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When I was an undergrad student, the first application that was given to me of the construction of the fundamental group was the non-retraction lemma : there is no continuous map from the disk to the circle that induces the identity on the circle. From this lemma, you easily deduce the Brouwer fixed point Theorem for the circle.

This was (for me) one of this "WOOOOW" moments where you realize that abstract constructions and some seemingly innocuous functorial lemmas may yield striking results (especially as I knew a quite long and complicated proof of Brouwer Theorem in dimension 2 before taking this topology class).

I was wondering if there exist (+ reference if they do) similarly "cute" applications of the construction of the étale fundamental group in Algebraic Geometry. Of course "cute" is not well-defined and may vary for each one of us, but existence of fixed points for the Frobenius morphism would I find especially cute. Any other relatively elementary result related to algebraic geometry over fields of positive characteristic will be appreciated!

Edit : I am obviously curious of any application of the étale fundamental group endowed with the aforementioned "WOOOW feeling". However, I'd be really interested in examples I could explain to smart grad students who are taking a first (but relatively advanced) course in Algebraic Geometry.

Libli
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    I would argue that your example is rather about the power of functoriality than the fundamental group itself as the (co)homology functor seems better suited when $n>2$. As Brian Conrad puts it, "before functoriality, people lived in caves" :-) Other than that, I definitely share the WOW-sentiment :-) – M.G. Mar 10 '21 at 02:12
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    Did you try Morishita's book on analogies between knots and pimes? https://www.springer.com/de/book/9781447121572?utm_campaign=3_pier05_buy_print&utm_content=en_08082017&utm_medium=referral&utm_source=google_books#otherversion=9781447121589 A shorter available version is on arXiv: https://arxiv.org/abs/0904.3399 – D.-C. Cisinski Mar 10 '21 at 09:05
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    @Denis-CharlesCisinski : Thanks a lot. I wasn't aware of this book, neither of the theory linking knots and primes. I must admit I am little bit lazy to read the whole book. Do you suggest a specific chapter that would answer to my question? – Libli Mar 10 '21 at 11:37
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    @Libli Chapter 4 on Legendre symbols is nice: they are identified with the analogue of linking numbers (Prop. 4.4). You have to read the part on classical reading number to appreciate it though. That is not an application per se (hence I did not post that an answer), but I find this cute! – D.-C. Cisinski Mar 10 '21 at 13:25
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    For many familiar objects (Abelian varieties...) the etale fundamental group can be identified with a more classical, familiar invariant (the Tate module...). Would you count applications of the Tate module for instance, as applications of the etale fundamental group? – Asvin Mar 10 '21 at 14:44
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    @Asvin : why not! If I can easily make the link between étale maps and the Tate modules. I must admit I am more inclined to think about complex geometry and I am starting to think again about all these incarnations of the (étale) fundamental group for this course I am teaching. But a cute and easily explainable application of the Tate module to obstruction of rational solutions would certainly be a fact I will be delighted to teach and explain – Libli Mar 10 '21 at 14:53
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    One thing that wowed me when I first saw it was that you could bound the size of the endomorphism ring of your elliptic curve by studying it's action on the Tate module. You could also try doing this with the tangent space but mod p, the tangent space is defined over a finite field so the endomorphisms can't act faithfully but the Tate module works! – Asvin Mar 10 '21 at 15:49
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    But really, the tate module is so pervasive in arithmetic geometry that pretty much any big result about abelian varieties/jacobians will feature it prominently (Mazur's theorem, Herbrand-Ribet,Faltings'...) but maybe these are too sophisticated? A more elementary example might be the theorem about good reduction - an elliptic curve has good redn at a prime iff the galois action on the tate module is unramified at that prime! – Asvin Mar 10 '21 at 15:52
  • If you are happy with conjectures, Grothendieck's section conjecture is another nice candidate. It says that the rational points of hyperbolic curves should be in bijection with splittings of a fundamental short exact sequence involving fundamental groups. – Asvin Mar 12 '21 at 18:14
  • @Asvin : interesting. In some concrete examples, is it any easier to find some splitting of these exact sequences than the corresponding rational point? Or at least, is there a way to prove existence of such splittings, again in some concrete examples, without knowing they correspond to rational points in the first place? – Libli Mar 12 '21 at 23:11
  • I am not sure and by all accounts, it looks to be a really hard conjecture. Until very recently (?), the only cases where we knew the conjecture was when there were no rational points to begin with and then one could prove there were no sections so maybe it's not really an exact fit to your question. Something else in a similar vein is the https://en.wikipedia.org/wiki/Neukirch%E2%80%93Uchida_theorem – Asvin Mar 12 '21 at 23:34
  • For a positive result though, it is known that the truth of a section conjecture implies an algorithm to determine if a curve has a rational point. See the answer to https://mathoverflow.net/questions/179664/why-is-the-section-conjecture-important – Asvin Mar 12 '21 at 23:35

3 Answers3

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Using the étale fundamental group one can construct an injective group homomorphism

$\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \hookrightarrow \operatorname{Out}(\widehat{F_2})$

which is canonical in the sense that there are no choices involved in its construction (once the algebraic closure is fixed), $\operatorname{Out}$ refers to the outer automorphism group and $\widehat{F_2}$ to the profinite completion of the free group on two letters.

