# Do the singular matrices form a topological manifold

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.

• "Singularity" is generally used to describe a point where the rules that hold everywhere else in a neighborhood of the point fail to hold at the point itself. So in the context of topological manifolds, a singularity would exactly be an isolated point with no neigborhood homeomorphic to $\Bbb R^n$. That isn't a cop out. That is the exact description of the condition of concern. – Paul Sinclair Dec 6 '15 at 23:55
• @PaulSinclair, why would it have to be an isolated point? The singularity at $0$ of $y^2=x^3$ given the subspace topology induced from $\mathbb{R}^2$ is not isolated I don't think. – Thoth Dec 7 '15 at 0:06
• As far as feeling it is a cop out definition I guess because it's a negative definition saying that something doesn't exist, which might make it difficult to check if the property holds. – Thoth Dec 7 '15 at 0:12
• How is it not isolated? It is only at $0$ that this curve does not have a neighborhood homeomorphic to $\Bbb R$. The term "singularity" can be used for higher dimensional anomalies, I suppose, but if it disconnects neighborhoods, the term "boundary" would be more appropriate. – Paul Sinclair Dec 7 '15 at 3:17
• @PaulSinclair, isolated (at least in the topological sense) means there is an open set which contains only the point, which as far as I can tell isn't true for $y^2=x^3$. – Thoth Dec 7 '15 at 3:26

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 $V \backslash\{0\}$.

• I'm pushing the limits of my knowledge trying to understand this answer, could you explain why the cone over the torus is given by the quotient topology $T\times[0,\infty)/T\times\{0\}$? Wouldn't this just be equal to $[0,\infty)$? It seems to me it should instead just be $T\times[0,\infty)$. – Thoth Dec 14 '15 at 0:08
• @Toth: when $r=0$ you only get one point, not the full torus. Think of a simpler example, the $3$ dimensional space with coordinates spherical coordinates $(r, \phi, \tau)$. Well $r \in [0,\infty)$ but when $r=0$ you only get one point. So the space is not $S \times [0, \infty)$, but a quotient of that. – Orest Bucicovschi Dec 14 '15 at 0:14
• Ohh!! I get it! – Thoth Dec 14 '15 at 0:16
• I'm afraid I don't know enough topology to to follow your argument once you introduce the path in $T\times(0,\epsilon)$. So I understand that any ball in $\mathbb{R}^3$ with a point removed is simply connected (any path can be deformed into any other), and I understand that this path you've introduced travels strictly on positive radii tori, but is there any way you could dumb down or spell out your discussion of its first coordinate being not null-homotopic and how this implies that $V\backslash\{0\}$ is not simply connected? – Thoth Dec 14 '15 at 2:18

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 $n$.

• thanks for expanding on Orangeskid's answer, if I could give you both the bounty I would. I don't fully follow what you wrote but I can sort of 'see' it, I'll probably return to your answer and try to fully understand it once I have some more geometry and topology under my belt. – Thoth Dec 17 '15 at 8:20