Birational invariants and fundamental groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:54:37Z http://mathoverflow.net/feeds/question/33021 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33021/birational-invariants-and-fundamental-groups Birational invariants and fundamental groups Tom Goodwillie 2010-07-22T22:53:16Z 2010-07-24T21:01:02Z <p>In pondering <a href="http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities" rel="nofollow">this</a> MO question and people's efforts to answer it, and recalling also something that I learned in my youth about using Morse theory ideas to prove some results of Lefschetz in the complex case, I seem to have learned two things -- things that I suppose are absorbed in the cradle by those who study algebraic geometry as opposed to learning it by osmosis -- or maybe I haven't got them quite right. Anyway:</p> <p>(1) Blowing up a surface at a smooth point does not change the fundamental group, and (therefore ?) the fundamental group of a smooth projective surface is a birational invariant.</p> <p>(2) Surfaces in $\mathbb P^3$ are simply connected, and more generally for a set $X\subset\mathbb P^n$ defined by a single homogeneous equation of degree $>0$ the pair $(\mathbb P^n,X)$ is at least $(n-1)$-connected. That is, the relative homotopy groups and therefore the relative homology groups vanish up through dimension $n-1$.)</p> <p>(This is all over the complex numbers.) </p> <p>Is this correct? And, taking off from (1), what are some other simple statements about invariants from homotopy theory that are birational invariants? And what are the first things to know about birational invariants that do not come from topology?</p> <p>EDIT I wish I could accept more than one answer.</p> http://mathoverflow.net/questions/33021/birational-invariants-and-fundamental-groups/33024#33024 Answer by algori for Birational invariants and fundamental groups algori 2010-07-22T23:46:53Z 2010-07-23T01:39:00Z <ol> <li><p>True since topologically blowing up a point is taking a connected sum with $\overline{ \mathbf{P}^2(\mathbf{C})}$.</p></li> <li><p>True for smooth surfaces (generally, for smooth projective hypersurfaces, but it suffices to prove this for surfaces; I will try to give more details a bit later). If $X$ is a smooth hypersurface of $\mathbf{P}^n$, then moreover the homology map induced by the inclusion is an iso in degree $\leq n-2$ and is surjective in degree $n-1$ (this follows from the Lefschetz hyperplane theorem as given e.g. in Griffiths-Harris, chapter 1).</p></li> </ol> <p>upd Here is a sketch of the proof of the fact that smooth projective hypersurfaces are simply connected. If we have a positive holomorphic line bundle $L$ on a complex analytic manifold $M$ (i.e. a line bundle whose Chern class is represented by $-i$ times a positive linear combination of $dz_j\wedge d\bar{z}_j$'s) and a smooth section $s$ of $L$ then we can equip the bundle with a hermitian metric so that the curvature will be $-2\pi i$ times the "positive" representative of the Chern class. We can then consider the function $\log |s|^2$ on the complement of a tubular neighborhood of the zero locus $V$ of $s$. A local calculation shows that at each singular point the Hessian of this function has at least $\dim M$ negative eigenvalues. I.e. $M$ is homotopy equivalent to $V$ with some cells of dimension $\geq n$ attached to it. For more details see Griffiths-Harris, chapter 1, Lefschetz hyperplane theorem.</p> http://mathoverflow.net/questions/33021/birational-invariants-and-fundamental-groups/33061#33061 Answer by damiano for Birational invariants and fundamental groups damiano 2010-07-23T07:43:41Z 2010-07-23T07:43:41Z <p>I would like to mention one more homotopy invariant of smooth projective varieties that is also a birational invariant. If $X$ is a smooth projective variety, then the torsion subgroup $T(X)$ of ${\rm H}^3(X,\mathbb{Z})$ is a birational invariant. This is explained in the beautiful paper "Some elementary examples of unirational varieties which are not rational" by Artin-Mumford, which starts exactly outlining a "homotopical" approach to showing that there exist unirational threefolds that are not rational. Their homotopic criterion is that rational varieties $X$ have trivial group $T(X)$ and they construct unirational varieties with non-trivial two-torsion in such group, showing that these are examples of unirational, non rational varieties.</p> <p>Finally, for surfaces the quantity $K^2+\rho$ is a birational invariant, where $K^2$ is the self intersection of the canonical divisor class on $X$ and $\rho$ is the rank of the N\'eron-Severi group of $X$). This is not too deep, as it is immediate knowing that a birational map of surfaces is a composition of blow ups and blow downs of smooth points (which for surfaces is quite easy!) and the above quantity is obviously invariant under a blow up. The reason for mentioning it, is that the N\'eron-Severi group can be defined in terms of Hodge theory, which, while not entirely homotopical, certainly has a homotopical feel to it. Moreover, many birational invariants are defined using Hodge structures, so it seemed useful to point it out!</p> http://mathoverflow.net/questions/33021/birational-invariants-and-fundamental-groups/33071#33071 Answer by Francesco Polizzi for Birational invariants and fundamental groups Francesco Polizzi 2010-07-23T09:45:58Z 2010-07-24T21:01:02Z <p>The paper of Keum and Zhang </p> <p>"Fundamental groups of open K3 surfaces, Enriques surfaces and Fano 3-folds", Journal of Pure and Applied Algebra vol. 170</p> <p>provides some answers to Tom's question "What happens to the fundamental group of a singular variety when removing the singular points?"</p> <p>It appears that the fundamental group of the smooth part can be quite complicate also in simple situations. For instance, one of the results in the paper is the following:</p> <p>"THEOREM. Let X be a $K3$ surface with at worst Du Val singularities (then X is still simply connected), and let $X^0$ be its smooth part. The number $c$=#(Sing $X$) is bounded by $16$, and if $c=16$ then $\pi_1(X^0)$ is infinite".</p> <p>So, given for instance a quartic surface $X \subset \mathbb{P}^3$ with $16$ nodes (a Kummer surface) we have </p> <p>$\pi_1(X)={1}$, but $\pi_1(X^0)$ is infinite! </p> <p>This follows from the fact that $X^0$ has an étale $\mathbb{Z}_2$-cover $Y^0 \to X^0$, where $Y^0$ is an Abelian surface minus 16 points.</p> <p>For smaller values of $c$, the group $\pi_1(X^0)$ is finite, but not trivial in general. </p> <p>In higher dimension, there is the following</p> <p>"CONJECTURE. Let $V$ be a $\mathbb{Q}$-Fano $n$-fold. Then the topological fundamental group $\pi_1(V^0)$ of the smooth part $V^0$ of $V$ is finite". </p> <p>(a normal variety $V$ with at worst log terminal singularities is $\mathbb{Q}$-Fano if, by definition, the anti-canonical divisor $−K_V$ is $\mathbb{Q}$-Cartier and ample).</p>