# Classification of $C^*$ algebras whose all non scalar elements have disconnected spectrum

To what extent have all unital $C^*$ algebras $A$ with the following property been classified? Is there a simple $C^*$ algebra with this property? Does $C(K)$ satisfy this property, where $K$ is an arbitrary compact extremally disconnected space?

The scalar elements, are the only elements with connected spectrum.

## 4 Answers

In the commutative case $A=C(K)$, this is equivalent to asking for which compact Hausdorff spaces there are no continuous surjections $K\to[0,1]$, since the spectrum of $f\in C(K)$ is its image. A compact Hausdorff space $K$ has this property iff $K$ is scattered: that is, iff every nonempty subset of $K$ has an isolated point.

Indeed, suppose $K$ is not scattered. Then $K$ contains a nonempty subset $P\subseteq K$ with no isolated points, which we may assume to be closed (take its closure). If $P$ has a nontrivial connected subset $Q$, then any function in $C(Q)$ has connected image, and thus any nonconstant function on $Q$ gives a continuous surjection $Q\to[0,1]$. Otherwise, $P$ is totally disconnected, and so since $P$ is perfect it surjects onto the Cantor set. Since there is a continuous surjection from the Cantor set to $[0,1]$, there is a continuous surjection $P\to [0,1]$.

In either case, we get a continuous surjection from a closed subspace of $K$ to $[0,1]$. This function then extends to a continuous surjection $K\to[0,1]$.

Conversely, suppose there exists a continuous surjection $f:K\to[0,1]$. Note that if $\mathcal{C}$ is a chain of closed subsets of $K$ on which $f$ is surjective, then $f$ is also surjective on $\bigcap\mathcal{C}$, since for each $x\in[0,1]$ the sets $S\cap f^{-1}(\{x\})$ for $S\in\mathcal{C}$ are closed and have the finite intersection property. So by Zorn's lemma, there exists a minimal closed subset $P\subset K$ on which $f$ is surjective. This subset $P$ cannot have an isolated point $x$, since then $P\setminus\{x\}$ would be closed but $f$ would be surjective on $P\setminus\{x\}$ since its image must be closed in $[0,1]$ and contain all but at most one point of $[0,1]$. Thus $K$ is not scattered.

Some examples of compact Hausdorff scattered spaces include all compact Hausdorff spaces of cardinality $<2^{\aleph_0}$, one-point compactifications of discrete spaces, and successor ordinals. A closed subset of $\mathbb{C}$ is scattered iff it is countable, and it follows easily that if $K$ is compact Hausdorff and scattered, then actually every element of $C(K)$ has countable spectrum.

Finally, to address your last question, if $K$ is infinite, compact, and extremally disconnected, then $K$ is not scattered. Indeed, in that case $K$ is the Stone space of an infinite complete Boolean algebra $B$. Since $B$ is infinite, it contains a countably infinite set of disjoint nonzero elements $\{b_0,b_1,b_2,\dots\}$. We may assume the join of all of these elements is $1$ (if not, add the complement of the join as one more element in the set). There is then an embedding of Boolean algebras $\mathcal{P}(\mathbb{N})\to B$ sending $S\subseteq\mathbb{N}$ to $\bigvee_{n\in S}b_n$. This dually gives a continuous surjection from $K$ to the Stone space $\beta\mathbb{N}$ of $\mathcal{P}(\mathbb{N})$. There exists a continuous surjection $\beta\mathbb{N}\to[0,1]$ (map $\mathbb{N}$ to a countable dense subset of $[0,1]$), and thus there exists a continuous surjection $K\to[0,1]$.

I don't really know anything about the noncommutative case, but apparently there is a notion of a "scattered $C^*$-algebra" which there is a lot of literature on and which I would guess is equivalent to your condition.

