The Noether Inequality for Algebraic Threefolds

.... Meng Chen School of Mathematical Sciences, Fudan University 2018.09.08

P1. Surface geography Back ground Two famous inequalities c 2 1 Miyaoka-Yau inequality Noether inequality O χ(o)

P1. Surface geography Back ground Two famous inequalities c 2 1 Miyaoka-Yau inequality Noether inequality O χ(o) Further classification the geometry of M c 2 1,c 2.

Threefold geography P2. The strategy for threefold geography X: a minimal 3-fold of general type

Threefold geography P2. The strategy for threefold geography X: a minimal 3-fold of general type Main birational invariants: K 3 X, χ(o X), q(x) = h 1 (O X ), q 2 = h 2 (O X ), p g (X) = h 3 (O X ).

Threefold geography P2. The strategy for threefold geography X: a minimal 3-fold of general type Main birational invariants: K 3 X, χ(o X), q(x) = h 1 (O X ), q 2 = h 2 (O X ), p g (X) = h 3 (O X ). Main task: find optimal relations among above birational invariants.

Threefold geography P3. The possible Noether type inequality The 3-dimensional analogy of Miyaoka-Yau inequality: Neither K 3 X?χ(O X) is possible, as < χ(o X ) < +, nor is K 3 X?p g(x) possible.

Threefold geography P3. The possible Noether type inequality The 3-dimensional analogy of Miyaoka-Yau inequality: Neither K 3 X?χ(O X) is possible, as < χ(o X ) < +, nor is K 3 X?p g(x) possible. Seek for the Noether type inequality: K 3 X a p g(x) b, a, b Q >0. K 3 X The Noether inequality O p g (X)

Threefold geography P4. The simplest case: Gorenstein minimal 3-folds X minimal, the Cartier index r X 1. X is Gorenstein r X = 1 {smooth minimal 3-folds} {Grenstein minimal 3-folds} {arbitrary minimal 3-folds}

Threefold geography P5. The Noether inequality for Gorenstein minimal 3-folds Kobayashi (1992): an infinite series of examples of canonically polarized 3-folds satisfying K 3 X = 4 3 p g(x) 10 3.

Threefold geography P5. The Noether inequality for Gorenstein minimal 3-folds Kobayashi (1992): an infinite series of examples of canonically polarized 3-folds satisfying K 3 X = 4 3 p g(x) 10 3. M. Chen (2004): K 3 X 4 3 p g(x) 10 3 polarized 3-folds. for canonically Catanese-Chen-Zhang (2006): K 3 X 4 3 p g(x) 10 3 for smooth minimal 3-folds of general type.

The main statement P6. General case: non-gorenstein minimal 3-folds May always assume p g (X) 2, since K 3 X > 0. So φ K X is non-trivial.

The main statement P6. General case: non-gorenstein minimal 3-folds May always assume p g (X) 2, since K 3 X > 0. So φ K X is non-trivial. Set up for φ 1 = φ KX. X Γ g π X ----------- f φ KX s Σ

P7. Notations The main statement K X = M + Z, S a generic irreducible element of M.

P7. Notations The main statement K X = M + Z, S a generic irreducible element of M. The canonical dimension d X = dim(γ).

P7. Notations The main statement K X = M + Z, S a generic irreducible element of M. The canonical dimension d X = dim(γ). π (K X ) M + E, E an effective Q-divisor.

P7. Notations The main statement K X = M + Z, S a generic irreducible element of M. The canonical dimension d X = dim(γ). π (K X ) M + E, E an effective Q-divisor. When d X 2, S M, K 3 X (π (K X ) S M S ) + (π (K X ) S E S ).

P8. The main statement The main statement. Theorem.. Let X be a minimal projective 3-fold of general type. Assume that one of the following holds: d X 2; or d X = 1 and K X is not composed with a rational pencil of (1, 2)-surfaces; or d X = 1, K X is composed with a rational pencil of (1, 2)-surfaces, and K X is weakly free. Then the inequality K 3 X 4 3 p g(x) 10 3 holds....

