Re: Determine the set of the conjugators between two affine unimodular matrix groups, whose linear parts are the same finite group belonging to the subgroup of GLNZ.

[Hongyi Zhao <hongyi.zhao@gmail.com>]

On Sun, Jan 8, 2023 at 12:44 PM Hongyi Zhao <hongyi.zhao@gmail.com> wrote:
>
> Hi here,
>
> I have two affine unimodular matrix groups whose generators are as follows:
>
> gens2=[
> [  -1, 0, 0, 0 ; 0, 1, 0, 1; 0, 0, 1, 0; 0, 0, 0, 1 ],
> [ -1, 0, 0, 0; 0, 1, 0, -1/2; 0, 0, -1, 0; 0, 0, 0, 1 ],
> [ -1, 0, 0, 0; 0, -1, 0, 1/2; 0, 0, 1, 0; 0, 0, 0, 1 ],
> [ 1, 0, 0, 1/2; 0, 1, 0, 1/2; 0, 0, 1, 1/2; 0, 0, 0, 1 ] ]
>
> gens3=[
> [ 1, 0, 0, 0; -1, -1, 0, 0; 0, 0, 1, -1/2; 0, 0, 0, 1 ],
> [ -1, 0, 0, 0; 0, -1, 0, 1/2; 1, 0, 1, 0; 0, 0, 0, 1 ],
> [ 1, 0, 0, 0; -1, -1, 0, 0; -1, 0, -1, -1/2; 0, 0, 0, 1 ],
> [ 1, 0, 0, -1; 0, 1, 0, 1; 0, 0, 1, 1; 0, 0, 0, 1 ] ]
>
> # Sorry here once more, I really don't know what's corresponding
> syntax for this in PARI/GP:
> grp2=Group(gens2)
> grp3=Group(gens3)
>
> Here, the 3-by-3 matrices corresponding to linear parts of the above
> two set of generators form the same finite group G, which is a
> subgroup in GLNZ.
>
> I want to determine the set of the conjugators between these two groups, a.k.a.,
>
> conjs = { c | grp2 ^ c = grp3  }
>
> Where, the linear part of c is an element of the normalizer of G, and
> the translation part of c is a rational vector which has the following
> form:
>
> [ x, y, z, 1 ]
>
> If there's no such conjugator, how can I make the decision quickly?

At present, I have the following ideas on solving this problem, but it
is still a very complicated thing to implement:

1. Find the set of the affine normalizers of these two groups, denoted
by an2 and an3, respectively.
2. Find the set of the permissible affine conjugators, and then
compute the difference with an2 and an3. In this way, the affine
conjugators which are not normalizers of these two groups, denoted by
ann2 and ann3, respectively, will be identified.
3. Compute the intersection of ann2 and ann3 ^ -1.

Regards,
Zhao