Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $(X,\Sigma,\mu)$ is a measured set with respect to $\sigma$-algebra $\Sigma$. Suppose that $L^\infty(X,\mu)$ is the set of all $\mu$-equal bounded $\Sigma$-measurable functions on $X$. Indeed equally, one may say that $L^\infty(X,\mu)$ is the dual of $L^1(X,\mu)$. What is the spectrum of $(L^\infty(X,\mu),\|\cdot\|_\infty)$ as a Banach (C^*) algebra? Thank you very much.

share|improve this question
1  
It doesn't directly affect your question, but I have a feeling that $L^\infty$ is only the dual of $L^1$ under certain mild conditions on your space and your sigma-algebra. –  Yemon Choi Jun 8 '12 at 6:45
1  
Why the vote to close? –  Johannes Ebert Jun 8 '12 at 7:56
1  
I've had a rough night, and might be missing something, but are you looking for the topological space which would be the Gelfand spectrum of the abelian $C*-$algebra $L^\infty$? If that's the case you should look for completely discontinuous spaces on Google, or have a look at Kadison-Ringrose, the part on VonNeumann algebras. –  Amin Jun 8 '12 at 8:26
2  
Yes, a hyperstonean space will be the spectrum. For example, if $X=\mathbb Z$ with the counting measure, then the spectrum is $\beta Z$ (the Stone-Cech compactification). –  Yulia Kuznetsova Jun 8 '12 at 8:31
1  
@Matthew: It's not a "standard fact"... it's a mathematical object. The OP is simply asking for extra information/intuition about that mathematical object. –  André Henriques Jun 8 '12 at 11:33

1 Answer 1

up vote 7 down vote accepted

Here is a description of the spectrum of $L^\infty([0,1];\mu)$ for an arbitrary Borel measure $\mu$ on $[0,1]$.

Consider the following poset, which I call $P$ :

• The objects of $P$ are decompositions $\mathbf X=\{X_1,\ldots, X_n\}$ of $[0,1]$ into finitely many $\mu$-measurable sets $[0,1]=X_1\cup X_2\cup\ldots\cup X_n$,  $X_i\cap X_j=\emptyset$. Two decompositions $\mathbf X$ and $\mathbf Y$ are declared equal is there exists a permutation $\sigma$ such that $X_i=Y_{\sigma(i)}$ up to a $\mu$-measure zero set.
• The partial order on $P$ is given by refinement: $\mathbf X \prec \mathbf Y$ if
$Y_1=X_1\cup\ldots \cup X_{n_1}$, $Y_2=X_{n_1+1}\cup\ldots \cup X_{n_2}$, $\ldots$ (up to permutation and $\mu$-measure zero sets)

Note that the poset $P$ is filtered: given a finite set $\mathbf X_1, \mathbf X_2, \ldots, \mathbf X_n$ of elements of $P$, there is always a common refinement, i.e., an element $\mathbf X\in P$ such that $\mathbf X\prec \mathbf X_i\\,\forall i$

Given a $\mu$-measurable subset $X\subset [0,1]$, let me denote by $|X|_ \mu \subset [0,1]$ the $\mu$-adherence of $X$: $$ |X|_ \mu:=\{x\in [0,1]: \forall \varepsilon>0\quad \mu(X\cap B_{x,\varepsilon})>0\}, $$ where $B_{x,\varepsilon}$ denotes the ball of radius $\varepsilon$ around the point $x$. Given $\mathbf X=\{X_1,\ldots,X_n\} \in P$, we also write $|\mathbf X|_ \mu$ for the disjoint union $$|\mathbf X|_ \mu:=|X_1|_ \mu\sqcup\ldots\sqcup|X_n|_ \mu.$$

Note that if $\mathbf X \prec \mathbf Y$, then there is a natural projection map $|\mathbf X|_ \mu \twoheadrightarrow |\mathbf Y|_ \mu$.

Given the above preliminaries, the spectrum of $L^\infty([0,1];\mu)$ is given by the inverse limit of the functor $P\to Top, \mathbf X\mapsto |\mathbf X|_ \mu$:

$$Spec\big(L^\infty([0,1];\mu)\big) =\quad \underset{\mathbf X\in P}{\underset\leftarrow\lim} |\mathbf X|_ \mu$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.