The main statement P9. Preparation: surfaces admitting a genus 2 fibration. Proposition.. Let S be a smooth projective surface of general type and T a smooth complete curve. Suppose that f : S T is a fibration of which the general fiber C is of genus 2. Assume that p g (S) 3 and K S nc + G for some effective integral divisor G on S and a positive integer n. Then... vol(s) 8 (n 1). 3

The main statement P10. Preparation: Kawamata extension theorem. Theorem.. Let V be a smooth variety and D a smooth divisor on V. Assume that K V + D Q A + B where A is an ample Q-divisor and B is an effective Q-divisor such that D Supp(B). Then the natural homomorphism H 0 (V, O V (m(k V + D))) H 0 (D, O D (mk D )). is surjective for all m 2...

The main statement P11. Preparation: comparison result. Corollary.. Let X be a minimal projective 3-fold of general type and D a semi-ample Weil divisor on X. Let π : W X be a resolution and S, a semi-ample divisor on W, is assumed to be a smooth surface of general type. Assume that λπ D S is Q-effective for some positive rational number λ. Then π (K X + λd) S σ K S0 Q-effective on S, where σ : S S 0 is the contraction onto the minimal. model S 0... is

The main statement P12. (Preparation): weak positivity Let X be a smooth projective variety and F a torsion-free coherent sheaf on X. We say that F is weakly positive on X if there exists some Zariski open subvariety U X such that for every ample invertible sheaf H and every positive integer α, there exists some positive integer β such that (S αβ F) H β is generated by global sections over U, which means that the natural map H 0 (X, (S αβ F) H β ) O X (S αβ F) H β is surjective over U. Here (S k F) denotes the reflexive hull of the symmetric product S k F.

The main statement P13. (Preparation): Viehweg-Campana weak positivity. Theorem.. Let g : Y Z be a surjective morphism between smooth projective varieties and D a reduced divisor on Y with simple normal crossing support. Then g O Y (m(k Y/Z + D)) is torsion free and weakly positive. for every integer m > 0...

The main statement P14. (Preparation): log canonical threshold The log canonical threshold of D with respect to (X, B) is defined by lct(x, B; D) = sup{t 0 (X, B + td) is lc}.

The main statement P14. (Preparation): log canonical threshold The log canonical threshold of D with respect to (X, B) is defined by lct(x, B; D) = sup{t 0 (X, B + td) is lc}. Definition of global log canonical threshold ( glct ) : Let Y be a normal projective variety with at worst klt singularities such that K Y is nef and big. The glct of Y is defined as the following: glct(y) = inf{lct(y; D) 0 D Q K Y } = sup{t 0 (Y, td) is lc for all 0 D Q K Y }.

P15. The glct of (1,2)-surfaces The main statement (Chen-Chen-Jiang) Let S be a minimal (1, 2)-surface (i.e. K 2 S = 1, p g(s) = 2). Then glct(s) 1 13.

P15. The glct of (1,2)-surfaces The main statement (Chen-Chen-Jiang) Let S be a minimal (1, 2)-surface (i.e. K 2 S = 1, p g(s) = 2). Then glct(s) 1 13. (János Kollár) glct(1, 2) 1 10. (optimal)

P15. The glct of (1,2)-surfaces The main statement (Chen-Chen-Jiang) Let S be a minimal (1, 2)-surface (i.e. K 2 S = 1, p g(s) = 2). Then glct(s) 1 13. (János Kollár) glct(1, 2) 1 10. (optimal) Example of Kollár: Consider the pair S := ( x 7 y 3 +y 10 +z 5 +t 2 = 0 ) P(1, 1, 2, 5) and := (y = 0). It is easy to check that S has a unique singular point, at (1:0:0:0), and it has type E 8. Thus S is a projective surface with Du Val singularities, K S = O S (1) is ample, K 2 S = 1 and h 0( S, O S (2K S ) ) = 4. Furthermore, lct(s, ) = 1 10.

