RATIONALLY INEQUIVALENT POINTS ON HYPERSURFACES IN P n

RATIONALLY INEQUIVALENT POINTS ON HYPERSURFACES IN P n XI CHEN, JAMES D. LEWIS, AND MAO SHENG Abstract. We prove a conjecture of Voisin that no two distinct points on a very general hypersurface of degree 2n in P n are rationally equivalent. Contents 1. Introduction 2 Conventions 3 2. Relative cycle map 3 3. Positivity of the sheaf of holomorphic N-forms 6 3.1. A key lemma 6 3.2. Sheaf of holomorphic N-forms 8 3.3. Global generation of TX 2 (1) 11 3.4. Differential map dσ 13 3.5. Criterion for two fixed sections 14 3.6. Criterion for two varying sections 20 4. Hypersurfaces of degree 2n + 2 in P n+1 23 4.1. Versal deformation of the Fermat hypersurface 23 4.2. A basis for W X,b 25 4.3. An observation on L λ 27 4.4. The space H 0 (I Z (1)) Span J d 1 28 4.5. Special case 30 4.6. Generic case 39 5. Notes on algebraic invariants 48 5.1. (Generalized) normal functions 50 5.2. The invariants 51 5.3. Example 5.5 revisited 52 References 53 Date: January 10, 2018. Research of the first two authors is partially supported by Discovery Grants from the Natural Sciences and Engineering Research Council of Canada. Research of the third named author is partially supported by National Natural Science Foundation of China (Grant No. 11471298, No. 11622109, No. 11626253) and the Fundamental Research Funds for the Central Universities. 1

2 XI CHEN, JAMES D. LEWIS, AND MAO SHENG 1. Introduction In [V1] and [V2], C. Voisin proved the following ([V1, Theorem 3.1] and [V2, Theorem 0.6]) Theorem 1.1 (C. Voisin). Let X be a very general complete intersection in P n+k of type (d 1, d 2,..., d k ). If (d i 1) 2n + 2, no two distinct points on X are Q-rationally equivalent. If (n, k, d 1 ) = (2, 1, 6), there are at most countably many points on X that are Q-rationally equivalent to a fixed point p for all p X. The main purpose of this note is to generalize this result in two directions. First, we will make a minor improvement by replacing rational equivalence by Roĭtman s Γ-equivalence [R1]: fixing a smooth projective curve Γ and two points 0 Γ, for every algebraic cycle ξ Z k (X Γ) with supp(ξ) flat over Γ, the fibers ξ 0 and ξ of ξ over 0 and are Γ-equivalent, written as ξ 0 Γ ξ. We will prove Theorem 1.2. For a fixed smooth projective curve Γ with two fixed points 0, no two distinct points on a very general complete intersection X in P n+k of type (d 1, d 2,..., d k ) are Γ-equivalent over Q if (d i 1) 2n + 2, Second, we will try to find the optimal bound for d i where the result holds. Our most optimistic expectation is Conjecture 1.3. For a very general complete intersection X P n+k of type (d 1, d 2,..., d k ) and every point p X, (1.1) dim R X,p,Γ 2n k (d i 1) where R X,p,Γ = {q p X : N(p q) Γ 0 for some N Z + } and Γ is a fixed smooth projective curve with two fixed points 0. Note that R X,p,Γ is a locally noetherian scheme. The case (d i 1) = n + 1 follows from Roĭtman s generalization of Mumford s famous theorem ([Mu], [R1] and [R2]). Of course, Voisin proved (1.2) dim R X,p,P 1 2n + 1 i=1 k (d i 1) for (d i 1) 2n + 2 or (n, k, d 1 ) = (2, 1, 6). Theorem 1.2 shows that (1.1) holds for (d i 1) 2n + 2. If our conjecture holds, R X,p,Γ = when (d i 1) 2n + 1. So the boundary case is (d i 1) = 2n + 1. For example, it is expected that R X,p,Γ = for a very general sextic surface X P 3. Voisin s theorem shows that dim R X,p,P 1 = 0 for such surfaces X. This boundary case is quite challenging, even only for sextic surfaces. We claim the following: i=1

RATIONALLY EQUIVALENT POINTS 3 Theorem 1.4. No two distinct points are Γ-equivalent over Q on a very general hypersurface X P n+1 of degree 2n+2 for a fixed smooth projective curve Γ with two fixed points 0. That is, (1.1) holds for k = 1 and d 1 = 2n + 2. Note that the bound d 2n + 2 is optimal for hypersurfaces of degree d in P n+1 : For a general hypersurface X of degree d 2n + 1 in P n+1, there exist two lines L 1 and L 2 in P n+1 such that each L i meets X at a unique point p i with p 1 p 2. Conventions. We work exclusively over C. Indeed, any two points on a variety over F p are rationally equivalent over Q. 2. Relative cycle map Voisin s proof consists of two major components. One is relative cycle map. For a relative Chow cycle Z CH n hom (X/B) for a smooth projective family π : X B of relative dimension n, if AJ n (Z b ) = 0 under the Abel- Jacobi map on each fiber X b, one can define some infinitesimal invariant δz H n (R n π Q). This invariant can be defined in a Hodge-theoretical way as in [V2]. Please see 5 for a comprehensive treatment along this line. Here we take a different approach: we define δz directly by (2.2) (see below) and then we prove δz is invariant under rational equivalence. This has the advantage of being elementary: no Hodge theory is involved in the definition of δz. In addition, we will obtain for free that δz is invariant under Γ-equivalence. Another advantage of this approach is that δz is well defined for an arbitrary flat family π : X B without any extra hypotheses on X/B. Definition 2.1. Let π : X B be a flat and surjective morphism of relative dimension n from X onto a smooth variety B of dim B = N. For a multi-section Z X, we define (2.1) δz Hom(π ( N Ω X ), N Ω B ) = Hom(π Ω N X, K B ) as follows: (2.2) δz = Tr Z/B (dσ) : π Ω N X dσ (π σ) Ω N Z = (π σ) K Z Tr Z/B K B where Tr Z/B is the trace map and σ : Z X is the embedding. We can easily extend δ to the free abelian group Z n (X/B) of algebraic cycles Z of pure codimension n in X whose support supp(z) is flat over B. For Z = m i Z i with Z i multi-sections of π, we let δz = m i δz i. Remark 2.2. The definition (2.2) of δz might need some further explanation. The differential map dσ is usually dσ : σ Ω N X ΩN Z. In (2.2), it is the

4 XI CHEN, JAMES D. LEWIS, AND MAO SHENG composition of dσ and (π σ) : (2.3) π Ω N X π (Ω N X O Z) (π σ) Ω N Z π (σ σ Ω N X ). The trace map Tr Z/B can be defined for π ( m Ω Z ) m Ω B under a generically finite map π : Z B. Obviously, it is well defined outside of the ramification locus of π. Since every meromorphic differential form in m Ω B is regular if it is regular in codimension 1, it suffices to show that the image of a differential m-form on Z under the trace map can be extended to a regular m-form on B in codimension 1 [K, Proposition 5.77, p. 185]. Moreover, the trace map is well defined for B normal if we follow the convention to define Ω B to be the sheaf of differential forms regular in codimension 1. However, Tr Z/B cannot be defined for π (Ω m Z ) Ω m B when m 2, which is the reason why Mumford s argument cannot be generalized using pluri-canonical forms. For Z Z n (X/B) and a morphism f : B B, we clearly have the commutative diagram (2.4) f Ω N X Ω N X f (δz) δ(f Z) f K B K B where X = X B B and we also use f to denote the map X X. Lemma 2.3. Let π : X B be a flat and projective morphism of relative dimension n from X onto a smooth variety B of dim B = N and let Z be a cycle in Z n (X/B). If π Ω N X is locally free and Z b Γ 0 for all b B, then δz = 0, where Γ is a fixed smooth projective curve with two fixed points 0. Proof. Since π Ω N X is locally free, δz = 0 if and only if δz = 0 at a general point of B. Using a Hilbert scheme argument, we can find a dominant and generically finite morphism f : B B and a cycle Y Z n (X Γ) such that supp(y ) is flat over B Γ and Y 0 Y = f Z, where X = X B B, Y t is the fiber of Y over t Γ and f Z is the pullback of Z under f : X X. Obviously, δz = 0 if δf Z = 0 by (2.4) and the fact that π Ω N X is locally free. To simplify our notations, we replace (X, B) by (X, B ). For every t Γ, Y t Z n (X/B) and thus it induces a map (2.5) γ : Γ Hom(π Ω N X, K B )

