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Gelfand-Naimark structure theorem for $C^* $ algebras gives a canonical isometric * isomorphism between any commutative unital $C^* $ algebra $A$ and the algebra of continuous complex-valued functions on $A$^. This is the spectrum (or structure space) of $A$, i.e. the non-zero multiplicative linear continuous functionals with the topology of pointwise convergence (alias weak*), which is compact and hausdorff. Apart from the easy case $A = C(X)$, with $X$ compact hausdorff, for which $A$^ is $X$ itself, there are a lot of non trivial and not immediately visible examples of spectra, for example:

If $X$ loc. compact hausdorff $A = C_b(X)$ (continuous and bounded functions with uniform topology) is a $C^*$ algebra. If X is non compact then A^ cannot be $X$ and is in fact $\beta X$, the Stone-Cech compactification of $X$.

If $X$ is loc. compact hausdorff and you take $C_0(X)$, then you get another compactification of $X$.

If instead you simply take $C(X)$ for $X$ compact non-hausdorff you get a natural "hausdorfization" of $X$.

I'm particularly interested in the existence of other constructions which can be described by gelfand theory as above. I mean to associate functorially to each space (in an appropriate subcategory of Top, maybe not full) a $C^*$ algebra and then to look at its spectra.

A related question: what are the spectra of $L^\infty(R)$, and similar algebras (maybe $L^\infty(G)$, G loc. compact group with haar measure)?

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up vote 7 down vote accepted

The spectrum of $L^\infty(R)$ is the hyperstonean space associated with the measurable space R. More information can be found in Takesaki's Theory of Operator Algebras I, Chapter III, Section 1, available here:

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Yes, but in a sense, this is just a way of getting out of difficulty by naming a hard-to-understand object, right? I am not saying this is useless or that one can't say a lot about this space, but it doesn't really give you a handle on what the spectrum is. (Though, as Bill Clinton famously said, it depends on what the meaning of the word “is” is.) – Harald Hanche-Olsen Nov 11 '09 at 15:55
Well, you can say a lot about this space. Bounded functions on it are in bijective correspondence with equivalence classes of bounded functions on the original measurable space; clopen sets are in bijective correspondence with equivalence classes of measurable sets etc. etc. etc. From the viewpoint of topology this space is weird because it is extremally disconnected, therefore to understand what this space really is you need to use quite different methods from what you are used to. – Dmitri Pavlov Nov 11 '09 at 17:23
After all, the category of commutative von Neumann algebras is contravariantly equivalent to the category of hyperstonean spaces and hyperstonean maps between them, and this category is equivalent to the category of measurable spaces and measurable maps between them. Thus understanding hyperstonean spaces is the same thing as understanding measure theory. There is a complete classification of measurable spaces: Every measurable space is a coproduct of points and real lines. Thus the spectrum of $L^\infty(R)$ is the only non-trivial interesting example of a measurable space. – Dmitri Pavlov Nov 11 '09 at 17:27
Hmm. I'll add hyperstonean spaces to the list of things I want to learn more about some day. Thanks. – Harald Hanche-Olsen Nov 11 '09 at 18:38

For $L^\infty(X)$, the spectrum is the Stone space of the algebra of measurable sets mod null sets. This is because a character is determined by what it does on characteristic functions because their span is dense.

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Your question (especially the first part) is a bit vague, but I'll shoot: A very nice example is provided by Carleson's corona theorem, stating that the unit disk is dense in the spectrum of the Hardy space $H^\infty$ (the bounded holomorphic functions on the unit disk).

As for the spectra of $L^\infty$, I don't think you can ever come up with a concrete example of a character on this space. You actually need the axiom of choice to prove that the spectrum is nonempty. Likewise with the points of the spectrum of $H^\infty$ outside the unit disk.

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I believed you could describe the spectrum of the algebra of translation invariant bounded operators on L^2(R^n), which is isomorphic to L^\infty. – Gian Maria Dall'Ara Nov 11 '09 at 15:21
That isomorphism provides the most concrete representation of the algebra you mention, via multiplication of the Fourier transform. I don't see how this helps in describing the spectrum, though of course maybe it's just my ignorance showing. – Harald Hanche-Olsen Nov 11 '09 at 15:47

I think, if $X$ is locally compact and Hausdorff, then the spectrum of $C_0(X)$ is just $X$.

You can get the one point compactification of $X$ by looking at the spectrum of the unitisation of $C_0(X)$. This is the vector space $C_0(X) \oplus \mathbb C$ with the unique $C^*$-norm. (Just embed it into $C_b(X)$ for example).

The spectrum of $L^\infty(G)$ will in general be very large: I don't know any "nice" way of describing it.

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I really meant to add a unit to $C_0(X)$, otherwise it is not a unital $C^* $ algebra and you cannot expect a compact spectrum, and hence the compactification of $X$ – Gian Maria Dall'Ara Nov 11 '09 at 13:32

The Gelfand representation also works for non-unital commutative C^*-algebras. In this case, it establishes a category equivalence to the category of locally compact Hausdorff spaces with proper maps (implemented by C_0(.) and the spectrum). Hence Matthew's comment, the spectrum of C_0(X) is just X.

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For another equivalent construction, you can take the set of primitive ideals and endow it with jacobson (or hull-kernel) topology and this space is homeomorphic to the spectrum for commutative things.

This is also the usual definition for the spectrum of a non-commutative $C^*$-algebras. It is analogous to the construction of spectra in algebraic geometry.

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Really? My dim recollection is that there should be many more prime ideals, see – Yemon Choi Jul 8 '13 at 8:55
prime $\neq$ primitive. Thanks. – Marc Palm Jul 8 '13 at 9:12

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