I have been thinking about polar decomposition of $C^*$-algebras. I could not find a proper reference where it says: Let $A$ be a $C^*$-algebra, and $b$ an invertible element of $A$ with modulus $\lvert b\rvert$. Why is it true that $b$ equals $w\lvert b\rvert$ for some unitary $w$? Is there some algebraic version of the polar decomposition that we have for $C^*$-algebras? By "algebraic version" I mean that we have only rings with involution, but without norm.
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ To get the modulus $|b| = \sqrt{b^\ast b}$ of $b$, you need to be able to take positive square roots of positive definite elements, presumably through some sort of functional calculus on your $\ast$-algebra more general than the polynomial functional calculus. Setting up such a functional calculus, however, is precisely where a topology on your $\ast$-algebra will likely enter the picture. $\endgroup$– Branimir ĆaćićCommented Aug 19, 2016 at 20:08
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Let $w=b|b|^{-1}$. It is then obvious that $w^*w=1$, and, since $w$ is invertible, this implies that $w^*=w^{-1}$, so $w$ is unitary. It is also clear that $b=w|b|$.
How algebraic is this argument? In a *-algebra one does not necessarily have a notion of absolute value, but let us suppose that for your given $b$, you may find a self-adjoint element, called $|b|$, which satisfies $|b|^2=b^*b$. Then $|b|$ is necessarily invertible and the above argument applies, providing your desired unitary $w$.
