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 $$ \|[xx^*,yy^*]\| > \gamma? $$
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$\begingroup$ I think you want another _0 in the definition of $A$. (Can't edit that myself.) $\endgroup$– RasmusCommented Oct 14, 2015 at 8:04
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$\begingroup$ Thanks. Hopefully that was the only reason no one has answered my question and now the answers will flow. $\endgroup$– Aaron TikuisisCommented Oct 18, 2015 at 9:46
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$\begingroup$ I'm not too sure about that. $\endgroup$– RasmusCommented Oct 18, 2015 at 9:52
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1 Answer
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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 $\|[a,u^*yy^*u]\|\geq 1/4$.
answered Oct 24, 2015 at 11:18
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$\begingroup$ How does this relate to the free product in Aaron's question? $\endgroup$ Commented Oct 24, 2015 at 11:31
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1$\begingroup$ 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. $\endgroup$ Commented Oct 24, 2015 at 11:48
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$\begingroup$ 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.) $\endgroup$ Commented Oct 24, 2015 at 11:50
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4$\begingroup$ 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. $\endgroup$ Commented Oct 24, 2015 at 12:47
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$\begingroup$ 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). $\endgroup$ Commented Oct 24, 2015 at 18:39