8
$\begingroup$

It is clear that commutative C*-algebras correspond to locally compact Hausdorff spaces. And locally compact Hausdorff spaces are completely regular. Now, does the complete regularity statement have some kind of generalization to general C*-algebras. In particular, I am looking for a statement using pure states of the C*-algebra. So something of the form:

If $\tau$ is a pure state on the C*-algebra and if $T$ is a collection of pure states satisfying some conditions (this should correspond to the closed set condition in commutative cases), then there exists an element $a$ of the $C^*$-algebra such that $\tau(a) = 1$ and $\sigma(a)=0$ for each $\sigma\in T$.

Is is a statement like this true? Are there other interesting generalizations of complete regularity?

$\endgroup$
1
  • $\begingroup$ [deleted some earlier comments, which arose from me misreading the question] $\endgroup$
    – Yemon Choi
    Sep 4 '15 at 19:52
12
$\begingroup$

You may be looking for the noncommutative Urysohn lemma: if $A$ is a C*-algebra and $p,q \in A^{**}$ are projections, $p \leq q$, with $p$ compact and $q$ open, then there is a positive element $x \in A$ such that $p \leq x \leq q$. It is due to Akemann, The general Stone-Weierstrass problem, J. Funct. Anal. 4 (1969), 277-294. Or just Google "noncommutative Urysohn lemma".

$\endgroup$
4
  • $\begingroup$ Thanks a lot Nik. This is helpful and I will most definitely look into this. But I was looking more at something that uses pure states. Is there some obvious relation between pure states and projections in $A^{**}$ that I am missing? $\endgroup$ Oct 25 '15 at 14:57
  • $\begingroup$ Yes, there is a relation, specifically the map taking projections p in A** to the set of states Φ with Φ(p)=0 is a bijection between open projections and weak*-closed faces of the state space (each of which is the closed convex hull of its extreme points, i.e. pure states). A projection p in A** is closed iff 1-p is open and compact iff there is also some positive a in A with p<=a (so if A is unital closed=compact). In particular, pure states correspond to minimal compact projections. See Pedersen's "C*-Algebras and their Automorphism Groups" or the papers by Akemann et al mentioned there. $\endgroup$ Oct 25 '15 at 15:28
  • $\begingroup$ I think you can just say that the pure states correspond to the minimal projections in $A^{**}$. $\endgroup$
    – Nik Weaver
    Oct 26 '15 at 3:13
  • $\begingroup$ True, I just thought the other terms should be explained too. Incidentally, one might hope that a truly non-commutative Urysohn lemma would apply to projections that do not commute, leading me to ask this question. $\endgroup$ Oct 29 '15 at 10:19
5
$\begingroup$

Nik's answer seems perfectly suited to what the OP was looking for, but there are indeed other generalizations of (complete) regularity to C*-algebras, as shown in Rosický's "Multiplicative lattices and C*-algebras" Cahiers Top. Geom. Diff. Cat. tome 30 no 2 (1989), 95-110. To paraphrase part of that paper, note first that regularity can be stated in terms of open and closed sets, without reference to points. Specifically, a topological space $X$ is regular iff, for every open $O\subseteq X$, $$O=\bigcup\{N:N\text{ is open and }\overline{N}\subseteq O\}.$$ So a natural question to ask is whether, for any C*-algebra $A$ and open projection $q\in A^{**}$, $$q=\bigvee\{p\in A^{**}:p\text{ is an open projection and }\overline{p}\leq q\}?$$ The answer is yes. To see this, take any $a\in A^1_+$ with $a\leq q$ and note that $\overline{a_{(\epsilon,1]}}\leq a_{[\epsilon,1]}\leq a_{(0,1]}\leq q$, for any $\epsilon>0$, where $a_S$ denotes the spectral projection of $a$ in $A^{**}$ corresponding to $S\subseteq\mathbb{R}$. The result now follows because $q$ being open means $$q=\bigvee\{a_{(0,1]}:a\in A^1_+\text{ and }a\leq q\}.\\$$

As $\overline{N\cup M}=\overline{N}\cup\overline{M}$, for open subsets $N$ and $M$, above we actually have a directed union. On the other hand, we can have $\overline{p\vee r}\neq\overline{p}\vee\overline{r}$ for open projections $p$ and $r$, as in Example II.6 in Akemann's paper referred to by Nik, which also shows that the supremum above is not always directed. However, we might ask:

Can the open projections above be replaced by a directed subset?

In general I do not know. If $A$ is separable then the answer is yes, as then every open $p$ will be of the form $a_{(0,1]}$, for some $a\in A^1_+$, and $a_{(\epsilon,1]}$ is directed, for $\epsilon>0$. More generally, the answer is yes if $A$ has an "almost idempotent" approximate unit $(a_\lambda)\subseteq A^1_+$, i.e. such that $a_\lambda a_\gamma=a_\lambda$ whenever $\lambda<\gamma$. Wondering if, in fact, all C*-algebras have such an approximate unit led me to post this question.

Regardless, in general we can still get a weaker iterated form of directedness. Specifically, note that $(a+b)_{(0,1]}=a_{(0,1]}\vee b_{(0,1]}$, for all $a,b\in A^1_+$, so every open projection $r$ is a directed supremum of open projections $q$, which are all themselves directed supremums of open projections $p$ with $\overline{p}\leq q$.

Also note the relation $\overline{N}\subseteq O$ can be expressed purely in terms of open sets as $$\text{there exists open $M$ disjoint from $N$ such that }X=M\cup O.$$ Wondering if, for open projections $p,q\in A^{**}$, $\overline{p}\leq q$ can also be expressed similarly in terms of open projections just led me to post this question. In fact, it is really relation 2. of that question, used with iterated directness as above, that Rosický considers as the appropriate notion of regularity for C*-algebras in the paper mentioned above.

For complete regularity one replaces $\overline{N}\subseteq O$ with the statement that there exist open $(O_r)$ for $r\in[0,1]$ (or $[0,1]\cap\mathbb{Q}$, as in Rosický's paper) such that $N\subseteq O_0$, $O_1\subseteq O$ and $\overline{O}_s\subseteq O_t$ whenever $s<t$. Again, the analogous statement for complete regularity holds for open projections in C*-algebras although again it is not clear if the supremum can always be replaced with a directed supremum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.