Date: Mon, 07 Apr 1997 21:59:21 +0400
From: Laurent BADOUD
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 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