As for your example of a "WOOOOW" moment, this statement no longer contains the étale fundamental group in its statement, even though it's vital for the construction.

The fact that the absolute Galois group of the rationals is canonically a subgroup of the outer automorphisms of a (profinite-) free group is completely non-obvious. (Try to prove it from scratch...)

One can try to determine the image of this map. This leads to the profinite Grothendieck-Teichmueller group, which sometimes is conjectured to be isomorphic to $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

The étale fundamental group enters because one considers étale coverings of the projective line minus three points (i.e. the affine line minus two points). Over an algebraically closed field of characteristic zero, this has the étale fundamental group $\widehat{F_2}$ (think of the loops around two of the removed points as its generators). But now study the projective line minus three points over the rationals instead of their algebraic closure, this makes the Galois group of the rationals enter.

The projective line minus three points enters because by a construction due to Belyi, every algebraic curve which can be defined over a finite extension of the rationals can be realized as an étale covering of the projective line minus three points (this is an if and only if). The idea is to modify a map to the projective line so that its ramification gets concentrated in only three points.

You will find the rest of the story for example in surveys by Leila Schneps on the Grothendieck-Teichmueller theory.

Glorfindel
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    This is certainly a very nice example and I agree I ave absolutely no idea on how I may possibly prove this result from scratch. I was howeer thinking of applications I may explain to grad students for which I teach a first (slightly advanced) course in Algebraic Geometry. I will edit my question. – Libli Mar 10 '21 at 07:36
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    which sometimes is conjectured to be isomorphic to Gal(Q¯¯¯¯/Q) - sometimes conjectured, as in sometimes it's conjectured to not be isomorphic also? – Kimball Mar 10 '21 at 18:01
  • Is there a nice application of this embedding which only talks to algebraic extensions of $\mathbb{Q}$? – Martin Brandenburg Jul 01 '21 at 08:48
  • I am confused. Isn't the action of $Gal(\bar{\bf Q}/{\bf Q})$ the obvious one? Since $\sigma(0)=0$ and $\sigma(1)=1$, it acts on $\bar{\bf Q}-{0,1}$ and thus also on the space of loops, say piecewise linear ones if we want to stay in $\bar{Q}$, and that space is the free group (up to conjugacy because we miss a basepoint). Take a completion if needed. – coudy Dec 13 '21 at 20:14
  • @Kimball there doesn't seem to be very good evidence that this is the case; see the papers/surveys of Leila Schneps on this topic. – user149000 Dec 30 '21 at 03:14
  • @coudy not even close... you aren't staying in $\overline{\mathbb{Q}}$ even if you restrict to piecewise linear loops, the action of this Galois group is totally independent of the Euclidean topology on $\mathbb{C}$ (which you need for continuity), even if all that were true there's no reason homotopic loops should remain homotopic, etc... – user149000 Dec 30 '21 at 03:31
  • The action is actually quite easy to describe if you accept the homotopy exact sequence for etale fundamental groups, which splits up $\pi_1(X)$ (where $X$ is the rational projective line minus 3 points) into its geometric component, which is just the profinite completion of $F_2$, and its arithmetic part, which is the Galois group. The existence of a rational point on $X$ gives a section by functoriality, so there's a map back from the Galois group to $\pi_1(X)$. You use this to conjugate $\widehat{F_2}$, so you get an outer action. – user149000 Dec 30 '21 at 03:36
  • @Libli in my opinion this is definitely something you can teach in a first-year grad course in AG if you're planning to introduce etale $\pi_1$. The result may seem quite out-of-the-blue at first glance, but the proof isn't very advanced. It follows quite readily from the homotopy exact sequence, which is definitely a key feature of etale $\pi_1$ you would probably want to include. All of this and more is also included in Tamas's book (in Gabriel's answer) which, despite Martin's points, I would still heartily recommend. – user149000 Dec 30 '21 at 03:41
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I think a fantastic application of the étale fundamental group is in extensions of the Chabauty-Coleman(-Kim) method. The original idea was that we could bound the rational points on a curve $C$ by embedding it into the Jacobian and under good enough conditions (on the rank of the Jacobian compared to it's dimension), this leads to a proof of finiteness.

Well, what happens if these conditions aren't met? The idea is that we can replace the Jacobian (which can be thought of as coming from the abelian quotient of the fundamental group) with more non abelian analogues. This goes under the name of Chabauty-Coleman-Kim and Jennifer Balakrishnan, for instance, has lots of recent work on this topic.

