Introduction
Let X X X X X X X X X Q ⊂ X Q \subset X Q ⊂ X meromorphically convex hull is
Q ^ M ( X ) = { x ∈ X : every complex hypersurface of X through x meets Q } , \widehat{Q}_{M}(X) = \{x \in X : \text{every complex hypersurface of } X \text{ through } x \text{ meets } Q\}, Q M ( X ) = { x ∈ X : every complex hypersurface of X through x meets Q } , and Q Q Q meromorphically (or hypersurface ) convex if Q ^ M ( X ) = Q \widehat{Q}_{M}(X) = Q Q M ( X ) = Q [3 Prop. 1.2] , this is precisely the hull of Q Q Q X X X
In [2 ] it was proved that a sublevel set K = { h ≤ 0 } K = \{h \le0\} K = { h ≤ 0 } h h h hypersurface ) convex on a Stein manifold X X X d d c h dd^{c}h d d c h X X X X X X Ω \Omega Ω [ Ω ] ∈ H 2 ( X , R ) [\Omega] \in H^{2}(X,\mathbb{R}) [ Ω ] ∈ H 2 ( X , R ) H 2 ( X , Z ) → H 2 ( X , R ) H^{2}(X,\mathbb{Z}) \to H^{2}(X,\mathbb{R}) H 2 ( X , Z ) → H 2 ( X , R ) h h h U ⊃ K U \supset K U ⊃ K d d c h dd^{c}h d d c h ω \omega ω X X X ω \omega ω X X X ω = d d c h \omega= dd^{c}h ω = d d c h K K K [5 Criterion 1.6] stated, without proof, that such convexity of K K K [5 ] ) is equivalent to the extension of d d c h dd^{c}h d d c h h h h K K K X X X
Theorem 1. Let X X X
H 2 ( X , Z ) ≅ F ⊕ T , (1) H_{2}(X,\mathbb{Z}) \cong F \oplus T, \tag*{(1)} H 2 ( X , Z ) ≅ F ⊕ T , (1) where F F F T T T U ⊂ X U \subset X U ⊂ X h : U → R h : U \to\mathbb{R} h : U → R K = { h ≤ 0 } K = \{h \le0\} K = { h ≤ 0 } U U U d d c h dd^{c}h d d c h ω \omega ω X X X N N N d d c ( N h ) dd^{c}(Nh) d d c ( N h ) Ω H \Omega_{H} Ω H X X X K K K X X X
Note that the decomposition (1) always holds if H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) § 2 for details. Further, by [1 ] , a Stein manifold of dimension n n n n n n dim C X = 2 \dim_{\mathbb{C}} X = 2 dim C X = 2 H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) § 4.3 shows that (1) is not required if the image of H 2 H_{2} H 2 K ^ O ( X ) \widehat{K}_{\mathcal{O}(X)} K O ( X ) K K K H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) d d c h dd^{c}h d d c h
But in general, as the next theorem shows, the topological hypothesis (1) in Theorem 1 cannot be dropped. We construct a Stein manifold X X X H 2 ( X , Z ) ≅ Q H_{2}(X,\mathbb{Z}) \cong\mathbb{Q} H 2 ( X , Z ) ≅ Q H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) F ⊕ T F \oplus T F ⊕ T Im H 2 ( X , Z ) → H 2 ( X , R ) = 0 \operatorname{Im} H_{2}(X,\mathbb{Z}) \to H_{2}(X,\mathbb{R}) = 0 Im H 2 ( X , Z ) → H 2 ( X , R ) = 0 X X X Lemma 8 ). Together with K K K Λ ⊂ int K \Lambda\subset\operatorname{int} K Λ ⊂ int K X X X X X X
