Amalgamated free products of n-slender groups RIMS Set Theory - - PowerPoint PPT Presentation

Amalgamated free products

• f n-slender groups

RIMS Set Theory Workshop 2010 Waseda University Jun Nakamura

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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

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1.Specker’theorem and slender groups

E.Specker(1950) ℎ : ℤ𝜕 → ℤ a homomorphism. ℤ𝜕 ℤ𝑛 ℤ

• ℎ

∃ℎ ∃𝑛 𝑞𝑛 ℎ = ℎ ∘ 𝑞𝑛 𝑞𝑛:projection. ℎ(𝑦) =

𝑛−1

∑

𝑗=0

𝑦(𝑗)ℎ(𝑓𝑗) 𝑓𝑗:i-th component is 1 , other components are all zero.

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𝑦 =

∑

𝑗<𝜕

𝑦(𝑗)𝑓𝑗 =

𝑛−1

∑

𝑗=0

𝑦(𝑗)𝑓𝑗 +

∑

𝑛≤𝑗<𝜕

𝑦(𝑗)𝑓𝑗 ℎ(𝑦) = ℎ(

𝑛−1

∑

𝑗=0

𝑦(𝑗)𝑓𝑗) + ℎ(

∑

𝑛≤𝑗<𝜕

𝑦(𝑗)𝑓𝑗) =

𝑛−1

∑

𝑗=0

𝑦(𝑗)ℎ(𝑓𝑗) + ℎ(

∑

𝑛≤𝑗<𝜕

𝑦(𝑗)𝑓𝑗) =

𝑛−1

∑

𝑗=0

𝑦(𝑗)ℎ(𝑓𝑗) ℎ factors through a finitely generated free abelian group ℤ𝑛. ℎ(𝑦) is determined by only finite components of 𝑦.

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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𝜕 ℤ𝑛 𝑇

• ℎ

∃ℎ ∃𝑛 𝑞𝑛 ℎ = ℎ ∘ 𝑞𝑛 A slender group 𝑇 satisfies Specker’theorem. ℤ is a typical example of slender groups.

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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 𝕂𝑞 for any prime 𝑞.

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2.Non-commutative Specker’theorem and n-slender groups G.Higman (1952) Let 𝐺 be a free group and ℎ : × ×

𝑜<𝜕ℤ𝑜 → 𝐺

a homomorphism. × ×

𝑜<𝜕ℤ𝑜

∗𝑗<𝑛ℤ𝑗 𝐺 ℎ ∃ℎ ∃𝑛 𝑞𝑛

• ℎ = ℎ ∘ 𝑞𝑛

𝑞𝑛: canonical projection × ×

𝑜<𝜕ℤ𝑜 is the free complete product of copies of ℤ.

It is isomorphic to the fundamental group of the Hawaiian earring.

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n-slenderness was introduced by K.Eda in 1992. A group 𝑇 is n-slender if 𝐻 satisfies the following diagram. × ×

𝑜<𝜕ℤ𝑜

∗𝑗<𝑛ℤ𝑗 𝑇 ℎ ∃ℎ ∃𝑛 𝑞𝑛

• ℎ = ℎ ∘ 𝑞𝑛

A n-slender group satisfies non-commutative Specker’theorem. ℤ is also a good example of n-slender groups.

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Theorem(K.Eda) Let 𝐵 be an abelian group. 𝐵 is slender if and only if, 𝐵 is n-slender. Theorem(K.Eda) Let 𝐻𝑗(𝑗 ∈ 𝐽) be n-slender. Then, the free product ∗𝑗∈𝐽𝐻𝑗 and the restricted direct product ∏𝑠

𝑗∈𝐽 𝐻𝑗 = {𝑦 ∈ ∏ 𝑗∈𝐽 𝐻𝑗∣{𝑗 ∈

𝐽∣𝑦(𝑗) ∕= 𝑓} is finite } are n-slender.

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There is a characterization of n-slender groups using fundamen- tal groups. Theorem(K.Eda) 𝜌1(𝑌, 𝑦) is n-slender if and only if, for any homomorphism ℎ : 𝜌1(ℍ, 𝑝) → 𝜌1(𝑌, 𝑦), there exists a continuous map 𝑔 : (ℍ, 𝑝) → (𝑌, 𝑦) such that ℎ = 𝑔∗ where 𝑔∗ is the induced homomorphism.

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We can rephrase Higman’s theorem in topological terms as follows: Let ℎ be a homomorphism from 𝜌1(ℍ, 𝑝) to 𝜌1(𝕋1). Then, there exists a continuous map 𝑔 : ℍ → 𝕋1 such that ℎ = 𝑔∗.