The following might be useful. For a C*-algebra $A$, the following statements are equivalent:

1) $A$ is scattered;

2) the spectrum of any self-adjoint element in $A$ is countable

3) all maximal commutative C*-subalgebras of $A$ have a scattered Gelfand spectrum;

4) the Gelfand spectrum of each commutative C*-subalgebra of $A$ is a Stone space;

5) every commutative C*-subalgebra of $A$ is generated by its projections;

6) every C*-subalgebra $C\subseteq A$ is AF (but possibly non-separable, i.e. there is a directed set of finite-dimensional C*-subalgebras of $C$ whose union is dense in $C$);

7) Every separable commutative C*-subalgebra of $A$ is AF;

8) Every C*-subalgebra of $A$ has real rank zero;

9) Every positive functional on $A$ can be written as a countable sum of pure functionals;

10) every non-degenerate representation of $A$ is unitarily equivalent with a subrepresentation of a sum of irreducible representations;

11) The enveloping von Neumann algebra of $A$ is isomorphic to $\prod_{i\in I}B(H_i)$

12) $A$ does not have a commutative C*-subalgebras whose Gelfand spectrum is homeomorphic to $[0,1]$;

Scattered C*-algebras were defined in the articles "Scattered Algebras" and "Scattered Algebras II" by Jensen, where the equivalence of 9,10 and 11 have been proven. Kusuda proved in "C*-algebras in which every C*-subalgebra is AF the equivalence of all statements involving (commutative) C*-subalgebras. Lin proved in "The structure of Quasi-Multipliers of C*-algebras" that scattered C*-algebras are AF.

An example of a non-commutative scattered C*-algebra is $K(H)$, the compact operators on some Hilbert space $H$. An example of a unital non-commutative scattered C*-algebra is $K(H)+1_H\mathbb{C}$.

• $K(H)$ doesn't work as an example: $\left( \begin{smallmatrix} 0 & 1 \\ 0 & 0\end{smallmatrix} \right)$ (for $H=\mathbb C^2$) is an element with connected spectrum. (I think Ali points out a general version of this in his answer.) – Christian Remling Jul 3 '17 at 17:48
• You are right. I edited my post, but as it is now it is actually quite irrelevant, since his algebras must be the commutative scattered C*-algebras. – Bert Lindenhovius Jul 3 '17 at 22:18

If $K$ is compact, extremally disconnected and infinite, then for every compact subset $D\subseteq\mathbf{C}$ there exists $a\in C(K)$ with spectrum $\sigma(a)=D$.

[edit: Previously, $K$ was only assumed totally disconnected. As pointed out by Eric Wofsey, this is not enough.]

Proof: Choose a sequence $(x_k)_k$ in $D$ such that $\{x_k:k\geq 1\}$ is dense in $D$. Using that $K$ is totally disconnected and infinite, choose pairwise disjoint, closed and open subsets $A_k\subseteq K$ with $K=\bigcup_{k=1}^\infty A_k$. Let $a\colon K\to\mathbf{C}$ be the function that takes value $x_k$ on $A_k$, for each $k$. One checks that $a$ belongs to $C(K)$.

The spectrum of $a$ contains each $x_k$. Since the spectrum is closed, it contains $D$. One checks that the spectrum of $a$ is also contained in $D$. It follows that $\sigma(a)=D$.

• Typically it will not be possible to find such an $a$ that is continuous. Try it in the case $K=\mathbb{N}\cup\{\infty\}$, for instance: if the sequence $(x_k)$ does not converge, then you can't extend continuously to $\infty$. – Eric Wofsey Jul 3 '17 at 7:23
• Yes, you are right. So one needs a "bigger" space. I guess extremally disconnected (as was actually the original question) would do? – Hannes Thiel Jul 3 '17 at 7:43

Regarding the non commutative case, I just realized that a non commutative $C^*$ algebra has a non scalar element $1+a$ with connected spectrum $\{1\}$ where $a$ is a non zero nilpotent element. According to a theorem of Kaplanski, a non commutative algebra possess a non zero nilpotent element.

Quasinilpotent elements of group C-star algebras