Do the singular matrices form a topological manifold

Question

So the definition of manifold I'm using is that of a topological manifold (a topological space with an atlas of homeomorphisms to $\mathbb{R}^n$).

I have two related questions:

1. Is the set of singular matrices (over $\mathbb{R}$) of dimension $n$ a manifold? My understanding is that it cannot be if it has the subspace topology inherited from $\mathbb{R}^{n^2}$, since in this case it's given by the surface $\det(A)=0$, which has a singularity at $0$. I'm aware of the similar questions here and here, but the answers given seem to be for smooth manifolds and not topological manifolds.

2. My second question is what exactly is the topological definition of a singularity? Since a topological space being a manifold is a property of the topology itself, there should be a purely topological definition of a singularity. The best I can come up with is that a singular point in a topological space is a point such that no open set which contains it can be homeomorphic to $\mathbb{R}^n$, however this definition seems like a bit of a cop out.

Update

So as far as I can tell my definition of a topological singularity is correct. However as to whether the surface $\det(A)=0$ is a topological manifold or not, I still couldn't say.

Considering the determinant as a potential submersion of manifolds $\mathbb{R}^{n^2}\rightarrow\mathbb{R}$, we see by looking at its derivative that it fails to be a submersion whenever the adjoint (or equivalently the cofactor) matrix is identically zero. And since the adjoint of a matrix being zero implies the matrix is singular, this will be some subset of the singular matrices. So it is these matrices which prevent us from using the regular level set theorem to conclude that the singular matrices form a smooth manifold, however it doesn't mean they aren't a topological manifold.

The set of points in $\mathbb{R}^{n^2}$ which are singularities is thus given by the polynomial surface $$\{\det(A_{ij})=0\; :\; 1\leq i,j\leq n\}$$ where $A_{ij}$ is the sub-matrix of $A$ obtained by deleting the $i^{th}$ row and $j^{th}$ column.

An example of a singular curve which does admit a $C^0$ structure is $y^2-x^3=0$, via the parametrization $t\mapsto (t^2,t^3)$.

An example of a singular curve which doesn't admit a $C^0$ structure is $xy=0$, since any open set containing the origin has 4 disconnected components when the origin is removed, while removing a point in $\mathbb{R}^k$ results in at most 2 disconnected components.

Answer 1

I will consider only the first non-trivial case of $2\times 2$ real matrices with determinant $0$: $\{ (a,b,c,d) \ | \ a d - b c = 0\}$. Write $ a = x+y$, $d = x-y$, $b = z+ t$, $c = z - t$ and get $$X \colon \{ (x,y,z,t) \ | \ x^2 - y^2 = z^2 - t^2\}$$ or $\{ (x,y,z,t) \ | x^2 + t^2 = y^2 + z^2 \}$. Notice that $X$ is homeomorphic to the cone over the torus by the map $(r, \phi, \tau) \mapsto (r \cos \theta, r \cos \tau, r \sin \tau, r\sin \theta)$ that is $X \simeq T \times [0, \infty) / T \times \{0\}$.

Let's show that no open neighborhood of $(0,0,0,0)$ in $X$ is homeomorphic to a 3-dimensional ball. It's enough to do that for $C(T) \colon = T \times [0, \infty) / T \times \{0\}$ ( with the point $0$ the image of $T \times \{0\}$). If it had such a neigborhood $V$ then $V \backslash \{0\}$ would be simply connected. Let's show that for no neighborhood $V$ of $0$ in $C(T)$, $V \backslash\{0\}$ is simply connected. Indeed, every such neighborhood must contain $T \times (0, \epsilon)$ for some $\epsilon > 0$. Consider a path in $T \times (0, \epsilon)$ whose first component is not null-homotopic. Then the image of this path in $C(T) \backslash \{ 0\}$ is not null-homotopic, so it's also not null-homotopic in $V \backslash\{0\}$.

Answer 2

This post is to expand the details in Orangeskid's argument.

What we have is some open set $U = V \setminus \{0\}$. Hypothetically, it's homeomorphic to $B^3 \setminus \{0\}$, and hence simply connected.

Now, one picks a map $\varphi: S^1 \to T^2 \times (0,\varepsilon)$ that's not null-homotopic. Because $\pi_1(T^2 \times (0,\varepsilon)) \cong \Bbb Z^2$, there are quite a lot. (Concretely: $T^2 = S^1 \times S^1$, so pick the inclusion $\varphi(x) = (x,1,\varepsilon/2)$. That this is not null-homotopic can be proved in a first algebraic topology course after learning that the identity map $S^1 \to S^1$ is not null-homotopic.)

If $U$ was simply connected, then this map would be null-homotopic in $U$. That is, there would be a map $\tilde \varphi: D^2 \to U$ that extended $\varphi$ (in the sense that $\tilde \varphi = \varphi$ on $S^1 \subset D^2$.) Because $U \subset T^2 \times (0,\infty)$, it would be null-homotopic in $T^2 \times (0,\infty)$ as well. But that's precisely the trouble! What an algebraic topologist would say is that the map $T^2 \times (0,\varepsilon) \to T^2 \times (0,\infty)$ is a homotopy equivalence, hence induces an isomorphism on all homotopy groups. Let's be concrete.

Consider the map $f: T^2 \times (0,\infty) \to T^2 \times (0,\infty)$ given by $f(x,t) = (x,\varepsilon/2)$. Now consider $f\tilde \varphi$. This is a map whose image lies in $T^2 \times (0,\varepsilon)$ and that agrees with $\varphi$ on the boundary circle $S^1$. So it is a null-homotopy of our map $S^1 \to T^2 \times (0,\varepsilon)$; this contradicts our choice of $\varphi$. So there was no such $U$ after all!

This exact same argument proves that the cone on any non-simply connected manifold is not a manifold; with some more tools one can prove that the cone on a manifold whose homology is not that of $S^n$ is not a manifold; and then putting these together and the Whitehead theorem and the Poincare conjecture (which is now a theorem topologically in all dimensions) one concludes that the cone on a manifold $M$ is only a manifold if $M = S^n$ for some $n$.