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I recently came across this question on MathOverflow asking if there are any polynomials {P} of two variables with rational coefficients, such that the map {P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}} is a bijection. The answer to this question is almost surely “no”, but it is remarkable how hard this problem resists any attempt at rigorous proof. (MathOverflow users with enough privileges to see deleted answers will find that there are no fewer than seventeen deleted attempts at a proof in response to this question!)

On the other hand, the one surviving response to the question does point out this paper of Poonen which shows that assuming a powerful conjecture in Diophantine geometry known as the Bombieri-Lang conjecture (discussed in this previous post), it is at least possible to exhibit polynomials {P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}} which are injective.

I believe that it should be possible to also rule out the existence of bijective polynomials {P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}} if one assumes the Bombieri-Lang conjecture, and have sketched out a strategy to do so, but filling in the gaps requires a fair bit more algebraic geometry than I am capable of. So as a sort of experiment, I would like to see if a rigorous implication of this form (similarly to the rigorous implication of the Erdos-Ulam conjecture from the Bombieri-Lang conjecture in my previous post) can be crowdsourced, in the spirit of the polymath projects (though I feel that this particular problem should be significantly quicker to resolve than a typical such project).

Here is how I imagine a Bombieri-Lang-powered resolution of this question should proceed (modulo a large number of unjustified and somewhat vague steps that I believe to be true but have not established rigorously). Suppose for contradiction that we have a bijective polynomial {P: {\bf Q} \times {\bf Q} \rightarrow {\bf Q}}. Then for any polynomial {Q: {\bf Q} \rightarrow {\bf Q}} of one variable, the surface

\displaystyle  S_Q := \{ (x,y,z) \in \mathbb{A}^3: P(x,y) = Q(z) \}

has infinitely many rational points; indeed, every rational {z \in {\bf Q}} lifts to exactly one rational point in {S_Q}. I believe that for “typical” {Q} this surface {S_Q} should be irreducible. One can now split into two cases:

  • (a) The rational points in {S_Q} are Zariski dense in {S_Q}.
  • (b) The rational points in {S_Q} are not Zariski dense in {S_Q}.

Consider case (b) first. By definition, this case asserts that the rational points in {S_Q} are contained in a finite number of algebraic curves. By Faltings’ theorem (a special case of the Bombieri-Lang conjecture), any curve of genus two or higher only contains a finite number of rational points. So all but finitely many of the rational points in {S_Q} are contained in a finite union of genus zero and genus one curves. I think all genus zero curves are birational to a line, and all the genus one curves are birational to an elliptic curve (though I don’t have an immediate reference for this). These curves {C} all can have an infinity of rational points, but very few of them should have “enough” rational points {C \cap {\bf Q}^3} that their projection {\pi(C \cap {\bf Q}^3) := \{ z \in {\bf Q} : (x,y,z) \in C \hbox{ for some } x,y \in {\bf Q} \}} to the third coordinate is “large”. In particular, I believe

  • (i) If {C \subset {\mathbb A}^3} is birational to an elliptic curve, then the number of elements of {\pi(C \cap {\bf Q}^3)} of height at most {H} should grow at most polylogarithmically in {H} (i.e., be of order {O( \log^{O(1)} H )}.
  • (ii) If {C \subset {\mathbb A}^3} is birational to a line but not of the form {\{ (f(z), g(z), z) \}} for some rational {f,g}, then then the number of elements of {\pi(C \cap {\bf Q}^3)} of height at most {H} should grow slower than {H^2} (in fact I think it can only grow like {O(H)}).

I do not have proofs of these results (though I think something similar to (i) can be found in Knapp’s book, and (ii) should basically follow by using a rational parameterisation {\{(f(t),g(t),h(t))\}} of {C} with {h} nonlinear). Assuming these assertions, this would mean that there is a curve of the form {\{ (f(z),g(z),z)\}} that captures a “positive fraction” of the rational points of {S_Q}, as measured by restricting the height of the third coordinate {z} to lie below a large threshold {H}, computing density, and sending {H} to infinity (taking a limit superior). I believe this forces an identity of the form

\displaystyle  P(f(z), g(z)) = Q(z) \ \ \ \ \ (1)

for all {z}. Such identities are certainly possible for some choices of {Q} (e.g. {Q(z) = P(F(z), G(z))} for arbitrary polynomials {F,G} of one variable) but I believe that the only way that such identities hold for a “positive fraction” of {Q} (as measured using height as before) is if there is in fact a rational identity of the form

