If I imagine that (the self-adjoint part of) a C\*-algebra $A$ represents the algebra of observables of some quantum system, then [certain](http://arxiv.org/abs/1412.2177) [perspectives](http://dx.doi.org/10.1007/s00220-009-0865-6) on algebraic quantum theory would ask me to imagine that each (maximal) commutative C\*-subalgebra $C \subseteq A$ provides a (maximal) "classical snapshot" of this quantum system. Gelfand duality yields that $C \cong C(X)$ for the compact Hausdorff space $X = \mathrm{Spec}(C)$, so I would picture $X$ as a classical state space that (maximally) "approximates" the would-be quantum state space corresponding to $A$.


I would like to know how different these spaces $X$ can be as the maximal commutative $\*$-subalgebra $C \subseteq A$ varies. Specifically, can it happen that these have different cardinalities?


I'm interested in the particular case $A = B(H)$ for a separable Hilbert space $H$ and two of its well-known masas: the continuous one $C \cong L^\infty[0,1] \subseteq A$ and discrete one $D \cong \ell^\infty(\mathbb{N}) \subseteq A$. Thus I ask:



> 
> **Q:** Is there a bijection between the Gelfand spectra $\mathrm{Spec}(C)$ and $\mathrm{Spec}(D)$ for the continuous and discrete masas $C,D \subseteq B(H)$?
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> 
> 


It's possible to describe these spectra in more explicit terms using Boolean algebra. Note that each of these masas is an (A)W\*-algebra. By a combination of Gelfand and Stone dualities (see [section 2 of this paper](http://arxiv.org/abs/1212.5778) for a bit more detail), the spectrum of a commutative AW\*-algebra $K$ is the [Stone space](http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras) of the complete Boolean algebra $\mathrm{Proj}(K)$ of projections in $K$, whose points are the ultrafilters of $\mathrm{Proj}(K)$.


The continuous masa $C \cong L^\infty[0,1]$ has $\mathrm{Proj}(C)$ isomorphic to the Boolean algebra of measurable subsets of $[0,1]$ modulo the null sets. I have just learned through the magic of Wikipedia that this is called the [random algebra](http://en.wikipedia.org/wiki/Random_algebra); I will denote it by $B$.


The discrete masa $D \cong \ell^\infty(\mathbb{N})$ has $\mathrm{Proj}(D)$ isomorphic to the power set Boolean algebra $2^\mathbb{N}$. (Note that an ultrafilter on the Boolean algebra $2^\mathbb{N}$ is alternatively referred to as an ultrafilter on the set $\mathbb{N}$.)


Thus my question is equivalent to:



> 
> **Q':** Is there a bijection between the sets of ultrafilters on the random algebra $B$ and the power set algebra $2^\mathbb{N}$?
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> 
> 


I am aware that $\mathrm{Spec}(D)$ has spectrum homeomorphic to the Stone-Cech compactification $\beta\mathbb{N}$ of the discrete space $\mathbb{N}$, and that this space has various properties that depend on set-theoretic assumptions. Now that I know what the random algebra is called, I see that it bears a relationship to forcing. Thus I can imagine that the answer to my question could be independent of ZFC. Nevertheless, as I am not asking exactly what the cardinality of this spectrum is, but whether it is in bijection with some other (possibly complicated) spectrum, I have an ounce of hope that this can indeed be decided in ZFC.


(By the way, the [classification](http://en.wikipedia.org/wiki/Abelian_von_Neumann_algebra#Classification) of the possible masas of $B(H)$ implies that, if the answer to my question is affirmative, then the spectra of all masas of $B(H)$ are in bijection with one another.)



Comparing cardinalities of the spectrum of two masas in $B(H)$
**Yes**. The spectra of $\ell\_\infty$ and $L\_\infty$ have the same cardinality, namely $2^{\mathfrak{c}}$.


Indeed, every infinite, compact $F$-space space (in particular, an extremely disconnected compact space such as the spectrum of $L\_\infty$) contains a copy of $\beta \mathbb{N}$ (which happens to be the spectrum of $\ell\_\infty$). This is 14N(5) in



> 
> L. Gillman and M. Jerison, *Rings of continuous functions*, van Nostrand Reinhold, New York, 1960.
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> 
> 


and it is quite elementary. The [Hausdorff–Pospíšil theorem](https://math.stackexchange.com/questions/83526/the-set-of-ultrafilters-on-an-infinite-set) implies that $|\beta\mathbb{N}|=2^{\mathfrak{c}}$. Thus, $2^{\mathfrak{c}}\leqslant |{\rm spec}\, L\_\infty|$.


I *claim* that the cardinality of ${\rm spec}\, L\_\infty$ cannot be bigger than $2^{\mathfrak{c}}$. Indeed, $L\_\infty^\*$ is a bidual of a separable Banach space, hence by [Goldstine's theorem](http://en.wikipedia.org/wiki/Goldstine_theorem), it is separable in the weak\*-topology. [Every separable space has cardinality at most $2^{\mathfrak{c}}$](http://en.wikipedia.org/wiki/Separable_space#Cardinality). Thus 
$|{\rm spec}\, L\_\infty|\leqslant |L\_\infty^\*|\leqslant 2^{\mathfrak{c}}.$


