Cohomological properties of maximal pro-p Galois groups that are preserved under profinite completion

Theorem 14.

Let G be a pro-p group acting trivially on F p such that G has hereditary the vanishing n-Massey product property. Then so does G ^ .

Proposition 15.

Let G be an abstract group and consider F p as a trivial G ^ -module. Then G ^ satisfies the n-vanishing Massey product property if and only if for every abstract homomorphism φ : G → U n + 1 ⁢ ( F p ) ¯ there exists an abstract homomorphism ψ : G → U n + 1 ⁢ ( F p ) such that for every element g ∈ G and 1 ≤ i ≤ n , [ ψ ⁢ ( g ) ] i , i + 1 = [ φ ⁢ ( g ) ] i , i + 1 .

Proof.

( ⇒ ) Let φ : G → U n + 1 ⁢ ( 𝔽 p ) ¯ be an abstract homomorphism. By definition of the profinite completion it can be lifted to a continuous homomorphism φ ¯ : G ^ → U n + 1 ⁢ ( 𝔽 p ) ¯ . By assumption, there exists a twisted (continuous) solution ψ : G ^ → U n + 1 ⁢ ( 𝔽 p ) . Let i : G → G ^ denote the natural homomorphism associated to the profinite completion. Then ψ ∘ i : G → U n + 1 ⁢ ( 𝔽 p ) is an abstract twisted solution for G.

( ⇐ ) Let φ : G ^ → U n + 1 ⁢ ( 𝔽 p ) ¯ be a continuous homomorphism. Then φ ∘ i is an abstract homomorphism G → U n + 1 ⁢ ( 𝔽 p ) ¯ . By assumption, G admits an abstract twisted solution ψ : G → U n + 1 ⁢ ( 𝔽 p ) . We claim that its continuous lifting ψ ¯ : G ^ → U n + 1 ⁢ ( 𝔽 p ) is a twisted solution. Indeed, for every 1 ≤ i < n , the maps [ ψ ¯ ] i , i + 1 and [ φ ] i , i + 1 which go to a Hausdorff space, identify on the dense subset i ⁢ ( G ) , thus they are equal. ∎

Lemma 16.

A profinite group G satisfies hereditarily the n-vanishing Massey product property if and only if every open subgroup of G satisfies the n-vanishing Massey product property.

Proof.

Let H ≤ c G and let φ : H → U n + 1 ⁢ ( 𝔽 p ) ¯ be a continuous homomorphism. By Lemma 6 there exists an open normal subgroup O ⊴ c G such that φ can be lifted to a continuous homomorphism φ ¯ : O → U n + 1 ⁢ ( 𝔽 p ) ¯ . Let ψ : O → U n + 1 ⁢ ( 𝔽 ) p be a “twisted” solution to the corresponding embedding problem. Then ψ | H is a twisted solution for H. ∎

Proof of Theorem 14.

Let U be a finite index subgroup of G and φ : U → U n + 1 ⁢ ( 𝔽 p ) ¯ an abstract homomorphism. By Proposition 15, in order to prove that U ^ satisfies the n-vanishing Massey product property it is enough to prove that φ admits an abstract twisted solution. For every finite subset D = { x 1 , … , x n } of U, look at the restriction of φ to H D = 〈 D 〉 ¯ , the closed subgroup of G generated by D. Since H D is finitely generated, by the Nikolov–Segal Theorem, φ | H D is continuous. So, since H D is a closed subgroup of G, by assumption, it admits a twisted solution ψ D . It is easy to see that if D ⊆ E , then any twisted solution from H E induces a twisted solution from H D by restriction of the domain, and that the union of twisted solutions is a twisted solution. Thus, by Proposition 11 there exists an abstract twisted solution from U. In conclusion, U ^ ≅ U ¯ satisfies the n-fold vanishing Massey product property. Since by Remark 5 these are all the open subgroups of G ^ , by Lemma 16 we are done. ∎

Definition 17 ([24]).

Let G be an oriented pro-p group. Then G is called a 1-cyclotomic oriented pro-p group if the induced module structure on ℤ p satisfies that the natural map

is surjective, for all closed subgroups H ≤ c G and for every natural number n.

Corollary 18.

Let F be a field containing a primitive root of unity of order p and let G F ⁢ ( p ) be its maximal pro-p Galois group. Then equipped with the arithmetical orientation, G F ⁢ ( p ) is a 1-cyclotomic oriented pro-p group.

Remark 19.

Since G acts trivially on 𝔽 p , we have H 1 ⁢ ( H , ℤ p / p ⁢ ℤ p ) = Hom ⁡ ( H , ℤ p / p ⁢ ℤ p ) .

Theorem 20.

Let G be a 1-cyclotomic oriented pro-p group. Then so is G ^ .

Observation 21.

