By Mikhail Borovoi

During this quantity, a brand new functor $H^2_{ab}(K,G)$ of abelian Galois cohomology is brought from the class of attached reductive teams $G$ over a box $K$ of attribute $0$ to the class of abelian teams. The abelian Galois cohomology and the abelianization map$ab^1:H^1(K,G) \rightarrow H^2_{ab}(K,G)$ are used to offer a functorial, virtually specific description of the standard Galois cohomology set $H^1(K,G)$ whilst $K$ is a host box.

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We will need the following fundamental results on Galois cohomology over local and global fields. 1 ([Kn1], [Kn3]). Let G be a simply connected group over a nonarchimedian local field K. Then H 1 (K, G) = 1. Another proof of this result appeared in [Br-T]. 2. Let K be a number field. A K-group G is said to satisfy the Hasse principle if X(G) := ker[H 1 (K, G) → Π H 1 (Kv , G)] = 0. 3 (Kneser-Harder-Chernousov). For any semisimple simply connected group G over a number field K, the map H 1 (K, G) → Π H 1 (Kv , G) ∞ is bijective.

Let T be a K-torus. Suppose that there is a place v0 of K such that T is anisotropic over Kv0 . Then X2 (K, T ) := ker[H 2 (K, T ) → Π H 2 (Kv , T )] = 0. v∈V Proof: See[Ha1], II, p. 2, Thm. 7, p. 3. 3 (Harder). Let G be a K-group. Let Σ ⊂ V be a finite set of places of K. For any v ∈ Σ let Tv ⊂ GKv be a maximal torus. Then there exists a maximal torus T ⊂ G such that TKv is conjugate to Tv under G(Kv ) for any v ∈ Σ. 3. 4. Let G be a semisimple simply connected K-group. Let j: T → G be a maximal torus of G such that for every v ∈ V∞ the torus TKv is fundamental in GKv .

This result is essentially due to Kottwitz [Ko3]. Proof: It suffices to find a maximal torus T ⊂ G such that the map 1 H 1 (K, T ) → Hab (K, G) = H1 (K, T (sc) → T ) is surjective. Let T be a fundamental torus of G; then T (sc) is a fundamental torus of Gsc . 3, we see tht for such T the map H 1 (K, T ) → 1 Hab (K, G) is surjective. The theorem is proved. 1. 4 is bijective. 17 any fiber of ab1G comes from H 1 (K, ψ Gsc ) for some cocycle ψ ∈ Z 1 (K, G). 1 H 1 (K, ψ Gsc ) = 1. Hence the map ab1G is injective.