direct summand in M(Q_{1})(d_{1})[2d_{1}], and N_{2} is indecomposable direct summand in M(Q_{2})(d_{2})[2d_{2}], for some d_{1}, d_{2}. Suppose α ∈ Hom(N_{1}, N_{2}) and β ∈ Hom(N_{2}, N_{1}) are such that the map: \overline{deg} ∘ β ∘ α : CH(N_{1}|\overline{k}) → ℤ/2 is nonzero. Then N_{1} ≃ N_{2}.

Proof

Statement follows from Theorem A.14 , taking into account that M(Q_{i})(d_{i})[2d_{i}] is a direct summand in M(Q′_{i}), where q′_{i} := q_{i} ⊥ d_{i} · ℍ, by [13], Proposition 1 (Theorem A.0 ).

Theorem A.16 ([15], Lemma 3.26).

Suppose Q_{1}, Q_{2} be some smooth projective quadrics over k, and

α ∈ Hom(M(Q_{1})(d_{1})[2d_{1}], M(Q_{2})(d_{2})[2d_{2}]),
β ∈ Hom(M(Q_{2})(d_{2})[2d_{2}], M(Q_{1})(d_{1})[2d_{1}])

be such morphisms, that the composition

\overline{deg} ∘ β ∘ α : CH_{r}(M(Q_{1})(d_{1})[2d_{1}]|\overline{k}) → Z/2

is nonzero for some r. Then there exist indecomposable direct summands N_{1} of M(Q_{1})(d_{1})[2d_{1}], and N_{2} of M(Q_{2})(d_{2})[2d_{2}], such that N_{1} ≃ N_{2}, and ℤ(r)[2r] is a direct summand in N_{1}|\overline{k}.

Proof

Lemma A.16.1 .

Suppose N and L be indecomposable direct summands of M(Q) in normal form, and γ ∈ Hom(N, L) be such map, that the composition \overline{deg} ∘ j_{L} ∘ γ : CH(N|\overline{k}) → Z/2 is nonzero. Then N ≃ L.

Proof

Let \overline{deg}∘j_{L}∘γ : CH_{r}(N|\overline{k}) → Z/2 is nonzero for some r. In particular, CH_{r}(N|\overline{k}) ≠ 0 ≠ CH_{r}(L|\overline{k}). By Theorem A.12 , either N ≃ L, or r = dim(Q)/2 (since only for such r, rank(CH_{r}(M(Q)|\overline{k})) = 2), and rank(CH_{r}(N|\overline{k})) = rank(CH_{r}(L|\overline{k})) = 1.

If N is not isomorphic to L, then, in the notations of Definition A.7 , N and L are of types (2), or (3). If L is of type (3), then \overline{deg} ∘ j_{L} : CH_{r}(L|\overline{k}) → Z/2 is zero (since in such case CH_{r}(L|\overline{k}) is generated by the class of h^{r}). This is, clearly, not the case, so L is of type (2). Since L is not isomorphic to N, and N,L are indecomposable direct summands in normal form, we have by Theorem A.9 : N is of type (3). But then the generator h^{r} of CH_{r}(N|\overline{k}) is defined over k. That means: the map \overline{deg} ∘ j_{L} ∘ γ : CH_{r}(N) → Z/2 is nonzero. This implies that on Q there exists r-dimensional cycle of odd degree (namely, j_{L}∘γ(h^{r})). By Sublemma 1.3.1 , Q is hyperbolic. Then N and L must be Tate-motives. Since rank(CH_{r}(N|\overline{k})) = rank(CH_{r}(L|\overline{k})) = 1, we have: N ≃ ℤ(r)[2r] ≃ L. Contradiction. So, N ≃ L.

Lemma A.16.2 .

Let N is a direct summand of M(Q_{2})(d_{2})[2d_{2}], and the maps: α ∈ Hom(M(Q_{1})(d_{1})[2d_{1}], N), β ∈ Hom(N, M(Q_{1})(d_{1})[2d_{1}]) are such that the composition \overline{deg} ∘ β ∘ α : CH(M(Q_{1})(d_{1})[2d_{1}]|\overline{k}) → Z/2 is nonzero. Then there exists direct summand L of M(Q_{1})(d_{1})[2d_{1}] isomorphic to N.

Proof

Let M(Q_{1})(d_{1})[2d_{1}] = ⊕_{a∈Λ_{1}}L^{a}, where L^{a} are indecomposable direct summands in normal form (by Theorem A.12 , we can always find such decompositions). Let α_{a} := α ∘ j_{L^{a}} ∈ Hom(L^{a}, N), and β_{c} := φ_{L^{c}} ∘ β. We have: β ∘α = ∑_{a,c∈Λ_{1}}(p_{L^{c}} ∘β ∘α∘p_{L^{a}}) = ∑_{a,c∈Λ_{1}}(j_{L^{c}} ∘β_{c} ∘α_{a} ∘φ_{L^{a}}).