By the universal property of profinite completion, the orientation of G can be lifted to an orientation of G ^ .

Lemma 22.

Let G be an oriented pro-p group. Then G is 1-cyclotomic if and only if the induced module structure on Z p satisfies that the natural map

is surjective, for all open subgroups U ≤ o G and for every natural number n.

Proof.

Let n be a natural number, H a closed subgroup of G and f ∈ H 1 ⁢ ( H , ℤ p / p ) = Hom ⁡ ( H , ℤ p / p ) . By Lemma 6, any homomorphism from H to ℤ p / p can be lifted to a homomorphism from some open subgroup U of G. By assumption, there exists a crossed homomorphism α ∈ H 1 ⁢ ( U , ℤ p ⁢ ( 1 ) / p n ) which projects on this homomorphism. So the restriction of α to H will do. ∎

Proof of Theorem 20.

First we lift the orientation of G to G ^ .

Let n be a natural element and α ∈ H 1 ⁢ ( U ^ , 𝔽 p ) . We shall find an element f ^ ∈ H 1 ⁢ ( U ^ , ℤ p ⁢ ( 1 ) / p n ) such that the natural homomorphism H 1 ⁢ ( U ^ , ℤ p ⁢ ( 1 ) / p n ⁢ ℤ p ⁢ ( 1 ) ) → H 1 ⁢ ( H , 𝔽 p ) sends f ^ to α.

Let us look at α | U . This is an element in H abs 1 ⁢ ( U , 𝔽 p ) ≅ Hom abs ⁡ ( U , 𝔽 p ) . Now let H be a finitely generated closed subgroup of G that is contained in U and look at α | H . By the Nikolov–Segal Theorem, H is strongly complete and thus α | H ∈ H cont 1 ⁢ ( H , 𝔽 p ) . Hence, since G is cyclotomic, there exists an element β H ∈ H cont 1 ⁢ ( H , ℤ p ⁢ ( 1 ) / p n ) which projects over α | H .

Recall that an element in H cont 1 ⁢ ( H , ℤ p ⁢ ( 1 ) / p n ) is in fact an equivalence class of continuous crossed homomorphisms H → ℤ p ⁢ ( 1 ) / p n . However, since H acts trivially on 𝔽 p , every element in H 1 ⁢ ( H , 𝔽 p ) is a singleton. Thus, if f H is a primitive of β H in C 1 ⁢ ( H , 𝔽 p ) , then the π ∘ f H = α | H , where π : ℤ p ⁢ ( 1 ) / p n → ℤ p ⁢ ( 1 ) / p ≅ 𝔽 p is the natural projection.

The number of continuous crossed homomorphisms from a finitely generated profinite group to a finite module is finite since it depends only on the values of the generators. In addition, if f H is a crossed homomorphism H → ℤ p ⁢ ( 1 ) / p n satisfying π ∘ f H = α | H and K ≤ H , then f H | K is a crossed homomorphism H → ℤ p ⁢ ( 1 ) / p n satisfying π ∘ f H | K = α | K . Hence, we have a projective system of finite sets, so its inverse limit is nonempty. Observe that the inverse limit is a crossed homomorphism f : ⋃ X : → ℤ p ( 1 ) / p n satisfying π ∘ f = ⋃ H ∈ X α | H , where X is the set of all closed subgroups of G which are contained in U. By Lemma 4 f is in fact a crossed homomorphism U → ℤ p ⁢ ( 1 ) / p n satisfying π ∘ f = α | U . Now using the cohomological goodness for n = 1 , there is a continuous crossed homomorphism f ^ ∈ C 1 ⁢ ( U ^ , ℤ p ⁢ ( 1 ) / p n ) whose restriction to U is equivalent to f. Thus ( π ∘ f ^ ) | U ∼ π ∘ f = α | U . Again, since every element in H abs 1 ⁢ ( U , 𝔽 p ) is a singleton, we have ( π ∘ f ^ ) | U = α | U . We conclude that π ∘ f ^ and α, being two continuous functions to an Hausdorff space which identify on a dense subset, are equal. ∎

Remark 23.

Notice that sequence (1) is always a complex.

Theorem 24.

Let G be a pro-p group, hereditarily of p-absolute Galois type. Then so is G ^ .

Lemma 25.

Let G be a pro-p group. Then G is hereditarily of p-absolute Galois type if and only if every open subgroup of G is of p-absolute Galois type.

Proof.

