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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.

神宮寺春姫
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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.

Paul Frost
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  • 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 01 '19 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 01 '19 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 03 '19 at 03:15
  • 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 https://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 03 '19 at 08:29
  • I see. Thank you so much. – 神宮寺春姫 Apr 04 '19 at 16:05
  • $\mathbb{Q}_p$ and their finite extensions also locally compact. In fact, except for the ones with discrete topology (every field with discrete topology is uninterestingly locally compact), these are all. – CO2 Apr 08 '21 at 15:09