\displaystyle  P( f_0(z), g_0(z) ) = z

for some rational functions {f_0,g_0} with rational coefficients (in which case we would have {f = f_0 \circ Q} and {g = g_0 \circ Q}). But such an identity would contradict the hypothesis that {P} is bijective, since one can take a rational point {(x,y)} outside of the curve {\{ (f_0(z), g_0(z)): z \in {\bf Q} \}}, and set {z := P(x,y)}, in which case we have {P(x,y) = P(f_0(z), g_0(z) )} violating the injective nature of {P}. Thus, modulo a lot of steps that have not been fully justified, we have ruled out the scenario in which case (b) holds for a “positive fraction” of {Q}.

This leaves the scenario in which case (a) holds for a “positive fraction” of {Q}. Assuming the Bombieri-Lang conjecture, this implies that for such {Q}, any resolution of singularities of {S_Q} fails to be of general type. I would imagine that this places some very strong constraints on {P,Q}, since I would expect the equation {P(x,y) = Q(z)} to describe a surface of general type for “generic” choices of {P,Q} (after resolving singularities). However, I do not have a good set of techniques for detecting whether a given surface is of general type or not. Presumably one should proceed by viewing the surface {\{ (x,y,z): P(x,y) = Q(z) \}} as a fibre product of the simpler surface {\{ (x,y,w): P(x,y) = w \}} and the curve {\{ (z,w): Q(z) = w \}} over the line {\{w \}}. In any event, I believe the way to handle (a) is to show that the failure of general type of {S_Q} implies some strong algebraic constraint between {P} and {Q} (something in the spirit of (1), perhaps), and then use this constraint to rule out the bijectivity of {P} by some further ad hoc method.

In 1946, Ulam, in response to a theorem of Anning and Erdös, posed the following problem:

Problem 1 (Erdös-Ulam problem) Let {S \subset {\bf R}^2} be a set such that the distance between any two points in {S} is rational. Is it true that {S} cannot be (topologically) dense in {{\bf R}^2}?

The paper of Anning and Erdös addressed the case that all the distances between two points in {S} were integer rather than rational in the affirmative.

The Erdös-Ulam problem remains open; it was discussed recently over at Gödel’s lost letter. It is in fact likely (as we shall see below) that the set {S} in the above problem is not only forbidden to be topologically dense, but also cannot be Zariski dense either. If so, then the structure of {S} is quite restricted; it was shown by Solymosi and de Zeeuw that if {S} fails to be Zariski dense, then all but finitely many of the points of {S} must lie on a single line, or a single circle. (Conversely, it is easy to construct examples of dense subsets of a line or circle in which all distances are rational, though in the latter case the square of the radius of the circle must also be rational.)

The main tool of the Solymosi-de Zeeuw analysis was Faltings’ celebrated theorem that every algebraic curve of genus at least two contains only finitely many rational points. The purpose of this post is to observe that an affirmative answer to the full Erdös-Ulam problem similarly follows from the conjectured analogue of Falting’s theorem for surfaces, namely the following conjecture of Bombieri and Lang:

Conjecture 2 (Bombieri-Lang conjecture) Let {X} be a smooth projective irreducible algebraic surface defined over the rationals {{\bf Q}} which is of general type. Then the set {X({\bf Q})} of rational points of {X} is not Zariski dense in {X}.

In fact, the Bombieri-Lang conjecture has been made for varieties of arbitrary dimension, and for more general number fields than the rationals, but the above special case of the conjecture is the only one needed for this application. We will review what “general type” means (for smooth projective complex varieties, at least) below the fold.

The Bombieri-Lang conjecture is considered to be extremely difficult, in particular being substantially harder than Faltings’ theorem, which is itself a highly non-trivial result. So this implication should not be viewed as a practical route to resolving the Erdös-Ulam problem unconditionally; rather, it is a demonstration of the power of the Bombieri-Lang conjecture. Still, it was an instructive algebraic geometry exercise for me to carry out the details of this implication, which quickly boils down to verifying that a certain quite explicit algebraic surface is of general type (Theorem 4 below). As I am not an expert in the subject, my computations here will be rather tedious and pedestrian; it is likely that they could be made much slicker by exploiting more of the machinery of modern algebraic geometry, and I would welcome any such streamlining by actual experts in this area. (For similar reasons, there may be more typos and errors than usual in this post; corrections are welcome as always.) My calculations here are based on a similar calculation of van Luijk, who used analogous arguments to show (assuming Bombieri-Lang) that the set of perfect cuboids is not Zariski-dense in its projective parameter space.

