$\mathbb{P}^n$ is simply connected - MathOverflow

$\mathbb{P}^n$ is simply connected

2011-04-19T15:18:29Z

In his chapter about Hurwitz' theorem for curves, Hartshorne shows that $\mathbb{P}^1$ is simply connected, i.e. every finite étale morphism $X \to \mathbb{P}^1$ is a finite disjoint union of $\mathbb{P}^1$s. In an exercise the reader is invited to show that $\mathbb{P}^n$ is simply connected, using the result for $\mathbb{P}^1$.

I have no idea how to do this. Perhaps someone can give a hint? There are closed immersions $\mathbb{P}^1 \to \mathbb{P}^n$, along which we may pull back a finite étale morphism, but the trivializations don't have to coindice ... perhaps we can resolve this using cohomology theory? I'm a bit confused since $\mathbb{P}^n$ is $n$-dimensional, but this is in Hartshorne's chapter about curves. I don't want to use the more advanced material of SGA.

Answer by BS for $\mathbb{P}^n$ is simply connected

2011-04-19T17:05:45Z

You can induct on $n$. Let $f:X\to\mathbb{P}^n$ be finite and étale.

If $H$ is a hyperplane in $\mathbb{P}^n$, there is a trivialization $\phi:f^{-1}(H)\simeq H\times F$, for a finite $F$, by the induction hypothesis.

If $L$ is any line in $\mathbb{P}^n$, $f^{-1}(L)$ is a finite disjoint union of $\mathbb{P}^1$'s, and you can label the components by elements of $F$ using the trivialization at any point of $L\cap H$ (in case $L\subset H$, otherwise there is only one).

Now any fiber $f^{-1}(x)$, $x\in \mathbb{P}^n$, is identified with $F$ through the labeling of the components of $f^{-1}(L)$, for any line $L$ through $x$ (this doesn't depend on the line through $x$, their space being connected).

Answer by Angelo for $\mathbb{P}^n$ is simply connected

2011-04-19T18:37:32Z

There is somewhere a theorem in Hartshorne's book saying that an ample divisor on a normal projective connected scheme of dimension at least 2 is connected. Now proceed by induction on $n$. If there is a non-trivial étale cover $X \to \mathbb P^n$, consider the inverse image of a hyperplane $\mathbb P^{n-1}$; this is connected, since the pullback of an ample divisor by a finite map is ample, and this give the required inductive step.

Answer by Sándor Kovács for $\mathbb{P}^n$ is simply connected

2011-04-19T18:41:58Z

We may assume that $n\geq 2$. Let $f:X\to \mathbb P^n$ be a finite étale morphism where $X$ is connected and $H\subset \mathbb P^n$ a hyperplane. Then $f^*H$ is an ample divisor on $X$ and hence connected. By induction, then the restriction $f^*H\to H$ is an isomorphism, so $\deg f=1$ and $f$ is an isomorphism.

EDIT added previously silently assumed assumption that $X$ is connected.

Answer by Sándor Kovács for $\mathbb{P}^n$ is simply connected

2011-04-20T05:47:56Z

I meant to add that there are other interesting ways to think about this issue. These do not conform to the request of a simple proof, but seem relevant to mention.

Every rationally connected smooth variety is simply connected (at least over $\mathbb C$) this is a result of Kollár-Miyaoka-Mori and Campana independently.

Hartshorne's conjecture, proved by Mori says that $\mathbb P^n$ is the only smooth projective variety whose tangent bundle is ample. This allows for a simple proof that $\mathbb P^n$ is simply connected: Let $f:X\to \mathbb P^n$ be a finite étale morphism and assume that $X$ is connected. Then clearly $X$ is smooth and projective and furthermore it follows that $\Omega_X\simeq f^*\Omega_{\mathbb P^n}$ and hence the tangent bundle of $X$ is also ample. By Mori's theorem it is then isomorphic to $\mathbb P^n$. However, $\mathbb P^n$ does not admit unramified self-maps of degree $d>1$ (because the induced map on the Picard group would be multiplication by $d$ and then it would imply that $\deg K_{\mathbb P^n}=0$), so $f$ has to be an isomorphism.

Answer by ACL for $\mathbb{P}^n$ is simply connected

2011-04-20T07:41:44Z

(From SGA I, Exposé XI).

You can prove it using the following two facts:

1) A product of simply connected proper varieties is simply connected (SGA I, X, 1.7). (*)

2) The fundamental group -- so in particular, being simply connected -- is a birational invariant of proper regular varieties (SGA I, X, 3.4).

(*) I do not know whether the properness is necessary here; it is required for the more general computation of the fundamental group of a product: in positive characteristic, one of the factors needs to be proper.

Answer by Emerton for $\mathbb{P}^n$ is simply connected

2011-04-20T13:06:17Z

Here is a sketch of an argument which directly uses simple connectedness of $\mathbb P^1$, and is related to the simple connectedness of rationally connected smooth varieties mentioned by Sandor in one of his answers.

The idea is to treat the $\mathbb P^1$s in $\mathbb P^n$ as analogous to arcs in a topological space, and to make a lifting argument (just as one does in the basic topological theory of covering spaces).

Let $Y \to \mathbb P^n$ be a finite etale map. Fix a base points $x \in \mathbb P^n$ and a point $y \in Y$ lying over $X$. If $x' \in \mathbb P^n \setminus {x}$, there is a unique line $L$ joining $x$ and $x'$. The preimage of $L$ is a disjoint union of curves $L'$, each mapping isomorphically to $L$ (by simple connectedness of $\mathbb P^1$), and we can choose a unique $L'$ containing $y$. Now let $y'$ be the point of $L'$ lying over $x'$.

The map $x' \mapsto y'$ (and of course mapping our original point $x$ to $y$) gives a section to the given map $Y\to \mathbb P^n$, which is what we wanted.

Added: Here is one explanation of why the map $x' \mapsto y'$ is algebraic. Let $\pi:Y \to \mathbb P^n$ be our given etale map. First note that $x' \mapsto \pi^{-1}(L)$ (where $L$ is the line joining $x$ and $x'$, as above) is a morphism from $\mathbb P^n \setminus {x}$ to the Hilbert scheme of $Y$. Now picking out the connected component $L'$ of $\pi^{-1}(x')$ containing $y$ is a morphism from our given locally closed subset of the Hilbert scheme to the Hilbert scheme, and so altogether we see that $x' \mapsto L'$ is a morphism. Finally, mapping $L'$ to $x'$ (which can be described as forming the intersection $L' \cap \pi^{-1}(x')$) is again a morphism. So altogether we have a section $\mathbb P^n \setminus{x} \to Y$. One way to show that this extends as a section over all of $\mathbb P^n$ (by sending $x$ to $y$) is just to repeat the whole process for a different choice of $x$, and glue the two resulting sections.

Answer by Lars for $\mathbb{P}^n$ is simply connected

2011-04-20T14:45:06Z

Let me give another answer, even though it does not fit into Hartshorne's context:

Show that $\pi_1(\mathbb{P}^n)$ has to be abelian.

Use Kummer-Theory to relate coverings to torsion in $Pic (\mathbb{P}^n)=\mathbb{Z}$, see e.g. Milne's Etale Cohomology, Prop 4.11. This implies that there are no nontrivial étale coverings of degree prime to the base characteristic.

Then use Artin-Schreier theory to relate the rest of the coverings to $\Gamma(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})/(F-1)\Gamma(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})=0$ , and $H^1(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})^F=0$, where $F$ is the Frobenius, see e.g. Milne's, Prop 4.12.