polar decomposition in von Neumann algebras

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

Proof :

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

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

Suppose $S\in\mathcal{M}^{\prime}$. 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 $\overline{Ran\;R}$.

As $R$ is self-adjoint, $\overline{Ran\;R}^{\perp}=KerR$, 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 $V\in\mathcal{M}^{\prime\prime}=\mathcal{M}\;\square$

Title polar decomposition in von Neumann algebras PolarDecompositionInVonNeumannAlgebras 2013-03-22 17:28:54 2013-03-22 17:28:54 asteroid (17536) asteroid (17536) 7 asteroid (17536) Result msc 47A05 msc 46L10