Mobius transformation proving equivalence class

[d2j2003]

Homework Statement

I have to show that if there is a mobius transformation p such that m=p°n°p$^{-1}$
forms an equivalence class.

Homework Equations

need to show that aRa, if aRb then bRa, and if aRb and bRc then aRc

The Attempt at a Solution

well.. for aRa I somehow need to show that m=p°m°p$^{-1}$ right? so if we just say m=p°p°m°p$^{-1}$ °p$^{-1}$ then this is of the correct form... if we say (p°p)=q then m=q°m°q$^{-1}$ which is of the correct form so mRm am I moving in the right direction?

Thanks in advance

 

[mighty2000]

for any help!

let me offer some guidance on how to approach this problem.

First, let's define our terms. A Mobius transformation is a function that maps the extended complex plane (including infinity) to itself. It is represented by the formula p(z) = (az+b)/(cz+d), where a, b, c, and d are complex numbers and ad-bc ≠ 0. This transformation can be composed with other Mobius transformations to form a new transformation.

Now, let's look at the given equation m = p°n°p^{-1}. This means that m is the result of applying the Mobius transformation p to the result of applying the Mobius transformation n to the inverse of p. In other words, m is in the equivalence class of p.

To show that this forms an equivalence class, we need to demonstrate the three properties listed in the homework equations. Let's start with aRa. This means that m = p°m°p^{-1}. We can rewrite this as p(m) = (p°p)(m)(p^{-1}°p^{-1}), which simplifies to p(m) = q(m)q^{-1}, where q = p°p. This shows that m is in the equivalence class of q, which is the composition of two Mobius transformations and is itself a Mobius transformation.

Next, let's consider the property aRb. This means that if aRb, then bRa. In this case, if m is in the equivalence class of p, then there exists a Mobius transformation q such that m = q°p°q^{-1}. In order for b to be in the equivalence class of p, we need to show that there exists a Mobius transformation r such that b = r°p°r^{-1}. This can be achieved by letting r = q^{-1}. Then, b = q^{-1}°(q°p°q^{-1})°q = (q^{-1}°q)°p°(q^{-1}°q) = p. This shows that b is in the equivalence class of p.

Finally, we need to show the property aRb and bRc then aRc. This means that if a is in the equivalence class of p and b is in the equivalence class of p, then c is also in the equivalence class of p. In other words,