Timeline for Conjugacy in $GL(n,\mathbb Z)$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2011 at 19:10 | comment | added | user6976 | @Wox: Simultaneous conjugacy is much harder than simple conjugacy because the centralizer of a matrix may be infinite, so if you do what you suggest you need to check infinitely many matrices. In fact of course neither Grunewald-Segal nor Sarkisjan do that. They prove a version of the Hasse local-to-global principle (look at their papers!). Note that solvability of simultaneous conjugacy implies solvability of isomorphism problem for nilpotent (even virtually polycyclic) groups and for commutative semigroups. Both problems are very difficult. | |
| Oct 5, 2011 at 16:03 | comment | added | Wox | @HW: I only need to consider finite subgroups of $GL(n,\mathbb Z)$. | |
| Oct 5, 2011 at 14:37 | comment | added | HJRW | Wox - also, if you are interested in infinite subgroups generated by more than one element, then checking conjugacy between a finite set of matrices is not going to be enough. | |
| Oct 5, 2011 at 12:45 | comment | added | Sam Nead | @Wox - Sarkisjan solves the following problem: Given two lists of matrices $\{A_i\}_{i=1}^n$ and $\{B_i\}_{i=1}^n$ is there a single matrix $P$ so that for all $i$, $P A_i P^{-1} = B_i$? This answers your question, because you can take $n = 1$. Note, however, that the $n=1$ case does not solve the $n > 1$ case. This is because the conjugating matrix $P$ need not be unique. | |
| Oct 5, 2011 at 8:29 | comment | added | Wox | @Mark: in the end, this is exactly what I need. But isn't that just an extension of the similarity-of-two-matrices-problem? Try all pairs of matrices, one from each set (or group in my case). If you find a pair that conjugates in $GL(n,\mathbb Z)$, you can check if the groups conjugate by the same similarity transform P. If you can't find such P or if you can't find conjugate pairs at all, then the two groups don't conjugate. Or am I missing something? | |
| Oct 4, 2011 at 16:51 | comment | added | user6976 | @Sam: Your answer is misleading. Sarkisjan (and independently Grunewald-Segal) solved the much harder problem of multiple conjugacy. | |
| Oct 4, 2011 at 15:26 | comment | added | Igor Rivin | A good reference, but Grunewald's solution is not effective (as he himself points out in arxiv.org/pdf/0801.3011 ) | |
| Oct 4, 2011 at 15:20 | history | answered | Sam Nead | CC BY-SA 3.0 |