Geometry dictates arithmetic

Transcription

1 Geometry dictates arithmetic Ronald van Luijk February 21, 2013 Utrecht

2 Curves Example. Circle given by x 2 + y 2 = 1 (or projective closure in P 2 ).

3 Curves Example. Circle given by x 2 + y 2 = 1 (or projective closure in P 2 ). Definition. Genus of a smooth projective curve C over Q is the genus of C(C). g = 0 g = 1 g 2

4 Genus 0 (0, 1) ( 4 5, 3 5) (1, 0) P 2 C : x 2 + y 2 = 1

5 Genus 0 (0, 1) ( 1, 0) (1, 0) P 2 C : x 2 + y 2 = 1

6 Genus 0 (0, 1) y = t(x + 1) ( 1, 0) (1, 0) P 2 C : x 2 + y 2 = 1

7 Genus 0 (0, 1) y = t(x + 1) ( 1, 0) ( ) 1 t 2 2t, 1+t 2 1+t 2 (1, 0) P 2 C : x 2 + y 2 = 1

8 Genus 0 (0, 1) y = t(x + 1) ( 1, 0) ( ) 1 t 2 2t, 1+t 2 1+t 2 (1, 0) Theorem. If a conic over Q has a rational point, then it has infinitely many. P 2 C : x 2 + y 2 = 1

9 Genus 0 (0, 1) y = t(x + 1) ( 1, 0) ( ) 1 t 2 2t, 1+t 2 1+t 2 (1, 0) Theorem. If a conic over Q has a rational point, then it has infinitely many. Theorem. If a conic D over Q has a rational point, then there is an isomorphism P 1 (C) D(C), so the genus of D is 0. P 2 C : x 2 + y 2 = 1

10 Genus 0 (0, 1) y = t(x + 1) ( 1, 0) ( ) 1 t 2 2t, 1+t 2 1+t 2 (1, 0) Theorem. If a conic over Q has a rational point, then it has infinitely many. Theorem. If a conic D over Q has a rational point, then there is an isomorphism P 1 (C) D(C), so the genus of D is 0. Theorem. Any curve of genus 0 over Q is isomorphic to a conic. P 2 C : x 2 + y 2 = 1

11 Genus 0 (0, 1) y = t(x + 1) ( 1, 0) ( ) 1 t 2 2t, 1+t 2 1+t 2 (1, 0) Theorem. If a conic over Q has a rational point, then it has infinitely many. Theorem. If a conic D over Q has a rational point, then there is an isomorphism P 1 (C) D(C), so the genus of D is 0. Theorem. Any curve of genus 0 over Q is isomorphic to a conic. P 2 C : x 2 + y 2 = 1 Theorem. If a curve of genus 0 over Q has a rational point, then it is isomorphic to P 1 and it has infinitely many rational points.

12 Genus 0 ( 11 61, 60 ) 61 N C (B) = 150 number of ( a c, b ) c with a, b, c B P 2 C : x 2 + y 2 = B

13 Genus 0 N C (B) = 150 number of ( a c, b ) c with a, b, c B P 2 C : x 2 + y 2 = 1 50 N C (B) 4 π B B

14 Genus 0 N C (B) = 150 number of ( a c, b ) c with a, b, c B P 2 C : x 2 + y 2 = 1 50 N C (B) 4 π B Theorem. The number N D (B) of rational points on a conic D grows linearly with the height B (or is zero) B

15 Genus 1 (elliptic)

16 Genus 1 (elliptic) Q P E : y 2 = x 3 15x + 19

17 Genus 1 (elliptic) Q P E : y 2 = x 3 15x + 19

18 Genus 1 (elliptic) Q P P + Q E : y 2 = x 3 15x + 19

19 Genus 1 (elliptic) Fact. E(k) is an abelian group! Q P P + Q E : y 2 = x 3 15x + 19

20 Genus 1 (elliptic) Fact. E(k) is an abelian group! Theorem (Mordell-Weil). For any elliptic curve E over Q, the group E(Q) is finitely generated Here: rank= 1, and Z = E(Q) = (3, 1) E : y 2 = x 3 15x + 19

21 Genus 1 (elliptic) ( , ) N E (B) = number of ( a c, b ) c with a, b, c B E : y 2 = x 3 15x N E (B) γ log B γ = log B

22 Genus 1 (elliptic) ( , ) N E (B) = number of ( a c, b ) c with a, b, c B Theorem. For any elliptic curve E over Q with r = rank E(Q), we have N E (B) c(log B) r/ N E (B) γ log B γ = log B

23 Genus g 2 Examples. y 2 = f (x) with f separable of degree 2g + 2. smooth projective plane curve of degree d 4 with g = 1 2 (d 1)(d 2).

