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S Sep 24, 2015 at 9:15 history bounty ended CommunityBot
S Sep 24, 2015 at 9:15 history notice removed CommunityBot
S Sep 16, 2015 at 8:04 history bounty started Andrei Smolensky
S Sep 16, 2015 at 8:04 history notice added Andrei Smolensky Draw attention
Sep 13, 2015 at 21:44 comment added Pablo @algorithmshark even though your comment has been upvoted by many, there seems to be a delicate point: the bound you present is one which allows the (elementary) matrices to appear in arbitrary order in the product, and it clearly disagrees with the definition given in the question. It is clear that a bound (for my definition) can be deduced from the bound you gave, but it may become highly nonoptimal. Reading the proof referenced by Ian Agol though, I get an impression that the order of elementary operations is independent of the matrix in question, so the bound is probably correct.
Sep 13, 2015 at 6:09 history edited Pablo CC BY-SA 3.0
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Sep 13, 2015 at 6:08 comment added Pablo @NoamD.Elkies this is a very interesting argument!!! Let me please increase $m$ again.
Sep 13, 2015 at 5:48 history edited Pablo CC BY-SA 3.0
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Sep 12, 2015 at 23:18 comment added Noam D. Elkies Sorry, my answer to "do you have a good reference for X?" is almost always "no"... This is the kind of thing where it feels easier to figure it out directly than to look it up in the literature.
Sep 12, 2015 at 22:52 comment added Andrei Smolensky @Noam, do you have a good reference for such things (the interplay between $SL_n(\mathbb{Z})$ and $SL_n(\mathbb{Z}_p)$)?
Sep 12, 2015 at 20:36 comment added Noam D. Elkies You're welcome. I see that you've now changed "$m=2$" to "$m=5$". In fact I think any $m<8$ can be excluded by a $p$-adic dimension argument: for any $p$, each $\langle g_i \rangle$ is contained in the union of finitely many one-dimensional $p$-adic patches, so their product has dimension at most $m$, and thus cannot exhaust ${\rm SL}_3({\bf Z})$ beause this group's closure in ${\rm SL}_3({\bf Z}_p)$ has dimension $8$.
Sep 12, 2015 at 20:30 history edited Pablo CC BY-SA 3.0
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Sep 12, 2015 at 20:28 comment added Pablo @NoamD.Elkies This is great! I had some other argument in mind, regarding the possible eiganvalues of a product $g^ih^j$.
Sep 12, 2015 at 19:48 comment added Noam D. Elkies $m=2$ is clearly impossible: you can't even generate the invertible matrices mod 2 this way, because each $\left|\langle g_i\rangle\right| \leq 7$ and $7 \times 7 < 168$.
Sep 12, 2015 at 15:01 history edited Neil Strickland CC BY-SA 3.0
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Sep 12, 2015 at 14:46 history edited Pablo CC BY-SA 3.0
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Sep 12, 2015 at 11:59 comment added Andrei Smolensky Another exposition of the elementary proof can be found in a paper by Adian and Mennicke worldscientific.com/doi/abs/10.1142/s0218196792000220
Sep 11, 2015 at 20:14 comment added Ian Agol Section 4.1 gives the proof. It uses Dirichlet's theorem on primes in arithmetic progressions. perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf
Sep 11, 2015 at 16:49 comment added algorithmshark The Wikipedia article cites a 1984 paper by Carter and Keller, Elementary expressions for unimodular matrices. They determine that for $G = \mathrm{SL}_n(\mathbb Z)$ ($n > 2$) you can take elementary matrices for all the $g_i$ and get an upper bound $m \le \frac12(3n^2-n) + 36$.
Sep 11, 2015 at 16:05 history asked Pablo CC BY-SA 3.0