Homotopy type of intersection of complement of hyperplanes in projective space.

Let $U_i = \{x=(x_0 :… :x_n) \in \mathbb{P}^n(\mathbb{C}); x_i \neq 0 \}$ be the usual trivialization of the complex projective space. I have been trying to compute the homotopy type of all the possible intersections $I_{i_1, …, i_m} = U_{i_1} \cap … \cap U_{i_m}$. I don't want to know exactly the value of the homotopy groups. Actually I just want to find an already known object (for instance, spheres, projective spaces, grasmannians, ...) such that $I_{i_1, …, i_m}$ is diffeomorphic to it.

I think that $\pi_0 (I_{i_1, …, i_m})$ is always trivial (but I'm not sure). Furthermore, $I_{0, …, n}$ is empty. However, even after some hours playing with this, I could not find any information about higher homotopy groups (including $\pi_1$).

So my questions are:

1) What's the homotopy type of $I_{i_1, …, i_m}$?

2)Is $I_{i_1, …, i_m}$ diffeomorphic to an already known object (for instance, spheres, grasmmanians, …)?

• Actually $I_{0,\dots, n}$ is not empty. For example, $[1, \dots, 1] \in I_{0,\dots, n}$. – Michael Albanese Jul 6 '15 at 17:26

Let $\{j_1, \dots, j_{n-m+1}\} = \{0, \dots, n\}\setminus\{i_1, \dots, i_m\}$ where $j_1 < j_2 < \dots < j_{n-m+1}$. Then there is a homeomorphism (even biholomorphism) $L : \mathbb{CP}^n \to \mathbb{CP}^n$ given by

$$L([x_0, \dots, x_n]) = [x_{i_1}, \dots, x_{i_m}, x_{j_1}, \dots, x_{j_{n-m+1}}].$$

Note that $L|_{I_{i_1,\dots,i_m}}$ is a homeomorphism from $I_{i_1,\dots,i_m}$ to $I_{0,\dots,m-1}$, so it is enough to determine the homotopy/homeomorphism type of $I_{0,\dots,m-1}$.

There is a map continuous $\varphi : I_{0, \dots, m-1} \to (\mathbb{C}^*)^{m-1}\times\mathbb{C}^{n-m+1}$ given by

$$\varphi([x_0, \dots, x_n]) = \left(\frac{x_1}{x_0}, \dots, \frac{x_{m-1}}{x_0}, \frac{x_m}{x_0}, \dots, \frac{x_n}{x_0}\right)$$

with continuous inverse $\varphi^{-1} : (\mathbb{C}^*)^{m-1}\times\mathbb{C}^{n-m+1} \to I_{0, \dots, m-1}$ given by

$$\varphi^{-1}(y_1, \dots, y_n) = [1, y_1, \dots, y_n].$$

Therefore $U_{0,\dots,m-1}$ is homeomorphic (again, even biholomorphic) to $(\mathbb{C}^*)^{m-1}\times\mathbb{C}^{n-m+1}$, and hence homotopy equivalent to $(S^1)^{m-1}$. This generalises the familiar situation of $\mathbb{CP}^1$ where $U_0\cap U_1$ is biholomorphic to $\mathbb{C}^*$ (topologically it is a two-sphere with two antipodal points removed), and is therefore homotopy equivalent to a circle.

• Thanks for your answer. I've just found out this by myself and I was going to delete the question, but since you answered…I need to sleep before asking questions here :) – user40276 Jul 6 '15 at 23:56