Let H be a closed subgroup of G and χ ∈ H 1 ⁢ ( H , 𝔽 p ) . Recall that since G acts trivially on 𝔽 p , it follows that H 1 ⁢ ( H , 𝔽 p ) ≅ Hom ⁡ ( H , 𝔽 p ) . Thus, by Lemma 6, χ can be lifted to some open subgroup U ≤ o G . Let ρ ∈ H 1 ⁢ ( H , 𝔽 p ) be an element such that χ ∪ ρ = 0 . By Lemma 6 again there is an open subgroup U ′ such that ρ can be lifted to U ′ . Note that χ ∪ ρ = 0 means that there is some f ∈ C 1 ⁢ ( H , 𝔽 p ) such that ∂ ⁡ ( f ) = χ ∪ ρ . Since f is a continuous function from a profinite space to a finite space, it projects through some finite quotient of H (see [25, Proposition 1.1.16 (a)]). That is, there exist a finite group A, a continuous homomorphism f ′ : H → A and a continuous function f ′′ : A → 𝔽 p such that f = f ′′ ∘ f ′ . So again, there exists a lifting of f ′ , and thus of f, to an open subgroup U ′′ of G. By defining O = U ∩ U ′ ∩ U ′′ , we get that O is an open subgroup of G to which both χ and ρ can be lifted. Denote these lifting by χ ¯ and ρ ¯ , and notice that χ ¯ ∪ ρ ¯ is trivial. By assumption,

for some ρ ′ ∈ H 1 ⁢ ( O ∩ ker ⁡ ( χ ) , 𝔽 p ) . Taking

will do.

For the second exactness, let α ∈ H 2 ⁢ ( H , 𝔽 p ) such that Res ker ⁡ ( χ ) H ⁡ ( α ) = 0 . By the same arguments, there exists some open subgroup O ≤ o G containing H such that α and χ can be lifted to it, and α ¯ , the lifting of α, is trivial. By assumption, α ¯ = χ ¯ ∪ ρ for some ρ ∈ H 1 ⁢ ( O , 𝔽 p ) , when χ ¯ denotes the lifting of χ to O. Then Res H O ⁡ ( ρ ) = α ′ proving the exactness in H 2 ⁢ ( H , 𝔽 p ) . ∎

Remark 26.

Lemma 25 has also been proven in [6, Proposition 5.10].

Lemma 27.

Let K be a group and let L , H be subgroups such that L has finite index in K. Assume further that [ H : H ∩ L ] = [ K : L ] . Then for every K-module M the following diagram is commutative:

Proof.

Choose a right transversal h 1 , … , h n of L ∩ H in H. Then h 1 , … , h n is a right transversal for H in K too. Recall that given a primitive f ∈ C 1 ⁢ ( L , M ) of an element α ∈ H 1 ⁢ ( L , M ) , Cor K L ⁡ ( α ) is represented by the function

(see [22, p. 46]). By substituting an element h ∈ H , the commutativity follows. ∎

Proof of Theorem 24.

Using Remark 5, let U ^ be an open subgroup of G ^ , for some finite-index subgroup U of G, and χ ∈ H 1 ⁢ ( U ^ , 𝔽 p ) a continuous homomorphism. We will prove the exactness of the following series:

Exactness at H 1 ⁢ ( U ^ , F p ) . Let ρ ∈ H 1 ⁢ ( U ^ , 𝔽 p ) be such that χ ∪ ρ = 0 . Let us look at the restrictions of ρ and χ to U, these are not-necessarily-continuous homomorphisms U → 𝔽 p .

First assume that χ | U = 0 . Since χ is continuous and U is dense in U ^ , we conclude that χ = 0 . Hence ker ⁡ ( χ ) = U ^ and for every ρ ∈ H 1 ⁢ ( U ^ , 𝔽 p ) , ρ = Cor U ^ ⁡ ( ρ ) .

Now assume that χ | U ≠ 0 and let x ∈ U be such that χ ⁢ ( x ) ≠ 0 . By the Nikolov–Segal Theorem, for every finitely generated closed subgroup H of G that is contained in U, Res H U ⁡ ( χ ) and Res H U ⁡ ( ρ ) are continuous homomorphisms. Moreover, taking the restriction of f to H, for f ∈ C 1 ⁢ ( U ^ , 𝔽 p ) such that ∂ ⁡ ( f ) = χ ∪ ρ implies that Res H U ⁡ ( χ ) ∪ Res H U ⁡ ( ρ ) = 0 . So by assumption,

for some ρ ′ ∈ H 1 ⁢ ( ker ⁡ ( Res H ⁡ ( χ ) ) , 𝔽 p ) . Since H is finitely generated and H ∩ ker ⁡ ( χ ) is of finite index in H, it follows that H ∩ ker ⁡ ( χ ) is open in H. Thus, by [25, Corollary 2.5.5] H ∩ ker ⁡ ( χ ) is finitely generated too. Moreover, if H ≤ K are finitely generated closed subgroups of G that are contained in U and that contain x, then using Lemma 27 and the fact that [ H : H ∩ ker ( χ ) ] = p = [ K : K ∩ ker ( χ ) ] , for all ρ K ′ such that Res K U ⁡ ( ρ ) = Cor K ker ⁡ ( χ ) ∩ K ⁡ ( ρ K ′ ) one has