RATIONALLY EQUIVALENT POINTS 5 by γ(t) = δy t. More precisely, since Y is flat over B Γ, we have (2.6) δy Hom(ε Ω N+1 X Γ, K B Γ) = Hom(η 1π Ω N+1 X η 1π Ω N X η 2K Γ, η 1K B η 2K Γ ) Hom(η 1π Ω N X η 2K Γ, η 1K B η 2K Γ ) = Hom(η 1π Ω N X, η 1K B ) where ε, η 1 and η 2 are the projections ε : X Γ B Γ, η 1 : B Γ B and η 2 : B Γ Γ, respectively. Clearly, γ(t) is the restriction of δy to the point t Γ. It follows that γ is a morphism. And since Γ is projective, it must be constant. Therefore, δz = δy 0 δy = 0. We are done. So, to show that σ 1 (b) Γ σ 2 (b) over Q at a general point b B for two sections σ i : B X of X/B, we only need to find s H 0 (U, π Ω N X ) satisfying (2.7) (dσ 1 )σ 1s (dσ 2 )σ 2s 0 over some open dense subset U B. The existence of such s is guaranteed if H 0 (X b, Ω N X ) is imposed independent conditions by σ i(b) for b B general. This observation leads to the following: Proposition 2.4. Let π : X B be a smooth and projective morphism from X onto a smooth variety B of dim B = N. Suppose that H 0 (X b, Ω N X ) is imposed independent conditions by all pairs of distinct points p q X b for b B general. Then R Xb,p,Γ = for b B very general and all p X b, where Γ is a fixed smooth projective curves with two fixed points 0. More generally, { R Xb,p,Γ q X b : q p and {p, q} does not impose independent (2.8) } conditions on H 0 (X b, Ω N X) for b B very general. Here we say that a closed subscheme Z X or its ideal sheaf I Z O X imposes independent conditions on a coherent sheaf F or its global sections H 0 (F) (resp. a linear series D H 0 (F)) on X if H 0 (F) H 0 (F O Z ) (resp. D H 0 (F O Z )) is surjective. Proof of Proposition 2.4. Suppose that there are a pair of points p q on a general fiber X b such that p Γ q over Q and {p, q} imposes independent conditions on H 0 (X b, Ω N X ). By a base change and shrinking B to an affine variety, we may assume that there exists two disjoint sections P and Q X of π : X B such that m(p b Q b ) Γ 0 for some m Z + and all b B, h 0 (X b, Ω N X ) is constant for all b B and H 0 (Ω N X ) is imposed independent conditions by P Q.

6 XI CHEN, JAMES D. LEWIS, AND MAO SHENG Since P Q imposes independent conditions on H 0 (Ω N X ) and ΩN X free, the map is locally (2.9) H 0 (Ω N X I σq P ) H 0 (σq ΩN X ) is a surjection, where σ P and σ Q : B X are the embeddings of P and Q to X, respectively. Combining (2.9) with the pullback map of σ Q : B X on differentials, we have a composition of two surjections (2.10) H 0 (Ω N X I σq P ) H 0 (σq ΩN X ) dσ Q H0 (Ω N B ) where dσ P and dσ Q are the pullback maps induced by σ P and σ Q on the differentials, respectively. Therefore, there exists s H 0 (Ω N X ) such that (2.11) σ P s = 0 and (dσ Q )σ Qs 0. It follows that (2.12) δz, s = (dσ P )σ P s (dσ Q )σ Qs = (dσ Q )σ Qs 0 for Z = P Q. On the other hand, δz = 0 by Lemma 2.3. Contradiction. The above argument shows that no irreducible component of { S X,Γ = (b, p, q) : b B and p q X b satisfy that p Γ q over Q (2.13) and {p, q} imposes independent conditions } on H 0 (X b, Ω N X) dominates B via the projection ξ : S X,Γ B. Note that S X,Γ is a locally noetherian subscheme of X B X. Therefore, for b B\ξ(S X,Γ ) very general, (2.8) holds. Remark 2.5. Note that the right hand side (RHS) of (2.8) is a subscheme that does not depend on the choice of the triple (Γ, 0, ). 3. Positivity of the sheaf of holomorphic N-forms 3.1. A key lemma. Let us first review some basic notions on global generation and very ampleness of coherent sheaves. A coherent sheaf V on a variety X is globally generated (resp. very ample) if the map H 0 (V ) H 0 (V O Z ) is surjective for all 0-dimensional subschemes Z X of length h 0 (O Z ) = 1 (resp. 2), i.e., V is imposed independent conditions by all 0-subschemes of length 1 (resp. 2). More generally, we say that a linear series D H 0 (V ) is globally generated (resp. very ample) if the map D H 0 (V O Z ) is surjective for all 0-dimensional subschemes Z X of length h 0 (O Z ) = 1 (resp. 2). The hypothesis in Proposition 2.4 that Ω N X O X b is imposed independent conditions by two distinct points is a weak version of very-ampleness, which is technically easier to treat and suffices for our purpose. We call V weakly very ample if H 0 (V ) is imposed independent conditions by all pairs of two distinct points on X.

RATIONALLY EQUIVALENT POINTS 7 Let us go through some basic facts on these notions: A quotient of a globally generated (resp. (weakly) very ample) coherent sheaf is also globally generated (resp. (weakly) very ample). More generally, if a coherent sheaf V on a variety X is imposed independent conditions by a 0-dimensional subscheme Z X, so is a quotient Q of V. For coherent sheaves V and W on a variety X, if V is globally generated and W is imposed independent conditions by a 0-dimensional subscheme Z X, then V W is imposed independent conditions by Z. In particular, if V is globally generated and W is globally generated (resp. (weakly) very ample), V W is also globally generated (resp. (weakly) very ample); if V is globally generated (resp. (weakly) very ample), so are V N, Sym N V and N V for all N 1. Let η (3.1) 0 U V W 0 be a short exact sequence of coherent sheaves on a variety X, if the map η Γ : Γ(V ) Γ(W ) induced by η is surjective and both U and W are imposed independent conditions by a 0-dimensional subscheme Z X, the same is true for V. Thus, if η Γ is surjective and both U and W are globally generated (resp. (weakly) very ample), V is also globally generated (resp. (weakly) very ample). Here we write Γ(A) = H 0 (A). Basically, if we have a short exact sequence (3.1), the global generation (resp. very-ampleness) of V implies that of W ; the global generation (resp. very-ampleness) of U and W implies that of V under the extra hypothesis that η Γ is surjective. The hard question is how to tell whether a 0- dimensional scheme Z imposes independent conditions on U if it does on V. The following key lemma gives us a criterion for that. Lemma 3.1. Let (3.2) 0 A 1 A 2 A 3 α 1 α 2 α 3 η 0 B 1 B 2 B 3 be a commutative diagram of sheaves over a topological space X whose rows are left exact. Suppose the map α 2 Γ is surjective. Then the map α 1 Γ is surjective if and only if (3.3) η(ker(α 2 Γ)) = η Γ(A 2 ) ker(α 3 Γ).

