- #1
nonequilibrium
- 1,439
- 2
By the well-known Whitney embedding theorem, any manifold can be embedded in [itex]\mathbb R^n[/itex].
You might have also heard the Nash embedding theorem, which basically says that this is still true for Riemannian manifolds (i.e. now we demand the metric is induced from [itex]\mathbb R^n[/itex]).
So fine, any Riemannian manifold can be seen as a submanifold of [itex]\mathbb R^n[/itex]. But my question is: in how many ways can one do this?
For example, embedding a sphere in [itex]\mathbb R^3[/itex], it doesn't matter where we do it (translation symmetry) or how we orient it (rotational symmetry). I.e. the only freedom in embedding it is the symmetry of [itex]\mathbb R^3[/itex]. So an alternative formulation of the question is: are there isometric submanifolds of [itex]\mathbb R^n[/itex] which are not the same up to rotation and translation?
You might have also heard the Nash embedding theorem, which basically says that this is still true for Riemannian manifolds (i.e. now we demand the metric is induced from [itex]\mathbb R^n[/itex]).
So fine, any Riemannian manifold can be seen as a submanifold of [itex]\mathbb R^n[/itex]. But my question is: in how many ways can one do this?
For example, embedding a sphere in [itex]\mathbb R^3[/itex], it doesn't matter where we do it (translation symmetry) or how we orient it (rotational symmetry). I.e. the only freedom in embedding it is the symmetry of [itex]\mathbb R^3[/itex]. So an alternative formulation of the question is: are there isometric submanifolds of [itex]\mathbb R^n[/itex] which are not the same up to rotation and translation?