```From:           greg@math.math.ucdavis.edu (Greg Kuperberg)
Date:           10 Oct 1996 22:08:39 GMT
Newsgroups:     sci.math.research
Subject:        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/
```