Direct proof that trace equivalence of Mobius transformations implies conjugacy

Question

I've seen it stated that two nontrivial Mobius transformations with the same trace (up to sign) are conjugate; now clearly this follows from the typical exhaustive classification of Mobius transformations by their trace and consequent class representative, but is there any short and direct way of proving this result using e.g. representation theory of $\text{PGL}(2,ℂ)$?

Answer 1

If two elements of $GL(2,\Bbb C)$ are conjugate, then they're corresponding equivalence classes are conjugate in $PGL$. It therefore suffices to show that for any $A,B \in GL$: if they have the same trace, then there $\alpha_1A$ is conjugate to $\alpha_2 B$ for some $\alpha_1,\alpha_2 \in \Bbb C$.

First, we may multiply $A$ and $B$ by scalars so that WLOG, they have determinant $1$. Let $\lambda_1,\lambda_2$ denote the eigenvalues of $A$. We know that $\lambda_1\lambda_2 = 1$ and $tr(A) = \lambda_1 +\lambda_2$. Note that the function $\phi: \lambda \mapsto \lambda + \frac 1{\lambda}$ is one-to-one over $U = \{\lambda : |\lambda| \geq 1; Arg(\lambda) \in [0,\pi)\}$ Note also that WLOG, we may multiply $A$ by $\pm 1$ so that $A$ necessarily has an eigenvalue in $U$.

Case 1: Suppose first that $tr(A) \neq 2$. It follows that $\lambda_1 \neq 1$, which means that $\lambda_1 \neq \lambda_2$. If we apply the same process to $B$ and find $tr(B) = tr(A)$, then $B$ must have the same distinct eigenvalues as $A$, which means that it is conjugate to $A$.

Case 2: Suppose that $tr(A) = 2$. We find that the eigenvalues of $A$ are necessarily $1,1$. There are two conjugacy classes of $GL$ corresponding to these eigenvalues, namely those of the Jordan form matrices $$ \pmatrix{1&0\\0&1}, \quad \pmatrix{1&1\\0&1} $$ the first matrix corresponds to the trivial case, so we discount it. It follows that if $A$ and $B$ come from non-trivial Mobius transformations with trace $2$, their matrices are conjugate to the second matrix above, which means that they are conjugate to each other.