11

Let $X$ and $Y$ be topological spaces, and consider the compact-open topology on $C(X,Y)$, which is generated by open sets of the form $$\{\text{continuous }f\colon X\to Y:f(K)\subseteq U\}\text{ for compact }K\subseteq X\text{ and open }U\subseteq Y.$$

Without assuming that $Y$ is a metric space, are there conditions on $X$ and $Y$ that imply that the compact open topology on $C(X,Y)$ is locally compact?

One related result of particular importance is the fact that $\mathrm{Hom}(G,T)$ is locally compact, where $G$ is a locally compact Hausdorff abelian group, and $T$ is the circle group. This is important for Pontryagin duality, because it is one part of knowing that the Pontryagin dual of a locally compact Hausdorff abelian group is still a locally compact Hausdorff abelian group.

For this related result, the standard proof uses equicontinuity and Arzela-Ascoli. But this approach only works if $Y$ is a metric space.

Thomas Browning
  • 4,541
  • 1
    You may find you'll get a better answer rate if you expand on some of these points. What is the compact open topology on $C(X, Y)$? If it's not too onerous, you could also expand on how Pontryagin duality is useful. The fewer things that people have to run and look up, the more people you'll hook into thinking about your problem. – Theo Bendit Feb 24 '22 at 22:19
  • Maybe you should explain what Pontryagin duality has do to with your highlighted question. – Tyrone Feb 24 '22 at 22:35
  • 2
    I added some more explanation. Let me know if you have other suggestions. – Thomas Browning Feb 24 '22 at 22:41
  • 1
    Are you sure that $C(G,T)$ is locally compact? But certainly this is true for the subspace of all homomorphisms $G \to T$. – Paul Frost Feb 25 '22 at 00:34
  • Oh, good point. I'm not actually sure that $C(G,T)$ is locally compact. I know that $\widehat{G}$ is a closed subspace of $C(G,T)$, and I assumed that's how local compactness of $\widehat{G}$ is derived, but I don't know. – Thomas Browning Feb 25 '22 at 00:43
  • @ThomasBrowning It is not true. See Tyrone's comment to my answer. – Paul Frost Feb 25 '22 at 14:32

2 Answers2

3

This is not a complete answer, but gives some useful information. Let us assume that $Y$ is Hausdorff.

  1. Necessary condition: If $C(X,Y)$ is locally compact, then $Y$ is locally compact.
    It is well-known that $j : Y \to C(Y,X), j(y) =$ constant map with value $y$, is an embedding. If $Y$ is Hausdorff, then $j$ is a closed embedding which proves 1.

  2. Sufficient condition: If $Y$ is locally compact and $X$ is a finite discrete space, then $C(X,Y)$ is locally compact.
    This is true because $C(X,Y)$ is homeomorphic to a finite product of copies of $Y$.

  3. If $Y$ is locally compact non-compact and $X$ is an infinite discrete space, then $C(X,Y)$ is not locally compact.
    This is true because $C(X,Y)$ is homeomorphic to an infinite product of copies of $Y$.

  4. If $X$ is an infinite compact Hausdorff space, $C(X,\mathbb R)$ is not locally compact.
    It is well-known that for compact Hausdorff $X$ the compact open topology on $C(X,\mathbb R)$ agrees with the $\sup$-norm topology. Thus $C(X,\mathbb R)$ is a normed linear space which is locally compact if and only if it is finite-dimensional. But for an infinite $X$ the space $C(X,\mathbb R)$ is not finite-dimensional. To see this, let $i : A \hookrightarrow X$ be the inclusion of a closed subspace $A$. It induces a linear map $i^* : C(X,\mathbb R) \to C(A,\mathbb R), i^*(f) = f \mid_A$. By the Tietze extension theorem this map is onto. But for each $n$ the set $X$ contains finite subsets $A_n$ with $n$ elements, thus we get linear surjections $C(X,\mathbb R) \to C(A_n,\mathbb R) \approx \mathbb R^n$.