Since the set of continuous homomorphisms from a finitely generated profinite group to a finite group is finite, there is an element ρ ′ ∈ H abs 1 ⁢ ( ker ⁡ ( χ | U ) , 𝔽 p ) which is an inverse limit of all these homomorphisms. This follows from the fact that U equals the union of all finitely generated closed subgroups H that are contained in it and contain x and thus ker ⁡ ( χ | U ) = ⋃ ker ⁡ ( Res H U ⁡ ( χ ) ) . Moreover, it is easy to see that ρ = Cor U ker ⁡ ( χ ) ∩ U ⁡ ( ρ ′ ) . Since H abs 1 ⁢ ( U , 𝔽 p ) = Hom abs ⁡ ( U , 𝔽 p ) , it follows that ρ ′ can be lifted to a continuous homomorphism ρ ′ ^ : ker ⁡ ( χ ) → 𝔽 p . The only thing left to show is that

For that choose a transversal { u 1 , … , u n } of ker ⁡ ( χ | U ) in U. By [25, Proposition 3.2.2] this is also a transversal of ker ⁡ ( χ ) in U ^ . Recall that for each x ∈ U ^ ,

where u i ⁢ x ~ denotes the representative of the coset of u i ⁢ x from the chosen set of transversal. An analog of this interpretation holds for Cor U ker ⁡ ( χ ) ∩ U ⁡ ( ρ ′ ) . So ρ and Cor U ^ ker ⁡ ( χ ) ⁡ ( ρ ′ ^ ) identify on the dense subset U, and hence they are equal.

Exactness at H 2 ⁢ ( U ^ , F p ) . Let α ∈ H 2 ⁢ ( U ^ , 𝔽 p ) be such that

Its restriction to U is an element in the second abstract cohomology of U whose restriction to ker ⁡ ( χ | U ) is trivial. Since α is continuous, there is some finite continuous homomorphic image A of U ^ such that α splits through the natural projection U ^ → A (recall that for the continuous cohomology groups, H n ( U ^ , 𝔽 p ) ) ≅ lim → H n ( A , 𝔽 p ) when A runs over the set of finite continuous homomorphic images of U ^ (see [25, Proposition 6.5.5]). That is, α = Inf U ^ ⁡ ( α A ) , where α A ∈ H 2 ⁢ ( A , 𝔽 p ) . Thus Res U U ^ ⁡ ( α ) is also inflated from A, and hence so is Res H ⁡ ( α ) for every finitely generated closed subgroup H of G that is contained in U. By the Nikolov–Segal Theorem we get that the reduced homomorphism H → A is continuous, so Res H ⁡ ( α ) is a continuous cocycle. The same holds for Res H U ⁡ ( χ ) . So by assumption there is some ρ H ∈ H cont 1 ⁢ ( H , 𝔽 p ) such that Res H U ⁡ ( χ ) ∪ ρ H = Res H U ⁡ ( α ) . Observe that this is an equation of equivalence classes. That is, if we identify α with some primitive of α in C 2 ⁢ ( U ^ , 𝔽 p ) , then there exists some f H ∈ C cont 1 ⁢ ( H , 𝔽 p ) such that Res H U ⁡ ( χ ) ∪ ρ H + δ ⁢ ( f H ) = Res H U ⁡ ( α ) . Obviously, if K ≤ H , then Res K H ⁡ ( χ ) ∪ ρ H | K + δ ⁢ ( f H | K ) = Res K U ⁡ ( α ) . Moreover, the number of continuous homomorphisms from a finitely generated profinite group to a finite group is finite. In addition, for every continuous homomorphism ρ H from H to 𝔽 p , the number of f H ∈ C cont 1 ⁢ ( H , 𝔽 p ) satisfying Res H U ⁡ ( χ ) ∪ ρ H + δ ⁢ ( f H ) = Res H U ⁡ ( α ) is finite as is can be determined by the values of finite set of generators. Taking the inverse limit of the finite sets of such pairs ( ρ H , f H ) , we obtain a cocycle ρ ∈ H abs 1 ⁢ ( U , 𝔽 p ) such that Res U U ^ ⁡ ( χ ) ∪ ρ = Res U U ^ ⁡ α . Since ρ is in fact a homomorphism, it can be lifted to a continuous homomorphism ρ ^ ∈ Hom ⁡ ( U , 𝔽 p ) = H 1 ⁢ ( U , 𝔽 p ) . So the only thing left to show is that χ ∪ ρ ^ = α . But this follows from the injectivity of the second cohomological map H cont 2 ⁢ ( U ^ , 𝔽 p ) → H abs 2 ⁢ ( U , 𝔽 p ) described in [26, Section 2.6]. ∎