I'm looking at some messy data, and I found by accident that some properties seem preserved under some transformations of a subset of the variables.
After some work I think I found a base set of transformation from which I can build up all the special transformations. I'd like to figure out what group this is, so I can read more about it and hopefully better understand it.
All values are integers, and the last symmetry is parameterized by an integer $t$: $$ \begin{eqnarray} (a,b,c,d) &\rightarrow& (-a,-b,-c,-d) \\ (a,b,c,d) &\rightarrow& (a,-b,c,-d) \\ (a,b,c,d) &\rightarrow& (d,c,b,a) \\ (a,b,c,d) &\rightarrow& (a,b + at,c + 2bt + a t^2,d + 3ct + 3bt^2 + at^3) \\ \end{eqnarray} $$
Can someone help me identify what group this is?
EDIT: I made some progress, but can't quite pull it all together.
I rewrote this in linear algebra. And I decided it might be a bit more natural to apply #2 to #3 for a different choice of a basis transformation (for reasons discussed below): $$ \begin{eqnarray} (a,b,c,d) &\rightarrow& (d,-c,b,-a) \\ \end{eqnarray} $$ Okay, we can view the transformation as 4x4 matrices, for example the first one: $$ \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} -a \\ -b \\ -c \\ -d \end{bmatrix} $$ And I can now see that the last parameterized symmetry is just the $t=1$ case as a generator $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ t & 1 & 0 & 0 \\ t^2 & 2t & 1 & 0 \\ t^3 & 3t^2 & 3t & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \end{bmatrix}^t % {{1, 0, 0, 0},{1,1,0,0},{1,2,1,0},{1,3,3,1}}^n $$ So the four generators are: $$ \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} , \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} , \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix} , \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \end{bmatrix} $$ The first is just the negative identity matrix, and so commutes with the others. Therefore the full group just has this as a product that is easily factored out, so we can focus on the subgroup without it.
When viewed as blocks of 2x2 matrices, all the non-zero blocks have determinant 1 except for the second generator (which has non-zero blocks of determinant -1). So that is somehow a bit "different", and if this turns out to be a known group I'm guessing there is a "special" version of it without that matrix.
So ... a "special" linear group over the integers with two generators ... SL(2,Z) ?
If that is it, I'm really close. But I'm just guessing. I still don't quite see how it all fits together.