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(A related question is this On the fundamental group of hypersurfaces).

Let $X$ be a simply connected projective complex manifold of dimension at least 3. Let $Y\subset X$ be a smooth hypersurface. What is $\pi_1(Y)$ then? If $Y$ is a hyperplane section, then $\pi_2(X,Y)=0$ by the Lefschetz theorem, hence $\pi_1(Y)=0$. The question is, can we say anything about the fundamental group of a hypersurface without assuming ampleness?

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2 Answers 2

up vote 6 down vote accepted

I think the answer is no. Take any smooth projective surface $S$ in $\mathbb{P}^N$, and choose a generic projection to $\mathbb{P}^3$. The image is a singular surface $S'$ (with a double curve, some triple points etc.). There exists a birational morphism $P\rightarrow \mathbb{P^3}$ such that the strict transform of $S'$ in $P$ is isomorphic to $S$. Then $P$ is simply-connected, while $S$ is arbitrary.

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You are probably right. But I am not sure I understand why the strict transform is isomorphic to $S$, not only birationally equivalent. –  Alex Gavrilov Jul 2 at 6:52
    
Even if it is only birationally equivalent, its $\pi_1$ is isomorphic to that of $S$, hence is arbitrary. –  abx Jul 2 at 8:51

I think that it is hard to say something is such a generality, as the following example shows.

Take $X= \mathbb{P}^1 \times \mathbb{P}^2$ and let $C \subset \mathbb{P}^2$ be a smooth curve of genus $g$. Set $Y = \mathbb{P}^1 \times C$.

Then $\pi_1(Y) = \pi_1(C)$. The latter group is trivial for $g=0$, it is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$ for $g=1$ and it is a infinite nonabelian group for $g \geq 2$.

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