On projective manifolds with semi-positive holomorphic sectional - - PowerPoint PPT Presentation

On projective manifolds with semi-positive holomorphic sectional curvature Shin-ichi Matsumura ( )

Tohoku University (mshinichi-math@tohoku.ac.jp, mshinichi0@gmail.com)

Taipei Conference on Complex Geometry

December, 2019

This talk is organized as follows: Section 1: Introduction Section 2: Main results Section 3: Strategy of proof Section 4: Details of proof Section 5: Related topics HSC:=holomorphic sectional curvature BSC:=holomorphic bisectional curvature Promise: all (bi)sectional curvature in this talk is holomorphic one (not real one).

This talk is organized as follows: Section 1: Introduction Section 2: Main results Section 3: Strategy of proof Section 4: Details of proof Section 5: Related topics HSC:=holomorphic sectional curvature BSC:=holomorphic bisectional curvature Promise: all (bi)sectional curvature in this talk is holomorphic one (not real one).

This talk is organized as follows: Section 1: Introduction Section 2: Main results Section 3: Strategy of proof Section 4: Details of proof Section 5: Related topics HSC:=holomorphic sectional curvature BSC:=holomorphic bisectional curvature Promise: all (bi)sectional curvature in this talk is holomorphic one (not real one).

This talk is organized as follows: Section 1: Introduction Section 2: Main results Section 3: Strategy of proof Section 4: Details of proof Section 5: Related topics HSC:=holomorphic sectional curvature BSC:=holomorphic bisectional curvature Promise: all (bi)sectional curvature in this talk is holomorphic one (not real one).

This talk is organized as follows: Section 1: Introduction Section 2: Main results Section 3: Strategy of proof Section 4: Details of proof Section 5: Related topics HSC:=holomorphic sectional curvature BSC:=holomorphic bisectional curvature Promise: all (bi)sectional curvature in this talk is holomorphic one (not real one).

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Section 1 Introduction

• (X, g) : compact K¨

ahler manifold of dimension n

• Rg: curvature tensor defined by

Rg(x, y, z, u) :=

• Θg(TX)(x, y)(z), u
• g ∈ C

for tangent vectors x, y, z, u ∈ TX.

• BSCg: defined by

BSCg(v, w) := Rg(v, v, w, w)

• |v|2

• HSCg: defined by

HSCg(v) := Rg(v, v, v, v)

• |v|4

Problem (Naive question) What is a relation between the geometry of X and positivity (or negativity) of BSCg.

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Frankel conjecture prove by Siu-Yau’80, Hartshorne conjecture proved by Mori’ 79) (1) If BSCg is positive (that is, BSCg(v, w) > 0 for any non-zero vectors v, w ∈ TX at any point of X), then X ∼ = Pn. (2) If TX is ample, then the same conclusion holds. Ampleness of TX is an algebraic counterpart of BSCg > 0. Theorem (Igusa ’55) TFAE. (1) ∃g: BSCg ≡ 0 (2) ∃g: Rg ≡ 0 (3) X is a torus up to finite ´ etale cover, namely, there exists a torus T s.t. T

− − − − − − → X

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Theorem (Howard-Smyth-Wu’81) If BSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. BSCgF ≥ 0 and RicgF >q 0. Moreover, we obtain Xuniv ∼ = F × Cm(biholo and isometric) Theorem (Mok’88) A manifold F (which corresponds to the fiber of φ : X → Y ) satisfying that BSCgF ≥ 0 and RicgF >q 0 is always a Hermitian symmetric space.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Problem (Main problem in this talk) (1) If HSCg > 0, then what can we say for X?? (HSC version of Mok’s result). (2) If HSCg ≥ 0, then can we obtain a structure thm?? (HSC version of Howard-Smyth-Wu’s result). Motivation:

• HSCg determines Rg (in particular BSCg),

namely, if HSCg = HSCh, then Rg = Rh.

• In this sense, HSCg is a more primitive object.
• However, a relation between positivity of BSC and HSC is

unknown and still mysterious.

