Why is a branched line in R2 not a topological manifold?

In summary, the branched line in R2 is not a topological manifold because there does not exist a chart at the point of branching. A chart is necessary to prove that the line is a manifold.
  • #1
CSteiner
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0
Is there a topologist out there that wants to explain why exactly a branched line in R2 is not not a topological manifold? I know it's because there doesn't exist a chart at the point of branching, but I don't understand why not. I'm just starting to self study this, so go easy on me :).
 
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  • #2
CSteiner said:
but I don't understand why not.
How do you suggest such a chart is constructed?
 
  • #3
Well why don't we form a chart (U, x) where U is an element of the subset topology of the standard topology on R2. Then U can be a subset of of the branch containing the point of branching with open endpoints on each of the three branches. x can then map each of these points to a point in R2. This mapping seems to be invertable, and continuous in both directions.
 
  • #4
But it is not a mapping from an open subset of R^2.
 
  • #5
Orodruin said:
But it is not a mapping from an open subset of R^2.
Why not? Sorry if this seems like a stupid question, but I'm only just now learning (and trying to understand) the definitions of these things. Open means that it belongs the standard topology on R2, right? Why doesn't it?
 
  • #6
Because it is not a union of balls. This is the very definition of the standard topology.
 
  • #7
CSteiner said:
Is there a topologist out there that wants to explain why exactly a branched line in R2 is not not a topological manifold? I know it's because there doesn't exist a chart at the point of branching, but I don't understand why not. I'm just starting to self study this, so go easy on me :).

Do you know the concept of "connected"?
 
  • #8
micromass said:
Do you know the concept of "connected"?
Not rigorously.
 
  • #9
You should learn about that then. Learn about "cut points" too. With that you can prove your OP.
 
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  • #10
micromass said:
You should learn about that then. Learn about "cut points" too. With that you can prove your OP.
Will do, thanks for pointing me in the right direction.
 

FAQ: Why is a branched line in R2 not a topological manifold?

1. What is a topological manifold?

A topological manifold is a topological space that locally resembles Euclidean space. In other words, it is a space that can be mapped onto a subset of Euclidean space using a continuous and invertible function.

2. Why is a branched line not a topological manifold in R2?

A branched line is not a topological manifold in R2 because it fails to meet the requirement of being locally homeomorphic to Euclidean space. This is because at the branching point, the space is not locally homeomorphic to a line in R2.

3. What does it mean for a space to be locally homeomorphic?

Being locally homeomorphic means that for every point in the space, there exists a neighborhood around that point that is equivalent to a neighborhood in Euclidean space. This means that the local properties of the space are the same as those of Euclidean space.

4. Can a branched line be a topological manifold in a different dimension?

Yes, a branched line can be a topological manifold in a different dimension. For example, a branched line in R3 can be a topological manifold because at the branching point, it can be locally homeomorphic to a line in R2.

5. What are the implications of a space not being a topological manifold?

If a space is not a topological manifold, it means that it does not have the nice properties of a smooth and continuous space. This can make it difficult to analyze and work with mathematically. Additionally, topological manifolds have important applications in physics, so a space not being a topological manifold can limit its usefulness in this field.

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