Computing the number of path-connected components of a subset of $\mathcal{M}_n(\mathbb{C})$

Question

For positive integers $n,m$, is there a way of computing the number of path-connected components of $\{A\in\mathcal{M}_n(\mathbb{C})\colon\,A^m=I_n\}$?

Answer 1

The solutions to $A^m = I_n$ will be all diagonalizable matrices with eigenvalues satisfying $\lambda^m = 1$. Two matrices will be contained within the same path component if and only if they have the same multiset of eigenvalues.

To show that two matrices with distinct eigenvalue sets are in distinct path components, it suffices to note that the function that takes a matrix to its characteristic polynomial is continuous.

To show that two matrices with the same eigenvalue lie in the same path component, it helps to note that $S:[0,1] \to \mathcal M_2(\Bbb C)$ given by $$ S(t) = \pmatrix{t & 1-t\\t-1 & t} $$ is such that $$ S(0)\pmatrix{\lambda_1 \\ & \lambda_2} S(0)^{-1} = \pmatrix{\lambda_1 \\ & \lambda_2}\\ S(1)\pmatrix{\lambda_1 \\ & \lambda_2}S(1)^{-1} = \pmatrix{\lambda_2 \\ & \lambda_1} $$

Answer 2

Given a multiset of $n$ complex numbers, the set of all diagonalizable matrices which has that multiset as spectrum is connected: indeed, it is the orbit of any one diagonal matrix with that spectrum under the action of $GL_n(\mathbb C)$ on $M_n(\mathbb C)$ by conjugation — we are using here the fact that the group $GL_n(\mathbb C)$ is connected.

Now, the set of solutions of the equation $A^m=I$ is a disjoint union of such conjugacy classes, so we need only count the corresponding multisets.