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Is it true that every smooth rational variety X is simply connected? How is the proof? Would it be still true if X has mild (for example orbifold) singularities?

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up vote 27 down vote accepted

Yes! (I assume it was implicit in your question that the variety be projective?)

More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book ("Higher dimensional algebraic geometry").

Alternatively, take a look at Debarre's Bourbaki talk ("Varietes rationnellement connexes"). The idea is that a rationally connected variety has no holomorphic forms, so by Hodge theory the structure sheaf $\mathcal{O}_X$ is acyclic, implying $\chi(X,\mathcal{O}_X) = 1$. If $f : Y \to X$ is a connected etale cover, then $Y$ is again rationally connected, so by the same argument $\chi(Y,\mathcal{O}_Y) = 1$, so $\deg{f} = 1$. So $X$ is simply connected.

In positive characteristic the Hodge theory fails, so the argument as such doesn't stand, but you may still get the simple connectedness as a consequence of the fibration theorem of Graber-Harris-Starr-de Jong. See the mentioned Bourbaki expose.

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By the way, the nodal curve ${y^2 = x^3}$ is not simply connected, so the result does not extend to singular rationally connected varieties. – Vesselin Dimitrov Jan 31 '13 at 19:36
@Tim. The topological fundamental group of any rationally chain connected projective variety is a priori finite by Chow variety / Hilbert scheme methods: a degree of a certain Chow variety of curves gives an upper bound on the size of the fundamental group. Combined with the arguments in Debarre's article (which are originally due to Koll\'ar), this implies simple connectedness of the fundamental group of a smooth, projective rationally connected variety. – Jason Starr Jan 31 '13 at 19:39
@Vesselin: The curve $y^2-x^3$ is cuspidal, not nodal, although your point is correct. In my comment, the variety should be unibranch. – Jason Starr Jan 31 '13 at 20:24
Indeed, I intended to write $y^2 = x^3 + x^2$ (for a nodal rational curve). Thanks for the correction! – Vesselin Dimitrov Jan 31 '13 at 20:29
Mohammad: being algebraically simply connected means that there are no non-trivial connected etale covers. Such covers are finite, while in topology one makes no finiteness condition on the covers. SeeÉtale_fundamental_group – Tim Perutz Jan 31 '13 at 23:10

In positive characteristic, the (étale) fundamental group of a rationally connected variety is finite (Kollár, Inventiones Math., 1993).

And this happens: Let us assume that the base field has characteristic $p$, where $p\neq 5$ and $p\not\equiv 1\pmod 5$. Then, the hypersurface with equation $X_0^5+\dots+X_3^5=0$ in $\mathbf P^3$ is unirational; so is its quotient by the obvious action of $\mu_5$ (a surface of general type known as the Godeaux surface), which is therefore an unirational variety with fundamental group $\mathbf Z/5$.

However, Ekedahl has proved that this group is prime to the characteristic. I discussed that in my Bourbaki Seminar talk, « Points rationnels et groupes fondamentaux, applications de la cohomologie $p$-adique », Astérisque 294, p. 125-146.

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Also using Raynaud's work on the Abhyankar conjecture combined with Koll\'ar's method, in characteristic $p$, any $p$-group acting on a smooth, projective, separably rationally connected variety has a fixed point. This is stronger than the $p$-part of the fundamental group being trivial, since that only implies there is a point with nontrivial stabilizer, not that there is a fixed point (i.e., point whose stabilizer is the entire group). – Jason Starr Jan 31 '13 at 20:19
I wasn't aware of that, thanks! But may I ask for details? – ACL Feb 1 '13 at 12:06
@ACL: I am not an expert on Abhyankar's conjecture, but I believe that Raynaud proved that for every algebraically closed field $k$ of characteristic $p$, and for every "quasi-p-group" $G$, there exists a finite, flat morphism of smooth, pointed curves over $k$, $f:(C,q)\to (B,p)$ that is a $G$-torsor over $B\setminus\{p\}$ and that is totally ramified over $p$, i.e., the inverse image of $\{p\}$ equals $\{q\}$. Given an action of $G$ on a smooth, projective, separably rationally connected variety $X$, form the diagonal action of $G$ on $C\times X$. contd. – Jason Starr Feb 1 '13 at 18:17
contd. Denote the quotient of this $G$-action by $C\times X \to Y$. The morphism $f\circ \text{pr}_C:C\times X \to B$ is $G$-invariant, hence factors through a morphism $\pi:Y\to B$. This is a morphism that satisfies the hypotheses of the "rationally connected fibration theorem". Thus there exists a section. The inverse image of this section in $C\times X$ is the image of a $G$-equivariant section of $\text{pr}_C$, i.e., the graph of a $G$-equivariant morphism $s:C\to X$. In particular, the image of $q$ is a $G$-fixed point. This is "Koll\'ar's method". Ekedahl's method might work too... – Jason Starr Feb 1 '13 at 18:21
I believe that the Godeaux surface is a quotient of your quintic surface, not the quintic itself. – René Feb 3 '13 at 10:54

Every proper, normal, rational variety over an algebraically closed field is simply connected. Dmitri explains this via the smooth case and resolution of singularities, but this is true in any characteristic: see SGA 1, XI, Cor. 1.2.

