Timeline for Periodic mapping classes of the genus two orientable surface
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2009 at 17:36 | vote | accept | janmarqz | ||
| Dec 8, 2009 at 17:34 | vote | accept | janmarqz | ||
| Dec 8, 2009 at 17:36 | |||||
| Dec 8, 2009 at 17:33 | vote | accept | janmarqz | ||
| Dec 8, 2009 at 17:33 | |||||
| Dec 5, 2009 at 18:49 | comment | added | Ryan Budney | 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. | |
| Dec 5, 2009 at 18:49 | comment | added | Ryan Budney | 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 $\sum_i p_i/q_i = 0$. | |
| Dec 5, 2009 at 17:25 | comment | added | janmarqz | do you mean $\chi(F)=n[\chi(B)-m+\sum_i{q_i}^{-1}]$, right? | |
| Dec 4, 2009 at 6:22 | comment | added | Ryan Budney | I think you'll have to "get your hands dirty" a little bit but once you do I hope you'll like my explanation. | |
| Dec 4, 2009 at 6:03 | comment | added | janmarqz | 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... | |
| Dec 4, 2009 at 4:34 | history | edited | Ryan Budney | CC BY-SA 2.5 |
qualifiers
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| Dec 4, 2009 at 4:29 | history | answered | Ryan Budney | CC BY-SA 2.5 |