Why should I believe the Mordell Conjecture? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:50:13Z http://mathoverflow.net/feeds/question/56011 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture Why should I believe the Mordell Conjecture? Barinder Banwait 2011-02-19T20:37:43Z 2011-07-07T23:31:20Z <p>It was Faltings who first proved in 1983 the Mordell conjecture, that a curve of genus 2 or more over a number field has only finitely many rational points.</p> <p>I am interested to know why Mordell and others believed this statement in the first place. What intuition is there that the statement must hold? Without reference to any proof, why should the conjecture 'morally' be true? Supposing one had to give a colloquium (to a general mathematically literate audience) on this, how could one convince them without going into details of heights or étale cohomology?</p> <p>Answers I'm not looking for will be of the form "You fool, because it's been proved already", or even "Read Faltings' proof". </p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56013#56013 Answer by Sándor Kovács for Why should I believe the Mordell Conjecture? Sándor Kovács 2011-02-19T21:17:25Z 2011-02-22T03:22:40Z <p><em>Disclaimer</em> In the spirit of the question I am not trying to be very precise below. The key word is <em>motivation</em>. :)</p> <h2>Reason #1 (simple, but lame)</h2> <p>Fermat's Last Theorem says that a curve of that kind of a particular form has <em>no</em> rational points. One could consider Mordell's conjecture as a weak form of that. This is probably not very convincing but perhaps a good way to start a colloquium. A derivative of this idea is that you could look at a lot of examples and make this conjecture.</p> <h2>Reason #2 (hyperbolicity and rational curves)</h2> <p>Another, perhaps more convincing idea is that there is a trichotomy of curves given by $g=0$, $g=1$, and $g\geq 2$. If you look at topological, geometric, arithmetic properties of these curves, their properties align very strongly with these classes. </p> <p>Having genus at least $2$ in other words means that the curve is of general type. From a differential geometric point of view that means (mostly) negative curvature or (roughly) being <em>hyperbolic</em>. A compact complex analytic space is hyperbolic if and only if there are no non-constant holomorphic maps to the space from $\mathbb C$. Algebraically this roughly means that there are no rational curves or abelian varieties in the space. This is of course not entirely true, but we have </p> <p><strong>Lang's Conjecture</strong> Let $X$ be a variety of general type, then there exists a proper closed subvariety $Z\subsetneq X$ such that any rational curve or abelian variety contained in $X$ is contained in $Z$.</p> <p>The existence of rational curves in the geometric setting is very closely related to the existence of rational points in the arithmetic setting. So, it is a natural idea that if a space has properties that suggest that it does not contain many rational curves then if it is defined over a number field then it should not contain many rational points.</p> <p>In fact there is an analogue of Lang's conjecture for number fields and rational points which reduces to the Mordell conjecture for curves:</p> <p><strong>Lang's conjecture</strong> (number field version) Let $X$ be a variety of general type defined over a number field $F$. Then the set of $F$-rational points of $X$ is not dense in $X$.</p> <p>Then at this point one (who is more qualified to do this than I) could explain how this is taken further by Vojta (see Matt's comment below). Then one gets all kinds of interesting conjectures estimating various heights. The motivation for these come from Nevanlinna theory and analogies between results on value distribution of meromorphic functions and Diophantine approximation.