Since the composition \overline{deg} ∘ β ∘ α : CH(M(Q_{1})(d_{1})[2d_{1}]|\overline{k}) → Z/2 is nonzero, we have: for some a, c, the composition \overline{deg} ∘ j_{L^{c}} ∘ β_{c} ∘ α_{a} : CH(L^{a}|\overline{k}) → Z/2 is nonzero.

By Lemma A.16.1 , there exists an isomorphism ψ : L^{c} → L^{a}. Take: u := α_{a} ∘ ψ ∈ Hom(L^{c}, N), and v := β_{c} ∈ Hom(N, L^{c}). We have: \overline{deg} ∘ j_{L^{c}} ∘ v ∘ u : CH_{r}(L^{c}|\overline{k}) → Z/2 is nonzero. Since the composi-

tion CH_{r}(L^{c}|\overline{k}) →^{j_{L^{c}}} CH_{r}(M(Q_{1})(d_{1})[2d_{1}]|\overline{k}) →^{\overline{deg}} Z/2 either coinsides with \overline{deg} : CH_{r}(L^{c}|\overline{k}) → Z/2, or is zero, we get: \overline{deg}∘v∘u : CH_{r}(L^{c}|\overline{k}) → Z/2 is nonzero. By Theorem A.15 , L^{c} ≃ N.

Let M(Q_{2})(d_{2})[2d_{2}] = ⊕_{b∈Λ_{2}}N_{2}^{b}, where N^{b} are indecomposable direct summands. Let α^{b} := φ_{N^{b}} ∘ α, and β^{b} := β ∘ j_{N^{b}}. We have: β ∘ α = ∑_{b∈Λ_{2}}(β ∘ p_{N^{b}} ∘ α) = ∑_{b∈Λ_{2}}(β_{b} ∘ α_{b}).

So, if \overline{deg} ∘ β ∘ α : CH_{r}(M(Q_{1})(d_{1})[2d_{1}]|\overline{k}) → Z/2 is nonzero, then for some b ∈ Λ_{2}, the map: \overline{deg} ∘ β_{b} ∘ α_{b} : CH_{r}(M(Q_{1})(d_{1})[2d_{1}]|\overline{k}) → Z/2 is nonzero. By Lemma A.16.2 , there exists some indecomposable direct summand L of M(Q_{1})(d_{1})[2d_{1}] isomorphic to N^{b}. Clearly, rank(CH_{r}(N^{b}|\overline{k})) ≠ 0, so, ℤ(r)[2r] is a direct summand in N^{b}|\overline{k}.

Let N be a direct summand in M(Q). We define: a(N) := min(s| CH_{s}(N|\overline{k}) ≠ 0), b(N) := max(s| CH_{s}(N|\overline{k}) ≠ 0), and c(N) := dim(Q) − b(N).

Let k = F_{0} ⊂ F_{1} ⊂ ⋯ ⊂ F_{h} be generalized splitting tower of M.Knebusch for Q.

We have the following application of Theorem A.16 :

Theorem A.17 ([16], Corollary 2).

Suppose Q is smooth projective quadric of dimension m, and i_{W} (q|_{F_{t}}) ⩽ i, j < i_{W} (q|_{F_{t+1}}). Let N be indecomposable direct summand of M(Q), such that a(N) = i. Then in M(Q) there exists direct summand isomorphic to N(j − i)[2j − 2i].

Proof

Sublemma A.17.1 .

Let N be a direct summand in M(Q), and π : M(Q) → ℤ be the map induced by the natural projection Q → Spec(k). Then the map

N ⊗ M(Q) →^{id⊗π} N has a section ν : N → N ⊗ M(Q).

Proof

It is easy to see that the composition: N →^{j_{N}} M(Q) →^{Δ} M(Q) ⊗ M(Q) →^{φ_{N} ⊗id} N ⊗ M(Q) gives desired section.

Sublemma A.17.2 .

Let i < j ⩽ m/2. Let N be a direct summand of M(Q), s.t. a(N) = i. Suppose over k(Q) there exists map α̃ : N(j − i)[2j − 2i]|_{k(Q)} → M(Q)|_{k(Q)}, s.t. the composition \overline{deg} ∘ α̃ : CH_{j}(N(j − i)[2j − 2i]|\overline{k(Q)}) → Z/2 is nonzero. Then there exists map α : N(j − i)[2j − 2i] → M(Q), s.t. the map \overline{deg} ∘ α : CH_{j}(N(j − i)[2j − 2i]|\overline{k}) → Z/2 is nonzero.

