# Infinitely many solutions of a diophantine equation

If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor.

Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric).

Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space.

The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).

Update: To make the question more concrete, consider one of the easiest (comparing to the other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $m\ne 0$ the equation $P(x,y)=m$ has only finitely many integer solutions. It is Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this.

Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have $|(x/y-a)|\cdot |(x/y-b)|\cdot |(x/y-c)|=O(1)/|y^3|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $|x/y-a|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $|x/y-a|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction.

The question is then: is there an asymptotic geometry proof of the Thue theorem.

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Great question! – HJRW Jan 7 '11 at 17:38

Revised

Interesting question.

Here's a thought:

You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms $x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times the order 2 group generated by $-1$.

If you take a submonoid that uses only one prime, it is quasi-isometric to a quotient of the hyperbolic plane by an action of $\mathbb Z$, which is multiplication by $p$ in the upper half-space model. To see this, place a dots labeled by integer $n$ at position $(n*p^k, p^k)$ in the upper half plane, for every pair of integers $(n,k)$, and connect them by horizontal line segments and by vertical line segments whenever points are in vertical alignment. The quotient of upper half plane by the hyperbolic isometry $(x,y) \rightarrow (p*x, p*y)$ has a copy of the Cayley graph for this monoid. This is also quasi-isometric to the 1-point union of two copies of the hyperbolic plane, one for negative integers, one for positive integes. It's a fun exercise, using say $p = 2$. Start from 0, and recursively build the graph by connecting $n$ to $n+1$ by one color arrow, and $n$ to $2*n$ by another color arrow. If you arrange positive integers in a spiral, you can make a neat drawing of this graph (or the corresponding graph for a different prime.) The negative integers look just the same, but with the successor arrow reversed.

If you use several primes, the picture gets more complicated. In any case, one can take rescaled limits of these graphs, based at sequence of points, and get asymptotic cones for the monoid. The graph is not homogeneous, so there is not just one limit.

Another point of view is to take limits of $\mathbb Z$ without rescaling, but with a $k$-tuple of constants $(n_1, \dots , n_k)$. The set of possible identities among polynomials in $k$ variables is compact, so there is a compact space of limit rings for $\mathbb Z$ with $k$ constants. Perhaps this is begging the question: the identitites that define the limits correspond to diophantine equations that have infinitely many solutions. Rescaling may eliminate some of this complexity.

A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to $\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on some asymptotic cone for the affine monoid for $\mathbb Z$.

With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest.

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@Bill: Something is wrong with your comment, I think there is a TeX error there. I did manage to read the first line though. The Cayley graph of a monoid is a directed graph and one cannot go against the arrows (unlike in the group case). So the distance is not quite defined. – Mark Sapir Jan 7 '11 at 1:09
@Bill: How to define an asymptotic cone of a monoid? What to rescale and by which factors? I was thinking about Vojta's proof of Faltings' theorem and previous, easier, results of Thue, Siegel, Roth and many others. They deduce finiteness of the number of solutions from some diophantine approximation results saying that certain numbers cannot be approximated very well by rational numbers. Perhaps these results can have some geometric interpretation. Continuous fractions do appear naturally in the study of the action of PSL2 on H2 also there are results of Margulis and others. – Mark Sapir Jan 7 '11 at 2:17
@Bill: The problem is also how to define the scaling factors. Rescaling in the group theoretic case implies that the group $\Gamma$ acts on the cone so that every point of the cone is moved by at least one of the generators by distance 1. Hence the action does not have a global fixed point. If the scaling constants are not carefully chosen, this property will fail. If the distance is not symmetric (as in a directed graph), then the distance may be infinitity between some points. In particular, a map may take a point to a point which is infinitely far away. But one cannot rescale by $\infty$. – Mark Sapir Jan 7 '11 at 2:24
This was actually the first comment. I removed it because I thought it was the cause of the TeX problem. How to define an asymptotic cone of a monoid? What to rescale and by which factors? I was thinking about Vojta's proof of Faltings' theorem and previous, easier, results of Thue, Siegel, Roth and many others. They deduce finiteness of the number of solutions from some diophantine approximation results saying that certain numbers cannot be approximated very well by rational numbers. Continuous fractions do appear naturally in the study of the action of $PSL_2$ on $\mathbb{H}^2$. – Mark Sapir Jan 7 '11 at 2:26
@Mark: For a monoid, I don't see the problem with using the symmetric distance function on its Cayley graph, just thinking of it as a space. The algebraic description of distance is more complicated, but OK. But the quasigeometry of the monoid of affine endomorphisms of Z is more tangled than I first imagined, so I think the big issue is getting a good enough picture of the rescaled limits to be able to say something interesting. Unfortunately I don't know enough the work of Vojta, Faltings, Siegel, Roth or others to have a good feel for what might be helpful. – Bill Thurston Jan 7 '11 at 4:16