Quadratic families of elliptic curves and degree 1 conic bundles

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Quadratic families of elliptic curves and degree 1 conic bundles János Kollár Princeton University joint with Massimiliano Mella

Elliptic curves E := ( y 2 = a 3 x 3 + a 2 x 2 + a 1 x + a 0 ) A 2 xy Major problem 1: Find all solutions. (over Q or number fields or...) Major problem 2: Are there infinitely many solutions? Weak variant: There are many (a 3,..., a 0 ) K 4 for which E has infinitely many solutions.

Families of elliptic curves a i (t) K[t] polynomials family of elliptic curves E t := ( y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) )

Families of elliptic curves a i (t) K[t] polynomials family of elliptic curves E t := ( y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) ) nontrivial family: at least two of the curves E t are smooth, elliptic and not isomorphic to each other over K. Equivalent to: discriminant is not identically 0 and not all the a i (t) are multiples of the same square.

Main Corollary K a number field, a i (t) K[t] polynomials of degree 2, nontrivial family. Then E t := ( y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) ) has infinitely many solutions for about all possible t K (arranged by height).

Previous work 1. If a i have degree 1: easy (linear equation for t) 2. Degree 2 case: R. Munshi studied the case when the a i are multiples of each other (+ CM case) 3. If a i have degree 3: probably not true (K3 surfaces)

Geometry enters We focus on the algebraic surface T := ( y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) ) A 3 xyt

Geometry enters We focus on the algebraic surface T := ( y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) ) A 3 xyt Theorem K field of characteristic 2 a i (t) K[t] polynomials of degree 2 giving a nontrivial family. T is unirational over K. unirational: there is a dominant map φ : P 2 T, so φ ( P 2 (K) ) gives many K-points on T.

Conic bundles Projection to x-axis (using deg a i = 2) T := ( y 2 = b 2 (x)t 2 + b 1 (x)t + b 0 (x) ) A 3 xyt where the b i (x) K[x] are cubics. T A 1 x is a conic bundle.

Conic bundles Projection to x-axis (using deg a i = 2) T := ( y 2 = b 2 (x)t 2 + b 1 (x)t + b 0 (x) ) A 3 xyt where the b i (x) K[x] are cubics. T A 1 x is a conic bundle. Compactify: S P 1, generic fibers are smooth, rational. Max Noether (1870): birational to P 1 P 1 over C. Corollary: S is rational after a finite degree field extension K /K (In our case, degree 2 7 7!)

Minimal conic bundles S P 1 such that every fiber is a conic; general fibers: smooth conics, special fibers: conjugate pairs of lines. Main invariant: δ(s) = number of singular fibers Degree: ( K 2 S) = 8 δ(s)

Minimal conic bundles S P 1 such that every fiber is a conic; general fibers: smooth conics, special fibers: conjugate pairs of lines. Main invariant: δ(s) = number of singular fibers Degree: ( K 2 S) = 8 δ(s) Arithmetic gets harder as δ(s) increases Del Pezzo cases: 0 δ(s) 7 Boundary case: δ(s) = 8 ( families of g = 1 curves) Hard cases: δ(s) 9 ( families of g 2 hyperelliptic curves)

Our surface T T = ( y 2 = b 2 (x)t 2 + b 1 (x)t + b 0 (x) ) A 3 xyt where the b i (x) K[x] are cubics. Singular fibers: at roots of b 2 1 4b 2 b 0 : 6 singular fibers at infinity: need to blow up/down to get that singular iff a 3 (t) is not a square. δ(t ) = 7 so T is a degree 1 conic bundle.

Theorem K field of characteristic 2 π : S P 1 conic bundle with δ(s) 7, then S is unirational over K S(K).

Theorem K field of characteristic 2 π : S P 1 conic bundle with δ(s) 7, then S is unirational over K S(K). Segre (1951), Manin (1966): δ(s) 5 δ(s) = 6: Manin knew many cases δ(s) = 7: there is always a K-point since (K 2 S ) = 1.

