MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A composition operator $C\_T : C(X) \to C(Y)$ with $T \in C(Y, X)$ is defined by $C\_T f := f \circ T, f \in C(X)$.

I read in the book about Composition Operators by Singh and others that a nontrivial algebra homomorphism $A : C(X) \to C(Y)$ is a composition operator (meaning there is a $T$ with $A = C\_T$) if $A(\overline{f}) = \overline{A(f)}$ holds for all $f \in C(X)$. This is true for $X$, $Y$ compact Hausdoff spaces. The proof is not difficult if one uses the isometric isomorphism $j(X) = M(C(X))$ ($j$ mapping $X$ into the space of dirac functionals, $M$ being the spectrum of the algebra $C(X)$).

Is this still true if $X, Y$ are hemicompact k-spaces?

If not can you give a counterexample?

Def.: A topological space $X$ is hemicompact if there is a sequence $(K_n)$ of compact sets in $X$ with $\bigcup_n K_n = X$ and $K_n \subset K_{n+1}$ for all natural $n$ and if for any compact $K$ in $X$ there is an $n$ with $K \subset K_n$.

Def.: A topological space $X$ is a k-space if every subset intersecting each compact subset in a closed set is itself closed.

EDIT: As was rightfully pointed out I forgot to mention that $A$ has to be an algebra homomorphism. I have corrected this now and added the definitions of hemicompact and k-space.

share|cite|improve this question
You should probably define all of your terms. – Qiaochu Yuan Dec 2 '09 at 17:32
You must have missed something out: as written, couldn't I pick some (real-valued) measure $\mu$ on X and define $A:C(X)\rightarrow C(Y)$ by $A(f) = \mu(f) 1$. Then $A(\overline{f}) = \overline{A(f)}$ but $A$ would only be induced by a composition operator if $\mu$ were a point mass at $x_0$, with $T(y)=x_0$ for all $y\in Y$. If $A$ is also a homomorphism, then it's fine (and, AFAIK, you don't actually need it to be a $*$-homomorphism...) What is a hemicompact k-space? – Matthew Daws Dec 2 '09 at 20:05
Seeing as my general topology is, despite Kelley's injunction, pretty shoddy - what exactly is the interest in the particular hypotheses you've chosen? For instance, is it the case that these conditions (hemicompact k-space) are the weakest which allow one to detect properties of the space X from the algebra C_R(X), in some sense? Or are you just choosing some general conditions because they seem interesting to you? – Yemon Choi Dec 3 '09 at 3:18

In this other question on mathoverflow, Eric Wofsey says that for any topological space $X$, the maximal ideals of $C(X)$ correspond to the points of the Stone-Cech compactification $\beta X$. He then says that if $C(X)/I$ is isomorphic to $\mathbb{C}$, then every continuous function on $X$ extends continuously to that point in $\beta X$. My intuition is that you'll get what you want if you can construct a proper continuous function from $X$ to the real numbers; as usual proper means that the inverse image of any compact set is compact. I don't know that you would need conditions on $Y$. I also don't know whether your conditions on $X$ yield such a function, but they look similar.

(This is not meant as a complete answer, but it is something.)

share|cite|improve this answer

For hemicompact k-space $X$ the space of continuous homomorphisms of algebra $C(X)$ to ℂ is $X$ (up to the obvious isomorphism). The proof can be found, for example, in H. Goldmann "Uniform Frechet Algebras". Then the same construction as for compact spaces give you the map $T$.

share|cite|improve this answer
Ok, thanks. I looked it up. That is a very neat trick and it answers my question, but if we drop the k-space hypothesis the original approach doesn't work anymore. I would still like to see a counterexample or alternative proof in this case. – santker heboln Dec 4 '09 at 23:39
I don't remember if there are examples of hemicompact spaces (without k-space propery) such that $M(C(X))\neq X$ (I don't have Goldmann's book at home). Here, $M(C(X))$ is the space of all continuous homomorphisms of the algebra $C(X)$ to ℂ. It seems, that $C(M(C(X)))=C(X)$ (is it?). Then try $Y=M(C(X))$ and identity maps between $C(X)$ and $C(Y)$. – Oleg Eroshkin Dec 5 '09 at 1:29

I'll risk making this a post, not a commment.

I think the real numbers $\mathbb R$ are a hemicompact $k$-space. Certainly $\mathbb R = \bigcup_n [-n,n]$ and if $K\subseteq\mathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $K\subseteq\mathbb R$ has closed intersection with all compacts, then by looking at sequences, it's easy to see that $K$ is closed.

But $\mathbb R$ is not compact, so I guess you really mean to look at $C^b(\mathbb R)$, the algebra/space of all bounded continuous functions. Is that right? If not, then it's a whole new ball game (as $C(\mathbb R)$ the space of all continuous functions is not a Banach space).

But if so, then $C^b(\mathbb R)$ has character space $\beta\mathbb R$, and we can apply Jonas's construction: just pick a point $w\in\beta\mathbb R\setminus \mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(\mathbb R)\rightarrow\mathbb C$ which is not a composition operator.

Edit: Yes, the original question was about all continuous functions on X, not just the bounded ones. My mistake...

share|cite|improve this answer
No I don't mean C^b(X). C(X) is not a Banach space, it's a Frechet space. – santker heboln Dec 4 '09 at 14:29
Santker's questions is much easier for $C_b(X)$. The answer would then be no if $\beta X$ has any points that are not in $X$. – Greg Kuperberg Dec 4 '09 at 18:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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