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) The 8th China-Japan-Korea International Symposium on Ring Theory Nagoya University, Japan, Mon. Aug. 26, 2019 *Partially supported by Grants-in-Aid for Scientific Research under grant no. 17K05207

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. ▶ 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.

Two-edge coloured graphs A two-edge colored graph is a simple graph each of whose edges colored with one of two different colors. V = { v 1 , v 2 , …, v n } is a vertex set, E and F are two edge sets; E = { e 1 , e 2 , …, e m } F = { f 1 , f 2 , …, f m } v 1 e 7 e 1 v 2 A cycle in the graph is called f 1 an alternating cycle if its edges f 3 f 4 belong alternatively to E and F . v 3 For example, f 1 e 3 f 2 e 5 f 3 e 7 v 4 e 5 e 3 f 2 v 5

SR-graphs A two-edge colored graph S = ( V , E , F ) is an SR-graph if every component of G = ( V , E ) is a complete graph. In an SR-graph, we call an alternating cycle an SR-cycle; v 1 for example, f 1 e 3 f 2 e 5 f 3 e 7 . e 7 e 1 v 2 f 1 f 3 f 4 v 3 v 4 e 5 e 3 f 2 Complete graphs v 5 𝑤 6 E = { e 1 , e 2 , …, e m } I ( G ) = {𝑤 3 , 𝑤 6 } F = { f 1 , f 2 , …, f m }

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

Let G be a group which has a nonabelian free subgroup. In this case, G has always a free subgroup of infinite rank: 𝑏 1 , 𝑏 2 , 𝑐 1 , 𝑐 2 , ⋯ . Let 𝐵 = 𝑏 1 + 𝑏 2 and 𝐶 = 𝑐 1 + 𝑐 2 . Suppose, to the contrary, that 𝐵𝑌 + 𝐶𝑍 = 0 for some 𝑌, 𝑍 ∈ 𝐿𝐻 ∗ . Since 𝑌, 𝑍 ∈ 𝐿𝐻 ∗ , they are expressed as follows: 𝑌 = σ 𝑦∈𝑇 𝑌 𝛽 𝑦 𝑦 , 𝑍 = σ 𝑧∈𝑇 𝑍 𝛾 𝑧 𝑧 , where 𝛽 𝑦 , 𝛾 𝑧 ∈ 𝐿 ∖ 0 , 𝑇 𝑌 = 𝑇𝑣𝑞𝑞 𝑌 and 𝑇 𝑍 = 𝑇𝑣𝑞𝑞 𝑍 . Since 𝐵𝑌 + 𝐶𝑍 = 0, we have ෍ 𝛽 𝑦 (𝑏 1 𝑦 + 𝑏 2 𝑦) + ෍ 𝛾 𝑧 (𝑐 1 𝑧 + 𝑐 2 𝑧) = 0. 𝑦∈𝑇 𝑌 𝑧∈𝑇 𝑍 We would like to regard these elements 𝑏 𝑗 𝑦 and 𝑐 𝑗 𝑧 as vertices. Because of that, we need to distinguish all these elements even if for 𝑗 ≠ 𝑘, 𝑏 𝑗 𝑦 = 𝑏 𝑘 𝑦′ , 𝑐 𝑗 𝑧 = 𝑐 𝑘 𝑧′ or 𝑏 𝑗 𝑦 = 𝑐 𝑘 𝑧 in G . So we define the vertex set and two edge set as follows: 𝑊 = 𝑏 𝑗 , 𝑦 , 𝑐 𝑗 , 𝑧 | 𝑗 = 1, 2, 𝑦 ∈ 𝑇 𝑌 , 𝑧 ∈ 𝑇 𝑍 𝐹 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = [𝑥] in G }, where 𝑤 = 𝑏𝑦 if 𝑤 = 𝑏, 𝑦 . 𝐺 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = 𝑏 1 , 𝑦 , 𝑥 = (𝑏 2 , 𝑦) or 𝑤 = 𝑐 1 , 𝑧 , 𝑥 = (𝑐 2 , 𝑧) }