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Many things about wild algebraic topology can be re- duced to the Hawaiian earring and how the homomorphic image of the fundamental group

• f the Hawaiian earring can detect a point in the space

in question. It is due to the non-commutative Specker phenomenon.

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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.

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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.

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Main Theorem If, 𝐻 = ∗

𝑉{𝐻𝑗 : 𝑗 ∈ 𝐽} is a free product of 𝐻𝑗 with amalgamated

subgroup 𝑉 and satisfies (1), (2); (1)∀𝑕 ∈ ∪

𝑗∈𝐽 𝐻𝑗 ∖ 𝑉 (𝑕2 /

∈ 𝑉) (2)∀𝐼 ≤ 𝐻 (𝐼 ⊆ 𝐷1 → ∃𝑋 ∈ 𝐻∃𝑗 ∈ 𝐽(𝐼 ≤ 𝑋 −1𝐻𝑗𝑋)) then, for any homomorphism ℎ from × ×

𝑜<𝜕ℤ𝑜 to 𝐻 , there exists

an natural number 𝑂 such that ℎ[× ×𝑂≤𝑜<𝜕ℤ𝑜] ≤ 𝑋 −1𝐻𝑗𝑋 for some 𝑋 ∈ 𝐻 and 𝑗 ∈ 𝐽.

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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.

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How to apply the main theorem It is well known that 𝜌1(𝑁𝑕) =< 𝑦1, ⋅ ⋅ ⋅ , 𝑦2𝑕∣[𝑦1, 𝑦2] ⋅ ⋅ ⋅ [𝑦2𝑕−1, 𝑦2𝑕] >. 𝐻0 =< 𝑦1 > ∗ < 𝑦2 >, 𝑉0 is the subgroup of 𝐻0 generated by [𝑦1, 𝑦2], 𝐻1 =< 𝑦3 > ∗ ⋅ ⋅ ⋅ ∗ < 𝑦2𝑕 > and 𝑉1 is the subgroup of 𝐻1 gener- ated by [𝑦3, 𝑦4] ⋅ ⋅ ⋅ [𝑦2𝑕−1, 𝑦2𝑕]. We amalgamate 𝑉0 and 𝑉1 according to the isomorphism which maps [𝑦1, 𝑦2] to ([𝑦3, 𝑦4] ⋅ ⋅ ⋅ [𝑦2𝑕−1, 𝑦2𝑕])−1. Cleary, such an amalgamated free product 𝐻0 ∗

ℤ 𝐻1 is equal to

𝜌1(𝑁𝑕).

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𝜌1(𝑂𝑕) =< 𝑦1, ⋅ ⋅ ⋅ , 𝑦𝑕∣𝑦1𝑦1 ⋅ ⋅ ⋅ 𝑦𝑕𝑦𝑕 >. 𝐻0 =< 𝑦1 > ∗ < 𝑦2 >, 𝑉0 is the subgroup of 𝐻0 generated by 𝑦1𝑦1𝑦2𝑦2, 𝐻1 =< 𝑦3 > ∗ ⋅ ⋅ ⋅ ∗ < 𝑦𝑕 > and 𝑉1 is the subgroup of 𝐻1 generated by 𝑦3𝑦3 ⋅ ⋅ ⋅ 𝑦𝑕𝑦𝑕. We amalgamate 𝑉0 and 𝑉1 according to the isomorphism which maps 𝑦1𝑦1𝑦2𝑦2 to (𝑦3𝑦3 ⋅ ⋅ ⋅ 𝑦𝑕𝑦𝑕)−1. Such an amalgamated free product 𝐻0 ∗

ℤ 𝐻1 is equal to 𝜌1(𝑂𝑕).

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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.

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Propotision 1.(K.Eda) If 𝑕𝜇(𝜇 ∈ Λ) are elements of × ×

𝑗∈𝐽𝐻𝑗 such that {𝜇 ∈ Λ ∣ 𝑚𝑗(𝑕𝜇) ∕= 0}

is finite for all 𝑗 ∈ 𝐽, then there exists a natural homomorphism 𝜒 :× ×𝜇∈Λℤ𝜇 → × ×𝑗∈𝐽𝐻𝑗 via 𝜀𝜇 → 𝑕𝜇 (𝜇 ∈ Λ) where 𝜀𝜇 is 1 of ℤ𝜇. 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 𝑓𝑗 to 𝑕𝑗 for any 𝑗 < 𝜕.

