Are there simple proofs of some concrete special cases of Faltings's theorem? Any help would be appreciated.

1$\begingroup$ I guess you would like to see a simple proof of some special case of Faltings's theorem (which is not what you wrote). $\endgroup$ – GH from MO Jan 5 '18 at 11:57

1$\begingroup$ I updated your question, I hope it is ok. $\endgroup$ – GH from MO Jan 5 '18 at 12:24

1$\begingroup$ this might be what you are looking for: mathoverflow.net/a/202800/11260 $\endgroup$ – Carlo Beenakker Jan 5 '18 at 12:24

2$\begingroup$ @CarloBeenakker Vojta's proof (especially the Bombieri simplification) is definitely more elementary than Faltings original proof, but it is still not simple, and it covers the full theorem. I think that the OP wants, say, a single curve $C$ for which there is a simple proof that $C(K)$ is finite for all number fields $K$. In fact, before Faltings work, there were no such examples proven. OTOH, if the OP just wants some curves $C/\mathbb Q$ for which it is possible to prove, in a relatively easy way, that $C(\mathbb Q)$ is finite, that can be done, e.g., by the method of Dem'janenko. $\endgroup$ – Joe Silverman Jan 5 '18 at 12:29

1$\begingroup$ Thank everyone for your comment, a special case which keep the outline or the approach of Faltings proof is better. $\endgroup$ – XL _At_Here_There Jan 5 '18 at 12:35
Based on the OP's comment clarifying his question, I fear that the answer is no, there are no concrete special cases in which one can follow the approach of Faltings' proof that yield any significant simplifications. Faltings' proof is very indirect. First one uses rational points in $C(K)$ to construct coverings of $C$ that have good reduction outside a certain finite set of primes $S$. (This idea is, I believe, due to Parshin.) Taking Jacobians yields abelian varieties with good reduction outside $S$. One can, in principle, do this for a specific curve fairly concretely. But now one is reduced to Faltings proof of the Shafarevich conjecture, that there are only finitely many (suffices to do principally polarized) abelian varieties of a given dimension having good reduction outside $S$. And the proof of that is via reversing an argument of Tate to show it suffices to prove the isogeny conjecture, which gives a Galoistheoretic interpretation of isogenies between abelian varieties. And the proof of the isogeny conjecture is sufficiently complicated that I won't try to summarize it here. Anyway, bottom line is it does not appear that applying Faltings' ideas to a single curve would simplify the argument very much.
Having said that, there are more elementary methods that can prove that is $C(K)$ finite in some cases. There's the method of Dem'janenko: If there are independent maps $f_1,...,f_n:C\to E$ for some elliptic curve $E$, and if $n>\text{rank } E(K)$, then $C(K)$ is finite. This is done via a height calculation. It was generalized by Manin, who applied it to towers of modular curves $X_1(p^k)$. There is the method of Chabauty (strengthened to what is now usually called the ChabautyColeman method): Let $C/K$ and $J=\text{Jac}(C)$. If $\text{genus}(C)>\text{rank }J(K)$, then $C(K)$ is finite. The proof is via $p$adic analytic methods. Coleman made the method quite precise (using Coleman integration), so in many cases it allows one to actually compute $C(K)$. Note that Faltings original proof, and the VojtaBombieriFaltings alternative proof via Diophantine approximation, are ineffective.
Recently Lawrence and Venkatesh gave a proof that uses Faltings' setup but looks at variation of padic Galois representations in a family of algebraic varieties. Neither simpler nor a concrete special case, though.

1$\begingroup$ If it's neither simpler not a concrete special case, how is this an answer? This would make a fine comment, but not an answer on its own $\endgroup$ – Wojowu Sep 13 '19 at 22:10
