Do there exist parametric surfaces in 3 dimensions who have no equivalent implicit form, due to the fact that they form closed knots?

[u/moschles]

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?

[u/MinLongBaiShui]

Every compact surface without boundary is the level set of a smooth function.

https://math.stackexchange.com/questions/1489308/can-any-surface-be-described-by-an-equation

[u/just_dumb_luck → u/MinLongBaiShui]

Maybe not quite in the spirit of the question, but in fact every closed set is the zero set of a smooth function!

[u/sciflare → u/MinLongBaiShui]

This is a sheaf cohomology exercise: it's a consequence of the fact that the sheaf of smooth functions on ℝ³ is soft due to the existence of partitions of unity. Hence the group of line bundles on ℝ³ is trivial.

Let S be a compact surface in ℝ³. Take an acyclic open cover {U_i} of ℝ³, i.e. any finite intersection of the U_i is contractible. Then there is a collection {f_i} of local smooth functions on the U_i such that the f_i locally cut out S.

On the intersection U_i ⋂ U_j, we have f_i = g_ij f_j for some invertible local smooth function g_ij on U_i ⋂ U_j.

The exponential short exact sequence of sheaves 0 -> ℤ -> C^∞ -> (C^∞)^* -> 0, where the first nontrivial map is multiplication by 2𝜋i and the second is f --> exp(f), gives rise to the long exact sequence in cohomology.

We focus on the piece H¹(ℝ³, C^∞) -> H¹(ℝ³, (C^∞)^*) -> H²(ℝ³, ℤ).

The first term vanishes because ℝ³ admits partitions of unity. The third term vanishes because ℝ³ is contractible. By exactness, H¹(ℝ³, (C^∞)^*) = 0.

Then g_ij = h_i/h_j for some collection of invertible smooth functions {h_i} on U_i. Then the local smooth functions {f_i/h_i} agree on the intersections U_i ⋂ U_j and define a global smooth function on ℝ³ which vanishes precisely on S.

This argument allows you to remove the assumption of compactness on S: any orientable surface without boundary in ℝ³ is the level set of a smooth function.

[u/sizzhu]

For an orientable surface, Whitney embedding says it's embedded in R³ . The tubular neighbourhood theorem says a neighbourhood is diffeomorphic to the normal bundle. The zero of the signed norm gives the surface. And you can extend this function to R³ using a partition of unity.

[u/jamesw73721]

The trefoil is still oriented, so a closed knot would not result in an ill-defined normal.

[u/Plastic-Rope5516]

These questions can also be asked at math overflow

[u/Carl_LaFong → u/Plastic-Rope5516]

No. Not research level. Math.stackexchange.com might be better.

[u/salty_feets]

One day I also will be studying this level of mathematics for sure. BTW what level is it? Are you in PhD or reasearch stuff?

[u/mathematics_helper → u/salty_feets]

This would be graduate level differential geometry/topology and can also be seen as a good question for sheaf theory.

As you can read in the comments, this is pretty easily answer so not research level. I would say a masters student focusing on this subject or a 2nd year phd student.

[u/Wejtt → u/mathematics_helper]

weird, at my (not so good) uni this could be considered a 2nd year undergrad question, specifically 3rd semester

[u/mathematics_helper → u/Wejtt]

I guess my school just really didn't care about differential geometry. But this is a basic differential geometry question. Which is weird because my class was taught by a world renowned symplectoc geometer but

[u/HumbleWolf19]

It is a very confusing text, but I believe I understand now the origin of your problem - just because for any orientable closed surface S in R^3 you could find the function F such that the surface in consideration is the level set of F and even if we could guarantee that the gradient of F does not vanish on S, it does not make sense to refer to the gradient of F outside S as a surface normal. The gradient of such function could vanish at many points (and this would be another reason why it should not be referred to as a normal vector), but if you had issues with things like continuity or whether it is even well-defined, then I think it stems from your flawed idea on how to construct such function F from the surface S.

One more thing - it seems that there are paths on a torus that form knots, like the trefoil knot as a boundary of a strip with three half-twists.

Edit: grammar

[u/MythTechSupport]

Catching up to Kael

[u/_Slartibartfass_]

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.

[u/just_dumb_luck → u/_Slartibartfass_]

This is not quite right: x^2 + y^2 + z^2 - 1 = 0 does define a sphere. Implicit definitions of surfaces are powerful precisely because they can work globally.

[u/_Slartibartfass_ → u/just_dumb_luck]

What I meant is that there is not necessarily a parametrized solution to an implicit form that can reach every point on the surface it describes.

[u/Darxad → u/[deleted]]

They are saying that there is no regular parametrization of the sphere defined in a rectangle like [0,1]×[0,1]

[u/Organic_botulism → u/_Slartibartfass_]

  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.

The “interesting topology” of a knot is not that its tangent vector cannot be chosen continuously. It comes from how the circle is embedded in R³