Clearly the etale fundamental group of $\mathbb{P}^1_{\mathbb{C}} \setminus \{a_1,...,a_r\}$ doesn't depend on the $a_i$'s, because it is the profinite completion of the topological fundamental group. Does the same hold for when I replace $\mathbb{C}$ by a finite field? How about an algebraically closed field of positive characteristic?

(note that I'm talking about the full $\pi_1$ and not the prime-to-$p$ part)

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    $\begingroup$ Maybe you are aware that the fundamental group of a projective curve of genus $g>1$ does depend on moduli. See, e.g., this paper of Saidi: empslocal.ex.ac.uk/people/staff/ms220/Site/Publications_files/… I don't know the answer to your question but my guess is that it will depend on the $a_i$'s $\endgroup$ Dec 27, 2010 at 23:34
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    $\begingroup$ I don't know the answer, but it follows from Abhyankar's conjecture, proved by Raynaud and Harbater, that the finite quotients of the fundamental groups in the algebraically closed case are the same, which suggests that the fundamental groups might be isomorphic. $\endgroup$
    – Angelo
    Dec 28, 2010 at 9:16

2 Answers 2


It is a result of Tamagawa that for two affine curves $C_1, C_2$ over finite fields $k_1,k_2$ any continuous isomorphism $\pi_1(C_1)\rightarrow \pi_1(C_2)$ arises from an isomorphism of schemes $C_1\rightarrow C_2$. Hence, if $\pi_1( \mathbb{P}^1\setminus\{a_1,\ldots, a_r\})$ were independent of the choice of the $a_i$, then the isomorphism class of the schemes $\mathbb{P}^1\setminus\{a_1,\ldots, a_r\}$ would be independent of the choice of $a_1,\ldots,a_r$.

Tamagawa's result is Theorem 0.6 in this paper:

The Grothendieck conjecture for affine curves, A Tamagawa - Compositio Mathematica, 1997 http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=298922

In the case of an algebraically closed field, the answer is also that the fundamental group depends on the choice of the points that are being removed. Again by a theorem by Tamagawa: If $k$ is the algebraic closure of $\mathbb{F}_p$, and $G$ a profinite group not isomorphic to $(\hat{\mathbb{Z}}^{(p')})^2\times \mathbb{Z}_p$, then there are only finitely many $k$-isomorphism classes of smooth curves $C$ with fundamental group $G$ (the restriction on $G$ excludes ordinary elliptic curves).

This can be found in

Finiteness of isomorphism classes of curves in positive characteristic with prescribed fundamental groups, A Tamagawa - Journal of Algebraic Geometry, 2004

  • $\begingroup$ There is a lot of interesting information here! Note though that I think you are missing the word "isomorphism" at the end of the first line. $\endgroup$ Dec 28, 2010 at 11:36

No --- given two triples of Q-rational points, there is an automorphism of the projective line over Q carrying one to the other.

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    $\begingroup$ I don't see how this answers anything... You just claimed that in the case r=3 and the field=Q (notice that I required positive characteristic) the a_i's don't matter (which, btw, would be a positive answer in that case and not a negative one). $\endgroup$ Dec 28, 2010 at 6:09
  • $\begingroup$ @MakhalanDuff, regarding your parenthetical remark: the question in the title and the body are not the same $\endgroup$
    – user74900
    May 24, 2018 at 18:57

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