Which is the function that the zero-set of which is Möbius strip?

Question

Many surfaces can be expressed by implicit functions. As described later, the spherical surface, the side surface of the cylinder, and the torus can be expressed by implicit functions. However, I have never seen the implicit function of Möbius strip.

As will be explained later, the parametrization of the Torus and Möbius strip seems very similar form.

【My Question】:

1. Can Möbius strip (or "Joint-around-portion removed version of Möbius strip"†) be expressed by implicit function?
2. If possible, which is the function that the zero-set of which is Möbius strip (or "Joint-around-portion removed version of Möbius strip" †) ?

†. Here, the "Joint-around-portion removed version of Möbius strip" represent the surface created by removing a narrow closed set near the joint from Möbius strip so as to be diffeomorphic to closed rectangle. For example, with ε> 0 as a small fixed constant, replace the domain of u in Box 2 with $0 \le u \le 2\pi - \epsilon$

I do not know much about whether the manifold, which is not orientable, can be expressed as a zero set of some functions. But, above-mantioned "Joint-around-portion removed version of Möbius strip" is topologically the closed rectangle (therefore closed and orientable) and 'almost' Möbius strip.

Many surfaces can be expressed by implicit functions. The spherical surface is the zero set of following ${f}_{suf}$, $${f}_{suf}(x,y,z)={x}^{2} + {y}^{2} +{z}^{2} -1 $$ The side-surfaxe of cylinder is the zero set of following $f_sils$, and $$f_{sils}={x}^{2} + {y}^{2} -1$$

When $R \ge r >0 $, the Torus defined by following parametrization is the zero set of following ${f}_{torus,R,r}$ $${f}_{torus,R,r} = \left(\sqrt{x^2 + y^2}-R\right)^2 + z^2 = r^2 .$$

The parametrization of above-mentioned Torus is:

Box.1 (A Parametrization of Torus, quoted with minor modification from the wikipedia) $$\begin{align} x(\theta, \varphi) &= (R + r \cos \theta) \cos{\varphi}\\ y(\theta, \varphi) &= (R + r \cos \theta) \sin{\varphi}\\ z(\theta, \varphi) &= r \sin \theta \end{align}$$ where

• θ, φ are angles which make a full circle, so that their values start and end at the same point,
• R is the distance from the center of the tube to the center of the torus,
• r is the radius of the tube.
• The constant value $R$ and $r$ shall be $R \ge r >0 $

The parametrization of Torus looks to me very similar to that of Möbius strip.

As described in Wikipedia, one way to represent the Möbius strip as a subset of three-dimensional Euclidean space is using following parametrization:

Box.2 (A Parametrization of Möbius strip quoted from the Wikipedia with minor modification.) $$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u$$ $$y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u$$ $$z(u,v)= \frac{v}{2}\sin \frac{u}{2}$$ where,

• $0 \le u< 2\pi$, and,
• $-1 \le v\le 1$.

This creates a Möbius strip of width 1 whose center circle has radius 1, lies in the $xy$-plane and is centered at $(0, 0, 0)$. The parameter $u$ runs around the strip while $v$ moves from one edge to the other.

The parametrization of the Torus and Möbius strip look same to me (If $R = 1$, $r=v/2$　,the essential difference is what is considered as a variable and what is considered as a constant.) Therefore, I think that Möbius strip(or almost part of Möbius strip) be expressed implicitly, by following fuctuion.

$${f}_{torus,1,v/2} = \left(\sqrt{x^2 + y^2}-1\right)^2 + z^2 = (\frac{v}{2} )^2 $$

Try the slimier manner of 1578756

First plug in $\;z=z(u,v) = \frac{v}{2}\sin(u/2) $ to ${f}_{torus,1,v/2}(x,y,z)$ to get

$$\left(1-\sqrt{x^2+y^2}\right)^2+{(v/2)}^2\sin^{2} (u/2)={(v/2)}^{2} ,$$

which is $$\left(1-\sqrt{x^2+y^2}\right)^{2}={(v/2)}^{2}(1-\sin^{2}(v/2)).$$

The last expression can also be written as $$\left(\sqrt{x^2+y^2}-1\right)^2={(v/2)}^{2}(1-\sin^{2}(v/2))={(v/2)}^{2}(\cos^{2}(v/2)).$$

Clear away squares and rearrange to obtain $$\sqrt{x^2+y^2}=1+(v/2)\cos\theta,$$ or $$\sqrt{x^2+y^2}=1-(v/2)\cos\theta,$$

and from this you get $$\;x^2+y^2=(1+(v/2)\cos\theta)^2,$$ or $$\;x^2+y^2=(1-(v/2)\cos\theta)^2,$$ or

Perhaps it seems to be necessary to distinguish between positive and negative values of v / 2. But I think this seems to be a natural. I forcibly separated the Mobius strip into front-side and back-side. It seems difficult to obtain θ geometrically unlike the torus.

So, I'll divide the question of implicit function of catted out version of Mobius strip, later.

P.S.
I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions. English review is also welcomed.

Answer 1

For any closed $C\subset \Bbb R^n$, we can define the distance-from-$C$ function as $f\colon\Bbb R^n\to\Bbb R$ given by $$f(x):=\inf\{\,\|x-y\|:y\in C\,\}.$$ Then $f(x)=0\iff x\in C$.

If this $f$ is not smooth enough for your purposes, replace it with the $g\circ f$ where $g(t)=\begin{cases}e^{-1/t}&t\ne0\\0&t=0\end{cases}$.

However, it is not possible to find any continuous implicit function for the Möbius strip such that the values "cross" through $0$ with a sign change: If the function takes negative and positive values, you can always find a path between points with different signs that avoids the Möbius strip (in fact, you can do so by moving very close to the strip until you reappear at its "other" side), and by the IVT, some point along any such path must have function value $=0$.