Fundamental group of the space $\Bbb CP^n-p(X)$

Question

Let $p : \Bbb C^{n+1} \to \Bbb CP^n$ be the standard quotient map, let $X=\{(z_0, ..., z_n ) \in \Bbb C^{n+1} ~|~ z_0^2 + ... +z_n^2=0 \}$, and consider the space $p(X)$. I want to compute the fundamental group of the space $\Bbb CP^n-p(X)$. To do this, I first want to know how does the space $p(X)$ look like, but I have no idea. Any hints?

Answer 1

First consider $V:=\{(z_0,\ldots,z_n)\in\mathbb{C}^{n+1}:z_0^2+\cdots+z_n^2=1\}$ and verify that the standard projection restricted to $V$ becomes a 2-sheeted covering map (actually it is sufficient to notice that $p\mid_V$ is a quotient map, but in this particular case it can be easily seen by noticing that $V$ is a cover).

To be specific, let an arbitary point $[z_0:\cdots:z_n]\in\mathbb{C}P^n$ be given. WLOG assume $z_0=1$. For sufficiently small $\epsilon>0$,

$\qquad U_\epsilon:=\{\,[1:w_1:\cdots:w_n]:(w_1,\ldots,w_n)\,\text{is in the $\epsilon$-neighborhood of}\,(z_1,\ldots,z_n)\,\text{in}\,\mathbb{C}^n.\}$

is a well-defined neighborhood of $[1:z_1:\cdots:z_n]$ both in $\mathbb{C}P^n$ and $\mathbb{C}P^n-p(X)$. Then $(p\mid_V)^{-1}(U_\epsilon)$ has two connected components $\widetilde{U_\epsilon}^{(1)}$ and $\,\widetilde{U_\epsilon}^{(2)}=-\widetilde{U_\epsilon}^{(1)}$ that evenly cover $U_\epsilon$, again provided that you chose sufficiently small $\epsilon$ (Check this by constructing explicit homeomorphisms from $U_\epsilon$ to its sheets!).

Now you can understand $\mathbb{C}P^n-p(X)$ via its cover $V$. Embed $V$ in $\mathbb{R}^{2(n+1)}$ instead of $\mathbb{C}^{n+1}$ to see that

$\qquad V=\{(\vec{x},\vec{y})\in\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}: |\vec{x}|^2=|\vec{y}|^2+1\quad\text{and}\quad\vec{x}\cdot\vec{y}=0\}$.

Construct an explicit homeomorphism from $V$ to

$\qquad TS^{n}=\{(\vec{x}',\vec{y}')\in\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}:|\vec{x}'|^2=1\quad \text{and}\quad \vec{x}'\cdot\vec{y}'=0\}$, the tangent bundle over $S^{n}$.

Since you're looking for the fundamental group, you can deformation retract $TS^{n}$ into an even simpler space, namely $S^{n}$; just send every tangent vector $\vec{y}'$ to $(1-t)\vec{y}'$ as $t\in [0,1]$.

To mimic $\,\mathbb{C}P^n-p(X)\,$ we shall quotient $TS^{n}$ and $S^{n}$ by the identification

$\qquad\mathbf{v}\sim\mathbf{-v}\quad\text{for every}\,\mathbf{v}.$

Observe that $\mathbb{C}P^n-p(X)\simeq TS^{n}/_{\sim}$ since they are quotients obtained from homeomorphic spaces by identifying (essentially) same sets of points.

The deformation retraction $TS^{n}\rightarrow S^{n}$ suggests our final deformation retraction

$\qquad TS^{n}/_{\sim}\rightarrow S^{n}/_{\sim}\simeq\mathbb{R}P^n$.

$\require{AMScd}$ \begin{CD} V @>\simeq>> TS^{n} @>d.r.>> S^n\\ @V p VV @VV quot. V @VV quot. V\\ \mathbb{C}P^n-p(X) @>>\simeq> TS^{n}/_{\sim} @>d.r.>>\mathbb{R}P^n \end{CD}

('d.r.' for deformation retractions, 'quot.' for quotient maps)

As it turns out, the fundamental group of $\;\mathbb{C}P^n-p(X)\;$ is that of $\;\mathbb{R}P^n\;$, to wit: $\mathbb{Z}\;(n=1), \mathbb{Z}_2\;(n\geq 2)$.

Answer 2

I don't know how to do this with algebra, but the space is diffeomorphic to the tangent bundle on $\mathbb RP^n$ (see my question and the answer here), and hence homotopic to $\mathbb RP^n$, so its fundamental group is $\mathbb Z/2$.

For a different approach: I think the suggestion in the comments is also good, since for a manifold of dimension at least three, you can remove a point from your space, and pass to affine coordinates, solving the questiion there (since the fundamental group will be unchanged, which is proven by seifert van kampen.)