fundamental groups of curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:39:11Z http://mathoverflow.net/feeds/question/57197 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57197/fundamental-groups-of-curves fundamental groups of curves Igor Rivin 2011-03-03T03:02:25Z 2011-03-03T17:34:21Z <p>I saw the following statement made without proof in a paper of Bogomolov and Tschinkel:</p> <p>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 $X.$</p> <p>I was wondering if someone could supply a reference, and perhaps some idea of what the most general version of this statement was...</p> http://mathoverflow.net/questions/57197/fundamental-groups-of-curves/57201#57201 Answer by J.C. Ottem for fundamental groups of curves J.C. Ottem 2011-03-03T03:26:17Z 2011-03-03T17:34:21Z <p>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'.</p> http://mathoverflow.net/questions/57197/fundamental-groups-of-curves/57267#57267 Answer by ACL for fundamental groups of curves ACL 2011-03-03T17:18:08Z 2011-03-03T17:18:08Z <p>There is also an algebraic proof which works over fields of positive characteristic.</p> <p>See, e.g., Lemma 5.1 of the paper <a href="http://www.numdam.org/item?id=ASENS_1983_4_16_2_305_0" rel="nofollow">Zariski's conjecture and related problems</a> by Madhav Nori (Annales scientifiques de l'École Normale Supérieure, Sér. 4, 16 no. 2 (1983), p. 305-344).</p> <p>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, $\deg(f)=1$. </p> <p>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 $$\begin{vmatrix} (B+R)^2 &amp; (B+R)\cdot B \cr B\cdot (B+R) &amp; B^2\end{vmatrix}$$ 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$.</p> <p>This proof generalizes to the so-called Ramanujam lemma according to which an effective divisor on a surface which is big and nef is numerically connected (doesn't decompose as the sum of two nonzero effective divisors with 0-intersection), hence connected. In our case, the existence of the section implies that$f^*A\$ is not numerically connected; it is however big and nef because this property is stable under finite pull-back.</p> <p>See also the paper of J-B. Bost, <a href="http://www.numdam.org/item?id=ASENS_1999_4_32_2_241_0" rel="nofollow">Potential theory and Lefschetz theorems for arithmetic surfaces</a> (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.</p>