𝑊 = 𝑏 𝑗 , 𝑦 , 𝑐 𝑗 , 𝑧 | 𝑗 = 1, 2, 𝑦 ∈ 𝑇 𝑌 , 𝑧 ∈ 𝑇 𝑍 𝐹 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = [𝑥] in G }, where 𝑤 = 𝑏𝑦 if 𝑤 = 𝑏, 𝑦 . 𝐺 = {𝑤𝑥 | 𝑤, 𝑥 ∈ 𝑊; 𝑤 ≠ 𝑥, 𝑤 = 𝑏 1 , 𝑦 , 𝑥 = (𝑏 2 , 𝑦) or 𝑤 = 𝑐 1 , 𝑧 , 𝑥 = (𝑐 2 , 𝑧) } ෍ 𝛽 𝑦 (𝑏 1 𝑦 + 𝑏 2 𝑦) + ෍ 𝛾 𝑧 (𝑐 1 𝑧 + 𝑐 2 𝑧) = 0, Since 𝑦∈𝑇 𝑌 𝑧∈𝑇 𝑍 all elements of 𝐻 in this equation are cancelled each other. Hence (𝑑 1 , 𝑨 1 ) ∈ {𝑏 1 , 𝑏 2 } × 𝑇 𝑌 , ∃ 𝑑 2 , 𝑨 2 ∈ {𝑏 1 , 𝑏 2 } × 𝑇 𝑌 ) ∪ ({𝑐 1 , 𝑐 2 } × 𝑇 𝑍 , 𝑑 1 𝑨 1 = 𝑑 2 𝑨 2 , If 𝑑 2 = 𝑏 𝑗 (resp. 𝑑 2 = 𝑐 𝑗 ), then for 𝑗 ≠ 𝑘 , 𝑏 𝑘 𝑨 2 (resp. 𝑐 𝑘 𝑨 2 ) exists in the above expression, and so v 1 = 𝑑 1 𝑨 1 , 𝑑 1 = 𝑏 𝑗 ∃𝑑 3 ∈ {𝑏 1 , 𝑏 2 , 𝑐 1 , 𝑐 2 } with 𝑑 3 ≠ 𝑑 2 and ∃ 𝑑 4 , 𝑨 3 ∈ {𝑏 1 , 𝑏 2 } × 𝑇 𝑌 ) ∪ ({𝑐 1 , 𝑐 2 } × 𝑇 𝑍 , v 2 = 𝑑 2 𝑨 2 , 𝑑 3 𝑨 2 = 𝑑 4 𝑨 3 , 𝑑 2 = 𝑏 𝑗 or 𝑐 𝑗 ⋮ v 3 = 𝑑 3 𝑨 2 𝑑 𝑛 𝑨 𝑚 = 𝑑 𝑛+1 𝑨 1 , −1 𝑑 2 𝑑 3 −1 𝑑 4 ⋯ 𝑑 𝑛 −1 𝑑 𝑛+1 = 1, v m = 𝑑 𝑛 𝑨 𝑚 𝑑 1 v 4 = 𝑑 4 𝑨 3 where 𝑑 𝑗 ∈ {𝑏 1 , 𝑏 2 , 𝑐 1 , 𝑐 2 } and 𝑑 𝑗 ≠ 𝑑 𝑗+1 . Since {𝑏 1 , 𝑏 2 , 𝑐 1 , 𝑐 2 } is a free basis, this implies a contradiction. ∎ We have thus seen that 𝐵 = 𝑏 1 + 𝑏 2 , 𝐶 = 𝑐 1 + 𝑐 2 ⟹ 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. ▶ 𝐺 is amenable ⟹ ∀𝐵, 𝐶 ∈ 𝐿𝐻 ∗ , ∃𝑌, 𝑍 ∈ 𝐿𝐻 ∗ , 𝐵𝑌 = 𝐶𝑍 for any 𝑌, 𝑍 ∈ 𝐿𝐻 ∗ . ▶ Is Thompson’s group 𝐺 is amenable?

Thompson’s group F −1 𝑦 𝑘 𝑦 𝑗 = 𝑦 𝑘+1 , for 𝑗 < 𝑘 . F = 𝑦 0 , 𝑦 1 , ⋯ , 𝑦 𝑗 , ⋯ | 𝑦 𝑗 = 𝑦 0 , 𝑦 1 [𝑦 0 𝑦 1 −1 , 𝑦 0 −1 𝑦 1 𝑦 0 , [𝑦 0 𝑦 1 −1 , 𝑦 0 −2 𝑦 1 𝑦 0 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 We considered the following SR-graph . We set here 𝐵 = 𝑏 1 + 𝑏 2 + 𝑏 3 and 𝐶 = 𝑐 1 + 𝑐 2 + 𝑐 3 . 𝑏 1 𝑦 1 𝑏 2 𝑦 1 −1 and 𝐶 1 = 𝑐 1 𝑐 2 −1 . Let 𝐵 1 = 𝑏 1 𝑏 2 𝑐 1 𝑧 3 𝑐 1 𝑧 1 −1 = 1 −1 𝐶 1 𝐵 1 𝐶 1 𝐶 1 𝐵 1 𝑐 3 𝑧 3 𝑏 3 𝑦 1 We have to choose 𝐵 1 and 𝐶 1 −1 ≠ 1 so as to 𝐵 1 𝐶 1 𝐶 1 𝐵 1 −1 𝐶 1 𝑐 3 𝑧 1 𝑐 2 𝑧 3 𝑐 2 𝑧 1 Generally, it may happen that 𝑐 3 𝑧 2 𝑏 1 𝑦 2 ±𝛾 1 ⋯ 𝐵 1 ±𝛾 𝑛 = 1 𝑐 1 𝑧 2 ±𝛽 1 𝐶 1 ±𝛽 𝑛 𝐶 1 𝐵 1 According to our method, we have to 𝑏 2 𝑦 2 choose 𝐵 1 and 𝐶 1 so as to 𝑏 3 𝑦 2 ±𝛾 1 ⋯ 𝐵 1 ±𝛾 𝑛 ≠ 1 𝑐 2 𝑧 2 ±𝛽 1 𝐶 1 ±𝛽 𝑛 𝐶 1 𝐵 1