Geometry for δ(s) = 5, 6, 7: Weak del Pezzo K S is semiample (with few of exceptions) maps by K S or 2K S :

Geometry for δ(s) = 5, 6, 7: Weak del Pezzo K S is semiample (with few of exceptions) maps by K S or 2K S : δ(s) = 5: S P 3 as a cubic (with a line)

Geometry for δ(s) = 5, 6, 7: Weak del Pezzo K S is semiample (with few of exceptions) maps by K S or 2K S : δ(s) = 5: S P 3 as a cubic (with a line) δ(s) = 6: S P 2 double cover, deg 4 ramification

Geometry for δ(s) = 5, 6, 7: Weak del Pezzo K S is semiample (with few of exceptions) maps by K S or 2K S : δ(s) = 5: S P 3 as a cubic (with a line) δ(s) = 6: S P 2 double cover, deg 4 ramification δ(s) = 7: K S is a pencil but 2K S gives S Q P 3 double cover, Q: quadric cone, degree 2 3 ramification. Bertini involution: τ B : S S over Q

Classification of degree 1 conic bundles I K S pencil with a unique base point p S(K) π : S P 1 F : fiber of π containing p.

Classification of degree 1 conic bundles I K S pencil with a unique base point p S(K) π : S P 1 F : fiber of π containing p. (General case) F smooth. Thus F = P 1 over K. Good! (Pf: Conic rational over K iff it has a K-point.)

Classification of degree 1 conic bundles I K S pencil with a unique base point p S(K) π : S P 1 F : fiber of π containing p. (General case) F smooth. Thus F = P 1 over K. Good! (Pf: Conic rational over K iff it has a K-point.) (Special case) F singular. Thus p is the unique singular point of F.

Classification of degree 1 conic bundles I K S pencil with a unique base point p S(K) π : S P 1 F : fiber of π containing p. (General case) F smooth. Thus F = P 1 over K. Good! (Pf: Conic rational over K iff it has a K-point.) (Special case) F singular. Thus p is the unique singular point of F. Two pencils: double cover σ : S P 1 P 1

Classification of degree 1 conic bundles II σ : S P 1 P 1 double cover (General case) In suitable coordinates: y 2 = x 4 + a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) where deg a i 2 and deg a 3 = 2.

Classification of degree 1 conic bundles II σ : S P 1 P 1 double cover (General case) In suitable coordinates: y 2 = x 4 + a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) where deg a i 2 and deg a 3 = 2. (Special case) In suitable coordinates: y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t), where deg a i 2 and deg a 3 = 2.

Classification of degree 1 conic bundles II σ : S P 1 P 1 double cover (General case) In suitable coordinates: y 2 = x 4 + a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) where deg a i 2 and deg a 3 = 2. Bertini involution σ-involution

Classification of degree 1 conic bundles II σ : S P 1 P 1 double cover (General case) In suitable coordinates: y 2 = x 4 + a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t) where deg a i 2 and deg a 3 = 2. Bertini involution σ-involution (Special case) In suitable coordinates: y 2 = a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t), where deg a i 2 and deg a 3 = 2. Bertini involution = σ-involution

Unirationality in general case Lemma (Enriques criterion) A conic bundle S P 1 is unirational iff there is a multi-section s : P 1 S.

Unirationality in general case Lemma (Enriques criterion) A conic bundle S P 1 is unirational iff there is a multi-section s : P 1 S. We have P 1 = F S a fiber; Bertini involution τ B σ-involution; τ B (F ) S is a (degree 8) multi-section; S is unirational.

Unirationality in general case Lemma (Enriques criterion) A conic bundle S P 1 is unirational iff there is a multi-section s : P 1 S. We have P 1 = F S a fiber; Bertini involution τ B σ-involution; τ B (F ) S is a (degree 8) multi-section; S is unirational. Problems in the Special Case: F is not rational and even if we find a rational fiber F 0, then τ B (F 0 ) = F 0.

