aa r X i v : . [ m a t h . N T ] N ov THE OPTIMALITY OF THE BOUNDED HEIGHT CONJECTURE
Evelina Viada
In this article we show that the Bounded Height Conjecture is optimal in thesense that, If V is an irreducible variety in a power of an elliptic curve with emptydeprived set, then all open subsets of V do not have bounded height. The BoundedHeight Conjecture is known to hold. We also present some examples and remarks.1. introduction This work concerns principally the optimality of the Bounded Height Conjecture,stated by Bombieri, Masser and Zannier [2] and proven by Habegger [3]. In section2, we clarify the assumption on the varieties, understanding such a hypothesisgeometrically and from different points of view. We give some examples, to makesure that certain situations can occur. In section 3, we prove the optimality of theBounded Height Conjecture. In the final section we present some further remarksand possible open questions.Denote by A an abelian variety over Q of dimension g . Consider on A ( Q ) acanonical height function. Denote by || · || the induced semi-norm. For ε ≥
0, wedenote O ε = { ξ ∈ A ( Q ) : || ξ || ≤ ε } . Consider a proper irreducible algebraic subvariety V of dimension d embedded in A , defined over Q . We say that:- V is transverse, if V is not contained in any translate of a proper algebraicsubgroup of A .- V is weak-transverse, if V is not contained in any proper algebraic subgroupof A .Given an integer r with 1 ≤ r ≤ g and a subset F of A ( Q ), we define the set S r ( V, F ) = V ( Q ) ∩ [ cod B ≥ r B + F where B varies over all abelian subvarieties of A of codimension at least r and B + F = { b + f : b ∈ B, f ∈ F } . Note that S r +1 ( V, F ) ⊂ S r ( V, F ) . We denote the set S r ( V, A
Tor ) simply by S r ( V ), where A Tor is the torsion of A . Fora subset V ′ ⊂ V , we denote S r ( V ′ , F ) = V ′ ∩ S r ( V, F ) . It is natural to ask: ‘For which sets F and integers r , has the set S r ( V, F ) boundedheight or is it non-Zariski dense in V ?’ Evelina Viada, Universit´e de Friboug Suisse, P´erolles, D´epartement de Math´ematiques,Chemin du Mus´ee 23, CH-1700 Fribourg, Switzerland, [email protected]. Supported by the SNF (Swiss National Science Foundation). Mathematics Subject classification (2000): 11G50, 14H52, 14K12.Key words: Height, Elliptic curves, Subvarieties.
Sets of this kind, for r = g , appear in the literature in the context of the Mordell-Lang, of the Manin-Mumford and of the Bogomolov Conjectures. More recentlyBombieri, Masser and Zannier [1] have proven that:For a transverse curve C in a torus,i. The set S ( C ) has bounded height,ii. The set S ( C ) is finite.They investigate for the first time, intersections with the union of all algebraicsubgroups of a given codimension. This opens a vast number of conjectures forsubvarieties of semi-abelian varieties.Most naively, one could risk the following:For V a transverse subvariety of A ,i. S d ( V ) has bounded height,ii. S d +1 ( V ) is non-Zariski dense in V .We will show that i. is a too optimistic guess.Several problems rise for varieties. A proper Zariski closed subset of a curve hasbounded height. In general, a proper Zariski closed subset of a variety does not havebounded height, however it is still a ‘small’ set. So one shall say, that outside ananomalous Zariski closed subset of V , the points we consider have bounded height.Bombieri, Masser and Zannier introduced the anomalous set. Hardest is to showthat it is closed. Definition 1.1 ([2] Definition 1.1 and 1.2) . An irreducible subvariety X of V isanomalous if it has positive dimension and lies in a coset H of A satisfying dim H ≤ n − dim V + dim X − . The deprived set V oa is what remains of V after removing all anomalous subvari-eties. For tori, they prove
Theorem 1.1 ([2] Theorem 1.4.) . The deprived set V oa is a Zariski open of V . Then, they state the following conjecture for tori and ε = 0. Conjecture 1.1 (Bounded Height Conjecture) . Let V be an irreducible variety in A of dimension d . Then, there exists ε > such that S d ( V oa , O ε ) has boundedheight. We remark that in all known effective proofs, the bound for the height of S d ( V oa )is independent of the field of definition of V . Then, a set F of bounded height doesnot harm.For transverse curves in a torus [1] and in a product of elliptic curves [7], Conjec-ture 1.1 is effectively proven. In a preprint P. Habegger [3] deals with subvarietiesof an abelian variety A defined over the algebraic numbers. He shows: Theorem 1.2 (Habegger [3]) . For V an irreducible subvariety of A , Conjecture1.1 holds. In the first instance we analyze several geometric properties which are differentfor varieties, but they all collapse to the transversal condition for curves.
