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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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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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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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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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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