Amalgamated free products of n-slender groups RIMS Set Theory Workshop 2010 Waseda University Jun Nakamura 1
1.Specker’theorem and slender groups 2.Non-commutative Specker’theorem and n-slender groups 3.Amalgamated free products of n-slender groups 4.Problems about n-slender groups 2
1.Specker’theorem and slender groups E.Specker(1950) ℎ : ℤ 𝜕 → ℤ a homomorphism. ℤ 𝜕 ℤ � ℎ �� � � � � � ∃ 𝑛 𝑞 𝑛 � � � � ∃ ℎ � � � ℤ 𝑛 𝑛 − 1 ∑ ℎ = ℎ ∘ 𝑞 𝑛 𝑞 𝑛 :projection. ℎ ( 𝑦 ) = 𝑦 ( 𝑗 ) ℎ ( 𝑓 𝑗 ) 𝑗 =0 𝑓 𝑗 :i-th component is 1 , other components are all zero. 3
𝑛 − 1 ∑ ∑ ∑ 𝑦 = 𝑦 ( 𝑗 ) 𝑓 𝑗 = 𝑦 ( 𝑗 ) 𝑓 𝑗 + 𝑦 ( 𝑗 ) 𝑓 𝑗 𝑗<𝜕 𝑗 =0 𝑛 ≤ 𝑗<𝜕 𝑛 − 1 ∑ ∑ ℎ ( 𝑦 ) = ℎ ( 𝑦 ( 𝑗 ) 𝑓 𝑗 ) + ℎ ( 𝑦 ( 𝑗 ) 𝑓 𝑗 ) 𝑗 =0 𝑛 ≤ 𝑗<𝜕 𝑛 − 1 ∑ ∑ = 𝑦 ( 𝑗 ) ℎ ( 𝑓 𝑗 ) + ℎ ( 𝑦 ( 𝑗 ) 𝑓 𝑗 ) 𝑗 =0 𝑛 ≤ 𝑗<𝜕 𝑛 − 1 ∑ = 𝑦 ( 𝑗 ) ℎ ( 𝑓 𝑗 ) 𝑗 =0 ℎ ( 𝑦 ) is determined by only finite components of 𝑦 . ℎ factors through a finitely generated free abelian group ℤ 𝑛 . 4
Slenderness was introduced by J.̷ Lo´ s. An abelian group 𝑇 is slender, if 𝑇 satisfies the following diagram. ℎ : ℤ 𝜕 → 𝑇 a homomorphism. ℤ 𝜕 𝑇 � ℎ �� � � � � � ∃ 𝑛 𝑞 𝑛 � � � � ∃ ℎ � ℎ = ℎ ∘ 𝑞 𝑛 � � ℤ 𝑛 A slender group 𝑇 satisfies Specker’theorem. ℤ is a typical example of slender groups. 5
Theorem (L.Fuchs) Direct sums of slender groups are slender. Theorem (R.J.Nunke) the characterization of slender groups. An abelian group is slender if and only if, it is torsion-free and contains no copy of ℚ , ℤ 𝜕 , or 𝑞 -adic integer group 𝕂 𝑞 for any prime 𝑞 . 6
2.Non-commutative Specker’theorem and n-slender groups G.Higman (1952) Let 𝐺 be a free group and ℎ : × × 𝑜<𝜕 ℤ 𝑜 → 𝐺 a homomorphism. × × 𝐺 𝑜<𝜕 ℤ 𝑜 � ℎ �� � � � � � ∃ 𝑛 𝑞 𝑛 � � � � ∃ ℎ � � � ∗ 𝑗<𝑛 ℤ 𝑗 ℎ = ℎ ∘ 𝑞 𝑛 𝑞 𝑛 : canonical projection 𝑜<𝜕 ℤ 𝑜 is the free complete product of copies of ℤ . × × It is isomorphic to the fundamental group of the Hawaiian earring. 7
n-slenderness was introduced by K.Eda in 1992. A group 𝑇 is n-slender if 𝐻 satisfies the following diagram. × × 𝑇 𝑜<𝜕 ℤ 𝑜 � ℎ �� � � � � � ∃ 𝑛 𝑞 𝑛 � � � � ∃ ℎ � ℎ = ℎ ∘ 𝑞 𝑛 � � ∗ 𝑗<𝑛 ℤ 𝑗 A n-slender group satisfies non-commutative Specker’theorem. ℤ is also a good example of n-slender groups. 8
Theorem (K.Eda) Let 𝐵 be an abelian group. 𝐵 is slender if and only if, 𝐵 is n-slender. Theorem (K.Eda) Let 𝐻 𝑗 ( 𝑗 ∈ 𝐽 ) be n-slender. Then, the free product ∗ 𝑗 ∈ 𝐽 𝐻 𝑗 and the restricted direct product ∏ 𝑠 𝑗 ∈ 𝐽 𝐻 𝑗 = { 𝑦 ∈ ∏ 𝑗 ∈ 𝐽 𝐻 𝑗 ∣{ 𝑗 ∈ 𝐽 ∣ 𝑦 ( 𝑗 ) ∕ = 𝑓 } is finite } are n-slender. 9
