[考古] 19世纪的几何学

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Nineteenth Century Geometry

• 5. The differential geometry of Riemann
• 6. Lie Groups
  • Supplement: A modern formulation of Riemann's theory
    
  
  

Lie sphere geometry
spherical wave transformations
Laguerre transformations

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The problem of Apollonius

The incidence of cycles in Lie sphere geometry provides a simple solution to the problem of Apollonius.^[3] This problem concerns a configuration of three distinct circles (which may be points or lines): the aim is to find every other circle (including points or lines) which is tangent to all three of the original circles. For a generic configuration of circles, there are at most eight such tangent circles.

The solution, using Lie sphere geometry, proceeds as follows. Choose an orientation for each of the three circles (there are eight ways to do this, but there are only four up to reversing the orientation of all three). This defines three points [x], [y], [z] on the Lie quadric Q. By the incidence of cycles, a solution to the Apollonian problem compatible with the chosen orientations is given by a point [q] ∈ Q such that q is orthogonal to x, y and z. If these three vectors are linearly dependent, then the corresponding points [x], [y], [z] lie on a line in projective space. Since a nontrivial quadratic equation has at most two solutions, this line actually lies in the Lie quadric, and any point [q] on this line defines a cycle incident with [x], [y] and [z]. Thus there are infinitely many solutions in this case.

If instead x, y and z are linearly independent then the subspace V orthogonal to all three is 2-dimensional. It can have signature (2,0), (1,0), or (1,1), in which case there are zero, one or two solutions for [q] respectively. (The signature cannot be (0,1) or (0,2) because it is orthogonal to a space containing more than one null line.) In the case that the subspace has signature (1,0), the unique solution q lies in the span of x, y and z.

The general solution to the Apollonian problem is obtained by reversing orientations of some of the circles, or equivalently, by considering the triples (x,ρ(y),z), (x,y,ρ(z)) and (x,ρ(y),ρ(z)).

Note that the triple (ρ(x),ρ(y),ρ(z)) yields the same solutions as (x,y,z), but with an overall reversal of orientation. Thus there are at most 8 solution circles to the Apollonian problem unless all three circles meet tangentially at a single point, when there are infinitely many solutions.