This shows that it is quite difficult to get a locally compact $C(X,Y)$.

Paul Frost
  • 87,968
  • Thanks, this is helpful for intuition. Seems like compactness of $Y$ might be important? – Thomas Browning Feb 25 '22 at 00:42
  • 1
    Suppose $X$ is compact. Then an open inclusion $V\subseteq Y$ induces an open embedding $C(X,V)\subseteq C(X,Y)$. Apply this to the embedding $\mathbb{R}\cong(0,1)\subseteq[0,1]$ to exhibit $C(X,\mathbb{R})$ as an open subspace of $C(X,I)$. Now suppose $Y$ is a Hausdorff space containing a nontrivial path. By the Hahn–Mazurkiewicz Theorem $Y$ contains an embedded arc., and this induces a closed embedded $C(X,I)\subseteq C(X,Y)$. In particular, if $C(X,Y)$ is locally compact, then so is $C(X,I)$, and hence also $C(X,\mathbb{R})$. Thus $X$ is a finite space. – Tyrone Feb 25 '22 at 03:17
  • 1
    @Tyrone Your comment should be an official answer. – Paul Frost Feb 25 '22 at 08:06
  • I agree. Tyrone's comment settles this question pretty well I think. For example, if neither $C(T,T)$ or $C(\mathbb{Z}_p,T)$ is locally compact, then there's really not much hope at all. – Thomas Browning Feb 25 '22 at 16:30
  • @ThomasBrowning If $\mathbb Z_p$ denotes the finite cylic group of order $p$, then $C(\mathbb Z_p,T)$ is compact. $C(T,T)$ is not locally compact. – Paul Frost Feb 25 '22 at 17:15
  • No, I mean the $p$-adics (a profinite group). – Thomas Browning Feb 25 '22 at 17:17
1

In general $C_K(X,Y)$ (the continuous functions from $X$ to $Y$ in the compact-open topology) will rarely be locally compact (see the other answer and its comments).

But I found in my copy of the Encyclopedia of General Topology (h-1, topological groups, by Dikranjan) facts on the special case of $Y=\Bbb T$ (the circle group ) and its subspace the characters (not all continuous maps). I believe that Abstract Harmonic Analysis (Vol. 1) by Hewitt and Ross has all the proofs. One defines the dual of a group $G$ as:

$$\hat{G} = \{\chi: G \to \Bbb T\mid \chi \text{ a continuous group homomorphism }\} \subseteq C_K(G,\Bbb T)$$

in its subspace topology (so with relativised subbase elements).

  • $G$ is compact iff $\hat{G}$ is discrete.
  • $G$ is discrete iff $\hat{G}$ is compact.
  • If $G$ is LC (locally compact Hausdorff) and Abelian (in total: LCA), then $\hat{G}$ is LC.

The last fact is what you're after. And in that case ($G$ LCA) the Pontryagin-van Kampen duality theorem tells us that $$\omega_G: G \to \hat{\hat {G}}\,;\, \omega_G(g)(\chi) = \chi(g)$$ is a topological group isomorphism.

But that is all quite specific for this subspace and seemingly not a consequence of general facts about the topology of $C_K(X,Y)$ in general. This question also asks about the duality facts. Tao has online notes with the proofs too, IIRC. The question refers to them... These notes also seem useful and do use Arzela-Ascoli for local compactness, so general facts on $C_K(X,Y)$, really.

Henno Brandsma
  • 250,824
  • Yes, I am aware of these properties of the Pontryagin dual, but I was hoping for something more general. Thanks. – Thomas Browning Feb 25 '22 at 16:27
  • 1
    @ThomasBrowning the large (continuous function) space has much worse properties than the dual as it’s subspace. It’s a subtle interplay between the algebra and the topology. – Henno Brandsma Feb 25 '22 at 18:08