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Let $K$ be a compact-Hausdorff space and $\mu$ be a regular Borel measure. We have four important space to consider: the $C^*$ algebra of continuos functions $(C(K),||\cdot ||_{\infty})$, the abstract Von Neumann algebra of bounded measurable functions $(L^{\infty}(K),||\cdot||_{\infty})$, the space of integrable functions $(L^1(K),||\cdot||_1)$ (its predual) and the Hilbert space of square integrable functions $(L^2(K),||\cdot||_2)$.

Q1: $L^{\infty}$ depends only on the sets of measure zero $N$, can $L^1$ and $L^2$ (or $L^p$ in general) be defined only with $N$? Are there any restrictions on $N$ similar to $\mu$ to be regular?

We can recover $K$ from $A=C(K)$, and then also the spaces $W=L^{\infty}(K)$, $W_*=L^1(K)$ and $H=L^2(K)$.

Q2: Given a $C^*$ algebra $A$, can we associate a Hilbert space $H$ and a Von Neumann algebra $W$ so that in the case of $A=C(K)$, then $H=L^2(K)$, $W=L^{\infty}(K)$ and $W_*=L^1(K)$?

Q3: If so, what about associating to $A$, Banach spaces $B_p$ such that if $A=C(K)$ then $B_p=L^p(K)$?

Edit: Maybe in Q2 we have to consider a generalization of $N$, and for $B_p$ a generalization of a regular measure.

alexpglez98
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A lot of serious research went into these questions, so I'll have to restrict myself mostly to giving some keywords. Feel free to ask for more details or references for concrete points.

Q1: Abstractly (that is, up to isometric isomorphism), the spaces $L^1(K)$ and $L^2(K)$ only depend on $N$. You can show this by writing down an explicit isomorphism using the Radon-Nikodym density of two measures with the same collection of measure zero sets. But more abstractly, $L^1(K)$ is the predual of $L^\infty(K)$ (as you wrote yourself) and $L^2(K)$ is the underlying Hilbert space of the standard form. Both are uniquely determined (up to isometric isomorphism) by the von Neumann algebra. The correct condition on $N$ is that the measure algebra be localizable.

But this is in some sense not the complete picture. The way $C(K)$ "sits" inside $L^1(K)$ and $L^2(K)$ does depend on the measure, that is, for different measures $\mu,\nu$ with the same collection of sets of measure zero, there is not necessarily a $\ast$-automorphism $\alpha\colon C(K)\to C(K)$ that is isometric with respect to the $L^1$- or $L^2$-norm.

Q2: As you noted, the von Neumann algebra depends on the choice of $N$. However, given a unital $C^\ast$-algebra $A$ and a state $\omega\colon A\to\mathbb{C}$, you can form the GNS Hilbert space $H_\omega$ and the weak closure of $A$ in the GNS representation. In the case when $A=C(K)$, every state is of the form $\omega=\int\cdot\,d\mu$ for some regular $\mu$ and you obtain $L^2(K)$ and $L^\infty(K)$ from this construction. I think it is hard to find a good analog of $N$ in the $C^\ast$-algebra picture because measurable sets don't fall in the realm of topology, but measure theory.

Q3: Again, von Neumann algebras seem more suitable here. Given a von Neumann algebra $M$ and a normal faithful state (or, more generally, norma semi-finite faithful weight) $\omega$ on $M$, one can form the noncommutative $L^p$-spaces $L^p(M,\omega)$. As it turns out, these spaces do not depend (up to isometric isomorphism) on the choice of $\omega$, although once again the interpolation scale does.

There are several different constructions of $L^p(M,\omega)$. The one that is probably easiest to describe uses interpolation. The space $L^2(M,\omega)$ is simply the GNS Hilbert space associated with $\omega$, which comes (if $\omega$ is a state) with a natural inclusion map $\Lambda_\omega\colon M\to L^2(M,\omega)$. Thus $(L^2(M,\omega),M)$ forms an interpolation couple and one defines $L^p(M,\omega)=(L^2(M,\omega),M)_{2/p}$ for $p\in (2,\infty)$. Moreover, $L^\infty(M,\omega)=M$, $L^1(M,\omega)=M_\ast$ and $L^p(M,\omega)=L^q(M,\omega)^\ast$ for $p\in (1,2)$ and $1/p+1/q=1$.