Proof

We have natural group homomorphism θ : Hom_{Chow(k)}(N(j − i)[2j − 2i]⊗M(Q), M(Q)) → Hom_{Chow(k(Q))}(N(j −i)[2j −2i]|_{k(Q)}, M(Q)|_{k(Q)}), which corresponds to the natural map ϑ_{2} : CH^{m+i−j}(Q × Q × Q) → CH^{m+i−j}(Q × Q|_{k(Q)}) (notice, that Hom_{Chow(k)}(N(j − i)[2j − 2i] ⊗ M(Q), M(Q)) is naturally a direct summand in CH^{m+i−j}(Q × Q × Q), and Hom_{Chow(k(Q))}(N(j − i)[2j − 2i]|_{k(Q)}, M(Q)|_{k(Q)}) is naturally a direct summand in CH^{m+i−j}(Q × Q|_{k(Q)})). θ is surjective, since ϑ_{2} is. So, there exists γ ∈ Hom_{Chow(k)}(N(j − i)[2j − 2i] ⊗ M(Q), M(Q)), s.t. θ(γ) = α̃. Define α := γ ∘ ν(j − i)[2j − 2i].

Since M(Q)|\overline{k}, and N|\overline{k} are direct sums of Tate-motives, we have: CH_{i}(N ⊗ M(Q)|\overline{k}) = ⊕_{s+t=i} CH_{s}(N|\overline{k}) ⊗ CH_{t}(M(Q)|\overline{k}) = CH_{i}(N|\overline{k}) ⊗ CH_{0}(M(Q)|\overline{k}). In particular, id⊗π : CH_{i}(N ⊗M(Q)|\overline{k}) → CH_{i}(N|\overline{k}) is an isomorphism. Hence, ν(j − i)[2j − 2i] : CH_{j}(N(j − i)[2j − 2i]|\overline{k}) → CH_{j}(N(j − i)[2j − 2i] ⊗ M(Q)|\overline{k}) is an isomorphism too. So, taking l_{i} as a generator of CH_{i}(N|\overline{k}) = CH_{j}(N(j − i)[2j − 2i]|\overline{k}) (notice, that 0 ⩽ i < m/2), we have: ν(j−i)[2j−2i](l_{i}) = l_{i}⊗l_{0}. By the definition of θ, \overline{deg}(γ(l_{i} ⊗ l_{0})) = \overline{deg}(θ(γ)(l_{i})) = \overline{deg}(α̃(l_{i})) ≠ 0. So, the composition \overline{deg} ∘ α : CH_{j}(N(j − i)[2j − 2i]|\overline{k}) → Z/2 is nonzero.

Sublemma A.17.3 .

Let i_{W} (q|_{F_{t}}) ⩽ i < j < i_{W} (q|_{F_{t+1}}). Let N be a direct summand of M(Q), s.t. a(N) = i. Then there exists map α : N(j − i)[2j − 2i] → M(Q), s.t. the map \overline{deg}∘α : CH_{j}(N(j−i)[2j−2i]|\overline{k}) → Z/2 is nonzero.

Proof

We can use induction on dim(q). If dim(q) ⩽ 1, then there is nothing to prove.

Suppose the statement is proven for all forms of dimension < dim(q). Changing k by k(Q), and using Sublemma A.17.2 , we can assume that i_{W} (q) > 0.

Then, by Theorem A.0 , M(Q) = ⊕_{0⩽s<i_{W} (q)}(ℤ(s)[2s] ⊕ ℤ(m − s)[2m − 2s]) ⊕ M(Q_{an})(i_{W} (q))[2i_{W} (q)], and End(M(Q)) = ×_{0⩽s<i_{W} (q)} (End(ℤ(s)[2s])×End(ℤ(m−s)[2m−2s]))×End(M(Q_{an})(i_{W} (q))[2i_{W} (q)]). This implies, that if N is a direct summand of M(Q), then N = ⊕_{0⩽s<i_{W} (q)}(N_{s} ⊕ N_{m−s}) ⊕ N_{an}, where N_{an} is a direct summand of M(Q_{an})(i_{W} (q))[2i_{W} (q)], and N_{t} is a direct summand of ℤ(t)[2t] (i.e. either ℤ(t)[2t], or 0).

We have two cases: 1) 0 ⩽ i < i_{W} (q); 2) i_{W} (q) ⩽ i < m/2.