Edit (adding in more details): The idea behind the original proof is simple: Suppose $C$ is a curve of genus $\geq 2$, $K$ is a number field and we want to prove that $C(K)$ is finite (Mordell's conjecture). For simplicity of notation, let me assume $K = \mathbb Q$

Under the Abel Jacobi embedding, we can think of $C$ as living inside the Jacobian ( with $g$ it's dimension) $$C(\mathbb Q) \subset J(\mathbb Q_p) \cong \mathbb Z_p^g\times\text{finite group}$$ If we consider the p-adic closure of the Mordell-Weil group (of rank $r$), $\overline{J(\mathbb Q)} \subset J(\mathbb Q_p)$ we get a lattice of rank $\leq r$ inside $\mathbb Z_p^g$ and let us suppose that $r+1 \leq g$. On the other hand $C(\mathbb Q_p)$ is a one dimensional p-adic curve and we have the inclusion: $$C(\mathbb Q) \subset C(\mathbb Q_p) \cap \overline{J(\mathbb Q)}$$ and we expect this intersection to be finite because we have a one dimensional thing intersection something of codimension at least 1. This can be made rigorous and even gives an effective bound under the assumption that $r + 1 \leq g$.

That's reasonable straightforward at least in concept but unfortunately, the restriction that $r < g$ is pretty severe and often won't hold. The really cool idea is that we can reinterpret this method in terms of data intrinsic to the curve $C$, in particular in terms of it's étale fundamental group. This paper by David Corwin gives a much better/complete exposition but let me try my best nevertheless.

We can reinterpret the Mordel-Weil group $J(\mathbb Q)\otimes\mathbb Z_p$ as living inside the cohomology group $H^1(\mathbb Q, T_p)$ where $T_p$ is the p-adic Tate module (ie, the abelian quotient of the fundamental group) and we identify which subgroup it corresponds to in terms of data intrinsic to $C$.What we need to do next is rewrite the entire Chabauty-Coleman method in terms of data that is intrinsic to $K$ (without reference to an Abelian variety) and then we can replace $T_p = \pi^{et}_C/[\pi^{et}_C,\pi^{et}_C]$ by a larger quotient by going deeper in the central filtration and it gets technically a lot more complicated but I hope that gives a general flavour of the idea.

Asvin
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    This looks really great, and exactly in the vein of results I am looking for. Could you give some more details in your answer? Facts accessible to a beginnign grad student whose taking a first (but long) course in AG? – Libli Mar 10 '21 at 19:57
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    I gave it my best shot! – Asvin Mar 10 '21 at 23:58
  • Thanks a lot for the details. This looks really interesting and I like a lot this approach the Mordell onjecture. Unfortunately, it seems to be out of the scope of my course which is designed for students having a first (though relatively advanced) course in AG. – Libli Mar 11 '21 at 23:51
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    Possibly a stupid question, but by "codimension larger than 1", did you mean "codimension at least 1"? – stupid_question_bot Mar 22 '21 at 04:27
  • That's right! Thanks for catching the error. @stupid_question_bot – Asvin Mar 22 '21 at 04:28
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The inverse Galois problem asks whether every finite group appears as the Galois group of some Galois extension of $\mathbb{Q}$. A well-known variant is the regular inverse Galois problem:

Let $G$ be a finite group. There is necessarily a Galois extension $K/\mathbb{Q}(t)$, regular over $\mathbb{Q}$, whose Galois group is isomorphic to $G$? Such an extension is said to be regular if there is no subextension of the form $L(t)$, where $L/\mathbb{Q}$ is non-trivial.

It is true that a positive answer to this question implies a positive answer to the real inverse Galois problem. This problem is still open as of today, but we can give an affirmative answer under some conditions on $G$ by exploring the relations between the Galois group and the étale fundamental group of the projective line.

Surprisingly, this is not that hard! A nice discussion about this may be found in the book Galois Groups and Fundamental Groups by Tamás Szamuely.

Gabriel
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  • I cannot recommend the mentioned book "Galois Groups and Fundamental Groups", though. It contains several inaccuracies, and most notably the definition of a sheaf is wrong (not just inaccurate). – Martin Brandenburg Jul 01 '21 at 10:16
  • @MartinBrandenburg having read it from the first to the last page, I agree that it has some inaccuracies, mainly because the author tries to do "an exposition" of some hard stuff to people that don't yet have the needed background. That doesn't always works (who can learn scheme theory in less than 10 pages, for example?) but this book has some stuff that is hard to find in elsewhere. (But what is the problem with its definition of a sheaf btw?) – Gabriel Jul 01 '21 at 19:06
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    The "definition" restricts to non-empty open subsets (at least in my edition), which then also implies that the sheaf condition only applies to covers with non-empty intersections. In particular, $F(U \sqcup V) = F(U) \times F(V)$ may fail with that "definition". There is a general theme in mathematics that some authors want to "simplify" the definitions by omitting "non-interesting" cases (e.g. to define Banach algebras to be non-trivial, or graphs to be non-empty), which is a mistake; a long elaboration can be found here. – Martin Brandenburg Jul 01 '21 at 22:28
  • ... and also at this wonderful mathoverflow thread: https://mathoverflow.net/questions/45951/interesting-examples-of-vacuous-void-entities – Martin Brandenburg Jul 01 '21 at 22:34