Identifying a group involving transformation of 4 integers
$begingroup$
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.
group-theory integers combinatorial-group-theory
asked Jan 13 at 4:01
PineDoorsPineDoors
335
$endgroup$
add a comment |
$begingroup$
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.
group-theory integers combinatorial-group-theory
asked Jan 13 at 4:01
PineDoorsPineDoors
335
$endgroup$
$begingroup$
Why do you assume this is a group?
$endgroup$
– Don Thousand
Jan 13 at 4:13
$begingroup$
@DonThousand Because I can compose those base transformations to get another valid transformation.
$endgroup$
– PineDoors
Jan 13 at 4:18
$begingroup$
Which properties are preserved?
$endgroup$
– zoli
Jan 13 at 6:38
$begingroup$
Just a few small observations: the orders of the generators are $2,2,4,infty$. Next, the first generator is the square of the third one, so we can throw it out. Next, the second generator acts by inversion on the last two, so the subgroup generated by the last two is normal of index at most $2$, so we can focus on that.
$endgroup$
– verret
Jan 14 at 1:49
add a comment |
$begingroup$
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.
group-theory integers combinatorial-group-theory
asked Jan 13 at 4:01
PineDoorsPineDoors
335
$endgroup$
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.
group-theory integers combinatorial-group-theory
group-theory integers combinatorial-group-theory
asked Jan 13 at 4:01
PineDoorsPineDoors
335
asked Jan 13 at 4:01
PineDoorsPineDoors
335
edited Jan 13 at 6:01
PineDoors
asked Jan 13 at 4:01
PineDoorsPineDoors
335
asked Jan 13 at 4:01
PineDoorsPineDoors
335
asked Jan 13 at 4:01
PineDoorsPineDoors
335
335
$begingroup$
Why do you assume this is a group?
$endgroup$
– Don Thousand
Jan 13 at 4:13
$begingroup$
@DonThousand Because I can compose those base transformations to get another valid transformation.
$endgroup$
– PineDoors
Jan 13 at 4:18
$begingroup$
Which properties are preserved?
$endgroup$
– zoli
Jan 13 at 6:38
$begingroup$
Just a few small observations: the orders of the generators are $2,2,4,infty$. Next, the first generator is the square of the third one, so we can throw it out. Next, the second generator acts by inversion on the last two, so the subgroup generated by the last two is normal of index at most $2$, so we can focus on that.
$endgroup$
– verret
Jan 14 at 1:49
add a comment |
$begingroup$
Why do you assume this is a group?
$endgroup$
– Don Thousand
Jan 13 at 4:13
$begingroup$
@DonThousand Because I can compose those base transformations to get another valid transformation.
$endgroup$
– PineDoors
Jan 13 at 4:18
$begingroup$
Which properties are preserved?
$endgroup$
– zoli
Jan 13 at 6:38
$begingroup$
Just a few small observations: the orders of the generators are $2,2,4,infty$. Next, the first generator is the square of the third one, so we can throw it out. Next, the second generator acts by inversion on the last two, so the subgroup generated by the last two is normal of index at most $2$, so we can focus on that.
$endgroup$
– verret
Jan 14 at 1:49
$begingroup$
Why do you assume this is a group?
$endgroup$
– Don Thousand
Jan 13 at 4:13
$begingroup$
Why do you assume this is a group?
$endgroup$
– Don Thousand
Jan 13 at 4:13
$begingroup$
@DonThousand Because I can compose those base transformations to get another valid transformation.
$endgroup$
– PineDoors
Jan 13 at 4:18
$begingroup$
@DonThousand Because I can compose those base transformations to get another valid transformation.
$endgroup$
– PineDoors
Jan 13 at 4:18
$begingroup$
Which properties are preserved?
$endgroup$
– zoli
Jan 13 at 6:38
$begingroup$
Which properties are preserved?
$endgroup$
– zoli
Jan 13 at 6:38
$begingroup$
Just a few small observations: the orders of the generators are $2,2,4,infty$. Next, the first generator is the square of the third one, so we can throw it out. Next, the second generator acts by inversion on the last two, so the subgroup generated by the last two is normal of index at most $2$, so we can focus on that.
$endgroup$
– verret
Jan 14 at 1:49
$begingroup$
Just a few small observations: the orders of the generators are $2,2,4,infty$. Next, the first generator is the square of the third one, so we can throw it out. Next, the second generator acts by inversion on the last two, so the subgroup generated by the last two is normal of index at most $2$, so we can focus on that.
$endgroup$
– verret
Jan 14 at 1:49
add a comment |
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$begingroup$
Why do you assume this is a group?
$endgroup$
– Don Thousand
Jan 13 at 4:13
$begingroup$
@DonThousand Because I can compose those base transformations to get another valid transformation.
$endgroup$
– PineDoors
Jan 13 at 4:18
$begingroup$
Which properties are preserved?
$endgroup$
– zoli
Jan 13 at 6:38
$begingroup$
Just a few small observations: the orders of the generators are $2,2,4,infty$. Next, the first generator is the square of the third one, so we can throw it out. Next, the second generator acts by inversion on the last two, so the subgroup generated by the last two is normal of index at most $2$, so we can focus on that.
$endgroup$
– verret
Jan 14 at 1:49