1) In this case, N_{i} = ℤ(i)[2i], since CH_{i}(N|\overline{k}) ≠ 0. Since 0 ⩽ i < i_{W} (q), we have: j < i_{W} (q), and cycle h^{i} × l_{j} ∈ CH^{m+i−j}(Q × Q|\overline{k}) is defined over k, so it gives us the map α ∈ Hom(M(Q)(j − i)[2j − 2i], M(Q)) = CH^{m+i−j}(Q × Q). We can take l_{i} ∈ CH_{i}(Q|\overline{k}) = CH_{j}(M(Q)(j − i)[2j − 2i]|\overline{k}) as a generator of CH_{j}(N(j − i)[2j − 2i]|\overline{k}). Clearly, α(l_{i}) = l_{j}. So, the composition \overline{deg} ∘ α : CH_{j}(N(j − i)[2j − 2i]|\overline{k}) → Z/2 is nonzero.

2) Since CH_{j}(ℤ(s)[2s]) = 0, for j ≠ s, we have: CH_{j}(N(j − i)[2j − 2i]|\overline{k}) = CH_{j}(N_{an}(j−i)[2j−2i]|\overline{k}). Taking i′ := i−i_{W} (q), j′ := j−i_{W} (q), q′ = q_{an}, and N′ := N_{an}(−i_{W} (q))[−2i_{W} (q)]. Clearly, N′, q′, i, j satisfy the conditions of Theorem A.17 . Since i_{W} (q) > 0, we have: dim(q′) < dim(q). By inductive assumption, there exists α′ ∈ Hom(N′(j′ − i′)[2j′−2i′], M(Q′)), s.t. the composition \overline{deg}∘α′ : CH_{j′}(N′(j′−i′)[2j′− 2i′]|\overline{k}) → Z/2 is nonzero. Let ψ : M(Q′)(i_{W} (q))[2i_{W} (q)] → M(Q), and ρ : N → N′(i_{W} (q))[2i_{W} (q)] be natural “embedding” and “projection”.

Take: α := ψ ∘ α′(i_{W} (q))[2i_{W} (q)] ∘ ρ(j − i)[2j − 2i] ∈ Hom(N(j − i)[2j − 2i], M(Q)).

Since ψ and ρ act isomorphically on CH_{j}, and \overline{deg} ∘ ψ = \overline{deg} : CH_{j}(M(Q′)(i_{W} (q))[2i_{W} (q)]|\overline{k}) → Z/2, we have: the composition \overline{deg} ∘ α : CH_{j}(N(j − i)[2j − 2i]) → Z/2 is nonzero.

Sublemma A.17.4 (cf. [15], Lemma 4.5).

Let i_{W} (q|_{F_{t}}) ⩽ i < j < i_{W} (q|_{F_{t+1}}). Let N be indecomposable direct summand of M(Q), s.t. a(N) = i. Then there exists direct summand of M(Q), isomorphic to N(j − i)[2j − 2i].

Proof

Let Δ_{h^{j−i}} ∈ CH^{m+j−i}(Q × Q) be the cycle given by the diagonal embedding of the (j −i)-codimensional plane section. Denote corresponding element of Hom(M(Q), M(Q)(j − i)[2j − 2i]) = CH^{m+j−i}(Q × Q) as ε_{j−i}. Define: β := φ_{N}(j − i)[2j − 2i] ∘ ε_{j−i} ∈ Hom(M(Q), N(j − i)[2j − 2i]).

Clearly, ε_{j−i}(l_{j}) = l_{i} ∈ CH_{i}(M(Q)|\overline{k}) = CH_{j}(M(Q)(j − i)[2j − 2i]|\overline{k}), and it is a generator of the later group.

Since i < m/2, and CH_{i}(N|\overline{k}) = ℤ, the map φ_{N}(j − i)[2j − 2i] : CH_{j}(M(Q)(j − i)[2j − 2i]|\overline{k}) → CH_{j}(N(j − i)[2j − 2i]|\overline{k}) is an isomorphism. So, the map β : CH_{j}(M(Q)|\overline{k}) → CH_{j}(N(j − i)[2j − 2i]|\overline{k}) is surjective.

By Sublemma A.17.3 , we also have a map: α ∈ Hom(N(j − i)[2j − 2i], M(Q)), s.t. the composition \overline{deg}∘α : CH_{j}(N(j−i)[2j−2i]|\overline{k}) → Z/2 is nonzero. By above, the composition \overline{deg}∘α∘β : CH_{j}(M(Q)|\overline{k}) → Z/2 is nonzero. By Lemma A.16.2 , M(Q) has a direct summand isomorphic to N(j − i)[2j − 2i].

Sublemma A.17.5 (cf. [16], proof of Statement).

Let Q be anisotropic quadric of dimension m, s.t. i_{1}(q) = 1. Let N be direct summand of M(Q), s.t. CH_{m}(N|\overline{k}) ≠ 0. Then CH_{0}(N|\overline{k}) ≠ 0.

Proof

Suppose it is not the case. So, CH_{m}(N|\overline{k}) ≠ 0 and CH_{0}(N|\overline{k}) = 0.