MaoWao
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  • I think $L^1(K,\mu)$ can be identified with the Radon measures with bounded variation which are absolutely continuos to $\mu$. But for $L^2(K)$ I think is more complicated: we can identify it with the equivalence pairs $[f,\nu]$, where $\nu$ is a positive measure equivalent to $\mu$ and $f\in L^2(\nu)$, and $(f,\nu)\simeq (f',\nu')$ if $f\sqrt{d\nu/d\nu'}=f'$. Does for every set of zero sets $N$ exist a positive measure having exactly these null sets? Why the correct condition is localizable? – alexpglez98 May 03 '21 at 14:00
  • The problem of finding the non-commutative analog of $N$ is that if $x\in A$ is positive and $\omega(x)=0$, it doesn't follow that $x=0$. Given a state you can think of $I={x\in A ; | ; \omega(x^x)=0 }$ as the ideal of null functions a.e., after consider that there could be non zero "continuos functions" that are zero a.e. Is $A/I$ a $C^$-algebra in a canonical way? It is obvious why $L^2(M,\omega)$ should be the completion $A/I$ with $<x,y>=\omega(x^*y)$. – alexpglez98 May 03 '21 at 14:21
  • Sorry, I didn't noticed you wrote that $\omega$ should be faithful so that $I={0}$. Is there a way to work with non faithful states to define $L^p(M,\omega)$? And, what is an interpolation couple? I think you are thinking of these https://en.wikipedia.org/wiki/Interpolation_space#The_setting_of_interpolation, but I don't know how it works – alexpglez98 May 03 '21 at 14:52
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    That depends on what you mean by set of zero sets. If $N\in \mathcal{F}$ is a $\sigma$-ideal and $(X,\mathcal{F},N)$ is localizable, then there exists a localizable measure $\mu$ on $\mathcal{F}$ such that $N={A\in \mathcal{F}\mid \mu(A)=}$. Localizability is the correct condition here because $L^\infty(X,\mathcal{F},N)$ is a von Neumann algebra if and only if $(X,\mathcal{F},N)$ is localizable. – MaoWao May 03 '21 at 16:14
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    If $\omega$ is not faithful, then one can always restrict it to the corner $eMe$, where $e$ is the support projection of $\omega$ (so the definition would be $L^p(M,\omega)=L^p(eMe,\omega(e\cdot e))$. This is akin to restricting a measure to its support. For this reason one usually deals only with faithful states/weights. – MaoWao May 03 '21 at 16:16
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    In general, the set $I$ is only a left and not a right ideal, so $A/I$ is not a $C^\ast$-algebra. But $L^2(M,\omega)$ is indeed defined this way, this is exactly the GNS construction. – MaoWao May 03 '21 at 16:18
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    Yes, interpolation couples are the ones you linked to. There are also other constructions for noncommutative $L^p$-spaces by Haagerup and Connes, but they are not easier to describe. Everything is much simpler if the state is a trace, that is, $\omega(xy)=\omega(yx)$. This is of course always the case for noncommutative von Neumann algebras, but not every noncommutative von Neumann algebra admits a normal faithful trace. – MaoWao May 03 '21 at 16:21
  • Where can I read a proof that given a (localizable) $\sigma$-ideal $F$ there exist a measure $\mu$ that vanish on $F$? Did you find a good book to read all of this? I only know the book of Connes (but I haven't read yet), but I think it isn't very easy to follow. – alexpglez98 May 03 '21 at 16:43
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    When it comes to measure theory, Fremlin's book are the place to go (and they certainly contain the result on localizable measure algebras). But I don't think you can read them as anything but references. A lot of the stuff from this answer should be covered by Takesaki's books on operator algebras. Also this preprint might interest you: https://arxiv.org/abs/2005.05284 – MaoWao May 03 '21 at 16:58
  • Two lasts questions. What is the universal enveloping Von Neumman algebra of $C(K)$? The intuition tells me that is the space of bounded functions, so in the case of having a measure $\mu$, the ideal $I$ are the zero functions a.e. and the quotient is $L^{\infty}(K,\mu)$. In general, does there exist any regular measure $\mu$ such that the canonical map $C(K)\longrightarrow L^{\infty}(K,\mu)$ is injective? (Is that the case of Haar measures?) – alexpglez98 May 03 '21 at 17:16
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    The second dual of $C(K)$ is discussed in this question: https://math.stackexchange.com/questions/74875/dual-space-of-the-space-of-finite-measures. In short, there does not seem to be a nice description of it. The canonical map $C(K)\to L^\infty(K,\mu)$ is injective whenever $\mu$ has full support. This is indeed the case for the Haar measure on a (locally) compact group. – MaoWao May 03 '21 at 22:43