# Sphere Eversion

Is it possible to gradually turn a sphere inside out without ripping, creasing, or crushing it? If the sphere is a solid sphere, the answer is obviously no. However, if the sphere is allowed to pass through itself, like a ghost, then the answer is yes! In mathematical terminology, we're asking whether or not there exists a regular homotopy of immersions connecting the identity map with multiplicaiton by -1.

This fact isn't obvious, and when Stephen Smale told his advisor, Raoul Bott about his discovery that such an eversion was possible, Bott responded by saying that the result was obviously false! Eventually Bott was eventually convinced that the result was correct, but he was frustrated by the lack of a concrete description of such an eversion.

The interactive visualization above allows you to see what one family of sphere eversions looks like. The sphere shown above is blue on the outside and golden on the inside. After the eversion takes place the sides and colors are reversed.

The first step corresponds to molding a sphere into a bowl with an inner and outer layer. It is obvious that this can be done. It can even be done with a physical ball.

The next step is a subtle puckering of the top of the bowl. As we do so lips form at the top of the bowl as if it is about to whistle. Also we see the inner bowl bulge out and pass through the outer bowl in some areas to form a symmetric pattern.

The twist is the middle step of the transformation, and the one that most strains the imagination. The twist creates arms at the upper lip of the bowl that fold and unfold over each other, while the base of the bowl barely moves at all. At the end of the twist the inner bowl has become the outer bowl and vice versa.

Next, we simply unpucker and reinflate the bowl. We get our sphere back, only now it is inside out!

To create this visualization I followed the paper Analytic Eversion of the Sphere using Ruled Surfaces by Adam Bednorz and Witold Bednorz.

The sphere is parametrized by two angles $$\phi$$ and $$\theta$$ with $$\phi$$ in the interval $$[-\pi, \pi]$$ and $$\theta$$ in the interval $$[-\pi / 2 , \pi / 2]$$.

The mapping in the visualization above is given by $$(\theta, \phi) \mapsto (x'', y'', z'')$$, with

\begin{aligned} x &=& p \sin((n-1) \phi) + t(1-\lambda + \lambda \cos^n \theta) \cos \phi - \lambda \omega \sin \theta \sin \phi / \cos^n \theta \\ y &=& p \cos((n-1) \phi) + t(1-\lambda + \lambda \cos^n \theta) \sin \phi - \lambda \omega \sin \theta \cos \phi / \cos^n \theta \\ z &=& \lambda(\omega \sin \theta (\sin (n\phi) - qt)/ \cos^n \theta - (t/n) \cos (n\phi)) - (1-\lambda)\eta^{1+\kappa}t |t|^{2\kappa} \sin \theta / \cos^{2n}\theta \end{aligned}

\begin{aligned} x' &=& x(\xi + \eta(x^2 + y^2))^{-\kappa} \\ y' &=& y(\xi + \eta(x^2 + y^2))^{-\kappa} \\ z' &=& z/(\xi + \eta(x^2 + y^2)) \end{aligned}

\begin{aligned} x'' &=& x' e^{\gamma z'} / (\alpha + \beta(x'^2 + y'^2)) \\ y'' &=& y' e^{\gamma z'} / (\alpha + \beta(x'^2 + y'^2)) \\ z'' &=& \frac{\alpha - \beta(x'^2 + y'^2)}{\alpha + \beta(x'^2 + y'^2)} \frac{e^{\gamma z'}}{\gamma} - \gamma^{-1} \frac{\alpha - \beta}{\alpha + \beta} \end{aligned}

To obtain these equations, I combined several family of equations in the paper so that all steps of the eversion could be viewed in one go.

Each integer $$n>2$$ gives a distinct eversion, and increases the number of arms that are created in the pucker phase.

The eversion consists of several steps which correspond to various groups of parameters in the equations.

Throughout the eversion, $$\omega=2$$, $$\eta=1$$, $$\gamma=2\sqrt{\alpha\beta}$$, and $$\kappa=(n-1)/2n$$.

At the beginning of the eversion, we have $$t=-1.5$$, $$p=0$$, $$q=0.66$$, $$\xi=0$$, $$\alpha=0.01$$, $$\beta=1$$, and $$\lambda=0$$

In the bowl step, $$\lambda$$ goes linearly from 0 to 1, $$\xi$$ goes from 0 to 1, $$\alpha$$ goes from 0 to 1, and $$\beta$$ goes from 1 to 0.12.

In the pucker step, $$p$$ goes from 0 to 1 and $$q$$ goes from 1 to 0.66.

In the twist step, $$t$$ goes linearly from -1.5 to 1.5.

If you select wireframe mode, you can look closely at each of the triangular elements and see that throughout the eversion they are smoothly and gently stretched, translated, and rotated into their final position, without getting smashed down to a line or a point, and without getting bent, torn, punctured, or kinked.

In the visualization below, I've handed over full control over all of the parameters. Note that the mapping does not give a smooth immersion for all possible combinations of the parameters.