properties of the poset of self adjoint elements in a $C^*$ algebra It is well known that the set of self-adjoint elements in a $C^*$ algebra naturally forms a poset. I assume that the general properties of this poset are known. Can anybody point me to a reference where the basic properties of that poset are established?
In particular, I'm interested to know when the poset will be a lattice, a complete lattice, and when the joins/meets behave nicely with respect to the $C^*$ algebra operations. 
 A: To compile the comments and add a little bit of info beyond leslie townes's answer here:


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*In order for the self-adjoint elements $A_{\rm sa}$ of a unital $C^\ast$-algebra $A$ to be a Riesz lattice in its natural order it is necessary (and sufficient) that $A$ be commutative. 
The non-trivial direction is Theorem 1 in S. Sherman's Order in Operator Algebras, American Journal of Mathematics
Vol. 73, No. 1 (Jan., 1951), pp. 227-232.
For a nice elaboration on these ideas see leslie townes's answer to C*-algebras as Banach lattices?

*If $A$ is a unital and commutative $C^\ast$-algebra, then $A = C(K,\mathbb{C})$ where $K = \sigma(A)$ is the spectrum of $A$ which is a compact Hausdorff space. The subspace of self-adjoint elements can then be identified with $A_{\rm sa} = C(K,\mathbb{R})$. This always is a lattice with respect to the pointwise order. Of course, the order and algebra structures interact as nicely as possible.

*Nakano proved in Über das System aller stetigen Funktionen auf einem topologischen Raum, Proc. Imp. Acad. Volume 17, Number 8 (1941), 308-310 the following two results. See also Stone, Boundedness properties in function-lattices, Canad. J. Math. 1(1949), 176-186: 


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*Let $X$ be a completely regular space. If the space of bounded and continuous functions $C_b(X,\mathbb{R})$ is an order-complete Riesz lattice, then the space $X$ is extremally disconnected, i.e., the closure of every open set is open.

*If $X$ is an extremally disconnected topological space, then $C_b(X,\mathbb{R})$ is an order-complete Riesz lattice.


Since Nakano's paper is in German and this is a very special case of Stone's related arguments (which are vastly more general and therefore a bit convoluted), here's an outline of the proof:


*

*Suppose $C_b(X,\mathbb{R})$ is order-complete and let $U \neq \emptyset$ be open. Consider the set $F(U)$ of continuous functions $f \geq \chi_{U}$. Since $C_b(X,\mathbb{R})$ is order-complete, there exists a continuous function $g \geq \chi_{U}$ such that $g \leq f$ for all $f \in F(U)$. We are going to prove that $g = \chi_{\overline{U}}$. As $g$ is continuous, this will imply that $\overline{U}$ is clopen, hence $X$ is extremally disconnected.   
Because $g \geq \chi_{U}$ and $g$ is continuous, we must have $\overline{U} \subset \{x : g(x) \geq 1\}$. If $g(x_0) \gt 1$ for some $x_0 \in \overline{U}$, then there is an open neighborhood $V$ of $x_0$ such that $g(v) \gt 1 + \varepsilon$ for all $v \in V$. Since $X$ is completely regular, we can choose a continuous $h \colon X \to [0,1]$ with support in $V$ and $h(x_0) = 1$. Then $g- \varepsilon h \geq \chi_U$, so  $h \in F(U)$. However $h \lt g$ contradicts $g \leq f$ for all $f \in F(U)$. Similarly one shows $g(x) = 0$ for all $x \in X \setminus \overline{U}$, so $g = \chi_{\overline{U}}$ as claimed earlier.

*(Sketch)
Suppose that $X$ is extremally disconnected. Let $F \subset C_b(X,\mathbb{R})$ be bounded above. Set $g(x) = \sup\{f(x): f \in F\}$. Then $g$ is lower semicontinuous and hence $U_\alpha = \{x \in X : g(x) \gt \alpha\}$ is open for all $\alpha \in \mathbb{R}$. Let $h(x) = \sup{\{\alpha : x \in \overline{U_\alpha}\}}$. Observe that $h(x) \geq g(x)$. Moreover, $\{x \in X : h(x) \geq \alpha\} = \overline{U_\alpha}$. Since $X$ is extremally disconnected, we know that $\overline{U_\alpha}$ is clopen so that $h$ is both upper and lower semicontinuous, hence continuous. It is not difficult to see that $h$ is the least upper bound for $F$.


*Here's a simple fact related to 1. As usual with Riesz lattices the proof is somewhat lengthy to carry out from scratch; it is proved as Theorem 140.10 on page 658 of Luxemburg-Zaanen, Riesz Spaces, Vol II:  
Suppose $A$ is a Riesz lattice with an associative algebra structure. Suppose further that order and multiplication are compatible in the sense that 


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*$a \geq 0$ and $b \geq 0$ implies that $ab \geq 0$

*$a \wedge b = 0$ implies $(ac) \wedge b = 0 = (ca) \wedge b$.
Algebras with these two properties are called $f$-algebras in the literature.
If the order on $A$ is Archimedean (i.e., $na \leq b$ for all $n \in \mathbb{N}$ implies $a \leq 0$) then $A$ must be commutative.
