Timeline for What fraction of n x n invertible integer matrices contain at least one unit?
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Sep 16, 2012 at 15:56 | vote | accept | Vidit Nanda | ||
| Sep 15, 2012 at 21:34 | comment | added | Gerhard Paseman | I suspect the distribution of your matrices is far from uniform. I appreciate your motivation, but am presently too ignorant to understand it. I recommend a follow up question which describes the source and the process in more detail (and if I am to understand it, pitching it at a bright undergraduate level would not insult me, and MathOverflow would survive it), and poses the question as to why your process seems as successful as it does. Gerhard "Or Let These Responses Satisfy" Paseman, 2012.09.15 | |
| Sep 15, 2012 at 15:07 | answer | added | Igor Rivin | timeline score: 2 | |
| Sep 15, 2012 at 8:21 | answer | added | Denis Chaperon de Lauzières | timeline score: 3 | |
| Sep 15, 2012 at 6:08 | comment | added | Vidit Nanda | Gerhard: I tried to reduce one of the spheres 50 times. Only 4 times the reductions were imperfect. Once, the reduced complex had a 3x3 boundary matrix with highest entry 17, twice 4x4 matrices with highest entries 45 and 84, and finally one 5x5 matrix with highest entry 172. | |
| Sep 15, 2012 at 2:53 | answer | added | smoked salmon sandwiches | timeline score: 10 | |
| Sep 15, 2012 at 2:22 | answer | added | Greg Martin | timeline score: 3 | |
| Sep 15, 2012 at 2:16 | answer | added | Will Jagy | timeline score: 3 | |
| Sep 15, 2012 at 1:41 | comment | added | Gerhard Paseman | Can you say more about these situations? I don't mind being wrong on the above, but 80% seems high for anything but small n and small moduli. If you give the range of moduli and number of samples, I might get a better feel for what is going on. Gerhard "Doesn't Know What To Feel" Paseman, 2012.09.14 | |
| Sep 14, 2012 at 23:26 | comment | added | Will Jagy | Sorry, it's always something. I actually had your resume on the screen and thought, maybe I should print this out so I get the first name right when asking about leading the host of the dead through the skies. But then I didn't and put in some randomized memory construction without realizing. Sigh. Now I'm going with Vidit. Because the CV says so. Also raising whirlwinds. | |
| Sep 14, 2012 at 23:02 | comment | added | Vidit Nanda | Will: maybe, but I am not Dilip. | |
| Sep 14, 2012 at 22:58 | comment | added | Will Jagy | Meanwhile, with a number of for loops in a fast language, you could evaluate $r_2(m)$ for some large $m$ and $r_3(m)$ for medium large $m.$ | |
| Sep 14, 2012 at 22:48 | comment | added | Vidit Nanda | Gerhard: quite the opposite, actually. I am finding that over 80% of the time I do in fact end up in the perfect situation with only two cells. This surprised me also, which is why I want the question answered as a control experiment: if GL_n(Z) elements typically lack units, then the computations are not just "getting lucky". | |
| Sep 14, 2012 at 22:09 | comment | added | Gerhard Paseman | ON further reflection, A cannot be just any set. For your situation though, I imagine that matrices with no units will show up distressingly often in GL_n(Z). Gerhard "Ask Me About System Design" Paseman, 2012.09.14 | |
| Sep 14, 2012 at 22:05 | comment | added | Vidit Nanda | Will: thank you, that makes sense at least for $n = 2$. | |
| Sep 14, 2012 at 21:58 | comment | added | Gerhard Paseman | Thanks Vel and Will, I wasn't reading carefully. I think you will find that for any small set A and larger superset B (possibly symmetric about 0), that many elements of your set will have entries coming from B-A, and that will represent more than 99% (say) of those that have entries coming from B. I think B needs to be not much larger than A for this too hold. Gerhard "Ask Me About System Design" Paseman, 2012.09.14 | |
| Sep 14, 2012 at 21:53 | comment | added | Will Jagy | I meant that I am demanding that the product of the two diagonal elements be my $k,$ but if both $m, m-1 < k$ and $m$ is the bound on element absolute values, $1 \cdot m < k$ and both diagonal elements must be larger than $1.$ So it should probably read $k \geq m+1.$ For $n=2$ we can simply switch the diagonal and the off-diagonal. Meanwhile, we must have $k \leq m^2$ here. The numerator and denominator in your $r_2(m)$ can thus be separately estimated pretty well, I think, with sums over my $1 \leq k \leq m^2,$ where $k \leq m$ and $m < k \leq m^2$ behave differently. Number theory. | |
| Sep 14, 2012 at 21:31 | comment | added | Vidit Nanda | Will: what do you mean by "if $k \geq m+2$ no $1$'s" in your first comment? | |
| Sep 14, 2012 at 21:11 | comment | added | Vidit Nanda | Gerhard: this is a theorem when the determinant is a unit (which it must be in the general linear group over Z): see [here][1] for instance. [1]: math.stackexchange.com/questions/19528/… | |
| Sep 14, 2012 at 21:07 | comment | added | Gerhard Paseman | also, by doing determinant invariant row operations, you can easily produce rows with no units for each row that has a unit. I say for n large, you will find more matrices avoiding small numbers with unit determinant than matrices with small numbers and unit det. Gerhard "Ask Me About Small Determinants" Paseman, 2012.09.14 | |
| Sep 14, 2012 at 21:06 | comment | added | Will Jagy | @Gerhard, must be, the determinant needs to be $\pm 1$ for $GL_n.$ Also the limit is clearly $0$ if not. | |
| Sep 14, 2012 at 21:00 | comment | added | Gerhard Paseman | Are the inverses also supposed to be integer matrices? Gerhard "Ask Me About System Design" Paseman, 2012.09.14 | |
| Sep 14, 2012 at 20:52 | comment | added | Will Jagy | seems tractable for $n=2.$ Demand $a_{11} a_{22} = k, \; a_{12} a_{21} = k-1,$ all positive, why not? Then something about numbers of divisors of $k,k-1$ no larger than $m.$ If $k \geq m+2$ no $1$'s. I do see that you want $n$ large, but if the answer for $n=2$ is $0,$ while the answer for $n=1$ is $1,$ that is persuasive. | |
| Sep 14, 2012 at 20:31 | history | asked | Vidit Nanda | CC BY-SA 3.0 |