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Math thread (need a brake from prep moving city) - my first one ever so be kind!
"Geometry determines Arithmetic" S. Lang
What does it mean? Let's check it out!
1/n
"Geometry determines Arithmetic" S. Lang
What does it mean? Let's check it out!
1/n
I'll focus on the easiest case. Consider a polynomial f(X,Y) with (say) integral coefficients. We are interested in the solution set S: f(X,Y) = 0 with both X and Y rationals (or integers). Q: is there a way to say something not trivial about S depending on properties of f?
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2/n
A first guess could be that the (total) degree of f plays a role. This is sort of true: the larger the degree the smaller the set S. But there's a catch: this is only true generically. Namely there are polys of high degree with a lot of rational sols (and viceversa).
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3/n
So what do we do? We need some better invariant that helps us describe the solution set. The idea is to consider the geometric object associated to f = 0, namely the variety/manifold that is the set of *complex* solutions of the original equation.
4/n
4/n
Since we restricted to two variables this geometric object is (almost always - singularities aside) an (open subset of a) Riemann surface R_f.
We have a natural invariant of R_f, namely its genus g. Turns out that the value of g determines the behavior of the set S.
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We have a natural invariant of R_f, namely its genus g. Turns out that the value of g determines the behavior of the set S.
5/n
Various results and theorems (some highly non-trivial) show that if g <= 1 then S is either empty or (Zariski) dense in R_f. OTOH if g>=2 then S is always finite.
So at least qualitatively g "determines" the arithmetic.
In fact we can give a more precise tricothomy:
6/n
So at least qualitatively g "determines" the arithmetic.
In fact we can give a more precise tricothomy:
6/n
If g = 0 then R_f is the Riemann Sphere (with a metric of positive const curvature). If there is a rational solution to f(X,Y) = 0 then there are infinitely many and they are dense in R_f. Moreover there is always a solution up to looking in a finite (deg 2) extension of \\Q.
7/n
7/n
In this case we say that the solution set S is "potentially" dense. This means that the set is dense once we consider solutions in a *finite* extension of \\Q. E.g. f = x^2 + y^2 + 1 has empty sol set but the set of \\Q(i) sols is dense.
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8/n
If g = 1 then R_f is a torus (with a flat metric). Here there are 3 possibilities: either S is empty, or S is finite or it is dense. However the set of solutions is always potentially dense (i.e. there is always an extension that contains a point of infinite order)
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9/n
In this case, up to a finite extension k of \\Q, R_f is an elliptic curve of positive rank over k, and the set S has the structure of a finitely generated abelian group - this is the Mordell(-Weil) Theorem.
10/n
10/n
Finally If g>=2 then R_f is a connected sum of g tori, i.e. R_f has g holes (with a metric of negative constant curvature). Here the set S is always finite, even when looking at solutions defined over an extension. This is Faltings' Theorem, aka Mordell's Conjecture.
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11/n
Lang's insight is that a similar behavior should be true in any dimension (i.e. for solutions sets of polynomials in any number of variables). Of course one would need more invariants in this case and one enters the world of conjectures.
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12/n
Already in dimension 2 we know very few particular cases where we can say non trivial things about the solution set. The story becomes even more intriguing when looking at the analogy with complex analysis (Vojta's dictionary) but that's it for now.
13/13 n=13
13/13 n=13