Property ( S n ) . We say that V satisfies Property ( S n ) if, for all morphism φ : A → A such that dim φ ( A ) ≥ d + n , dim φ ( V ) = d. We simply say Property ( S ) for ( S ) . HE OPTIMALITY OF THE BOUNDED HEIGHT CONJECTURE 3
In some sense Property ( S ) is natural. Property ( S n ) implies Property ( S n +1 )and also implies transversality. For curves, transverse implies Property ( S ).Habegger and R´emond (see lemma 3.3) show that property (S) is equivalent tothe assumption V oa = ∅ . Then, one can easily reformulate the Bonded HeightConjecture in terms of Property ( S ), avoiding the notion of deprived set. Conjecture 1.2 (Bounded Height Conjecture) . Let V be an algebraic subvariety of A defined over Q . Suppose that V satisfies Property ( S ) . Then, there exists ε > and a non-empty open subset V e of V such that S d ( V e , O ε ) has bounded height. One could hope to relax the assumption of Property S on the variety. Could itbe sufficient to assume, as we do for curves, that V is transverse? What abouta product of varieties which do satisfy Property S ? In section 3, we prove thattheorem 1.2 is optimal for subvarieties of a power of an elliptic curves E g . Theorem 1.3.
Let V be a subvariety of E g of dimension d . Suppose that V doesnot satisfy Property ( S ) (or equivalently that V oa = ∅ ). Then, for every non-emptyZariski open subset U of V the set S d ( U ) does not have bounded height. The proof is constructive. A fundamental point is to associate to a non-torsionpoint of E ( Q ) a Zariski dense subgroup of E n .A natural rising question is to investigate the height for larger codimension ofthe algebraic subgroup. Let Γ be a subgroup of A ( Q ) of finite rank. We denoteΓ ε = Γ + O ε . Conjecture 1.3.
Let V be an irreducible algebraic subvariety of A of dimension d , defined over Q . Then there exists ε > and a non-empty Zariski open subset V e of V such that: i. If V is weak-transverse, S d +1 ( V e , O ε ) has bounded height. ii. If V is transverse, S d +1 ( V e , Γ ε ) has bounded height. In some cases Conjecture 1.3 is proven. For Γ = 0 or V weak-transverse but nottransverse, the method used for the proofs is based on a Vojta inequality. Thismethod is not effective. It gives optimal results for curves (see [6] Theorem 1.5and [8] Theorem 1.2). On the contrary, for varieties of dimension at least two ahypothesis stronger than transversality is needed. Part i. of the following theoremis proven by R´emond [4] Theorem 1.2 and [5]. Whereas, Part ii. is proven by theauthor [9] Theorem 1.5. Theorem 1.4.
Let V be an irreducible subvariety of E g of dimension d , definedover Q . Let p be a point in E s ( Q ) not lying in any proper algebraic subgroup of E s . Assume that V satisfies (1) dim( V + B ) = min(dim V + dim B, g ) for all abelian subvarieties B of E g . Then there exists a non-empty Zariski opensubset V e of V and ε > such that: i. S d +1 ( V e , Γ ε ) has bounded height, ii. S d +1 ( V e × p, O ε ) has bounded height. In Lemma 4.1, we will see that the assumption (1) is equivalent to Property ( S ).Finally we give some examples of varieties satisfying Property ( S ) and of varietieswhich do not satisfy Property ( S ) but for which Conjecture 1.3 holds.To conclude we remark that, if one knows that, for r ≥ d + 1 and V transverse, theset S r ( V e , Γ ε ) has bounded height, then [9] Theorem 1.1 implies that S r ( V e , Γ ε )is not Zariski dense in V . If Γ has trivial rank, it is sufficient to assume V weak-transverse. This makes results on heights particularly interesting. Acknowledgments:
I kindly thank the Referee for his accurate and nice suggestions.
