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.
Best Answer
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.