The main statement P16. A pencil of surfaces on a 3-fold over a curve. Proposition.. Assume that there exists a resolution π : W X such that W admits a fibration structure f : W Γ onto a smooth curve Γ. Denote by F a general fiber of f and F 0 the minimal model of F. Assume that..1 there exists a π-exceptional prime divisor E 0 on W such that (π (K X ) F E 0 F ) > 0, and..2 π (K X ) Q bf + D for some rational number b > 0 and an effective Q-divisor D on W. Then b. 2.. glct(f 0 ). (Application of Kollár s connectness lemma )

P17. When is K X weakly free! The main statement. Corollary.. Let X be a minimal projective 3-fold of general type such that K X is composed with a pencil of (1, 2)-surfaces. Assume that one of the following holds:..1 K X is composed with an irrational pencil; or..2 K X is composed with a rational pencil and p g (X) 21; or..3 X is Gorenstein. Then there exists a minimal projective 3-fold Y, being birational to X,. such that Mov K Y is base point free...

P18. Proof: The case d X 2 Main steps of the proof d X = 3, Kobayashi K 3 X 2p g(x) 6.

P18. Proof: The case d X 2 Main steps of the proof d X = 3, Kobayashi K 3 X 2p g(x) 6. d X = 2, f : X Γ, C the general fiber of f. d X = 2 and g(c) 3: K 3 X (π K X S S S ) = a(π K X S C) (p g (X) 2)(π K X S C). The comparison inequality implies (π K X S C) 2. Hence K 3 X 2p g(x) 4.

Main steps of the proof P19. Proof:The case d X = 2 and g(c) = 2 K S nc +, n 2(p g (X) 2), S admits a pencil of genus 2. Hence vol(s) 8 (n 1). 3

Main steps of the proof P19. Proof:The case d X = 2 and g(c) = 2 K S nc +, n 2(p g (X) 2), S admits a pencil of genus 2. Hence vol(s) 8 (n 1). 3 The extension theorem implies K 3 X 1 4 vol(s) 4 3 p g(x) 10 3.

P20. Proof: The case d X = 1 Main steps of the proof f : X Γ is the induced fibration. The surface theory implies: (1) F is a (1, 1)-surface; (2) F is a (1, 2)-surface; (3) F is of other types, i.e. K 2 2 and p g (F) > 0. When F is of Type (3): K 2 F 0 2, the comparison inequality implies K 3 X a(π (K X ) F ) 2 where a p g (X) 1. a 3 (a + 1) 2 K2 F 0 2a3 (a + 1) 2 > 2a 4 2p g(x) 6

Main steps of the proof P21. Proof: The case d X = 1 and F a (1, 1)-surface The main idea is to study the bicanonical restriction: dim V 2 P 2 (F) = 3. H 0 (2K X ) ν 2 V 2 H 0 (F, 2K F ).

Main steps of the proof P21. Proof: The case d X = 1 and F a (1, 1)-surface The main idea is to study the bicanonical restriction: dim V 2 P 2 (F) = 3. H 0 (2K X ) ν 2 V 2 H 0 (F, 2K F ). dim V 2 = 3 φ 2 is generically finite of degree 4 K 3 X 2p g(x) 4. dim V 2 = 2 dim φ 2 (X) = 2. 4K 3 X (π (K X ) S 2 S 2 ) = a 2 (π (K X ) S2 C ) 2a a + 1 (P 2(X) 2) 2a a + 1 ( 1 2 K3 X + 3p g(x) 5).

Main steps of the proof P22. Proof: The case d X = 1 and F a (1, 1)-surface Thus it follows that K 3 X 6a 3a + 4 p g(x) 10a 3a + 4 = 2p g (X) 6 + 8a 8p g(x) + 24 3a + 4 > 2p g (X) 6.

Main steps of the proof P23. Proof: The case d X = 1 and F a (1, 1)-surface dim V 2 = 1, φ 2 and φ 1 induce the same fibration. One has 2K 3 X a 2(π (K X ) F ) 2 a 3 2 (a 2 + 2) 2 K2 F 0 = a 2 4 + 12a 2 + 16 (a 2 + 2) 2 > P 2 (X) 5 1 2 K3 X + 3p g(x) 8, which gives K 3 X > 2p g(x) 16 3.

