On Thompson's group F and its group algebra Tsunekazu Nishinaka* - - PowerPoint PPT Presentation

On Thompson's group F and its group algebra

Tsunekazu Nishinaka* (University of Hyogo)

*Partially supported by Grants-in-Aid for Scientific Research under grant no. 17K05207

The 8th China-Japan-Korea International Symposium on Ring Theory Nagoya University, Japan, Mon. Aug. 26, 2019

▶ Finally, in order to be able to investigate the group algebra of this group, we use a ‘directed’ two edge-colored graph and improve our method. In this talk, ▶ we first introduce an application of (undirected) two edge-colored graphs to group algebras of groups which have non-abelian free subgroups.

We have used these graphs to study primitivity of group algebras of non-Noetherian groups, where generally a ring R is right primitive if it has a faithful irreducible right R-module . Our method using two edge-colored graphs seems to be effective to investigate a group algebra if its group has non-abelian free subgroups. But there exist some non-Noetherian groups with no non-abelian free subgroups; for example Thompson’s group F and a free Burnside group of large exponent.

▶ Next we introduce briefly Thompson’s group F and consider amenability of it.

A cycle in the graph is called an alternating cycle if its edges belong alternatively to E and F. For example, f1 e3 f2 e5 f3 e7 Two-edge coloured graphs E = {e1, e2, …, em } F = {f1, f2, …, fm }

v1 v2 v3 v4 e1 f1 e3 v5 f2 e5 f3 e7 f4

A two-edge colored graph is a simple graph each of whose edges colored with one of two different colors. V = { v1, v2, …, vn } is a vertex set, E and F are two edge sets;

SR-graphs E = {e1, e2, …, em } F = {f1, f2, …, fm } A two-edge colored graph S = (V, E, F ) is an SR-graph if every component of G = (V, E) is a complete graph.

I(G)= {𝑤3, 𝑤6}

v1 v2 v3 v4 e1 f1 e3 v5 f2 e5 f3 e7 f4

𝑤6

In an SR-graph, we call an alternating cycle an SR-cycle; for example, f1 e3 f2 e5 f3e7 . Complete graphs

An introduction to an application of SR-graph theory We begin with the following simple problem. Find elements 𝐵, 𝐶 ∈ 𝐿𝐻∗ such that 𝐵𝑌 + 𝐶𝑍 ≠ 0 for any 𝑌, 𝑍 ∈ 𝐿𝐻∗. Problem 1 Let G be a group and KG the group algebra of G over a field K. If G has a non-abelian free subgroup, then we can find this kind of elements. We denote 𝐿𝐻 ∖ {0} , the non-zero elements in 𝐿𝐻, by 𝐿𝐻∗.

Since 𝑌, 𝑍 ∈ 𝐿𝐻∗, they are expressed as follows: 𝑌 = σ𝑦∈𝑇𝑌 𝛽𝑦𝑦, 𝑍 = σ𝑧∈𝑇𝑍 𝛾𝑧𝑧, Since 𝐵𝑌 + 𝐶𝑍 = 0, we have ෍

𝑦∈𝑇𝑌

𝛽𝑦(𝑏1𝑦 + 𝑏2𝑦) + ෍

𝑧∈𝑇𝑍

𝛾𝑧(𝑐1𝑧 + 𝑐2𝑧) = 0. We would like to regard these elements 𝑏𝑗𝑦 and 𝑐𝑗𝑧 as vertices. Because of that, we need to distinguish all these elements even if for 𝑗 ≠ 𝑘, 𝑏𝑗𝑦 = 𝑏𝑘𝑦′, 𝑐𝑗𝑧 = 𝑐

𝑘𝑧′ or 𝑏𝑗𝑦 = 𝑐 𝑘𝑧 in G.