8 XI CHEN, JAMES D. LEWIS, AND MAO SHENG Proof. This follows from the diagram 0 0 E F (3.4) 0 Γ(A 1 ) Γ(A 2 ) G 0 η α 1 α 2 α 3 0 Γ(B 1 ) Γ(B 2 ) Γ(B 3 ) 0 and the snake lemma, where E = ker(α 2 Γ), F = η Γ(A 2 ) ker(α 3 Γ) and G = η Γ(A 2 ). Of course, this lemma can be formulated and proved in abelian categories for left exact functors. A typical way to apply the above lemma is the following: if U, V and W are locally free in (3.1) and V is globally generated, then U is globally generated if and only if (3.5) η Γ(V I p ) = η Γ(V ) Γ(W I p ) for all p X. 3.2. Sheaf of holomorphic N-forms. The other component of Voisin s proof is the positivity of the sheaf of holomorphic N-forms. More precisely, we are considering the global generation and very-ampleness of the sheaf N Ω X = Ω N X when restricted to a general fiber of a family π : X B over B of dim B = N. Voisin proved [V1, Proposition 3.4] and [V2, Corollary 1.2]: Theorem 3.2 (C. Voisin). Let X B P n+k be a versal family of complete intersections of type (d 1, d 2,..., d k ) in P n+k over a smooth variety B of dim B = N. Then for a general point b B, (3.6) Ω N X π K 1 B = TX n K X/B is globally generated on X b if (d i 1) 2n + 1 and very ample on X b if (d i 1) 2n + 2, where π is the projection X B, T X = Ω X is the holomorphic tangent bundle of X and TX n = n T X. Let us go over Voisin s proof of the above theorem. The key fact is that T X (1) is globally generated [V2, Proposition 1.1]: Theorem 3.3 (Clemens). For a versal family X Y = B P of complete intersections in P = P n+k over a smooth variety B, T X (1) = T X O X (1) is globally generated on a general fiber X b, where O P (1) is the hyperplane bundle on P.

RATIONALLY EQUIVALENT POINTS 9 This theorem was originally due to Herbert Clemens [C]. We will give a proof following closely the argument of Lawrence Ein in [E1] and [E2]. Proof of Theorem 3.3. We have the so-called adjunction sequence η (3.7) 0 T X T Y O X N X 0 associated to X Y, where N X is the normal bundle of X in Y. By (3.3) in Lemma 3.1, T X (1) O Xb is globally generated if (3.8) T Y (1) O Xb is globally generated and (3.9) η Γ b (T Y (1) I p ) = η Γ b (T Y (1)) Γ b (N X (1) I p ) where we use the notation Γ b for Γ b (F) = H 0 (X b, F). Obviously, (3.8) follows immediately from the fact that (3.10) T P (1) and O P (1) are globally generated, while (3.9) follows if we can prove that the map (3.11) H 0 (X b, T Y (1) I p ) H 0 (X b, N X (1) I p ) H 0 (N Xb (1) I p ) is surjective for all p X b, where N Xb is the normal bundle of X b in P. The surjectivity of the map (3.11) comes from the surjectivity of two maps (3.12) H 0 (N Xb ) H 0 (O Xb (1) I p ) H 0 (N Xb (1) I p ) and (3.13) H 0 (X b, T P ) T B,b H 0 (X b, T Y ) H 0 (N Xb ) via the diagram (3.14) H 0 (X b, T Y ) H 0 (O Xb (1) I p ) H 0 (X b, T Y (1) I p ) H 0 (N Xb ) H 0 (O Xb (1) I p ) H 0 (N Xb (1) I p ), where T B,b is the holomorphic tangent space of B at b and the map (3.13) is induced by the Kodaira-Spencer map of the family X/B Y/B. Finally, (3.12) follows from the fact that (3.15) H 0 (O P (d)) H 0 (O P (1) I p ) H 0 (O P (d + 1) I p ) is surjective for all p P and d 0 and (3.13) is a consequence of the hypothesis that X/B is versal.

10 XI CHEN, JAMES D. LEWIS, AND MAO SHENG To put it in a nutshell, the global generation of T X (1) O Xb comes down to three easy-to-verify facts (3.10), (3.12) and (3.13). Thus, we can put Theorem 3.3 in a more general setting: Theorem 3.4. Let P be a smooth projective variety, X be a smooth closed subvariety of Y = B P that is flat over a smooth variety B and let L be a line bundle on the fiber X b of X/B over a point b B. Suppose that (3.16) T P L and L are globally generated on X b, (3.17) H 0 (N Xb ) H 0 (L I p ) H 0 (N Xb L I p ) is surjective for all p X b and the Kodaira-Spencer map (3.18) T B,b H 0 (N Xb )/H 0 (X b, T P ) is surjective. Then T X L is globally generated on X b. In addition, (3.19) H 1 (X b, T X L) = H 1 (X b, T X L I p ) = 0 for all p X b if H 1 (X b, T P L) = H 1 (X b, L) = 0. Note that the map (3.18) is the Kodaira-Spencer map associated to the family X/B Y/B, as given in the following diagram (3.20) T B,b 0 H 0 (T Xb ) H 0 (X b, T P ) H 0 (N Xb ) H 1 (T Xb ). The surjectivity of (3.18) simply says that B dominates the versal deformation space of X b P. Once we have the global generation of T X (1), Theorem 3.2 follows easily from the fact ( ) (3.21) TX n K X/B = n (T X (1)) O X (di 1) (2n + 1). Indeed, we can put Theorem 3.2 in a more general form as 3.4: Theorem 3.5. Under the same hypotheses of Theorem 3.4, TX n K X/B is globally generated (resp. very ample) on X b if K Xb L n is globally generated (resp. very ample). Of course, combining Proposition 2.4 and Theorem 3.5, we arrive at the following: Theorem 3.6. Under the same hypotheses of Theorem 3.4, we assume that (3.16), (3.17) and (3.18) hold and K Xb L n is very ample for b B general and n = dim X b. Then R Xb,p,Γ = for b B very general and all p X b, where Γ is a fixed smooth projective curve with two fixed points 0. This implies Theorem 1.2.

RATIONALLY EQUIVALENT POINTS 11 3.3. Global generation of TX 2 (1). In order to prove part of Conjecture 1.3, e.g., that no two points on a very general sextic surface are rationally equivalent, we need to show that Ω N X π K 1 ( ) B = TX n O X (di 1) (n + 1) (3.22) ( ) = TX(n) n O X (di 1) (2n + 1) is imposed independent conditions by two distinct points on a general fiber X b when (d i 1) 2n + 1. Namely, we need improve Voisin s theorem 3.3 to show that TX n (n) is weakly very ample. Of course, if this is true for TX m(m) for some m n, it is true for T X n (n). So we conjecture Conjecture 3.7. Let X B P n+k be a versal family of complete intersections of type (d 1, d 2,..., d k ) in P n+k over a smooth variety B of dim B = N. Then for a general point b B, H 0 (X b, TX 2 (2)) is imposed independent conditions by all pairs of points p q X b. Voisin actually had a stronger conjecture [V2, Question 2.1]: Conjecture 3.8. Let X B P n+k be a versal family of complete intersections of type (d 1, d 2,..., d k ) in P n+k over a smooth variety B of dim B = N. Then for a general point b B, T 2 X (1) = ( 2 T X ) O X (1) is globally generated on X b if X b is of general type. Clearly, Voisin s conjecture implies that TX 2 (2) = T X 2 (1) O X(1) is very ample on X b and hence our conjecture 3.7. In addition, it implies that Ω N X is globally generated when (d i 1) 2n. Unfortunately, both of the above conjectures fail. Basically, we are considering whether TX m L is imposed independent conditions by a 0-dimensional subscheme Z X b for a line bundle L. Using Lemma 3.1 again, we can obtain the following criterion: Theorem 3.9. Let Y be a smooth projective family of varieties over a smooth variety B, X be a smooth closed subvariety of Y that is flat over B, L be a line bundle on X b for a point b B and Z be a 0-dimensional subscheme of X b. Suppose that (3.23) H 0 (X b, T m Y L) is imposed independent conditions by Z. Then H 0 (X b, TX m L) is imposed independent conditions by Z if and only if (3.24) η m Γ(T m Y L I Z ) = η m Γ(T m Y L) Γ(T m 1 Y N X L I Z ), where η m : TY m O X T m 1 Y N X is the map m (3.25) η m (ω 1 ω 2... ω m ) = k=1( 1) k+1 η(ω k ) i k induced by η : T Y O X N X with N X the normal bundle of X in Y. ω i

