I have two unital $C^*$ algebras $A$ and $B$. Let $X\subset A$ and $Y\subset B$ be such that span $X$ and span $Y$ are dense * subalgebras of $A$ and $B$ respectively. I have a map $\Psi: Span ~X \to Span ~Y$ which is a *-homomorphism. Also I have a map $\Phi: Span ~Y \to Span ~X $ such that $\Phi\circ \Psi=id$. Can I conclude that $\Psi$ extends to an isomorphism between $A$ and $B$?
Asked
Active
Viewed 116 times
1
-
3Consider the group algebra $X=Y=\mathbb CG$ for a discrete, non-amenable group $G$, let $A=C^_{max}G$, $B=C^rG$, and $\Phi=\Psi=id{\mathbb CG}$. – Aweygan Sep 26 '20 at 19:32
1 Answers
2
No, because $\Psi$ may fail to be a contraction. Let $A=B=C[0,2]$, and $X=\mathbb C[x]$, $Y=\mathbb C[x^3]$. Both are dense by Stone-Weierstrass. Let $\Psi:X\to Y$ be given by $(\Psi f)(x)=f(x^3)$. Then $\Psi$ is an invertible $*$-isomorphism. And, with $f(x)=x^n$, $$\frac{\|\Psi f\|}{\|f\|}=\frac{2^{3n}}{2^n}=4^n.$$ So $\Psi$ is unbounded.
Martin Argerami
- 205,756