# Polar decomposition of functionals

Posted on 13th March 2019

A well-known fact from the basic theory of von Neumann algebras is the polar decomposition of normal functionals: given $$\newcommand{\ip}[2]{\langle #1, #2 \rangle}$$ $$\varphi\in M_{*}$$ there is $$\omega\in M_{*}^+$$ and a partial isometry $$v\in M$$ with $$\varphi = v\omega$$. If $$v^* v$$ is equal to the support of $$\omega$$, then this decomposition is unique, and we write $$|\varphi|$$ for $$\omega$$.

I am mostly following Takesaki's book here; but the material is also nicely presented in a the MSc thesis of Zwarich. Neither of these sources quite gives a correct proof (IMHO) so I thought I would record here the main steps in proving existence.

We shall need two lemmas, which are Chapter III, Lemma 3.2 and Lemma 4.1 in Takesaki, and 2.12.1 and 3.6.1 in Zwarich.

Claim 1: Let $$A$$ be a $$C^{*}$$-algebra and let $$\omega\in A^*$$. If there is $$a\in A^+$$ with $$\|a\|\leq 1$$ and $$\ip{a}{\omega} = \|\omega\|$$, then $$\omega\geq 0$$

Claim 2: Let $$M$$ be a von Neumann algebra, and let $$\omega\in M_*$$. If $$e\in M$$ if a projection with $$\|e\omega\| = \|\omega\|$$ then $$e\omega =\omega$$.

We can now proceed with the existence part of the proof. By Hahn-Banach, there is $$a\in M$$ of norm one with $$\|\varphi\| = \ip{a}{\varphi}$$. Let $$a^{*} = u|a^{*} |$$ be the polar decomposition of $$a^{*}$$. Thus $$u^{*} a^{*} = |a^{*} |$$ and $$e=uu^{*}$$ is the projection onto the closure of the image of $$a^{*}$$. Define $$\omega = u^{*} \varphi$$.

Then $$u\omega = uu^{*} \varphi = e \varphi$$. Also $$ea^{*} = u|a^{*} | = a^{*}$$ so $$ae = a$$. Thus we have that $] \|\varphi\| = \ip{a}{\varphi} = \ip{ae}{\varphi} = \ip{a}{e\varphi} \leq \|e\varphi\| \leq \|\varphi\|,$ and so there is equality throughout, which implies that $$\|e\varphi\| = \|\varphi\|$$. By Claim 2, $$\varphi = e\varphi = u\omega$$.

We now observe that $\|\varphi\| = \ip{a}{\varphi} = \ip{|a^{*} |u^{*} }{\varphi} = \ip{|a^{*} |}{\omega} \leq \|\omega\| = \|u^{*} \varphi\| \leq \|\varphi\|,$ so again we have equality, and hence $$\ip{|a^{*} |}{\omega} = \|\omega\|$$. From Claim 1 we conclude that $$\omega$$ is positive.

Finally, as $$u^{*} u \omega = u^{*} u u^{*} \varphi = u^{*} \varphi = \omega$$ we have that $$u^{*} u \geq s(\omega)$$, the support projection of $$\omega$$. Setting $$v = us(\omega)\in M$$ we have that $$v$$ is a partial isometry, because $v^{*} v v^{*} = s^{*} u^{*} u s s^{*} u^{*} = s^{*} s s^{*} u^{*} = s u^{*} = v^{*},$ where I have written $$s=s(\omega)$$. Also $$v\omega = us(\omega)\omega = u\omega = \varphi$$, and finally $$v^{*} v = s u^{*} u s = s = s(\omega)$$ as required.

Writing out $$C^{*}$$-algebra theory in markdown is a pain...