Is a Full-Cone in E^3 a Topological Manifold?

[cianfa72]

TL;DR

Full-cone topological space but not topological manifold

Hello,

consider a full-cone (let me say a cone including bottom half, upper half and the vertex) embedded in $E^3$. We can endow it with the topology induced by $E^3$ defining its open sets as the intersections between $E^3$ open sets (euclidean topology) and the full-cone thought itself as subset of $E^3$. This way it has topological space structure.

Nevertheless I believe it is not a topological manifold because any neighborhood of the vertex cannot be homeomorphic to $E^3$ or one of its open subset.

Is that correct ? Thanks

 

[Orodruin]

The space that the cone embedded in $E^3$ is locally homeomorphic to everywhere but at the apex is $E^2$, not $E^3$.

It is true that the cone is not locally homeomorphic to either $E^2$ or $E^3$ at the apex.

 

[cianfa72]

The space that the cone embedded in $E^3$ is locally homeomorphic to everywhere but at the apex is $E^2$, not $E^3$.

yes, sure.

 

[fresh_42]

Summary: Full-cone topological space but not topological manifold

Hello,

consider a full-cone (let me say a cone including bottom half, upper half and the vertex) embedded in $E^3$. We can endow it with the topology induced by $E^3$ defining its open sets as the intersections between $E^3$ open sets (euclidean topology) and the full-cone thought itself as subset of $E^3$. This way it has topological space structure.

Nevertheless I believe it is not a topological manifold because any neighborhood of the vertex cannot be homeomorphic to $E^3$ or one of its open subset.

Is that correct ? Thanks

The cone (one half) is homeomorphic to a ball or a cube. It is topologically a compact subset of $\mathbf{E}^3$. Inner points are locally homeomorphic to $\mathbf{E}^3$, boundary points to $\mathbf{E}^2$.

The vertex is no specific point from a topological point of view, except that it is on the boundary.

The cone is no analytic manifold, but it is a topological manifold, or a compact, connected subset of one to be precise.

 

[Orodruin]

The cone (one half) is homeomorphic to a ball or a cube. It is topologically a compact subset of $\mathbf{E}^3$. Inner points are locally homeomorphic to $\mathbf{E}^3$, boundary points to $\mathbf{E}^2$.

The vertex is no specific point from a topological point of view, except that it is on the boundary.

The cone is no analytic manifold, but it is a topological manifold, or a compact, connected subset of one to be precise.

He is considering a double cone, joined at the apex. I.e., style light-cone:

 

[lavinia]

Summary: Full-cone topological space but not topological manifold

Hello,

consider a full-cone (let me say a cone including bottom half, upper half and the vertex) embedded in $E^3$. We can endow it with the topology induced by $E^3$ defining its open sets as the intersections between $E^3$ open sets (euclidean topology) and the full-cone thought itself as subset of $E^3$. This way it has topological space structure.

Nevertheless I believe it is not a topological manifold because any neighborhood of the vertex cannot be homeomorphic to $E^3$ or one of its open subset.

Is that correct ? Thanks

Have you come up with a proof?

 

[cianfa72]

Have you come up with a proof?

Basically removing a point from an $\mathbb R^2$ disk we get a connected set. Yet the set attained removing the apex from a double cone is not.

 

[WWGD]

Yes, I think in general an n-manifold must have an n- or (n-1)-dimensional cutset. A 1-ball ( interval) can be separated by a single point., a 2-ball must be separated by a line(segment), etc.