Let $A$ be a $C^*$-algebra, and $S,T\in A$ be normal. If $ST=TS$, then must $S^*T=TS^*$?

Question

I wish to show a kind of "simultaneous diagonalizability" statement for commuting normal operators $S,T\in\mathscr{L}(H)$ over a complex separable Hilbert space $H$. However, I've run into the hitch that even if $S$ and $T$ commute, I don't see why $S^*$ and $T$ should. So is it true that $ST=TS$ implies $S^*T=TS^*$?

Answer 1

The proof of this which I have seen goes like this. First a quick lemma:

Lemma: Let $S,T \in A$. Then, $ST=TS$ commute if and only if $e^{zS}T=Te^{zS}$ for all $z \in \mathbb{C}$.

Proof: The forward implication follows from the series formula for $e^{zS}$. For the reverse implication, apply $\frac{d}{dt}|_{t=0}$ to both sides of $e^{tS}T=Te^{tS}$.

Now the main result:

Fuglede's Theorem: Let $S,T \in A$ with $S$ normal. Then $ST=TS$ if and only if $S^*T=TS^*$.

Proof: It suffices to prove one direction. For notational convenience, I'll actually assume $S^*T=TS^*$. We need to show $e^{zS}Te^{-zS}=T$ for all $z \in \mathbb{C}$. We have $$e^{zS}Te^{-zS} = e^{zS} e^{-(z S)^*} T e^{(z S)^*} e^{-zS}=e^{zS-(zS)^*} T e^{-zS + (zS)^*}$$ where the first equality uses $S^*T=TS^*$ and the lemma, and the second uses normality of $S$. But now note that $e^{zS-(zS)^*}$ (and its inverse) is unitary (since $zS-(zS)^*$ is anti-self-adjoint), and it follows that the right hand side is bounded (with norm identically equal to $\|T\|$). On the other hand, the left hand side is a holomorphic function of $z$, so the result follows from an appropriate version of Liouville's theorem.

Added: I found the place where I saw this argument. See Lemma 6.1 in these lecture notes of John Roe.

Answer 2

This result is not very trivial at all. It is true, and this result is known as Fuglede's theorem (in fact, Fuglede's theorem only requires that $S$ is a normal operator).