So we define the vertex set and two edge set as follows: 𝑊 = 𝑏𝑗, 𝑦 , 𝑐𝑗, 𝑧 | 𝑗 = 1, 2, 𝑦 ∈ 𝑇𝑌, 𝑧 ∈ 𝑇𝑍 𝐺 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = 𝑏1, 𝑦 , 𝑥 = (𝑏2, 𝑦) or 𝑤 = 𝑐1, 𝑧 , 𝑥 = (𝑐2, 𝑧) } 𝐹 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = [𝑥] in G }, where 𝑤 = 𝑏𝑦 if 𝑤 = 𝑏, 𝑦 . Let G be a group which has a nonabelian free subgroup. In this case, G has always a free subgroup of infinite rank: 𝑏1, 𝑏2, 𝑐1, 𝑐2, ⋯ . Let 𝐵 = 𝑏1 + 𝑏2 and 𝐶 = 𝑐1 + 𝑐2. Suppose, to the contrary, that 𝐵𝑌 + 𝐶𝑍 = 0 for some 𝑌, 𝑍 ∈ 𝐿𝐻∗. where 𝛽𝑦, 𝛾𝑧 ∈ 𝐿 ∖ 0 , 𝑇𝑌= 𝑇𝑣𝑞𝑞 𝑌 and 𝑇𝑍 = 𝑇𝑣𝑞𝑞 𝑍 .

෍

𝑦∈𝑇𝑌

𝛽𝑦(𝑏1𝑦 + 𝑏2𝑦) + ෍

𝑧∈𝑇𝑍

𝛾𝑧(𝑐1𝑧 + 𝑐2𝑧) = 0, Hence (𝑑1, 𝑨1) ∈ {𝑏1, 𝑏2} × 𝑇𝑌, ∃ 𝑑2, 𝑨2 ∈ {𝑏1, 𝑏2} × 𝑇𝑌) ∪ ({𝑐1, 𝑐2} × 𝑇𝑍 , 𝑑1𝑨1 = 𝑑2𝑨2, If 𝑑2 = 𝑏𝑗(resp. 𝑑2 = 𝑐𝑗), then for 𝑗 ≠ 𝑘, 𝑏𝑘𝑨2 (resp. 𝑐

𝑘𝑨2) exists in the above expression, and so

𝑑3𝑨2 = 𝑑4𝑨3, ∃𝑑3 ∈ {𝑏1, 𝑏2, 𝑐1, 𝑐2} with 𝑑3 ≠ 𝑑2 and ∃ 𝑑4, 𝑨3 ∈ {𝑏1, 𝑏2} × 𝑇𝑌) ∪ ({𝑐1, 𝑐2} × 𝑇𝑍 , 𝑑𝑛𝑨𝑚 = 𝑑𝑛+1𝑨1, ⋮ 𝑑1

−1𝑑2𝑑3 −1𝑑4 ⋯ 𝑑𝑛 −1𝑑𝑛+1 = 1,

where 𝑑𝑗 ∈ {𝑏1, 𝑏2, 𝑐1, 𝑐2} and 𝑑𝑗 ≠ 𝑑𝑗+1. Since {𝑏1, 𝑏2, 𝑐1, 𝑐2} is a free basis, this implies a contradiction. ∎

v1=𝑑1𝑨1, 𝑑1 = 𝑏𝑗 v2=𝑑2𝑨2,

𝑑2 = 𝑏𝑗 or 𝑐𝑗

v3=𝑑3𝑨2 v4=𝑑4𝑨3 vm=𝑑𝑛𝑨𝑚 𝑊 = 𝑏𝑗, 𝑦 , 𝑐𝑗, 𝑧 | 𝑗 = 1, 2, 𝑦 ∈ 𝑇𝑌, 𝑧 ∈ 𝑇𝑍 𝐺 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = 𝑏1, 𝑦 , 𝑥 = (𝑏2, 𝑦) or 𝑤 = 𝑐1, 𝑧 , 𝑥 = (𝑐2, 𝑧) } 𝐹 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = [𝑥] in G }, where 𝑤 = 𝑏𝑦 if 𝑤 = 𝑏, 𝑦 . Since all elements of 𝐻 in this equation are cancelled each other. We have thus seen that 𝐵 = 𝑏1 + 𝑏2, 𝐶 = 𝑐1 + 𝑐2 ⟹ 𝐵𝑌 + 𝐶𝑍 ≠ 0 for any 𝑌, 𝑍 ∈ 𝐿𝐻∗.

