If $a,b$ are commuting normal elements of a $C^*$-algebra, then does $a$ commute with $b^*$?

Question

If $a,b$ in a $C^*$-algebra, such that $aa^*=a^*a,bb^*=b^*b$, and $ab=ba$.

Then do we have $ab^*=b^*a$ and $a^*b=ba^*$?

Answer 1

This is true, and this result is known as Fuglede's theorem. Here is a proof outline:

(STEP 1) If $A$ is a unital $C^*$-algebra and $a,b \in A$, show that the map $$f: \mathbb{C}\to A: \lambda \mapsto e^{i\lambda b}a e^{-i\lambda b}$$ is differentiable with $f'(0) =i(ba-ab).$

(STEP 2) Let $A$ be an arbitrary $C^*$-algebra and $a\in A$ normal and $b \in A$ an element that commutes with $a$. Let $\widetilde{A}$ be the unitisation of $A$ and define $$f: \mathbb{C}\to \widetilde{A}: \lambda \mapsto e^{i \lambda a^*}be^{-i \lambda a^*}.$$ Show that $\|f(\lambda)\| = \|b\|$ for all $\lambda \in \mathbb{C}$ and invoke's Liouville's theorem from functional analysis to deduce that $f$ is necessarily constant (hint: Liouville's theorem applies only to analytic functions that take values in $\mathbb{C}$, so you will have to compose with functionals on $\widetilde{A}$ and use that functionals separate the points).

(STEP 3) Use step 2 to conclude.