Punctured complex projective space

Question

Let $\mathcal{P}\mathbb{C}^{n}$ be the complex projective space of $\mathbb{C}^{n+1}$, and let $B=\{\mathbf{e}_{1},\cdots,\mathbf{e}_{n+1}\}$ be a basis in $\mathbb{C}^{n+1}$.

I would like to understand what happens to $\mathcal{P}\mathbb{C}^{n}$ if I remove the points in it corresponding to the basis vectors in $B$.

In the case $\mathbb{C}^{2}$ I know that $\mathcal{P}\mathbb{C}^{1}\cong S^{2}$ and, from the quantum mechanical picture of $\mathcal{P}\mathbb{C}^{1}$ as the Bloch sphere, I know that the points in $\mathcal{P}\mathbb{C}^{1}$ corresponding to two basis vectors in $\mathbb{C}^{2}$ are just antipodal points. Consequently, removing these points from $\mathcal{P}\mathbb{C}^{1}$ I get a cilinder $S^{1}\times\mathbb{R}$.

Unfortunately, this visually-inspired procedure does not work in higher dimensions, and I do not know how to even start to face the problem, therefore, I appreciate any comment, suggestion or reference.

Thank You.

EDIT

I thought of something.

Let $\mathbf{E}_{j}$ be the rank-one projector associated to $\mathbf{e}_{j}\in B$, let $\mathbf{H}=\sum_{j}\nu_{j}\mathbf{E}_{j}$ be a self-adjoint operator on $\mathbb{C}^{n+1}$ such that the one-parameter group of unitary operators $\mathbf{U}_{\tau}:=\exp(-\imath\tau\mathbf{H})$ on $\mathbb{C}^{n+1}$ generated by $\mathbf{H}$ is a closed subgroup of the unitary group $U(n+1)$. Therefore, $\mathbf{U}_{\tau}$ is an action of the circle group $U(1)\cong S^{1}$ on the complex projective space $\mathcal{P}\mathbb{C}^{n}$. The fixed points of this action are just the points corresponding to the elements of $B$.

Let $\mathcal{P}\mathbb{C}^{n}_{*}$ denotes the complex projective space without the fixed points of the action. Since $U(1)\cong S^{1}$ is a compact group, its action on $\mathcal{P}\mathbb{C}^{n}_{*}$ is proper. Furthermore, it is free by construction. This means that the orbit space $\mathcal{P}\mathbb{C}^{n}_{*}/U(1)\equiv M$ is a differential manifold, the canonical projection $\pi:\mathcal{P}\mathbb{C}^{n}_{*}\rightarrow M$ is a surjection, and $\left(\mathcal{P}\mathbb{C}^{n}_{*}\,;M\,;\pi\,;U(1)\right)$ is a $U(1)$-principal bundle.

At this point, if the bundle is trivial, we have that $\mathcal{P}\mathbb{C}^{n}_{*}\cong M\times U(1)$, however, I am not able to go further.

Answer 1

Since all points on equivalent on $\mathbb{C}P^n$, WLOG, let $p = [1,0,0,\cdots]$. Then I claim $\mathbb{C}P^n-p$ homotopy retracts to $\mathbb{C}P^{n-1}$.

To see this explicitly, define the homotopy to be $$[a,b,c,\cdots]\mapsto [(1-t)a,b,c,\cdots].$$

or

To see this abstractly, the projection $q:\mathbb{C}P^n-p\to \mathbb{C}P^{n-1}$ defined by $$ q:[a,b,c,\cdots]\mapsto [b,c,\cdots] $$ is a fibration with fiber $\mathbb{C}$ which is contractible. The LES of the fibration says that $q$ is a weak homotopy equivalence, by Whitehead's theorem this is a homotopy equivalence.