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The following claim is a well-known consequence of the Whitney-Graustein theorem:

Claim. It does not exist $H\colon\mathbb{S}^1\times[0,1]\overset{C^1}{\rightarrow}\mathbb{R}^2$ such that for all $t\in [0,1]$, $H(\cdot,t)\colon\mathbb{S}^1\rightarrow\mathbb{R}^2$ is an immersion, $H(\cdot,0)=(\cos(2\pi\cdot),\sin(2\pi\cdot))$ and $H(\cdot,1)=(\cos(2\pi \cdot),-\sin(2\pi\cdot))$.

enter image description here

In other words, it is impossible to perform a circle eversion in the plane, namely it is impossible to continuously and regularly change the orientation of the circle while sticking to the plane.

However, I want to illustrate that it is possible to realize the circle eversion in the $3$-dimensional space.

The idea is to thicken the circle into a cylinder, perform a $\pi$-twist on the cylinder in order to put its inside out and finally to retract the everted cylinder onto its equatorial circle.

My main concern is to graphically represent the above process using a mathematical software, e.g. SageMath. I tried in vain to write down explicit formulas for it and here I am stuck. Please note that the following homotopy did not seem to do any good:

$$\forall x\in\mathbb{S}^1\times [-1,1],\forall t\in [0,1],H(x,t)=\frac{x}{\|x\|^{2t}}.$$

Any enlightenment will be greatly appreciated!

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  • $\begingroup$ Helpful: m.youtube.com/watch?v=R_w4HYXuo9M $\endgroup$ – Michael Hoppe Dec 22 '16 at 20:05
  • $\begingroup$ @MichaelHoppe I already know this movie by Thurston. Please note that I actually want to perform a circle eversion in the $3$d-space not a sphere eversion. In the full movie such an animation can be seen, but it did not help me to reproduce it on my own. $\endgroup$ – C. Falcon Dec 22 '16 at 20:15
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    $\begingroup$ What you're describing looks unnecessarily hard. Represent the circle as being in the $xy$-plane and rotate around the $x$-axis. $\endgroup$ – user98602 Dec 22 '16 at 21:49
  • $\begingroup$ @MikeMiller I guess I totally overlooked the problem! You are totally right, I am grateful for your input! $\endgroup$ – C. Falcon Dec 22 '16 at 21:56
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As pointed out by Mike Miller in the comments, the process I described is unnecessarily complex. In order to perform the circle eversion in the $3$-dimensional space, it suffices to assume that the circle lies in the $(xOy)$ plane and then gradually operate a rotation of angle $\pi$ and axe $x$ on it. This is achieved by the following : $$H\colon\left\{\begin{array}{ccc} \mathbb{S}^1\times[0,1]& \rightarrow & \mathbb{R}^3\\ (x,t) & \mapsto & (\cos(2\pi x),\cos(\pi t)\sin(2\pi x),\sin(\pi t)\sin(2\pi x)) \end{array}\right..$$

enter image description here

As a bonus, here are some pictures on how to unravel a figure-eight knot:

enter image description here enter image description here enter image description here

And just for fun, here is the unraveling in real life:

enter image description here enter image description here

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