If $A$ is a commutative, unital $C^*$-algebra, then $A=C(X)$ for some compact, $T_2$ space $X$. Then $a=f,x=g$ and $y=h$ are continuous functions on $X$. Then there is $x_0\in X$ such that $\lVert f\rVert=|f(x_0)|$. Take $\phi=\hat{x_0}$. Then $\phi(xa^*ay)=g(x_0)|f(x_0)|^2h(x_0)=\lVert f\rVert^2\phi(gh)=\lVert a\rVert^2\phi(xy)$.

I'm doubtful about the non-commutative case here. If $x,y=1$, then the we consider the (commutative) $C^*$-algebra generated by $a^*a$, and then extend to whole of $A$ by Hahn-Banach. But here $x,y,a^*a$ may not commute with each other, so the $C^*$-algebra generated by $x,y,a^*a$ may not be commutative. Therefore, I cannot apply the above argument.

Can anyone help me with the non-commutative case? Thanks for your help in advance.


1 Answer 1


It is not true in general. The problem arises in a sense from the fact that in a non-commutative environment it is possible to have a product of non-invertible elements be invertible.

Let $A=B(H)$. Fix an orthonormal basis $\{e_n\}$, and let $\{E_{kj}\}$ the corresponding matrix units. Let $$ x=\sum_kE_{k,k+1},\qquad y=\sum_kE_{k+1,k},\qquad a=E_{11}. $$ Then $\|a\|=1$, and $$ x(1-a^*a)y=\sum_{h,k,j}E_{h,h+1}E_{k+1,k+1}E_{j+1,j}=\sum_kE_{kk}=I. $$ This means that $\phi(x(1-a^*a)y)=1$ for any state $\phi$, which means that $$ 1+\phi(xa^*ay)=\|a\|^2\,\phi(xy) $$ for any state $\phi$.

  • $\begingroup$ I'm a bit curious why the computation $x(1-a^*a)y=1$ fails if $H$ is finite dimensional $\endgroup$ Oct 25, 2023 at 21:30
  • $\begingroup$ In finite dimension the equality $x(1-a^*a)y=1$ implies that $1-a^*a$ is invertible. This forces $\|a\|<1$ (because $1$ cannot be an eigenvalues of $a^*a$), contrary to what was assumed. $\endgroup$ Oct 25, 2023 at 21:54
  • $\begingroup$ So it's a matter of being stably finite, interesting. Nice answer Martin! $\endgroup$ Oct 26, 2023 at 6:14

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