There is a characterization of n-slender groups using fundamen- tal groups. Theorem(K.Eda) 𝜌 1 ( 𝑌, 𝑦 ) is n-slender if and only if, for any homomorphism ℎ : 𝜌 1 ( ℍ , 𝑝 ) → 𝜌 1 ( 𝑌, 𝑦 ) , there exists a continuous map 𝑔 : ( ℍ , 𝑝 ) → ( 𝑌, 𝑦 ) such that ℎ = 𝑔 ∗ where 𝑔 ∗ is the induced homomorphism. 10
We can rephrase Higman’s theorem in topological terms as follows: Let ℎ be a homomorphism from 𝜌 1 ( ℍ , 𝑝 ) to 𝜌 1 ( 𝕋 1 ) . Then, there exists a continuous map 𝑔 : ℍ → 𝕋 1 such that ℎ = 𝑔 ∗ . 11
Many things about wild algebraic topology can be re- duced to the Hawaiian earring and how the homomorphic image of the fundamental group of the Hawaiian earring can detect a point in the space in question. It is due to the non-commutative Specker phenomenon. 12
Theorem (K.Eda) Let 𝑌 and 𝑍 be a one-dimensional Peano continua which are not semi-locally simply connected at any point. Then, 𝑌 and 𝑍 are homeomorphic if and only if, the fundamental groups of 𝑌 and 𝑍 are isomorphic. Theorem (K.Eda) Let 𝑌 and 𝑍 be one-dimensional Peano continua. If the fundamental groups of 𝑌 and 𝑍 are isomorphic, then 𝑌 and 𝑍 are homotopy equivalent. 13
3.Amalgamated free products of n-slender groups We try a generalization of the theorem that free prod- ucts of n-slender groups are n-slender. We partially succeed and find new n-slender groups. 14
Main Theorem If, 𝐻 = ∗ 𝑉 { 𝐻 𝑗 : 𝑗 ∈ 𝐽 } is a free product of 𝐻 𝑗 with amalgamated subgroup 𝑉 and satisfies (1) , (2); 𝑗 ∈ 𝐽 𝐻 𝑗 ∖ 𝑉 ( 2 / (1) ∀ ∈ ∪ ∈ 𝑉 ) (2) ∀ 𝐼 ≤ 𝐻 ( 𝐼 ⊆ 𝐷 1 → ∃ 𝑋 ∈ 𝐻 ∃ 𝑗 ∈ 𝐽 ( 𝐼 ≤ 𝑋 − 1 𝐻 𝑗 𝑋 )) then, for any homomorphism ℎ from × × 𝑜<𝜕 ℤ 𝑜 to 𝐻 , there exists × 𝑂 ≤ 𝑜<𝜕 ℤ 𝑜 ] ≤ 𝑋 − 1 𝐻 𝑗 𝑋 for an natural number 𝑂 such that ℎ [ × some 𝑋 ∈ 𝐻 and 𝑗 ∈ 𝐽 . 15
Corollary 1. The fundamental group of the closed orientable sur- face 𝑁 of genus is n-slender. Corollary 2. The fundamental group of the closed non-orientable surface 𝑂 of genus ≥ 4 is n-slender. 16
How to apply the main theorem It is well known that 𝜌 1 ( 𝑁 ) = < 𝑦 1 , ⋅ ⋅ ⋅ , 𝑦 2 ∣ [ 𝑦 1 , 𝑦 2 ] ⋅ ⋅ ⋅ [ 𝑦 2 − 1 , 𝑦 2 ] > . 𝐻 0 = < 𝑦 1 > ∗ < 𝑦 2 > , 𝑉 0 is the subgroup of 𝐻 0 generated by [ 𝑦 1 , 𝑦 2 ], 𝐻 1 = < 𝑦 3 > ∗ ⋅ ⋅ ⋅ ∗ < 𝑦 2 > and 𝑉 1 is the subgroup of 𝐻 1 gener- ated by [ 𝑦 3 , 𝑦 4 ] ⋅ ⋅ ⋅ [ 𝑦 2 − 1 , 𝑦 2 ]. We amalgamate 𝑉 0 and 𝑉 1 according to the isomorphism which maps [ 𝑦 1 , 𝑦 2 ] to ([ 𝑦 3 , 𝑦 4 ] ⋅ ⋅ ⋅ [ 𝑦 2 − 1 , 𝑦 2 ]) − 1 . Cleary, such an amalgamated free product 𝐻 0 ∗ ℤ 𝐻 1 is equal to 𝜌 1 ( 𝑁 ). 17
