Continuous group representation Suppose you have a topological
group $G$ 
, a normed $k$-
vector space $V$ and a group homomorphism $\rho:G\longrightarrow GL(V)$. How do you define the topology on $GL(V)$ to make this
map continuous? 
Many thanks
 A: In principle this is standard: what we really want is that $G\times V\rightarrow V$ by $g\times v\rightarrow \rho(g)(v)$ is (jointly) continuous. This correctly suggests taking the "strong operator topology" on (continuous) operators $V\rightarrow V$, namely, the seminorms $p_v(T)=|Tv|$ for $v\in V$, Banach, say.
With the uniform operator norm topology on endomorphisms of $V$, rarely will $G\rightarrow GL(V)$ be continuous. This is not a pathology, but a natural occurrence: already with $G$ being the circle acting on $V=L^2(G)$, the map $G\rightarrow GL(V)$ is not continuous when $GL(V)$ is given the uniform norm operator topology.
Edit: the topology is given by the collection of all such seminorms. The resulting topology on continuous operators is not Banach, etc. That is, there's no "open ball of radius $\epsilon$ centered at $T$". A basis for the open nbds of given $T$ are indexed by finite lists of vectors $v_1,\ldots,v_n$, small $\epsilon>0$, and requiring $|(T-S)v_i|<\epsilon$ for $i=1,\ldots,n$.
A: If $V$ is a normed real (or complex) vector space, then prove that $GL(V)$ is a normed real (or complex) vector space with the following norm:
$\left\|T\right\|=\sup_{\left\|v\right\|=1}\left\|Tv\right\|$
for $T\in GL(V)$. (We use the same notation, $\left\|\cdot\right\|$, for both the norm on $V$ and in the definition of the norm on $GL(V)$ using the norm on $V$; I hope this doesn't cause confusion.)
I hope this helps!
Edit: It appears that I didn't read your question as carefully as I should have and that you are looking for a topology on $GL(V)$ such that a given group homomorphism $\rho:G\to GL(V)$ is continuous. The norm on $GL(V)$ that I defined above is the natural norm induced by the given norm on $V$.
I'm keeping this answer up anyway in case it's useful.