By Sublemma A.17.1 , the natural projection N ⊗ M(Q) → N has a section ν. So, π ∘ ν = id_{N}.

Let β_{0} ∈ Hom(ℤ(m)[2m], M(Q)) = CH_{m}(Q) be the morphism corresponding to the “generic cycle” on Q. Let u := (φ_{N} ⊗ id) ∘ (β_{0} ⊗ id) ∈ Hom(M(Q)(m)[2m], N ⊗ M(Q)), v := Δ^{∨} ∘ (j_{N} ⊗ id) ∈ Hom(N ⊗ M(Q), M(Q)(m)[2m]), where Δ^{∨} ∈ Hom(M(Q × Q), M(Q)) is dual to the “diagonal embedding” Δ via duality: CH^{∗}(A × B) ≃ CH^{∗}(B × A).

Since j_{N} ∘ φ_{N} ∘ β_{0} = β_{0} (since CH_{m}(N|\overline{k}) = ℤ, and hence j_{N} ∘ φ_{N} : CH_{m}(Q|\overline{k}) → CH_{m}(Q|\overline{k}) is identity map), we have: v ∘ u = id ∈ End(M(Q)(m)[2m]). So, N ⊗ M(Q) = M(Q)(m)[2m] ⊕ X. Let φ_{X} ∈ Hom(N ⊗ M(Q), X), and j_{X} ∈ Hom(X, N ⊗ M(Q)) be the corresponding projection and embedding. I.e.: φ_{X} ∘ j_{X} = id_{X}, j_{X} ∘ φ_{X} + u ∘ v = id_{N⊗M(Q)}, and φ_{X} ∘ u = 0, v ∘ j_{X} = 0.

In particular, π ∘ ν = (π ∘ j_{X}) ∘ (φ_{X} ∘ ν) + (π ∘ u) ∘ (v ∘ ν). So, there exist maps α_{1} : N → M(Q)(m)[2m], β_{1} : M(Q)(m)[2m] → N, and α_{2} : N → X, β_{2} : X → N, s.t. β_{1} ∘ α_{1} + β_{2} ∘ α_{2} = id_{N}.

If \overline{deg} ∘ β_{1} ∘ α_{1} : CH_{m}(N|\overline{k}) → Z/2 is nonzero, then, by Theorem A.16 , we have a direct summand N′ of N, which is isomorphic to a direct summand of M(Q)(m)[2m], and CH_{m}(N′) ≠ 0. Since CH_{m}(Q) → CH_{m}(Q|\overline{k}) is an isomorphism, and N′ is a direct summand of M(Q), we have: CH_{m}(N′) → CH_{m}(N′|\overline{k}) is an isomorphism. Since N′ is also a direct summand in M(Q)(m)[2m], and CH_{m}(N′) = Z, we have: CH_{m}(N′) = CH_{m}(M(Q)(m)[2m]), and CH_{m}(M(Q)(m)[2m]) → CH_{m}(M(Q)(m)[2m]|\overline{k}) is an isomorphism. This implies that on Q there is a 0-cycle of degree 1. By Springer’s Theorem, Q is isotropic - a contradiction. So, \overline{deg} ∘ β_{1} ∘ α_{1} : CH_{m}(N|\overline{k}) → Z/2 is zero.

Consider E = k(Q). We have: q|_{E} = ℍ ⊕ q′, where q′ is anisotropic (since i_{1}(q) = 1). By [13], Proposition 1 (Theorem A.0 ), M(Q|_{E}) = ℤ ⊕ M(Q′)(1)[2] ⊕ ℤ(m)[2m]. And then N|_{E} = N_{an} ⊕ ℤ(m)[2m], where N_{an} is a direct summand of M(Q′)(1)[2] (since CH_{m}(N|\overline{k}) = Z, and CH_{0}(N|\overline{k}) = 0). Moreover, the corresponding map j_{ℤ(m)[2m]→N|_{E}} : ℤ(m)[2m] → N|_{E} coinsides with (φ_{N}∘β_{0})|_{E}. Then the map j_{ℤ(m)[2m]→N|_{E}}⊗ id_{M(Q)} : M(Q)(m)[2m]|_{E} → N ⊗ M(Q)|_{E} coinsides with u|_{E}. This implies that the complimentary direct summand N_{an} ⊗ M(Q) (in (N ⊗ M(Q))|_{E}) is isomorphic to X|_{E}. Notice that N_{an} ⊗ M(Q) is a direct summand in M(Q′)(1)[2] ⊗ M(Q). So, α_{2}|_{E} and β_{2}|_{E} give us maps α′_{2} : N|_{E} → M(Q′ × Q)(1)[2], and β′_{2} : M(Q′ × Q)(1)[2] → N|_{E}, s.t. β′_{2} ∘ α′_{2} = (β_{2} ∘ α_{2})|_{E}.

