Fundamental group of spaces of diagonalizable matrices

Your post is very interesting, but it contains quite a lot of different questions. I’ll answer the second part, which concerns matrices of finite order. It seems to me there are a few minor misconceptions here. Afterwards, we can probably discuss the first part about matrices with a simple spectrum.

Let $B\subset M_n(\mathbb K)$ be the set of matrices of finite order, with $\mathbb K=\mathbb C$ or $\mathbb R$.

• Over $\mathbb C$: $B$ is the disjoint union of conjugacy classes of diagonalizable matrices whose eigenvalues lie in $\mu_\infty$ (the roots of unity). These classes are indexed by multiplicity functions $m:\mu_\infty\to\mathbb N$ with finite support and $\sum m_{\zeta\in\mu_\infty}(\zeta)=n$. Each class is a connected homogeneous manifold $$ GL_n(\mathbb C)/\prod_{\zeta} GL_{m(\zeta)}(\mathbb C). $$ Hence $B$ has countably many path-connected components and is not totally disconnected.

• Over $\mathbb R$: $B$ is the disjoint union of conjugacy classes determined by the dimensions of the $+1$- and $-1$-eigenspaces and by the multiplicities of conjugate pairs $(\zeta,\overline\zeta)$ of complex roots of unity. Each class is a connected homogeneous manifold.
  Again, there are countably many components, so $B$ is not totally disconnected.

Regarding the hypothesis about simply diagonalizable matrices.

Fix $n=3$ and $\mathbb K = \mathbb C$. Call the set of simply diagonalizable matrices $A$ as in your question. We can use the fact that a simply diagonalizable matrix is completely determined by its eigenvalues and corresponding eigenspaces. The space of all one-dimensional subspaces of $\mathbb C^3$ is the complex projective plane $\mathbb {CP}^2$, which parameterizes the possible eigenspaces.

Let $B$ be the space of three complex eigenvalues and three elements of $\mathbb {CP}^2$ representing the eigenspaces, where the three eigenvalues are distinct and the three eigenspaces are linearly independent:

$$ B = \left\{ \begin{array}{l|l} \begin{align} (& \lambda_1 \in \mathbb C, \\ &v_1 \in \mathbb{CP}^2, \\ &\lambda_2 \in \mathbb C, \\ &v_2 \in \mathbb{CP}^2, \\ &\lambda_3 \in \mathbb C, \\ &v_3 \in \mathbb{CP}^2 ) \end{align} & \begin{array}{l} \lambda_1, \lambda_2, \lambda_3 \text{ are distinct}, \\ v_1, v_2, v_3 \text{ are linearly independent} \\ \text{(as subspaces of $\mathbb C^3$)} \end{array} \end{array} \right\} \subset (\mathbb C \times \mathbb{CP}^2)^3 . $$

This is an open subspace of $(\mathbb C \times \mathbb{CP}^2)^3$. There is a map $B \to A$ which sends an element of $B$ to the matrix with the prescribed eigenvalues and eigenspaces. Since the components of $B$ are ordered but the eigenvalues/eigenspaces of a matrix are unordered, each element of $A$ has $3!=6$ elements of $B$ in its preimage under this map. This map is a 6-fold covering of $A$.

We can reorder the factors of $B \subset (\mathbb C \times \mathbb{CP}^2)^3$ and view it as

$$ B \subset \mathbb{C}^3 \times (\mathbb{CP}^2)^3 $$

where the same restriction of the eigenvalues and eigenspaces all being distinct from each other applies, and since there is no constraint involving both a factor of $\mathbb C$ and a factor of $\mathbb{CP}^2$ this is actually a product space. For a topological space $S$, denote by $\mathrm{Conf}(S, m)$ the space of ordered tuples of $m$ distinct points in $S$. Then

$$B = \mathrm{Conf}(\mathbb{C}, 3) \times \text{something},$$

where the "$\text{something}$" represents the data about the eigenspaces. So we can compute the fundamental group of $B$ as

$$ \begin{align} \pi_1(B) &= \pi_1(\mathrm{Conf}(\mathbb{C}, 3)) \times \pi_1(\text{something}) \\ &= \pi_1(\mathbb C) \times \pi_1(\mathbb C \setminus \mathbf 1) \times \pi_1(\mathbb C \setminus \mathbf 2) \times \pi_1(\text{something}) \\ &= \pi_1(\mathbb C) \times \pi_1(\mathbb C \setminus \mathbf 1) \times \pi_1(\mathbb C \setminus \mathbf 2) \times \text{something} \\ &= \mathbb Z \times (\mathbb Z \ast \mathbb Z) \times \text{something} \end{align} $$

where $C \setminus \mathbf 1$ and $C \setminus \mathbf 2$ mean the complex plane with one or two points removed respectively. This means, at least, that $\mathbb Z \times (\mathbb Z \ast \mathbb Z)$ is a subgroup of $\pi_1(B)$.

