Topology of non-degenerate $n\times n$ Hermitian matrices .. where I guess by topology I mean its homology / homotopy groups. Here "degenerate" means having repeated eigenvalues. This is interesting because it defines the space that can be explored via adiabatic quantum computation.
I believe it's fundamental group (windings) will be trivial, because the space is 3 dimensions more than degenerate Hermitian matrices, and I expect there to be no links then. I'm not sure how to prove this though, or how this works out for higher groups.
Edit:
I've been thinking on this and I have what I think is significant progress to the solution. I'm posting it as an answer (that I won't accept) for now, I'm hoping someone else more versed in algebraic topology can give me the relevant terms I would need to search to complete this!
Each matrix is fully specified by its eigenvectors and eigenvalues. It's clear that we can deform the eigenvalues freely as long as we preserve their order; e.g. if I have eigenvectors $v_1,\,v_2,\,v_3$ as sorted by their eigenvalues, I can continuously change my matrix to have eigenvalues $0,\,1,\,2$. This is a deformation retract and doesn't significantly change anything about the space. This leaves the problem as exploring the space of ordered normal eigenvectors.
First consider the real symmetric (as opposed to Hermitian) matrices, because they're a bit easier to visualize, to me at least. And for even more intuition, let's take just $\mathbb{R}^3$. I could first considering "choosing" a direction for my largest eigenvector; normalized, it can be anywhere on $S_{n-1} = S_2$ in this case. But since an eigenvector in one direction is equivalent to negative, the space is really $\mathbb{R}P^2$. Then I have to select an orientation for my second-largest-by-value eigenvector, which comes from the plane left over, normal to the first eigenvector. This is $\mathbb{R}P^1 \cong S_1$ by a similar argument. The third vector is then determined fully.
This indicates to me that full space should be some sort of product of $\mathbb{R}P^2$ and $S_1$, but where the motion in $\mathbb{R}P^2$ somehow... modifies? the $S_1$ component. I don't know how to describe this more rigorously. Having done a lot more group theory than topology, this feels sort of like a semi-direct product. A friend of mine suggested this might be a fiber bundle, but neither of us were confident in that after some thought. My thought is that it's similar in some sense to how $S_2$, $S_1\times S_1$, the Klein bottle, and $\mathbb{R}P^2$ all contain two independent copies of $S_1$, and so all somehow look a little bit like a product. I don't know what this operation is, so I'll write it as $\otimes$ for now.
The extension to the full problem then is that, for a general $n\times n$ real symmetric matrix, the space looks like $\mathbb{R}P^{n-1}\otimes\mathbb{R}P^{n-2}\dots \mathbb{R}P^{2}\otimes \mathbb{R}P^{1}$. I believe that for a Hermitian matrix, where the only difference is that the eigenvectors can be complex and they're still considered equivalent up to an arbitrary (now complex) factor, it would be $\mathbb{C}P^{k}$ instead in each term.
So I'm wondering if (1) the above logic seems sound, (2) this $\otimes$ operation (which clearly has some degrees of freedom) is something well-defined, (3) anyone knows where I could find information on the different ways to $\otimes$ projective spaces together.
 A: Call this space $E$. Let $\pi : E \to B$ be the map sending an element of $E$ (a Hermitian matrix with distinct eigenvalues) to its set of eigenvalues. $B$ here is the configuration space $\text{Conf}_n(\mathbb{R})$ of $n$ distinct unordered points in $\mathbb{R}$, which is contractible; it can be described as a certain open subset of $\mathbb{R}^n$, and is homeomorphic to $\mathbb{R}^n$. 
$\pi$ is a fiber bundle. The fiber $F$ over a point in $B$ is the space of possible eigenvectors a Hermitian matrix with fixed eigenvalues can have. This is, equivalently, the space of ordered $n$-tuples of orthogonal $1$-dimensional subspaces of $\mathbb{C}^n$ (the ordering being fixed by the ordering of the corresponding eigenvalues). The unitary group $U(n)$ acts transitively on this space with stabilizer the maximal torus $U(1)^n$, so
$$F \cong U(n)/U(1)^n$$
is the flag variety of $U(n)$. (Another way to see this is that given a flag in $\mathbb{C}^n$ we can write down an ordered $n$-tuple of orthogonal $1$-dimensional subspaces by using Gram-Schmidt, and vice versa.)
Since $B$ is contractible, the inclusion of any fiber $F$ into $E$ is a homotopy equivalence, and so we get that $E$ is homotopy equivalent to the flag variety of $U(n)$. Much is known about the topology of flag varieties; for example, their rational cohomology is known. 
