Are the coordinate axes a 1d- or 2d-differentiable manifold?

In summary, the set D, which is a subset of R^2 with subspace topology, is not a manifold. It does not satisfy the requirement of having a neighborhood homeomorphic to an open subset of Euclidean space for every point. Additionally, removing the origin would result in 4 disconnected components, making it different from any surface or 1-manifold. However, if the origin is removed, it can be considered a topological manifold with 4 connected components, each being a 1-manifold and globally homeomorphic to the Reals.
  • #1
Delong66
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Suppose $$ D=\{ (x,0) \in \mathbb{R}^2 : x \in \mathbb{R}\} \cup \{ (0,y) \in \mathbb{R}^2 : y \in \mathbb{R} \}$$ is a subset of $$\mathbb{R}^2 $$ with subspace topology. Can this be a 1d or 2d manifold?
Thank you!
 
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  • #2
The set D is not a manifold. Every point in a manifold must have a neighbourhood that is homeomorphic to an open subset of a Euclidean space. The point (0,0) in set D has no such neighbourhood, as any open set containing (0,0) has an intersection of the two lines in it, and neither 1D nor 2D Euclidean space has an open subset consisting of such an intersection.
 
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Likes Orodruin
  • #3
And removal of a single point, the origin, would disconnect it into 4 components, unlike any surface or 1-manifold. It's clearly not a differentiable ( if it was a manifold ), as its tangent space is not defined at the origin. It's not even a manifold with boundary, as no neighborhood of the origin is homeomorphic to a (subspace) neighborhood of the upper half plane .
 
  • #4
WWGD said:
And removal of a single point, the origin, would disconnect it into 4 components, unlike any surface or 1-manifold. It's clearly not a differentiable ( if it was a manifold ), as its tangent space is not defined at the origin. It's not even a manifold with boundary, as no neighborhood of the origin is homeomorphic to a (subspace) neighborhood of the upper half plane .
If you remove the origin, I believe it is a topological manifold.
 
  • #5
jbergman said:
If you remove the origin, I believe it is a topological manifold.
Indeed, 4 lines, each a 1-manifold, globally homeomorphic to the Reals. A manifold with 4 connected components.
 

FAQ: Are the coordinate axes a 1d- or 2d-differentiable manifold?

What is a differentiable manifold?

A differentiable manifold is a mathematical concept used to describe spaces that are locally similar to Euclidean spaces, but may have more complex global structures. It is a generalization of the concept of a smooth curve or surface in traditional geometry.

What do we mean by "1d- or 2d-differentiable manifold"?

This refers to the dimensionality of the manifold. A 1-dimensional differentiable manifold is a curve, while a 2-dimensional differentiable manifold is a surface. In general, a d-dimensional differentiable manifold is a space that locally resembles d-dimensional Euclidean space.

How do we determine if the coordinate axes are a differentiable manifold?

We can determine if the coordinate axes are a differentiable manifold by checking if they satisfy the definition of a differentiable manifold. This includes being locally homeomorphic to Euclidean space, having a smooth structure, and being Hausdorff and second-countable.

Why is it important to know if the coordinate axes are a differentiable manifold?

Knowing if the coordinate axes are a differentiable manifold is important in many areas of science and mathematics, such as differential geometry, topology, and physics. It allows us to study and understand the properties of these spaces and apply them to real-world problems.

What are some applications of differentiable manifolds?

Differentiable manifolds have many applications in various fields, including computer graphics, robotics, image processing, and machine learning. They are also used in physics to describe the geometry of spacetime in general relativity and in the study of dynamical systems and chaos theory.

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