Timeline for Sums of inverse determinants over matrices
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2013 at 15:57 | vote | accept | Anton Menshov | ||
| Jun 4, 2013 at 15:05 | answer | added | David E Speyer | timeline score: 1 | |
| Jun 3, 2013 at 12:52 | comment | added | Daniel Loughran | Out of interest, there is a related zeta function which is very well-behaved. Namely, the set of matrices in $M_n(\mathbb{Z})$ up to unimodular equivalence is naturally in bijection with the set of all sublattices of $\mathbb{Z}^n$. The associated zeta function $\sum_{ \Lambda \subset \mathbb{Z}^n} (\mathrm{det \Lambda})^{-s}$ is well behaved and has a nice expression in terms of products of the Riemann zeta function. | |
| Jun 3, 2013 at 12:48 | comment | added | Daniel Loughran | One natural thing to do in this setting is to introduce a zeta function like $Z_n(s)=\sum_{A \in M_n(\mathbb{Z})}|\mathrm{det} A|^{-s}$. Unfortunately this zeta function is not very well behaved and does not converge for any $s \in \mathbb{C}$ due to there being infinitely many unimodular matrices. Perhaps you can consider a zeta function in your setting which takes into account the condition $||A|| \leq r$. | |
| Jun 3, 2013 at 0:14 | answer | added | David E Speyer | timeline score: 4 | |
| Jun 2, 2013 at 9:09 | comment | added | user18180 | A lower bound for the problem can be found from Example 1.6 in "DENSITY OF INTEGER POINTS ON AFFINE HOMOGENEOUS VARIETIES" by Duke, Rudnick, and Sarnak, Vol. 71, Duke, July 1993. This gives a lower bound of form $mr^{n^2−n}$, where m is a constant, by estimating the number of matrices in $SLn(\mathbb{Z})$, with bounded coefficients. I would guess that this could be also used for further analysis. A related question: mathoverflow.net/questions/76839/… Related work: arxiv.org/abs/1111.6289 | |
| Jun 1, 2013 at 23:01 | answer | added | Sungjin Kim | timeline score: 4 | |
| Jun 1, 2013 at 13:51 | comment | added | Gerhard Paseman | For 4r greater than log(2^n), you can restrict the off diagonal entries to beng nonzero, but replace upper triangular in Davide Giraudo's suggestion by "comb" patterns of mostly zero rows and columns to get a factor of 2^(n-1). As r grows, the loss by replacing 2r+1 by 2r becomes negligible comapred to the 2^(n-1) gain. Because the denomnator of your limit is not (2r+1)^(n^2), I suspect your limit does not go to 0. Gerhard "Ask Me About Integer Matrices" Paseman, 2013.06.01 | |
| Jun 1, 2013 at 10:32 | history | asked | Anton Menshov | CC BY-SA 3.0 |