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Let G be an non-Abelian locally compact group. What is the set af all multiplicative functionals of L1(G)? (When G is abelian the answer is the dual group)

flag	asked Apr 21 2011 at 2:31 saghar 23●2

Are you looking for the dual group of the abelianization? – S. Carnahan♦ Apr 21 2011 at 7:52

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It is the set of all multiplicative functionals of $L^1(G)$ – Yemon Choi Apr 21 2011 at 8:46

Anatoly Kochubei · Accepted Answer · 2011-04-21 05:15:03Z UTC

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Multiplicative functionals on the group algebra correspond to one-dimensional unitary representations of the group. In the non-commutative case such (nontrivial) representations can be absent or constitute a small part of the set of all irreducible unitary representations. See, for example,

M. A. Naimark, Normed rings, Noordhof, Groningen, 1970.

answered Apr 21 2011 at 5:15

Anatoly Kochubei
3,052●1●4●9

	Just to expand on the word "nontrivial" - there is of course at least one multiplicative functional on $L^1(G)$, namely the augmentation character $f\mapsto \int_G f$. The point is that there may not be any others. – Yemon Choi Apr 21 2011 at 8:46
	(... apart from zero, obviously) – Yemon Choi Apr 21 2011 at 22:59

multiplicative functionals

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