Gerard Westendorp
Oct12-06, 04:56 AM
Lately I was thinking about some matrices, and I wasn't quite sure how
this fits in with what I've learned so far about group theory.
Normally, when you want to rotate 2 dimensional space, you might start
with a basis of 2 orthogonal vectors:
Y
^
|
|
-----> X
You can do a 90 degree rotation, and turn (X,Y) into (Y,-X). You can do
this operation using a matrix acting on the vector (X,Y):
( X ) => ( 0 1 ) ( X )
( Y ) ( -1 0 ) ( Y )
This can be turned into a continuous group that has the above discrete
group as a subgroup:
( cos(a) sin(a) )
( -sin(a) cos(a) )
Now I was looking at a space in which UP is not quite the same as
-(DOWN), and LEFT not quite the same as -(RIGHT). It doesn't matter at
this stage why they are different, we will just write them a formally
distinct terms. In a specific coordinate system, our basis looks like this:
UP
^
|
|
|
LEFT <----------------> RIGHT
|
|
|
V
DOWN
A 90 degree rotation of this basis could be described by a matrix (A)
acting on a 4-component vector:
(RIGHT') ( 0 0 0 1 ) (RIGHT)
(UP' ) = ( 1 0 0 0 ) (UP )
(LEFT' ) ( 0 1 0 0 ) (LEFT )
(DOWN' ) ( 0 0 1 0 ) (DOWN )
It seems pretty straight forward to rotate this space by multiples of
90 degrees. But what about rotating by other angles? It turns out that
we there is a continuous group that has the above matrix as a discrete
subgroup. The matrix form is:
(a b c d)
(d a b c)
(c d a b)
(b c d a)
a, b, c and d can be expressed as weighted sums of the eigenvalues, and
the eigenvalues are:
C1 = cos(1a) + i* sin(1a)
C2 = cos(2a) + i* sin(2a)
C3 = cos(3a) + i* sin(3a)
C4 = cos(4a) + i* sin(4a)
In this case, our group seems isomorphic to the ordinary 2D rotation
group. But in 3D, you could also take a discrete subgroup of the
rotation group, say the octahedron. You could look at this subgroup as a
discrete subgroup of the rotation group in 3D, but you could also write
it as a matrix acting on the vertices. For an octagon with 6 vertices,
this would give a group of 6X6 matrices acting on
(RIGHT')
(UP' )
(LEFT' )
(DOWN' )
(FRONT )
(BACK )
Again, we can turn this into a continuous group of 6X6 matrices.
My question is, is this just a representation of SO(3)? Or am I making
another group here?
Gerard
this fits in with what I've learned so far about group theory.
Normally, when you want to rotate 2 dimensional space, you might start
with a basis of 2 orthogonal vectors:
Y
^
|
|
-----> X
You can do a 90 degree rotation, and turn (X,Y) into (Y,-X). You can do
this operation using a matrix acting on the vector (X,Y):
( X ) => ( 0 1 ) ( X )
( Y ) ( -1 0 ) ( Y )
This can be turned into a continuous group that has the above discrete
group as a subgroup:
( cos(a) sin(a) )
( -sin(a) cos(a) )
Now I was looking at a space in which UP is not quite the same as
-(DOWN), and LEFT not quite the same as -(RIGHT). It doesn't matter at
this stage why they are different, we will just write them a formally
distinct terms. In a specific coordinate system, our basis looks like this:
UP
^
|
|
|
LEFT <----------------> RIGHT
|
|
|
V
DOWN
A 90 degree rotation of this basis could be described by a matrix (A)
acting on a 4-component vector:
(RIGHT') ( 0 0 0 1 ) (RIGHT)
(UP' ) = ( 1 0 0 0 ) (UP )
(LEFT' ) ( 0 1 0 0 ) (LEFT )
(DOWN' ) ( 0 0 1 0 ) (DOWN )
It seems pretty straight forward to rotate this space by multiples of
90 degrees. But what about rotating by other angles? It turns out that
we there is a continuous group that has the above matrix as a discrete
subgroup. The matrix form is:
(a b c d)
(d a b c)
(c d a b)
(b c d a)
a, b, c and d can be expressed as weighted sums of the eigenvalues, and
the eigenvalues are:
C1 = cos(1a) + i* sin(1a)
C2 = cos(2a) + i* sin(2a)
C3 = cos(3a) + i* sin(3a)
C4 = cos(4a) + i* sin(4a)
In this case, our group seems isomorphic to the ordinary 2D rotation
group. But in 3D, you could also take a discrete subgroup of the
rotation group, say the octahedron. You could look at this subgroup as a
discrete subgroup of the rotation group in 3D, but you could also write
it as a matrix acting on the vertices. For an octagon with 6 vertices,
this would give a group of 6X6 matrices acting on
(RIGHT')
(UP' )
(LEFT' )
(DOWN' )
(FRONT )
(BACK )
Again, we can turn this into a continuous group of 6X6 matrices.
My question is, is this just a representation of SO(3)? Or am I making
another group here?
Gerard