Profinite genus of the fundamental groups of compact flat manifolds with holonomy group of prime order

To what extent can a finitely generated residually finite group Γ be determined by its profinite completion?

Let Γ be an 𝑛-dimensional Bieberbach group with the holonomy group 𝐺 of prime order.

1. If all indecomposable summands of Z ⁢ G -module 𝑀 have ℤ-rank p - 1 except one trivial summand of ℤ-rank 1, then | g ⁢ ( Γ ) | = | C 2 ∖ H ⁢ ( Q ⁢ ( ζ ) ) | , where C 2 is the group of order 2 acting on H ⁢ ( Q ⁢ ( ζ ) ) by invertion.

2. Otherwise, | g ⁢ ( Γ ) | = | G ⁢ ( p ) ∖ H ⁢ ( Q ⁢ ( ζ ) ) | .^[1]

Let Γ be an 𝑛-dimensional Bieberbach group with the holonomy group 𝐺 of prime order. Then | g ⁢ ( Γ ) | = 1 if and only if | G | ⩽ 19 .

Let 𝑋 be an 𝑛-dimensional compact flat manifold with the holonomy group of prime order. If n ⩽ 21 , then 𝑋 is determined among all 𝑛-dimensional compact flat manifolds by the profinite completion of its fundamental group.

Let 𝑀 be a free abelian group of rank 𝑛 and 𝐺 a group of prime order 𝑝. Let Γ 1 and Γ 2 be 𝑛-dimensional Bieberbach groups that are extensions of 𝑀 by 𝐺. If M 1 and M 2 are 𝐺-modules induced by the action of Γ 1 and Γ 2 on 𝑀, respectively, then

1. Γ 1 ≅ Γ 2 if and only if either 𝐺-modules M 1 and M 2 satisfy hypothesis (ii) of Theorem 1.1 and are isomorphic, or M 1 and M 2 satisfy hypothesis (i) of Theorem 1.1 and are isomorphic up to a twist of 𝐺 by inversion,

2. Γ ^ 1 ≅ Γ ^ 2 if and only if M ^ 1 ≅ M ^ 2 as 𝐺-modules.

Let Γ be an 𝑛-dimensional Bieberbach group with translation subgroup 𝑀. Then M ^ is the unique open normal, torsion-free maximal abelian subgroup of Γ ^ .

Proof

There is a one-to-one correspondence between the set of all finite index normal subgroups of Γ, and the set of all open normal subgroups of Γ ^ (see [21, Proposition 3.2.2]). Therefore, the lemma follows from Proposition 2.14 together with [21, Theorem 4.7.10]. ∎

The profinite completion of a Bieberbach group is a torsion-free profinite group.

Proof

Let Γ be a Bieberbach group defined by the exact sequence

where 𝑀 is the translation subgroup and 𝐻 the holonomy group. From the above discussion, we have

Suppose that Γ ^ has an element of finite order. Then Γ ^ has an element of prime order. Since M ^ is torsion-free, there is an element h ∈ H of prime order 𝑝 such that π ^ - 1 ⁢ ( ⟨ h ⟩ ) is not torsion-free. Note that π ^ - 1 ⁢ ( ⟨ h ⟩ ) is an open subgroup of Γ ^ containing M ^ such that π ^ - 1 ⁢ ( ⟨ h ⟩ ) / M ^ ≅ ⟨ h ⟩ . Now, by [21, Proposition 3.2.2], there is a subgroup Γ p of Γ containing 𝑀 such that

Then Γ ^ p is not torsion-free, a contradiction in view of [21, Theorem 4.7.10]. Therefore, Γ ^ is a torsion-free profinite group. ∎

Let Γ 1 and Γ 2 be 𝑛-dimensional Bieberbach groups with translation subgroups M 1 and M 2 and holonomy groups G 1 and G 2 , respectively. If φ : Γ ^ 1 → Γ ^ 2 is an isomorphism, then there are isomorphisms ϕ : M ^ 1 → M ^ 2 and ψ : G 1 → G 2 such that the following diagram commutes:

(3.1)

Proof

By Lemma 3.1, we have φ ⁢ ( M ^ 1 ) = M ^ 2 . We define 𝜙 to be the restriction of 𝜑 to M ^ 1 . Since M ^ 1 , M ^ 2 are profinite free abelian groups of the same rank, ϕ : M ^ 1 → M ^ 2 is an isomorphism. Thus, 𝜑 induces an isomorphism ψ : G 1 → G 2 such that diagram (3.1) commutes. ∎

Let

be Z ⁢ G -lattices. Then M ^ ≅ M ^ ′ as 𝐺-modules if and only if a = a ′ , b = b ′ , and c = c ′ .

