Unitary and central in a C* algebra

Question

Is it true that if an element of a unital $C^*$-algebra is unitary and commutes with all other elements, then it is a scalar multiple of the identity?

Answer 1

Let $A$ be a unital $C^*$-algebra, and $Z(A)$ the center of $A$. Then $Z(A)$ is a $C^*$ sub-algebra of $A$ and a commutative unital $C^*$-algebra by itself. $Z(A)$ is therefore isomorphic to $C(\Delta)$ where $\Delta$ is the maximal-ideal-space/space-of-homomorphisms of $Z(A)$. $\Delta$ is a compact Hausdorff space (with the weak-* topology from the Banach-space dual of $Z(A)$.) The unitaries of $Z(A)$ correspond to functions from $\Delta$ to the unit circle $S^1 = \{z: |z|=1\}\subseteq\Bbb{C}$, and they are unitaries of $A$ as well. Since $\Delta$ is compact-Hausdorff, it is normal and by Urysohn's lemma completely normal. It follows that if $\Delta$ has more than $1$ point, there are non-constant continuous functions from $\Delta$ to $\Bbb{R}$ and therefore to $S^1$. Therefore, the unitaries of $Z(A)$ are all scalar if and only if $\Delta$ is a one-point space if and only if $Z(A) = \Bbb{C}$.

Obviously not every $C^*$-algebra has $Z(A)=\Bbb{C}$, but there are such. If they also happen to be von-Neumann algebras, they're called factors. For example $M_n(\Bbb{C})$ and the algebra of bounded operators on an infinite-dimensional separable Hilbert space are (type I) factors. There are also $C^*$-algebras with $Z(A)=\Bbb{C}$ that are not von-Neumann. These can be constructed for instance from ergodic actions of groups on topological spaces.

Answer 2

This is false in any C$^*$-algebra with non-trivial centre.

The simplest example is to take $u=(1,-1)$ in $\mathbb C^2$.