EVELINA VIADA preliminaries Let E be an elliptic curve defined over a number field. All statement in the in-troduction become trivially verified for a zero-dimensional variety. In the followingwe avoid this case. Let V be an irreducible algebraic subvariety of E g of dimension0 < d < g defined over Q .We fix on E ( Q ) the canonical N´eron-Tate height function. We denote by || · || theinduced semi-norm on E ( Q ). For x = ( x , . . . , x g ) ∈ E g ( Q ), we denote || x || = max i || x i || . For ε ≥
0, we define O ε = { ξ ∈ E g ( Q ) : || ξ || ≤ ε } . The height of a non-empty set S ⊂ E g ( Q ) is the supremum of the heights of itselements. The degree of S is the degree (possibly ∞ ) of the field of definition ofthe points of S .The ring of endomorphism End( E ) is isomorphic either to Z (if E does not haveC.M.) or to an order in an imaginary quadratic field (if E has C.M.). We consideron End( E ) the hermitian scalar product h· , ·i induced by C and denote by | · | the associated norm. Note that the metric does not depend on the embedding ofEnd( E ) in C .We denote by M r,g (End( E )) the module of r × g matrices with entries in End( E ).For F = ( f ij ) ∈ M r,g (End( E )), we define | F | = max ij | f ij | . We identify a morphism φ : E g → E r with a matrix in M r,g (End( E )).Let B be an algebraic subgroup of E g of codimension r . Then B ⊂ ker φ B fora surjective morphism φ B : E g → E r . Conversely, we denote by B φ the kernel ofa surjective morphism φ : E g → E r . Then B φ is an algebraic subgroup of E g ofcodimension r .If φ : E g → E g ′ is a surjective morphism, we can complement φ and definean isogeny f : E g → E g such that f (ker φ ) = 0 × E g − g ′ and π f = φ , where π : E g → E g ′ is the natural projection on the first g ′ coordinate. More precisely;recall that every abelian subvariety of E g of dimension n is isogenous to E n . Thenker φ is isogenous to E g − g ′ , let i be such an isogeny. Let (ker φ ) ⊥ be an orthogonalcomplement of ker φ in E g . Then E g ′ is isogenous to (ker φ ) ⊥ . Let j : E g ′ → (ker φ ) ⊥ be such an isogeny. Define the isogeny f : E g → E g x → ( φ ( x ) , i ( x − j ( φ ( x ))) . This f has the wished property.Let us state a classical: Lemma 2.1.
For every algebraic subvariety X of E g of dimension d there existsa projection on d coordinates such that the restriction to X is dominant.Proof. Let d be the maximal integer such that the restriction of π : E g → E d to X is surjective. If d ≥ d , nothing has to be shown. Suppose that d < d. Without loss of generality, suppose that π projects on the first d coordinates.For d < i ≤ g , we define π i : E g → E d +1 to be the projection π i ( x , . . . , x g ) → ( x , . . . , x d , x i ) . Let i π : E d +1 → E g be the immersion such that π i · i π = id E d .We denote by X i = i π · π i ( X ) ⊂ E g . By maximality of d we see that dim X i = d .Furthermore X is the fiber product of X i over π ( X ) = π ( X i ). Then d = dim X =dim( X d +1 × π ( X ) · · · × π ( X ) X g ) = d , which contradicts d > d . (cid:3) HE OPTIMALITY OF THE BOUNDED HEIGHT CONJECTURE 5
We show an easy application.
Lemma 2.2. If V does not satisfy Property ( S ) then there exists a surjective mor-phism φ : E g → E d such that < dim φ ( V ) < d .Proof. If V does not satisfy Property ( S ), then there exists a surjective morphism φ : E g → E d such that dim φ ( V ) < d . If dim φ ( V ) >
0, nothing has to be shown.If dim φ ( V ) = 0, Lemma 2.1 gives a morphism r : E g → E such that the restrictionto X is surjective. Replace the first row of φ by r . (cid:3) The Bounded Height Conjecture and its optimality
In the following we first show that the set S d ( V ) is dense in V .We then ask if Property ( S ) is necessary to show that S d ( V ) has bounded height.We give here a positive answer. Meanwhile we try to understand the geometricaspect of Property ( S ).An easy example of a variety which does not satisfy Property ( S ) is a split variety V × V × · · · × V n with the V i ⊂ E g i . It is natural to ask if only this kind of productvarieties do not satisfying Property ( S ). This is not the case, as Lemma 3.2 andExample 3.3 show. Definition 3.1.