𝑏 1 𝑦 1 𝑏 2 𝑦 1 𝑐 1 𝑧 3 𝑐 1 𝑧 1 𝑐 3 𝑧 3 𝑏 3 𝑦 1 𝑐 3 𝑧 1 𝑐 2 𝑧 3 𝑐 2 𝑧 1 𝑐 3 𝑧 2 𝑏 1 𝑦 2 𝑐 1 𝑧 2 𝑏 2 𝑦 2 𝑏 3 𝑦 2 𝑐 2 𝑧 2 We replace an undirected SR-graph with a directed one.

𝑏 1 𝑦 1 𝑏 2 𝑦 1 𝑐 1 𝑧 3 𝑐 1 𝑧 1 𝑐 3 𝑧 3 𝑏 3 𝑦 1 𝑐 3 𝑧 1 𝑐 2 𝑧 3 𝑐 2 𝑧 1 𝑐 3 𝑧 2 𝑏 1 𝑦 2 𝑐 1 𝑧 2 𝑏 2 𝑦 2 𝑏 3 𝑦 2 𝑐 2 𝑧 2 In this graph, an SR-cycle means a cycle along the direction of arrows.

𝑏 1 𝑦 1 𝑏 2 𝑦 1 𝐵𝑌 + 𝐶𝑍 = 0 𝑐 1 𝑧 3 𝑐 1 𝑧 1 𝑐 3 𝑧 3 𝑏 3 𝑦 1 𝑐 3 𝑧 1 𝑐 2 𝑧 3 𝑐 2 𝑧 1 𝑏 3 𝑦 3 𝑐 3 𝑧 2 𝑏 2 𝑦 3 𝑏 1 𝑦 2 𝑐 1 𝑧 2 𝑏 1 𝑦 3 𝑏 2 𝑦 2 𝑏 3 𝑦 2 𝑐 2 𝑧 2 −1 and 𝐶 3 = 𝐶 3 𝐶 1 −1 , 𝐵 2 = 𝑏 2 𝑏 3 −1 , 𝐶 1 = 𝑐 1 𝑐 2 −1 . Let 𝐵 1 = 𝑏 1 𝑏 2 𝐵 1 𝐶 1 𝐶 1 𝐵 2 𝐵 1 𝐶 3 = 1; We have only to choose 𝐵 1 𝐶 1 𝐶 1 𝐵 2 𝐵 1 𝐶 3 ≠ 1.

A result : If 𝐺 satisfies ∃𝑏 𝑗 , 𝑐 𝑗 ∈ 𝐺 ∖ {1} (𝑗 = 1,2,3) such that 𝑣 1 𝑣 2 ⋯ 𝑣 𝑜 = 1 −1 , 𝑏 2 𝑏 3 −1 , 𝑏 3 𝑏 1 −1 , 𝑐 1 𝑐 2 −1 , 𝑐 2 𝑐 3 −1 , 𝑐 3 𝑐 1 −1 }. for 𝑣 𝑗 ∈ {𝑏 1 𝑏 2 −1 , 𝑣 𝑗+1 = 𝑑 𝑙 𝑑 𝑚 −1 , ∃𝑗, 𝑣 𝑗 = 𝑑 𝑘 𝑑 𝑙 ⟹ where 𝑑 𝑗 ∈ {𝑏 1 , 𝑏 2 , 𝑏 3 , 𝑐 1 , 𝑐 2 , 𝑐 3 }, then two elements 𝐵 = 𝑏 1 + 𝑏 2 + 𝑏 3 and 𝐶 = 𝑐 1 + 𝑐 2 + 𝑐 3 satisfy ∀𝑌, 𝑍 ∈ 𝐿𝐻 ∗ , 𝐵𝑌 + 𝐶𝑍 ≠ 0. In particular, 𝐺 is not amenable.