From:greg@math.math.ucdavis.edu (Greg Kuperberg)Date:10 Oct 1996 22:08:39 GMTNewsgroups:sci.math.researchSubject:Conceptual proof that inversion sends circles to circles

A while ago, it occured to me that if you understand the Mobius group well enough, you can prove with an absolute minimum of computation that inversion in a circle sends circles to (generalized) circles. Does this argument appears in the literature? The argument goes as follows: Consider the set of complex lines in C^2; it is parametrized by the complex numbers and infinity. The set of linear transformations on C^2 is a group action on this set; let the corresponding permutation group of the set be G. Since [[a,b],[c,d]] sends (1,z) to (a+bz,c+dz), it thus sends the slope z to (c+dz)/(a+bz). If a linear transformation preserves the lines (z,0) and (0,z), it must be diagonal, and if it preserves a third line also, it must be proportional to the identity. As any pair of lines in C^2 is equivalent to any other, we conclude that if an element of G fixed three points, it is the identity. Extend G to G-hat by adding complex conjugation. In G-hat, there are two elements which fix any given triple of points. Indeed, we can say what the other one is: Inversion in the (generalized) circle through the three points. To know this, we only need to know that all inversions in circles are in G-hat. Inversion in the unit circle is simply z -> 1/z-bar, and by conjugating by elements of G we can make inversion in any other circle. So we now have a characterization of generalized circles: They are the collection of fixed sets of elements of G-hat with at least three points. Any element of G-hat must permute this collection. In particular, z -> 1/z-bar does. -- /\ Greg Kuperberg greg@math.ucdavis.edu / \ \ / Recruiting or seeking a job in math? Check out my Generic Electronic \/ Job Application form, http://www.math.ucdavis.edu/~greg/geja/