From: rong@math.gwu.edu (Yongwu Rong)
Newsgroups: sci.math.research,sci.math
Subject: Re: A 3-manifold question
Date: 22 Aug 1994 14:15:06 GMT

In article <32obpe$i2j@cronkite.seas.gwu.edu>,
>
>This question is more or less equivalent to:  determine all
>periodic maps up to conjugacy on a given surface.
>


I tried to post the following last week but it didn't work, so I am
giving another shot.   Y.R.      8/22

%%%%%%%%%%   text written on 8/17

After some email contact with Charlie Frohman and Dan Asimov, plus some 
patient work last night, I now believe I have a solution to my question.
Below is a complete (I hope) list of periodic homeomorphisms of
the genus 2 surface. There are nine such maps up to some equivalence 
relations such as conjugacy, or taking a power that is relatively 
prime to the order. Here F denotes the genus 2 surface, f: F -> F 
is the periodic homeomorphism of F with order d, and G is the quotient
orbifold.

1. d=2, G = T(2,2),   f is a \pi rotation of F along the "short axis".
              f has two fixed points.
2. d=2, G = S(2,2,2,2,2,2), f is a \pi rotation along the "long axis".
              f has 6 fixed points, and f is often called the mutation map.
3. d=3, G = S(3,3,3,3),  to see f, consider two concentric spheres connected
              by 3 tubes that are 120 degree apart. The resulting surface
              is F.  Now rotate F 120 degree gives the map f with order 3.
              f has 4 fixed points.
4. d=4, G = S(2,2,4,4), to see f, consider the standard 8-gon representation 
              of F by identify pairs of edges. A 90 degree rotation of the
              8-gon induces the order 4 map f.
              f has 2 fixed points, and 2 periodic orbits each of length 2.
5. d=5, G = S(5,5,5), cut the above 8-gon into two pentagons, and rotate each
              by a 72 degree rotation, inducing f.
              f has 3 fixed points.
6. d=6, G = S(3,6,6),  I don't know what is the best way to describe f, but G
              is a 2-fold quotient of the S(3,3,3,3) (d=3 case). To see this
              put the 4 cone points in S(3,3,3,3) symmetrically in East, South,
              West, North.  Now rotation S(3,3,3,3) 180 degree along the
              East-West axis gives S(3,6,6).  The order 6 map f on F should 
              be the lift of the involution on F.
              f has 2 fixed points, and 1 periodic orbits of length 2.
7. d=6, G = S(2,2,3,3),  f is the product of the f in S(3,3,3,3) (d=3 case) and
              the mutation (d=2, S(2,2,2,2,2,2) case).  The mutation can be
              seen in the "two-concentric-spheres-plus-3-tubes" representation
              of F as follows: simply interchange the inner sphere with the
              outer one, and flip the 3 tubes.
              f has no fixed points, but has a few periodic orbits.
8. d=8, G = S(2,8,8),  f is induced by a 45 degree rotation of the 8-gon,
               (identified by the word: a b c d a^-1 b^-1 c^-1 d^-1)
              f has 2 fixed points, and 1 periodic orbits of length 4.
9. d=10, G = S(2,5,10), f is the product of the order 5 map and the mutation.
              f has 1 fixed point, and 1 periodic orbit of length 2, and one
              with length 5.

  There are a lot of relations among these maps. For example the square of
  an order 6 map is the order 3 map. 

The main idea in making the list is: first using the Rieman-Hurwitz formula
to eliminate possibilities of G, then remove the cone points and get genuine
covering from F - disks to G - disks. Covering space theory says this is
determined by representations from pi_1 of G - disks to the symmetric group
S_d.  Also note that covering transformation group is a cyclic group 
generated by f.  All these eliminates more possibilities of G and gives
the description of the covering transformation.


- Yongwu Rong, George Washington University