</p> <h2>Reason #3 (Shafarevich's Conjecture, de Franchis's Theorem)</h2> <p>Faltings proved the Mordell Conjecture using Parshin's trick and proving Shafarevich's Conjecture. These are the following:</p> <p><strong>Shafarevich's Conjecture</strong> Let $F$ be either a number field or the function field of a curve defined over an algebraically closed field of characteristic $0$. Let $R\subset F$ be a subring; if $F$ is a number field, let $R$ be the ring of integers, if $F$ is a function field, let $R$ be the coordinate ring of a smooth affine curve whose fraction field is $F$. Let $\Delta\subset {\rm Spec} R$ be a finite set of primes. A smooth projective curve over $F$ is called <em>admissible</em> (with respect to $\Delta$) if it has good reduction outside $\Delta$ and in the case $F$ is a function field it is also non-isotrivial. Then the Shafarevich conjecture is that for a fixed $F$ and $\Delta$ there exist only finitely many admissible curves of a fixed genus$\geq 2$.</p> <p><strong>Parshin's Trick</strong> Shafarevich's conjecture implies Mordell's Conjecture.</p> <p><em>The idea of Parshin's trick</em>: Mordell's conjecture says that if $X$ is a smooth projective curve over $F$ of genus at least $2$ then there are only finitely many sections of the structure map $X\to {\rm Spec F}$. Parshin's idea is that for any such section one can obtain a new curve by a cover that's ramified exactly over the section. One proves that this can be done so the genus of the resulting curve is bounded in terms of the initial data. </p> <p>This way one associates a set of curves with bounded genus to the set of sections of a fixed curve and concludes that Shafarevich's conjecture implies Mordell's.</p> <p>Truth be told, one also needs to prove that this association is at worst finite-to-one in the sense that there cannot be infinitely many sections producing the same (abstract) covering curve. But this follows from de Franchis's (classical) theorem:</p> <p><strong>de Franchis's Theorem</strong> Let $C$ and $D$ be smooth curves of genus at least $2$. Then there exist only finitely many dominant rational maps $D\to C$.</p> <p>So, at this point you might say: OK, but why should Shafarevich's conjecture be true?</p> <p>Let's try to answer that. I will concentrate on the function field case for two reason. Mostly because that is what I know a little about and second because I claim that for motivation it is good enough if you know why it should be true in that case. The number field case ought to behave similarly. </p> <p>What does Shafarevich's conjecture mean in terms of moduli? One could say that the function field version is equivalent to the following:</p> <p><strong>Shafarevich's Conjecture 2.0</strong> (function field case) Let $B$ be a smooth curve. Then there exist only finitely many non-trivial maps $B\to \mathscr M_g$ where $\mathscr M_g$ is the moduli stack of smooth projective curves of genus $g$.</p> <p>But this is "just" a higher dimensional log version of de Franchis's Theorem: $B$ and ${\mathscr M}_g$ are of <em>log general type</em>, so we expect only finitely many non-trivial maps between them (remembering that $\dim B=1$).</p> <p><em>Remark</em> I think this is a pretty good reason to believe Mordell's conjecture <em>now</em>. Obviously this was not known to those who believed it before the (late) sixties.</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56014#56014 Answer by George Lowther for Why should I believe the Mordell Conjecture? George Lowther 2011-02-19T21:59:56Z 2011-02-20T15:55:09Z <p>I'm not an expert here, but reading J.S.Milne's notes on elliptic curves (available from his <a href="http://www.jmilne.org/math/Books/index.html" rel="nofollow">website</a>), I noticed the following statement.</p> <blockquote> <p>ASIDE 2.4 There is a heuristic explanation for Mordell’s conjecture. Let $C$ be a curve of genus $g\ge1$ over $\mathbb{Q}$, and assume that $C(\mathbb{Q})\not=\emptyset$. It is possible to embed $C$ into another projective variety $J$ of dimension $g$ (its <em>jacobian</em> variety). The jacobian variety $J$ is an abelian variety, i.e., it has a group structure, and a generalization of Mordell’s theorem (due to Weil) says that $J(\mathbb{Q})$ is finitely generated. Hence, inside the $g$-dimensional set $J(\mathbb{C})$ we have the countable set $J(\mathbb{Q})$ and the (apparently unrelated) one-dimensional set $C(\mathbb{C})$. If $g > 1$, it would be an extraordinary accident if the second set contained more than a finite number of elements from the first set.