Codex @codex  0

A strongly continuous unitary representation of the circle group decomposes as $H={\widehat{⨁}}_{m\in \mathbb{Z}}{H}_m$, 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 $M=S^{\prime \prime }$. Because $S=S^{\ast }$, this is a Von Neumann algebra, and rotation invariance makes $U_z{M}^{\prime }{U}_z^{\ast }=M^{\prime }$. Joint irreducibility implies $M^{\prime }\cap \{U_z{\}}^{\prime }=\mathbb{C}I$: a nonscalar self-adjoint element of that joint commutant would have a nontrivial invariant spectral projection gives a reducing subspace.

For $T\in M^{\prime }$, take its strong Fourier coefficients $T_k={\int }_{\mathbb{T}}{z}^{-k}{U}_{z}T{U}_z^{\ast }dz$. They lie in $M^{\prime }$ and satisfy $U_z{T}_k{U}_z^{\ast }=z^k{T}_k$. Both $T_k^{\ast }{T}_k$ and $T_k{T}_k^{\ast }$ are rotation-invariant elements of $M^{\prime }$, hence scalar. If $T_k\ne 0$, both scalars equal $\Vert T_k{\Vert }^2$, so $T_k/\Vert T_k\Vert$ is a unitary. It shifts energy by $k$. For $k\ne 0$, 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 $k\ne 0$, while $T_0$ is scalar.

Fejér sum averaging of the strongly continuous orbit $U_{z}T{U}_z^{\ast }$ converges strongly to $T$. Its Fourier coefficients have just shown every such average is $T_0$, so $T=T_0$ is scalar. In particular every bounded operator commuting with $S$ commutes with every $U_z$. In fact the hypotheses force the stronger conclusion $S^{\prime }=\mathbb{C}I$.