Manin obstruction to strong approximation for homogeneous spaces

More documents

Recommendations

Info

48 Mikhail Borovoi and Cyril Demarchew ∈ S be a nonarchimedean place, then the map β w is bijective. It followsthat the map µ w : H 1 (k w , H) → M Γ is surjective (because µ w = cor w ◦ β w ).Since w is nonarchimedean, we have H 1 (k w , G) = 1 (because G is simply connected),hence ker[H 1 (k w , H) → H 1 (k w , G)] = H 1 (k w , H). It follows that themap µ w : ker[H 1 (k w , H) → H 1 (k w , G)] → M Γ is surjective. Now it is clear thatthe map µ S = ∑ v∈S µ v is surjective.We prove the proposition. We must prove that the localization mapX(k)/G(k) → X(A S )/G(A S )is surjective. In the language of Galois cohomology, we must prove that the localizationmapker[H 1 (k, H) → H 1 (k, G)] → ⊕ v /∈Sker[H 1 (k v , H) → H 1 (k v , G)]is surjective.Let ξ S = (ξ v ) ∈ ⊕ v /∈S ker[H1 (k v , H) → H 1 (k v , G)]. Set s := ∑ v /∈S µ v(ξ v ) ∈π 1 (H) Γ . Since the map µ S is surjective, there exists an element ξ S in the product∏v∈S ker[H1 (k v , H) → H 1 (k v , G)] such that ∑ v∈S µ v(ξ v ) = −s. Setξ := (ξ S , ξ S ) ∈ ⊕ v∈Ωker[H 1 (k v , H) → H 1 (k v , G)] ⊂ ⊕ v∈ΩH 1 (k v , H),then µ(ξ) = ∑ v∈Ω µ v(ξ v ) = s + (−s) = 0. By Proposition 8.2 there exists aclass ξ 0 ∈ H 1 (k, H) with image ξ in ⊕ v∈Ω H1 (k v , H). Since the Hasse principleholds⊕for G, we have ξ 0 ∈ ker[H 1 (k, H) → H 1 (k, G)]. Since the image of ξ 0 inv∈Ω H1 (k v , H) is (ξ S , ξ S ), we see that the image of ξ 0 in ⊕ v /∈S H1 (k v , H) is ξ S .Thus ξ S lies in the image of ker[H 1 (k, H) → H 1 (k, G)].Theorem 8.4. Let G be a simply connected k-group over a number field k, andlet H ⊂ G be a connected geometrically character-free subgroup (i.e. H <strong>to</strong>r =1). Set X := H\G. Let S be a finite set of places of k containing at least onenonarchimedean place v 0 . Assume that G ss (k) is dense in G ss (A S ). Then X hasstrong approximation away from S in the following sense. Let x = (x v ) ∈ X(A)and let U S ⊂ X(A S ) be any open neighbourhood of the A S -part x S of x. Thenthere exists a k-point x 0 ∈ X(k) ∩ U S . Moreover, one can ensure that for v ∈Ω ∞ ∩ S the points x 0 and x v lie in the same connected component of X(k v ). Moreprecisely, there exists y v0 ∈ X(k v0 ) such that the point x ′ ∈ X(A) defined byx ′ v 0:= y v0 and x ′ v := x v for v ≠ v 0 belongs <strong>to</strong> the closure of the set X(k).G(k S ) inX(A) for the adelic <strong>to</strong>pology.Proof. Set Σ := {v 0 }. We denote by x Σ ∈ X(A Σ ) and x S ∈ x(A S ) the correspondingprojections of x. By Proposition 8.3 applied <strong>to</strong> the finite set of placesΣ, there exists a k-point x ′ 0 ∈ X(k) ∩ x Σ .G(A Σ ). Let y v0 := (x ′ 0) v0 ∈ X(k v0 )