</p> </blockquote> <p>Now, I'm not completely convinced about quite how unrelated the one-dimensional set $C(\mathbb{C})$ and the countable set $J(\mathbb{C})$ are. More generally, Lang's conjecture (now a theorem) says that if $X$ is a subvariety of an abelian variety $A$ and $\Gamma$ is a finite rank subgroup of $A(\mathbb{C})$, then the Zariski closure of $X(\mathbb{C})\cap\Gamma$ is a union of finitely many translates of subvarieties of $A$. In particular, if $X$ does not contain a translate of a nontrivial abelian subvariety of $A$, then $X(\mathbb{C})\cap\Gamma$ is finite (I'm working from the statement of Lang's conjecture given in Theorem F.1.1.1 of <em>Diophantine Geometry: An Introduction</em> by Hindry and Silverman). So, it does seem that subvarieties and finitely generated subgroups of abelian varieties have an intersection which is small.</p> <hr> <p>I'm also going to add an elementary heuristic argument to try and answer James Borger's question above, "<em>Why should a generic polynomial $f(x,y)\in\mathbb{Q}[X,Y]$ of degree $d\ge4$ have only finitely many solutions in $\mathbb{Q}$?</em>". This is something I have thought about before, and can at least come up with a rough heuristic argument why this should be the case. Setting $g(x,y,z)=f(x/z,y/z)z^d$, which is a homogeneous polynomial in $\mathbb{Q}[x,y,z]$, the question is equivalent to asking why $g$ should have only finitely many coprime solutions in $\mathbb{Z}^3$.</p> <p>The probability of a randomly chosen triple $(x,y,z)\in\mathbb{Z}^3$ being coprime is $\zeta(3)^{-1}$. So, in the ball $B_R$ of radius $R$ centered about the origin in $\mathbb{R}^3$, we expect about $\frac43\pi R^3\zeta(3)^{-1}=O(R^3)$ such triples. We want to know how many satisfy $g(x,y,z)=0$. Choosing any $a\gg1$, we can instead ask how many coprime triples $(x,y,z)\in\mathbb{Z}^3$ satisfy $\vert g(x,y,z)\vert\le a$. Letting $V$ be the surface $\{{\bf x}\in\mathbb{R}^3\colon g({\bf x})=0\}$, we can approximate $\vert g({\bf x})\vert$ to leading order as $$\vert g({\bf x})\vert\approx \Vert\nabla g({\bf x})\Vert d({\bf x},V),$$ where $d({\bf x},V)$ is the distance of a point ${\bf x}\in\mathbb{R}^3$ from $V$. This gives the following estimate for the volume of the set ${\bf x}\in B_R$ with $\vert g({\bf x})\vert\le a$ in terms of an integral over the surface $V$, $$2a\int_{V\cap B_R}\frac{1}{\Vert \nabla g({\bf x})\Vert}\,d{\bf x}.$$ As $\nabla g$ is homogeneous of degree $d-1$, this can be rewritten as a line integral over the intersection of $V$ with the unit sphere $S^2$, $$2a\int_0^R\int_{V\cap S^2}\frac{1}{r^{d-1}\Vert\nabla g({\bf x})\Vert}\,r d{\bf x}dr\approx 2a \frac{R^{3-d}}{3-d}\int_{V\cap S^2}\frac{1}{\Vert\nabla g({\bf x})\Vert}\,d{\bf x}$$ for $d\not=3$ (I'm ignoring the part of the integral near the origin in $\mathbb{R}^3$, as $\nabla g=0$ there, and the estimate of the volume becomes very inaccurate). Multiplying by $\zeta(3)^{-1}$ gives the "expected number" of coprime integer solutions $\Vert{\bf x}\Vert\le R$. Note that this tends to infinity as $R\to\infty$ for $d &lt; 3$. If $d=3$, then we should replace the $R^{3-d}/(3-d)$ term by $\log R$, so it still tends to infinity. On the other hand, for $d\ge4$, we have a finite volume. So, we should expect the number of coprime solutions ${\bf x}\in\mathbb{Z}^3$ with $\vert g({\bf x})\vert\le a$ to be finite when $d\ge4$ and probably infinite when $d\le3$.</p> <p>That's only part of the story though. We should also look at the probability that $g({\bf x})=0$ is satisfied modulo prime powers $p^r$. If this probability is about $p^{-r}$, as you might expect, then the heuristic argument above seems good. If the probability is much more than $p^{-r}$ (as is the case if we don't restrict to coprime triples), then the argument above only gives a lower estimate and, if the probability is much less than $p^{-r}$, then we expect fewer solutions than the argument above would suggest.</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56015#56015 Answer by Sam Lichtenstein for Why should I believe the Mordell Conjecture? Sam Lichtenstein 2011-02-19T22:05:17Z 2011-02-20T00:11:35Z <p>One reason to believe Mordell's conjecture is Chaubaty's proof of it (from 1940ish) in the case where the Mordell-Weil rank of the Jacobian is less than the genus of the curve. This basically builds on the idea from Milne's notes mentioned in George Lowther's answer, but instead uses the $p$-adic Lie group structure of the points of the Jacobian in a non-archimedean completion of the number field. You can read an explanation of the method in section 6 of <a href="http://dl.dropbox.com/u/1589321/Faltings%20seminar/L02.pdf" rel="nofollow">these notes</a> by Brian Conrad.