24 Genus g 2 Examples. y 2 = f (x) with f separable of degree 2g + 2. smooth projective plane curve of degree d 4 with g = 1 2 (d 1)(d 2). Theorem ( Mordell Conjecture by Faltings, 1983). Any curve over Q with g 2 has only finitely many rational points.

25 Genus g 2 Examples. y 2 = f (x) with f separable of degree 2g + 2. smooth projective plane curve of degree d 4 with g = 1 2 (d 1)(d 2). Theorem ( Mordell Conjecture by Faltings, 1983). Any curve over Q with g 2 has only finitely many rational points. Conclusion. The higher the genus, the lower the number of rational points.

26 Differentials Definition. Let X be a smooth projective variety with function field k(x ). Then Ω k(x )/k is the k(x )-vectorspace of differential 1-forms, generated by {df : f k(x )} and satisfying d(f + g) = df + dg, d(fg) = fdg + gdf, da = 0 for a k.

27 Differentials Definition. Let X be a smooth projective variety with function field k(x ). Then Ω k(x )/k is the k(x )-vectorspace of differential 1-forms, generated by {df : f k(x )} and satisfying d(f + g) = df + dg, d(fg) = fdg + gdf, da = 0 for a k. Proposition. We have dim k(x ) Ω k(x )/k = dim X. Example. For curve C : y 2 = f (x) we have 2ydy = f (x)dx in Ω k(c)/k.

28 Holomorphic differentials on curves Definition. For a point P on a smooth projective curve C with local parameter t P k(c) and a differential ω Ω k(c)/k, we write ω = f P dt P ; then ω is holomorphic at P if f P has no pole at P. Example. Curve C : y 2 = f (x) with f separable of degree d 3. Then is holomorphic everywhere. ω = 1 y d(x c) = 1 y dx = 2 f (x) dy

29 Holomorphic differentials on curves Definition. For a point P on a smooth projective curve C with local parameter t P k(c) and a differential ω Ω k(c)/k, we write ω = f P dt P ; then ω is holomorphic at P if f P has no pole at P. Example. Curve C : y 2 = f (x) with f separable of degree d 3. Then is holomorphic everywhere. ω = 1 y d(x c) = 1 y dx = 2 f (x) dy Definition. Set Ω C/k = {ω Ω k(c)/k : ω holom. everywhere}. Proposition. We have g = dim k Ω C/k.

30 Holomorphic differentials in general Recall. If X smooth, projective, then dim k(x ) Ω k(x )/k = dim X. Fact. If V is a vector space with dim V = n, then dim n V = 1.

31 Holomorphic differentials in general Recall. If X smooth, projective, then dim k(x ) Ω k(x )/k = dim X. Fact. If V is a vector space with dim V = n, then dim n V = 1. Definition (unconventional notation for (dim X )-forms). Set Ω X /k = {ω dim X Ω k(x )/k : ω holom. everywhere}.

32 Holomorphic differentials in general Recall. If X smooth, projective, then dim k(x ) Ω k(x )/k = dim X. Fact. If V is a vector space with dim V = n, then dim n V = 1. Definition (unconventional notation for (dim X )-forms). Set Ω X /k = {ω dim X Ω k(x )/k : ω holom. everywhere}. Definition For a k-basis (ω 0, ω 1,..., ω N ) of Ω X /k, we get f i k(x ) such that ω i = f i ω 0. The Kodaira dimension κ(x ) of X is 1 if dim k Ω X /k = 0, or the dimension of the image of the map X A N, P (f 1 (P), f 2 (P),..., f N (P)).

33 Holomorphic differentials in general Recall. If X smooth, projective, then dim k(x ) Ω k(x )/k = dim X. Fact. If V is a vector space with dim V = n, then dim n V = 1. Definition (unconventional notation for (dim X )-forms). Set Ω X /k = {ω dim X Ω k(x )/k : ω holom. everywhere}. Definition (Wrong: use tensor powers of dim X Ω k(x )/k.) For a k-basis (ω 0, ω 1,..., ω N ) of Ω X /k, we get f i k(x ) such that ω i = f i ω 0. The Kodaira dimension κ(x ) of X is 1 if dim k Ω X /k = 0, or the dimension of the image of the map X A N, P (f 1 (P), f 2 (P),..., f N (P)).