Problem 1 is strongly connected with amenability of groups. ∃𝐵, 𝐶 ∈ 𝐿𝐻∗, ∀𝑌, 𝑍 ∈ 𝐿𝐻∗, 𝐵𝑌 + 𝐶𝑍 ≠ 0 ⟹ 𝐻 is not amenable. ▶ 𝐻 has a non-abelian free subgroup ⟹ 𝐻 is not amenable. ▶(Definition) 𝐻 is amenable if for 𝑄 𝐻 = 𝑇 𝑇 ⊆ 𝐻}, ∃𝜈: 𝑄 𝐻 → [0, 1] such that 1.𝜈 𝐻 = 1.

• 2. If 𝑇 and 𝑈 are disjoint subsets of 𝐻, then 𝜈 𝑇 ∪ 𝑈 = 𝜈 𝑇 + 𝜈(𝑈)
• 3. If 𝑇 ∈ 𝑄 𝐻 and 𝑕 ∈ 𝐻, then 𝜈 𝑕𝑇 = 𝜈 𝑇 .

▶Finite groups and abelian groups are amenable. ▶The Burnside group 𝐶(𝑛, 𝑜) is not amenable if 𝑛 > 1 and 𝑜 is enough large. ▶Is Thompson’s group 𝐺 is amenable? ▶ 𝐺 is amenable⟹ ∀𝐵, 𝐶 ∈ 𝐿𝐻∗, ∃𝑌, 𝑍 ∈ 𝐿𝐻∗, 𝐵𝑌 = 𝐶𝑍 for any 𝑌, 𝑍 ∈ 𝐿𝐻∗.

Thompson’s group F F = 𝑦0, 𝑦1, ⋯ , 𝑦𝑗, ⋯ | 𝑦𝑗

−1𝑦𝑘𝑦𝑗 = 𝑦𝑘+1, for 𝑗 < 𝑘 .

= 𝑦0, 𝑦1 [𝑦0 𝑦1

−1, 𝑦0 −1𝑦1𝑦0 , [𝑦0 𝑦1 −1, 𝑦0 −2𝑦1𝑦0 2] .

▶ 𝐺 is non-noetherian ▶ 𝐺 has no non-abelian free subgroups. ▶ 𝐺 is a torsion free group and includes a free subsemigroup. ▶ ∃𝑈 ⊃ 𝐺 such that 𝑈 is simple.

We need to improve our graph theory so as to be effective for Thompson group F; generally for a non-Noetherian group with no free subgroup.

• 5. Improvement on SR-graph theory

Let 𝐵1 = 𝑏1𝑏2

−1 and 𝐶1 = 𝑐1𝑐2 −1.

𝐵1𝐶1𝐶1𝐵1

−1𝐶1 −1 = 1

Generally, it may happen that 𝐵1

±𝛽1𝐶1 ±𝛾1 ⋯ 𝐵1 ±𝛽𝑛𝐶1 ±𝛾𝑛 = 1

We have to choose 𝐵1 and 𝐶1 so as to 𝐵1𝐶1𝐶1𝐵1

−1𝐶1 −1 ≠ 1

According to our method, we have to choose 𝐵1 and 𝐶1 so as to 𝐵1

±𝛽1𝐶1 ±𝛾1 ⋯ 𝐵1 ±𝛽𝑛𝐶1 ±𝛾𝑛 ≠ 1

We considered the following SR-graph. We set here 𝐵 = 𝑏1 + 𝑏2 + 𝑏3 and 𝐶 = 𝑐1 + 𝑐2 + 𝑐3. 𝑏1𝑦1 𝑏2𝑦1 𝑏3𝑦1 𝑏1𝑦2 𝑏2𝑦2 𝑏3𝑦2 𝑐2𝑧1 𝑐1𝑧1 𝑐3𝑧1 𝑐1𝑧3 𝑐1𝑧2 𝑐2𝑧3 𝑐2𝑧2 𝑐3𝑧3 𝑐3𝑧2

