Date: Mon, 07 Apr 1997 21:59:21 +0400
From: Laurent BADOUD <lbadoud@iprolink.ch>
Newsgroups: sci.math
Subject: Group - Mobius

=46rom Laurent Badoud     lbadoud@iprolink.ch

Do the two functions :z -> 1/z   and  z -> z + 1 generate the whole
Mobius group with integer coefficients ? =

If so, given an element  f(z) = (az +b)/(cz +d) with ad - bc = 1 or -=1
how do you find the shortest "word" that generates f ? If not, what
numerical invariant characterises the elements generated by the two
latter functions ?

Laurent Badoud Gen=E8ve

==============================================================================

From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Group - Mobius
Date: 8 Apr 1997 20:03:43 GMT

In article <334935F9.771D@iprolink.ch>,
Laurent BADOUD  <lbadoud@iprolink.ch> wrote:

>Do the two functions :z -> 1/z   and  z -> z + 1 generate the whole
>Mobius group with integer coefficients ? =
>
>If so, given an element  f(z) = (az +b)/(cz +d) with ad - bc = 1 or -=1
>how do you find the shortest "word" that generates f ? If not, what
>numerical invariant characterises the elements generated by the two
>latter functions ?

First let me comment that the Moebius group may be described either as
PGL(2,C) or PSL(2,C): given complex coefficients, one may normalize to
assume that the determinant is 1. With other coefficient rings, one must
be more careful: it seems you apparently intend to ask about
PGL(2,Z), given your  f  above, but one could equally well ask for
all invertible functions  f(z)=(az+b)/(cz+d)  with  a,b,c,d, integral;
this is really  PGL(2,Q), which is certainly a larger group, including
e.g. the functions  f(z)=2z.

The subgroup PSL(2,Z) is freely generated by the elements F(z)=-1/z of
order 2 and G(z)=-1/(z+1) of order 3. (Note that  FG  is the second element
K  mentioned in the original post). Thus every element of  PSL(2,Z)  is a
_unique_ word in  F's and  G's (at least, it is unique if there are no
substrings  FF  or  GGG).

The group PGL(2,Z) contains this as a subgroup of index 2; the
element  H(z) = 1/z  of order 2 lies outside  PSL(2,Z). Thus every element
of  PGL(2,Z) may be uniquely written either as a (reduced) word in  F  and G
or as such a word times H. This applies in particular to words in
H and K: they fall into the two classes according as the number of
H's is  even or odd. To perform the reduction you need to know that
HK=FG^2H and HK^(-1)=GFH, so that in a long word of  H's , K's, and
K^(-1)'s, one may "move all the H's to the right" and convert to a
long string of G's and F's with either  H  or  1  at the very right.

Conversely, any string of  G's and F's may be converted to H's and K's
since  G  is the commutator  [K,H]=K^(-1)HKH  and then of course
F=FGG^(-1)=K[H,K]=KHK^(-1)HK (I hope you are surprised to see that
this means  N(z)=-z  can be expressed using only  H  and  K  !)

This group  PSL(2,Z) is known as the modular group. Discussions occur
throughout the mathematical literature: in group theory, complex
analysis, Riemann surfaces, elliptic curves, and number theory. See
for example Apostol's book on modular functions.

dave