When is the set of positive elements in a C* algebra a totally ordered set

Question

As the title suggests I was wondering if there are any properties that ensure a C*-algebra has all positive elements comparable to each other. ( Recall $a\leq b$ if $b-a$ is a positive element in the C* algebra)

Answer 1

Here is a partial answer:

Let $A$ be a $C^*$-algebra. Assume that there is a projection $p$ such that $p \not \in \{0,1\}$. Then $p$ and $1-p$ are not comparable : indeed, the spectrum of $p - (1-p)$ is $\{\pm 1\}$ so $p - (1-p)$ is not positive, and the same goes for $1-p - p = 1 - 2p$.

EDIT:

Claim: It never happens if the algebra is unital and not $\mathbb{C}$.

Let $A$ be a unital $C^*$-algebra of dimension greater or equal than $2$. Then there is $a \in A$ such that the subalgebra generated by $a$ isn't isomorphic to $\mathbb{C}$.

[EDIT2: We can assume that $a$ is normal. Indeed, if both $a+a^*$ or $a-a^*$ are elements of $\mathbb{C}$, then $a$ is as well. So, $a+a^* \not \in \mathbb{C}$ or $a-a^* \not \in \mathbb{C}$, and each of these possibilities gives a normal element of $A \setminus \mathbb{C}$.]

Let us denote by $B$ the closure of this subalgebra. Then $B$ is commutative, since $a$ is assumed to be normal, so it is isomorphic to $C_0(X)$ where $X$ is a compact space with at least two points, $x$ and $y$. Since $X$ is normal, there is a positive continuous function $f$ such that $f(x) = 0$ and $f(y) = 1$, and there is a positive continuous function $g$ such that $g(x) = 1$ and $g(y) = 0$. Then $f$ and $g$ are not comparable.