Proof

Our proof starts with the observation that

If M ^ ≅ M ^ ′ as 𝐺-modules, then the pro-𝑝 components of M ^ and M ^ ′ are isomorphic as Z p ⁢ G -modules. Therefore, the necessity follows from Remark 2.8 together with Propositions 2.5 and 2.6.

On the other hand, by Remark 2.8, the statement holds for pro-𝑝 components of M ^ and M ^ ′ . Hence, by Proposition 2.9, M q ≅ N q for all primes 𝑞, and therefore, M ^ ≅ M ^ ′ as 𝐺-modules. ∎

Let Γ 1 and Γ 2 be 𝑛-dimensional Bieberbach groups with holonomy group 𝐺 and translation subgroups M 1 and M 2 , respectively. Then Γ 1 ≅ Γ 2 if and only if M 1 ≅ ( M 2 ) ψ as 𝐺-modules, where ψ ∈ Aut ⁢ ( G ) .

Proof

Let φ : Γ 1 → Γ 2 be an isomorphism. By Proposition 2.14, M 1 and M 2 are the unique normal, maximal abelian subgroups of Γ 1 and Γ 2 , respectively. Hence, φ ⁢ ( M 1 ) = M 2 . Therefore, 𝜑 induces an isomorphism ψ : Γ 1 / M 1 → Γ 2 / M 2 , i.e., ψ ∈ Aut ⁢ ( G ) because Γ 1 / M 1 ≅ G ≅ Γ 2 / M 2 . Define ϕ := φ | M 1 . By definition of 𝜙 and 𝜓, the following diagram commutes:

Thus,

for x ∈ G , m ∈ M 1 , and α ∈ Γ 1 with π 1 ⁢ ( α ) = x . Therefore, M 1 ≅ ( M 2 ) ψ as 𝐺-modules.

Conversely, suppose there is ψ ∈ Aut ⁢ ( G ) such that M 1 ≅ ( M 2 ) ψ as 𝐺-modules. Then M 1 is semi-linearly isomorphic to M 2 . Therefore, by Propositions 2.13, 2.17, and 2.18, we have Γ 1 ≅ Γ 2 . ∎

Proof of Theorem 1.4

(i) This follows from Proposition 3.5 and by Remark 2.19.

(ii) “Only if”. From Lemma 3.3, we see there are isomorphisms ϕ : M ^ 1 → M ^ 2 and ψ : G → G such that the following diagram is commutative:

Then ϕ : M ^ 1 → M ^ 2 is a 𝐺-module homomorphism that is an isomorphism by Lemma 3.3.

Conversely, by Lemma 2.15, M 1 = M n - 1 ⊕ Z and M 2 = M n - 1 ′ ⊕ Z ′ such that Γ 1 = M n - 1 ⋊ C 1 and Γ 2 = M ′ n - 1 ⋊ C 2 , where C 1 , C 2 contain ℤ and Z ′ as subgroups of index 𝑝 and act on M n - 1 , M ′ n - 1 as 𝐺. Hence, M ^ 1 = M ^ n - 1 ⊕ Z ^ , M ^ 2 = M ^ n - 1 ′ ⊕ Z ^ ′ and Γ ^ 1 = M ^ n - 1 ⋊ C ^ 1 and Γ ^ 2 = M ^ n - 1 ′ ⋊ C ^ 2 , where C ^ 1 , C ^ 2 act on M ^ n - 1 , M ^ n - 1 ′ as 𝐺. By Lemma 3.4 and Proposition 2.5, we get that the profinite completions of 𝐺-modules M ^ n - 1 , M ^ n - 1 ′ are isomorphic (because they are isomorphic as Z p ⁢ G -modules for all primes 𝑝). Hence, Γ ^ 1 is isomorphic to Γ ^ 2 . ∎

Let Γ be an 𝑛-dimensional Bieberbach group whose holonomy group has prime order. Then there is a one-to-one correspondence between isomorphism classes of Γ ^ and triples ( a , b , c ) from decomposition (2.1) with a > 0 .

