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If I imagine that (the self-adjoint part of) a C*-algebra $A$ represents the algebra of observables of some quantum system, then certain perspectives 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 $C,D \subseteq B(H)$?

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 for a bit more detail), the spectrum of a commutative AW*-algebra $K$ is the Stone space 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; 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 $2^\mathbb{N}$?

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 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.)

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  • $\begingroup$ I don't understand your last statement. Isomorphic in what category? $\endgroup$
    – Yemon Choi
    Commented Feb 26, 2015 at 19:31
  • $\begingroup$ If I understand your question correctly, it does not really have anything to do with MASAs. YOu are merely comparing the Gelfand spectra of the two Banach algebras $L^\infty[0,1]$ and $\ell^\infty({\bf N})$ and asking if the two spectra have the same cardinality -- is that correct? $\endgroup$
    – Yemon Choi
    Commented Feb 26, 2015 at 19:33
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    $\begingroup$ @YemonChoi, I meant in the category of sets. I'll edit accordingly in a moment. And you're right about my question; but the only reason that I would dream to ask if these spectra have the same cardinality is that they both occur as masas of the same C*-algebra. $\endgroup$
    – Manny Reyes
    Commented Feb 26, 2015 at 19:34

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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.

and it is quite elementary. The Hausdorff–Pospíšil theorem 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, it is separable in the weak*-topology. Every separable space has cardinality at most $2^{\mathfrak{c}}$. Thus $|{\rm spec}\, L_\infty|\leqslant |L_\infty^*|\leqslant 2^{\mathfrak{c}}.$

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  • $\begingroup$ Thank you! This answer is so nice, especially the use of (pre)duals, that I now suspect a similar argument could be made for the spectrum of masas of $B(H)$, independent of the dimension of $H$. $\endgroup$
    – Manny Reyes
    Commented Feb 26, 2015 at 20:37
  • $\begingroup$ @Manny Reyes, for non-separable spaces the situation is trickier but I guess that the only types of masas are the following: $\ell_\infty(\lambda)$ and $L_\infty(\{0,1\}^\lambda)$ and certain $\ell_\infty$-sums of them ($\lambda$ is the dimension of the Hilbert space) but at the end of the day the spectra will have the same cardinality. $\endgroup$
    – Tomasz Kania
    Commented Feb 26, 2015 at 20:42
  • $\begingroup$ I was thinking that one might be able to work even without resorting to an explicit description. If $C \subseteq B(H)$ is a masa, then the predual $C_*$ will be a homomorphic image of $B(H)_*$ (trace class operators), giving upper bounds on $C_*$ and then presumably on $C^* = (C_*)^{**}$. If this upper bound coincides with $|\mathrm{Spec}(\ell^\infty(\dim(H))| = 2^{2^{\dim(H)}}$, and if one can similarly embed $\beta \dim(H)$ into $\mathrm{Spec}(C)$, then we'd have effectively the same proof. $\endgroup$
    – Manny Reyes
    Commented Feb 26, 2015 at 20:47
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    $\begingroup$ No, in general you cannot embed $\beta \lambda$ into ${\rm spec}\, L_\infty(\{0,1\}^{\lambda})$. This fails for all uncountable $\lambda$. (You will find this in old papers of H. P. Rosenthal.) $\endgroup$
    – Tomasz Kania
    Commented Feb 26, 2015 at 20:51
  • $\begingroup$ OK, that's no good. Can we get an injective $*$-homomorphism $\ell^\infty(\lambda) \hookrightarrow L^\infty([0,1]^\lambda)$, perhaps? (We could construct a normal $*$-homomorphism if there are $\lambda$-many orthogonal projections in the latter algebra that sum to 1.) This would give a surjection of spectra in the opposite direction. $\endgroup$
    – Manny Reyes
    Commented Feb 26, 2015 at 21:01

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