Proposition 44.3.2.label Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, $x: D(x) \to H$ be a closed densely defined operator, $x = u|x|$ be its polar decomposition, and $E: \cb_{\real}\to B(H)$ be the resolution of the identity for $|x|$, then the following are equivalent:
- (1)
$x$ is affiliated with $M$.
- (2)
$u \in M$ and $E(B) \in M$ for all $B \in \cb_{\real}$.
Proof. (1) $\Rightarrow$ (2): Let $v \in U(M')$, then for any $\xi, \eta \in D(x^{*}x)$,
so $x^{*}x$ is affiliated with $M$. As such, $(x^{*}x-i)(x^{*}x+i)^{-1}\in M'' = M$ by the Bicommutant Theorem. By construction of the spectral measure, $E(B) \in M$ for all $B \in \cb_{\real}$, and $f(x^{*}x)$ is affiliated with $M$ for all $f \in B^{\infty}([0, \infty); \complex)$. In particular, $|x|$ is affiliated with $M$ as well.
As such, for any $v \in U(M')$, $uv|x| = u|x|v = xv = vx = vu|x|$, so $vu = uv$ on $\ol{|x|(H)}$. On the other hand, for any $\xi \in \ker(|x|)$, $|x|v\xi = v|x|\xi = 0$, so $v(\ker(|x|)) \subset \ker(|x|)$ and $\ker(uv) \supset \ker(|x|)$. As $H = \ker(|x|) \oplus \ol{|x|(H)}$, $uv = vu$ on $H$.$\square$
Post a Comment