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Jul 27, 2013 at 0:54 comment added Ian Agol The rank problem is unsolvable among word-hyperbolic groups: ams.org/mathscinet-getitem?mr=1246477
Sep 27, 2011 at 11:37 comment added user6976 @Igor: See my update 3.
Sep 27, 2011 at 11:36 history edited user6976 CC BY-SA 3.0
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Sep 27, 2011 at 11:30 comment added user6976 @Igor: For $n\ge 28$ you only need 2. See books.google.com/…
Sep 27, 2011 at 10:28 comment added Igor Rivin Thanks! I am pretty sure that the smallest known generating sets for $SL(n, \mathbb{Z})$ itself has $3$ elements...
Sep 27, 2011 at 9:24 history edited user6976 CC BY-SA 3.0
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Sep 27, 2011 at 9:23 comment added user6976 @HW: You are right of course, one just needs to check if $\delta=1, 2,...$.
Sep 27, 2011 at 7:48 comment added HJRW Mark - Papazoglou's algorithm will find a constant $\delta$ given a presentation of a word-hyperbolic group.
Sep 27, 2011 at 3:28 comment added Andy Putman I'm not sure what the correct upper bound should be, and would be very interested in knowing the answer.
Sep 27, 2011 at 2:54 history edited user6976 CC BY-SA 3.0
corrected the first part of the answer
Sep 27, 2011 at 2:52 comment added user6976 @Andy: Thanks! Is $n^2-1$ the upper bound for all the lattices in $SL_n$? I will correct the answer.
Sep 27, 2011 at 2:15 comment added Andy Putman You need more than $2$ generators for lattices in $SL_n(\mathbb{R})$. The abelianizations of the level $p$ congruence subgroups of $SL_n(\mathbb{Z})$ have rank $n^2-1$ (see the paper "On the homology and cohomology of congruence subgroups" by Lee-Szczarba), so you need at least $n^2-1$ generators to generate them.
Sep 26, 2011 at 23:54 history answered user6976 CC BY-SA 3.0