If α′_{2}∘j_{ℤ(m)[2m]→N|_{E}} ∈ Hom(ℤ(m)[2m], M(Q′×Q)(1)[2]) = CH_{m−1}(Q′× Q) is represented by the cycle A, and φ_{N|_{E}→ℤ(m)[2m]}∘β′_{2} ∈ Hom(M(Q′× Q)(1)[2], ℤ(m)[2m]) = CH^{m−1}(Q′ × Q) is represented by the cycle B, then the composition (φ_{N|_{E}→ℤ(m)[2m]} ∘ β′_{2}) ∘ (α′_{2} ∘ j_{ℤ(m)[2m]→N|_{E}}) ∈ End(ℤ(m)[2m]) = Z is given by the degree of the 0-cycle A ∩ B ∈ CH_{0}(Q′×Q). Since Q′ is anisotropic, this number is even, by Springer’s Theorem. Since j_{ℤ(m)[2m]→N|_{E}} and φ_{N|_{E}→ℤ(m)[2m]} are isomorphisms on CH_{m}, we have: the composition \overline{deg}∘β′_{2}∘α′_{2} = \overline{deg}∘β_{2}∘α_{2} : CH_{m}(N|\overline{k}) → Z/2 is zero. Since also \overline{deg} ∘ β_{1} ∘ α_{1} : CH_{m}(N|\overline{k}) → Z/2 is zero, we get a contradiction with: β_{1} ∘ α_{1} + β_{2} ∘ α_{2} = id_{N}.

Sublemma A.17.6 .

Let Q be anisotropic quadric, and L is indecomposable direct summand in M(Q), s.t. a(L) = 0. Then for any subquadric P ⊂ Q, s.t dim(P) = dim(Q) − i_{1}(q) + 1, we have:

1) M(P) contains a direct summand isomorphic to L;

2) p|_{k(Q)} and q|_{k(P)} are isotropic.

Proof

For any field extension E/k, we have: p|_{E} is isotropic if and only if q|_{E} is. In particular we get 2). So, we have rational (algebro-geometric) maps f : Q → P, and g : P → Q. Let α ∈ CH^{dim(P)}(Q × P) = Hom(M(Q), M(P)), and β ∈ CH^{dim(Q)}(P × Q) = Hom(M(P), M(Q)) be closures of the graphs of f and g, respectively. Clearly, α(l_{0}) = l_{0}, and β(l_{0}) = l_{0}. So, the composition \overline{deg} ∘ β ∘ α : CH_{0}(Q|\overline{k}) → Z/2 is nonzero. By Theorem A.16 , M(P) contains a direct summand isomorphic to L (notice that if M is indecomposable direct summand of M(Q), s.t. CH_{0}(M) ≠ 0, then M ≃ L, by Theorem A.9 ).

Sublemma A.17.7 .

Let Q be anisotropic quadric, and L is indecomposable direct summand in M(Q), s.t. a(L) = 0. Then there exists subquadric P ⊂ Q, s.t.:

1) i_{1}(p) = 1;

2) M(P) contains a direct summand isomorphic to L;

3) p|_{k(Q)} and q|_{k(P)} are isotropic.

Proof

Use induction on the dimension of Q. The case of dim(Q) = 0 is trivial. Suppose the statement is true for all quadrics of dimension < dim(Q). Consider P from Sublemma A.17.6 . Either i_{1}(q) = 1, in which case statement is trivial, or dim(P) < dim(Q). Then there exists P′, s.t. P′ satisfy 1) and 2), and p′|_{k(P)}, p|_{k(P′)} are isotropic. Since we also have: p|_{k(Q)}, q|_{k(P)} are isotropic, we get: P′ satisfy 3) (since if q_{2}|_{k(q_{1})}, q_{3}|_{k(q_{2})} are isotropic, then q_{3}|_{k(q_{1})} is).

Sublemma A.17.8 .

Let Q be anisotropic quadric of dimension m, and L is indecomposable direct summand in M(Q), s.t. a(L) = 0. Then c(L) = i_{1}(q) − 1.

Proof

Let P be quadric from Sublemma A.17.7 . Then L is also a direct summand in M(P). By Sublemma A.17.5 , b(L) = dim(P). In particular, by [13], Proposition 1 (Theorem A.0 ), L|_{k(P)} contains ℤ and ℤ(dim(P))[2 dim(P)] as direct summands.

