# Continuity of representation of topological group

First, We set notations as follows.
$$G$$ : topological group , $$k$$ : field , $$V$$ : linear topological space over $$k$$ ,
$$\mathrm{Map}(V,V)$$ : Set of all continuous maps from $$V$$ to $$V$$
$$\mathrm{Aut}_k (V)$$ : Set of all homeomorphism from $$V$$ to $$V$$

We give compact-open topology to $$\mathrm{Map}(V,V)$$ and its subpace topology to $$\mathrm{Aut}_k (V)$$ .
Let $$\rho : G \rightarrow \mathrm{Aut}_k (V)$$ be a group homomorphism between topological spaces.

Then , are following conditions equivalent $$???$$

$$(1)$$ $$\rho$$ is a continuous map between topological spaces.
$$(2)$$ $$G \times V \rightarrow V , (g,x) \mapsto \rho(g)(x)$$ is a continuous map.

• Is $V$ finite-dimensional? What is known about $k$? – Paul Frost Mar 29 at 12:49
• Yes, I assume $V$ is finite dimensional. But, I don't impose a condition on $k$. – 神宮寺春姫 Mar 31 at 12:33

For spaces $$X,Y$$ let $$C(X,Y)$$ denote the set of continuous functions $$X \to Y$$. This set endowed with the compact-open will be denoted by $$Y^X$$. There is a canonical function $$E : C(X \times Y,Z) \to C(X,Z^Y)$$ where for $$f : X \times Y \to Z$$ we define $$E(f) : X \to Z^Y$$ by $$E(f)(x) : Y \to Z, E(f)(x)(y) = f(x,y)$$. This function is known as the exponential correspondence. See any book on general topology treating function spaces. There are also a number of contributions in this forum, for example When is the exponential law in topology discontinuous? (search for "exponential law").
The function $$E$$ is trivially injective, but surjectivity requires to assume that $$Y$$ is locally compact.
For your question this means that (2) implies (1). The converse is true under the additional assumption that $$V$$ is locally compact. Thus, if $$k$$ is a locally compact topological field, then (1) and (2) are equivalent because $$V \approx k^n$$ is locally compact.
• When I set $n= dim_k V$ , I can get a isomorphism $f:V \rightarrow k^n$. But, I cannot figure out $f$ is a homeomorphism. How do we get the homeomorphism $V \approx k^n$ $??$ – 神宮寺春姫 Apr 1 at 20:50
• Ah, you are right, there is a gap. I had in mind $k = \mathbb{R}, \mathbb{C}$ and finite fields. But I have no idea which other locally compact fields may exist and whether $V \approx k^n$ topologically. Therefore the most general solution seems to be to assume that $V$ is locally compact. – Paul Frost Apr 1 at 22:25
• I see. Can you tell me a reference which contains the proof in the case $k= \mathbb{R}$ $?$ – 神宮寺春姫 Apr 3 at 3:15
• It is a standard result that on each finite-dimensional $\mathbb R$-vector space $V$ there exists a unique Hausdorff topology making it a topological vector space. See any book on functional analysis and math.stackexchange.com/q/445547. Note that $\mathbb R ^n$ with the product topology is the standard model of such a space. This implies $V \approx \mathbb R ^n$ topologically. – Paul Frost Apr 3 at 8:29