# A Two cones connected at their vertices do not form a manifold

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1. Jan 10, 2017

### 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?

2. Jan 10, 2017

### Orodruin

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

3. Jan 10, 2017

### Heisenberg1993

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

4. Jan 10, 2017

### Orodruin

Staff Emeritus
It doesn't look like either. There is no smooth map from an open set of R nor R2 that maps to that region.

5. Jan 10, 2017

### Staff: Mentor

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:

@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.)

6. Jan 10, 2017

### lavinia

- What is your definition of "manifold"?
- What do you mean by "look like"?

7. Jan 10, 2017

### lavinia

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.

Last edited: Jan 11, 2017
8. Jan 10, 2017

### Staff: Mentor

Yes, I indeed emphasized the wrong property as the bijection is the crucial point.

9. Jan 11, 2017

### lavinia

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. Jan 11, 2017

### Orodruin

Staff Emeritus
Fine, as a physicist, I mainly just consider smooth manifolds.

11. Jan 12, 2017

### 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?

12. Jan 12, 2017

### 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.

13. Jan 12, 2017

### micromass

Staff Emeritus
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. Jan 12, 2017

### lavinia

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. Jan 12, 2017

### 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 $R^3$.

Last edited: Jan 13, 2017