If $X$ is a smooth projective variety of dimension at least $3$ over $\mathbb{C}$, Lefschetz's Hyperplane theorem says that for every hyperplane section $H$ $$\pi^1(H)\to\pi^1(X)$$ is an isomorphism, where $\pi^1$ can be taken to be the topological fundamental group or the étale one (as it is the profinite completion of the topological).

The same holds for the étale fundamental group over any field in any characteristic (using Leff conditions, e.g. SGA 2, XII Cor. 3.5), when $X$ is a smooth projective variety and $H$ is a regular hyperplane section.

Over the complex numbers, though, more is true: if $X$ is any smooth quasi-projective variety (of dimension at least $3$) and $H$ is a general hyperplane section, then $$\pi^1(H)\to\pi^1(X)$$ is again an isomorphism, as far as I know this is proven using Morse Theory (see e.g. Sec. 5.1 in Goresky and MacPherson "Stratified Morse Theory").

My question is:

Could one expect a similar statement in positive characteristic? Is there a counterexample?

Note that the tame part of the fundamental group should not pose any problem, as there is a Lefschetz theorem for it (once we have a good compactification).

EDIT: I always thought that generic is commonly used for "at the generic point", general for "every rational point of something containing an open" and very general for "every rational point of something dense" (in this case, in the projective space parametrizing hyperplanes). Hence by general I meant for every hyperplane in some open of the parametrizing space.

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    $\begingroup$ Surjectivity for the algebraic fundamental group (the only one defined in positive characteristic) follows from Bertini's theorem as proved, for instance, in Jouanolou's book. $\endgroup$ – Jason Starr Feb 9 '16 at 12:47
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    $\begingroup$ You could be interested in my paper arxiv.org/abs/1203.2595. I will possibly think whether its methods can give an answer to your question. Would a statement like "the morphism of $\pi_1$s is surhective with the kernel being a pro-p-group" (where p is the characteristic of the base field) satisfy you? $\endgroup$ – Mikhail Bondarko Feb 9 '16 at 17:46
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    $\begingroup$ Dear Jason, I'm confused about your comment due to Will 's answer. Would you be so kind as to provide the exact reference? $\endgroup$ – Giulia Feb 10 '16 at 6:57
  • $\begingroup$ Dear Mikhail, thank you for the link. If it is not injective, I do expect the kernel to be a pro-p group : as said there is a Lefschetz theorem for the tame part, in particular for the prime-to-p part, hence all that can go wrong lives in a subgroup whose prime-to-p quotient is zero. $\endgroup$ – Giulia Feb 10 '16 at 7:07
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    $\begingroup$ @Giulia: "I do expect the kernel to be a pro-$p$ group: as said there is a Lefschetz theorem for the tame part, in particular for the prime-to-$p$ part, hence all that can go wrong lives in a subgroup whose prime-to-$p$ quotient is zero." This is not quite correct. The tame fundamental group is the quotient of the fundamental group by the normal subgroup generated by all $p$-subgroups. However, it can happen that this normal subgroup is not itself a $p$-group, e.g., if $n> p>2$, then the alternating group $\mathfrak{A}_n$ is generated by $p$-cycles. $\endgroup$ – Jason Starr Feb 10 '16 at 14:42

This won't work for arbitrary smooth quasiprojective varieties. Take $X = \mathbb A^n$, then certainly any $H = \mathbb A^{n-1}$,.

But $\pi_1(\mathbb A^{n-1}) \to \pi_1(\mathbb A^n)$ is not an isomorphism in characteristic $p$ because there are nontrivial finite etale coverings of $\mathbb A^1$, which when pulled back to $\mathbb A^n$ become nontrivial coverings of $\mathbb A^n$ that trivialize when restricted to $\mathbb A^{n-1}$.

  • $\begingroup$ Dear Will, this implies that the map is not surjective. Do you have an idea about the likelihood of injectivity? Thank you. $\endgroup$ – Giulia Feb 10 '16 at 6:55
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    $\begingroup$ @Giulia. Will is correct, but also I am (sort of) correct. Let $X\subset \mathbb{P}^n$ be a quasi-projective variety over an algebraically closed field $k$. By Bertini's theorem, for every connected, finite, etale cover $Y\to X$, there exists a dense, Zariski open subset $U\subset (\mathbb{P}^n_k)^\vee$ such that for every $[H] \in U(k)$, also $Y\times_{\mathbb{P}^n_k} H \to X\times_{\mathbb{P}^n_k} H$ is a connected, finite, etale cover. Thus, for the geometric generic point $[H]$ of $(\mathbb{P}^n_k)^\vee$, the induced map $\pi^1(X\times_{\mathbb{P}^n_k} H) \to \pi^1(X)$ is surjective. $\endgroup$ – Jason Starr Feb 10 '16 at 12:06

I am posting my comments above as an answer, together with a reference (per the OP's request). The difference between Will's answer and my answer results from the ambiguous term "general". Only the OP can specify what is meant by "general" in the question. In this answer, I interpret "general" as meaning "the property is true for the geometric generic point".

Let $k$ be an algebraically closed field. Let $X$ be an integral $k$-scheme of dimension $\geq 2$, and let $f:X\to \mathbb{P}^n_k$ be a finite type morphism that is unramified on a dense, open subscheme of $X$. Denote by $(\mathbb{P}^n_k)^\vee$ the projective space parameterizing hyperplanes in $\mathbb{P}^n_k$. In particular, for every algebraically closed field extension $\kappa/k$, there is a natural bijection between the hyperplanes $H\subset \mathbb{P}^n_\kappa$ and the $k$-morphisms $[H]:\text{Spec}(\kappa) \to (\mathbb{P}^n_k)^\vee$.

Let me restate Corollary 2.2 of the following (which, in turn, depends on Théorème 4.10 and 6.10 of Jouanolou's "Théorèmes de Bertini et applications").

Graber, Tom; Starr, Jason Michael
Restriction of sections for families of abelian varieties.
A celebration of algebraic geometry, 311–327,
Clay Math. Proc., 18, Amer. Math. Soc., Providence, RI, 2013.

For every generically finite morphism $g:Y\to X$ that admits no rational section, there exists a dense, Zariski open subscheme $U\subset (\mathbb{P}^n_k)^\vee$ such that for every algebraically closed field $\kappa$ and for every geometric point $[H]:\text{Spec}(\kappa) \to U$, also $Y\times_{\mathbb{P}^n_k} H \to X\times_{\mathbb{P}^n_k} H$ admits no rational section.

Now let $K$ denote the algebraic closure of the function field of $(\mathbb{P}^n_k)^\vee$ as an extension of $k$, and denote by $[H]:\text{Spec}(K)\to (\mathbb{P}^n_k)^\vee$ the corresponding $k$-morphism. Then for every connected, finite, etale morphism $g:Y\to X$, also the morphism $Y\times_{\mathbb{P}^n_k} H \to X\times_{\mathbb{P}^n_k} H$ is connected, finite and etale. Therefore, the induced homomorphism, $$\pi^1(X\times_{\mathbb{P}^n_k} H) \to \pi^1(X),$$ is surjective.


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