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An introduction to the classification of amenable C*-algebra.

page 140 Lemma 3.5.1 Let $x\in A$ with the polar decomposition $x=u|x|$ in $A''$ and $B=\overline{x^*Ax}$. Then $ub\in B$ for every $b\in B$.

$A''$ refers to the enveloping C*-algebra of $A$, the weak closure of $A$ in $B(H)$ where $A$ is universally represented. But these are not important.

Let $A$ be $B(l^2(\mathbb N))$ and let $x$ be the shift operator such that $x(e_j)=e_{j-1}$. Then $\overline {x^*Ax}=B(0\oplus l^2(\mathbb N^+))$ (regarding $B(0\oplus l^2(\mathbb N^+))$ as a subalgebra of $B(l^2(\mathbb N))$). Since $x$ is a partial isometry itself so $x=x|x|$ and $|x|$ is the projection onto $0\oplus l^2(\mathbb N^+)$. However, $x=x|x|\not\in \overline{x^*Ax}$ since no element in $\overline{x^*Ax}=\overline{|x|A|x|}$ has range larger than $0\oplus l^2(\mathbb N^+)$.

Am I wrong, or is the book wrong?

Sui
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  • Bear in mind that I really know next to nothing about $C^$ algebras, but why are you expecting $x \in x^ A x$? The lemma says $ub \in B$ for every $b \in B$. If $e$ is the identity map in $A = B(\ell^2(\Bbb{N}))$, then $e \notin \overline{x^Ax}$ for the same reason you stated, so I don't see why $x = xe$ ought to be in $\overline{x^ A x}$? Or am I missing something? – user824599 Sep 14 '20 at 06:15
  • @user824599 maybe i should write more details. There is a fact that every C* algebra contains an approximate unit and $\overline {x^*Ax} $ is actually called the hereditary C* algebra generated by $x^x$. Approximate is a net of elements $f_\lambda$ such that $f_\lambda x\to x$ holds for every $x$. This guarantees that $x^ f_\lambda x\to x^x=|x|^2$ is always in $x^Ax$. Then by the existence of square root, $|x|\in x^Ax$. By hypothesis, $x|x|=x\in x^Ax$. – Sui Sep 14 '20 at 07:29
  • @user824599 and in the example i gave, the projection $|x|$ onto $0\oplus B(l^2(\mathbb N^+))$ is in $x^Ax$ which makes $x|x|\in x^Ax$ according to the lemma. – Sui Sep 14 '20 at 07:33
  • I think Lin wanted the statement of that Lemma to say "Then $ubu^* \in B$ for every $b \in B$" rather than "$ub \in B$". At least, when he uses it in the proof of (3.5.3) in the next page, it seems that this is what he has in mind. – Ruy Sep 17 '20 at 21:15
  • @Ruy Do you mean $ubu^∗∈B$ for every $b∈B$ ? Suppose you are right, note that $x^Ax=|x|A|x|$ and $uBu^=xAx^$, it is still not true, since $x^Ax= xAx^*$ does not always hold. – Sui Sep 18 '20 at 01:30
  • @Ruy The counterexample is still the same, let $x$ be the left shift operator then $x|x|x^=id$ which is not in $x^Ax$. – Sui Sep 18 '20 at 01:34
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    You are absolutely right and I am sorry for having misled you. I guess the whole point is to try to salvage (3.5.1) in such a way that the application in (3.5.3) becomes lawful. For that purpose it would be enough to show that if $b\in B$ then $ubu^$ is in $A$ but I think in fact Lin's proof may be modified to show that if $b\in B$ then $ubu^$ is in $\overline{xAx^*}$. – Ruy Sep 18 '20 at 01:56

2 Answers2

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It is my impression that Lin wanted his Lemma to say:

Lemma 3.5.1. Let $x\in A$ with the polar decomposition $x=u|x|$ in $A''$. Also let $B_1=\overline{x^*Ax}$ and $B_2=\overline{xAx^*}$. Then $uB_1u^*\subseteq B_2$.

Here is a proof of this result broken into Lemmas, each of which might have some interest in itself.

Lemma 1. $x = \lim_n x(x^*x)^{1/n}$.

Proof. Compute $\Vert x - x(x^*x)^{1/n}\Vert ^2$ using the C*-identity $\Vert y\Vert ^2 = \Vert y^*y\Vert $.

Lemma 2. For every $\alpha >0$ one has that $u(x^*x)^\alpha = (xx^*)^\alpha u\in A$.

Proof. Since both sides vanish on the kernel of $x$ (seen in any given faithful Hilbert space representation), it is enough to show that they agree on $$ \text{Ker}(x)^\perp = \overline{\text{Ran}(x^*)} = \overline{ \text{Ran}(|x|)}. $$ We have $$ u(x^*x)^\alpha |x| = u|x| (x^*x)^\alpha = x (x^*x)^\alpha = (xx^*)^\alpha x = (xx^*)^\alpha u|x|. $$

This proves the identity in the statement, so let us now prove that $u(x^*x)^\alpha \in A$.

Approximate the function $f(t)=t^{2\alpha}$ on the spectrum of $|x|$ by a polynomial $p$ without constant term and hence we may write $p(t) = t q(t)$ for some other polynomial $q$. Then $$ u(x^*x)^\alpha = u |x|^{2\alpha} \sim u p(|x|) = u |x| q(|x|) = x q(|x|) \in A. $$ QED.

Lemma 3. $ux^*\in \overline{xA}$.

Proof. $$ ux^* = \lim_n u(x^*x)^{2/n}x^* = \lim_n (xx^*)^{1/n}u(x^*x)^{1/n}x^* \in \overline{xA}. $$ QED

Therefore the result we need, namely $$ ux^*Axu^* \subseteq \overline{xAx^*} $$ follows easily.

Ruy
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  • By the way, do you know how to prove 1.11.44? I am having trouble on it... – Sui Sep 18 '20 at 03:12
  • Yes. Lets go over to where you asked this question. – Ruy Sep 18 '20 at 03:30
  • Sorry for bothering you again... I have had trouble showing $b_1=bz_1b\geq 0$ in Lemma 3.5.8. Could you please give me some hint about it? I don't see why $g(|x|)$ and $d_1$ are commutative. – Sui Sep 20 '20 at 06:03
  • I think Huaxin made another mistake. If you'd like to post this as a new question I'll have more room to describe a counter-example. – Ruy Sep 20 '20 at 15:22
  • Lol ok let me post a new question. I hadn't thought it would be a mistake. – Sui Sep 20 '20 at 15:32
  • https://math.stackexchange.com/questions/3833542/ here I have posted it. – Sui Sep 20 '20 at 16:00
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Yes, the book is wrong. It was supposed to be $ub \in A$ for all $b \in B$.

See also Farah's book Corollary 1.6.13.

Junqi Yang
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