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In a book that I am reading, Polygon Mesh Processing (page 1, last paragraph), the authors say this:

[...] implicit definition is only available for planar curves, i.e., $\mathcal{C} = \{x \in \mathbb{R}^2 | F(x)=0\}$ with $F:\mathbb{R}^2 \to \mathbb{R}$.

in the context of defining parametric and implicit surface representations for curves. On the other hand, the authors specify no such restriction of planar curves when talking about parametric surfaces.

What I fail to reason, or get an intuition for is why we are unable to define a 3D non-planar curve in implicit representation, while being able to come up with a parametric definition of the same?

Harsh
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  • What is their definition of an implicit definition? The answer will depend on that. – Moishe Kohan Feb 10 '22 at 17:48
  • If you happen to know the Serret-Frenet differential system, there is a very interesting representation called the intrinsic representation describing under the form of an equation $f(\kappa,\tau)=0$ where $\kappa(s)$ is the curvature and $\tau(s)$ is the torsion of the curve, for example $\kappa = a \tau$ defines a helix. Caution: an intrinsic equation defines a curve up to rigid motion (for example a horizontal ans a vertical helix have the same intrinsic equation. – Jean Marie Feb 10 '22 at 18:17
  • In that book an Implicit curve is the "zero" level set of a function of two variables. The level set of a function of n variables lives in n-dimensional space - always. If you've f(x,y,z)=0 your level set is a surface that lives in 3D space and so on. – Mauricio Cele Lopez Belon Feb 11 '22 at 07:14
  • I also think they require the function f to be differentiable. – Mauricio Cele Lopez Belon Feb 11 '22 at 07:29
  • @MauricioCeleLopezBelon Then every closed subset of $R^n$ is the zero level set of an infinitely differentiable function. – Moishe Kohan Feb 11 '22 at 18:11

1 Answers1

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  1. Every closed subset of ${\mathbb R}^n$ is the zero level set of an infinitely differentiable function, see here. In particular, according to the definition you gave in a comment, every curve (with the mild restriction that it is a closed subset, that is also necessary) admits an implicit description.

  2. However, likely, the authors of your test are simply sloppy with their language and in the definition of an implicit description of a subset $C$ as the zero-level set of a function $f: {\mathbb R}^n \to {\mathbb R}$, they also (implicitly) require that $0$ is a regular value of the function $f$, meaning that $$ \nabla f(x)\ne {\mathbf 0}, \forall x\in C. $$ The implicit function theorem will imply that the set $C$ will be a manifold of dimension $n-1$. Hence, $C$ cannot be a (nonempty) curve unless $n=2$.

Moishe Kohan
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