# topologies on $C(X,Y)$

Let $X,Y$ be two topological spaces. One can define various topologies on $C(X,Y),$ the space of continuous functions from $X$ to $Y.$ Let us take the group $G=Hom(X),$ the homeomorphisms of $X.$ Let $G_1$ be a subgroup of $G.$

Now, given $G_1$ and a $f\in C(X,Y),$ I am looking for "useful" nontrivial topologies $\tau$ on $C(X,Y)$ so that the set $\left\{f\circ h\colon h\in G_1\right\}$ has some nice properties with respect to $\tau.$

For example, when $X=Y=\mathbb{R}$ and $G_1$ is the group ,say, generated by $x\mapsto 2x,$ I want to say that $\left\{f(x),f(2x),f(4x),\ldots; f(x/2),\ldots\right\}$ satisfy a topological property. Or, if $G_1=F$ and $f\colon I\rightarrow X$ then $F$ operates on $C(I,X),$ so , what I want is a topology $\tau$ on $C(X,Y)$ such that the embedding $F\hookrightarrow Hom(C(I,X))$ has nice properties. Thank you.

• what topologies have you looked up? Usually the compact open topology works very well. – Olivier Bégassat Jul 8 '11 at 4:06
• @Oliver: I couldn't find any other top to look up. So I asked this question to get a list of well known topologies on $C(X,Y).$ – categoryboy Jul 8 '11 at 6:01

The compact-open topology on $C(X,Y)$ has the virtue that it is complete (under various reasonable hypotheses on $X$ and $Y$).

A typical situation in which a group $G$ acts is that $G$ acts continuously on $X$. Without loss of generality, we can have $G$ acting on the right on $X$. There is the associativity $x(gh)=(xg)h$ for $x\in X$ and $g,h\in G$. Then $G$ acts on $f\in C(X,Y)$ by $gf(x)=f(gx)$.

One large class of examples is with $Y$ a real or complex vector space. Then $C(X,Y)$ is naturally a vector space, and for "representation theory" you'd want $C(X,Y)$ to have a topology so that $G \times C(X,Y) \rightarrow C(X,Y)$ is (jointly continuous).

It is easy to mess up this continuity requirement, even tho' one might imagine that it is somehow automatic. For example, with $G=X=Y={\mathbb R}$ and $C(X,Y)$ replaced by $L^\infty({\mathbb R})$, it is false. Similarly, for $X$ non-compact (and $Y=\mathbb R$), the plain sup-norm will cause the action of (big...) groups to fail to be continuous.

The simplest situation is when continuous _compactly_supported_ functions on $X$ are dense in the space of functions being considered. With $Y=\mathbb R$ and the uniform-convergence-on-compacts topology on $C(X,Y)$, this does hold, happily.

For $Y$ an abelian topological group, a similar situation holds, with similar caveats for $Y$ non-compact (like $\mathbb R$).

In summary, one almost always does want the compact-open topology on $C(X,Y)$, or else various things get weird.