TITLE: The conjugacy classes of diagonalizable $2 \times 2$ matrices can be identified with their eigenvalues, what about pairs?
QUESTION [14 upvotes]: For sake of simplicity, let's say that we live in $G = SL(2, \mathbb{C})$. Every conjugacy class of diagonalizable matrices $$[A] := \{gAg^{-1} \mid g \in G\}$$ can be identified with its set of eigenvalues $\{\lambda, \lambda^{-1}\}$. As such, the collection of diagonalizable matrices in $SL(2, \mathbb{C})$ can be identified with $\mathbb{C}^\times / \{z \sim z^{-1}\}$.
How about the conjugacy class of pair of matrices $$[A, B] = \{(gAg^{-1}, gBg^{-1}) \mid g \in G\}?$$ Can they similarly be identified with their eigenvalues in a natural way? What is the significance of this identification?

REPLY [2 votes]: Let me complete Ehud Meier's answer. One cannot distinguish one eigenvalue among $(\lambda,\lambda^{-1})$. In other words, one cannot distinguish between the projections $p_j$ and $q_j=1-p_j$. Therefore, $\alpha$ is only one among four relevant numbers. If $p_1q_2p_1=(1-\alpha)p_1$ is clear, it is less obvious that $q_1p_2q_1=(1-\alpha) q_1$.
Therefore one expect that the conjugacy classes are parametrized by the triples $(\alpha,\lambda,\mu)$, modulo the group generated by the involutions
$$(\alpha,\lambda,\mu)\mapsto(1-\alpha,\lambda^{-1},\mu)\quad\hbox{or}\quad(1-\alpha,\lambda,\mu^{-1}).$$