Let V ⊂ E g be a variety of dimension d . i. V is split if there exists an isogeny φ : E g → E g such that φ ( V ) = V × V with V i ⊂ E g i and g i = 0 , for i = 1 , .We say that V is non-split if the above property is not verified. ii. V is n -generically split if there exists an isogeny φ : E g → E g such that φ ( V ) is contained in a proper split variety W = W × W with W i ⊂ E g i and dim W < min( d, g − n ) . We say that V is n -generically non-split if it is not n -generically split.We simply say generically split for -generically split. Clearly generically non-split implies non-split. Note that non-split implies trans-verse. Indeed if V is not transverse, then there exists an isogeny φ : E g → E g suchthat φ ( V ) ⊂ p × E r . Set V = p and V = π ( φ ( V )), where π is the projection onthe last r coordinates.The following Lemma clarifies the equivalence between Property ( S n ) and the n -generically non-split property. Lemma 3.2.
A subvariety V ⊂ E g satisfies Property ( S n ) if and only if V is n -generically non-split.Proof. First suppose that V does not satisfy Property ( S n ). Then, there exists φ : E g → E d + n such that V = φ ( V ) has dimension d < d . Let f = (cid:0) φ φ ⊥ (cid:1) . Then f ( V ) ⊂ V × E g − d − n . Furthermore dim V < d = min( d, d + n − n ). Thus V is n -generically split.Secondly suppose V is n -generically non-split. Then, up to an isogeny, V iscontained in W = W × W with W i ⊂ E g i and dim W = d < min( d, g − n ).Consider the projection V of V on the first d + n coordinates.If g ≥ d + n , then V is contained in the projection of W . As dimensions cannotincrease by projection, dim V ≤ dim W < d .If g < d + n , then we have V ⊂ W × E d + n − g . Thus dim V ≤ d + d + n − g < d because d < g − n . So V does not satisfy Property ( S n ). (cid:3) EVELINA VIADA
It is then natural to give an example of a non-split variety which is genericallysplit, or equivalently which does not satisfy Property ( S ). Example 3.1.
Let us show at once that for a hypersurface, the notion of non-splitand generically non-split coincide.Let V be a non-split hypersurface in E d +1 . If V were generically split, then, foran isogeny φ , φ ( V ) would be contained in a proper split variety W × W . For adimensional argument φ ( V ) = W × W , contradicting the non-split assumption. Example 3.2.
In some sense, to give an example of a non-split but generically-splitvariety it is necessary to consider varieties of large codimension.In G nm it is easier to write equations. Consider the surface V in G m parameterizedby u and v , and given by the set of points ( u, u + 1 , u v + u, v + u + 1) . This issimply the envelope variety V of the irreducible plane curve C = ( u, u + 1) . Theenvelope is constructed as follows. To a point p ∈ C we associate the tangent line t p in p. Then V = ∪ p ∈ C ( p, t p ) . A property of the envelope is that it is not the fiberproduct of two varieties of positive dimension.The set V is an algebraic surface; let z , z , z , z be the variables, then V is thezero set of ( z = 5 z ( z − z ) + z ,z = z + 1 . The projection on the first two coordinates is exactly the curve C defined by z = z + 1 . Thus V does not satisfy Property ( S ) , however it is non-split. If, onthe contrary, V were split, then, for an isogeny φ , φ ( V ) = V × V . Since V is transverse, V i have positive dimension. In addition ker φ · V = φ − ( V ) × ker φ φ − ( V ) . Thus V = W × W ∩ W W where W i = V ∩ φ − ( V i ) . This contradictsthat V is not a fiber product. Example 3.3.
