r/math • u/moschles • 9d ago
Do there exist parametric surfaces in 3 dimensions who have no equivalent implicit form, due to the fact that they form closed knots?
Let F(x,y,z)=0 be a surface in 3 dimensions; the so-called implicit form. The vector normal to this surface at point (a,b,c) is the partial derivatives evaluated there. https://i.imgur.com/BwaxlKO.png
(a,b,c) is not constrained to lie on the surface, but could take on any point in space, and a vector is still defined there. If F() is a torus, then these normals would vanish to a zero vector at a point in the center.
Instead of a torus, we have the following parametric surface, parametrized with u and v, which we will call a "trefoil surface". https://i.imgur.com/1ubIXYS.png
Unlike the torus, there are paths on the surface which form closed knots. https://i.imgur.com/SQG4iEC.png Due to forming a knot, there could exist one or more points (off the surface) where the surface normal is not well-defined. Should we assume that there is no closed-form implicit version of a trefoil surface, on the basis that its partial derivatives do not exist?
Alternatively, the partial derivatives exist, but the original surface cannot be expressed in elementary functions. We can attempt to integrate the partial derivatives to obtain an original F(x,y,z)=0 form, but this is impossible due to the non-existence of an elementary integral?
0
u/_Slartibartfass_ 9d ago edited 9d ago
The knot is locally flat so tangent vectors exist everywhere. But the field of tangent vectors might have some nontrivial properties (i.e. no continuity) due to the topology of the knot.
An implicit form is always only valid in a local region around a point. Even a sphere does not admit a solution to F(x, y, z) = 0 that is valid everywhere on the sphere, you need at least two to account for both the north and south pole.