EDIT: As Dmitri and Vesselin observe, the "proof" in SGA1 is sketchy, to say the least. One can argue as follows: if $X$ is our variety, there is a birational morphism $U\to X$ where $U=\mathbb{P}^n\smallsetminus Z$ and $Z$ has codimension $\geq2$ in $\mathbb{P}^n$. By the purity theorem, $U$ is simply connected. So any étale covering of $X$ is generically trivial (because its pullback on $U$ is trivial), hence trivial since $X$ is normal.

In fact, this proves that if $X$ and $Y$ (both proper and normal) are birationally equivalent, and $Y$ is regular and simply connected, then $X$ is simply connected. But this does not answer Vesselin's final question.

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Dear Laurent I decided to check the reference and it looks to me that the proof of the fact is not really there. It is proven in two ways that projective spaces over algebraically closed fields are simply connected SGA 1. XI. Prop. 1.1, ( and then comes corollary 1.2 without an actual proof. It is just said there that the proof is the same as for projective space :). Could you indicate how to make this an actual proof? I am asking this because I want to see how to make a proof over C without Hironaka's resolution of singularities. – Dmitri Feb 4 '13 at 11:12
I had the same point as Dmitri. In expose X it is shown, as an application of the Zariski-Nagata purity theorem, that the algebraic fundamental group of a complete regular variety is birationally invariant. In particular, this certainly contains a proof of the simple connectedness of regular proper rational varieties over an alg. closed field. Statement XI, Cor. 1.2 simply refers to this birational invariance. Does the birational invariance extend to normal varieties? – Vesselin Dimitrov Feb 4 '13 at 11:57
@Vesselin: Is the affine line birational to the punctured line? If yes, whatever the characteristic, then you have two birational smooth curves over an alg. closed field whose fundamental groups are distinct. – ACL Feb 4 '13 at 15:25
The fundamental group is certainly not a birational invariant for normal proper varieties. For example, consider a cone over a curve of genus $>0$ (and its desingularisation). – ulrich Feb 4 '13 at 15:30
@ulrich: Thanks for the example! @ACL: I had in mind proper varieties. – Vesselin Dimitrov Feb 4 '13 at 15:39

I decided to rewrite the answer so it is more clear.

Smooth complex rational projective varieties are indeed simply-connected as is explained in the first answer to the question. My point is that from this fact it follows immediately that normal complex rational projective varieties are always simply-connected. Here is a proof.

Proof. Let $X$ be an normal complex projective variety and $X'\to X$ be a resolution of singularities. Then we have a homomorphism $\pi_1(X')\to \pi_1(X)$. I claim that is is surjective. In order to see this one just needs to know that every loop on $X$ can be lifted to $X'$ and this is true since all the fibres of $X'\to X$ are connected since $X$ is normal.

So if $X$ is rational then $X'$ is smooth rational and $\pi_1(X)=\pi_1(X')=0$

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Aren't you missing the words "rationally connected" in a couple of places? I know a few smooth complex projective varieties that are indeed not simply-connected... – Sándor Kovács Feb 4 '13 at 11:21
Sandor, thank you :), I completely agree with you, I added missing words. In fact I was meaning "rational complex projective varieties". I don't know what is the definition of rationally connected projective varieties in the case they are singular. For example, if you consider a cone over a genus $g>0$ curve, every to points can be connected by a two $\mathbb P^1$'s (through the center of the cone), but I don't think this variety should be called rationally connected... – Dmitri Feb 4 '13 at 12:02
A normal projective variety $X$ is (separably) rationally connected if there exists a finite-type integral scheme $T$ and a morphism $F : T \times \mathbb{P}^1 \to X$ such that the obvious morphism $T \times \mathbb{P}^1 \times \mathbb{P}^1 \to X \times X$ is dominant (and generically smooth). – Vesselin Dimitrov Feb 4 '13 at 12:14
Thank you Vesselin. The property of been rationally connected is a birational invariant, I guess? – Dmitri Feb 4 '13 at 12:35
Yes! So, in char. 0, $X$ is rationally connected iff. it has a smooth projective rationally connected (birational) model. But since resolution of singularities is not available in positive characteristic, this definition is the general one. In general, in any characteristic, a smooth separably rationally connected variety is simply connected. (And as noted above, all we can say about non-separably rationally connected varieties in char. $p > 0$ is that their fundamental group is finite of order prime to $p$). – Vesselin Dimitrov Feb 4 '13 at 12:56

In characteristic zero you do not neen rational. It is enough rationally connected.

Let $X$ be a smooth, projective, rationally connected variety over a field of characteristic zero. Then

  • Any finite étale cover of $X$ is trivial;
  • $X$ is simply connected.

You can find this, for instance, in O. Debarre "Higher-Dimensional Algebraic Geometry", Corollary 4.18.

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