Clarification about submanifold definition in ##\mathbb R^2##

In summary, the conversation discusses the concept of smooth curves and their images as smooth manifolds. It is mentioned that the map from ##\mathbb R## to ##\mathbb R^2## may or may not be a diffeomorphism onto its image, depending on the derivative being an injective linear map. The concept of smooth immersion is also touched upon in relation to this discussion.
  • #1
cianfa72
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TL;DR Summary
Clarification about submanifold definition and diffeomorphisms involved in ##\mathbb R^2##
Hi,
a clarification about the following: consider a smooth curve ##γ:\mathbb R→\mathbb R^2##. It is a injective smooth map from ##\mathbb R## to ##\mathbb R^2##. The image of ##\gamma## (call it ##\Gamma##) is itself a smooth manifold with dimension 1 and a regular/embedded submanifold of ##\mathbb R^2##.

Since it is a regular submanifold there is a global chart ##(\phi, \mathbb R^2)## such that ##\Gamma## is represented as ##(x,0),x∈\mathbb R##. Restricting such a chart to ##\Gamma## we get the manifold structure on it w.r.t. the inclusion map ##i:\Gamma ↪\mathbb R^2## is an embedding. So far so good.

My point is: is the above map ##γ:\mathbb R→\mathbb R^2## a diffeomorphism onto its image ? I believe the answer is positive.
##\Gamma## indeed is diffeomorphic to ##\mathbb R## by definition of chart. Now the composition ##g = \phi \circ \gamma## is a continuous injective map ##g: \mathbb R \rightarrow \mathbb R##. By virtue of Invariance of domain theorem ##g## is an homeomorphism hence the map ##\gamma## is actually a differentiable homeomorphism onto its image (i.e. a diffeomorphism onto the image).
 
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  • #2
I haven't done differential geometry in a while, so sorry if I say something wrong here. I think since ##g## isn't the identity map, you haven't proven that the inverse of ##\gamma## is differentiable.

Consider the map ##\gamma: x\to (x^3,0)##. Then the inverse is easy to compute as ##(z,0)\to z^{1/3}##. This is not differentiable. There are obviously choices of ##\phi:\Gamma\to \mathbb{R}## for which ##\phi \circ \gamma## is smooth (e.g. ##\phi(z,0)=z##), but this doesn't mean ##\gamma## is a diffeomorphism.
 
  • #3
Office_Shredder said:
I haven't done differential geometry in a while, so sorry if I say something wrong here. I think since ##g## isn't the identity map, you haven't proven that the inverse of ##\gamma## is differentiable.
Ah yes, that makes sense. Indeed from Invariance of domain theorem it follows that ##g## is a smooth homeomorphism ##g: \mathbb R \rightarrow \mathbb R## (i.e. it is a differentiable/smooth homeomorphism however we cannot claim the inverse map ##g^{-1}## is a differentiable/smooth map too).

So, from a general point of view, even though the image ##\Gamma## of a smooth curve ##\gamma## is a regular/embedded submanifold of ##\mathbb R^2##, the map ##\gamma## itself may or may not be a diffeomorphism onto its image ##\Gamma##.
 
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  • #4
I think ##\gamma## is locally a diffeomorphism with its image whenever its derivative is an injective linear map.
 
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  • #5
Office_Shredder said:
I think ##\gamma## is locally a diffeomorphism with its image whenever its derivative is an injective linear map.
Yes, that is the definition of smooth immersion: if ##\gamma## is an immersion from ##\mathbb R## in ##\mathbb R^2## then it is a local diffeomorphism.
 
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Related to Clarification about submanifold definition in ##\mathbb R^2##

1. What is a submanifold in ##\mathbb R^2##?

A submanifold in ##\mathbb R^2## is a subset of ##\mathbb R^2## that is locally diffeomorphic to a Euclidean space of lower dimension. In other words, it is a subset that looks like a curved or twisted version of a line, plane, or other higher-dimensional shape.

2. How is a submanifold different from a manifold?

A manifold is a topological space that is locally homeomorphic to Euclidean space. This means that every point on a manifold has a neighborhood that is homeomorphic to a Euclidean space. A submanifold is a subset of a manifold that also satisfies this property.

3. Can a submanifold in ##\mathbb R^2## have more than two dimensions?

No, a submanifold in ##\mathbb R^2## can only have two dimensions because it is a subset of ##\mathbb R^2##, which is a two-dimensional space. However, a submanifold in a higher-dimensional space, such as ##\mathbb R^3##, can have more than two dimensions.

4. How is a submanifold defined mathematically?

A submanifold is defined as a subset of a manifold that satisfies certain conditions. Specifically, it must be locally diffeomorphic to a Euclidean space of lower dimension, and its tangent space at each point must be a linear subspace of the tangent space of the larger manifold.

5. What is the importance of understanding submanifolds in ##\mathbb R^2##?

Understanding submanifolds in ##\mathbb R^2## is important in many areas of mathematics and physics. For example, it is crucial in the study of differential geometry, which has applications in fields such as relativity, fluid mechanics, and computer graphics. Submanifolds also play a role in optimization and control theory, as well as in the study of dynamical systems.

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