We now extend the previous example to a power of an elliptic curve E given by a Weierstrass equation in P . Consider the projection from E → P given by the projection of a point ( v : v : 1) ∈ E on the first coordinate ( v : 1) .Do the same on each factor to have a projection from E to (cid:0) P (cid:1) . The torus G m is naturally an open in (cid:0) P (cid:1) .Consider in G m the non-split but generically split V just constructed above. Takethe preimage of V on E and its closure V ′ . Then V ′ is generically split, howeverit is non-split. As above, suppose φ ( V ′ ) = V ′ × V ′ for an isogeny φ . Then thepreimage of φ − φ ( V ′ ) to ( P ) were a fiber product, contradicting that V is not afiber product. We remark, that there are also non-split transverse varieties which do not satisfyProperty ( S n ): One can extend this last example taking the envelope surface of atransverse curve in E n +2 .R´emond [5] theorem 1.9 and Habegger [3] corollary 2 prove that if V oa = ∅ then V does not satisfy property ( S ). Using the generically-split property we prove thereverse implication. Lemma 3.3.
A variety V does not satisfy property ( S ) if and only if V oa = ∅ Proof.
Suppose that V has dimension d and does not satisfy property ( S ). Bylemma 3.2, there exists an isogeny φ such that φ ( V ) ⊂ W × W with W i ⊂ E g i and dim W < min( d, g ). Then, the intersection of V with the cosets φ − ( x × E g )for x ∈ W are either empty of anomalous. In addition each point of V belongsto such an intersection. So V oa is empty. The reverse implication is proven byHabegger [3] corollary 2 using R´emond [5] theorem 1.9. (cid:3) HE OPTIMALITY OF THE BOUNDED HEIGHT CONJECTURE 7
The following lemma shows that in the Bounded Height Conjecture we can notexpect the set in the consequence to be non-dense. This lemma will also be usedin the proof of Theorem 1.3.
Lemma 3.4.
Let V be an irreducible subvariety of E g of dimension ≤ d < g .Then the set S d ( V ) \ S g ( V ) is dense in V .Proof. We shall distinguish two cases with regard to whether V is or not the trans-late of an abelian subvariety by a torsion point.Suppose V is not such a translate. Then, the Manin-Munford Conjecture, atheorem of Raynaud, ensures that the torsion S g ( V ) is not dense in V . Our claimis then equivalent to show that S d ( V ) is dense in V . Consider a surjective morphism(for example a projection) φ : E g → E d such that the restriction to V is dominant.Use lemma 2.1 to ensure the existence of such a morphism. Let E d Tor be the torsiongroup of E d . The preimage on V via φ of E d Tor is dense in V and it is a subset of S d ( V ).Suppose now that V is the translate of an abelian subvariety by a torsion point.Up to an isogeny, we can assume V = E d × p for p = ( p , . . . , p g − d ) ∈ E g − d Tor . Notethat, by Kronecker’s Theorem, for any x ∈ V , x + E d Tor is dense in V .Since p is a torsion point, (cid:0)(cid:0) E ( Q ) \ E Tor × { } d − × p (cid:1) + E d Tor (cid:1) ⊂ S d ( V ) \ S g ( V ) . In addition this set is dense in V , because E ( Q ) \ E Tor is non-empty (even densein E ). (cid:3) We now discuss the assumption of Property ( S ). In general, for V = V × V with dim V = d and dim V = d , we have S d ( V ) × S d ( V ) ⊂ S d ( V ). Could wehave equality if we assume, for example, that each factor satisfies Property ( S )?Similarly, does Conjecture 1.2 hold for such a product variety or for a non-splitvariety? The answer is negative.To simplify the formulation of the statements we characterize the sets which breakConjecture 1.2. Definition 3.5.
We say that a subset V u of V ( Q ) is densely unbounded if V u isZariski dense in V and for every non-empty Zariski open U of V the intersection V u ∩ U does not have bounded height. Equivalently V u is densely unbounded if, fora sequence { N } of positive reals going to infinity, the set V u [ N ] = { x ∈ V u : || x || > N } is Zariski dense in V .Proof of the equivalence of the definitions. Suppose that there exists a non-emptyopen U such that V u ∩ U has height bounded by N . The set Z = V \ U is a properclosed subset of V , and therefore not dense. So V u [ N + 1] ⊂ Z can not be dense.Suppose now that there exists an unbounded sequence { N } such that V u [ N ] isnot dense for some N . Then the Zariski closure Z of V u [ N ] is a proper closedsubset of V . So U = V \ Z is a non-empty open set such that V u ∩ U has heightbounded by N . (cid:3) Let us prove a preparatory lemma for the proof of Theorem 1.3.
