Manin obstruction to strong approximation for homogeneous spaces
Manin obstruction to strong approximation for homogeneous spaces
Manin obstruction to strong approximation for homogeneous spaces
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8 Mikhail Borovoi and Cyril Demarcheand if y ∈ Y (k), x = π(y), define Br 1,x (X, Y ) <<strong>strong</strong>>to</<strong>strong</strong>> beBr 1,x (X, Y ) := ker[x ∗ : Br 1 (X, Y ) → Br(k)] = {b ∈ Br(X) : π ∗ (b) ∈ Br 1,y (Y )} .We denote bythe evaluation map.〈, 〉: Br(X) × X(k) → Br(k): (b, x) ↦→ b(x)2.2. Be<strong>for</strong>e recalling the result of Sansuc, we give a few more definitions andnotations. Let A be an abelian category and F : Var/k → A be a contravariantfunc<<strong>strong</strong>>to</<strong>strong</strong>>r from the category of k-varieties <<strong>strong</strong>>to</<strong>strong</strong>> A . If X and Y are k-varieties, theprojections p X , p Y : X × k Y → X, Y induce a morphism in A (see [38], Section6.b):F (p X ) + F (p Y ): F (X) ⊕ F (Y ) → F (X × k Y )such thatF (p X ) + F (p Y ) = F (p X ) ◦ π X + F (p Y ) ◦ π Y , (3)where π X , π Y are the projections F (X) ⊕ F (Y ) → F (X), F (Y ) and the group lawin the right-hand side is the law in Hom(F (X) ⊕ F (Y ), F (X × k Y )).Let m: X × k Y → Y be a morphism of k-varieties. Assume that the morphismF (p X ) + F (p Y ) is an isomorphism. We define a mapϕ : F (Y ) → F (X × k Y ) → F (X) ⊕ F (Y ) → F (X)by the <strong>for</strong>mula(see [38], (6.4.1)).ϕ := π X ◦ (F (p X ) + F (p Y )) −1 ◦ F (m) (4)Lemma 2.3. Let F : Var/k → A be a contravariant func<<strong>strong</strong>>to</<strong>strong</strong>>r. Let X, Y be twok-varieties, m : X × k Y → Y be a k-morphism. Assume that:• F (Spec(k)) = 0.• F (p X ) + F (p Y ): F (X) ⊕ F (Y ) → F (X × k Y ) is an isomorphism.• There exists x ∈ X(k) such that the morphism m(x, .) : Y → Y is the identityof Y .Then F (m) = F (p X ) ◦ ϕ + F (p Y ): F (Y ) → F (X × k Y ).Proof. Consider the morphism x Y : Y → X × k Y defined by x. Then F (x Y ) ◦F (p X ) = 0, since the morphism p X ◦x Y : Y → X fac<<strong>strong</strong>>to</<strong>strong</strong>>rs through x : Spec(k) → Xand F (Spec(k)) = 0. Since p Y ◦ x Y = id Y , we have F (x Y ) ◦ F (p Y ) = id, and the