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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 $\pi_2(\mathbb{P}^n)\rightarrow \pi_1(\mathbb{C}^*)$.

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    $\begingroup$ You can use van Kampen's theorem to compute this, where the two opens $U$ and $V$ are the inverse images of the standard opens $\mathbb{C}$ in $\mathbb{CP}^1$ whose intersection is $\mathbb{C}^\times$. $\endgroup$ – Jason Starr Sep 28 '17 at 12:47
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    $\begingroup$ Anyway, the answer is that the fundamental group is isomorphic to $\mathbb{Z}/k\mathbb{Z}$. $\endgroup$ – Jason Starr Sep 28 '17 at 12:50
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    $\begingroup$ Now I see that $n$ is arbitrary. I thought the OP specified that $n$ should equal $2$. So you need more than two "standard opens" to cover $\mathbb{CP}^n$. However, it does not change the answer. You can remove a codimension $2$ linear subvariety from $\mathbb{CP}^n$, and its inverse image in $L^\times$, without changing the fundamental group. After doing that, there is a covering by two opens $\mathbb{C}^n$ whose intersection is $\mathbb{C}^\times \times \mathbb{C}^{n-1}$. So the same computation works. $\endgroup$ – Jason Starr Sep 28 '17 at 12:55
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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.

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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.

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