= Solution
A strongly continuous <unitary representation> of the <circle group> decomposes as $H=\widehat\bigoplus_{m\in\mathbb Z}H_m$, with $U_z\xi=z^m\xi$ on $H_m$. Its <positive energy unitary representation of the circle> condition is that the occurring integers are bounded below. Equivalently $U_{e^{it}}=e^{itL}$ with a <self-adjoint> generator bounded below. An integer character can shift the lowest weight to zero.
Let $M=S''$. Because $S=S^*$, this is a <Von Neumann algebra>, and rotation invariance makes $U_zM'U_z^*=M'$. Joint irreducibility implies $M'\cap\{U_z\}'=\mathbb CI$: a nonscalar <self-adjoint> element of that joint <commutant> would have a nontrivial invariant <spectral projection gives a reducing subspace>.
For $T\in M'$, take its strong <Fourier coefficients> $T_k=\int_{\mathbb T}z^{-k}U_zTU_z^*\,dz$. They lie in $M'$ and satisfy $U_zT_kU_z^*=z^kT_k$. Both $T_k^*T_k$ and $T_kT_k^*$ are rotation-invariant elements of $M'$, hence scalar. If $T_k\ne0$, both scalars equal $\|T_k\|^2$, so $T_k/\|T_k\|$ is a unitary. It shifts energy by $k$. For $k\ne0$, applying it or its adjoint to a nonzero lowest-energy <vector> produces a <vector> below the lowest energy, a contradiction. Thus $T_k=0$ for every $k\ne0$, while $T_0$ is scalar.
<Fejér sum> averaging of the strongly continuous orbit $U_zTU_z^*$ converges strongly to $T$. Its <Fourier coefficients> have just shown every such average is $T_0$, so $T=T_0$ is scalar. In particular \b[every <bounded operator> commuting with $S$ commutes with every $U_z$]. In fact the hypotheses force the stronger conclusion $S'=\mathbb CI$.
Back to article page
