Geometric intuition behind The Mordell Conjecture The Mordell Conjecture/Faltings Theorem says roughly that if $K$ is an algebraic number field and $X$ is an algebraic curve defined over $K$ of genus $g >1$ then the set of $K$-rational points $X(K)$ is finite.
Now in the cases when the genus is $g = 0$ or $g = 1$ we can have an infinite number of $K$-rational points in the curve, but then Mordell's conjecture says that when the genus goes on to the next higher value $g \geq 2$ then this can no longer happen and we can only have a finite number of $K$-rational points lying on the curve (if there are any).
The usual topological interpretation that one sees mentioned in the books when they talk about the genus is as the number of holes in the surface $X(\mathbb{C})$. Thus I was just wondering if there's some way to see intuitively or at least without having to resort to highly technical arguments using scheme theory why this happens?
To be more specific, is there a way to see intuitively how the fact that the surface $X(\mathbb{C})$ has more than one hole (when the genus is greater than one) prevents the corresponding algebraic curve from having an infinite number of $K$-rational points?

Note: If it helps to ease things at least a little bit, I would be more than happy with an argument for the case when $K = \mathbb{Q}$ the set of rational numbers. I'm aware that this is highly technical stuff so maybe I'm asking for too much, but it would be very interesting if there is an argument for this so I cannot resist asking.
Thank you very much in advance for any help. 
 A: You should certainly look at the answers to the MO question linked to above by Qiaochu.
In any case, even over $\mathbb Q$ there is no simple proof of Mordell's conjecture.
Here is one geometric way to think about the conjecture, which may not be emphasized in the linked MO answers:
A projective curve over $\mathbb Z$ admits a projective model over $\mathbb Z$, and by the valuative criterion, a $\mathbb Q$-valued point of this model is the same thing as a $\mathbb Z$-valued point.    This leads to the more general problem of thinking about $\mathbb Z$-valued points of curves over $\mathbb Z$.
The reason I write "more general" is because this problem now includes affine curves with coefficients in $\mathbb Z$ (in which case $\mathbb Z$-valued points are not the same as $\mathbb Q$-valued points; it is much more restrictive for a point in affine space to have integral coordinates than rational coordinates).
In this case one has Siegel's theorem mentioned above by Arturo.  This says that on an affine curve over $\mathbb Z$ of genus $> 0$ (and by genus here I mean genus of the projectivization) there are only finitely many rational points.  Of course, if $g > 1$ this is weaker than Faltings's theorem (which says the same thing even when we pass to the projectivization), but Siegel's theorem was earlier, and includes the case of
an elliptic curve with the point at infinity removed.  Note that one can also
replace $\mathbb Q$ by any number field $K$ and $\mathbb Z$ by its ring of integers, or in fact by any localization of the ring of integers obtained by inverting a finite
number of primes, i.e. at any ring of $S$-integers --- see e.g.
this formulation of Siegel's theorem.
This result in turn is related to the $S$-unit theorem, which says that if you consider the affine curve $\mathbb A^1 \setminus \{0,1\}$, there are only
finitely many $\mathcal O$-valued points whenever $\mathcal O$ is the ring of $S$-integers in any number field.  (To understand this, note that the
description of $\mathbb A^1\setminus \{0,1\}$ as an affine curves is the
set of $(x,y,z)$ such that $xy = (x-1)z = 1$.  Thus giving an $S$-integral
point of $\mathbb A^1\setminus \{0,1\}$ is the same as giving an $S$-unit
$x$ so that $x-1$, or equivalently $1-x$, is also an $S$-unit, which in turn
is the same as giving a pair of $S$-units satisfying $u+v =1$; this is how
the $S$-unit equation is usually phrased.)
What do (affine or projective) curves of genus $g > 1$, affine curves of
genus $g = 1$, and the curve $\mathbb A^1\setminus \{0,1\}$ (or indeed
$\mathbb P^1$ minus any collection of at least three points) have in common?
They are all hyperbolic curves.
So the basic geometric fact underlying Mordell's conjecture, Seigel's theorem, and the $S$-unit theorem, is that there are only finitely many integral (or even $S$-integral, for a given ring of $S$-integers) points on a hyperbolic curve.
There is an analogy (developed in the work of Vojta mentioned by Arturo above) between this situation in arithmetic and a certain phenomenon in complex function theory:
Recall Picard's theorem, which states that a holomorphic function on $\mathbb C$ missing more than one value is constant.  Another way to think of this is that it says that a holomorphic function from $\mathbb C$ into $\mathbb A^1\setminus \{0,1\}$ is constant.
More generally, the same is true for holomorphic maps of $\mathbb C$ into any hyperbolic Riemann surface.  (The usual proof is that one lifts to a map on
the universal cover, to get a holomorphic map from $\mathbb C$ to the unit disk,
which is constant by Liouville.)
To be more precise, then, there is an analogy between integral points of hyperbolic curves and holomorphic maps from $\mathbb C$ to hyperbolic curves.  If you want to see this analogy spelled out in more detail, you can look up some of the references on Vojta's work.
The idea that hyperbolic varieties should not contain many integral points was greatly extended by Lang, in analogy with corresponding results
in complex function theory.
