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Fix a set $X$ with right $G$-action. Give $X$ a topology $\tau$ and make $G$ a topological group. (These topologies need not make the action continuous).

We can define another topology $\tau'$ on $X$ as the largest topology making the action $(X,\tau) \times G \to (X,\tau')$ continuous. (This is also called the quotient topology on $X$ with respect to the action $(X,\tau) \times G \to X$.)

Note that if the $G$-action is continuous for $\tau$ then $\tau'= \tau$.

For example, if $X = \mathbb{R}$, $\tau$ is the discrete topology and $G$ is $(\mathbb{R}, +)$ with the usual topology acting on $X$ by addition, then $G \times X / \sim = \mathbb{R}$ with the usual topology (unless I am much mistaken).

More interesting examples exist, e.g. the Skorokhod topology (again unless I am mistaken).

This construction feels useful enough that it must be well known and have a name. Can anyone provide me with more information?

[EDIT: actually I don't think it's necessary that $G$ is a topological group, just that it's a group with a topology. Although it is probably necessary for inversion to be continuous at the identity and for multiplication to be continuous on $\{e\} \times G$.]

[EDIT: made the presentation clearer to address the existing comments, changed title]

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In your example with $\mathbb R$, the action of $G$ is discontinuous. – Sergei Ivanov Apr 20 '10 at 21:53
Yes, it's supposed to be. If the action of G is continuous then you don't get a new topology. See Brad Hannigan-Daley's answer below. – Tom Ellis Apr 21 '10 at 6:14

If the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous) then the resulting topology is $\tau$:

Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = \{(x,g):xg\in U\}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.

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Yes I know. This is why I wrote "whose point set $X$" is acted on, rather than "with a right $G$ action". Perhaps I should have made it clearer. – Tom Ellis Apr 21 '10 at 6:09
I've clarified that now. – Tom Ellis Apr 21 '10 at 6:10
N.B. by the definition of quotient topology, $G \times X / \sim$ is the largest topology on $X$ such that the product $G \times \tau \to X$ is continuous. – Tom Ellis Apr 21 '10 at 6:18
@Brad: it's more conceptual to use the universal property of the quotient. – Martin Brandenburg Apr 21 '10 at 6:28

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