𝑏1𝑦1 𝑏2𝑦1 𝑏3𝑦1 𝑏1𝑦2 𝑏2𝑦2 𝑏3𝑦2 𝑐2𝑧1 𝑐1𝑧1 𝑐3𝑧1 𝑐1𝑧3 𝑐1𝑧2 𝑐2𝑧3 𝑐2𝑧2 𝑐3𝑧3 𝑐3𝑧2 We replace an undirected SR-graph with a directed one.

𝑏1𝑦1 𝑏2𝑦1 𝑏3𝑦1 𝑏1𝑦2 𝑏2𝑦2 𝑏3𝑦2 𝑐2𝑧1 𝑐1𝑧1 𝑐3𝑧1 𝑐1𝑧3 𝑐1𝑧2 𝑐2𝑧3 𝑐2𝑧2 𝑐3𝑧3 𝑐3𝑧2 In this graph, an SR-cycle means a cycle along the direction of arrows.

𝑏1𝑦1 𝑏2𝑦1 𝑏3𝑦1 𝑏1𝑦2 𝑏2𝑦2 𝑏3𝑦2 𝑐2𝑧1 𝑐1𝑧1 𝑐3𝑧1 𝑐1𝑧3 𝑐1𝑧2 𝑐2𝑧3 𝑐2𝑧2 𝑐3𝑧3 𝑐3𝑧2 𝑏1𝑦3 𝑏2𝑦3 𝑏3𝑦3 Let 𝐵1 = 𝑏1𝑏2

−1, 𝐵2= 𝑏2𝑏3 −1, 𝐶1 = 𝑐1𝑐2 −1 and 𝐶3 = 𝐶3𝐶1 −1.

𝐵1𝐶1𝐶1𝐵2𝐵1𝐶3 = 1; We have only to choose 𝐵1𝐶1𝐶1𝐵2𝐵1𝐶3 ≠ 1. 𝐵𝑌 + 𝐶𝑍 = 0

for 𝑣𝑗 ∈ {𝑏1𝑏2

−1, 𝑏2𝑏3 −1, 𝑏3𝑏1 −1, 𝑐1𝑐2 −1, 𝑐2𝑐3 −1, 𝑐3𝑐1 −1}.

𝑣1𝑣2 ⋯ 𝑣𝑜 = 1 A result: If 𝐺 satisfies ∃𝑏𝑗, 𝑐𝑗 ∈ 𝐺 ∖ {1} (𝑗 = 1,2,3) such that ⟹ ∃𝑗, 𝑣𝑗= 𝑑

𝑘𝑑𝑙 −1 , 𝑣𝑗+1= 𝑑𝑙𝑑𝑚 −1,

where 𝑑𝑗 ∈ {𝑏1, 𝑏2, 𝑏3, 𝑐1, 𝑐2, 𝑐3}, then two elements 𝐵 = 𝑏1 + 𝑏2 + 𝑏3 and 𝐶 = 𝑐1 + 𝑐2 + 𝑐3 satisfy ∀𝑌, 𝑍 ∈ 𝐿𝐻∗, 𝐵𝑌 + 𝐶𝑍 ≠ 0. In particular, 𝐺 is not amenable.

[A-D, 2018] “Property Pnaive for acylindrically hyperbolic groups” To appear in Math Z [S, 2017] “Primitivity of group rings of non-elementary torsion-free hyperbolic groups”

• J. Algebra, Vol. 493, 438-443

[N, 2018] “Uncountable locally free groups and their group rings”

• J. group theory, Vol. 21(1), 101-105

[A-N, 2017] “Non-noetherian groups and primitivity of their group algebras”

• J. Algebra, Vol. 473, 221-246

References

Thank you!