fundamental groups of curves

2011-03-03T03:02:25Z

I saw the following statement made without proof in a paper of Bogomolov and Tschinkel:

If $X$ is an algebraic surface, and $C$ is an ample smooth curve in $X,$ then the fundamental group of $C$ surjects onto that of $X.$

I was wondering if someone could supply a reference, and perhaps some idea of what the most general version of this statement was...

Answer by J.C. Ottem for fundamental groups of curves

2011-03-03T03:26:17Z

There is a general Lefschetz hyperplane theorem for the homotopy groups of an ample divisor $D$ on a smooth complex variety $X$. Basically, this theorem says that the relative homotopy groups $\pi_i(X,D)$ are zero for all $i$ less than $\dim X$. In particular, the map $\pi_1(D)\to\pi_1(X)$ is an isomorphism for $\dim X\ge 3$ and surjective for $\dim X= 2$. There is a very nice proof of this theorem using Morse theory which can be found in Lazarsfeld's book 'Positivity in algebraic geometry'.

Answer by ACL for fundamental groups of curves

2011-03-03T17:18:08Z

There is also an algebraic proof which works over fields of positive characteristic.

See, e.g., Lemma 5.1 of the paper Zariski's conjecture and related problems by Madhav Nori (Annales scientifiques de l'École Normale Supérieure, Sér. 4, 16 no. 2 (1983), p. 305-344).

The proof of that Lemma is more general than what you need. It goes as follows. Let $X$ be a proper smooth algebraic surface and let $A\subset X$ be an ample curve. That $\pi_1(A)$ surjects onto $\pi_1(X)$ means the following: any finite étale cover $f:Y\to X$ which is split over $A$ is split. So if $Y$ is connected, $\deg(f)=1$.

Let us prove that if $A$ is ample; in fact, we only need $A$ big and nef. If $f$ has a section on $A$, one can write $f^{-1}(A)=B+R$ where $B\to A$ is an isomorphism, and $R$ is disjoint from $B$. The Hodge index theorem says that the intersection form restricted to the space generated by $B$ and $R$ has at most one +-sign. Since $(B+R)^2=(f^*A)^2=\deg(f) A^2>0$, it has exactly one +-sign, and the determinant $$\begin{vmatrix} (B+R)^2 & (B+R)\cdot B \cr B\cdot (B+R) & B^2\end{vmatrix} $$ is nonpositive. By the projection formula, one has $$(B+R)\cdot B=f^*A\cdot B=A\cdot f_*B=A^2.$$ Since $B$ and $R$ are disjoint, we obtain $B^2=A^2$. Then, $$ \deg(f)A^2 = (B+R)^2=B^2+R^2, $$ so $$R^2=(\deg(f)-1) A^2$. The above determinant is equal to $ (\deg(f)-1)A^2$. Since $A^2>0$, $\deg(f)\leq 1$.

This proof generalizes to the so-called Ramanujam lemma according to which an effective divisor on a surface which is big and nef is numerically connected (doesn't decompose as the sum of two nonzero effective divisors with 0-intersection), hence connected. In our case, the existence of the section implies that $f^*A$ is not numerically connected; it is however big and nef because this property is stable under finite pull-back.

See also the paper of J-B. Bost, Potential theory and Lefschetz theorems for arithmetic surfaces (Annales scientifiques de l'École Normale Supérieure, Sér. 4, 32 no. 2 (1999), p. 241-312) where this argument is explained for surfaces and adapted for arithmetic surfaces.

fundamental groups of curves - MathOverflow

fundamental groups of curves

Answer by J.C. Ottem for fundamental groups of curves

Answer by ACL for fundamental groups of curves