Isos of C*-algebras completely positive and contactive?

Question

I have the probably easy question, why every isomorphism between not necesarily unital C-algebras is completely positive and contractive. Can one in addition give examples of non contractive or non completely positive C-algebra morphisms, because I have no intuition to that. For example I somewhere read that every star-homomorphism is completely positive and hence every C*-algebra morphism is?

Answer 1

The question is indeed very easy.

1. Why every $*$-homomorphism between $C^*$-algebras is contractive:

Suppose first that $\phi:A\to B$ is a unital $*$-homomorphism between unital $C^*$-algebras, so $\phi(1)=1$. Let $a\in A$. It is an elementary thing from $C^*$-theory that, for self-adjoint elements (even for normal elements more generally) we have $\|x\|=\sup_{z\in\sigma(x)}|z|=\max_{z\in\sigma(x)}|z|=r(x)$, i.e. the norm is given by the spectral radius. Now let $a\in A$. We have $$\|\phi(a)\|^2=\|\phi(a^*a)\|=\|\phi(a)^*\phi(a)\|$$ and $\phi(a)^*\phi(a)$ is self-adjoint, so $\|\phi(a)^*\phi(a)\|=\sup_{z\in\sigma(\phi(a)^*\phi(a))}|z|$. But now one if $z\in\sigma(\phi(a)^*\phi(a))$, then $\phi(a^*a)-z1$ is non invertible. Thus $\phi(a^*a-z1_A)$ is non-invertible. This can only happen if $a^*a-z1_A$ is non-invertible in $A$, i.e. $z\in\sigma(a^*a)$. In other words, we have just shown that $\sigma_B(\phi(a^*a))\subset\sigma_A(a^*a)$. Thus $\sup_{z\in\sigma(\phi(a^*a))}|z|\leq\sup_{z\in\sigma(a^*a)}|z|=\|a^*a\|=\|a\|^2$ and combining the above, we have just shown that $\|\phi(a)\|\leq\|a\|$ for all $a\in A$.

This was under the assumption that $A,B$ and $\phi$ are all unital. For the general case, simply embed $A\subset\tilde{A}$, $B\subset\tilde{B}$ in the unitizations and consider the unitization of $\phi$, namely $\tilde{\phi}$ which is a unital $*$-homomorphism $\tilde{A}\to\tilde{B}$ and thus the previous can be applied. Since $\tilde{\phi}$ extends $\phi$, we have the desired result.

2. Why every $*$-homomorphism between $C^*$-algebras is completely positive.

First observe that every $*$-homomorphism between $C^*$-algebras is positive: indeed, if $\phi:A\to B$ is a $*$-homomorphism and $x\in A$ is positive, then $x$ is written as $x=a^*a$ for some $a\in A$. Thus $\phi(x)=\phi(a^*a)=\phi(a)^*\phi(a)$ which is positive in $B$, since every positive element is of the form $b^*b$.

Now to show that $\phi$ is completely positive, we must show that $\phi^{(n)}:M_n(A)\to M_n(B)$ is positive, where $\phi^{(n)}[a_{i,j}]=[\phi(a_{i,j})]_{i,j}$. But, by Gelfand's theorem we know that $M_n(A)$ and $M_n(B)$ are $C^*$-algebras. It is also trivial to check that, when $\phi$ is a $*$-homomorphism, the maps $\phi^{(n)}$ are also $*$-homomorphisms. By the work we did in the first paragraph, applied to the $*$-homomorphism of $C^*$-algberas $\phi^{(n)}:M_n(A)\to M_n(B)$, we have that $\phi^{(n)}$ is positive, as we wanted.