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.

  • $\begingroup$ Is $V$ finite-dimensional? What is known about $k$? $\endgroup$ – Paul Frost Mar 29 at 12:49
  • $\begingroup$ Yes, I assume $V$ is finite dimensional. But, I don't impose a condition on $k$. $\endgroup$ – 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.

  • $\begingroup$ 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 $V \approx k^n$ $??$ $\endgroup$ – 神宮寺春姫 Apr 1 at 20:50
  • $\begingroup$ 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. $\endgroup$ – Paul Frost Apr 1 at 22:25
  • $\begingroup$ I see. Can you tell me a reference which contains the proof in the case $k= \mathbb{R}$ $?$ $\endgroup$ – 神宮寺春姫 Apr 3 at 3:15
  • $\begingroup$ 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. $\endgroup$ – Paul Frost Apr 3 at 8:29
  • $\begingroup$ I see. Thank you so much. $\endgroup$ – 神宮寺春姫 Apr 4 at 16:05

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