Two cones connected at their vertices do not form a manifold

In summary, the double cone is not a manifold because it has a cut point of order 2 at its vertex, meaning that if the vertex is removed, the space becomes disconnected with 2 components. This property is preserved under homeomorphisms, leading to the conclusion that the double cone cannot be homeomorphic to an open set in the plane or the real line. While it can be given a smooth structure, it is not a smooth manifold as it is the image of a non-smooth embedding of a 2 disk into ##R^3##.
  • #1
Heisenberg1993
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Why is i that two cones connected at their vertices is not a manifold? I know that it has to do with the intersection point, but I don't know why. At that point, the manifold should look like R or R2?
 
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  • #2
Heisenberg1993 said:
At that point, the manifold should look like R or R2?
But it doesn't, so it is not a manifold.
 
  • #3
That's exactly my question, why it doesn't look like which (R or R2)?
 
  • #4
Heisenberg1993 said:
That's exactly my question, why it doesn't look like which (R or R2)?
It doesn't look like either. There is no smooth map from an open set of R nor R2 that maps to that region.
 
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  • #5
No matter how small you choose a neighborhood around the connected point, it always looks like two cones. So we could as well think of the whole thing to be homeomorph to an Euclidean space, that is two discs or balls sharing exactly one point. Now there is no continuous way to make one disc or ball out of it, because you lose uniqueness in this point. They would have to share at least a small disc.

Perhaps this helps:
https://www.physicsforums.com/threads/tangent-at-a-single-point.848275/#post-5319494

@micromass and @WWGD have been very, very patient with me as I once stepped in a similar trap. (Thanks, again. I haven't forgotten it.)
 
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  • #6
Heisenberg1993 said:
That's exactly my question, why it doesn't look like which (R or R2)?

- What is your definition of "manifold"?
- What do you mean by "look like"?
 
  • #7
fresh_42 said:
...Now there is no continuous way to make one disc or ball out of it, because you lose uniqueness in this point.

You can continuously map an open double cone onto an open interval in ##R##. You can then continuously map the interval onto an open subset of the plane.
 
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  • #8
lavinia said:
You can continuously ( in fact smoothly) map an open double cone onto an open interval in ##R##. You can then continuously map the interval onto an open subset of the plane.
Yes, I indeed emphasized the wrong property as the bijection is the crucial point.
 
  • #9
Orodruin said:
It doesn't look like either. There is no smooth map from an open set of R nor R2 that maps to that region.

There is no smooth map from an open set in the plane (or the real line) onto a region of the vertex of a single cone either. Yet a single cone is a manifold.
 
  • #10
lavinia said:
There is no smooth map from an open set in the plane (or the real line) onto a region of the vertex of a single cone either. Yet a single cone is a manifold.
Fine, as a physicist, I mainly just consider smooth manifolds.
 
  • #11
@Heisenberg1993 Is the OP satisfied with the answers or would he like to think about this some more - for instance how to prove that the double cone is not a manifold?
 
  • #12
The common vertex is a cut point of order 2, this means that if you take it away from the space, you end up with a disconnected space with 2 components. This is the reason for the failure of your space to be a manifold. You can easily prove that homeomorphisms preserve cut points of order n, and from this the result follows.
 
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  • #13
Orodruin said:
Fine, as a physicist, I mainly just consider smooth manifolds.

The single cone can be very easily be considered a smooth manifold. A double cone can never be a smooth manifold. That's the difference.
 
  • #14
micromass said:
The single cone can be very easily be considered a smooth manifold. A double cone can never be a smooth manifold. That's the difference.

Every 2 dimensional manifold has a smooth structure so smoothness is not the question. The double cone is not a manifold because there is no neighborhood of its vertex that is homeomorphic to an open set in the plane. Cruz Martinez gave a correct proof in post #12.
 
  • #15
While the cone is a 2- manifold and can be given a smooth structure, as a subset of ##R^3## I don't think that it is a smooth manifold. This is because it is the image of a non-smooth embedding of a 2 disk into ##R^3##.
 
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1. Why do two cones connected at their vertices not form a manifold?

This is because a manifold is a space that locally resembles Euclidean space, meaning it should be smooth and have a well-defined tangent space at every point. However, at the point where the two cones are connected, the surface becomes non-smooth and the tangent space is not well-defined, making it not a manifold.

2. Can two cones connected at their vertices still be considered a surface?

Yes, two cones connected at their vertices can still be considered a surface. However, it is not a manifold since it does not locally resemble Euclidean space.

3. What is the difference between a manifold and a non-manifold surface?

A manifold is a smooth surface that locally resembles Euclidean space, while a non-manifold surface is not smooth and does not have a well-defined tangent space at every point. A manifold can be described using mathematical equations, while a non-manifold surface cannot.

4. Are there any real-world examples of two cones connected at their vertices not forming a manifold?

Yes, a real-life example of this is the shape of a saddle. When viewed from above, it may look like two cones connected at their vertices, but the surface is not smooth and does not locally resemble Euclidean space.

5. How does the non-manifold property of two cones connected at their vertices affect its properties?

The non-manifold property of two cones connected at their vertices affects its properties by making it non-smooth and not well-defined at the point of connection. This can impact its curvature, area, and other geometric properties, making it difficult to analyze and describe mathematically.

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