We also remark that in a recent paper of Makhul and Shaffaf, the Bombieri-Lang conjecture (or more precisely, a weaker consequence of that conjecture) was used to show that if {S} is a subset of {{\bf R}^2} with rational distances which intersects any line in only finitely many points, then there is a uniform bound on the cardinality of the intersection of {S} with any line. I have also recently learned (private communication) that an unpublished work of Shaffaf has obtained a result similar to the one in this post, namely that the Erdös-Ulam conjecture follows from the Bombieri-Lang conjecture, plus an additional conjecture about the rational curves in a specific surface.

Let us now give the elementary reductions to the claim that a certain variety is of general type. For sake of contradiction, let {S} be a dense set such that the distance between any two points is rational. Then {S} certainly contains two points that are a rational distance apart. By applying a translation, rotation, and a (rational) dilation, we may assume that these two points are {(0,0)} and {(1,0)}. As {S} is dense, there is a third point of {S} not on the {x} axis, which after a reflection we can place in the upper half-plane; we will write it as {(a,\sqrt{b})} with {b>0}.

Given any two points {P, Q} in {S}, the quantities {|P|^2, |Q|^2, |P-Q|^2} are rational, and so by the cosine rule the dot product {P \cdot Q} is rational as well. Since {(1,0) \in S}, this implies that the {x}-component of every point {P} in {S} is rational; this in turn implies that the product of the {y}-coordinates of any two points {P,Q} in {S} is rational as well (since this differs from {P \cdot Q} by a rational number). In particular, {a} and {b} are rational, and all of the points in {S} now lie in the lattice {\{ ( x, y\sqrt{b}): x, y \in {\bf Q} \}}. (This fact appears to have first been observed in the 1988 habilitationschrift of Kemnitz.)

Now take four points {(x_j,y_j \sqrt{b})}, {j=1,\dots,4} in {S} in general position (so that the octuplet {(x_1,y_1\sqrt{b},\dots,x_4,y_4\sqrt{b})} avoids any pre-specified hypersurface in {{\bf C}^8}); this can be done if {S} is dense. (If one wished, one could re-use the three previous points {(0,0), (1,0), (a,\sqrt{b})} to be three of these four points, although this ultimately makes little difference to the analysis.) If {(x,y\sqrt{b})} is any point in {S}, then the distances {r_j} from {(x,y\sqrt{b})} to {(x_j,y_j\sqrt{b})} are rationals that obey the equations

\displaystyle (x - x_j)^2 + b (y-y_j)^2 = r_j^2

for {j=1,\dots,4}, and thus determine a rational point in the affine complex variety {V = V_{b,x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4} \subset {\bf C}^5} defined as

\displaystyle V := \{ (x,y,r_1,r_2,r_3,r_4) \in {\bf C}^6:

\displaystyle (x - x_j)^2 + b (y-y_j)^2 = r_j^2 \hbox{ for } j=1,\dots,4 \}.

By inspecting the projection {(x,y,r_1,r_2,r_3,r_4) \rightarrow (x,y)} from {V} to {{\bf C}^2}, we see that {V} is a branched cover of {{\bf C}^2}, with the generic cover having {2^4=16} points (coming from the different ways to form the square roots {r_1,r_2,r_3,r_4}); in particular, {V} is a complex affine algebraic surface, defined over the rationals. By inspecting the monodromy around the four singular base points {(x,y) = (x_i,y_i)} (which switch the sign of one of the roots {r_i}, while keeping the other three roots unchanged), we see that the variety {V} is connected away from its singular set, and thus irreducible. As {S} is topologically dense in {{\bf R}^2}, it is Zariski-dense in {{\bf C}^2}, and so {S} generates a Zariski-dense set of rational points in {V}. To solve the Erdös-Ulam problem, it thus suffices to show that

Claim 3 For any non-zero rational {b} and for rationals {x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4} in general position, the rational points of the affine surface {V = V_{b,x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4}} is not Zariski dense in {V}.

