We look at the Matrix group problems first -
1. For the group PSL(2,z) show that the identity element (e) = s^4 = t^3.
The identity element e =
[1 0]
[01]
While s =
(0 1)
(-1 0)
And t =
(0 -1)
(1 -1)
One therefore simply needs to multiply s x s x s x s in order to obtain e (using the rules of matrix multiplication) and t x t x t, to find e in the same.
2. For the group sl(2) show that:(a) [h.f] = h*f - f*h = -2f
Again, using the elements h and f as defined in the blog, we multiply:
h*f - f*h =
(0 0)
(-2 0)
And since f =
(0 0)
(1 0)
It means that the relation: h*f - f*h = -2f
(b) The "Casimir element", C, of sl(2) is defined according to: h^2/ 2 + h + 2f*e, find the element
By matrix multiplication, we determine: h^2/ 2 =
(½ 0)
(0 ½)
And adding the matrix h yields:
(1½ 0)
(0 -½)
Now, adding 2fe =
(0 0)
(0 2) yields:
C =
(1½ 0)
(0 1½)
1. For the group PSL(2,z) show that the identity element (e) = s^4 = t^3.
The identity element e =
[1 0]
[01]
While s =
(0 1)
(-1 0)
And t =
(0 -1)
(1 -1)
One therefore simply needs to multiply s x s x s x s in order to obtain e (using the rules of matrix multiplication) and t x t x t, to find e in the same.
2. For the group sl(2) show that:(a) [h.f] = h*f - f*h = -2f
Again, using the elements h and f as defined in the blog, we multiply:
h*f - f*h =
(0 0)
(-2 0)
And since f =
(0 0)
(1 0)
It means that the relation: h*f - f*h = -2f
(b) The "Casimir element", C, of sl(2) is defined according to: h^2/ 2 + h + 2f*e, find the element
By matrix multiplication, we determine: h^2/ 2 =
(½ 0)
(0 ½)
And adding the matrix h yields:
(1½ 0)
(0 -½)
Now, adding 2fe =
(0 0)
(0 2) yields:
C =
(1½ 0)
(0 1½)
3. Show that the Klein Viergruppe, V4, is Abelian.
This is easily shown by matrix multiplication for which we will find:
a*b = b*a =
(-1 0)
(0 -1)
And: a*c = c*a =
(-1 0)
(0 1)
And finally, b*c = c*b =
(1 0)
(0 -1)
Now the odd solutions to the lines and planes problems are as follows:
(1) Find the equation of the plane perpendicular to the vector N at (-3, -2, 4), and passing through the point P= (2, pi, -5)
Using the given triples, we obtain the equation as: -3x – 2y + 4z = P*N
Whence: P*N = [(-3)(2) + (2)(pi) + 4(-5)] = -26 + 2pi = -2(13 + pi)
So: -3x -2y +4z = -2(13 + pi)
Or reduced to:
-3x/2 – y + 2z = 13 + pi
(3) Find the cosine of the angle between the planes: x + y + z = 1 and x – y – z = 5
The respective vectors we need, from the coefficients of the equations are:
A = (1, 1, 1) and B = (1, -1, -1)
Then: cos(theta) = A*B/ [A]{B] =
{(1*1) + (1*(-1)) + (1)*(-1)} / {[1^2 + 1^2 + 1^2] [1^2 + (-1)^2 +(-1)^2]}
Or cos (theta) = -1 / {(3)^1/2 (3)^1/2} = -1/ 3
This is easily shown by matrix multiplication for which we will find:
a*b = b*a =
(-1 0)
(0 -1)
And: a*c = c*a =
(-1 0)
(0 1)
And finally, b*c = c*b =
(1 0)
(0 -1)
Now the odd solutions to the lines and planes problems are as follows:
(1) Find the equation of the plane perpendicular to the vector N at (-3, -2, 4), and passing through the point P= (2, pi, -5)
Using the given triples, we obtain the equation as: -3x – 2y + 4z = P*N
Whence: P*N = [(-3)(2) + (2)(pi) + 4(-5)] = -26 + 2pi = -2(13 + pi)
So: -3x -2y +4z = -2(13 + pi)
Or reduced to:
-3x/2 – y + 2z = 13 + pi
(3) Find the cosine of the angle between the planes: x + y + z = 1 and x – y – z = 5
The respective vectors we need, from the coefficients of the equations are:
A = (1, 1, 1) and B = (1, -1, -1)
Then: cos(theta) = A*B/ [A]{B] =
{(1*1) + (1*(-1)) + (1)*(-1)} / {[1^2 + 1^2 + 1^2] [1^2 + (-1)^2 +(-1)^2]}
Or cos (theta) = -1 / {(3)^1/2 (3)^1/2} = -1/ 3
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