The conjugacy classes of diagonalizable $2 \times 2$ matrices can be identified with their eigenvalues, what about pairs?

Question

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?

Answer 1

You have a geometric invariant theory question here: your space is the space of the pairs of matrices, and the action is given by conjugation. One possible way to deal with this is to study the ring of invariant functions. In case of one matrix it really boils down to the ring generated by the coefficients of the characteristic polynomial. In case of two matrices it is more complicated, and was studied by Proecsi in the paper "The invariants of n x n matrices". Basically the invariants are generated by polynomials of the form $tr(M_1M_2\ldots M_n)$ where $M_i=A$ or $B$. The relations between these polynomials are studied in Procesi's paper. Since you consider here only the two dimensional case, I would continue as follows: Lets say that both matrices are diagonalizable, and that $A$ has the eigenvalues $\lambda,\lambda^{-1}$ and $B$ has the eigenvalues $\mu,\mu^{-1}$. Then we can write the two dimensional vector space $V=\mathbb{C}^2$ as the direct sum $V_{\lambda}\oplus V_{\lambda^{-1}} = V_{\mu}\oplus V_{\mu^{-1}}$. Let $p_1$ be the projection onto $V_{\lambda}$ (with respect to the first decomposition) and let $p_2$ be the projection onto $V_{\mu}$ (with respect to the second decomposition). Then since the space is two dimensional, there will be a scalar $\alpha$ such that $p_1p_2p_1 = \alpha p_1$. I believe that $\alpha$, $\mu$ and $\lambda$ will determine the pair of matrices. You will have to mod out by a certain equivalence relation, but it is not so hard to write this equivalence relation explicitly.

Answer 2

It is fairly easy to do this directly. I consider the case that both $A$ and $B$ are individually diagonalizable, and not scalar. Note also that ${\rm SL}(2,\mathbb{C})$-conjugacy is equivalent to ${\rm GL}(2,\mathbb{C})$-conjugacy, since every matrix in ${\rm GL}(2,\mathbb{C})$ has a scalar multiple in ${\rm SL}(2,\mathbb{C})$. For a pair $(X,Y)$ to be $G$ conjugate to a pair $(U,V)$ we first need $X$ and $U$ to be $G$-conjugate, so we might as well assume that $X = U$, and that both are diagonal. In that case, $Y$ and $V$ need to be conjugate within $C_{G}(X)$. But $C_{G}(X)$ now consists of diagonal matrices. Hence $Y$ and $V$ need to be conjugate via a diagonal matrix.

Degenerate cases : If $U$ and $X$ are both the same scalar, then the pair $Y$ and $V$ simply need to be $G$-conjugate. If $X$ and $Y$ commute (neither scalar) and $U$ and $V$ commute (neither scalar), then if we assume wlog that $U = X$, then the pair $(U,V)$ is conjugate to $(U,Y)$ if and only if we also have $Y = V$.

Answer 3

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 $$(\alpha,\lambda,\mu)\mapsto(1-\alpha,\lambda^{-1},\mu)\quad\hbox{or}\quad(1-\alpha,\lambda,\mu^{-1}).$$