- #1
Pond Dragon
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In the following stackexchange thread, the answerer says that there is a Riemannian metric on [itex]\mathbb{R}[/itex] such that the isometry group is trivial.
http://math.stackexchange.com/questions/492892/isometry-group-of-a-manifold
This does not seem correct to me, and I cannot follow what he is saying. Could someone please explain this to me?
Thank you!
http://math.stackexchange.com/questions/492892/isometry-group-of-a-manifold
This does not seem correct to me, and I cannot follow what he is saying. Could someone please explain this to me?
Thank you!