<strong>Manin</strong> <strong>obstruction</strong> <strong>to</strong> strong approximation for homogeneous spaces 49and define x ′ ∈ X(A) as in the theorem. Then there exists g ∈ G(A) such thatx ′ 0.g = x ′ in X(A). Let U ⊂ X(A) be an open neighbourhood of x ′ . Since theorbit x ′ 0.G(A) ⊂ X(A) is open (because H is connected) and contains x ′ , we mayassume that U ⊂ x ′ 0.G(A).By assumption G(k).G(k S ) is dense in G(A). It follows that there exists g 0 ∈G(k) and g S ∈ G(k S ) such that x ′′ := x ′ 0.g 0 .g S belongs <strong>to</strong> U . Set x 0 := x ′ 0.g 0 ∈X(k), then x ′′ = x 0 .g S . We see that x ′′ ∈ X(k).G(k S )∩U . Therefore, we concludethat x ′ lies in the closure of X(k).G(k S ).Concerning the infinite places, for v ∈ Ω ∞ ∩ S we have x 0 ∈ x v .G(k v ), becausex ′ 0 ∈ x v .G(k v ). Since G is simply connected, the group G(k v ) is connected (see[34], Theorem 5.2.3), hence the image of x 0 in X(k v ) is contained in the connectedcomponent of x v in X(k v ).8.5. Proof of Theorem 1.7. To prove this theorem, we can follow the proof ofTheorem 1.4 <strong>to</strong> make reductions, so that we may assume the following:(i) G u = {1},(ii) H ⊂ G lin , i.e. H is linear,(iii) G ss is simply connected,(iv) X(G abvar ) is finite.(v) the homomorphism H <strong>to</strong>r → G sab is injective.Set Σ ′ := Ω ∞ ∪ {v 0 }. Let UXΣ′ ⊂ X(AΣ′ ) be an open neighbourhood of theprojection x Σ′ ∈ X(A Σ′ ) of x. Set U f Σ′X:= UX× X(k v 0). Let U X be the specialopen neighbourhood of x in X(A) defined by U f X . Set Y := Gsab /H <strong>to</strong>r , andconsider the canonical morphism ψ : X → Y . Set y := ψ(x) ∈ Y (A), then y isorthogonal <strong>to</strong> the group Br 1 (Y ) for the <strong>Manin</strong> pairing. Hence by [22], Theorem 4,there exists y 0 ∈ Y (k)∩ψ(U X ). Set X y0 := ψ −1 (y 0 ) ⊂ X and V := X y0 (A)∩U X .Then V is open and non-empty since y 0 ∈ ψ(U X ). As in the proof of Proposition7.4, we know that X y0 is a homogeneous space of the semisimple simply connectedgroup G ss = G sc , with connected character-free geometric stabilizers, and with a k-point. Therefore Theorem 8.4 implies that X y0 (k).G sc (k S ) ∩ V ≠ ∅. In particular,the set X(k).G sc (k S ) ∩ U X is non-empty. Set S ′ := S {v 0 }, Sf ′ := S′ ∩ Ω f ,U {v0}X:= UXΣ′× U X,∞, then it follows that the set X(k).G sc (k S ′) ∩ U {v0}X⊂.Gsc (k S ′f), we obtain easilyX(A {v0} ) is non-empty. Since U {v0}Xthat the set X(k).G sc (k S ′f) ∩ U {v0}Xproof of Theorem 1.7..Gsc (k S ′) = U {v0}X⊂ X(A {v0} ) is non-empty. This completes theAcknowledgements. The first-named author is grateful <strong>to</strong> the Tata Instituteof Fundamental Research, Mumbai, where a part of this paper was written, forhospitality and good working conditions. The second-named author thanks DavidHarari for many helpful suggestions. Both authors thank the referee for his/hercomments and suggestions.
Page 8: 8 Mikhail Borovoi and Cyril Demarch
Page 11: Manin obst
Page 14 and 15: 14 Mikhail Borovoi and Cyril Demarc
Page 20: 20 Mikhail Borovoi and Cyril Demarc
Page 25 and 26: Manin obst
Page 47: Manin obst

Manin obstruction to strong approximation for homogeneous spaces

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?