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L.s.,

I try to do some calculations on the Klein Quarctic curve, but there is a basic thing I don't know how to compute.

Let $\Gamma(7)$ denote the congruence subgroup of the modular group PSL(2, Z), that means

$ {\begin{pmatrix} a & b \\ c & d \end{pmatrix}}$

where a, b, c, d are integers,

$ad-cb = 1$

$a,d \equiv 1 \mod 7$

$ b,c \equiv 0 \mod 7$

My question now is, what are the generators of this group? (and how would you derive that, if possible!)

I unfortunately don't have a clue, and I can find it nowhere.

Any help would be greatly appreciated,

Willem

PS: this is not a homework question

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1 Answer 1

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There is a very beautiful algorithmic approach to finding generators of congruence subgroups of $PSL(2, \mathbf{Z})$ using "Farey symbols" (sequences of rational numbers specifying a fundamental domain of a special kind). There is a survey article by Chris Kurth and Ling Long describing the algorithm and the theory behind it; and recent versions of the Sage computer algebra system have an implementation of Kurth and Long's algorithm (mostly due to Hartmut Monien).

So in Sage you can do this:

masiao@fermat:~$ sage
┌────────────────────────────────────────────────────────────────────┐
│ Sage Version 6.1.1, Release Date: 2014-02-04                       │
│ Type "notebook()" for the browser-based notebook interface.        │
│ Type "help()" for help.                                            │
└────────────────────────────────────────────────────────────────────┘
sage: Gamma(7).generators()
[
[1 7]  [-48   7]  [-209   56]  [113 -35]  [-55  21]  [120 -49]
[0 1], [ -7   1], [ -56   15], [ 42 -13], [-21   8], [ 49 -20],

[ 15  -7]  [ 239 -140]  [113 -70]  [ 232 -161]  [-181  133]  [ 8 -7]
[ 28 -13], [  70  -41], [ 21 -13], [  49  -34], [ -49   36], [ 7 -6],

[-76 105]  [ 169 -238]  [ 43 -63]  [ 309 -490]  [ 134 -217]
[-21  29], [  49  -69], [ 28 -41], [  70 -111], [  21  -34],

[ 281 -476]  [-230  399]  [ 15 -28]  [-97 231]  [ 218 -525]
[  49  -83], [ -49   85], [  7 -13], [-21  50], [  49 -118],

[-279  763]  [ 22 -63]  [-118  399]  [  29 -112]  [-139  609]
[ -49  134], [  7 -20], [ -21   71], [   7  -27], [ -21   92],

[  36 -175]  [  43 -252]
[   7  -34], [   7  -41]
]

That command took less than 0.02 seconds to run. The theory actually guarantees that this is a free generating set, so these generators give an isomorphism between $\Gamma(7)$ and the free group on 29 generators.

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  • $\begingroup$ Thank you very much for your nice and quick answer! Would it be oké if I ask a follow-up question in the comments? Kind regards $\endgroup$ Feb 20, 2014 at 18:53
  • $\begingroup$ Sure, go ahead... $\endgroup$ Feb 20, 2014 at 21:12
  • $\begingroup$ I have a question regarding the last comment of your answer. If I run the command for the modular group SL(2,Z), one of the generators given by the algorithm is S = {{0,-1},{1,0}}. This one satisfies S^4=1, so how can the list be a free generating set ? $\endgroup$
    – Antoine
    Aug 14, 2016 at 14:23
  • $\begingroup$ I didn't claim that the list of generators returned was always a free generating set, only that it was so for Gamma(7). The group SL(2, Z) has torsion, so of course it cannot have a free generating set, as you have evidently noticed. What Sage returns for a general subgroup of SL(2, Z) is a generating set that is "as free as possible", in the sense that all the relations have a simple explicit form -- read the relevant entry in the Sage manual for a precise statement. $\endgroup$ Aug 20, 2016 at 7:25

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