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Proposition 2.(K.Eda) 𝑇 is n-slender if and only if, for any homomorphism ℎ : × ×

𝑜<𝜕ℤ𝑜 → 𝑇, the set {𝑜 < 𝜕∣ℎ(𝜀𝑜) ∕= 𝑓}

is finite.

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Lemma 1.(due to G.Higman.) Let 𝐺 be a free group, 𝑦𝑜, 𝑧𝑜 ∈ 𝐺 and 𝑔 ∈ 𝜕𝜕 such that 𝑔(𝑜) ≥ 𝑚(𝑦𝑜) + 2 and 𝑧𝑜 = 𝑦𝑜𝑧𝑔(𝑜)

𝑜+1 for any 𝑜 < 𝜕.

Then, there exists 𝑛 such that 𝑧𝑜 = 𝑓 for any 𝑜 ≥ 𝑛.

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Proof of Lemma 1. Assume not. Let 𝑜 be a natural number such that 𝑚(𝑧0) ≤ 𝑜 and 𝑧𝑜+1 ∕= 𝑓. 𝑚(𝑧𝑔(𝑜)

𝑜+1) ≥ 𝑚(𝑧𝑜+1) + 𝑔(𝑜) − 1 ≥ 𝑚(𝑧𝑜+1) + 𝑚(𝑦𝑜) + 1, 𝑧𝑜 = 𝑦𝑜𝑧𝑔(𝑜) 𝑜+1

𝑚(𝑧𝑜) ≥ 𝑚(𝑧𝑜+1) + 1. We can repeat the argument and find out 𝑚(𝑧0) ≥ 𝑜 + 1. It is a contradiction. □

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Now, we will prove the Higman’s theorem. × ×

𝑜<𝜕ℤ𝑜

∗𝑗<𝑛ℤ𝑗 𝐺 ℎ ∃ℎ ∃𝑛 𝑞𝑛

• ℎ = ℎ ∘ 𝑞𝑛

𝑞𝑛: canonical projection

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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(𝜀𝑜) ∕= 𝑓 for all 𝑜 < 𝜕. Let 𝑔 : 𝜕 → 𝜕 such that 𝑔(𝑜) = 𝑚(ℎ(𝜀𝑜)) + 2. We construct infinite words from 𝑔 and 𝜀𝑜.

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Firstly, we construct a tree using 𝑔. Let 𝑈𝑜 = {𝑡 ∈ 𝑜𝜕∣𝑡(0) = 0 ∧ 1 ≤ ∀𝑗 < 𝑜(𝑡(𝑗) < 𝑔(𝑗 − 1))} and 𝑈 = ∪

0<𝑜<𝜕 𝑈𝑜.

𝑈 is linear ordered by the lexicographical order. We define a word 𝑊0 : 𝑊0 → ∪

𝑜<𝜕 ℤ𝑜 such that

𝑊0 = 𝑈 and 𝑊0(𝑦) = 𝜀𝑜 iff 𝑦 ∈ 𝑈𝑜+1.

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𝜀2 ⋅ ⋅ ⋅ ... 𝜀1 . . . 𝜀2 ⋅ ⋅ ⋅ ... 𝑊0 ≡ 𝜀0 . . . 𝜀2 ⋅ ⋅ ⋅ ... 𝜀1 . . . 𝜀2 ⋅ ⋅ ⋅ ...

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𝜀𝑜+2 ⋅ ⋅ ⋅ ... 𝜀𝑜+1 . . . 𝜀𝑜+2 ⋅ ⋅ ⋅ ... 𝑊𝑜 ≡ 𝜀𝑜 . . . 𝜀𝑜+2 ⋅ ⋅ ⋅ ... 𝜀𝑜+1 . . . 𝜀𝑜+2 ⋅ ⋅ ⋅ ...

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By this construction, 𝑊𝑜 = 𝜀𝑜𝑊 𝑔(𝑜)

𝑜+1 for all 𝑜 < 𝜕.

Then, ℎ(𝑊𝑜) = ℎ(𝜀𝑜)ℎ(𝑊𝑜+1)𝑔(𝑜) for all 𝑜 < 𝜕. By lemma 1, there exists 𝑂 such that ∀𝑜 ≥ 𝑂(ℎ(𝑊𝑜) = 𝑓). It implies ℎ(𝜀𝑂) = 𝑓, which is a contradiction. □

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The key of this proof is lemma 1 and the words 𝑊𝑜. The idea is often used to show the n-slenderness and related things.