Lemma 3.6.
Let z be a non-torsion point in E ( Q ) . Let n be a positive integer.Define G z ,n = h z i n End( E ) . For N ∈ N , the set G z ,n [ N || z || ] = { p ∈ G z ,n : || p || > N || z ||} is Zariski dense in E n . As a consequence G z ,n is dense in E n . EVELINA VIADA
Proof.
Denote by Σ = h z i End( E ) the submodule of E generated by z . Then G z ,n = Σ n . Recall that Σ[ N || z || ] = { p ∈ Σ : || p || > N || z ||} . Then (Σ[ N || z || ]) n ⊂ G z ,n [ N || z || ]. As Σ[ N || z || ] is an infinite set, it is dense in E . Then (Σ[ N || z || ]) n is dense in E n .Note that G z ,n contains G z ,n [0], so it is also dense. (cid:3) We are ready to show the optimality of the Bounded Height Conjecture.
Proof of Theorem 1.3.
Suppose that V does not satisfy Property ( S ). We aregoing to construct a densely unbounded set of V which is a subset of S d ( V ).By Lemma 2.2, there exists a surjective morphism ψ : E g → E d such that 0 < dim ψ ( V ) < d . Denote V = ψ ( V ) and d = dim V . We can fix an isogeny andsuppose that ψ is the projection on the first d coordinates, thus V ⊂ V × E g − d .Let x ∈ V . Then x = ( x , x ) with x ∈ V and x ∈ E g − d . Consider x × W x = V ∩ ( x × E g − d ) . There exists an open dense subset U of V such that the algebraic variety W x isequidimensional of dimension d = d − d . Let V x be an irreducible componentof W x . By Lemma 2.1, there exists a projection π x : E g − d → E d such that therestriction(2) π x | V x : V x → E d is dominant and even surjective and therefore its fibers are generically finite.Consider V ⊂ E d . Since V is irreducible also V is. By Lemma 3.4, applied with V = V , d = d and g = d , the set S d ( V ) \ S d ( V ) is Zariski dense in V . Define V u = U ∩ ( S d ( V ) \ S d ( V )) . Then all points in V u are non-torsion and V u is a dense subset of V . By definitionof S d ( V ), if x ∈ V u ⊂ S d ( V ), then there exists φ : E d → E d of rank d suchthat(3) φ ( x ) = 0 . Let z k be a coordinate of x = ( z , . . . , z d ) such that || z k || = max i || z i || . Only thetorsion has norm zero. Since x is non-torsion, then || z k || > x ∈ V u we will construct a subset of x × V x which is, both,densely unbounded in x × V x and a subset of S d ( V ).We denote by φ = (0 , . . . , , ϕ k , , . . . ,
0) : E d → E d a morphism such that only the k -th column is non zero.For a positive integer N , we define F ( N ) := { φ = (0 , . . . , , ϕ k , , . . . ,
0) : E d → E d s . t . | φ | > N } and x × V ux ( N ) := { ( x , y ) ∈ ( x , V x ) s . t . ∃ φ ∈ F ( N ) with φ ( x ) = π x ( y ) } . We simply denote U x = V ux (1) . We want to show that x × U x is densely unbounded in x × V x .(a) - First we show that x × V ux ( N ) ⊂ ( x × U x )[ N || z k || ] . For ( x , y ) ∈ x × V ux ( N ) there exists φ ∈ F ( N ) such that φ ( x ) = π x ( y ). Thus, || y || ≥ || π x ( y ) || = || φ ( x ) || ≥ | φ ||| z k || > N || z k || . HE OPTIMALITY OF THE BOUNDED HEIGHT CONJECTURE 9
Whence ( x , y ) ∈ ( x × U x )[ N || z k || ].(b) - We now show that x × V ux ( N ) is dense in x × V x . Let ( a z k , . . . , a d z k ) ∈ G z k ,d [ N || z k || ] with a i ∈ End( E ). Then max i | a i | > N . Let φ be the morphismfrom E d to E d such that the k -th column of φ is the vector ϕ k = ( a , . . . , a d ) t and all other entries are zeros. Then φ ∈ F ( N ) and φ ( x ) = ( a z k , . . . , a d z k ).So, we have the inclusion G z k ,d [ N || z k || ] ⊂ [ φ ∈F ( N ) φ ( x ) . By Lemma 3.6, G z k ,d [ N || z k || ] is a Zariski dense subset of E d . Thus, also theset S φ ∈F ( N ) φ ( x ) is Zariski dense in E d . By (2) the map π x | V x is surjective.Then for any φ ∈ F ( N ) there exists y ∈ V x such that π x ( y ) = φ ( x ). Therefore