𝜌 1 ( 𝑂 ) = < 𝑦 1 , ⋅ ⋅ ⋅ , 𝑦 ∣ 𝑦 1 𝑦 1 ⋅ ⋅ ⋅ 𝑦 𝑦 > . 𝐻 0 = < 𝑦 1 > ∗ < 𝑦 2 > , 𝑉 0 is the subgroup of 𝐻 0 generated by 𝑦 1 𝑦 1 𝑦 2 𝑦 2 , 𝐻 1 = < 𝑦 3 > ∗ ⋅ ⋅ ⋅ ∗ < 𝑦 > and 𝑉 1 is the subgroup of 𝐻 1 generated by 𝑦 3 𝑦 3 ⋅ ⋅ ⋅ 𝑦 𝑦 . We amalgamate 𝑉 0 and 𝑉 1 according to the isomorphism which maps 𝑦 1 𝑦 1 𝑦 2 𝑦 2 to ( 𝑦 3 𝑦 3 ⋅ ⋅ ⋅ 𝑦 𝑦 ) − 1 . Such an amalgamated free product 𝐻 0 ∗ ℤ 𝐻 1 is equal to 𝜌 1 ( 𝑂 ). 18
We modify the proof of Higman’s theorem to prove the main theorem. Now, we explain a basic idea of the proof of Higman’s theorem and prepare some lemmas. 19
Propotision 1. (K.Eda) If 𝜇 ( 𝜇 ∈ Λ) are elements of × × 𝑗 ∈ 𝐽 𝐻 𝑗 such that { 𝜇 ∈ Λ ∣ 𝑚 𝑗 ( 𝜇 ) ∕ = 0 } is finite for all 𝑗 ∈ 𝐽 , then there exists a natural homomorphism 𝜒 : × × 𝜇 ∈ Λ ℤ 𝜇 → × × 𝑗 ∈ 𝐽 𝐻 𝑗 via 𝜀 𝜇 �→ 𝜇 ( 𝜇 ∈ Λ) where 𝜀 𝜇 is 1 of ℤ 𝜇 . Remark Let 𝑜 ∈ ℤ 𝜕 ( 𝑜 < 𝜕 ) such that { 𝑜 ∣ 𝑜 ( 𝑗 ) ∕ = 0 } is finite for any 𝑗 < 𝜕 . Then, we define a homomorphism 𝜒 : ℤ 𝜕 → ℤ 𝜕 by 𝜒 ( 𝑦 ) = 𝑗<𝜕 𝑦 ( 𝑗 ) 𝑗 which maps 𝑓 𝑗 to 𝑗 for any 𝑗 < 𝜕 . ∑ 20
Proposition 2. (K.Eda) 𝑇 is n-slender if and only if, for any homomorphism ℎ : × 𝑜<𝜕 ℤ 𝑜 → 𝑇 , the set { 𝑜 < 𝜕 ∣ ℎ ( 𝜀 𝑜 ) ∕ = 𝑓 } × is finite. 21
Lemma 1. (due to G.Higman.) Let 𝐺 be a free group, 𝑦 𝑜 , 𝑧 𝑜 ∈ 𝐺 and 𝑔 ∈ 𝜕 𝜕 such that 𝑔 ( 𝑜 ) ≥ 𝑚 ( 𝑦 𝑜 ) + 2 and 𝑧 𝑜 = 𝑦 𝑜 𝑧 𝑔 ( 𝑜 ) 𝑜 +1 for any 𝑜 < 𝜕 . Then, there exists 𝑛 such that 𝑧 𝑜 = 𝑓 for any 𝑜 ≥ 𝑛 . 22
Proof of Lemma 1. Assume not. Let 𝑜 be a natural number such that 𝑚 ( 𝑧 0 ) ≤ 𝑜 and 𝑧 𝑜 +1 ∕ = 𝑓 . 𝑚 ( 𝑧 𝑔 ( 𝑜 ) 𝑜 +1 ) ≥ 𝑚 ( 𝑧 𝑜 +1 ) + 𝑔 ( 𝑜 ) − 1 ≥ 𝑚 ( 𝑧 𝑜 +1 ) + 𝑚 ( 𝑦 𝑜 ) + 1, 𝑧 𝑜 = 𝑦 𝑜 𝑧 𝑔 ( 𝑜 ) 𝑜 +1 𝑚 ( 𝑧 𝑜 ) ≥ 𝑚 ( 𝑧 𝑜 +1 ) + 1. We can repeat the argument and find out 𝑚 ( 𝑧 0 ) ≥ 𝑜 + 1. It is a contradiction. □ 23
Now, we will prove the Higman’s theorem. × × 𝐺 𝑜<𝜕 ℤ 𝑜 � ℎ �� � � � � � ∃ 𝑛 𝑞 𝑛 � � � � ∃ ℎ � � � ∗ 𝑗<𝑛 ℤ 𝑗 ℎ = ℎ ∘ 𝑞 𝑛 𝑞 𝑛 : canonical projection 24
Proof of the Higman’s theorem. Assume a free group 𝐺 is not n-slender. By Proposition 1 and 2, there exists a homomorphism ℎ such that ℎ ( 𝜀 𝑜 ) ∕ = 𝑓 for all 𝑜 < 𝜕 . Let 𝑔 : 𝜕 → 𝜕 such that 𝑔 ( 𝑜 ) = 𝑚 ( ℎ ( 𝜀 𝑜 )) + 2. We construct infinite words from 𝑔 and 𝜀 𝑜 . 25
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