12 XI CHEN, JAMES D. LEWIS, AND MAO SHENG Proof. By the adjunction sequence (3.7), we obtain a left exact sequence (3.26) 0 TX m L T Y m L T m 1 N X L on X b. Since TY m L is imposed independent conditions by Z, we conclude the same for TX m L if and only if (3.24) holds by (3.3) in Lemma 3.1. Note that η m is actually the map in the generalized Koszul complex (3.27) m T Y O X η m η m m 1 T Y N X m 2 T Y Sym 2 N X... T Y Sym m 1 N X Sym m N X 0 of m T Y Sym N X induced by η. We are considering the very-ampleness of TX m (l) for X Y = B P for P = P r. By the Euler sequence Y (3.28) 0 O P O P (1) (r+1) T P 0 on P, we have the diagram 0 0 E 0 O X O X (3.29) 0 G X E Y O X N X 0 ξ 0 T X T Y O X N X 0 η where 0 0 (3.30) E Y = π BT B π P E with π B : Y B and π P : Y P the projections of Y onto B and P, respectively. Since T X is a quotient of G X, TX m (l) is imposed independent conditions by two distinct points on X b if GX m (l) is. Thus, we have the following easy corollary of Theorem 3.9: Corollary 3.10. Let P = P r, X be a smooth closed subvariety of Y = B P flat over B, and Z be a 0-dimensional subscheme of X b for a point b B. Suppose that (3.31) Z imposes independent conditions on O P (l).

RATIONALLY EQUIVALENT POINTS 13 Then Z imposes independent conditions on H 0 (X b, TX m (l)) if (3.32) ξ m Γ b (EY m (l) I Z ) = ξ m Γ b (EY m (l)) Γ b (E m 1 Y (l) N X I Z ), where ξ m : EY m O X E m 1 Y N X is the map induced by ξ. In addition, the converse holds if (3.33) H 1 (X b, T m 1 X (l)) = 0 and Z imposes independent conditions on H 0 (X b, T m 1 X (l)). Proof. This follows directly from the diagram (3.29) and Theorem 3.9. The converse follows from the exact sequence (3.34) 0 T m 1 X (l) Gm X (l) T m X (l) 0. For a versal family X of complete intersections, we have already proved (3.33) for m = 2 and l 1 by (3.19) in Theorem 3.4. So TX 2 (2) is weakly very ample if and only if (3.35) ξ 2 Γ b (E 2 Y (2) I Z ) = ξ 2 Γ b (E 2 Y (2)) Γ b (E Y (2) N X I Z ) for all Z = {p 1 p 2 } X b and TX 2 (1) is globally generated if and only if (3.36) ξ 2 Γ b (E 2 Y (1) I p ) = ξ 2 Γ b (E 2 Y (1)) Γ b (E Y (2) N X I p ) for all p X b. Unfortunately, neither (3.35) nor (3.36) holds for hypersurfaces by a direct computation, although we will not go through the details here as it is not the main purpose of this paper. 3.4. Differential map dσ. Since TX 2 (2) fails to be weakly very ample, we cannot apply Proposition 2.4 to show that no two points on a general sextic surface are Γ-equivalent. It is very likely that TX n (n) fails to be weakly very ample for n > 2 as well. So we are unable to prove Conjecture 1.3 for (di 1) = 2n + 1 in this way. A closer examination of the proof of Proposition 2.4 shows that we do not really need Ω N X to be weakly very ample on X b. We only need find s H 0 (U, π Ω N X ) satisfying (2.7). This is much weaker than the requirement that p 1 = σ 1 (b) and p 2 = σ 2 (b) impose independent conditions on H 0 (X b, Ω N X ) for b general. For one thing, (dσ 1)σ1 s (dσ 2)σ2 s = 0 imposes only one condition on Γ b (Ω N X ) = H0 (X b, Ω N X ). Let dσ i Γ b be the map induced by dσ i on Γ b (Ω N X ) as in (3.37) Γ b (T n X K X) Γ b (Ω N X ) dσ 1 dσ 2 Γ ( ( b Kσ1 (B)) Γb Kσ2 (B)). Clearly, (2.7) holds for some s H 0 (U, π Ω N X ) if (3.38) ker(dσ 1 Γ b ) ker(dσ 2 Γ b )

14 XI CHEN, JAMES D. LEWIS, AND MAO SHENG holds at a general point b B. More precisely, as long as (3.38) holds at a point b B such that h 0 (X t, Ω N X ) is locally constant for t in an open neighborhood of b, we can find a section s b Γ b (Ω N X ) with the property (3.39) (dσ 1 )s b (dσ 2 )s b 0 and this s b can be extended to a section s H 0 (U, π Ω N X ) over an open neighborhood U of b satisfying (2.7). Therefore, to show that σ i (b) are not Γ-equivalent over Q on a general fiber X b, we just have to prove (3.38). Let us formalize this observation in the following proposition: Proposition 3.11. Let X be a smooth projective family of varieties over a smooth variety B of dim B = N and let σ i : B X be two disjoint sections of X/B for i = 1, 2. Then σ 1 (b) and σ 2 (b) are not Γ-equivalent over Q on X b for b B general if (3.38) holds at a point b where h 0 (X t, Ω N X ) is locally constant in t. 3.5. Criterion for two fixed sections. To apply Proposition 3.11, we need an explicit description of the differential maps dσ i. They can be made very explicit if X Y = B P is a family of varieties in a projective space P passing through two fixed points p i P and σ i (b) p i for i = 1, 2. On the other hand, for an arbitrary family X Y with two sections σ i over B, we can always apply an automorphism λ B Aut(P ), after a base change, fiberwise to Y/B such that λ σ i (b) p i for two fixed points p i P ; thus, to test (3.38) for a general fiber X b of X/B, it suffices to test it for a general fiber X b of X/B, X = λ(x) and σi = λ σ i. Let us first consider families X B P with two fixed sections σ i (b) p i. To set it up, we let P = P r and fix two points p 1 p 2 in P. We let X Y = B P be a closed subvariety of Y that is flat over B with fibers X b containing p 1 and p 2 for all b B. We assume that X and B are smooth of dim X = N + n and dim B = N, respectively. We have two sections σ i : B X sending σ i (b) = p i for all b B and i = 1, 2. To state our next proposition on the differential map dσ, we need to introduce the filtration F Ω X associated to the fibration X/B. For a surjective morphism f : W B with B smooth, we have a filtration (3.40) Ω m W = F 0 Ω m W F 1 Ω m W... F m+1 Ω m W = 0 with Gr p F Ωm W = F p Ω m W F p+1 Ω m = f ( p Ω B ) m p Ω W/B W for Ω m W = m Ω W derived from the short exact sequence (3.41) 0 f Ω B Ω W Ω W/B 0. Note that F p is an exact functor.