34 Holomorphic differentials in general Recall. If X smooth, projective, then dim k(x ) Ω k(x )/k = dim X. Fact. If V is a vector space with dim V = n, then dim n V = 1. Definition (unconventional notation for (dim X )-forms). Set Ω X /k = {ω dim X Ω k(x )/k : ω holom. everywhere}. Definition (Wrong: use tensor powers of dim X Ω k(x )/k.) For a k-basis (ω 0, ω 1,..., ω N ) of Ω X /k, we get f i k(x ) such that ω i = f i ω 0. The Kodaira dimension κ(x ) of X is 1 if dim k Ω X /k = 0, or the dimension of the image of the map X A N, P (f 1 (P), f 2 (P),..., f N (P)). Proposition. For a curve C we get 1 g = 0 κ(c) = 0 g = 1 1 g 2

35 Varieties of general type In general, 1 κ(x ) dim X (complex X high κ(x )). Definition. We say that X is of general type if κ(x ) = dim X. ( many holom. differentials, canonical bundle is pseudo-ample )

36 Varieties of general type In general, 1 κ(x ) dim X (complex X high κ(x )). Definition. We say that X is of general type if κ(x ) = dim X. ( many holom. differentials, canonical bundle is pseudo-ample ) Conjecture (Lang). If X is a variety over Q that is of general type, then the rational points lie in a Zariski closed subset, i.e., a finite union of proper subvarieties of X.

37 Varieties of general type In general, 1 κ(x ) dim X (complex X high κ(x )). Definition. We say that X is of general type if κ(x ) = dim X. ( many holom. differentials, canonical bundle is pseudo-ample ) Conjecture (Lang). If X is a variety over Q that is of general type, then the rational points lie in a Zariski closed subset, i.e., a finite union of proper subvarieties of X. Corollary. Let X P 3 be a smooth, projective surface over Q of degree 5. Then the rational points are all contained in some finite union of curves.

38 Fano varieties Definition. A Fano variety is a smooth, projective variety X with ample anti-canonical bundle. We have κ(x ) = 1 and X is geometrically easy.

39 Fano varieties Definition. A Fano variety is a smooth, projective variety X with ample anti-canonical bundle. We have κ(x ) = 1 and X is geometrically easy. Conjecture (Batyrev-Manin). Suppose X over Q is Fano. Set ρ = rk Pic X. There is an open subset U X and a constant c with N U (B) cb(log B) ρ 1.

40 Fano varieties Definition. A Fano variety is a smooth, projective variety X with ample anti-canonical bundle. We have κ(x ) = 1 and X is geometrically easy. Conjecture (Batyrev-Manin). Suppose X over Q is Fano. Set ρ = rk Pic X. There is an open subset U X and a constant c with N U (B) cb(log B) ρ 1. This is proved in many cases for surfaces. False in higher dimension, but no counterexamples to lower bound.

41 Fano varieties Definition. A Fano variety is a smooth, projective variety X with ample anti-canonical bundle. We have κ(x ) = 1 and X is geometrically easy. Conjecture (Batyrev-Manin). Suppose X over Q is Fano. Set ρ = rk Pic X. There is an open subset U X and a constant c with N U (B) cb(log B) ρ 1. This is proved in many cases for surfaces. False in higher dimension, but no counterexamples to lower bound. Conclusion. The more complex a variety, the fewer rational points.

42 K3 surfaces Definition. A K3 surface over Q is a smooth, projective surface X with X (C) simply connected and with trivial canonical bundle. There is a unique holomorphic differential and we have κ(x ) = 0.

43 K3 surfaces Definition. A K3 surface over Q is a smooth, projective surface X with X (C) simply connected and with trivial canonical bundle. There is a unique holomorphic differential and we have κ(x ) = 0. Examples Smooth quartic surfaces in P 3. Double cover of P 2 ramified over a smooth sextic. Desingularization of A/ [ 1] for an abelian surface A.

44 Theorem (Tschinkel-Bogomolov). If rk Pic X 5, then there is a finite extension K of Q such that the K-rational points are Zariski dense on X, i.e., rational points are potentially dense on X.

