# polar decomposition

If we impose the further conditions that $1-u^{*}u$ is the projection to the kernel of $x$, and $\ker(h)=\ker(x)$, then $(h,u)$ is unique, and is called the polar decomposition of $x$. The operator $h$ will be $|x|$, the square root of $x^{*}x$, and $u$ will be the partial isometry, determined by

• $u\xi=0$ for $\xi\in\ker(x)$

• $u(|x|\xi)=x\xi$ for $\xi\in\mathscr{H}$.

If $x$ is a closed, densely defined unbounded operator on $\mathscr{H}$, the polar decomposition $(u,h)$ still exists, where now $h$ will be the unbounded  positive operator $|x|$ with the same domain $\mathscr{D}(x)$ as $x$, and $u$ still the partial isometry determined by

• $u\xi=0$ for $\xi\in\ker(x)$

• $u(|x|\xi)=x\xi$ for $\xi\in\mathscr{D}(x)$.

If $x$ is affiliated with a von Neumann algebra    $M$, both $u$ and $h$ will be affiliated with $M$.

Title polar decomposition PolarDecomposition 2013-03-22 16:01:54 2013-03-22 16:01:54 aube (13953) aube (13953) 10 aube (13953) Definition msc 47A05