Question

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.

Answer 1

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 $P\to Top, \mathbf X\mapsto |\mathbf X|_ \mu$:

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