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On amenability constants of Fourier algebras: new bounds and new examples

Yemon Choi, Mahya Ghandehari

Abstract

Let $G$ be a locally compact group. If $G$ is finite then the amenability constant of its Fourier algebra, denoted by ${\rm AM}({\rm A}(G))$, admits an explicit formula [Johnson, JLMS 1994]; if $G$ is infinite then no such formula for ${\rm AM}({\rm A}(G))$ is known, although lower and upper bounds were established by Runde [PAMS 2006]. Using non-abelian Fourier analysis, we obtain a sharper upper bound for ${\rm AM}({\rm A}(G))$ when $G$ is discrete. Combining this with previous work of the first author [Choi, IMRN 2023], we exhibit new examples of discrete groups and compact groups where ${\rm AM}({\rm A}(G))$ can be calculated explicitly; previously this was only known for groups that are products of finite groups with ``degenerate'' cases. Our new examples also provide additional evidence to support the conjecture that Runde's lower bound for the amenability constant is in fact an equality.

On amenability constants of Fourier algebras: new bounds and new examples

Abstract

Let be a locally compact group. If is finite then the amenability constant of its Fourier algebra, denoted by , admits an explicit formula [Johnson, JLMS 1994]; if is infinite then no such formula for is known, although lower and upper bounds were established by Runde [PAMS 2006]. Using non-abelian Fourier analysis, we obtain a sharper upper bound for when is discrete. Combining this with previous work of the first author [Choi, IMRN 2023], we exhibit new examples of discrete groups and compact groups where can be calculated explicitly; previously this was only known for groups that are products of finite groups with ``degenerate'' cases. Our new examples also provide additional evidence to support the conjecture that Runde's lower bound for the amenability constant is in fact an equality.

Paper Structure

This paper contains 21 sections, 30 theorems, 67 equations.

Key Result

Theorem 1.3

Let $G$ be a discrete group. Then there is a countable subgroup $\Gamma\leq G$ such that ${\rm AM}({\rm A}(G))={\rm AM}({\rm A}(\Gamma))$.

Theorems & Definitions (76)

  • Conjecture 1.2
  • Theorem 1.3: Countable saturation
  • Corollary 1.4
  • Corollary 1.5
  • Remark 1.6
  • Theorem 1.7
  • Theorem 1.8: Groups with two degrees of irreducible representations
  • Theorem 1.9
  • Definition 2.1
  • Remark 2.2
  • ...and 66 more