This is already very close to a claim that can be directly resolved by the Bombieri-Lang conjecture, but {V} is affine rather than projective, and also contains some singularities. The first issue is easy to deal with, by working with the projectivisation

\displaystyle \overline{V} := \{ [X,Y,Z,R_1,R_2,R_3,R_4] \in {\bf CP}^6: Q(X,Y,Z,R_1,R_2,R_3,R_4) = 0 \} \ \ \ \ \ (1)

 

of {V}, where {Q: {\bf C}^7 \rightarrow {\bf C}^4} is the homogeneous quadratic polynomial

\displaystyle (X,Y,Z,R_1,R_2,R_3,R_4) := (Q_j(X,Y,Z,R_1,R_2,R_3,R_4) )_{j=1}^4

with

\displaystyle Q_j(X,Y,Z,R_1,R_2,R_3,R_4) := (X-x_j Z)^2 + b (Y-y_jZ)^2 - R_j^2

and the projective complex space {{\bf CP}^6} is the space of all equivalence classes {[X,Y,Z,R_1,R_2,R_3,R_4]} of tuples {(X,Y,Z,R_1,R_2,R_3,R_4) \in {\bf C}^7 \backslash \{0\}} up to projective equivalence {(\lambda X, \lambda Y, \lambda Z, \lambda R_1, \lambda R_2, \lambda R_3, \lambda R_4) \sim (X,Y,Z,R_1,R_2,R_3,R_4)}. By identifying the affine point {(x,y,r_1,r_2,r_3,r_4)} with the projective point {(X,Y,1,R_1,R_2,R_3,R_4)}, we see that {\overline{V}} consists of the affine variety {V} together with the set {\{ [X,Y,0,R_1,R_2,R_3,R_4]: X^2+bY^2=R^2; R_j = \pm R_1 \hbox{ for } j=2,3,4\}}, which is the union of eight curves, each of which lies in the closure of {V}. Thus {\overline{V}} is the projective closure of {V}, and is thus a complex irreducible projective surface, defined over the rationals. As {\overline{V}} is cut out by four quadric equations in {{\bf CP}^6} and has degree sixteen (as can be seen for instance by inspecting the intersection of {\overline{V}} with a generic perturbation of a fibre over the generically defined projection {[X,Y,Z,R_1,R_2,R_3,R_4] \mapsto [X,Y,Z]}), it is also a complete intersection. To show (3), it then suffices to show that the rational points in {\overline{V}} are not Zariski dense in {\overline{V}}.

Heuristically, the reason why we expect few rational points in {\overline{V}} is as follows. First observe from the projective nature of (1) that every rational point is equivalent to an integer point. But for a septuple {(X,Y,Z,R_1,R_2,R_3,R_4)} of integers of size {O(N)}, the quantity {Q(X,Y,Z,R_1,R_2,R_3,R_4)} is an integer point of {{\bf Z}^4} of size {O(N^2)}, and so should only vanish about {O(N^{-8})} of the time. Hence the number of integer points {(X,Y,Z,R_1,R_2,R_3,R_4) \in {\bf Z}^7} of height comparable to {N} should be about

\displaystyle O(N)^7 \times O(N^{-8}) = O(N^{-1});

this is a convergent sum if {N} ranges over (say) powers of two, and so from standard probabilistic heuristics (see this previous post) we in fact expect only finitely many solutions, in the absence of any special algebraic structure (e.g. the structure of an abelian variety, or a birational reduction to a simpler variety) that could produce an unusually large number of solutions.

The Bombieri-Lang conjecture, Conjecture 2, can be viewed as a formalisation of the above heuristics (roughly speaking, it is one of the most optimistic natural conjectures one could make that is compatible with these heuristics while also being invariant under birational equivalence).

