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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?

flag	asked Mar 8 2012 at 0:17 Venus 11●2

MTS · Accepted Answer · 2012-03-08 01:44:36Z UTC

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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...

answered Mar 8 2012 at 1:44

MTS
5,306●1●9●27

	Just to add: the fact that $0$ compactifies this space is the Riemann Lebesgue lemma. – Phil Isett Mar 8 2012 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 2012 at 5:11
	Doesn't the usual proof that a character is continuous use that a character is unital? – Matthew Daws Mar 8 2012 at 8:14
	Yes, but can't you just extend the character naively to the unitalization of the algebra? – MTS Mar 8 2012 at 16:57
	@MTS: Yes! Sorry, silly question... – Matthew Daws Mar 8 2012 at 17:08

spectrum of Banach algebras

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