Interactive reciprocation and inversion

Tim Hutton - 2014-10-06 20:37:18+0000 - Updated: 2014-10-06 20:37:18+0000

Interactive reciprocation and inversion of a circle.

I'm trying to understand how the hyperbola in hyperbolic space arises. Why wouldn't a parabola work, for example? And for projecting the hyperbola onto the Poincaré disk, is there a nice operation like there is in the spherical geometry case, where circle inversion takes us from the sphere to the plane and vice versa?

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Tim Hutton - 2014-10-07 08:19:45+0000 - Updated: 2014-10-07 09:30:20+0000

I found this: "hyperbolic space is a hyperboloid of two sheets that may be thought of as a sphere of squared radius -1 or of radius i = sqrt(-1)"
http://www.math.brown.edu/~rkenyon/papers/cannon.pdf
Now this makes more sense to me. A hyperboloid is a sphere with negative curvature. This is exactly what we want for exploring plane tessellations with different curvatures.

I read elsewhere that the intersection of two sphere is a circle if they intersect, and a circle with imaginary radius if they don't intersect. That means that two non-intersecting spheres intersect at a hyperboloid? Interesting.

Tim Hutton - 2014-10-07 12:16:43+0000 - Updated: 2014-10-07 12:21:10+0000

Further reading:
http://en.wikipedia.org/wiki/Minkowski_space
The change in sign of the squared radius is expressed neatly by the metric signature of the space (x,y,z) changing from (+,+,+) for the sphere to (+,+,-) for the hyperboloid. Is that right?

Tim Hutton - 2014-10-07 14:11:41+0000 - Updated: 2014-10-07 14:17:45+0000

Yes, it seems right. As hoped, the formula for stereographic projection is exactly as normal, but the distance function now depends on the metric signature. I've put a demo and explanation here: http://timhutton.github.io/hyperplay/generalized_stereographic_projection.html

Tim Hutton - 2014-10-07 14:17:26+0000

So to answer my own questions: the original demo in this post, about reciprocation, was a complete red herring. Under the correct metric signature the hyperbola is very naturally the inverted form of the circle. And yes, there is a nice operation that does stereographic projection: it's the usual stereographic projection, only with the appropriate distance function for the metric signature. See the previous comment for the formulae.

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