45 Theorem (Tschinkel-Bogomolov). If rk Pic X 5, then there is a finite extension K of Q such that the K-rational points are Zariski dense on X, i.e., rational points are potentially dense on X. Question. Is there a K3 surface X over a number field with rk Pic X = 1 and rational points potentially dense?

46 Theorem (Tschinkel-Bogomolov). If rk Pic X 5, then there is a finite extension K of Q such that the K-rational points are Zariski dense on X, i.e., rational points are potentially dense on X. Question. Is there a K3 surface X over a number field with rk Pic X = 1 and rational points potentially dense? Question. Is there a K3 surface X over a number field with rk Pic X = 1 and rational points not potentially dense?

47 Theorem (Tschinkel-Bogomolov). If rk Pic X 5, then there is a finite extension K of Q such that the K-rational points are Zariski dense on X, i.e., rational points are potentially dense on X. Question. Is there a K3 surface X over a number field with rk Pic X = 1 and rational points potentially dense? Question. Is there a K3 surface X over a number field with rk Pic X = 1 and rational points not potentially dense? Question. Is there a K3 surface X over a number field K with X (K) neither empty nor dense?

48 K3 surfaces Theorem (Logan, McKinnon, vl). Take a, b, c, d Q with abcd (Q ) 2. Let X P 3 be given by ax 4 + by 4 + cz 4 + dw 4. If P X (Q) has no zero coordinates and P does not lie on one of the 48 lines (no two terms sum to 0), then X (Q) is Zariski dense.

49 K3 surfaces Theorem (Logan, McKinnon, vl). Take a, b, c, d Q with abcd (Q ) 2. Let X P 3 be given by ax 4 + by 4 + cz 4 + dw 4. If P X (Q) has no zero coordinates and P does not lie on one of the 48 lines (no two terms sum to 0), then X (Q) is Zariski dense. Question. Are the conditions on P necessary? Conjecture (vl) Every t Q can be written as t = x 4 y 4 z 4 w 4.

50 N U (B) N 13.5 log B S : x 3 3x 2 y 2 + 4x 2 yz x 2 z 2 + x 2 z xy 2 z xyz 2 + x + y 3 + y 2 z 2 + z 3 = 0 8 log B

51 K3 surfaces Conjecture (vl). Suppose X is a K3 surface over Q with rk Pic X C = 1. There is an open subset U X and a constant c such that N U (B) c log B.

52 Hasse principle Theorem (Hasse). Let Q P n be a smooth quadric over Q. Suppose that Q has points over R and over Q p for every p. Then Q(Q).

53 Hasse principle Theorem (Hasse). Let Q P n be a smooth quadric over Q. Suppose that Q has points over R and over Q p for every p. Then Q(Q). Proposition (Selmer). The curve C P 2 given by 3x 3 + 4y 3 + 5z 3 = 0 has points over R and over Q p for every p, but C(Q) =.

54 Brauer-Manin obstruction To every variety X we can assign the Brauer group Br X. Every morphism X Y induces a homomorphism Br Y Br X. For every point P over a field k we have Br(P) = Br(k).

55 Brauer-Manin obstruction To every variety X we can assign the Brauer group Br X. Every morphism X Y induces a homomorphism Br Y Br X. For every point P over a field k we have Br(P) = Br(k). Let X be smooth and projective. X (Q) Br(Q) v X (Q v ) φ v Br(Q v ) Q/Z

56 Brauer-Manin obstruction To every variety X we can assign the Brauer group Br X. Every morphism X Y induces a homomorphism Br Y Br X. For every point P over a field k we have Br(P) = Br(k). Let X be smooth and projective. X (Q) Br(Q) v X (Q v ) φ v Br(Q v ) Q/Z Corollary. If ( v X (Q v ) ) Br := φ 1 (0) is empty, then X (Q) =.

57 Brauer-Manin obstruction To every variety X we can assign the Brauer group Br X. Every morphism X Y induces a homomorphism Br Y Br X. For every point P over a field k we have Br(P) = Br(k). Let X be smooth and projective. X (Q) Br(Q) v X (Q v ) φ v Br(Q v ) Q/Z Corollary. If ( v X (Q v ) ) Br := φ 1 (0) is empty, then X (Q) =. Conjecture (Colliot-Thélène). This Brauer-Manin obstruction is the only obstruction to the existence of rational points for rationally connected varieties.