RATIONALLY EQUIVALENT POINTS 15 For π B : Y B with Y = B P, F p Ω m Y (3.42) F p Ω m Y = i p is simply that π BΩ i B π P Ω m i P and we have natural projections Ω m Y F p Ω m Y. Proposition 3.12. Let X Y = B P be a smooth projective family of varieties in a smooth projective variety P passing through a fixed point p P over a smooth variety B with the section σ : B X given by σ(b) = p for b B. Then the diagram (3.43) Ω N X Ω N+k Y det(n X ) Ω N+k+1 Y det(n X ) N X dσ Ω N σ(b) F N Ω N+k Y det(n X ) σ(b) F N Ω N+k+1 Y det(n X ) N X σ(b) commutes and has left exact rows, where N = dim B, k = dim Y dim X, det(n X ) = k N X and the vertical maps in the second and third columns are induced by the projections Ω Y F N Ω Y followed by the restrictions to σ(b). Proof. The rows of (3.43) are induced by Koszul complex (3.27) and hence left exact. We want to point out that the diagram (3.44) η Ω m Y Ω m+1 N X Y F l Ω m Y F l Ω m+1 Y N X does not commute in general. However, it commutes when we restrict the bottom row to σ(b). That is, we claim that the diagram (3.45) η Ω m Y Ω m+1 N X ρ m ρ m+1 F l Ω m η σ Y F l Ω m+1 σ(b) N X σ(b) commutes. Of course, this implies that the right square of (3.43) commute. Let (x 1, x 2,..., x r ) and (t 1, t 2,..., t N ) be the local coordinates of P and B, respectively. Let p = {x 1 = x 2 =... = x r = 0} and (3.46) X = {f 1 (x, t) = f 2 (x, t) =... = f k (x, t) = 0}. Then η is given by (3.47) η(ω) = (ω df 1, ω df 2,..., ω df k ). Y Y

16 XI CHEN, JAMES D. LEWIS, AND MAO SHENG Since p X b for all b B, we have f i (0, t) 0. Hence f i (3.48) t j = 0 x=0 for all t, i = 1, 2,..., k and j = 1, 2,..., N. It follows that (3.49) for all local sections ρ m+1 η(ω 1 ) = ρ m+1 (ω 1 df 1, ω 1 df 2,..., ω 1 df k ) = (ρ m+1 (ω 1 df 1 ), ρ m+1 (ω 1 df 2 ),..., ρ m+1 (ω 1 df k )) = 0 = η σ ρ m (ω 1 ) (3.50) ω 1 H 0 (U, i<l π BΩ i B π P Ω m i P ) H 0 (U, Ω m Y ), where U is an open subset of Y. Every ω H 0 (U, Ω m Y ) can be written as (3.51) ω = ω 1 + ω 2 with ω 1 given in (3.50) and ω 2 H 0 (U, F l Ω m Y ). It is clear that (3.52) ρ m+1 η(ω 2 ) = η σ ρ m (ω 2 ). Combining (3.49) and (3.52), we conclude that (3.53) ρ m+1 η(ω) = η σ ρ m (ω) and hence the diagram (3.45) commutes. It remains to prove that the left square of (3.43) commutes. Note that Ω N X can be identified with the image of the map (3.54) Ω N Y O θ X Ω N+k Y det(n X ) given by (3.55) θ(ω) = ω df 1 df 2... df k. By (3.48) again, we see that the diagram (3.56) Ω N Y O X dσ F N Ω N Y O X σ(b) θ Ω N+k Y det(n X ) F N Ω N+k Y det(n X ) σ(b) commutes. Thus, the diagram θ (3.57) Ω N Y O X Ω N X Ω N+k Y det(n X ) dσ dσ F N Ω N Y O X Ω N σ(b) σ(b) F N Ω N+k Y det(n X ) σ(b)

RATIONALLY EQUIVALENT POINTS 17 commutes. Setting m = n and L = K X in (3.26), we have (3.58) TX n K X TY n K X T n 1 K X N X η n Y Ω N X where det(n X ) = r n N X. Note that Ω N+r n Y det(n X ) Ω N+1+r n Y det(n X ) N X (3.59) F N Ω N+1 Y = π BΩ N B π P Ω P and F N Ω N+2 Y = π BΩ N B π P Ω 2 P. Combining (3.58), (3.43) and (3.59), we obtain commutative diagrams (3.60) TX n K X TY n K X T n 1 K X N X dσ i K σi (B) α n,i πp T P n K X σi (B) η n Y π P T n 1 P α n 1,i K X N X σi (B) with left exact rows for i = 1, 2. By the above diagram, we have (3.61) ker(dσ i Γ b ) = ker(α n,i Γ b ) ker(η n Γ b ) for i = 1, 2. Therefore, (3.38) is equivalent to (3.62) ker(α n,1 Γ b ) ker(η n Γ b ) ker(α n,2 Γ b ) ker(η n Γ b ). More explicitly, we can write Γ b (T n Y K X) as (3.63) Γ b (T n Y K X ) = Γ b (π P T n P K X ) j<n Then the kernel of α n,i Γ b is (3.64) ker(α n,i Γ b ) = Γ b (π P T n P K X ( p i )) j<n Γ b (π P T j P K X) T n j B,b. Γ b (π P T j P K X) T n j B,b for i = 1, 2, where K X ( p i ) = K X I pi for I pi the ideal sheaf of p i. So (3.62) is equivalent to (3.65) ker(η n Γ b ) (Γ b (π P T n P K X ( p 1 )) j<n ker(η n Γ b ) (Γ b (π P T n P K X ( p 2 )) j<n Γ b (π P T j P K X) T n j B,b ) Γ b (π P T j P K X) T n j B,b ) Combining it with Proposition 3.11, we obtain the following criterion:

18 XI CHEN, JAMES D. LEWIS, AND MAO SHENG Proposition 3.13. Let X Y = B P be a smooth projective family of n-dimensional varieties in a projective space P passing through two fixed point p 1 p 2 P over a smooth variety B. Then p 1 and p 2 are not Γ- equivalent over Q on X b for b B general if (3.65) holds at a point b where h 0 (X t, T n X K X) is locally constant in t. Remark 3.14. Since X B P is a family of varieties in P passing through p i, η(v) is a section in H 0 (N Xb ) vanishing at p i for all tangent vectors v T B,b and i = 1, 2. It follows that (3.66) η n n 1 2 i=1 j=0 Γ b (π P T j P K X) T n j B,b ker(α n 1,i Γ b ) = Γ b (π P T n 1 P K X N X ( p 1 p 2 )) n 2 j=0 Γ b (π P T j P K X N X ) T n 1 j B,b. Let us apply Proposition 3.13 to complete intersections in P = P n+k of type (d 1, d 2,..., d k ). When (d i 1) = 2n + 1, we have K X = O X (n). More general, let us consider a smooth projective family X Y = B P of varieties of dimension n in P with K X ( n) globally generated on each fiber X b. In this case, we have the following corollary of Proposition 3.13. Corollary 3.15. Let X Y = B P be a smooth projective family of n-dimensional varieties in a projective space P passing through two fixed point p 1 p 2 P over a smooth variety B and let W X,b be the subspace of Γ b (T P (1)) defined by { (3.67) W X,b = ω Γ b (T P (1)) : η(ω) η ( ) } Γ b (O(1)) T B,b, where the map η on Γ b (T P (1)) and Γ b (O(1)) T B,b are given by the diagram (3.68) Γ b (T Y (1)) η Γ b (N X (1)) Γ b (T P (1)) Γ b (O(1)) T B,b Suppose that there exists a point b B such that h 0 (X t, T n X K X) is constant for t in an open neighborhood of b, each point p i imposes independent conditions on both K Xb ( n) and T X (1) O Xb, i.e., the maps (3.69) Γ b (K X ( n)) K X ( n) O pi and (3.70) Γ b (T X (1)) T X (1) O pi

