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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.

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    Let's see. If no-one else posts an answer I'll do so in the next few days. Concerning the lattice of projections in C\*-algebras see Jonas Meyer's answer here: http://math.stackexchange.com/q/1151602012-11-13

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To compile the comments and add a little bit of info beyond leslie townes's answer here:

  1. 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?

  2. 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.

  3. 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:

    1. 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.
    2. 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:

    1. 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.

    2. (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 $F$.

  4. 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

    • $a \geq 0$ 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.

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    I don't know enough to say much about the structure of projections in C\*-algebras. Jonas Meyer's answer I mentioned in the comments should already give some good pointers, otherwise I'd suggest to ask a separate question about that.2012-11-15