Let $p$ be the 6-fold covering map $B \to A$. Since $B$ and $A$ are connected, p induces an injective homomorphism $p_{\#} : \pi_1(B) \to \pi_1(A)$. Since it is injective, it sends the $\mathbb Z \times (\mathbb Z \ast \mathbb Z)$ subgroup in $\pi_1(B)$ to a subgroup of $\pi_1(A)$ isomorphic to $\mathbb Z \times (\mathbb Z \ast \mathbb Z)$. So since the fundamental group of $A$ includes $\mathbb Z \times (\mathbb Z \ast \mathbb Z)$ as a subgroup, it can't be $\mathbb Z$.

Also note that the hypothesis fails for $n=1$ because every 1x1 matrix is simply diagonalizable.

Edit: I was wrong on the fibre, the mistake is now corrected. I thank @coiso for noticing it.

We can actually complete @chloe's anwser (by using similar techniques) and show the finiteness of $SD_n$, the set of simply diagonalisable matrices in the case $\mathbb{K} = \mathbb{R}$ if $n > 2$. In the beggining, we take any $\mathbb{K}$.

Let $$ \lambda : M \longmapsto \textrm{the set of eigenvalues of } M, $$ which is a continuous map from the set $\mathrm{Mat}_n(\mathbb{K})$ of $n \times n$ matrices with coefficients in $\mathbb{K}$ on the set $\mathbb{C}^n/\mathfrak{S}_n$ where the symmetric group $\mathfrak{S}_n$ acts on $\mathbb{C}^n/\mathfrak{S}_n$ by permutation of the coordinates (since we can't order eigenvalues).

By definition, $SD_n$ is exactly the set of $M \in \mathrm{Mat}_n(\mathbb{K})$ such that $\lambda(M) \in X$ where $X \subset \mathbb{K}^n/\mathfrak{S}_n$ is the set of pairwise distinct elements of $\mathbb{K}^n/\mathfrak{S}_n \subset \mathbb{C}^n/\mathfrak{S}_n$. Let us consider the restriction of $\lambda$, $$ \lambda : SD_n \longrightarrow X. $$

Let $x \in X$ that we can write as the equivalence class of some $(\lambda_1,\ldots,\lambda_n) \in \mathbb{K}^n/\mathfrak{S}_n$ with the $\lambda_i$ pairwise distinct. In this case, an element of the fibre of $\lambda$ over $x$ is a matrix with eigenvalues $(\lambda_1,\ldots,\lambda_n)$. Such a matrix $M$ is given by $M = PDP^{-1}$ where $D$ is diagonal whose entries are the $\lambda_i$ and $P$ is invertible. Since the $\lambda_i$ are pariwise disjoint, we can check that $P$ is unique up to a multiplication of the columns by non-zero scalars. Therefore, $$ F \cong \mathrm{GL}_n(\mathbb{K})/(\mathbb{K}^*)^n, $$ where $(\mathbb{K}^*)^n$ acts by multiplication of the columns.

Moreover, one can show that $\lambda$ is a fibre bundle thus a Serre fibration hence the exact sequence of homotopy groups (https://en.wikipedia.org/wiki/Homotopy_group), $$ \pi_2(X) \rightarrow \pi_1(F) \rightarrow \pi_1(SD_n) \rightarrow \pi_1(X) \rightarrow \pi_0(F). $$

The real case: let us assume now that $\mathbb{K} = \mathbb{R}$. In this case, thanks to the natural order on $\mathbb{R}$, for all elements $x \in X$, we can write $x$ uniquely as the class modulo permutation of $(\lambda_1,\ldots,\lambda_n)$ with $\lambda_1 < \cdots < \lambda_n$. Thus, $X$ is naturally homeomorphic to the set of all $(\lambda_1,\ldots,\lambda_n)$ strictly ordered. This subset of $\mathbb{R}^n$ is convex thus contractible. Since all fibred bundles over a contractible base are trivial, we even have a homeomorphism $SD_n \cong F \times X$ so we only need to study the topology of $F \cong \mathrm{GL}_n(\mathbb{R})/(\mathbb{R}^*)^n$.

Since multiplying the first column of a matrix $M$ by a scalar $a$ multiplies the determinant of $M$ by $a$, we see easily that $F \cong \mathrm{SL}_n(\mathbb{R})/(\mathbb{R}^*)^{n - 1}$ where $(\mathbb{R}^*)^{n - 1}$ acts on $\mathrm{SL}_n(\mathbb{R})$ by $$ (a_2,\ldots,a_n) \cdot (C_1,C_2,\ldots,C_n) = (a_2^{-1}\cdots a_n^{-1}C_1,a_2C_2,\ldots,a_nC_n). $$

