3D object inverted w.r.t. a sphere

[neelakash]

Few days back I posted a question here that dealt with inversion geometry. A point P inside a sphere can be inverted with respect to the spherical surface to another point P' outside. Center of the sphere O,P and P' are collinear and OPxOP'=r^2

I was wondering if it is possible to get a 3D object inverted with respect to a sphere. It means that given I have an object (a collection of points) outside the sphere. I know its shape, coordinates etc. Is it possible to calculate the shape inverted with respect to the sphere? After all, it is a collection of points for which inversion technique works elegantly. Does anyone know whether this has been done somewhere or not? To me, it looks that
for some regular objects, it may be found...but I am not sure how?

-Neel

 

[jasomill]

Sure it's possible — given a set of points, apply your definition to each point! One minor nit-pick: given an n-1 dimensional sphere in $$\mathbb{R}^n$$, inversion is only defined with respect to the complement of the center of the sphere. To extend to the center, we need to add a "point at infinity."

To see what's going on, let's try it with the simplest possible case: inversion in the unit circle in $$\mathbb{R}^2 \cup \{\infty\}$$. Since the reciprocal of a complex number is just "inversion followed by reflection about the real axis," we may as well work in $$\mathbb{C} \cup \{\infty\}$$ instead, then

$$i_C(z) = \frac{1}{z^*},$$

that is to say,

$$i_C(x+i y) = \frac{1}{x - i y} = \frac{x + i y}{x^2 + y^2},$$

which, as you explained, is the unique point on the line through z and the origin such that $$\left|z\right| \cdot \left|i_C(z)\right| = 1$$.

Now say we have a parameterized curve given by a map

$$\gamma: [0, 1] \to \mathbb{C}.$$

For simplicity, assume it doesn't pass through the origin. Then we invert it by passing the result through our inversion map: $$i_C \circ \gamma$$. For example, inverting the line segment

$$\gamma(t) = 2t - 1 + i$$

gives

$$i_C \circ \gamma(t) = \frac{1}{2t - 1 - i} = \frac{2t-1 + i}{4t^2 - 4t + 2}.$$

As you can http://www.wolframalpha.com/input/?.../(4t^2+-+4t++++2),1/(4t^2+-+4t++++2)}+t=0..1", this is a semicircle.

In case you're not familiar with complex numbers, the "y" coordinate is the imaginary part, that is, the bit being multiplied by $$i$$, while the "x" coordinate is the "real part," i.e., the rest. This is why I "simplified" the expression above in a way that makes it appear more complicated. Also, it'd probably look nicer in polar coordinates.

Working out similar formulas for spheres in $$\mathbb{R}^3$$ shouldn't be hard — the usual trick for this sort of thing is, work out the map for the simplest case (a sphere centered around the origin, say), then "translate" your formulas as appropriate. Taking spherical coordinates could also help.

Inversion is indeed an elegant technique — https://www.amazon.com/dp/069112745X/?tag=pfamazon01-20 (follow this link and watch the video!), which, among other things, have applications to complex analysis, non-Euclidean geometry, and special relativity.

Whether or not coordinate expressions of inversion are "elegant" is really a matter of what it is you're trying to invert — if you think of a line as a "circle of infinite radius," inversion "takes circles to circles," and there are especially nice relationships involving circles that intersect the inversion circle (as in the example above), or at right angles to it, and those passing through the origin, so your mileage may vary. Returning to the example above, I can't think of an easier way to find a rational parametrization of a circular arc.

The case with spheres (of any dimension!) is mostly the same.

Chapter 3 of Needham's http://usf.usfca.edu/vca/" has a short, but particularly well-written section on Möbius transformations in the complex plane.

In summary, inversive geometry is both beautiful and useful, and certainly worth learning about!

Cheers,
Jason