</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56016#56016 Answer by Emerton for Why should I believe the Mordell Conjecture? Emerton 2011-02-19T22:07:33Z 2011-03-26T00:40:29Z <p>If the curve $X$ (over the number field $k$) has no $k$-points at all, then Mordell's conjecture is true for $X$. Otherwise, if $O$ is a given $k$-point on $X$, we can get a map (the Albanese map) $X \to Jac(X)$ via $P \mapsto P - O$, embedding $X$ as a $k$-subvariety of $Jac(X)$.</p> <p>The Mordell--Weil theorem shows that $Jac(X)(k)$ is a finitely generated abelian group. Since $g \geq 2$ by assumption, the curve $X$ is of positive codimension in $Jac(X)$, and so it is not unreasonable to imagine that $X$ intersects $Jac(X)(k)$ in only finitely many points. (George Lowther's answer, posted while I was writing this, describes the same heuristic.)</p> <p>In fact, my understanding is that Weil proved the general form of the Mordell--Weil theorem (in his thesis, I think) precisely to try to implement this strategy and so prove Mordell's conjecture. Unfortunately, no-one has been able to implement this strategy quite as cleanly as the intuition above suggests, although it inspired an important approach, due to Chabauty, which established the Mordell conjecture in some non-trivial cases; see <a href="http://math.mit.edu/~poonen/papers/chabauty.pdf" rel="nofollow">this paper by McCallum and Poonen</a> for an explanation of Chabauty's method, and Coleman's strengthening of it.</p> <p>More recently, Minhyong Kim has developed an anabelian strengthening of Chabauty's method; see e.g. <a href="http://www.ucl.ac.uk/~ucahmki/alb.pdf" rel="nofollow">here</a> and <a href="http://www.ucl.ac.uk/~ucahmki/cm3.pdf" rel="nofollow">here</a>. The second paper (joint with Coates) gives a new proof of Mordell for Fermat curve's, among other examples.</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56024#56024 Answer by Stergios for Why should I believe the Mordell Conjecture? Stergios 2011-02-20T00:31:36Z 2011-02-20T00:41:13Z <p>You can also have a look at the following paper of Curt McMullen : <a href="http://www.math.harvard.edu/~ctm/papers/home/text/papers/fermat/fermat.pdf" rel="nofollow">http://www.math.harvard.edu/~ctm/papers/home/text/papers/fermat/fermat.pdf</a></p> <p>In his own words "This paper is an appreciation of some of the topological intuitions behind number theory."</p> <p>To moderators : How can I put my answer as a comment ? </p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56071#56071 Answer by S. Carnahan for Why should I believe the Mordell Conjecture? S. Carnahan 2011-02-20T16:14:42Z 2011-02-20T16:14:42Z <p>Here's a quick and dirty version of George Lowther's calculation that I learned from Bjorn Poonen. It is presented in a bit more generality in <a href="http://math.mit.edu/~poonen/slides/rational.pdf" rel="nofollow">the next-to-last slide of this talk</a>, so it is in a sense pre-masticated for colloquium use:</p> <p>Let $B$ be a large positive integer which we will allow to grow. The set of integer points in the box $[-B,B]^{\times 3}$ has size of order $B^3$. A homogeneous degree $d$ polynomial in 3 variables with integer coefficients (describing a degree $d$ plane curve over $\mathbb{Q}$) will take values of size about $B^d$ when inputs are taken from the box. If we assume the values are uniformly distributed, the expected number of zeroes in the box is $B^{3-d}$.</p> <p>There is a qualitative difference in expected number of zeroes based on the sign of the exponent $3-d$. This suggests that plane curves of degree more than 3 (namely those of genus at least 2) will only have "accidental" rational points.</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56072#56072 Answer by Denis Chaperon de Lauzières for Why should I believe the Mordell Conjecture? Denis Chaperon de Lauzières 2011-02-20T16:16:09Z 2011-02-20T16:16:09Z <p>I have a vague memory that Weil (or someone else) actually complained in print somewhere about people making conjectures unsupported by much evidence, and using Mordell's conjecture as an example, with words like "there is no evidence one way or another". If it is Weil, it is probably in his Collected Works, which I don't have handy to check. Obviously this was before Faltings's proof.