are surjective for i = 1, 2 and RATIONALLY EQUIVALENT POINTS 19 (3.71) { ω WX,b : ω(p 1 ) = 0 } { ω W X,b : ω(p 2 ) = 0 }. Then p 1 and p 2 are not Γ-equivalent over Q on X b for b B general. Proof. By (3.71), there exists ω W X,b such that ω(p i ) = 0 and ω(p 3 i ) 0 for i = 1 or 2. Without loss of generality, let us assume that ω 1 (p 1 ) = 0 and ω 1 (p 2 ) 0 for some ω 1 W X,b. It is easy to see that W X,b is the image of the projection from Γ b (T X (1)) to Γ b (T P (1)) via the diagram (3.72) Γ b (T X (1)) Γ b (T Y (1)) Γ b (N X (1)) η Γ b (T P (1)) where Γ b (T X (1)) can be identified with ker(η). In other words, for every ω W X,b, there exists τ Γ b (O(1)) T B,b such that η(ω +τ) = 0 and hence ω + τ Γ b (T X (1)). By (3.70), Γ b (T X (1)) generates the vector space T X (1) O p2. On the other hand, by the diagram T Xb (1) O p2 T X (1) O p2 T B,b (3.73) T Yb (1) O p2 T Y (1) O p2 T B,b T P (1) O p2 we see that the image of the projection T X (1) O p2 T P (1) O p2 is the same as the image of the map T Xb (1) O p2 T Yb (1) O p2 and thus has dimension n. Therefore, (3.74) dim{ω(p 2 ) : ω W X,b } = n. And since ω 1 (p 2 ) 0, we can find ω 2,..., ω n W X,b such that {ω j (p 2 )} are linearly independent. On the other hand, ω 1 (p 1 ) = 0 and hence {ω j (p 1 )} are linearly dependent. In other words, { ω1 (p 1 ) ω 2 (p 1 )... ω n (p 1 ) = 0 (3.75) ω 1 (p 2 ) ω 2 (p 2 )... ω n (p 2 ) 0. Let η(ω j + τ j ) = 0 for some τ j Γ b (O(1)) T B,b and j = 1, 2,..., n. Then (3.76) n (ω j + τ j ) s ker(η n Γ b ) j=1

20 XI CHEN, JAMES D. LEWIS, AND MAO SHENG for all s Γ b (K X ( n)). By (3.75), we have n ω j s Γ b (πp TP n K X ( p 1 )) and (3.77) j=1 n ω j s Γ b (πp TP n K X ( p 2 )) j=1 provided that s(p 2 ) 0. The combination of (3.76) and (3.77) yields (3.65). Since the validity of (3.71) is determined by the restriction of W X,b to Z = {p 1, p 2 }, we may let W X,b,Z be the subspace of H 0 (Z, T P (1)) given by (3.78) W X,b,Z = W X,b H 0 (O Z ) = {ω Z : ω Γ b (T P (1)) and η(ω) η ( ) } Γ b (O(1)) T B,b and reformulate (3.71) as (3.79) W X,b,Z H 0 (Z, T P (1) I p ) 0 for some p supp(z) = {p 1, p 2 }. 3.6. Criterion for two varying sections. So far we have obtained the key criterion, Corollary 3.15, for the Γ-equivalence of two fixed sections of X/B in the ambient space P. To apply it to two arbitrary sections of X/B, we need to use an automorphism λ Aut(Y/B) to move these two sections to two fixed points in P, as pointed out before. This line of argument leads to the following: Proposition 3.16. Let X Y = B P be a smooth projective family of n-dimensional varieties in a projective space P over the N-dimensional polydisk B = Spec C[[t j ]] and let σ i : B X be two disjoint sections of X/B with p i = σ i (b) at the origin b B for i = 1, 2. Let λ B Aut(P ) be an automorphism of Y preserving the base B, satisfying that λ b = id and λ(σ i (t)) p i for i = 1, 2 and all t B and given by x 1 (3.80) λ. = Λ x 1., x 0 x r where (x 0, x 1,..., x r ) are the homogeneous coordinates of P and Λ = Λ(t) is an (r + 1) (r + 1) matrix over C[[t j ]] satisfying Λ(0) = I. Let W X,b,Z,λ be the subspace of H 0 (Z, T P (1)) defined by { W X,b,Z,λ = ω Z + L λ (τ) : ω Γ b (T P (1)), τ Γ b (O(1)) T B,b, (3.81) } η(ω + τ) = 0 x 0 x r

RATIONALLY EQUIVALENT POINTS 21 for Z = {p 1, p 2 }, where L λ : πb T B,b T P O Z is the map given by /x ( ) (3.82) L λ = [ ] Λ T 0 x 0 x 1... x r /x 1 t j t j t=0.. /x r Suppose that h 0 (X t, TX n K X) is constant over B, K Xb ( n) and T X (1) O Xb are imposed independent conditions by each point p i for i = 1, 2, and (3.83) W X,b,Z,λ H 0 (Z, T P (1) I p ) 0 for some p supp(z). Then σ 1 (t) and σ 2 (t) are not Γ-equivalent over Q on X t for t B general. Proof. Note that W X,b,Z,λ = W X,b,Z if L λ = 0, i.e., σ i (t) p i. Let X = λ(x) Y = B P. Obviously, X is a smooth projective families of n-dimensional varieties in P over B passing through the two fixed points p 1 p 2. We define the map η : T Y O X N X and the space W X,b Γ b (T P (1)) for X Y = B P in the same way as η and W X,b. Note that since λ b = id, X b = X b and we may use Γ b ( ) to refer both H 0 (X b, ) and H 0 ( X b, ). Let us consider the commutative diagram: Γ b (T X (1)) Γ b (T Y (1)) Γ b (N X (1)) η (3.84) = (dλ) = (dλ) Γ b (T X(1)) Γ b (T Y (1)) Γ b (N X(1)) η π P, Γ b (T P (1)) As pointed out in the proof of Corollary 3.15, W X,b is simply the image of the projection from Γ b (T X (1)) to Γ b (T P (1)) when Γ b (T X (1)) is identified with the kernel of η : Γ b (T Y (1)) Γ b (N X (1)). The same holds for X. That is, W X,b is simply the image of the projection from Γ b (T X(1)) to Γ b (T P (1)) when Γ b (T X(1)) is identified with the kernel of η : Γ b (T Y (1)) Γ b (N X(1)). We may regard W X,b as the image of Γ b (T X (1)) under the map π P, (dλ) in the above diagram. Note that π P, (dλ) is not the same as the projection π P, : Γ b (T Y (1)) Γ b (T P (1)), i.e., (3.85) π P, (dλ) π P,. Indeed, we have (3.86) (dλ) (ω + τ) = (ω + L λ (τ)) + τ

22 XI CHEN, JAMES D. LEWIS, AND MAO SHENG for ω Γ b (T P (1)) and τ Γ b (O(1)) T B,b, where (3.87) Lλ : π B T B π P T P is a homomorphism induced by (dλ) : T Y T Y. Thus, (3.88) π P, (dλ) (ω + τ) = ω + L λ (τ) ω = π P, (ω + τ). It follows that W X,b = π P, dλ Γ b (T X (1)) (3.89) = { ω + L λ (τ) : ω Γ b (T P (1)), τ Γ b (O(1)) T B,b, η(ω + τ) = 0 }. We claim that L λ and W X,b,Z,λ are exactly the restrictions of L λ and W X,b to Z, respectively. Indeed, the differential map dλ : T Y T Y is given by ( ) (dλ) = x i x i ( ) (dλ) = ( ) + t j t L λ j t j (3.90) /x 0 = + [ ] Λ T /x 1 x 0 x 1... x r t j t j. /x r at b. Therefore, L λ is the restriction of L λ to Z and hence W X,b,Z = W X,b,Z,λ. In conclusion, the hypothesis (3.83) on W X,b,Z,λ translates to { } { } (3.91) ω W X,b : ω(p 1 ) = 0 ω W X,b : ω(p 2 ) = 0. Then by Corollary 3.15, σ 1 (t) and σ 2 (t) are not Γ-equivalent over Q on a general fiber X t of X/B. Remark 3.17. In the above proof, it is easy to see that ( ) ( ) ( ) ( ( (3.92) η = η and η = η x i x i t j t L λ j t j )). Since X t passes through p 1 and p 2, η(τ) vanishes at p i and hence L λ satisfies (3.93) η(l λ (τ)) = η(τ) for all τ T B,b. Z There is a more intrinsic way to define L λ : for every t B, we consider the line joining the two points σ i (t); we may regard σ i (t) as the image of two fixed points on P 1 mapped to this line and thus interpret L λ in terms of the deformation of this map P 1 P. We can put the above proposition in the following equivalent form.