x × V ux ( N ) is Zariski dense in x × V x .In view of Definition 3.5, part (a) and (b) above show that x × U x is a denselyunbounded subset of x × V x . In addition, by definition of x × U x , for every( x , y ) ∈ x × U x there exists φ : E d → E d such that(4) φ ( x ) = π x ( y ) . Consider ( x , y ) with x ∈ V u and y ∈ U x . By relations (3) and (4), the morphism φ = (cid:18) φ − φ π x (cid:19) : E g → E d , has rank equal to rk φ + rk π x = d + d and φ ( x , y ) = 0 . So ( x , y ) ∈ S d ( V ).Let W ux = [ V ux (1) = [ U x for V x varying over the irreducible components of W x . We conclude that the set [ x ∈ V u x × W ux ⊂ S d ( V )is densely unbounded in V . (cid:3) Final Remarks
It is then natural to investigate the height property for the codimension of thealgebraic subgroups at least d + 1. We expect that Conjecture 1.3 holds. Let ussay at once that the (weak)-transverse hypothesis is in general necessary, howeverit is not clear if it is sufficient.Theorem 1.4 is a special case of Conjecture 1.3. We show that the condition (1)coincides with Property ( S ). Compare the following lemma with [4] lemma 7.2. Lemma 4.1.
An irreducible variety V ⊂ E g satisfies Property ( S ) if and only if dim( V + B ) = min(dim V + dim B, g ) for all abelian subvarieties B of E g .Proof. Note that E g /B is isogenous to E g − dim B . Consider the natural projection π B : E g → E g − dim B . Then(5) dim π B ( V ) = dim( V + B ) − dim B. Denote by d the dimension of V . Suppose that V satisfies Property ( S ), we have- If g − dim B ≥ dim X , then dim π B ( X ) = d .- If g − dim B ≤ dim X , then dim π B ( X ) = g − dim B . Use (5) to deduce dim( V + B ) = min( d + dim B, g ).Suppose now that dim( V + B ) = min( d + dim B, g ) for all abelian subvarieties B of codimension d . Note that, if φ : E g → E d is a surjective morphism, then thezero component of ker φ is an abelian variety of codimension d . Relation (5) showat once that V satisfy Property ( S ). (cid:3) We observe that, for S d +1 ( V ), the natural analogue to Conjecture 1.2, is to assumeProperty ( S ). Property ( S ) is weaker than ( S ). There are even split varietieswhich satisfy Property ( S ).Potentially, the method used by Habegger to prove Theorem 1.2, extends to showthat, for V satisfying Property ( S ), there exists a non-empty open V e such that S d +1 ( V e ) has bounded height.However, neither such a statement nor Theorem 1.4 are optimal: transversality isexpected to be a sufficient assumption, as the following examples suggest. We givesimple examples of a transverse variety V of dimension d which does not satisfyProperty ( S ) or ( S ) but such that S d +1 ( V, Γ ε ) is non-Zariski dense. Example 4.1.
Let V be a variety in E d + n +1 of dimension d . Suppose that V satisfies Property ( S ) . If you like take a transverse curve. By Theorem 1.4 i., forevery Γ ′ of finite rank there exists ε > such that S d +1 ( V , Γ ′ ε ) has bounded height.By [9] Theorem 1.1, applied to V of dimension d , we obtain that there exists ε > such that: (1) S d +1 ( V , Γ ′ ε ) is non-Zariski dense in V .Let V = V × E d and g = d + d + n + 1 , then V is transverse in E g . Furthermore, V does not satisfy Property ( S n ) . Indeed dim V = d = d + d . The projection onthe first d + n coordinates is V × E d − which has dimension d − . Let Γ ⊂ E g be asubgroup of finite rank and let Γ ′ be its projection on the first d + n + 1 coordinates.By [9] Lemma 4.1 we obtain S d +1 ( V, Γ ε ) ⊂ S d +1 ( V , Γ ′ ε ) × E d . Then, using relation (1) above, S d +1 ( V, Γ ε ) is non-Zariski dense in V . Define Z = S d +1 ( V, Γ ε ) . Then S d +1 ( V \ Z, Γ ε ) is empty and so it also has bounded height. Example 4.2.