</p> <p>Most 'heuristic' explanations tend to suggest that failures of the Mordell conjecture should be rare, but don't really say why there shouldn't be any. For instance, one can give a pretty good heuristic that squares of integers don't exist (they have to be square modulo every prime, which means surviving infinitely many coin-tosses which are more or less independent).</p> <p>A last remark concerning JSE's answer: I remember from a lecture of Faltings that he was saying that he and/or a student of his were working hard at trying to generalize Kim's method to all number fields; off-topically, the following is -- as far as memory allows -- one of the pithy things he said: "Motivic cohomology is a bit like communism; in principle, it's great, but you can't compute anything").</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/56106#56106 Answer by Felipe Voloch for Why should I believe the Mordell Conjecture? Felipe Voloch 2011-02-20T21:05:58Z 2011-02-20T22:25:35Z <p>Mordell's reasoning might have been the following. If a curve of genus zero has a (smooth) point, then it has infinitely many. If a curve of genus one has two smooth points, it has a good chance of having infinitely many, as the difference will likely be of infinite order. Now, by Mazur's theorem, we know that seventeen points implies infinitely many for genus one. For higher genus, there is no such geometric reasons to get new points from old points, so a curve may have a million points and have no reason to have a million and one, so finiteness is reasonable.</p> <p>I mentioned above in a comment a reason why most plane curves of degree at least four have no point (now I see that Scott has reproduced the argument in an answer, actually it may not be the same argument, but it is similar). I have no idea if Mordell had a similar heuristic, but it's entirely possible that he did as the arguments don't use anything he didn't know.</p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/69714#69714 Answer by Minhyong Kim for Why should I believe the Mordell Conjecture? Minhyong Kim 2011-07-07T12:36:50Z 2011-07-07T23:31:20Z <p>Since summer is here, I will bore people a bit with my own ahistorical view of Mordell. The main point is that Serge Lang's strategy, mentioned by several people, is essentially correct. Nevertheless, we might add that</p> <p><em>you should believe the Mordell conjecture because most local principal bundles should not extend to global principal bundles.</em></p> <p>I've spoken about this in one form or another many times, but maybe this particular sentence hasn't received enough emphasis. So I will explain it. </p> <p>Jean-Benoit Bost has pointed out that this story is not entirely devoid of historical context, since Andre Weil's famous 1938 paper on vector bundles can be construed as giving motivation roughly of this nature.</p> <hr> <p>Recall that Lang's idea is to consider the diagram</p> <p>$$\begin{array}{ccc} X(F)&amp;\longrightarrow &amp; X(\mathbb{C}) \end{array}$$ $$\begin{array}{lcr} \downarrow \ \ \ \ &amp; &amp; \ \ \ \ \downarrow\end{array}$$ $$\begin{array}{ccc} J(F) &amp;\longrightarrow &amp; J(\mathbb{C})\ \end{array}$$</p> <p>for a smooth projective curve $X$ of genus $\geq 2$ with Jacobian $J$ and an embedding $F\hookrightarrow \mathbb{C}$ of the number field $F$ into the complex numbers. Lang suggested that</p> <p>$$X(\mathbb{C})\cap J(F)\subset J(\mathbb{C})$$</p> <p>should be finite. This is indeed plausible and turns out to be true after Faltings proof. Of course we are considering its status as motivation rather than corollary, following the original question posted.