Why is a two sheeted hyperboliod regular but two cone connected at their start point not

Question

Let $x^2-y^2-z^2=1$ be the equation for the two sheeted hyperboliod and $x^2+y^2-z^2=1$ If we say the connected cone is not regular because of the ‘pointy end’ at $z=0$ then if we rearrange the equation for the hyperboliod to $x^2-1=y^2+z^2$ Then at $x=1$ we get similar coordinates as the cones point except it has shifted $1$ in the $x$ direction and somehow this shape is now regular?

So I guess my question is if we say the two attached cones are not a regular surface then why is the two sheet hyperboliod (which is basically the same two cones but ‘pulled apart’) is now somehow regular?

Answer 1

Basically by the same reason why $\sqrt{1+x^2}$ is differentiable, whereas $\sqrt{x^2}(=\lvert x\rvert)$ is not: the surface $x^2=y^2+z^2$ is the union of the surfaces $x=\pm\sqrt{y^2+z^2}$ and you have a problem concerning differentiability when $(x,y,z)=(0,0,0)$. But you have no such problem with the surfaces $x=\pm\sqrt{1+x^2+y^2}$; each one of thm is well-behaved at any point, as far as differentiability is concerned.

Answer 2

The answer depends a bit on your definition of a "regular" surface; there seems to be no universally familiar definition.

Going by the definition that a surface is regular if it is everywhere locally diffeomorphic to a plane, then the reason that the standard hyperboloid is regular is that even at the "closest" points on the two sheets, at each point you could smoothly deform a small local neighborhood to map to a plane.

But the connected cones, at the origin, cannot be smoothly deformed in that way because a line through the origin, staying on one of the codes, has a sharp bend and cannot be smoothly deformed to any smooth curve on the plane. So there is no diffeomorphism and the definition of "regular" is not met.