• HSCg is closely related to the geometry of rational curves.
• For example, by Schwarz lemma generalized by Yau,

if HSCg < 0, then there is no rational curves in X.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Conjecture (Yau’s conjecture) If HSCg is positive, then X is (projective and) rationally connected. Definition (1) A curve R ⊂ X is called a rational curve if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R.

Definition (Reprint) (1) A curve R ⊂ X is called a rational curve (R-curve) if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R. Example (1) Any Fano manifolds (including Pn) are RC. (2) Any torus T has no rational curve. (3) X := P(E) → Y is always uniruled. (4) X := P(E) is not RC when Y is torus.

Definition (Reprint) (1) A curve R ⊂ X is called a rational curve (R-curve) if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R. Example (1) Any Fano manifolds (including Pn) are RC. (2) Any torus T has no rational curve. (3) X := P(E) → Y is always uniruled. (4) X := P(E) is not RC when Y is torus.

Definition (Reprint) (1) A curve R ⊂ X is called a rational curve (R-curve) if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R. Example (1) Any Fano manifolds (including Pn) are RC. (2) Any torus T has no rational curve. (3) X := P(E) → Y is always uniruled. (4) X := P(E) is not RC when Y is torus.

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Definition (Reprint) (1) A curve R ⊂ X is called a rational curve (R-curve) if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R. Example (1) Any Fano manifolds (including Pn) are RC. (2) Any torus T has no rational curve. (3) X := P(E) → Y is always uniruled. (4) X := P(E) is not RC when Y is torus.

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Definition (Reprint) (1) A curve R ⊂ X is called a rational curve (R-curve) if ∃f : P1 → X s.t. R = f (P1). (2) X is called rationally connected (RC) if ∀x, y ∈ X, ∃R ⊂ X s.t. x, y ∈ R. (3) X is called uniruled if ∀x ∈ X, ∃R ⊂ X s.t. x ∈ R. Example (1) Any Fano manifolds (including Pn) are RC. (2) Any torus T has no rational curve. (3) X := P(E) → Y is always uniruled. (4) X := P(E) is not RC when Y is torus.

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Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Conjecture (Reprint, Yau’s conjecture) If HSCg is positive, then X is projective and RC. Known Reults:

• Heier-Wong’15 solved Yau’s conjecture when X is projective.
• Yang’18 solved Yau’s conjecture in a general situation.

However

• Yang’s method is based on the maximal principle and it can not

be applied to the quasi-positive case.

• It is important to consider the quasi-positive case of Yau’s

conjecture from the viewpoint of a structure thm. Theorem (Thm A) If X is projective and satisfies P(X, g) = n (which is satisfied in the quasi-positive case), then X is RC.

• P(X, g) is an invariant defined by an analogue of the numerical

Kodaira dimension.

• The proof depends on Heier-Wong’s idea and RC positivity

(partial positivity) introduced by Yang.

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Theorem (Thm B) If X is projective and HSCg is non-negative, then ∃φ : X → Y satisfying that: (1) φ is locally trivial (all the fibers F are non-singular and biholo). (2) ∃ K¨ ahler metric gY on Y : RgY ≡ 0. (3) ∃ K¨ ahler metric gF on F s.t. HSCgF ≥ 0 and F is RC. Moreover, we obtain Xuniv ∼ = F × Cm (biholo and isometric)

• There are two bad points in Thm B, compared to

Howard-Smyth-Wu’s structure theorem.

• One is the assumption of X being projective.
• The other is the lack of “quasi-positivity” of HSCgF .

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

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Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

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Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

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Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

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Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

Step 1: MRC fibration

• ∃φ : X Y : MRC fibration (Campana, Koll´

ar-Miyaoka-Mori): (1) φ is a rational map whose general fiber is compact. (2) A general fiber F is RC. (3) ∄ horizontal rational curve at “general points”.

• For example, X := P(E) → Y is always an RC fibration, but it is

not necessarily an MRC fibration (consider Y = Pm).

• X := P(E) → Y is MRC when Y has no rational curve.

