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

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

5
This comment is a little late, since the question has been answered, but let me just say that the statement "There are n obvious closed immersions P^1→P^n, along which we may pull back a finite étale morphism, but they do not cover P^n" seems a bit strange to me. (There are a lot more than n ways to embed P^1 in P^n, and they all seem to me to be equally obvious.) Am I missing something? – Artie Prendergast-Smith Apr 20 2011 at 7:16

## 7 Answers

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.

-

2
This seems to be the most elementary approach, but still there are some details I don't understand. Namely, $x' \mapsto y'$ is first defined only as a set-theoretical map $\mathbb{P}^n \backslash x \to X \backslash x$. Why is it a morphism? And why can we extend it on $\mathbb{P}^n$? Why does it suffice to find a section? – Martin Brandenburg Apr 21 2011 at 8:22
@Martin : as in Sandor's answer, $Y$ must be silently assumed connected, so that a section is enough. Also, I have edited my answer, which seems to be a more cumbersome version of the same idea, but seemingly not needing extension. Does it seem more airtight to you ? I surmised that such simple geometric reasoning cannot let you out of algebraic geometry over anything, although I never felt so much assured with coverings in positive characteritic. – BS Apr 21 2011 at 15:47
There is no need to assume that $Y$ is connected (although of course it is harmless to do so): any section of a finite etale morphism over a connected base induces an isomorphism between the base and a connected component of the cover. Thus the statement that a connected scheme is simply connected is equivalent to the statement that any finite etale morphism admits a section. – Emerton Apr 21 2011 at 16:17
Also, the map extends to $\mathbb P^n$ just by sending $x$ to $y$. – Emerton Apr 21 2011 at 16:20
"any section of a finite etale morphism over a connected base induces an isomorphism between the base and a connected component of the cover." Why? Also, why is your set-map (yes I knew that we map $x$ to $y$) a morphism? – Martin Brandenburg Apr 21 2011 at 16:34
show 4 more comments

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

-

5
The theorem is III.7.9 in Hartshorne. – Dave Anderson Apr 19 2011 at 18:53

(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.

-
@ACL: I did some minor copyediting on your answer; I hope you don't mind. (I was having trouble with the sentence in 2), which I read as saying that the fundamental group itself was simply connected...) – Pete L. Clark Apr 20 2011 at 14:11
@Pete. Thanks a lot! – ACL Apr 21 2011 at 6:48

1
Nice! so 1) gives us that n-fold product of $P^1$'s is simply connected and since product variety is birational to $P^n$, we get the required result by 2). is this correct? – SGP Apr 24 2011 at 17:12

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.

-

@Sandor: You cannot conclude that $f$ is an isomorphism, just a finite disjoint union of isomorphisms.. – Martin Brandenburg Apr 19 2011 at 20:00
@Martin: sorry, I meant to say that you can assume at the start that $X$ is connected. – Sándor Kovács Apr 19 2011 at 21:21
How can we show that $f^* H$ is ample? And how the degree of $f$ is related with the degree of the restriction to $H$? – Martin Brandenburg Apr 22 2011 at 20:11
@Martin: The pull back of an ample divisor via a finite map is ample. This is an exercise in Hartshorne and can be proved using pretty much any characterization of ampleness. As for the degree: $f$ is unramified, so the degree is equal to the number of preimages of any (closed) point. This remains the same for any subvariety. – Sándor Kovács Apr 22 2011 at 22:56

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.

1

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

2

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.

-

What does the "at least over $\mathbb C$ mean in the first point? It is true over $\mathbb C$, but what does happen over other fields? I hate "at least"! :P – Mariano Suárez-Alvarez Apr 24 2011 at 17:23

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.

-

There is one thing which has always confused me with this arugment. Since H^1(X,Z) is essentially the abelianisation of the fundamental group, how are you ruling out the fact that the fundamental group might have trivial abelianisation i.e. is a perfect group?  – Daniel Loughran Apr 20 2011 at 15:39
Oops, you are of course perfectly right. In this particular case we can be saved though, if I am not mistaken: G_m^n is an open subscheme of $\mathbb{P}^n$,so $\pi_1(\mathbb{P}^n)$ is a quotient of the abelian group $\pi_1(G_m^n)$. – Lars Apr 20 2011 at 17:36
@Lars: so this works only in char. zero, otherwise $\pi_1(\mathbb{G}_m)$ is not abelian. – Laurent Moret-Bailly Apr 21 2011 at 6:40
Again oops, what was I thinking. I guess I don't know of a "easy" proof that $\pi_1(\mathbb{P}^n)$ is abelian then (as $\pi_1(\mathbb{A}_k^1)^{(p)}$ is free pro-p on $\#k$ generators). – Lars Apr 21 2011 at 10:04

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).

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 131, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9410375952720642, "perplexity_flag": "head"}