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In pondering this 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:

(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.

(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 $n-1$.)

(This is all over the complex numbers.)

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?

EDIT I wish I could accept more than one answer.

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Yes for (1) and (2). So far so good! Dimensions of spaces cooked out of the holomorphic tangent bundle -- specfically spaces of $p$-forms, and spaces of sections of tensors powers of $K=\det T_X^*$ (called plurigenera) -- are birational invariants that algebraic geometers love. These are not homotopy invariants in general. Hope that helps. –  Donu Arapura Jul 22 '10 at 23:51
    
And if a surface $S\subset\mathbb P^2$ has just one singular point $x$ then is there some rule for describing the fundamental group of $S-x$ in terms of the (formal isomorphism class of?) the singularity? If the thing is formally like a cone on a smooth plane curve, do you always get the fundamental group of that curve? –  Tom Goodwillie Jul 23 '10 at 0:25
    
Tom -- cones over curves are simply-connected as well. –  algori Jul 23 '10 at 1:43
    
I suspect Tom's questions were all concerning $S-x$. I don't have answer, however. –  Donu Arapura Jul 23 '10 at 2:00
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I've heard of it, that's all. –  Tom Goodwillie Jul 23 '10 at 2:28
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3 Answers

up vote 10 down vote accepted

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.

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!

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  1. True since topologically blowing up a point is taking a connected sum with $\overline{ \mathbf{P}^2(\mathbf{C})}$.

  2. 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).

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.

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In SGA1 it is proved that the fundamental group is a birational invariant of smooth projective varieties over any algebraically closed field. It rests on Zariski-Nagata purity, if I rememeber correctly. No need for deep results on factorization of birational maps, just that they're isoms away from codim 2 stuff. –  BCnrd Jul 23 '10 at 1:44
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Victor, indeed the argument in sga1 concerns the profinite group. But I think it should adapt to the topological one. The main point is that a proper birational map between smooth projective varieties is an isomorphism away from Zariski-closed sets of codimension > 1 in the source and target, so the issue is to check that the topological fundamental group is unaffected by removing such closed sets. That sounds quite reasonable, thinking in terms of real codimensions and topological loops. Is there a subtlety? (I haven't thought it through in a while, so maybe I miss something?) –  BCnrd Jul 23 '10 at 3:10
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BCnrd -- yes, removing a polyhedron $X$ of real codimension $>2$ from a smooth manifold does not affect the fundamental group for transversality reasons: not only we can push any loop away from $X$, but any homotopy between loops as well. –  algori Jul 23 '10 at 3:41
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BCnrd -- given a complex algebraic variety $X$ and a closed subvariety $Y$, the Lojasiewicz's theorem implies that one can triangulate $X$ so that $Y$ is a subpolyhedron. I think a similar result should hold for $X$ and $Y$ complex analytic (or even real analytic). In the smooth case there are pathologies: any closed subset of a smooth manifold is the zero locus of a smooth function. –  algori Jul 23 '10 at 4:55
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@Artie: whoops, I garbled what I meant to say. The correct statement is that since a rational map between smooth proper varieties is defined away from closed set of codim $> 1$, insensitivity of the topological fundamental group to removing any such closed sets (which algori justified in the topological case) implies that it's functorial in such dominant rational maps (hence takes such birat'l maps to isoms) provided any dense Zariski-open $U \subset X$ induces surjection on top. fund. groups. Surjectivity holds because loops push away from $X-U$ of real codim $\ge 2$. QED –  BCnrd Jul 23 '10 at 10:07
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The paper of Keum and Zhang

"Fundamental groups of open K3 surfaces, Enriques surfaces and Fano 3-folds", Journal of Pure and Applied Algebra vol. 170

provides some answers to Tom's question "What happens to the fundamental group of a singular variety when removing the singular points?"

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:

"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".

So, given for instance a quartic surface $X \subset \mathbb{P}^3$ with $16$ nodes (a Kummer surface) we have

$\pi_1(X)=\{1\}$, but $\pi_1(X^0)$ is infinite!

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.

For smaller values of $c$, the group $\pi_1(X^0)$ is finite, but not trivial in general.

In higher dimension, there is the following

"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".

(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).

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