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The polar decomposition of forms in a von Neumann algebra goes as follows - Let $\phi$ be a $\sigma$-weakly continuous form on a von Neumann algebra $M$. Then there exist a normal form $|\phi|$ and a partial isometry $v\in M$, such that $\phi(x) = |\phi|(xv)$, $\forall x\in M$.

The proof of this theorem can be found in any standard text of von Neumann algebras, eg. Stratila--Szido or Dixmier. They also give some uniqueness criterion, but here I only want to prove the existence part.

I am trying to give a different proof of this theorem as follows - For this, I assume the following theorem - Any $\sigma$-weakly continuous form on a von Neumann algebra $M\subset B(H)$ can be extended to a $\sigma$-weakly continuous form on $B(H)$.

Using this theorem, it follows that $\phi$ can be extended to a $\sigma$-weakly continuous form, say $\psi$ on $B(H)$. Then we can find $a\in B_1(H)$ such that $\psi(x) = tr(xa)$ $\forall x \in B(H)$. Let $a = v|a|$ be the polar decomposition of $a$. Then $\psi(x) = tr(xv|a|) = tr(|a|xv) = |\phi|(xv)$ $\forall x\in B(H)$. So, $\phi(x) = |\phi|(xv)$ $\forall x\in M$ where $|\phi|(x) = tr(|a|x)$ which is clearly a normal form. But I am unable to show that $v\in M$. Can anyone please help me with this? Thanks for any help.

Note - $B_1(H)$ refers to the set of trace-class operators.

Martin Argerami
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You cannot expect $v$ to be in $M$ with your approach.

Let $H=\mathbb C^2$, $$ M=\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix}:\ x,y\in\mathbb C\right\}, $$ and $\phi$ the linear functional given by $$ \phi\left(\begin{bmatrix}x&0\\0&y\end{bmatrix}\right)=x-y. $$ Then you can have $a=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, in which case $v=a\in M$. But your approach could also give you $$ a=\begin{bmatrix}1&1\\1&-1\end{bmatrix}, $$ in which case $$ |a|=\begin{bmatrix}2&0\\0&2\end{bmatrix},\ \ \text{ and }v=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}\not\in M. $$

Martin Argerami
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