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Wikipedia defines strongly continuous group action as follows:

A group action of a topological group G on a topological space X is said to be strongly continuous if for all x in X, the map g ↦ g.x is continuous with respect to the respective topologies.

My issue with this is that with this definition, every continuous group action is strongly continuous:

Suppose $G$ acts continuously on the space $X$. Then for all $x\in X$, the map $g\mapsto g\cdot x$ can be decomposed as $$G \xrightarrow{\;r_x\;} G\times X \xrightarrow{\;\alpha\;} X, $$ where $\alpha$ is multiplication and $r_x\colon g\mapsto (g,x)$. But $\alpha$ is continuous by definition, and it's clear that $r_x$ is continuous (if $U\times V$ is a cylinder set in $G\times X$, then $r_x^{-1}(U\times V)=U$, which is open). Thus, the map $g\mapsto g \cdot x$ is continuous for all $x\in X$. Hence, by the definition above, the action of $G$ on $X$ is strongly continuous.

Am I missing something?

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What is meant by respective topologies? –  Berci Feb 27 '13 at 17:18
@Berci You have a topological group $G$ and a topological space $X$. Both of which must have topologies. The action $\alpha : G \times X \to X$ goes from the topological space $G \times X$ to the topological space $X$. The respective topologies are the product topology on $G \times X$ and the topology on $X$. –  Fly by Night Feb 27 '13 at 17:51
See also: mathoverflow.net/questions/66394 –  Martin Apr 12 '13 at 17:08

1 Answer 1

up vote 8 down vote accepted

That's a feature, not a bug. Strong continuity is actually supposed to be a weaker condition than continuity (see also strong operator topology, which is weaker than the norm topology). For example, the action of $\mathbb{R}$ on $L^2(\mathbb{R})$ by translation is strongly continuous if $L^2(\mathbb{R})$ is given the norm topology but not continuous.

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