Main steps of the proof P24. Statement for the case: d X = 1 and F a (1, 2)-surface. Theorem.. Let X be a minimal projective 3-fold of general type with p g (X) 4. Assume that d X = 1 and K X is composed with a pencil of (1, 2)-surfaces. Moreover, assume that Mov K X is base point free. Then K 3 X. 4 3 p g(x) 10 3.... Corollary.. When d X = 1, F is a (1, 2)-surface and p g (X) 21, then... K 3 X 4 3 p g(x) 10 3.

. Main steps of the proof P25. Proof: The case d X = 1 and F a (1, 2)-surface Step 0. Overall setting. 88888888888888 W Σ f γ π 8 Φ f KX/Z Γ X Z β s

Main steps of the proof P26. Proof: The case d X = 1 and F a (1, 2)-surface Step 1. Let D = K X/Z and S Mov π (K D ) be a general member. We have S S ac where a h 0 (D) 2 p g (X) + 1 and C is a general fiber of the restricted fibration β S : S β(s) with p g (C) = 2. 4K 3 X + 16 = (π (K X + D) 2 π (D)) (π (K X + D) 2 S) (σ K S0 ) 2 8 (2a 3). 3 Condition: a p g (X) + 2 K 3 X 4 3 p g(x) 10 3.

. Main steps of the proof P27. Proof: The case d X = 1 and F a (1, 2)-surface Step 2. a = p g (X) + 1 Z is normal. Sub-step 2.1. We claim that Z = P(1, 1, a). Sub-step 2.2. Existence of a special resolution W, which factors through F a. W g F a π X f f P 1

Main steps of the proof P28. Proof: The case d X = 1 and F a (1, 2)-surface Sub-step 2.2. Continued... Denote g (σ 0 ) = B which is an effective Cartier divisor on W. We can write π K X + 2F S + E af + B + E for an effective Q-divisor E and K W = π K X + E π for an effective π-exceptional Q-divisor E π.

Main steps of the proof P29. Proof: The case d X = 1 and F a (1, 2)-surface Sub-step 2.3. Two distinguished components in B and E. The following statements hold:..1. there exists a unique π-exceptional prime divisor E 0 on W such that E 0 dominates F a. Moreover, (E 0 C) = 1 and coeff E0 E = coeff E0 E π = 1, where C is a general fiber of g;..2. there exists a unique prime divisor D 0 in B such that (D 0 E 0 F) = 1, coeff D0 B = 1, and (π (K X ) (B D 0 ) F) = 0.

. Main steps of the proof P30. Proof: The case d X = 1 and F a (1, 2)-surface Sub-step 2.4. Pseudo-effectivity of (3π K X (a 6)F) D0. By Viehweg, there is a resolutions ψ : Σ F a and a resolution W of W Fa Σ giving the commutative diagram: W g Σ π ψ W g F a such that every g -exceptional divisor is π -exceptional. We may assume that E 0 is smooth by taking further modification, where E 0 is the strict transform of E 0 on W.

Main steps of the proof P31. Proof: The case d X = 1 and F a (1, 2)-surface. Claim.. For any integer m > 0, there exists an integer c > 0 and an effective divisor D m cmk W/Fa + cme 0 + cg A such. that E 0 Supp(D m ).... Corollary...((3π K X (a 6)F) D 0 π K X ) 0...

Main steps of the proof P32. Proof: The case d X = 1 and F a (1, 2)-surface Sub-step 2.5. The main inequality for Step 2. ((3π K X (a 6)F) B π K X ) 0. Since (F B π K X ) = (F S π K X ) = (C π K X F ) = 1, hence Finally, we have (π K 2 X B) a 6 3 (F B π K X ) a 3 2. K 3 X = (π K 2 X (π K X + 2F)) 2 (π K 2 X (af + B)) 2 4a 3 4 = 4 3 p g(x) 8 3.

Main steps of the proof P33. Summary and open question Let X be a minimal 3-fold of general type. Then 1 1680, p g(x) = 0; 1 75, p g(x) = 1; 1 3, p g(x) = 2; K 3 X 1, p g (X) = 3; 2, p g (X) = 4; 4 3 p g 14 3, 5 p g(x) 20; 4 3 p g 10 3, p g(x) 21. Conjecture. K 3 X 4 3 p g(x) 10 3 holds.

P34 Main steps of the proof.... Thank you very much!