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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)

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Are you looking for the dual group of the abelianization? –  S. Carnahan Apr 21 '11 at 7:52
It is the set of all multiplicative functionals of $L^1(G)$ –  Yemon Choi Apr 21 '11 at 8:46
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1 Answer

up vote 4 down vote accepted

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.

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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 '11 at 8:46
(... apart from zero, obviously) –  Yemon Choi Apr 21 '11 at 22:59
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