</p> <p>In fact, the plausibility is strengthened when we replace the diagram above by a refinement $$\begin{array}{ccc} X(F)\ \ \ \ \ \ \ \ \ \ &amp;\longrightarrow &amp;\ \ \ \ \ \ \ X(F_v) \end{array}$$ $$\begin{array}{ccc} \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ &amp; &amp;\ \ \ \ \ \ \ \ \ \ \ \downarrow j \end{array}$$ $$\begin{array}{ccc} H^1_f(G_S, \pi_1(\bar{X}, b)) &amp;\stackrel{loc}{\longrightarrow} &amp; H^1_f(G_v, \pi_1(\bar{X}, b)) \end{array}$$ and try to prove that $$Im(j)\cap Im(loc) \subset H^1_f(G_v, \pi_1(\bar{X}, b))$$ is finite. Here, $$\pi_1(\bar{X}, b)$$ is some $\mathbb{Q}_p$-algebraic fundamental group of $\bar{X}$ with base-point $b\in X(F)$. The completion $F_v$ should be taken to have degree one over $\mathbb{Q}_p$. The $H^1$'s are moduli spaces of (locally constant) principal bundles for $\pi_1(\bar{X}, b)$, a global one$$H^1_f(G_S, \pi_1(\bar{X}, b))$$ consisting of principal bundles over some $Spec(O_F[1/S])$ and a local one $$H^1_f(G_v, \pi_1(\bar{X}, b)) \simeq \mathbb{A}^N_{\mathbb{Q}_p},$$ (almost naturally) isomorphic to affine space, consisting of principal bundles on $Spec(F_v)$ (satisfying some technical condition).</p> <p>Thus, we are replacing $\mathbb{C}$ by a non-Archimedean completion and the Jacobian (a moduli space of line bundles) by a moduli space of principal bundles. The vertical maps assign to a point $x$ the principal bundle of paths $$\pi_1(\bar{X};b,x).$$ This framework turns out to refine considerably the intuition that $J(F)$ and $X(\mathbb{C})$ have very different natures inside $J(\mathbb{C})$.</p> <hr> <p>Now, <em>why should the Mordell conjecture be true?</em></p> <p>There are two steps.</p> <p>A. The easy one: The map $j$, being a non-Archimedean period map, is highly transcendental, and maps $X(F_v)$ to a Zariski-dense compact analytic curve in $H^1_f(G_v, \pi_1(\bar{X}, b))$, which therefore meets any proper subvariety in finitely many points. One proves this by showing that certain transcendental functions on $X(F_v)$ (the coordinates of the map) are algebraically independent. Meanwhile, the localization map $$loc: H^1_f(G_S, \pi_1(\bar{X}, b))\longrightarrow H^1_f(G_v, \pi_1(\bar{X}, b)),$$ is algebraic, and hence, has constructible image in the Zariski topology. These are the different natures alluded to in the previous paragraph: $$\begin{array}{ccccc} H^1_f(G_S, \pi_1(\bar{X},b)) &amp;\stackrel{\scriptstyle \mbox{algebraic}}{\longrightarrow} &amp; H^1_f(G_v, \pi_1(\bar{X},b)) &amp; \stackrel{\scriptstyle \mbox{dense analytic}}{\longleftarrow}&amp; X(F_v)\end{array}$$</p> <p>Therefore, it suffices to show:</p> <p>B. The hard step: $Im(loc)$ is not Zariski dense, that is, the localization map is not dominant.</p> <p>Now, why should this be true? Well, the moduli space $$H^1_f(G_v, \pi_1(\bar{X}, b))$$ consists of <em>local</em> principal bundles, while $$H^1_f(G_S, \pi_1(\bar{X}, b))$$ is a moduli space of <em>global</em> principal bundles. So it makes sense that most local bundles should not extend to global ones when the group $\pi_1(\bar{X}, b)$ is sufficiently large and non-abelian. (Perhaps you will disagree...)</p> <p>These thoughts were actually inspired by Yang-Mills theory: We have something like local solutions to the Yang-Mills equation, that is, on a small annulus (or a handle-body) embedded in a Riemann surface (or a three-manifold). It seems they should not all extend to global solutions. Natural enough, but quite hard to prove in general. Galois cohomology, on the whole, appears to be harder than Yang-Mills theory. One motivation for posting this answer is the hope that some bright young person will have an idea.</p> <hr> <p>Added: As mentioned by Matthew Emerton, the strategy outlined above is an extension of Chabuaty's method, which quite likely inspired Lang's conjecture as well (according to one reading of the notes to <em>Fundamentals of Diophantine Geometry</em>). There, the analogue of $H^1_f(G_v, \pi_1(\bar{X},b))$ is $$T_eJ(F_v),$$ the Lie algebra of the Jacobian, while the role of the global moduli space is played by $$J(F)\otimes_{\mathbb{Z}}F_v.$$ My own feeling is that the 'local vs. global perspective' that emerges out of the principal bundle interpretation is somehow critical to understanding the Mordell conjecture, and constitutes a natural generalization not just of Chabauty's method, but of the arithmetic theory of curves of genus zero and one. In this sense, the essential motivation for Mordell's conjecture should not just be probabilistic, but something rather precise coming out of <em>class field theory.</em> It's fairly clear that this couldn't have been Mordell's reason for believing in it, but it is plausible, as mentioned above, that it was Weil's reason, in spite of his eventual non-committal assessment.</p>