Let V = V × V with V i a hypersurface in E d i +1 satisfying Property ( S ) . The projection on the first d = d + d coordinates shows that V does not satisfyProperty ( S ) . However V satisfies Property ( S ) . We are going to show that S d +1 ( V, F ) ⊂ S d ( V , F ) × S d ( V , F ) ∪ (cid:0) S d +1 ( V , F ) × V (cid:1) ∪ (cid:0) V × S d +1 ( V , F ) (cid:1) . (6) Let g = d + d + 2 . Let ( x, y ) ∈ S d +1 ( V, F ) with x ∈ V and y ∈ V . Then, thereexist φ : E g → E d +1 of rank d + 1 and ( f, f ′ ) ∈ F such that φ (( x, y ) − ( f, f ′ )) = 0 . Decompose φ = ( A | B ) with A : E d +1 → E d +1 and B : E d +1 → E d +1 . Then d + d + 1 = rk φ ≤ rk A + rk B. Note that rk A ≤ d + 1 and rk B ≤ d + 1 because of the number of columns. Thenone of the following cases occurs: (1) rk A = d or rk B = d , (2) rk A = d + 1 and rk B = d + 1 . HE OPTIMALITY OF THE BOUNDED HEIGHT CONJECTURE 11 (1) - If the rank of B is d then, with the Gauss algorithm, one finds an invertiblematrix ∆ ∈ Mat d +1 (End( E )) such that ∆ φ = (cid:18) ϕ ⋆ ϕ (cid:19) , with ϕ of rank d + 1 .If the rank of A is d then one finds an invertible matrix ∆ ∈ Mat d +1 (End( E )) such that ∆ φ = (cid:18) ϕ ⋆ ϕ (cid:19) , with ϕ of rank d + 1 .Then either x ∈ S d +1 ( V , F ) or y ∈ S d +1 ( V , F ) . So ( x, y ) ∈ (cid:0) S d +1 ( V , F ) × V (cid:1) ∪ (cid:0) V × S d +1 ( V , F ) (cid:1) . (2) - With the Gauss algorithm one can find two invertible matrices ∆ i ∈ Mat d +1 (End( E )) such that ∆ φ =( aI d +1 | l )∆ φ =( l ′ | bI d +1 ) with a, b ∈ End( E ) \ and I d +1 the identity matrix. The last d rows of ∆ φ showthat y ∈ S d ( V , F ) and the first d rows of ∆ φ show that x ∈ S d ( V , F ) . Thus ( x, y ) ∈ (cid:0) S d ( V , F ) × S d ( V , F ) (cid:1) .We now apply the inclusion (6) to the case of curves, and we deduce a non-densityresult for surfaces. Let V i = C i be transverse curve in E . By Theorem 1.2, thereexists ε > such that S ( C i , O ε ) has bounded height. In view of the BogomolovConjecture, a theorem of Ullmo, one can choose ε such that S ( C i , O ε ) is finite.Define F = O ε . Then, relation (6) implies that S ( C × C , O ε ) has bounded height.In addition C × C is transverse in E . Using [9] Theorem 1.1, we conclude that S ( C × C , O ε ) is non-Zariski dense.According to Theorem 1.2 and [9] Theorem 1.1, one can do similar considerationsfor hypersurfaces.
These last examples give evidence that the transverse or weak-transverse hypoth-esis is sufficient for Conjecture 1.3. Precisely, the idea is that if U is a dense subsetof V of bounded height, then the set U × V is densely unbounded in V × V , (thisis more or less what makes Property ( S ) necessary for Theorem 1.2). Instead if U is Zariski closed in V , then the set U × V is still Zariski closed in V × V .Could one extend the idea in the last examples to show that for the product ofvarieties satisfying Property ( S ) Conjecture 1.3 holds?This is not an easy matter; even the case of C × C for C transverse in E and C transverse in E remains open. References [1] E. Bombieri, D. Masser and U. Zannier,
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