WStarAlgebras have Unique Preduals

5 Stonean Spaces and Maximal Commutative Subalgebras

Recall that a compact Hausdorff space is Stonean if the closure of every open set is open. We note that Lemma 39 is only part of Sakai 1.3.1, since that is all we need for the proof of Theorem 45. If \(K\) is a topological space in which singletons are closed, and \(p\) is a projection in \(C(K)\), the spectrup of \(p\) is its range and is contained in \(\{ 0,1\} \). Furthermore, \(p^{-1}\{ 0\} =K\backslash p^{-1}\{ 1\} \), and so both of these sets are clopen, and identify connected components of \(K\). If \(K\) is Stonean, there are many such connected components:

Lemma 39

(Sak 1.3.1) Let \(K\) be a Stonean space. Then every positive self-adjoint element \(a\) in \(C(K)\) can be uniformly approximated by finite linear combinations of projections in \(C(K)\) having nonnegative coefficients.

Proof

Since \(a\) is self-adjoint, its spectrum is its range in \(\mathbb {R}_{\ge 0}\), and is therefore contained in \([0, \| a\| ]\), where \(\| a\| =\sup _{t\in K}|a(t)|\). Let \(\varepsilon {\gt}0\) and choose real numbers \(0{\lt}\lambda _1{\lt}\lambda _2{\lt}\ldots {\lt}\lambda _n {\lt} \| a\| +1\) so that \(|\lambda _{i+1}-\lambda _{i}|{\lt}\varepsilon \). Now let \(G_1= \overline{\{ t\in K | a(t){\lt}\lambda _1\} }\) and \(G_i=\overline{\{ t\in K \backslash \bigcup _{j=1}^{i-1}G_{i}|a(t){\lt}\lambda _i\} }\) for \(i{\gt}1\). These are pairwise disjoint, and \(\chi _{G_{i}}\) are each continuous projections. Furthermore, for all \(t\in K\),

\begin{align} |a(t)-\sum _{i=1}^{n}\lambda _i\chi _{G_{i}}(t)|{\lt}\varepsilon \end{align}

and therefore \(\| a-\sum _{i=1}^{n}\lambda _i\chi _{G_{i}}\| {\lt}\varepsilon \).

In Sakai, the pointwise order on \(C(K)\) in Proposition 41 isn’t specified beforehand in the book. Following \(C^{*}\)–algebra theory, \(a\ge 0\) in \(C(K)\) iff the range of \(a\) is contained in \(\mathbb {R}_{\ge 0}\), and \(b\le a\) iff \(b-a \ge 0\). The spectrum of a function in \(C(K)\) is just its range. So \(b\le a\) iff \(b(t)\le a(t)\) for all \(t\in K\). A net \((f_{\alpha })\) in \(C(K)\) is bounded if there is a positive constant that bounds the norms of all \(f_{\alpha }\). A net is said to be increasing if \(\alpha \le \beta \) implies \(f_{\alpha }\le f_{\beta }\) for all \(\alpha ,\beta \) in the associated directed set. We ought to note, here, that Sakai opts not to use net language (although he sort of does when introducing subscripts) since his directed set is identical to his net in Proposition 41. Recall that the support of \(f\in C(K)\) is \(\overline{\{ x\in K | f(x)\ne 0\} }\).

Lemma 40

Let \(K\) be a compact Hausdorff space, with \(U\subseteq K\) open, then the set of \(f\in C(K)\) supported on \(U\) such that \(0 \le f \le 1\) is a directed set w.r.t. the order on \(C(K)\).

Proof

The usual order is a partial order, and given \(f,g\in C(K)\) supported on \(U\) we have \(f\vee g \in C(K)\), supported on \(U\), where \(f\vee g(t)=\max \{ f(t),g(t)\} \).

Proposition 41

(Sakai 1.3.2) Let \(K\) be a compact Hausdorff space. Suppose every bounded increasing net of real valued, non-negative functions in \(C(K)\) has a least upper bound in \(C(K)\). Then \(K\) is Stonean.

Proof

Suppose for the sake of a contradiction that \(K\) isn’t Stonean, i.e. that there exists an open set \(U\) whose closure isn’t open, i.e. that \(\overline{U}^{c}\) isn’t closed. Consider the set of \(f\) in \(C(K)\) with support contained in \(U\) and such that \(0\le f\le 1\). By Lemma 29 and Lemma 40, this is a bounded, increasing net with respect to the order on \(C(K)\). For any point \(t\in U\), by Urysohn’s Lemma (compact Hausdorff Spaces are regular) there exists a continuous function \(g\in C(K)\) such that \(g(t)=1\) and \(g\) is zero on \(U^{c}\). It follows that if \(f_s\) is the least upper bound of the above net, then \(1=g(t)\le f_s(t)\), and therefore (since we can play this game for any \(t\in U\)) it must be that \(f_s(t)=1\) for any \(t\in U\). By continuity, \(f_s(t)=1\) for all \(t\in \overline{U}\). Now, since \(\overline{U}^{c}\subset U^{c}\), \(f_c\) is identically zero on \(U^{c}\). Since \(\overline{U}^{c}\) is not closed, it must have a limit point \(p\in \overline{U}\), and by continuity \(f_c(p)=0\) But \(f_c(p)=1\) by the above, which is a contradiction.

(Sak 1.7.5) If \(C\) is any maximal commutative \(C^*\)–subalgebra of the \(W^{*}\)–algebra \(M\), its spectrum space (maximal ideal space) is Stonean.

Proof

Suppose \((a_{\lambda })\) is a uniformly bounded, increasing net of positive elements in \(C\) with supremum \(a\). Given any unitary \(u\in C\), we have \(u^{*}\sup _{\lambda }a_{\lambda } u=\sup _{\lambda }u^{*}a_{\lambda }u\) and since every element in \(C\) is a linear combination of unitary elements from \(C\), we have that \(\sup _{\lambda }a_{\lambda }\in C\). The result follows from Proposition 41 after applying the Gelfand transform.