Unfortunately, {\overline{V}} contains some singular points. Being a complete intersection, this occurs when the Jacobian matrix of the map {Q: {\bf C}^7 \rightarrow {\bf C}^4} has less than full rank, or equivalently that the gradient vectors

\displaystyle \nabla Q_j = (2(X-x_j Z), 2(Y-y_j Z), -2x_j (X-x_j Z) - 2y_j (Y-y_j Z), \ \ \ \ \ (2)

 

\displaystyle 0, \dots, 0, -2R_j, 0, \dots, 0)

for {j=1,\dots,4} are linearly dependent, where the {-2R_j} is in the coordinate position associated to {R_j}. One way in which this can occur is if one of the gradient vectors {\nabla Q_j} vanish identically. This occurs at precisely {4 \times 2^3 = 32} points, when {[X,Y,Z]} is equal to {[x_j,y_j,1]} for some {j=1,\dots,4}, and one has {R_k = \pm ( (x_j - x_k)^2 + b (y_j - y_k)^2 )^{1/2}} for all {k=1,\dots,4} (so in particular {R_j=0}). Let us refer to these as the obvious singularities; they arise from the geometrically evident fact that the distance function {(x,y\sqrt{b}) \mapsto \sqrt{(x-x_j)^2 + b(y-y_j)^2}} is singular at {(x_j,y_j\sqrt{b})}.

The other way in which could occur is if a non-trivial linear combination of at least two of the gradient vectors vanishes. From (2), this can only occur if {R_j=R_k=0} for some distinct {j,k}, which from (1) implies that

\displaystyle (X - x_j Z) = \pm \sqrt{b} i (Y - y_j Z) \ \ \ \ \ (3)

 

and

\displaystyle (X - x_k Z) = \pm \sqrt{b} i (Y - y_k Z) \ \ \ \ \ (4)

 

for two choices of sign {\pm}. If the signs are equal, then (as {x_j, y_j, x_k, y_k} are in general position) this implies that {Z=0}, and then we have the singular point

\displaystyle [X,Y,Z,R_1,R_2,R_3,R_4] = [\pm \sqrt{b} i, 1, 0, 0, 0, 0, 0]. \ \ \ \ \ (5)

 

If the non-trivial linear combination involved three or more gradient vectors, then by the pigeonhole principle at least two of the signs involved must be equal, and so the only singular points are (5). So the only remaining possibility is when we have two gradient vectors {\nabla Q_j, \nabla Q_k} that are parallel but non-zero, with the signs in (3), (4) opposing. But then (as {x_j,y_j,x_k,y_k} are in general position) the vectors {(X-x_j Z, Y-y_j Z), (X-x_k Z, Y-y_k Z)} are non-zero and non-parallel to each other, a contradiction. Thus, outside of the {32} obvious singular points mentioned earlier, the only other singular points are the two points (5).

We will shortly show that the {32} obvious singularities are ordinary double points; the surface {\overline{V}} near any of these points is analytically equivalent to an ordinary cone {\{ (x,y,z) \in {\bf C}^3: z^2 = x^2 + y^2 \}} near the origin, which is a cone over a smooth conic curve {\{ (x,y) \in {\bf C}^2: x^2+y^2=1\}}. The two non-obvious singularities (5) are slightly more complicated than ordinary double points, they are elliptic singularities, which approximately resemble a cone over an elliptic curve. (As far as I can tell, this resemblance is exact in the category of real smooth manifolds, but not in the category of algebraic varieties.) If one blows up each of the point singularities of {\overline{V}} separately, no further singularities are created, and one obtains a smooth projective surface {X} (using the Segre embedding as necessary to embed {X} back into projective space, rather than in a product of projective spaces). Away from the singularities, the rational points of {\overline{V}} lift up to rational points of {X}. Assuming the Bombieri-Lang conjecture, we thus are able to answer the Erdös-Ulam problem in the affirmative once we establish

Theorem 4 The blowup {X} of {\overline{V}} is of general type.

This will be done below the fold, by the pedestrian device of explicitly constructing global differential forms on {X}; I will also be working from a complex analysis viewpoint rather than an algebraic geometry viewpoint as I am more comfortable with the former approach. (As mentioned above, though, there may well be a quicker way to establish this result by using more sophisticated machinery.)

I thank Mark Green and David Gieseker for helpful conversations (and a crash course in varieties of general type!).

Remark 5 The above argument shows in fact (assuming Bombieri-Lang) that sets {S \subset {\bf R}^2} with all distances rational cannot be Zariski-dense, and thus (by Solymosi-de Zeeuw) must lie on a single line or circle with only finitely many exceptions. Assuming a stronger version of Bombieri-Lang involving a general number field {K}, we obtain a similar conclusion with “rational” replaced by “lying in {K}” (one has to extend the Solymosi-de Zeeuw analysis to more general number fields, but this should be routine, using the analogue of Faltings’ theorem for such number fields).

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