## Wednesday, May 5, 2010

### Math Solutions

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)

(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