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Let $I$ is a compact topological space, $m$ is a positive regular Borel measure. Then $L^\infty(m)$ is a standard example of commutative $W^\ast$-algebra (von Neumann algebra), but it is also a commutative $C^\ast$-algebra, so it should be isomorphism with $C(I)$. I can not believe that $L^\infty(m)$ is isomorphic to $C(I)$, who can tell me what happens?

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Who says it is the same $I$? Take the simplest non-trivial example: $I = [0,1]$ with Lebesgue measure. Then $C(I)$ is separable while $L^{\infty}(I)$ isn't. The statement is that there exists a compact space $X$ such that $L^{\infty}(I) \cong C(X)$. Looking at the proof of the Gel'fand-Neumark theorem will tell you what $X$ you can take. It is huge (it can't be metrizable, as the space of continuous functions on a compact metrizable space is separable). It turns out that $X$ is an extremally disconnected space – t.b. Nov 12 '11 at 10:43
Another interesting example is the spectrum of the unital commutative $C^*$-algebra of bounded continuous functions $X\to\mathbb C$ for some locally compact non-compact Hausdorff space $X$. – Rasmus Nov 12 '11 at 11:53
@t.b.: Post as an answer? – Nate Eldredge Nov 12 '11 at 13:15
@Nate: sure, done. – t.b. Nov 12 '11 at 23:48
up vote 8 down vote accepted

Your notation indicates that you seem to assume that $L^{\infty}(m) = C(I)$. This can't be true, of course. Consider Lebesgue measure on $I = [0,1]$. Then $C(I)$ is a separable algebra while $L^{\infty}[0,1]$ is non-separable, so the two can't be isomorphic. Note also that $L^{\infty}(m)$ contains many functions that aren't equal to a continuous function almost everywhere.

Recall what the Gelfand representation theorem says: Given a commutative and unital $C^{\ast}$-algebra $A$, let $\hat{A} = \{\varphi:A \to \mathbb{C}\,:\,\varphi\text{ is a nonzero homomorphism}\}$ be its space of characters, which is a compact Hausdorff space when equipped with the weak$^{\ast}$-topology. The map $\operatorname{ev}: A \to C(\hat{A})$ given by $\operatorname{ev}_a (\varphi) = \varphi(a)$ is an (isometric) $\ast$-isomorphism.

Since von Neumann algebras are either finite-dimensional or non-separable, the space $\hat{A}$ is only metrizable in trivial cases, since $C(K)$ is separable for compact metrizable spaces $K$. It turns out that a commutative $C^\ast$-algebra $A$ is a von Neumann algebra if and only if its spectrum $\hat{A}$ is a hyperstonian space. In particular, the spectrum of a von Neumann algebra is extremally disconnected in the sense that the closure of an open set is open. This indicates that the spectrum of a commutative von Neumann algebra must be huge.

If the measure $m$ is non-atomic and $A = L^{\infty}(m)$, you don't even have a natural map $I \to \hat{A}$ (you can't evaluate an equivalence class of bounded measurable functions at a point $x \in I$), so the space $\hat{A}$ has little to do with the space $I$ itself (in fact, it only depends on the measure algebra of $(I,m)$). As I argued above, the natural map $C(I) \to L^{\infty}(m)$ is typically not surjective. This map need not be injective either (since the measure $m$ need not be everywhere supported).

Finally, as Rasmus points out, the Gelfand representation theorem can be used to construct the Stone–Čech compactification of a locally compact space $X$. Namely, let $A = C_b(X)$ be the commutative unital $C^{\ast}$-algebra of bounded continuous functions on $X$. Then $\hat{A} = \beta X$ together with the map that sends $x$ to the evaluation at $x$ can be shown to satisfy the universal property of the Stone–Čech compactification. This leads us to the simplest example of a non-discrete hyperstonian space: the Stone–Čech compactification $\beta\mathbb{N}$ of $\mathbb{N}$ which is the spectrum of $\ell^\infty(\mathbb{N})$ (you can get $\ell^{\infty}(\mathbb{N})$ as $L^{\infty}(m)$ by taking a measure on $I = [0,1]$ having countably many atoms whose weights sum up to $1$).

For all this, and much more, I recommend Pedersen's Chapter IV of Analysis Now for an efficient but accessible exposition. If you want to dig deeper, have a look at Volume 1 of Takesaki's Theory of operator algebras. Also, you may want to have a look at the Wikipedia article on abelian von Neumann algebras.

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Great answer. These results sort of show that in many "intuitive" senses, spaces like $\ell^{\infty}(\mathbb{N})$ aren't of the form $C(X)$, for the reason that our intuition for how topological spaces "ought to be" often imposes more structure/niceness than is present in the $X$'s one needs to resort to in the Gelfand-Naimark correspondence. (I am no set theorist, but I have heard that what you can prove about spaces like $\beta \mathbb{N}$ depends strongly on what you add to set theory beyond choice. There is a sense in which one cannot really "get one's hands on" these spaces.) – leslie townes Nov 13 '11 at 2:15
Thanks! I agree with what you say. Concerning your parenthetical remark: a good idea of what's possible under different assumptions is given in J. van Mills's contribution An introduction to $\beta\omega$ to the Handbook of set theoretic topology. – t.b. Nov 13 '11 at 2:59
Great!Thank you. – Strongart Nov 14 '11 at 7:03

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