4
$\begingroup$

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]

$\endgroup$
2
  • 2
    $\begingroup$ In your example with $\mathbb R$, the action of $G$ is discontinuous. $\endgroup$ Apr 20, 2010 at 21:53
  • $\begingroup$ 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. $\endgroup$
    – Tom Ellis
    Apr 21, 2010 at 6:14

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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. $\endgroup$
    – Tom Ellis
    Apr 21, 2010 at 6:09
  • $\begingroup$ I've clarified that now. $\endgroup$
    – Tom Ellis
    Apr 21, 2010 at 6:10
  • $\begingroup$ 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. $\endgroup$
    – Tom Ellis
    Apr 21, 2010 at 6:18
  • $\begingroup$ @Brad: it's more conceptual to use the universal property of the quotient. $\endgroup$ Apr 21, 2010 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.