Remark (1) MRC fibrations are not uniquely determined, but they are unique up to birational models of Y . (2) X is RC iff dim Y = 0. (3) X is not uniruled iff dim Y = dim X. (4) If MRC is non-trivial (that is, 0 < dim Y < dim X), then KY is always pseudo-effective (psef), namely, KY has a singular metric h such that Θh(KX) ≥ 0 (by the result of Graber-Harris-Starr).

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

• If HSCg ≡ 0, then X is torus up to ´

etale covers. We may assume: HSCg ≥ 0 and HSCg ̸≡ 0 Claim Then X is uniruled. In particular, KY is psef. Theorem (BDPP) X is not uniruled if and only if KX is psef. Proof. Assume KX is psef. Then 0 ≤

• X

c1(KX) ∧ ωn−1 = − 1 πn

• X

Scalg ωn. On the the hand, we have Scalg(p) =

• v∈P(TX,p)

HSC(v) dVFS. Hence Scalg is a quasi-positive function on X.

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 2: Splitting of TX

• Assume that MRC φ : X → Y is holo in Step 2 and Step 3.

In Step 2, we will prove the followings:

• φ : X → Y is smooth (that is, any fiber F is non-singular). In

particular, we have the exact sequence of the tangent bundles: 0 → TX/Y →TX → φ∗TY → 0. g gQ

• The sequence above admits the holo and orthogonal

decomposition: Namely, a bundle morphism ∃j : φ∗TY → TX satisfying that TX = TX/Y ⊕ j(φ∗TY ), TX/Y and j(φ∗TY ) is orthogonal w.r.t. g.

• a metric ∃gY on Y s.t. gQ = φ∗gY and RgY ≡ 0 .

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 3: Effect of the integrability of j(φ∗TY )

Check that the subbundle j(φ∗TY ) ⊂ TX is integrable. Then, by Ehresmann’s theorem, we obtain (E1) and (E2): (E1) X

• Xuniv ∼

= Cm × F

• p
• q

Y Yuniv ∼ = Cm

• (E2) The pull-back of the splitting TX = TX/Y ⊕ j(φ∗TY ) to

Xuniv coincides with the natural decomposition TXuniv = q∗TF ⊕ p∗TYuniv on Xuniv.

• Then all the conclusions in Thm B can be obtained.
• It remains to prove that MRC fibration can be chosen to be holo.

For this purpose, we use H¨

• ring’s result.

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Xo

Cs

X

• open ?

¢

Kkk

hol.hr#

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Step 4: Existence of holo MRC fibrations

Proposition (H¨

• ring ’07)

Let V ⊂ TX be a foliation whose general leaf is RC. Then V comes from the fibration. Namely, ∃φ : X → Y s.t. V = TX/Y .

• ∃X0 ⊂ X s.t. φ|X0 is holo and codim(X \ X0) ≥ 2.
• By the same way as in Step 2, we prove that φ is smooth on X0

and we show the following decomposition holds TX = TX/Y ⊕ j(φ∗TY ) on X0 (not X).

• By taking i∗ under the inclusion i : X0 → X, we obtain the

decomposition on the ambient space X TX = i∗(TX/Y ) ⊕ i∗(φ∗TY ) on X.

• Then we show that i∗(TX/Y ) is actually locally free, and it is a

foliation whose general leaf is RC.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

Claim (Key point) For simplicity, we assume that dim X = 2, dim Y = 1, and MRC φ : X → Y is smooth. Then we can prove that c1(KY ) = 0 and that TX splits.

• Our first goal: c1(KY ) = 0 (equivalently c1(φ∗TY ) = 0).
• For this purpose, we study the curvature ΘgQ of φ∗TY ,

where gQ is the quotient metric defined by 0 → (TX/Y , gS) →(TX, g) → (φ∗TY , gQ) → 0.

