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 $\pi\_2(\mathbb{P}^n)\rightarrow \pi\_1(\mathbb{C}^\*)$.



What is the fundamental group of $\mathcal O_{\mathbb P^n}(k)$ minus the zero section
The fibration $\mathbb{C}^\times\to L^\times\to\mathbb{P}^n$ can be "delooped" to a fibration $L^\ast \to\mathbb{P}^n\to{\rm B}\mathbb{C}^\times$ where the last map is the classifying map for the line bundle. Now we have ${\rm B}\mathbb{C}^\times\cong\mathbb{P}^\infty$, and we want to identify the map $\pi\_2(\mathbb{P}^n)\to\pi\_2(\mathbb{P}^\infty)\cong\pi\_1(\mathbb{C}^\times)$ induced by the classifying map for $\mathcal{O}(k)$. Note that the generator for $\pi\_2(\mathbb{P}^\infty)$ is given by the natural inclusion $\mathbb{P}^1\to\mathbb{P}^\infty$ classifying $\mathcal{O}(1)$. In particular, the map $\pi\_2(\mathbb{P}^n)\to\pi\_2(\mathbb{P}^\infty)$ classifying $\mathcal{O}(k)$ must be multiplication by $k$. Of course, the result is the same as in Jason Starr's comments. Just another funny way to see it.


