What is the fundamental group of the variety of singular matrices, without $0$?

Consider $S = V(det) \subset M_n(\mathbb{C})$. I want to know what the fundamental group of $S \setminus {0}$ is. Equivalently, I want to know the fundamental group of $V(det) \subset \mathbb{P}^{n^2 - 1}$. (Using the homotopy long exact sequence.)

In some cases it's easy - if $n = 2$, then the variety is a full rank (smooth) quadric hypersurface in $P^3$, and is thus isomorphic to $P^1 \times P^1$, which is simply connected. If $n= 3$, I already have no idea how to proceed. It is also non-smooth for $n \geq 3$.

The result is a variety with a lot of symmetry - conjugation by $GL_n(\mathbb{C})$, left and right multiplication by $GL_n(C) \times GL_n(C)$. There are finitely many orbits for the latter action, classified by rank, and the closure of each orbit is the union of the lower rank orbits. Maybe this is useful?

To put something at stake, I guess that it is simply connected for all $n$. What about the higher homotopy groups? (Is the limit with the natural inclusions as $n \to \infty$ contractible?)

This is thanks to Mohan's comment.

The Lefschetz hyperplane theorem states (among other things):

If $X$ is an $n$-dimensional projective variety in $P^N$, and $Y$ is a hyperplane section of $X$ so that $X \setminus Y$ is smooth, then for $k < n -1$, the map induced by inclusion $\pi_k(Y) \to \pi_k(X)$ is an isomorphism, and is surjective for $k = n - 1$.

In particular, we let $v : X = P^{n^2 - 1} \to P^N$ be the degree $n$ veronese, so that $V(det) \subset X$ becomes a hyperplane section with smooth complement. Then by the theorem the first $n^2 - 2$ homotopy groups agree with those of $P^{n^2 - 1}$.

The homotopy groups of $CP^M$ can be related to those of $S^{2M + 1}$ via the fibration $U(1) \to S^{2M + 1} \to CP^M$. In particular, for $2 < k < 2M + 1$, $\pi_k(CP^M) = \pi_k(S^{2M + 1}) = 0$, and we get $\pi_2(CP^M) = \mathbb{Z}$ and $\pi_1(CP^M) = 0$. (That $CP^M$ is simply connected can also been seen from its 2 skeleton being a sphere.)

So the variety of singular matrices is always simply connected, but $\pi_2$ is $\mathbb{Z}$ as long as $2 < n^2 - 2$. (In particular, this tells us nothing about $\pi_2(CP^1 \times CP^1)$. That we can compute with the Hurewicz map since it is $2 - 1$ connected, so $\pi_2 (CP^1 \times CP^1) = H_2(CP^1 \times CP^1) = Z^2$, the later equality using Kunneth.)

The homotopy groups are then trivial until $\pi_{n^2 - 2}$, where all we know from the Lefschetz theorem is that $\pi_{n^2 - 2} (V(det))$ surjects onto $\pi_{n^2 - 2}(CP^{n^2 - 1}) = \pi_{n^2 - 2}(S^{2n^2}) = 0$, which is no information. I computed this group to be $\mathbb{Z}^2$ for the case $n = 2$, so I guess in general they are non-trivial. (I can conjecture that $\pi_{2n^2- 2}(det(X)) = \mathbb{Z}^n$, but jumping on the most obvious pattern.)

The higher homotopy groups are presumably mysterious.