# Can a surface of revolution be built from a self-intersected curve?

I'm reading "Differential Geometry of Curves And Surfaces" of Manfredo Do Carmo. There's a point in his book about Surfaces of Revolution which confuses me a lot. Here is the part:

The part confuses me is the last part to show that x is a homeomorphism: "In fact, since $(f(v), g(v)$ is a parametrization of $C$, given $z$ and $x^2 + y^2 = (f(v))^2$, we can determine $v$ uniquely".

In my understanding, what he means is that because $(f(v), g(v))$ is a parametrization, then one pair of $(f(v), g(v))$ determines only one parameter $v$. But I think this is not true for a curve which intersects with itself. So if I'm right, how can we prove that $\mathsf{x}(u,v)$ is homeomorphism in general case. Most of the case for surface of revolution which I know, the generating curve is not self-intersected. Am I missing something? Please help me to clarify this. Thanks a lot

I am not sure if I understood you properly, ... whether self intersection is for the 3D curves on surface of revolution or for the meridian itself.By generating curve do you mean the meridian or the non-planar 3D space curve written on it?

(f(v),g(v)) is a parameterization determining a single unique point on a meridian through which any number of curves ( for example geodesics) can be made to pass. Like here in the first image

Consider the geodesics drawn on a torus

$$x = (\cos v + 2) \cos u ; y = (\cos(v) + 2) \sin u; z = \sin v; \,$$

in which varying $v$ draws the eccentric circle meridian and $u$ rotates it around x-axis.

For an arbitrary boundary condition draw geodesics on toroid surface. The geodesics criss cross infinitely many times creating many self intersections on a unique meridian. This can be made to happen by choosing initial constant point in position but varying inclination in the tangent plane.

Maybe you miss the distinction between self intersection among the 3D single parameter space curve on the surface and the uniquely defined meridian through parameter $v$. But I can be also wrong.Shall improve my answer on your reply.

• By self intersection, I mean the self intersection for meridian, because if you read the text from Carmo's book, the surface of revolution he uses is generated from a plane curve. So, you're right, $(f(v),g(v))$ will determine a unique point, but my question is that can $(f(v),g(v))$ determines an unique $v$ value (that's what I understand from Carmo last part)? It's great if you can help me clarify this, thanks! Commented Jan 31, 2016 at 11:19
• That is possible. Say the intersecting meridian is for Bernoulli's Lemniscate $x= sin(t)/(1+ sin^{2}(t)), y= cos(t) sin(t)/(1+ sin^{2}(t)) +1$ . It should be $parametrized$ . y = f(x) wont do. For a given $t$ , $(x,y)$ of meridian is found, but vice-versa, not possible always uniquely. There can be double roots. Commented Jan 31, 2016 at 14:02
• Yeah, I agree. So I think that part needs to clarify more, and to make the definition of surfaces of revolution unambiguous, we need to assume that the meridian is not self-intersected, like @Bombyx mori suggests Commented Jan 31, 2016 at 14:05
• The meridian can self intersect any number of times producing toroid annular rings, but parameterization is a must. Commented Jan 31, 2016 at 16:16

I think you are right that the example needs some work. A regular plane curve could have figure $8$ intersections with itself. In such cases the exterior and interior of the curve can be confusing. After rotation it is not hard to imagine the resulting surface, which tends to have a cusp-type singularity. It is very unlikely De Carmo meant this when writing an introductory textbook.

I think what De Carmo meant is a regular simple plane curve. If this hypothesis holds, then the curve is differeomorphic to $\mathbb{S}^{1}$, and the surface produced this way would be differeomorphic to $\mathbb{T}^{1}$.

• I agree with you about this. Thanks for making this clear and giving me a counter-example, but I want to hear more ideas, so I will accept your answer later. Is it ok? Commented Jan 30, 2016 at 8:20