lpetrich
Contributor
So nobody's bitten. Here are the answers.
* What determinant values can an orthogonal matrix have?
Multiplying the definition by M, we get M.transpose(M) = transpose(M).M = I
Now take the determinant. It's easy to show that (det(M))2 = 1, giving det(M) = +- 1
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* What are all the O(1) matrices? The SO(1) ones?
It is easy to show that a 1*1 matrix has only one element, and that that matrix's determinant is equal to that element. Thus,
O(1): { {{1}}, {{-1}} } -- Z2
SO(1): { {{1}} } -- the identity group, Z1
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* The O(2) matrices might seem to have 2*2 = 4 parameters, one for each matrix element. Can you show that their matrix elements can be specified with a smaller number of parameters?
We start with M = {{a11, a12}, {a21, a22}}
We find
det(M) = a11*a22 - a12*a21
transpose(M) = {{a11, a21}, {a12, a22}}
inverse(M) = {{a22, -a12}, {-a21, a11}}/det(M)
Set det(M) = s = +-1. Then
a22 = a11*s
a21 = - a12*s
(a11)2 + (a12)2 = 1
This suggests that we can use trigonometric functions:
For det +1: {{cos(a), -sin(a)}, {sin(a), cos(a)}}
For det -1: {{cos(a), sin(a)}, {sin(a), -cos(a)}}
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* What are all the finite subgroups of O(2)? Of SO(2)?
First, we prove a theorem about groups of orthogonal matrices with both determinant values: that there are as many negative-determinant ones as there are positive-determinant ones.
Start with sets P = {M(+)} the positive-determinant ones and N = {M(-)} the negative-determinant ones.
Is P -> N an injection?
Take an element of N, X, and multiply every element of P by it on one side. The result is a subset of N, since matrices' determinants multiply. Thus, P -> N is an injection, one-to-one.
Is P -> N a surjection?
Take the inverse of X, and multiply every element of N by it on one side. The result is a subset of P. Thus, P -> N is a surjection, onto.
Since both are true, P -> N is a bijection, a one-to-one correspondence, and every element of P can be identified with every element of N.
So one can specify all the elements of a group of orthogonal matrices by specifying all its positive-determinant ones, and by combining them with a negative-determinant one, if any are present, to specify the negative-determinant ones.
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Now for the main problem. Let us find the all-positive-determinant subgroups first, and then branch off to include negative-determinant ones. Thus, I find all finite subgroups of SO(2), and from them, all the finite subgroups of O(2) that are not in SO(2).
As I'd stated earlier, every element of SO(2) can be stated in form M(+,a) = {{cos(a), -sin(a)}, {sin(a), cos{a}} -- it can be pictured as a 2D rotation. These elements have a nice composition rule: M(+,a).M(+,b) = M(+,a+b), as can easily be verified from trigonometric identities.
We can strip away the matrix to get a (x) b = (a + b) mod 2π. We can get rid of the 2π factor by setting a = (2π)a'. That makes a' (x) b' = (a' + b') mod 1.
Let's see what form the a's are for a finite group. A finite group's subgroups must all be finite, as can easily be proved. So every element must generate a cyclic subgroup. This makes a = (2π)*m/n, where m is a nonnegative integer and n a positive one > n. Taking all the denominators in the a's, we find their least common multiple, and turn m/n into (m*(lcm/n)) / (lcm). All the a's now have only one denominator value.
Let's see about the numerators. There is a theorem that says that for any nonzero integers a and b, there exist integers m and n such that m*a - n*b = gcd(a,b). Doing this operation on the numerators of the a's, we eventually find the value 1, since they are all relatively prime to the denominator. In effect, we have an element a = (2π)*(1/N) that generates all the others, making all the a's equal to (2π)*(k/N) where k is a nonnegative integer less than N. Thus, all finite subgroups of SO(2) are cyclic ones: Z.
Turning to the subgroups with positive and negative determinants, the negative-determinant matrix can be expressed in the form M(-,a) = {{cos(a), sin(a)}, {sin(a), -cos(a)}}
We get an interesting multiplication table:
M(+,a).M(+,b) = M(+,a+b)
M(+,a).M(-,b) = M(-,a+b)
M(-,a).M(+,b) = M(-,a-b)
M(-,a).M(-,b} = M(+,a-b)
We find that M(+,a).M(-,b) = M(-,b).M(+,-a) where M(+,-a) is the inverse of M(+,a), and also that M(-,a).M(-,a) = I -- that the negative-determinant elements are their own inverses. Thus, the positive-and-negative cases are dihedral groups.
Finite subgroups:
SO(2): Cyc (~ Z)
O(2): Cyc, Dih
Positive-determinant elements of O: pure rotations
Negative-determinant elements of O: rotation-reflections or rotoreflections
* What determinant values can an orthogonal matrix have?
