Skip to main content
MathOverflow

Return to Question

added 3 characters in body; edited tags
Source Link
Ali Taghavi
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

A free action of $\mathbb{Z}/3\mathbb{Z}$ on a Riemannian manifold $(M, g)$ is called an equilateral action if for every $x\in M$ all three points of orbit of $x$ have the same distant todistance from each other.

What is an example of a Riemannian manifold which does not admit such an action but it already admit ana smooth free action by cyclic group of order $3$?

A free action of $\mathbb{Z}/3\mathbb{Z}$ on a Riemannian manifold $(M, g)$ is called an equilateral action if for every $x\in M$ all three points of orbit of $x$ have the same distant to each other.

What is an example of a Riemannian manifold which does not admit such an action but it already admit an smooth free action by cyclic group of order $3$?

A free action of $\mathbb{Z}/3\mathbb{Z}$ on a Riemannian manifold $(M, g)$ is called an equilateral action if for every $x\in M$ all three points of orbit of $x$ have the same distance from each other.

What is an example of a Riemannian manifold which does not admit such an action but it already admit a smooth free action by cyclic group of order $3$?

Source Link
Ali Taghavi
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

Riemannian manifolds which admit a smooth free $\mathbb{Z}/3\mathbb{Z}$ action but do not admit an equilateral triangle action

A free action of $\mathbb{Z}/3\mathbb{Z}$ on a Riemannian manifold $(M, g)$ is called an equilateral action if for every $x\in M$ all three points of orbit of $x$ have the same distant to each other.

What is an example of a Riemannian manifold which does not admit such an action but it already admit an smooth free action by cyclic group of order $3$?