TITLE: What is the fundamental group of $\mathcal O_{\mathbb P^n}(k)$ minus the zero section
QUESTION [8 upvotes]: Let $L^*$ be the total space of the line bundle $\mathcal{O}_{\mathbb{P}^n}(k)$ minus its zero section.
How can one compute the fundamental group of $L^*$?
For k = 0 the space $L^*$ is $\mathbb{P}^n \times \mathbb{C}^*$ hence $\pi_1(L^*) = \mathbb{Z}$.
For k=-1 the $L^*$ is $\mathbb{C}^{n+1} \setminus \{0\}$, therefore $\pi_1(L^*) = 0$.
What about the other $k$ ?
The long exact sequence of homotopy of a Serre fibration $\mathbb{C}^* \rightarrow L^* \rightarrow \mathbb{P}^n$ gives
$\pi_2(\mathbb{C}^*) = 0 \rightarrow \pi_2(L^*) \rightarrow \pi_2(\mathbb{P}^n) \simeq \mathbb{Z} \rightarrow \pi_1(\mathbb{C}^*)\simeq \mathbb{Z} \rightarrow  \pi_1(L^*) \rightarrow \pi_1(\mathbb{P}^n) = 0$.
So one needs to understand the map $\pi_2(\mathbb{P}^n)\rightarrow \pi_1(\mathbb{C}^*)$.

REPLY [2 votes]: Another way to see that the fundamental group is $\mathbb{Z}/k$ is using geometry of cyclic quotient singularities and it goes as follows. 
We consider the negative twists first. In this case the total space $\mathcal{O}_{\mathbb{P}^n}(-k)$ is the resolution of singularities of the quotient space $\mathbb{C}^{n+1} / G$ with $G=\mathbb{Z}/k$ acting diagonally by $k$-th roots of unity. To see this one computes the coordinate algebra of the quotient which is precisely the affine cone over the degree $k$ Veronese embedding of $\mathbb{P}^n$. Now the latter cone is resolved by a single blow up of its vertex and the resulting space is the total space $\mathcal{O}_{\mathbb{P}^n}(-k)$.
In particular the complement to the zero section of $\mathcal{O}(-k)$ is isomorphic to the punctured cone $(\mathbb{C}^{n+1} - 0) / G$, and so has fundamental group isomorphic to $G$ as $\mathbb{C}^{n+1} - 0$ is simply-connected. 
Now if the twist was positive, contracting zero section is impossible, but replacing $k$ by $-k$ makes a homeomorphic space, so it is $\mathbb{Z}/k$ in both cases; and the answer is $\mathbb{Z}$ for $k=0$ as the bundle is trivial then.