Elementary Differential Geometry: Cone Not a Surface - Exercise Problem

[paluskar]

cone as surface...

In Barrett O'Neill's Elementary Differential Geometry book...he says that the cone
M:$x^{2}$ + $y^{2}$=$z^{2}$ is not a surface in that
there exists a point p in M such that there exists no proper patch in M which can cover a neighbourhood of p in M
Intuitively I realize that this is point is the apex...here $\left(0,0,0\right)$
but how would the patches for the rest of points be??...and what goes wrong with the patch at the apex of the cone??
Note: A PATCH IS A 1-1 REGULAR FUNCTION X:D→$ℝ^{3}$..D open in $ℝ^{2}$
This is an exercise problem.

 

[jvicens]

In order for a surface to be well-defined, it must have a neighborhood around each point that can be represented by a patch (a 1-1 regular function from an open set in ℝ^{2} to ℝ^{3}). However, in the case of a cone, there is no proper patch that can cover the apex point (0,0,0) because any neighborhood around the apex will include the entire cone, making it impossible to define a 1-1 regular function on that neighborhood.

To understand this in more detail, let's consider a point on the cone that is not the apex. At this point, we can define a patch that maps a small neighborhood on the cone to a corresponding open set in ℝ^{2}. However, as we move closer to the apex, the neighborhood on the cone becomes larger and eventually includes the entire cone. This means that the patch we defined for the point on the cone is no longer 1-1, as multiple points on the cone are mapped to the same point in the open set in ℝ^{2}. This violates the definition of a patch and shows that the cone cannot be considered a surface.

In summary, the issue with the apex of the cone is that it cannot be covered by a proper patch, which is necessary for a surface to be well-defined. This is due to the fact that the apex is a singular point on the cone, where the neighborhood becomes too large to be represented by a 1-1 regular function.