Set $A:=C_0((0,1]) * C_0((0,1])$ (the free product C*-algebra), with canonical generators $a,b$ (positive contractions). Does there exists some $\gamma>0$ such that, for any $x,y \in A$ if $x^*x=a$ and $y^*y=b$ then $$ \|[xx^*,yy^*]\| > \gamma? $$

  • $\begingroup$ I think you want another _0 in the definition of $A$. (Can't edit that myself.) $\endgroup$ – Rasmus Oct 14 '15 at 8:04
  • $\begingroup$ Thanks. Hopefully that was the only reason no one has answered my question and now the answers will flow. $\endgroup$ – Aaron Tikuisis Oct 18 '15 at 9:46
  • $\begingroup$ I'm not too sure about that. $\endgroup$ – Rasmus Oct 18 '15 at 9:52

YES. (However, the answer for the same question for von Neumann algebras is NO.) I take $$A:=\lbrace f\in C([0,1],M_2) : f(0), f(1) \in D_2\rbrace.$$ Here $D_2\cong\ell_\infty^2$ is the diagonal. Let $$Q:=\mathrm{ev}_0\oplus\mathrm{ev}_1\colon A\to\ell_\infty^2\oplus\ell_\infty^2.$$ Let $$a=\left(\begin{matrix} 1 & 0\\ 0 & \frac{1}{2}\end{matrix}\right)\mbox{ (constant) and } b=\left(\begin{matrix} t & \sqrt{t(1-t)}\\ \sqrt{t(1-t)} & 1-t\end{matrix}\right)\mbox{ (projection)}.$$ Suppose $a=x^*x$ and $b=y^*y$. Then, $x=u|x|$ and $u\in A$ (unitary). One has $$\|[xx^*,yy^*]\|=\|[a,u^*yy^*u]\|.$$ Since $Q(A)$ is commutative, $u^*yy^*u$ is a {\bf projection} such that $Q(u^*yy^*u)=Q(b)=\mathrm{diag}(0,1)\oplus\mathrm{diag}(1,0)$. This implies $\|[a,u^*yy^*u]\|\geq 1/4$.

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  • $\begingroup$ How does this relate to the free product in Aaron's question? $\endgroup$ – Yemon Choi Oct 24 '15 at 11:31
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    $\begingroup$ I'm sure, coming from you Yemon, you know the answer, but here it is for the sake of others. Since the free product in my question is the universal C*-algebra generated by two commuting positive contractions, there is a canonical *-homomorphism from my $A$ to Taka's $A$, sending my $a,b$ to his $a,b$ respectively. As *-homomorphisms between C*-algebras are contractive, this shows that the answer to my question (in the statement, not in the title) is yes, as claimed. $\endgroup$ – Aaron Tikuisis Oct 24 '15 at 11:48
  • $\begingroup$ Great answer, Taka, thank you. Why is the answer no for von Neumann algebras? (I presume you mean the von Neumann algebraic free product of two diffuse abelian von Neumann algebras.) $\endgroup$ – Aaron Tikuisis Oct 24 '15 at 11:50
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    $\begingroup$ Thanks. In any von Neumann algebra $M$, $\inf\lbrace \|[a,ubu^*]\| : u\in U(M)\rbrace=0$. This follows by considering the masa generated by $a$ and by working on each type of von Neumann algebra separately. $\endgroup$ – Narutaka OZAWA Oct 24 '15 at 12:47
  • $\begingroup$ Thanks. A small comment about your initial answer. We could take a to be a constant projection $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, and then it becomes more obvious (to me at least) why $x$ has a polar decomposition (it follows from $A$ having stable rank one). $\endgroup$ – Aaron Tikuisis Oct 24 '15 at 18:39

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