Natural topology on a group acting on a topological space I'm wondering if there is a general way to give a topological structure to a group acting on a topological space. For example, the rotation group $SO(3)$ acts on the space $\mathbb{R}^3$, and it is normally given the subspace topology when you think of it as a a subspace of the collection of $3\times 3$ matrices. Is there a more natural way of arriving at this topology that is coordinate independent?
Here's what I am considering:
Suppose there is a group action by a group $G$ on a topological space $X$. Suppose that for $x\in X,$ $f_x:G\to X$ is a map so that $f_x(g)=g\cdot x.$ Define a topology on $G$ by the minimal topology such that for all $x\in X$, the map $f_x$ is continuous. Explicitly this would involve taking pre-images of open subsets of $X$ under $f_x$ for each $x\in X$ as a subbasis.
Basically I want the action $\cdot:G\times X \to X$ to be continuous. For this to even be possible, the action must already be continuous w.r.t. change in the $X$ component. Does anyone know a good way to do this?
 A: When $X, Y$ are two topological spaces there is a commonly used topology on the set of continuous functions $X \to Y$ called the compact-open topology. Under mild hypotheses ($X$ locally compact, which is true here) this describes the exponential object $Y^X$ in the category of topological spaces and so has a very pleasant universal property, namely that there is a natural bijection
$$\text{Hom}_{\text{Top}}(X, Z^Y) \cong \text{Hom}_{\text{Top}}(X \times Y, Z).$$
For example, a homotopy between two continuous functions $X \to Y$ is a map $[0, 1] \to Y^X$, the evaluation map $X \times Y^X \to Y$ is continuous, etc. So for $G$ a topological group and $X$ a locally compact space we have
$$\text{Hom}_{\text{Top}}(G, X^X) \cong \text{Hom}_{\text{Top}}(G \times X, X).$$
This implies in particular that if $G$ acts faithfully on $X$ and carries the subspace topology with respect to the inclusion $G \to X^X$, then the action map $G \times X \to X$ is continuous (although it doesn't imply that $G$ is a topological group with respect to this topology!). On the other hand, your proposal only asks for continuity in $G$ for fixed $x \in X$ and doesn't seem to me to get joint continuity.
So, what topology do we get when $SO(3)$ is topologized as a subspace of the space of maps $\mathbb{R}^3 \to \mathbb{R}^3$ with the compact-open topology? We can actually prove a more general fact about the compact-open topology on the space of linear maps between two finite-dimensional real vector spaces, as follows. First, recall that a finite-dimensional real vector space has a unique topology with respect to which addition and scalar multiplication are continuous (which agrees with the product / Euclidean topology if we pick a basis and an isomorphism to some $\mathbb{R}^d$). This is also the topology induced by any norm, and below I'll call it the Euclidean topology.

Proposition: Let $V, W$ be finite-dimensional real vector spaces (with Euclidean topologies). Consider the vector space $[V, W]$ of linear maps $V \to W$, topologized as a subspace of the space $\text{Hom}_{\text{Top}}(V, W)$ of continuous maps $V \to W$, with the compact-open topology. This topology agrees with the Euclidean topology.

Proof. We'll use the fact that when $Y$ is a metric space the compact-open topology on $Y^X$ is the topology of compact convergence. Pick norms $\| \cdot  \|_V, \| \cdot \|_W$ on $V, W$ (which you can take to be Euclidean if you prefer). The Euclidean topology on $[V, W]$ is induced by the operator norm
$$\| T \| = \sup_{\| v \|_V \le 1} \| T(v) \|_W.$$
We'll show that the topology of compact convergence, and hence the compact-open topology, is also the topology induced by the operator norm. The topology of compact convergence is, by definition, the topology in which a sequence $T_i$ of linear maps $V \to W$ converges iff it converges uniformly on compact subspaces of $V$. Since every compact subspace of $V$ is contained in a closed ball centered at the origin (which conversely is compact by the Heine-Borel theorem), equivalently $T_i$ converges compactly iff it converges uniformly on closed balls centered at the origin. And since linear maps are homogeneous, by scaling, $T_i$ converges compactly iff it converges uniformly on the closed ball of radius $1$ centered at the origin. But this is the same as converging with respect to the operator norm. $\Box$
(This argument is actually new to me and suggests the nice intuition that the compact-open topology is like a nonlinear generalization of the operator norm, at least if $X$ is locally compact.)

Corollary: The compact-open topology on $SO(3)$, thought of as a subspace of $\text{Hom}_{\text{Top}}(\mathbb{R}^3, \mathbb{R}^3)$, agrees with the Euclidean topology.

