Polar decomposition in a finite von Neumann algebra

Question

Suppose $\mathcal{M}$ is a finite von Neumann Algebra and $a \in \mathcal{M}$. Does it follow that we can write $a = |a|u$ where $u$ is a unitary operator. I know we can write $a = |a|u$ where $u$ is a partial isometry.

Answer 1

Note that that to have $a=|a|u$ you need to define $|a|=(aa^*)^{1/2}$. This is fine and it works, but it is more common to take the polar decomposition as $a=u|a|$ with $|a|=(a^*a)^{1/2}$. In any case the first decomposition applied to $a^*$ yields the second one for $a$, and viceversa.

The partial isometry $u$ can be extended to a unitary in any finite von Neumann algebra (not necessarily finite-dimensional).

Indeed, if $a=u|a|$ (with $|a|=(a^*a)^{1/2}$) is the polar decomposition, then $p=u^*u$ is the range projection of $a^*$, while $q=uu^*$ is the range projection of $a$. Now, in a finite von Neumann algebra, if the two projections $p,q$ are equivalent, then $1-p$ and $1-q$ are equivalent too. That is, there exists $w\in\mathcal M$ with $w^*w=1-p$, $ww^*=1-q$. Now $v=u+w$ is a unitary: $$ v^*v=(u^*+w^*)(u+w)=p+1-p+2\text{Re}\,w^*u=1+2\text{Re}\,w^*(1-q)qu=1, $$ with a similar computation to show that $vv^*=1$. We also have $vp=(u+w)p=up+wp=u+w(1-p)p=u$. So $$ a=u|a|=up|a|=vp|a|=v|a|. $$