# Are free positive operators equivalent to almost-commuting operators?

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 $$\|[xx^*,yy^*]\| > \gamma?$$

• I think you want another _0 in the definition of $A$. (Can't edit that myself.) – Rasmus Oct 14 '15 at 8:04
• Thanks. Hopefully that was the only reason no one has answered my question and now the answers will flow. – Aaron Tikuisis Oct 18 '15 at 9:46
• I'm not too sure about that. – 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 $\|[a,u^*yy^*u]\|\geq 1/4$.
• 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. – Aaron Tikuisis Oct 24 '15 at 11:48
• 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. – Narutaka OZAWA Oct 24 '15 at 12:47
• 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). – Aaron Tikuisis Oct 24 '15 at 18:39