Commuting operators and polar decomposition

Suppose that $V$ is an isometry and $X$ an arbitrary operator on a Hilbert space $H$. Let $X=U|X|$ be the polar decomposition for $X$.

If $VX=XV$, can I conclude that $VU=UV$?

Solution 1:

Edit: No, it does not hold in general. For some reason I always tended to use that $V$ commuted with $X^*$, which in general doesn't have to be true. A counterexample: Let $\mathcal{H}=\ell^2(\mathbb{N})$. Let $V$ be the right shift on $\mathcal{H}$, i.e. $V\delta_n=\delta_{n+1}$ where $\delta_n(m)=\left\{\begin{matrix} 1 & \mbox{if }n=m\\0 & \mbox{if }n \neq m\end{matrix}\right.$. Let $X=2+V$ (so $X\delta_n=2\delta_n+\delta_{n+1}$) then $XV=VX$. However, you can check that $X^*V \neq VX^*$. Now $X$ is invertible, so $U$ is unitary. But then $UV=VU \Leftrightarrow X^*V=VX^*$, which is false.

So the following only holds if $V$ commutes with $X^*$: Let $\mathcal{X}$ be the von Neumann algebra generated by $X$. One can show that in this case $U\in \mathcal{X}$. As clearly $V$ is an element of the commutant of $\mathcal{X}$ and $U \in \mathcal{X}$, we find that $VU=UV$. So I don't think you need the fact that $V$ is an isometry...