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Let $G$ is a locally compact group (non-Abelian)

Why $sp(L^1(G))$ , i.e. the set of all nonzero bounded multiplicative functionals on $L^1(G)$ is a locally compact group.

Even for any noncommutative Banach algebra A, why $sp(A)$ is a locally compact space?

Can you give me a reference?

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1 Answer 1

up vote 1 down vote accepted

If $A$ is a (noncommutative) Banach algebra, a character (i.e. a multiplicative linear functional) is automatically bounded. Then consider the set $\Sigma(A) \subseteq A^*$ of nonzero characters with the weak-$*$ topology inherited from $A^*$. Then $\Sigma(A) \cup \{ 0 \} $ is a weak-$*$ closed subset of the unit ball in $A^*$, which itself is weak-* compact. Thus $\Sigma(A) \cup \{ 0 \} $ is weak-$*$ compact, so $\Sigma(A)$ is locally compact.

Of course, for a general Banach algebra you may not have many characters, but that's a separate issue...

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Just to add: the fact that $0$ compactifies this space is the Riemann Lebesgue lemma. –  Phil Isett Mar 8 '12 at 4:45
    
Phil, I'm not sure what you mean. What is the Riemann Lebesgue lemma for a general Banach algebra? –  MTS Mar 8 '12 at 5:11
    
Doesn't the usual proof that a character is continuous use that a character is unital? –  Matthew Daws Mar 8 '12 at 8:14
    
Yes, but can't you just extend the character naively to the unitalization of the algebra? –  MTS Mar 8 '12 at 16:57
    
@MTS: Yes! Sorry, silly question... –  Matthew Daws Mar 8 '12 at 17:08

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