In particular, $F$ is connected as long as $n \geqslant 2$. The fibre bundle $(\mathbb{R}^*)^{n - 1} \rightarrow \mathrm{SL}_n(\mathbb{R}) \rightarrow F$ yields an exact sequence of homotopy groups $$ \pi_1((\mathbb{R}^*)^{n - 1}) \rightarrow \pi_1(\mathrm{SL}_n(\mathbb{R})) \rightarrow \pi_1(F) \rightarrow \pi_0((\mathbb{R}^*)^{n - 1}) \rightarrow \pi_0(F) = 1. $$ The base point can be chosen anywhere. Since the connected components of $(\mathbb{R}^*)^{n - 1}$ are simply connected, we obtain $$ 1 \rightarrow \pi_1(\mathrm{SL}_n(\mathbb{R})) \rightarrow \pi_1(F) \rightarrow 2^{n - 1} \rightarrow 1. $$ Here $2^{n - 1}$ has to be interpreted as a finite set of caridnality $2^{n - 1}$ (the set of connected components of $(\mathbb{R}^*)^{n - 1}$).

Recall that $$ \pi_1(\mathrm{SL}_2(\mathbb{R})) = \mathbb{Z}, \qquad \forall n > 2, \pi_1(\mathrm{SL}_n(\mathbb{R})) = \mathbb{Z}/2\mathbb{Z}. $$

In particular, if $n = 2$, $\pi_1(SD_n) = \pi_1(F)$ is countable infinite and if $n > 2$, $\pi_1(SD_n) = \pi_1(F)$ is finite of cardinality $2^n$. (A more complete description of this group can be found in @coiso's answer).

The complex case: In the complex case, $X$ is well-known to be connected with fundamental group $B_n$, the $n$th Braid group (https://en.wikipedia.org/wiki/Braid_group). Therefore, the above exact sequence of groups gives in particular $$ \pi_1(SD_n) \rightarrow B_n \rightarrow 0. $$

It doesn't give explicitely what $\pi_1(SD_n)$ is but it shows that it surjects onto $B_n$. In particular, it can't be isomorphic to $\mathbb{Z}$ as long as $n > 2$.

I've compiled my answers into a pdf (updated version). Highlights:

The space $\Theta_n$ of $n\times n$ simply diagonalizable real matrices has components

$$ \Theta_n=\bigsqcup_{r+2s=n} \Theta_{r,s} $$

where $\Theta_{r,s}$ is the simply diagonalizable real matrices with $r$ invariant lines and $s$ invariant planes. The component $\Theta_{n,0}$ has those matrices which are diagonalizable over $\Bbb R$ specifically.

It is a fiber bundle $G/T\to\Theta_{r,s}\to {\rm UConf}(s,{\cal H})$ where $G={\rm GL}(n,\Bbb R)$, $T$ is subgroup of block diagonal matrices with $r$ scalars followed by $s$ scalar multiples of $2\times 2$ rotation blocks, $\mathrm{UConf}(k,{\cal H})$ (unordered configuration space) is the space of sets of $k$ distinct points of the open upper half plane in $\Bbb C$. Using Orbit-Stabilizer, the Iwasawa decompositon $G=KAN$, and the Long Homotopy Exact Sequence, we can deduce:

Theorem. $\pi_1(\Theta_{r,s})$ is an extension of $\pi_1(G/T)$ (below) by either the braid group $B_s$ (if $r>0$) or its "alternating braid subgroup" (if $r=0$).

$$ \pi_1(G/T) \,\cong\, \begin{cases} \, \Bbb Z_2^{r-1} & s>0, r>0 \\ \, 0 & s>0, r=0 \\ \, \Bbb Z & s=0, r=2 \\ \, H_r & s=0, r>2 \end{cases} $$

Here $H_r$ is the subgroup of the spin group ${\rm Spin}(r)$ generated by $e_ie_j$ for $1\le i,j\le r$ (wrt a given orthonormal basis), subject to relations $e_i^2=1$ and $e_ie_j=-e_je_i$ ($i\ne j$). For example $H_3=Q_8$ is the quaternion group.

Theorem. The space $\Xi_n$ of all $n\times n$ simply diagonalizable complex matrices has $\pi_1(\Xi_n)=B_n$.

The space of $n\times n$ matrix solutions to $A^m=I$ is a representation variety whose components are orbits under conjugation. There is one orbit for each $n$-dim rep of $\Bbb Z_m$. Any rep can be encoded as a multiset of irreps (up to iso). The irreps of $\Bbb Z_m$ over both $\Bbb R$ and $\Bbb C$ are well-known. It is possible to translate "has dim $n$" and "has order $m$" each into a condition on irrep multiplicities.

Over $\Bbb C$, the orbits have fundamental group $\Bbb Z/g\Bbb Z$, where $g$ is the $\gcd$ of the dimensions of the irreps present in the given rep. Over $\Bbb R$, the fundamental group of the orbit is $\Bbb Z$ if $n=2$, else $\Bbb Z_2^{k-1}$ (where $k$ is the number of isoclasses of irreps in a given rep) if there is an odd-dim rep of multiplicity $>1$, else the aforementioned $H_k$ in the spin group.