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We modify lemma 1 for amalgamated free products. Lemma 2. Let 𝑦𝑜, 𝑧𝑜 ∈ ∗

𝑉{𝐻𝑗 : 𝑗 ∈ 𝐽}(𝑜 < 𝜕) , 𝑔 ∈ 𝜕𝜕 and 𝑧𝑜 = 𝑦𝑜𝑧𝑔(𝑜) 𝑜+1 for

any natural number 𝑜. If 1

2(∑ 𝑗≤𝑜 𝑚(𝑦𝑗) + 𝑜) + 1 ≤ 𝑔(𝑜) and 𝑚(𝑧𝑜+1) ≤ 𝑚(𝑧𝑔(𝑜) 𝑜+1) for any

natural number 𝑜, then there exists a natural number 𝑛 such that 𝑧𝑜 ∈ 𝐷1 for any 𝑜 ≥ 𝑛.

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Remark. 𝐷1 := {𝑊 −1𝑕𝑊 ∣𝑊 ∈ ∗

𝑉{𝐻𝑗 : 𝑗 ∈ 𝐽} ∧ 𝑕 ∈ ∪ 𝑗∈𝐽 𝐻𝑗}

𝐷2 := {𝑦𝑧∣𝑦, 𝑧 ∈ 𝐷1} It is clear that If 𝑦 ∈ 𝐷1, then 𝑚(𝑦𝑜) ≤ 𝑚(𝑦) and 𝑦𝑜 ∈ 𝐷1 for any 𝑜 ≥ 1.

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Proof of the main theorem.(rough sketch) We prove by contradiction. By Proposition 1 and some lemmas, there exists a homomorphism ℎ such that ℎ(𝜀𝑜) / ∈ 𝐷2 for any 𝑜 < 𝜕. Let 𝑔 ∈ 𝜕𝜕 be satisfying the assumption of lemma 2. We construct words 𝑊𝑜 (𝑜 < 𝜕) from 𝜀𝑜 and 𝑔 as before. ie; 𝑊𝑜 = 𝜀𝑜𝑊 𝑔(𝑜)

𝑜+1 for any 𝑜 < 𝜕.

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Applying lemma 2, we find 𝑂 such that ℎ(𝑊𝑂), ℎ(𝑊𝑂+1) ∈ 𝐷1. Since ℎ(𝑊𝑂) = ℎ(𝜀𝑂)ℎ(𝑊𝑂+1)𝑔(𝑂) and ℎ(𝜀𝑂) / ∈ 𝐷2, it is a contradiction.

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4.Problems of n-slender groups.

Question 1. Are finitely generated torsion-free groups n-slender ? It is true in abelian case. Because, such an abelian group is free by the fundamental the-

• rem of finitely generated abelian groups.

Especially, we are interested in 𝜌1(𝑂2) and 𝜌1(𝑂3).

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Question 2. Are amalgamated free products of n-slender groups also n-slender ? The main theorem answers only a part of this question. Can we drop the assumptions of the theorem ? For free products, it is proved. Theorem(K.Eda) Let ℎ be a homomorphism from × ×

𝑜<𝜕ℤ𝑜 to ∗𝑗∈𝐽𝐻𝑗.

Then, ℎ[× ×

𝑂≤𝑜<𝜕ℤ𝑜] ≤ 𝑋 −1𝐻𝑗𝑋 for some 𝑂 < 𝜕, 𝑋 ∈ ∗𝑗∈𝐽𝐻𝑗, 𝑗 ∈

𝐽.

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Question 3. What is a group theoretic characterization of n-slender groups ? Theorem(abelian case) 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 𝕂𝑞 for any prime 𝑞.

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References

[1]K.Eda, Free 𝜏-products and noncommutatively slender groups J.Algebra 148(1992) 243-263 [2]K.Eda, The fundamental groups of one-dimensional spaces and spatial homomorphisms Topology and its Applications 123(2002) 479-505 [3]K.Eda Atomic property of the fundamental groups of the Hawaiian earring and wild locally path-connected spaces to appear in Jour.Math.Soc.Japan. [4]K.Eda Homotopy types of one- dimensional Peano continua Fund.Math. 209(2010) 27-42 [5]P.C.Eklof and A.H.Mekler Almost free modules North-Holland, 1990. [6]L.Fuchs Infnite Abelian Groups Vol.2, Academic Press, New York. (1970) [7]M.Hall, Jr The theory of groups Chelsea Pub. Co., 1959 [8]G.Higman, Unrestricted free pruducts and varieties of topological groups 73-81 J.London Math.Soc.27(1952) [9]R.C. Lyndon and P.E. Schupp Combinatorial Group Theory Ergebn. Math. Grenzgeb. ,Springer(1977) 38