Since Q is anisotropic, P is also anisotropic. If dim(P) = 0, then rank(CH_{0}(L|\overline{k})) = 2, and hence, m = 0 (since L is a direct summand in M(Q)). In this case everything is evident. If b(L) = dim(P) > 0, then the map CH_{b(L)}(P) → CH_{b(L)}(P |\overline{k}) is surjective, and CH_{b(L)}(L|\overline{k}) = CH_{b(L)}(P |\overline{k}). So, the map CH_{b(L)}(L) → CH_{b(L)}(L|\overline{k}) is surjective. And the map \overline{deg} : CH_{b(L)}(L) → Z/2 is nonzero. If b(L) < m/2, then \overline{deg} = \overline{deg} ∘ j_{L} : CH_{b(L)}(L) → Z/2, and we get b(L)-dimensional cycle of odd degree on Q. By Sublemma 1.3.1 , Q is isotropic - contradiction. So, b(L) ⩾ m/2. Since L|_{k(P)} contains ℤ(b(L))[2b(L)] as a direct summand, L|_{k(Q)} also contains ℤ(b(L))[2b(L)] as a direct summand, by Sublemma A.17.7 (3). Then b(L) > m − i_{1}(q), by Lemma 1.3 (since b(L) ⩾ m/2).

By Sublemma A.17.4 , M(Q) contains direct summand isomorphic to L(i_{1}(q) − 1)[2i_{1}(q) − 2]. This implies: b(L) ⩽ dim(Q) − i_{1}(q) + 1. So, b(L) = dim(Q) − i_{1}(q) + 1, and c(L) = i_{1}(q) − 1.

Sublemma A.17.9 .

Let Q be anisotropic quadric of dimension m, and N is indecomposable direct summand in M(Q), s.t. a(N) < i_{1}(q). Let L be indecomposable direct summand in M(Q), s.t a(L) = 0. Then N ≃ L(a(N))[2a(N)].

Proof

We can assume that a(N) > 0. By Sublemma A.17.4 , L(a(N))[2a(N)] is isomorphic to a direct summand of M(Q) (since a(N) < i_{1}(q)). We have: CH_{a(N)}(N|\overline{k}) ≠ 0 ≠ CH_{a(N)}(L|\overline{k}). Consider L, L(a(N))[2a(N)], and N. Since a(N) > 0, we have: L is not isomorphic to L(a(N))[2a(N)]. By the same reason, L is not isomorphic to N (since a(L) ≠ a(N)). If L(a(N))[2a(N)] is not isomorphic to N, then N ⊕L⊕L(a(N))[2a(N)] is a direct summand of M(Q), by Theorem A.12 . Since rank(CH_{a(N)}(N⊕ L(a(N))[2a(N)]|\overline{k})) ⩾ 2, we get: a(N) = m/2, which implies: i_{1}(q) = m/2 + 1, and b(L) = c(L) = m/2 = a(N). In this case, CH_{a(N)}(L|\overline{k}) ≠ 0, and rank(CH_{a(N)}(N ⊕ L ⊕ L(a(N))[2a(N)]|\overline{k})) ⩾ 3 - contradiction.

Sublemma A.17.10 .

Let Q be anisotropic quadric of dimension m, and N is indecomposable direct summand in M(Q), s.t. either a(N), or c(N) is < i_{1}(q). Then a(N) + c(N) = i_{1}(q) − 1.

Proof

Changing N by N^{∨}, if nesessary, we can assume that a(N) ⩽ c(N), in particular, 0 ⩽ a(N) < i_{1}(q). By Sublemma A.17.9 , N ≃ L(a(N))[2a(N)]. In this case, a(N) + c(N) = a(L) + c(L) = c(L) = i_{1}(q) − 1.

Sublemma A.17.11 .

Let Q be anisotropic quadric of dimension m. Let N be indecomposable direct summand in M(Q). Then there exists 0 ⩽ t < h = height(q), s.t. for fields F_{t} ⊂ F_{t+1} from the generalized splitting tower of M.Knebusch for Q, we have: i_{W} (q|_{F_{t+1}}) > a(N), c(N) ⩾ i_{W} (q|_{F_{t}}), and a(N) + c(N) ⩽ i_{W} (q|_{F_{t}}) + i_{W} (q|_{F_{t+1}}) − 1.

