2016
DOI: 10.1017/s0017089516000252
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Dual Space and Hyperdimension of Compact Hypergroups

Abstract: We characterize dual spaces and compute hyperdimensions of irreducible representations for two classes of compact hypergroups namely conjugacy classes of compact groups and compact hypergroups constructed by joining compact and finite hypergroups. Also studying the representation theory of finite hypergroups, we highlight some interesting differences and similarities between the representation theories of finite hypergroups and finite groups. Finally, we compute the Heisenberg inequality for compact hypergroup… Show more

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Cited by 6 publications
(8 citation statements)
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References 24 publications
(37 reference statements)
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“…Proof It is known that Irr(G) as a strong hypergroup satisfies ( P 2 ), (look at [, Proposition 2.4]). Since ZA(G) is isometrically Banach algebra isomorphic to the hypergroup algebra of Irr(G), one can apply Theorem to the hypergroup trueĜ to finish the proof.…”
Section: Amenability Of Hypergroup Algebrasmentioning
confidence: 99%
See 2 more Smart Citations
“…Proof It is known that Irr(G) as a strong hypergroup satisfies ( P 2 ), (look at [, Proposition 2.4]). Since ZA(G) is isometrically Banach algebra isomorphic to the hypergroup algebra of Irr(G), one can apply Theorem to the hypergroup trueĜ to finish the proof.…”
Section: Amenability Of Hypergroup Algebrasmentioning
confidence: 99%
“…Proof. It is known that Irr( ) as a strong hypergroup satisfies ( 2 ), (look at [1,Proposition 2.4] Let be a compact group such that → ∞. Then is called a tall group.…”
Section: Corollary 52 Let Be a Non-discrete Compact Group Such Thatmentioning
confidence: 99%
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“…Every compact group is a trivial example of a compact hypergroup. Other essential and non-trivial examples are double coset hypergroups G//H asing from a Gelfand pair (G, H) for a compact group G and a closed subgroup H [27], conjugacy classes of compact Lie groups [46,11], countable compact hypergroups [16,11], Jacobi hypergroups [19,11], hypergroup joins [47] of compact hypergroups by finite hypergroups [6,11].…”
Section: Preliminariesmentioning
confidence: 99%
“…6 to get inequality(46) above. The condition(34) for β = 3 by choosing the sequence {µ χn } n∈N := {(1 − a)a −n } n∈N with µ χ 0 = 1 turns out to ben∈N 0 − a) 2 a −2n (1 − a) 3 a −3n = 1 + 1 1 − a n∈N a n = 1 − a + a 2 (1 − a) 2 = (1 − a) 2 − a (1 − a) 2 ,which is finite.…”
mentioning
confidence: 99%