Multiplying the definition by M, we get M.transpose(M) = transpose(M).M = I
Now take the determinant. It's easy to show that (det(M))2 = 1, giving det(M) = +- 1
-
* What are all the O(1) matrices? The SO(1) ones?
It is easy to show that a 1*1 matrix has only one element, and that that matrix's determinant is equal to that element. Thus,
O(1): { {{1}}, {{-1}} } -- Z2
SO(1): { {{1}} } -- the identity group, Z1
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* The O(2) matrices might seem to have 2*2 = 4 parameters, one for each matrix element. Can you show that their matrix elements can be specified with a smaller number of parameters?
We start with M = {{a11, a12}, {a21, a22}}
We find
det(M) = a11*a22 - a12*a21
transpose(M) = {{a11, a21}, {a12, a22}}
inverse(M) = {{a22, -a12}, {-a21, a11}}/det(M)
Set det(M) = s = +-1. Then
a22 = a11*s
a21 = - a12*s
(a11)2 + (a12)2 = 1
This suggests that we can use trigonometric functions:
For det +1: {{cos(a), -sin(a)}, {sin(a), cos(a)}}
For det -1: {{cos(a), sin(a)}, {sin(a), -cos(a)}}
-
* What are all the finite subgroups of O(2)? Of SO(2)?
First, we prove a theorem about groups of orthogonal matrices with both determinant values: that there are as many negative-determinant ones as there are positive-determinant ones.
Start with sets P = {M(+)} the positive-determinant ones and N = {M(-)} the negative-determinant ones.
Is P -> N an injection?
Take an element of N, X, and multiply every element of P by it on one side. The result is a subset of N, since matrices' determinants multiply. Thus, P -> N is an injection, one-to-one.
Is P -> N a surjection?
Take the inverse of X, and multiply every element of N by it on one side. The result is a subset of P. Thus, P -> N is a surjection, onto.
Since both are true, P -> N is a bijection, a one-to-one correspondence, and every element of P can be identified with every element of N.
So one can specify all the elements of a group of orthogonal matrices by specifying all its positive-determinant ones, and by combining them with a negative-determinant one, if any are present, to specify the negative-determinant ones.
-
Now for the main problem. Let us find the all-positive-determinant subgroups first, and then branch off to include negative-determinant ones. Thus, I find all finite subgroups of SO(2), and from them, all the finite subgroups of O(2) that are not in SO(2).
As I'd stated earlier, every element of SO(2) can be stated in form M(+,a) = {{cos(a), -sin(a)}, {sin(a), cos{a}} -- it can be pictured as a 2D rotation. These elements have a nice composition rule: M(+,a).M(+,b) = M(+,a+b), as can easily be verified from trigonometric identities.
We can strip away the matrix to get a (x) b = (a + b) mod 2π. We can get rid of the 2π factor by setting a = (2π)a'. That makes a' (x) b' = (a' + b') mod 1.
Let's see what form the a's are for a finite group. A finite group's subgroups must all be finite, as can easily be proved. So every element must generate a cyclic subgroup. This makes a = (2π)*m/n, where m is a nonnegative integer and n a positive one > n. Taking all the denominators in the a's, we find their least common multiple, and turn m/n into (m*(lcm/n)) / (lcm). All the a's now have only one denominator value.
Let's see about the numerators. There is a theorem that says that for any nonzero integers a and b, there exist integers m and n such that m*a - n*b = gcd(a,b). Doing this operation on the numerators of the a's, we eventually find the value 1, since they are all relatively prime to the denominator. In effect, we have an element a = (2π)*(1/N) that generates all the others, making all the a's equal to (2π)*(k/N) where k is a nonnegative integer less than N. Thus, all finite subgroups of SO(2) are cyclic ones: Z.
Turning to the subgroups with positive and negative determinants, the negative-determinant matrix can be expressed in the form M(-,a) = {{cos(a), sin(a)}, {sin(a), -cos(a)}}
We get an interesting multiplication table:
M(+,a).M(+,b) = M(+,a+b)
M(+,a).M(-,b) = M(-,a+b)
M(-,a).M(+,b) = M(-,a-b)
M(-,a).M(-,b} = M(+,a-b)
We find that M(+,a).M(-,b) = M(-,b).M(+,-a) where M(+,-a) is the inverse of M(+,a), and also that M(-,a).M(-,a) = I -- that the negative-determinant elements are their own inverses. Thus, the positive-and-negative cases are dihedral groups.
Finite subgroups:
SO(2): Cyc (~ Z)
O(2): Cyc, Dih
Positive-determinant elements of O: pure rotations
Negative-determinant elements of O: rotation-reflections or rotoreflections