RATIONALLY EQUIVALENT POINTS 23 Proposition 3.18. Let X Y = B P be a smooth projective family of n-dimensional varieties in a projective space P over a smooth variety B and let v : S = B P 1 Y be a closed immersion preserving the base B such that v O Y (1) = O S (1) and there are two fixed points p 1 p 2 on P 1 with v b (p i ) X b for all b B. Let W X,b,Z,λ be the subspace of H 0 (Z, vb T P (1)) defined by { W X,b,Z,λ = vb ω Z + L λ (vb τ) : ω Γ b(t P (1)), (3.94) τ Γ b (O(1)) T B,b, } η(ω + τ) = 0 for Z = {p 1, p 2 }, where L λ : πs,b T B,b vb T P O Z is the map induced by T S v T Y with π S,B the projection S B. Suppose that K Xb ( n) and T X (1) O Xb are imposed independent conditions by each point v b (p i ) for i = 1, 2 and (3.95) W X,b,Z,λ H 0 (Z, v b T P (1) I p ) 0 for some p Z and b B general. Then v b (p 1 ) and v b (p 2 ) are not Γ- equivalent over Q on X b for b B general. Note that the hypothesis v O Y (1) = O S (1) simply means that v maps S/B fiberwise to lines in P. Using Proposition 3.16 or 3.18, we obtain the following criterion for the Γ-inequivalence of all pairs of distinct points on X b. Corollary 3.19. Let X Y = B P be a smooth projective family of n-dimensional varieties in a projective space P over a smooth variety B and let W X,b,Z,λ be the subspace of H 0 (Z, T P (1)) defined by (3.81) for a 0-dimensional subscheme Z X b and L λ Hom(πB T B,b, T P O Z ). Suppose that K Xb ( n) and T X (1) O Xb are globally generated on X b and (3.83) holds for a general point b B, all pairs Z = {p 1, p 2 } of distinct points p 1 p 2 on X b, some p supp(z) and all L λ Hom(πB T B,b, T P O Z ) satisfying (3.93). Then no two distinct points on X b are Γ-equivalent over Q for b B very general. We believe that the above corollary will find application in the future. However, we will not use it to prove our main theorem 1.4; instead, we will apply Proposition 3.16 directly, i.e., apply it to families X B P n+1 of hypersurfaces of degree 2n + 2 in P n+1. In this case, both K X ( n) = O X and T X (1) are globally generated on X b if X/B is versal. So it suffices to verify (3.83), which we will carry out in the next section. 4. Hypersurfaces of degree 2n + 2 in P n+1 4.1. Versal deformation of the Fermat hypersurface. In this section, we are going to prove our main theorem 1.4 using the criteria developed in the previous section.

24 XI CHEN, JAMES D. LEWIS, AND MAO SHENG To start, let us choose a versal family of hypersurfaces in P n+1. Let X Y = B P be the family of hypersurfaces of degree d in P = P n+1 given by (4.1) F (x 0, x 1,..., x n, t f ) = x d 0 + x d 1 +... + x d n+1 + f J d t f f = 0, where (x 0, x 1,..., x n+1 ) are the homogeneous coordinates of P n+1, J d is the set of monomials in x i given by (4.2) J d = { x m 0 0 xm 1 1...xm n+1 n+1 : m 0, m 1,..., m n+1 N, m 0 + m 1 +... + m n+1 = d and m 0, m 1,..., m n+1 d 2 } and (t f ) are the coordinates of the affine space B = Span C J d = A N for ( ) d + n + 1 (4.3) N = h 0 (O P (d)) h 0 (T P ) 1 = (n + 2) 2. n + 1 We may regard X/B as a versal deformation of the Fermat hypersurface. At a general point b B, X/B is obviously versal, i.e., the Kodaira- Spencer map (4.4) T B,b H 0 (N Xb )/η(h 0 (X b, T P )) is an isomorphism, where η is the map in H 1 (T Xb ) η (4.5) 0 T X T Y O X N X 0. More explicitly, (4.4) is equivalent to saying { } F (4.6) Span x i Span J d = H 0 (N Xb ) = H 0 (X b, O(d)) x j for b B general. Let E = O P (1) n+2 be the Euler bundle on P. Then { } (4.7) H 0 (T P ) = H0 (E) = Span x i /(α) (α) x j by the Euler sequence (3.28) and ( ) (4.8) η = F ( ) and η x j x j t f for j = 0, 1, 2,..., n + 1 and f J d, where n+1 (4.9) α = i=0 x i. x i = F t f = f

RATIONALLY EQUIVALENT POINTS 25 We are going to show that no two distinct points on a very general fiber X b of X/B are Γ-equivalent over Q when d = 2n+2 6. To set it up, we fix a general point b B. Let us assume that there exist two disjoint sections σ i : B X in an analytic open neighborhood of b such that σ 1 (t) and σ 2 (t) are Γ-equivalent over Q for all t. We let λ B Aut(P ) be an automorphism of Y such that λ b = id and λ(σ i (t)) p i = σ i (b) for i = 1, 2 and let L λ be defined accordingly by (3.82). It comes down to the verification of (3.83). Definition 4.1. Let Z be a 0-dimensional scheme of length 2 in P = P n+1 with homogeneous coordinates (x 0, x 1,..., x n+1 ). We call Z generic with respect to the homogeneous coordinates (x i ) if (4.10) H 0 (O Z (1)) = Span{x j : j i} for every i = 0, 1,..., n + 1. Otherwise, we call Z special with respect to (x i ). We call Z very special with respect to (x i ) if (4.11) #{x i : x i H 0 (I Z (1))} = n = h 0 (O P (1)) 2 where I Z is the ideal sheaf of Z in P. Remark 4.2. Clearly, these notions depend on the choice of homogeneous coordinates of P. More generally, we can define these terms with respect to a basis of H 0 (L) for an arbitrary very ample line bundle L on P. When the choice of homogeneous coordinates is clear, we simply say Z is generic (resp. special/very special). Obviously, being very special implies being special. There always exist i j such that x i and x j span H 0 (O Z (1)) since O P (1) is very ample. Without loss of generality, we usually make the assumption that (i, j) = (0, 1), i.e., (4.12) H 0 (O Z (1)) = Span{x 0, x 1 }. Under the hypothesis of (4.12), Z is special if and only if (4.13) Span{x 0, x 1 } = H 0 (O Z (1)) Span{x 1, x 2,..., x n+1 }. Furthermore, by re-arranging x 2,..., x n+1, we may assume that there exists 1 a n + 1 such that (4.14) x 1,..., x a H 0 (I Z (1)) and x a+1,..., x n+1 H 0 (I Z (1)). Of course, Z is very special if and only if a = 1. We are considering two cases: with respect to (x j ), Generic case: Z = {σ 1 (b), σ 2 (b)} = {p 1, p 2 } is generic or Special case: Z = {σ 1 (b), σ 2 (b)} is special for all b B. 4.2. A basis for W X,b. For convenience, we identify the tangent space T B,b with Span J d. Then η(f) = f for all f Span J d. We start the proof of (3.83) by studying the space W X,b defined by (3.67). It has a basis given by:

26 XI CHEN, JAMES D. LEWIS, AND MAO SHENG Lemma 4.3. Let P = P n+1 and X Y = B P be the family of hypersurfaces in P given by (4.1) over B = Span J d for d 3. Then W X,b = { ω H 0 } (X b, E(1)) : η(ω) Span J d+1 (4.15) = Span { ω ijk : 0 i, j, k n + 1, i j and i, j k } has dimension ( ) n + 2 (4.16) dim W X,b = (n + 2) 2 for b = (t f ) in an open neighborhood of 0, where (4.17) with (4.18) c ijk = d 1 d! ω ijk = x i x j for i j k and x k ω iik = x 2 i c ijk x i x j for i k x k x i j i ( ) d F x d 2 = i x j x k { 2d 1 t f d 1 t f if i j = k if i j k for f = x d 2 i x j x k. Here we consider η as a map H 0 (E(1)) H 0 (O(d + 1)) given by (4.8). Proof. We have ( ) (4.19) η x i x j x k It is easy to check that F = x i x j = dx i x j x d 1 x k + f t f x i x j. k x k f J d (4.20) η(ω ijk ) = x i x j F x k Span J d+1 for i j k and (4.21) η(ω iik ) = x 2 i F x k j i d 1 d! ( ) d F x i x j xi d 2 x j x k F x i Span J d+1 for i k. Hence ω ijk W X,b for all i, j k. To show that {ω ijk : i j and i, j k} forms a basis of W X,b in an open neighborhood of 0, it suffices to verify this for b = 0: clearly, { } { } b=0 (4.22) ω ijk : i j and i, j k = x i x j : i j and i, j k x k is a basis of W X,0. Therefore, (4.15) and (4.16) follow. Clearly, W X,b is the image of W X,b under the map (4.23) H 0 (X b, E(1)) H 0 (X b, T P (1)).

RATIONALLY EQUIVALENT POINTS 27 More precisely, let ŴX,b be the lift of W X,b in H 0 (X b, E(1)). Then (4.24) Ŵ X,b = W X,b α H 0 (O(1)) where W X,b α H 0 (O(1)) = 0 because (4.25) Span J d+1 η ( α H 0 (O(1)) ) = Span J d+1 F H 0 (O(1)) = 0. 4.3. An observation on L λ. We observe the following: Lemma 4.4. Let P = P n+1 and X Y = B P be the family of hypersurfaces in P given by (4.1) over B = Span J d. For b B, a 0-dimensional subscheme Z X b of length 2 and L λ Hom(π B T B,b, T P O Z ), if (4.26) L λ (f) 0 for some f H 0 (I Z (1)) Span J d 1 Span J d, then (3.83) holds. Proof. Obviously, (4.26) holds for some f = lg with l H 0 (I Z (1)) and g J d 1. For each point p supp(z), we choose l p H 0 (O P (1)) such that l p (p) = 0 and l p H 0 (I Z (1)) and let (4.27) τ p = l p f l l p g H 0 (O Xb (1)) T B,b. Then η(τ p ) = 0 so L λ (τ p ) W X,b,Z,λ. Clearly, (4.28) L λ (τ p ) = l p L λ (f) ll λ (l p g) = l p L λ (f) since l H 0 (I Z (1)). Then by our choice of l p, L λ (τ p ) vanishes at p. If L λ (τ p ) 0, then (3.83) follows. Otherwise, (4.29) l p L λ (f) = 0. Since l p H 0 (I Z (1)), (4.29) implies that L λ (f) vanishes at all p supp(z). If Z consists of two distinct points, then we must have (4.30) L λ (f) = 0, which contradicts our hypothesis (4.26). If Z is supported at a single point p, then L λ (f) vanishes at p. Applying the same argument to τ q = l q f l l q g for some l q H 0 (O P (1)) satisfying l q (p) 0, we have (4.31) L λ (τ q ) = l q L λ (f) ll λ (l q g) = l q L λ (f) W X,b,Z,λ vanishing at p. Again, we have either (3.83) or (4.30) since l q (p) 0. Let us assume that (4.30) holds for all f H 0 (I Z (1)) Span J d 1. Otherwise, we are done by the above lemma. Then L λ : T B,b H 0 (Z, T P ) factors through (4.32) Span J d H 0 (I Z (1)) Span J d 1

28 XI CHEN, JAMES D. LEWIS, AND MAO SHENG and it can be regarded as a map (4.33) Span J d L λ H 0 H 0 (Z, T P ). (I Z (1)) Span J d 1 4.4. The space H 0 (I Z (1)) Span J d 1. Let us figure out the space (4.32). Obviously, (4.34) H 0 (I Z (1)) Span J d 1 Span J d H 0 (I Z (1)) H 0 (O P (d 1)). Furthermore, since H 0 (I Z (1)) H 0 (O P (d 1)) is the kernel of the map (4.35) ξ H 0 (O P (d)) Sym d H 0 (O Z (1)) we may write (4.34) as Sym d H 0 (O P (1)) (4.36) H 0 (I Z (1)) Span J d 1 Span J d ker(ξ). Actually, this inclusion is an equality for Z generic: Lemma 4.5. Let P = P n+1, J d be defined in (4.2) and Z be a 0-dimensional subscheme of P of length 2. If d 4 and Z is generic with respect to (x i ), then (4.37) H 0 (I Z (1)) Span J d 1 = Span J d H 0 (I Z (1)) H 0 (O P (d 1)) = Span J d ker(ξ). Or equivalently, H 0 (I Z (1)) Span J d 1 is the kernel of the map ξ (4.38) Span J d Sym d H 0 (O Z (1)). In addition, (4.39) Span J d ξ H 0 Sym d H 0 (O Z (1)) (I Z (1)) Span J d 1 is an isomorphism. Proof. To prove (4.37), it suffices to find a subset S J d such that (4.40) Span J d = H 0 (I Z (1)) Span J d 1 + Span(S) and (4.41) H 0 (I Z (1)) H 0 (O P (d 1)) Span(S) = 0. Let us assume (4.12). By (4.10), H 0 (O Z (1)) = Span{x 1, x 2,..., x n+1 } and hence there exists i 0, 1 such that (4.42) H 0 (O Z (1)) = Span{x 1, x i }.

RATIONALLY EQUIVALENT POINTS 29 Similarly, we have H 0 (O Z (1)) = Span{x 0, x 2,..., x n+1 } and hence there exists j 0, 1 such that (4.43) H 0 (O Z (1)) = Span{x 0, x j }. Then we let (4.44) S = { x d 3 0 x 3 i, x d 3 0 x 2 i x 1, x d 3 0 x i x 2 1, x d 3 0 x 3 1, x d 4 0 x 4 1,..., x 3 0x d 3 1, x 2 0x d 3 1 x j, x 0 x d 3 1 x 2 j, x d 3 1 x 3 j }. By (4.12), (4.42) and (4.43), for every k, (4.45) x k H 0 (I Z (1)) + Span{x 0, x 1 }, x k H 0 (I Z (1)) + Span{x 1, x i }, and x k H 0 (I Z (1)) + Span{x 0, x j }. Then (4.40) follows. To see (4.41), we just have to show that ker(ξ) Span(S) = 0, which is equivalent to (4.46) ξ (Span(S)) = Sym d H 0 (O Z (1)) since S = dim Sym d H 0 (O Z (1)) = d+1. Again it is easy to see from (4.12), (4.42) and (4.43) that ( ) ξ (Span(S)) = ξ Span{x d k 0 x k 1 : k = 0, 1,..., d} (4.47) = Sym d H 0 (O Z (1)). This also proves that (4.39) is an isomorphism. When Z is special, H 0 (I Z (1)) Span J d 1 is no longer the kernel of the map (4.38). Instead, we have the following result when Z is special but not very special. Lemma 4.6. Let P = P n+1, J d be defined in (4.2) and Z be a 0-dimensional subscheme of P of length 2. Suppose that d 4, Z satisfies (4.13) and {x 2,..., x n+1 } H 0 (I Z (1)). Then (4.48) Span J d ker(ξ) = H 0 (I Z (1)) Span J d 1 { + Span x d 2 0 x i (x j c j x 1 ) : i 1, j 2 } and x j c j x 1 H 0 (I Z (1)). Proof. We leave the verification of (4.48) to the readers.