Two cones connected at their vertices do not form a manifold

[Heisenberg1993]

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?

 

[Orodruin]

At that point, the manifold should look like R or R2?

But it doesn't, so it is not a manifold.

 

[Heisenberg1993]

That's exactly my question, why it doesn't look like which (R or R²)?

 

[Orodruin]

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.

 

[fresh_42]

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.)

 

[lavinia]

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"?

 

[lavinia]

...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.

 

[fresh_42]

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.

 

[lavinia]

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.

 

[Orodruin]

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.

 

[lavinia]

@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?

 

[Cruz Martinez]

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.

 

[micromass]

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.

 

[lavinia]

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.

 

[lavinia]

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 $R^3$.