Timeline for How boundedly generated is $SL_3(\mathbb{Z})$?
Current License: CC BY-SA 3.0
20 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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 |
edited body
|
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 |
added 1 character in body
|
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 |
edited body
|
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 |
LaTeX in title
|
Sep 12, 2015 at 14:46 | history | edited | Pablo | CC BY-SA 3.0 |
added 35 characters in body
|
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 |