I'm currently reading the book "A Course in Abstract Harmonic Analysis" of Folland and I oversee a point:
Let $\pi_1$ and $\pi_2$ be unitary representations of $G$, an intertwining operator for $\pi_1$ and $\pi_2$ is a bounded linear map $T: \mathbb H_{\pi_1} \to \mathbb H_{\pi_2}$ such that $T \pi_1(x)=\pi_2(x)T$ for all $x \in G$. The set of all such operators is denoted by $C(\pi_1,\pi_2)$.
For the proof of Schur's Lemma, we use the fact that of $T \in C(\pi)$ then $T^* \in C(\pi)$. Why do we have this? Is it because $C(\pi)$ is a $C^*$ algebra or what do I oversee?
If $T\in C(\pi),$ i.e. (I suppose) if $\pi$ is a unitary representation of $G$ and $T$ commutes with $\pi(x)$ for all $x\in G,$ then $T^*$ commutes with $\pi(x)^*=\pi(x)^{-1}=\pi(x^{-1})$ for all $x\in G,$ i.e. with $\pi(y)$ for all $y\in G,$ hence $T^*\in C(\pi).$
It is because of this that $C(\pi)$ is a $C^*$-algebra, and not the converse.