Let Γ be an 𝑛-dimensional Bieberbach group with holonomy group 𝐻. If Δ ∈ g ⁢ ( Γ ) , then Δ is an 𝑛-dimensional Bieberbach group with holonomy group 𝐻.

Proof

Let Γ be an 𝑛-dimensional Bieberbach group with translation subgroup 𝑀 and holonomy group 𝐻. Since Δ ⩽ Δ ^ and Δ ^ ≅ Γ ^ , it follows from Proposition 3.2 that Δ is torsion free. Then it remains to show that Δ has a normal, maximal abelian subgroup 𝑁 such that N ≅ Z n . By Lemma 3.1, M ^ is the unique open normal, maximal abelian subgroup of Γ ^ . Since Δ ^ ≅ Γ ^ , it follows by [21, Proposition 3.2.2] that there is a normal, torsion-free maximal abelian subgroup 𝑁 in Δ such that N ^ ≅ M ^ and Δ / N ≅ H . Then 𝑁 is a finitely generated torsion-free abelian group, i.e., N ≅ Z m . We claim that m = n . Indeed, for example, if n > m , we can construct a finite quotient ( Z / k ⁢ Z ) n of 𝑀 that cannot be quotient of 𝑁, for some non-zero integer 𝑘. Then N ^ is not isomorphic to M ^ , a contradiction (see [11]). Therefore, Δ is an 𝑛-dimensional Bieberbach group with the holonomy group 𝐻. ∎

Proof of Theorem 1.1

Let Γ 1 and Γ 2 be 𝑛-dimensional Bieberbach groups with holonomy group of prime order and translation subgroups M 1 and M 2 , respectively. By Lemma 3.3, we can assume that Γ 1 and Γ 2 have the same holonomy group 𝐺 of prime order 𝑝. As M 1 and M 2 are Z ⁢ G -modules, by Proposition 2.3,

Let Γ be a Bieberbach group whose holonomy group has prime order. If H ⁢ ( Q ⁢ ( ζ ) ) is non-trivial, then | g ⁢ ( Γ ) | > 1 .

It follows that the cardinality | g ⁢ ( Γ ) | of the genus of a Bieberbach group Γ with translation group 𝑀 and holonomy group 𝐺 of order 𝑝 is equal to the cardinality of the genus of the 𝐺-module 𝑀 for non-exceptional Γ and half of it if Γ is exceptional.

Proof of Corollary 1.2

This is a consequence of Theorem 1.1 together with Proposition 2.11. ∎

Proof of Corollary 1.3

From Proposition 2.20, we deduce that, for all n ⩽ 21 , no prime number p ⩾ 23 can occur as the order of an element in the holonomy group of an 𝑛-dimensional Bieberbach group. The result then follows from Corollary 1.2. ∎

If Γ = M ⋊ G , with 𝑀 free abelian of rank 𝑛 and 𝐺 of order 𝑝 acting non-trivially on 𝑀, then it is not a Bieberbach group. Nevertheless, using the fact that Γ and M 1 ⋊ G are isomorphic if and only if M ≅ M 1 as 𝐺-modules (up to a twist of 𝐺) and that the same holds for the profinite completions (see [12, Subsection 2.2]), one can deduce that the genus of Γ has the same cardinality as in Theorem 1.1 (ii) (see Proposition 3.5). Moreover, considering 𝐺 as a subgroup of Aut ⁢ ( M ) = GL ⁢ ( n , Z ) , one deduces from [12, Proposition 2.17] that the cardinality of the genus is exactly the number of conjugacy classes of subgroups of order 𝑝 of GL ⁢ ( n , Z ) in the conjugacy class of 𝐺 in GL ⁢ ( n , Z ^ ) .