Timeline for What is the permutation group generated by those three given morphisms of the affine space $\mathbb{F}_q^3$?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2015 at 18:58 | vote | accept | CommunityBot | ||
| Aug 6, 2015 at 20:00 | answer | added | Peter Mueller | timeline score: 3 | |
| Aug 6, 2015 at 16:20 | history | edited | user4324 | CC BY-SA 3.0 |
added 129 characters in body
|
| Aug 6, 2015 at 16:11 | history | edited | user4324 | CC BY-SA 3.0 |
added 428 characters in body
|
| Aug 6, 2015 at 16:03 | comment | added | Peter Mueller | Actually, the group has at least $q$ orbits, because the three generators leave invariant the sets $S_c=\{(x,y,z)|x^2+y^2+z^2-xyz=c\}$ for each $c$, and each of these sets is not empty. | |
| Aug 6, 2015 at 15:46 | comment | added | Nick Gill | @GeoffRobinson, Good point. The set you describe of size $3(q-1)$ breaks into orbits of size $6$ (provided $q>2$), and on these orbits these involutions generate a group of size $48$ I believe (this would need checking). So one needs to analyze the action on the remaining $q^3-3q+1$ points. | |
| Aug 6, 2015 at 14:54 | comment | added | Joe Silverman | @NickGill The Markoff equation has been much studied as an algebraic variety with an interesting automorphism group, especially in term of integer solutions. For example, every solution in positive integers to $x^2+y^2+z^2=xyz$ is obtained by starting with $(3,3,3)$ and applying automorphisms. Zagier has a beautiful paper in which he estimates the number of solutions of size at most $T$. The "Unicity Conjecture" says that if solutions are listed as $(a,b,c)$ with $a\ge b\ge c$, then no $a$ value appears more than once. (BTW, the classical equation is $x^2+y^2+z^2=3xyz$.) | |
| Aug 6, 2015 at 13:55 | comment | added | Geoff Robinson | Note that the action on non-zero vectors is not transitive. The triples with exactly two of $\{x,y,z \}$ equal to $0$ are invariant under the three given involutions. | |
| Aug 6, 2015 at 13:52 | comment | added | Nick Gill | BTW, I assume that @JoeSilverman's comment explains why the algebraic-geometry tag has been used? If you can give some more context, that would be nice. | |
| Aug 6, 2015 at 13:45 | comment | added | Nick Gill | In fact the permutations generate a subgroup of $S_{q^3-1}$, since $(0,0,0)$ is fixed. Do you know if the group they generate is transitive / primitive on non-zero vectors? It would also help to know the answer for $q=2,3,4,5$ for instance... In general if you pick permutations at random, you'll either get $A_n$ or $S_n$, so it would be worth seeing if this is the case for small $q$... | |
| Aug 6, 2015 at 13:41 | history | edited | Nick Gill | CC BY-SA 3.0 |
added a tag, and fixed a couple of typos.
|
| Aug 5, 2015 at 19:58 | comment | added | user4324 | Markoff equation... interesting. The action will depend on q (I have done some experiments). I found elements whose order in $\mathbb{F}_p^3$ is $p(p^2-1)/2$. | |
| Aug 5, 2015 at 19:57 | history | edited | user4324 | CC BY-SA 3.0 |
added 181 characters in body
|
| Aug 5, 2015 at 19:54 | comment | added | Joe Silverman | You might mention that these generate automorphisms of the affine variety defined by the Markoff equation $x^2+y^2+z^2=xyz$. Your group $\mathbb Z_2*\mathbb Z_2*\mathbb Z_2$ is (up to finite index) $\mathbb Z*\mathbb Z$, the free group on two generators. Do you have any reason to suspect that its action on $\mathbb F_q^3$ won't depend on $q$ in a reasonably random fashion? (Have you done some experiments?) If you do a Google search on "markoff equation finite field", you'll find lots of articles studying (generalized) Markoff equations over finite fields. | |
| Aug 5, 2015 at 19:46 | history | asked | user4324 | CC BY-SA 3.0 |