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Please, any information on the periodic mapping classes of the genus two orientable surface, $O_2$, will be greatly thanked. We had been studying the topological structure of 3d surface bundles and reintrepreting them as a circle bundles over orbifolds.

In the http://web.archive.org/web/20070316045651/http://www.smm.org.mx/SMMP/html/modules/Publicaciones/AM/Cm/35/artExp08.pdf -work you would like see the cases $O_1$, among $N_1$ and $N_2$, solved. Any feedback on the results and conjectures, some of them obviously false, will bring a lot of happiness :)

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2 Answers

up vote 4 down vote accepted

If you want to enumerate the finite-order automorphisms (up to conjugacy) I suggest the following exercise. The associated 3-manifold is Seifert fibred. So determine how the genus 2 surface is sitting in the Seifert manifold (horizontal incompressible surface).

This will give you a formula relating the various branch points of the monodromy to the Seifert data. Moreover, you should be able to go back-and-forth between the description of the Seifert-fibred space (unnormalized Seifert data, fibred over a genus 0, 1 or 2 surface) and the monodromy of the surface. So the classification of Seifert-fibred spaces basically gives you a dictionary for walking-through the finite-order automorphisms of a mapping class group.

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your answer induce me a headache... nah! it is a joke :), what I´m going to do is to trace your programme in my solved cases, after that I'll tell you, thanks! In the other hand as you might see professor Hatcher has enlighted us very sharp... –  janmarqz Dec 4 '09 at 6:03
I think you'll have to "get your hands dirty" a little bit but once you do I hope you'll like my explanation. –  Ryan Budney Dec 4 '09 at 6:22
do you mean $\chi(F)=n[\chi(B)-m+\sum_i{q_i}^{-1}]$, right? –  janmarqz Dec 5 '09 at 17:25
Yes, so $n$ is the order of the monodromy, $F$ is the horizontal incompressible surface, $B$ is the quotient of $F$ by the automorphism you're interested in, $m$ is the number of non-free orbits of the automorphism's action and the $q_i$'s are the orders of the stabilizers of points in the non-free orbits. In hatcher's notion if you have a seifert-fibred space $M[g,0;p_i/q_i]$ you have a horizontal incompressible surface if and only if the sum $\sum_i p_i/q_i = 0$. –  Ryan Budney Dec 5 '09 at 18:49
In this case your $n = LCM\{q_i : \forall i\}$ -- think about how the horizontal incompressible surface is sitting in the "model Seifert manifolds" (chapter 2 Hatcher's notes) to see this. –  Ryan Budney Dec 5 '09 at 18:49
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In the paper listed below there is a calculation of all the finite group actions on a genus 2 surface. There are 20 of them, with the groups ranging from order 2 to order 48. Nine of the actions are of cyclic groups, of orders 2,2,3,4,5,6,6,8,10 respectively. The paper also does the genus 3 case. The techniques are mostly algebraic. It is an interesting exercise to try to find nice geometric pictures of all the actions.

S.A.Broughton, Classifying finite group actions on surfaces of low genus, J.Pure Appl.Alg. 69 (1991), 233-270.

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Alternatively, Proposition 1.11 in Allen Hatcher's 3-manifolds notes available on his web-page, together with the formula on the next page relating the Euler characteristic of the horizontal surface to the euler characteristic of the base. This is what you need to get to the algebra Allen refers to. –  Ryan Budney Dec 4 '09 at 6:05
professor Hatcher: thank you very much... perhaps this scarse letters are to far from reflecting how happy I am, but believe me: I realy am!. My next work is to tell which SFS arise... –  janmarqz Dec 4 '09 at 6:10
Prof.Ryan: it greatly pleased me the arousal interest of my quizz, thanks again. –  janmarqz Dec 4 '09 at 6:14
Ah, the orbifold Euler-characteristic gonna rule the possibilities inter Faser-fläche und Zerlegungfläche... thank-x –  janmarqz Dec 5 '09 at 17:18
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