How is the action of $G$ on $G/H$ jointly continuous?

Question

Let $G$ be a separable topological group, and $H$ a closed subgroup. Then, according to Knapp (chapter 1, section 11), the action of $G$ on $G/H$ is jointly continuous. In this context, what does it mean for the action to be “jointly” continuous?

Answer 1

A group action $G\times X\to X$ is usually taken to be "jointly continuous" in the sense that it is ... um... :) ... continuous, as a map from $G\times X$ to $X$. Not merely continuous in $G$ for each fixed $x\in X$, etc.

(In particular, for group representations on topological vector spaces, e.g., Hilbert spaces, it is not the case that $G\to \mathrm{End}(V)$ is continuous when the endomorphisms are given the operator norm topology! For example, the action of $G=\mathbb R$ on $L^2(\mathbb R)$ by translation is not a continuous map $G\to \mathrm{End}(L^2(\mathbb R))$ with uniform operator norm topology, because, given $\varepsilon>0$, for every $\delta>0$, there is $f\in L^2$ of norm $1$ such that $|f-T_x f|\ge \sqrt{2}$, for $x=\delta/2$, and $T_x$ is translation. E.g., tents of $L^2$ norm $1$ with widths $\delta/2$. )

In Knapp's situation, the assertion is that $G\times G/H\to G/H$ is continuous.

The distinction does matter now and then, in some functional-analytic situations, where we do not have joint continuity. But, actually, often for $V\times W\to X$, separately continuous, when $V,W$ are Frechet spaces, joint continuity is a corollary of Baire category, etc. :)