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Most of the time one finds the proof for bounded operators on a Hilbert space, but Sakai in his book "C*-algebras and W*-agebras" gives a purely algbraic one, (Thm 1.12.1 p.27-28, partially at his link):

  • Instead of taking directly $\sqrt{a^*a}$ as the positive part he considers the sequence $h(n):= (a^*a+\frac{1}{n}1)^{1/2} $. (I guess the point is that this is invertible because it is strictly positive) Anyway, he also defines $a(n):=a h(n)^{-1}$ where $a$ is the element we want to "polar" decompose.
  • Of course $h(n)\rightarrow \sqrt{a^*a}$ uniformly, but we can reformulate this as $\sqrt{a^*a} \in h(n)+ \epsilon S $ for $n \geq n_0$ where $S$ is the unit "sphere" (understand ball) and $\epsilon>0$. Then multiplying by $a(n)$ gives $$\forall\ n\geq n_0,\quad a(n) \sqrt{a^*a}\in a+\epsilon S $$ because ($a(n)h(n)=a$ by construction) and one prooves that $\lVert a(n)\rVert\leq 1$
  • Applying a theorem saying that the unit ball in a von Neumann algebra is weakly compact, one obtains that $a(n) \sqrt{a^*a}$ has an accumulation point $b$ ( or very very explicitly $\frac{a(n) \sqrt{a^*a}-a}{\epsilon}\in S$). Because this is true for all $\epsilon$ we actually have the equality $$a = b \sqrt{a^*a}$$ which is the polar decomposition.
  • Now we want to check some claims that $b$ (with some adjustment...) is a partial isometry and that $b^*b$ and $bb^*$ are the support of $\sqrt{a^*a}$ and $\sqrt{aa^*}$, denoted $p$ and $q$. (Recall, the support $s$ of a self-adjoint element $c$ is the smallest projection such that $sc=c$ which is equivalent in that case to $cs=c$)

Here is the detail I don't get, he easily obtains $\sqrt{a^*a}(p-pb^*qbp)\sqrt{a^*a}=0 $ and then says

"Since $\lVert b\rVert\leq 1 $, we conclude that $p=pb^*qbp$."????

Of course with the interpretation that $p$ is the projection on the image of $\sqrt{a^*a}$ it works, but I don't get his argument. And also directly from the decomposition $a = b \sqrt{a^*a}$ and the property of a Banach norm we can already say that $\lVert b\rVert= 1$

Noix07
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  • My opinion is that it is more than an interpretation: let s be the support of a self adjoint element c and suppose we have cdc=0. Then sds=0. it seems that there is no condition on d. – Noix07 Jan 09 '14 at 17:49

2 Answers2

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The point is that $r=p-pb^*qbp$ is positive: $$ pb^*qbp\leq pb^*bp\leq p\|b\|^2p\leq p^2=p. $$ So you have $$ 0=\sqrt{a^*a}\,r\,\sqrt{a^*a}=(r^{1/2}\sqrt{a^*a})\,r^{1/2}\sqrt{a^*a}. $$ Then $r^{1/2}\sqrt{a^*a}=0$. Multiplying by $r^{1/2}$, we get $r\sqrt{a^*a}=0$. Now $p$ is the range projection of $\sqrt{a^*a}$, so we have $rp=0$. As $rp=r$, we get $r=0$.

Edit: here is a version that seems to avoid using that $b$ is contractive, and also seems to use material available up to that state in Sakai's book.

I will write $x=(a^*a)^{1/2}$, $r=p-pb^*qbp$. We know that $pr=rp=r$, and that $xrx=0$. If you look at definition 1.10.3 in the book, where $a$ there is $x$ here, you get that $xr$ is in $\mathscr L$, so $xr=xr(1-p)$, implying $xrp=0$. Then $xr=0$. Now take adjoint and repeat the process to get $r=0$.

Martin Argerami
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  • Actually it is that last step that I'm not sure how to proove: r\sqrt{aa}=0 implies rp=0. If that is always true, then we don't even need positivity, we do it directly for (\sqrt{aa}r)\sqrt{aa}=0 and then \sqrt{aa}r p=0, then prp=r=0. I want to proove it from the definition "the left (resp. right) support s of an element c is the smallest projection such that sc=c (resp. cs=c)" – Noix07 Jan 10 '14 at 12:23
  • There are several ways to see that $r|a|=0$ implies $rp=0$. The most natural to me is to multiply on the right by $|a|$ to deduce that $r f(|a|)=0$ for any polynomial $f$, and a fortiori for any bounded Borel function $f$. In particular, $rp=0$. In any case, I have added an edit to do it in a way suitable to Sakai's line of reasoning. – Martin Argerami Jan 10 '14 at 18:15
  • I actually had similar ideas which I made explicit below but still at some point a slight detail prevents me from concluding. In your reasoning I would say that xr in \mathcal{L} means there exists y such that xr=y(1-p), not directly xr(1-p)? – Noix07 Jan 10 '14 at 18:39
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    Notice that $\mathscr L$ is of the form $\mathscr M e$ for a projection $e$. Any $x\in\mathscr M e$ satisfies $x=xe$, since $e$ is a projection. Using the notation from your last comment, $xr=y(1-p)$, so $xr(1-p)=y(1-p)^2=y(1-p)=xr$. – Martin Argerami Jan 10 '14 at 19:04
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More explicit answer: let $c\in\mathfrak{M}$ be an arbitrary element in a von Neumann algebra, and $s$ its right support, $cs=c$. We want to proove $\forall\ d\in\mathfrak{M}, dc=0 \Rightarrow ds=0 $

  • It seems to be that it is very much related to the proposition 1.10.1 still in Sakai: there exists a unique projection $p$ such that a (ultraweakly?) closed left ideal can be written $I=\mathfrak{M}p $. Let's then consider the left ideal $K=\{x\in\mathfrak{M}, xc=0 \} $. By the proposition, there exists a projection such that $K=\mathfrak{M}p$
  • $dc=0 \Rightarrow \exists f\in\mathfrak{M}, d=fp\ $ then $ds=fps$.
  • One actually has a sufficient condition for it to vanish: in the definition 1.10.3 of left and right support Sakai seems to assume that $p=1-s_l $ where $s_l$ is the left support but in our case since $\sqrt{a^*a}$ is selft adjoint, left and right support coincide, $p=1-s$.

So if there is something to clarify, it is the link between the left/right support and the projections generating the left/right ideal of annihilators: the statement is probably

"$s$ left support of $c$ iff $K=\mathfrak{M}(1-s)$ "

and analogously for the right support and right annihilators.

  • We have obviously $K\supseteq\mathfrak{M}(1-s)$ : for any $d$, $\; d(1-s)c=d(1-s)sc=0$
  • At the same time the projection $p$ is by definition such that $\mathfrak{M}pc=0=pc $ Hence $(1-p)c=c$, which is similar to the definition of left support. But the later is smallest that satisfies such a condition, ie. $s \leq (1-p) $ and then $p \leq (1-s) $
  • But $p \leq (1-s) $, implies $p(1-s)=p $, cf. this, and then $K=\mathfrak{M}p=\mathfrak{M}p(1-s) \subseteq \mathfrak{M}(1-s)$
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Noix07
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