polar decomposition in von Neumann algebras

- Let $ℳ$ be a von Neumann algebra acting on a Hilbert space $H$ and $T\in ℳ$. If $T=VR$ is the polar decomposition for $T$ with $KerV=KerR$, then both $V$ and $R$ belong to $ℳ$.

Proof :

• As $ℳ$ is a $C^{*}$-algebra (http://planetmath.org/CAlgebra), it is known that $R=\sqrt{T^{*}T}$ belongs to $ℳ$. (proof will be added later)

• To see that $V$ also belongs to $ℳ$, by the double commutant theorem, it suffices to show that $V$ belongs to ${ℳ}^{\prime \prime }$ (the double commutant of $ℳ$).

  Suppose $S\in {ℳ}^{\prime }$. We intend to prove that $V$ commutes with $S$.

  For $x\in H$ we have that

  $$TSx=STx=SVRx$$

  and

  $$TSx=VRSx=VSRx$$

  So $SV$ and $VS$ agree on $\overline{RanR}$.

  As $R$ is self-adjoint, ${\overline{RanR}}^{⟂}=KerR$, and so it remains to show that $SV$ and $VS$ agree on $KerR$. Recall that, by hypothesis, $KerR=KerV$.

  Let $x\in KerR$. We have that $RSx=SRx=0$ and therefore

  $$S(KerR)\subseteq KerR=KerV$$

  and so we can conclude that $VS$ is identically zero in $KerR$.

  Clearly $SV$ is also identically zero on $KerR=KerV$.

  Thus $VS$ and $SV$ agree on $KerR$. Therefore $SV=VS$ and so $V\in {ℳ}^{\prime \prime }=ℳ\mathrm{\square }$