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Aug 24, 2021 at 18:08 answer added Steven Clontz timeline score: 2
Aug 23, 2021 at 8:35 comment added Alessandro Codenotti To add to Z.M.'s comment you don't need to assume $G$-invariance of the metric, if $d$ is not $G$-invariant it is another exercise to show that $d'(x,y)=\sum_g d(gx,gy)$ is an equivalent $G$-invariant metric. (This generalizes to actions of compact groups by integrating against the Haar measure along orbits)
Aug 22, 2021 at 20:18 comment added Sunrit Thank you so much!
Aug 22, 2021 at 16:03 review Close votes
Aug 27, 2021 at 3:04
Aug 22, 2021 at 12:46 comment added Tom Goodwillie I don't have a reference, but I think that what Z.M. asserts is a straightforward exercise.
Aug 22, 2021 at 8:08 comment added Alessandro Codenotti Another way to think about this is that the quotient is the orbit space with the Hausdorff metric
Aug 22, 2021 at 5:27 comment added Z. M Following Tom Goodwillie's comment, I think that the following is true: let $X$ be a metric space with a $G$-action where $G$ is a finite group and the metric is $G$-invariant. Then the orbit space $X_G$ is metrizable with $d([x],[y])=\min\{d(x,gy)\vert g\in G\}$ (check that "open" balls are really open and constitute a neighborhood basis). Maybe we could generalize this to compact groups $G$.
Aug 22, 2021 at 1:35 comment added Sunrit Hi @TomGoodwillie thank you for the response. Can you give some references/insight into why?
Aug 22, 2021 at 1:18 comment added Tom Goodwillie The topology on $\mathcal X/\sim$ given by the metric you describe is the same as the quotient topology.
Aug 21, 2021 at 22:57 review First posts
Aug 22, 2021 at 5:53
Aug 21, 2021 at 22:47 history asked Sunrit CC BY-SA 4.0