# What is the “spectrum of $L^1(G)$”?

If $G$ is a locally compact abelian group, what does "the spectrum of $L^1(G)$ mean?" This comes from Folland's A Course in Abstract Harmonic Analysis. As I understand it, $L^1(G)$ is the integrable functions (wrt Haar measure) on G. I always think of spectrum as the set of eigenvalues of an operator (or I guess the set of prime ideals if we're taking the spectrum of a ring). Can someone clarify for me the situation? Folland tells us on page 88 that the spectrum of $L^1(G)$ can be identified with the P-dual of $G$. What is going on?

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## 1 Answer

$L^1(G)$ is a commutative Banach algebra. As such, it has a spectrum: as a set, this is the set of its closed maximal ideals. It is also a topological space in a certain canonical way.

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This should be in the book somewhere... –  Mariano Suárez-Alvarez Sep 24 '12 at 1:07
Maximal ideals are already closed! –  Qiaochu Yuan Sep 24 '12 at 1:12
Right. Let's say it was for emphasis :-) –  Mariano Suárez-Alvarez Sep 24 '12 at 1:13
He defines the spectrum of an operator as the set of eigenvalues. On a commutative unital Banach algebra he defines it as the set of multiplicative functionals on the algebra. I guess this might be related to your definition. –  Thelonius Sep 24 '12 at 1:14
A multiplicative functional is determined by its kernel, which is a (closed!) maximal ideal of the algebra. This is done in section 1.2 of the 1st chapter of the book. –  Mariano Suárez-Alvarez Sep 24 '12 at 1:16
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