• As KY = T ∗

0 ≥

• X

ΘgQ ∧ ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV . Here {eh, ev} is o.n.b at a point in X s.t. eh (resp. ev) is a horizontal (resp. vertical) vector.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• Our goal:
• X ΘgQ(eh) ≥ 0 and
• X ΘgQ(ev) ≥ 0.
• By the inequality of Gauss-Codazzi type, we have

ΘgQ(v) = ⟨ΘgQ(v, v)eh, eh⟩gQ ≥ ⟨Θg(v, v)eh, eh⟩g.

• Hence we obtain ΘgQ(eh) ≥ 0 by applying it to v := eh.
• Let ωY be a K¨

ahler form on Y and t be a coord on Y . Then:

• X

ΘgQ(ev) dV =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ(ev) ω =

• y∈Y

ωY

• Xy

1 |φ∗dt|2

ΘgQ.

• By the definition ΘgQ = √−1∂∂ log |φ∗dt|2

can be expressed as

• Xy

1 |φ∗dt|2

ΘgQ =

• Xy

1 |φ∗dt|6

√ −1∂|φ∗dt|2 ∧ ∂|φ∗dt|2 ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

• In summary, we have

0 =

• X

c1(φ∗TY )∧ω =

• X

ΘgQ∧ω =

• X

ΘgQ(eh) dV +

• X

ΘgQ(ev) dV .

• Hence we obtain c1(φ∗TY ) = 0 (and also c1(TY ) = 0).
• The next our goal: ΘgQ = 0 (equivalently, ΘgQ is non-negative).
• Then HSCg(eh) = 0 by

0 =

• X

ΘgQ(eh) ≥

• X

HSCg(eh) ≥ 0.

• In particular, the horizontal vector eh is the minimizer of HSCg.

Proposition (Yang’18, Brendle’20, Brunebarbe–Klingler–Totaro’13) If a vector v ∈ TX is the minimizer of HSCg on TX,p, then we have BSCg(v, w) ≥ 0 for any w ∈ TX,p.

• Hence ΘgQ is non-negative (and thus ΘgQ = 0) by the inequality
• f Gauss-Codazzi type,

ΘgQ(w) = ⟨ΘgQ(w, w)eh, eh⟩gQ ≥ ⟨Θg(w, w)eh, eh⟩g ≥ 0.

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

RC-positivity and classification of surfaces

Definition (Yang’18) A hermitian vector bundle (E, h) is called RC-positive if ∀e ∈ E ∃v ∈ TX s.t. Rh(e, e, v, v) > 0. Theorem (BDPP, DPS’94, Yang’17, M-’13.) X is uniruled iff −KX := det TX is RC-positive. Conjecture (Yang’18) X is RC iff TX is RC-positive. Problem Can we classify all the surfaces with HSCg ≥ 0.

• By Thm B, X is (a) 2-dim torus, (b) ruled surface over 1-dim

torus, or (c) RC.

• The case (b) with HSCg ≥ 0 actually has BSCg ≥ 0.
• Hirzebruch surface Fn := P(O ⊕ O(n)) has HSCg > 0 (by

Hitchin’73).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

HSC and foliations

• From a fibration φ : X → Y , we constructed the subbundle

φ∗TY ⊂ TX. Also all the vectors v ∈ φ∗TY are truly flat, namely, Rg(v, x, y, z) = 0 for any vectors x, y, z ∈ TX

• Conversely, under the assumption of HSCg ≥ 0, can we construct

a fibration from a foliation: (a) Can we construct a “subbundle” F ⊂ TX consisting of “appropriate” truly flat vectors?? (b) Can we obtain the splitting TX = F ⊕ G?? (c) Can we prove G comes from fibration??

• The above strategy seems to remove the bad points in Thm B.
• Also it is interesting to apply the above strategy to HSCg ≤ 0.
• When HSCg ≤ 0, KX is nef. Hence it should be semi-ample if

the abundance conjecture is true.

• If the above strategy works, we may construct a fibration

φ : X → Y s.t. F = TX/Y that coincides with semi-ample fibration of KX. (cf. Heier-Lu-Wong-Fangyang).

Thank you for your attention!!