Week4Lecture3: Möbius transformatios, part I

[@shacharh5470]

Another way to show it's a bijcetion: You can define a mobius tra. using an invertible 2x2 matrix (a b c d), then show that composition of Mobious trs respects multiplication of matrices - if T is the transformation using matrix A and S using B then T after S is the transformation defined by (AB).
As a corrolary, a Mobius transformation is invertible - its inverse is the trans. defined with the inverse matrix - so it's a bijection.

[@xanderlewis]

That doesn't work - Möbius transformations aren't (always) linear so can't be represented by matrices in the way you describe.

[@shacharh5470]

​@@xanderlewisThat's not what I meant.
Forget that matrices represent linear transformations on a vector space.
I'm saying there is a grouop homomorphism between Mobius transformations (with composition) and GL(2,R).
Check it for yourself.
The matrix (a b; c d) is associated with the function f(z) -> (az+b)/(cz+d)

[@xanderlewis]

@@shacharh5470 OK! That makes a lot more sense and is a better way of putting it.

[@xanderlewis]

...and thanks for replying five years later. ;-)

[@Charky32]

@@shacharh5470 hey can you help with a question, Suppose that M is a 2 × 2 complex matrix of determinant 1, and that T_M maps the upper half plane onto itself. Show that M is a real matrix.

[@nilkarabulut3135]

Thank you so much

[@CornishMiner]

Right at the end, a division by zero of a non-zero number is defined as infinity. Not sure that it was made clear that in C ∪ {∞} z/0 = ∞ and z/∞ = 0, and that ∞/0 = ∞ and 0/∞ = 0, leaving 0/0 and ∞/∞ undefined.

[@personxy7443]

Teacher,why we called it linear?(A linear function is T(ax+b)=aT(x)+b,but it is not

[@formlessspace3560]

Trank you for this Video Petra

[@Kybeline]

Thanks for every video, Petra :)

[@esmirhodzic981]

thank yoou for so awesome videos :D really helpful

[@ganeshdilli2699]

Too good