Multiplication maps (a side direction) E elliptic curve: we have m E : x m x. S with elliptic pencil: these glue to m S : S S. Compute: let F S be a smooth fiber. Then either: F consists of 6-torsion points or: m 2 (F ) or m 3 (F ) is a multi-section. Corollary: If S has a K-point that is not 6-torsion then S is unirational. Main problem: Can p be the only K-point?

Historic note: Secant map for cubic surfaces F K P 3 cubic surface over K, p 1, p 2 F : let l be the line through them, l F = {p 1, p 2, q}, we get φ : F F F given by φ(p 1, p 2 ) = q; descends to Sym 2 F F.

Historic note: Secant map for cubic surfaces F K P 3 cubic surface over K, p 1, p 2 F : let l be the line through them, l F = {p 1, p 2, q}, we get φ : F F F given by φ(p 1, p 2 ) = q; descends to Sym 2 F F. Working with conjugate point pairs gives the following: Proposition. Let L/K be a degree 2 field extension. Weil restriction gives a dominant rational map φ L/K : R L/K (F L ) F K. Corollary. If we have u : P 2 L F L then we get φ L/K (u) : P 4 K F K.

Reinterpreting the secant map F P 3 cubic surface over K, q = unique base point of K S ( p 1 p 2 ).

Reinterpreting the secant map F P 3 cubic surface over K, q = unique base point of K S ( p 1 p 2 ). Concrete geometric question: Find class of surfaces X such that for n 1 there are linear systems B n such that for general {p 1,..., p n } X B n ( p 1 p n ) has a unique base point.

Reinterpreting the secant map F P 3 cubic surface over K, q = unique base point of K S ( p 1 p 2 ). Concrete geometric question: Find class of surfaces X such that for n 1 there are linear systems B n such that for general {p 1,..., p n } X B n ( p 1 p n ) has a unique base point. Abstract question: For all n 1 find φ n : Sym n X X

Reinterpreting the secant map F P 3 cubic surface over K, q = unique base point of K S ( p 1 p 2 ). Concrete geometric question: Find class of surfaces X such that for n 1 there are linear systems B n such that for general {p 1,..., p n } X B n ( p 1 p n ) has a unique base point. Abstract question: For all n 1 find φ n : Sym n X X Meta Conjecture Only for conic bundles with δ = 7.

Multi-secant map for δ(s) = 7 S: conic bundle with δ(s) = 7 K S : canonical class, F : fiber class. Theorem For general {p 1,..., p n } S K S + nf ( 2p 1 2p n ) is a pencil with a unique base point.

Corollary S: conic bundle with δ(s) = 7. Then multi-secant map is defined on an open subset of S n combining with the Weil restriction gives φ : P 1290240 S. (exponent = 2n for n = 7! 2 7 )

More economical method? Step 1. Pick x, y K at random. Solve for t to get conjugate pair r 1, r 1 S.

More economical method? Step 1. Pick x, y K at random. Solve for t to get conjugate pair r 1, r 1 S. Step 2. q, q : projection of r 2 2r 1, r 2 2r 1 to A 2 xt

More economical method? Step 1. Pick x, y K at random. Solve for t to get conjugate pair r 1, r 1 S. Step 2. q, q : projection of r 2 2r 1, r 2 2r 1 to A 2 xt Step 3. Look for the unique curve C in 4K S + 7F ( p 8q 8q )

More economical method? Step 1. Pick x, y K at random. Solve for t to get conjugate pair r 1, r 1 S. Step 2. q, q : projection of r 2 2r 1, r 2 2r 1 to A 2 xt Step 3. Look for the unique curve C in 4K S + 7F ( p 8q 8q ) Birational preimage of the new F after elementary transformation at q, q plus Bertini involution.

there are polynomials f, g, F, G, H of degrees 8,6,17,17,26 C is given by x = f /g, t = F /G, y = H/(gG) multiplicity 8 at q, q, multiplicity 1 at p and ( H/(gG) ) 2 = a3 (F /G)(f /g) 3 + + a 0 (F /G).

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