Mobius transformation (isomorphism)

[lavster]

Im having difficulty understanding this satement - can someone please explain it to me...

let M be the class of mobius transformations

M is isomorphic to GL2/Diag isomorphic to SL2/Id, where GL2 is the group of non-degenerate matrices of size 2 x 2 with complex entries, SL2 = A in GL2 : detA = 1, Diag is the group of non-zero diagonal 2 x 2 matrices and Id is the identity.

I know what a mobius transform is and that the matrix of its coefficients cannot be zero, i know that isomorphism = one to one correspondence... but i don't understand this statement at all - why are we exclusing the diagonal matrices... surely we should be excluding the matrices with det 0?

thanks

 

[Deveno]

actually (my memory is a bit foggy on these) i think that M is isomorphic to GL2(C)/(CI), where the matrices in CI are of the form:

[k 0]
[0 k], for some non-zero complex number, k.

we're not "excluding" the matrices kI, we're treating them as if they all were the identity matrix. if m =

[a b]
[c d] represents the mobius transformation f(z) = (az+b)/(cz+d), then m' =

[ka kb]
[kc kd] represents the transformation g(z) = (kaz+kb)/(kcz+kd) = k(az+b)/(k(cz+d)) = f(z), the same transformation as f.

so we need to take the matrices m "mod a scalar".

 

[lavster]

actually (my memory is a bit foggy on these) i think that M is isomorphic to GL2(C)/(CI), where the matrices in CI are of the form:

[k 0]
[0 k], for some non-zero complex number, k.

we're not "excluding" the matrices kI, we're treating them as if they all were the identity matrix. if m =

[a b]
[c d] represents the mobius transformation f(z) = (az+b)/(cz+d), then m' =

[ka kb]
[kc kd] represents the transformation g(z) = (kaz+kb)/(kcz+kd) = k(az+b)/(k(cz+d)) = f(z), the same transformation as f.

so we need to take the matrices m "mod a scalar".

how do you know we are treating them as the identity matrix? i thought eg R/{0} meant all the real numbers excluding zero, so why is it different here? and what does "mod a scalar" mean? mod = modulo arithmetic? (Im not very good at this kind of thing - it confuses me!)

thanks :)

 

[Deveno]

GL2(C) is the general linear group of degree 2 (2x2 invertible matrices), over the field C.

in a group like GL2(C), you ignore the additive structure, and deal just with the multiplication. to have a group structure, every element must have an inverse (this is the same as requiring that ad - bc is non-zero in the 2x2 case).

the technical term for GL2(C)/(CI) is a "quotient" group. this is like modulo arithmetic, but more general. with integers, we say that a ≡ b (mod n) if a - b = kn, for some integer k. here, the operation is "+" (we can write a - b as a+(-b)).

with matrix multiplication, we have a different operation, under this operation the inverse of B is B^-1, and the parallel of a - b is AB^-1.

now the set of all integer kn (multiples of n) form what is called a subgroup of the integers (any sum of multiples of n is a multiple of n, and the additive inverse of a multiple of n is also a multiple of n. also, 0, the additive identity, is a multiple of n,0 = 0n). so the condition: a - b = kn, is really the condition a - b is in the subgroup nZ. this has the effect of setting all multiples of n congruent to 0, all numbers of the form n+1 congruent to 1, etc.

by direct analogy, the matrices kI, form a multiplicative group:

(kI)(k'I) = (kk')I

I = 1I,

(kI)^-1 = (1/k)I

we can use this to define an equivalence of matrices: A ~ B if AB^-1 = kI.

this means that we are regarding A and B as "the same" if they differ by just a scalar multiple. this is what is meant by "mod a scalar", we are treating kA =(kI)A, the same as A.

this is a similar notation as A\B = {x in A, but not in B}, so A\{0} means {the non-zero elements of A}, but the meaning is entirely different.

the meaning of the statement that M is isomorphic to GL2(C)/(CI), is just that there is a 1-1
correspondence between the transformations f(z) = (az+b)/(cz+d) and matrices of the form:

[ka kb]
[kc kd] , k non-zero in C, ad-bc non-zero.

by multiplying a matrix

[a b]
[c d] (with ad-bc non-zero) by 1/(ad-bc) (to which it is equivalent, since 1/(ad-bc) is a scalar) we obtain a matrix of determinant 1, which is what SL2(C) is, the invertible matrices of determinant 1 (this is another "subgroup" of GL2(C)).

 

[lavster]

Ah... perfect! thanks so much! :)