Proof

Let k = F_{0} ⊂ F_{1} ⊂ ⋯ ⊂ F_{h} be the generalized splitting tower for Q. Let 1 ⩽ t < h be the minimal number such that N|_{F_{t+1}} contains either ℤ(a(N))[2a(N)], or ℤ(b(N))[2b(N)] as a direct summand (since q|_{F_{h}} is hyperbolic, and CH_{a(N)}(N|\overline{k}) ≠ 0, such t exists). Changing N by N^{∨}, if nesessary, we can assume that N|_{F_{t+1}} contains ℤ(a(N))[2a(N)] as a direct summand (notice, that a(N) + c(N) will not change under this procedure). Since N|_{F_{t}} does not contain Tate-motives, we have: a(N), c(N) ⩾ i_{W} (q|_{F_{t}}). This shows that N|_{F_{t}} is a direct summand in M((Q|_{F_{t}})_{an})(i_{W} (q|_{F_{t}}))[2i_{W} (q|_{F_{t}})]. Let N′ be indecomposable direct summand of N|_{F_{t}}, s.t. CH_{a(N)}(N′|\overline{F_{t}}) ≠ 0. Then a(N′) + c(N′) = 2i_{W} (q|_{F_{t}})+i_{1}((q|_{F_{t}})_{an})−1, by Sublemma A.17.10 . Since a(N′) = a(N), and c(N′) ⩽ c(N), we get: a(N) + c(N) ⩽ i_{W} (q|_{F_{t}}) + i_{W} (q|_{F_{t+1}}) − 1.

We have the following easy corollary of Sublemma A.17.11 .

Corollary A.17.12 ([16], Statement).

Let Q be anisotropic quadric of dimension m. Let N be indecomposable direct summand in M(Q).

Then there exists 0 ⩽ t < h, s.t. i_{W} (q|_{F_{t+1}}) > a(N), c(N) ⩾ i_{W} (q|_{F_{t}}), and a(N) + c(N) = i_{W} (q|_{F_{t}}) + i_{W} (q|_{F_{t+1}}) − 1.

Proof

Let l := i_{W} (q|_{F_{t+1}}) − 1 − a(N). By Sublemma A.17.4 , N(l)[2l] is isomorphic to a direct summand M of M(Q). Clearly, a(M) = a(N) + l = i_{W} (q|_{F_{t+1}}) − 1, and c(M) = c(N) − l. Since a(M) ⩾ i_{W} (q|_{F_{t}}), we have, by Sublemma A.17.11 , c(M) ⩾ i_{W} (q|_{F_{t}}). This implies: a(N) + c(N) ⩾ i_{W} (q|_{F_{t}}) + i_{W} (q|_{F_{t+1}}) − 1. Combined with Sublemma A.17.11 , this gives required result.

Now we can prove Theorem A.17 . Let Q, i, j, N be as in Theorem A.17 . Clearly, i = a(N). If j > i, then everything is contained in Sublemma A.17.4 .

Let (i − j) > 0. Let F = F_{r} be a field in the generalized splitting tower of M.Knebusch, given by Sublemma A.17.11 . Then i_{W} (q|_{F_{r+1}}) > a(N) ⩾ i_{W} (q|_{F_{r}}). By conditions of Theorem A.17 , i_{W} (q|_{F_{r+1}}) > a(N)− (i − j) ⩾ i_{W} (q|_{F_{r}}). Also, i_{W} (q|_{F_{r+1}}) > c(N) ⩾ i_{W} (q|_{F_{r}}). On the other hand, from Sublemma A.17.11 it follows that i_{W} (q|_{F_{r+1}}) > c(N) + (i − j) ⩾ i_{W} (q|_{F_{r}}). That means that the pair (i′, j′) := (c(N), c(N) + (i − j)) satisfy the conditions of Sublemma A.17.4 . But c(N) = a(N^{∨}). By Sublemma A.17.4 , in M(Q) there exists a direct summand L, isomorphic to N^{∨}(i − j)[2i − 2j]. Then L^{∨} will be isomorphic to N(j − i)[2j − 2i].

Theorem A.17 is proven.

We proved all the results we use in the main part of the paper. Let us just show simple consequence of Sublemma A.17.10 .

Corollary A.18 ([16], Corollary 3).

1) If q_{1}|_{k(Q_{2})} and q_{2}|_{k(Q_{1})} are isotropic, then dim(q_{1})−i_{1}(q_{1}) = dim(q_{2})− i_{1}(q_{2}).

2) If P ⊂ Q is a subquadric, s.t. p|_{k(Q)} is isotopic, then codim(P ⊂ Q) < i_{1}(q).

3) In the situation of 2), i_{1}(p) = i_{1}(q) − codim(P ⊂ Q).

Proof

1) Since q_{1}|_{k(Q_{2})} and q_{2}|_{k(Q_{1})} are isotropic, by Theorem A.14 , there exist isomorphic direct summands N_{1} of M(Q_{1}), and N_{2} of M(Q_{2}), s.t. a(N_{i}) = 0. By Sublemma A.17.10 , dim(N_{i}) := b(N_{i}) − a(N_{i}) = dim(Q_{i}) − a(N_{i}) − c(N_{i}) = dim(Q_{i}) − i_{1}(q_{i}) + 1. Since N_{1} ≃ N_{2}, we get the equality.

2) and 3) Follow from 1), taking into account that i_{1}(p) ⩾ 1.

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