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It is well know that the genus three non orientable surface, N3, has only periodic and reducible auto-homeomorphisms, meanwhile the surface N4 is the first non orientable surface with pseudo Anosov maps. Also, recently profesor B.Szepietowski gave the MCG presentation of N4, from where, I calculated that there are seven periodic mapping classes. The question is: are there more?

Curiously, the torus and N3 show also seven periodic mapping classes each but we would like to understand better why N3 loses pseudo Anosov maps which contrast with the fact that N3 is got from the 2-torus via an one-point blow up...

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Maybe if you provided a couple of references, more people would find the question interesting. –  Mariano Suárez-Alvarez Nov 30 '09 at 20:18
    
Do you mean "The mapping class group of N_4 has at least 7 distinct conjugacy classes of periodic elements. Are there any more?"? –  Sam Nead Nov 30 '09 at 20:54
    
R.C. Penner. A construction of pseudo-Anosov homeomorphisms, Trans. Amer. Math. Soc., 310 (1988) No 1, 179-197, for the results in the existence of pA maps. In the other hand Szepietowski is easy to find in the Front-for-the-arXiv –  janmarqz Nov 30 '09 at 20:56
    
Sam, yes! it seems to me that the Szepietowski presentation produce at least seven automorphisms in the aut(MCG)... –  janmarqz Nov 30 '09 at 20:59
    
oops I mean Ab(MCG(N_4))... –  janmarqz Nov 30 '09 at 21:01
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1 Answer

up vote 2 down vote accepted

Just to lend some context to the above question: the mapping class group of the two-torus is naturally isomorphic to GL(2, Z). If we restrict to orientation preserving homeomorphism the mapping class group is SL(2, Z). The periodic mapping classes (isotopy classes of homeomorphisms) are exactly those with trace less than two in absolute value. (Hmm, and +/- Id, I guess!) Now we need to count the number of conjugacy classes of periodic elements. There should be a cool algebraic way to do this. (Perhaps it would help to give a purely algebraic proof that the order of torsion is at most 6?)

I think that there is a geometric way to do this: every periodic element occurs as the symmetry of some flat torus (= parallelogram with opposite sides identified). All tori have have the hyperelliptic symmetry, corresponding to rotation by 180 degrees about any point. (These maps lie in the mapping class of the negative identity.) Other symmetries:

Rombic tori have a reflection symmetry as do rectangular tori.
The square torus has a rotation by 90 degrees.
The hexagonal torus has a rotation by 60 degrees.

So I count:
1. the identity, Id
2. the hyperelliptic = -Id = rotation by 180
3. rotation by 90
4. rotation by 60
5. rotation by 120
6. the reflection [[-1,0],[0,1]] (reflection in an axis) and
7. the reflection [[0,1],[1,0]] (exchange axes).

You can prove that the last two are distinct algebraically. Perhaps the lack of 45 degree rotation is a geometric proof.

Now, we could perform similar geometric tricks to obtain symmetries of $N_4$ and get at least all of the rotations... [Edit: For example, it is possible to build a copy of $\rm{Sym}_4$ by placing the cross-caps at the vertices of a tetrahedron.]

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You can easily check that a finite subgroup of SL(2,Z) maps injectively to SL(2, Z_3) by the obvious map, and you can also easily check that elements of SL(2,Z_3) have orders 1, 2, 3, 4, or 6. –  Mariano Suárez-Alvarez Dec 4 '09 at 7:19
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Alternatively, the minimum polynomial of an element of $GL(2,Z)$ of finite order $n$ must be divisible by the minimal polynomial of a primitive $n$th root of unity, which has degree $\phi(n)$. This is at most $2$ iff $n\in\{1,2,3,4,6\}$ –  Mariano Suárez-Alvarez Dec 4 '09 at 7:25
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