This question relates to one on topology and C^*-algebras that was asked two days ago, namely at the link: C*-algebras with no nontrivial endomorphisms link text

Let D be the closed unit disk in the plane. Let C(D) be the unital ring of continuous complex-valued functions on D. Then, C(D) is naturally a Banach algebra with pointwise addition and multiplication as the ring operations. Furthermore, the "star-operation" on elements of C(D) can be defined by: $g*(x)$ to be the complex conjugate (pointwise) of g(x), any x in D, and for any function g in C(D).

The question in reference 1 above was related to injective star-endomorphisms of some $C*$ -algebras. Without saying so, I think the famous commutative Gelfand-Naimark theorem and the Gelfand representation figured "in the background", because of the interplay of commutative $C*$-algebras and topology on compact spaces ... If $\alpha$ is an injective star-morphism of C(D) to itself, is it possible for $\alpha$(C(D)) to be a proper (i.e. a `strict') star-sub-algebra of the $C*$-algebra C(D) ?

If so, I cannot find such a special *-morphism $\alpha$, hence my question.

  • 1
    $\begingroup$ Am I missing something? Can't you take $\alpha(f)= f(g)$ where $g$ is a continuous surjection from $D$ onto $D$ that is not injective? $\endgroup$ Oct 18 '12 at 15:18

Assume $D$ lies in $\mathbb{R}^2$ and define $f:D\rightarrow D$ by $f(x):=2x$ if $\|x\|\leq 1/2$ and $f(x):=\frac{x}{\|x\|}$ if $\|x\|\geq 1/2$. Then, $f$ is onto and continuous, but it is not injective. It is easy to see that $f^\ast :C(D)\rightarrow C(D)$ is an injective $^\ast$-homomorphism, but it is not onto.

  • $\begingroup$ Thanks, Vahid Shirbisheh. I'm beginning to understand your answer. I'm not familiar with the definition or naming of $f*$ , given $f$ . I have heard of pull-back maps, but I don't know what they are. The same goes for so-called "push-forward" maps. David Bernier $\endgroup$ Oct 18 '12 at 21:24
  • $\begingroup$ $f^\ast$ is defined by $f^\ast (g) (x):= g(f(x))$ for every $g\in C(D)$ and $x\in D$. $\endgroup$
    – user23860
    Oct 19 '12 at 6:24
  • $\begingroup$ Thanks. $f*$ is obtained by "pre-composing" through the map $f: D -> D$. It's clear to me now. $\endgroup$ Oct 19 '12 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.