Theorem 2. There exist a Stein manifold X X X H 2 ( X , Z ) ≅ Q H_{2}(X,\mathbb{Z}) \cong\mathbb{Q} H 2 ( X , Z ) ≅ Q U ⊂ X U \subset X U ⊂ X h : U → R h: U \to\mathbb{R} h : U → R K = { h ≤ 0 } K = \{h \leq0\} K = { h ≤ 0 } U U U Λ ⊂ int K \Lambda\subset\operatorname{int} K Λ ⊂ int K
(i) d d c h dd^{c}h d d c h ω \omega ω X X X
(ii) The torus Λ \Lambda Λ ω \omega ω ω ∣ Λ = 0 \omega|_{\Lambda} = 0 ω ∣ Λ = 0
(iii) For every λ > 0 \lambda> 0 λ > 0 Ω \Omega Ω X X X d d c ( λ h ) dd^{c}(\lambda h) d d c ( λh ) [ Ω ] ≠ 0 [\Omega] \neq0 [ Ω ] = 0 H 2 ( X , R ) H^{2}(X,\mathbb{R}) H 2 ( X , R ) N N N d d c ( N h ) dd^{c}(Nh) d d c ( N h ) X X X
(iv) The meromorphically convex hull Λ ^ M ( X ) \widehat{\Lambda}_{M(X)} Λ M ( X ) A A A ∂ A ⊂ Λ \partial A \subset\Lambda ∂ A ⊂ Λ A ⊄ K A \not\subset K A ⊂ K Λ \Lambda Λ K K K
The theorem shows that the equivalence between meromorphic convexity of K K K d d c h dd^{c}h d d c h [5 Criterion 1.6] , does not hold on arbitrary Stein manifolds.
Concerning (ii) and (iv): a classical result of Duval and Sibony [7 ] states that a smooth compact totally real submanifold of C n \mathbb{C}^{n} C n C n \mathbb{C}^{n} C n C n \mathbb{C}^{n} C n [9 ] extended this to Stein manifolds: a smooth compact totally real submanifold of a Stein manifold X X X X X X [3 Theorem 3.2] it was shown that isotropy with respect to an exact Kähler form d d c φ dd^{c}\varphi d d c φ C n \mathbb{C}^{n} C n [7 ] . By (ii) and (iv), the torus Λ \Lambda Λ [5 Criterion 1.8(a)] . By Guedj’s theorem, Λ \Lambda Λ X X X Λ \Lambda Λ X X X
The construction in Theorem 2 uses the following ingredients (see Figure 1 in § 5 ): a Stein manifold Y Y Y
Figure 1. Schematic picture of Λ ⊂ X \Lambda\subset X Λ ⊂ X arg w \arg w arg w Λ \Lambda Λ Γ \Gamma Γ h 0 h_0 h 0 π − 1 ( 0 ) \pi^{-1}(0) π − 1 ( 0 ) C 0 , C 1 C_0,C_1 C 0 , C 1 A A A
Hom ( H 1 ( Y , Z ) , Z ) = Hom ( H 2 ( Y , Z ) , Z ) = 0 and H 1 ( Y , Z ) ⊗ Q ≠ 0. \operatorname{Hom}(H_{1}(Y,\mathbb{Z}),\mathbb{Z}) = \operatorname{Hom}(H_{2}(Y,\mathbb{Z}),\mathbb{Z}) = 0 \quad\text{and}\quad H_{1}(Y,\mathbb{Z}) \otimes\mathbb{Q} \neq0. Hom ( H 1 ( Y , Z ) , Z ) = Hom ( H 2 ( Y , Z ) , Z ) = 0 and H 1 ( Y , Z ) ⊗ Q = 0. Such a manifold Y H Y_{H} Y H [12 Prop. 2] ; it is the complement in the unit bidisc of a sequence of smooth analytic curves (see also [6 Example 1] ). Our construction is motivated by this example but is done differently: our Y Y Y C 5 \mathbb{C}^{5} C 5 R 5 \mathbb{R}^{5} R 5 Y ∩ R 5 Y \cap\mathbb{R}^{5} Y ∩ R 5 Γ \Gamma Γ H 1 ( Y , Q ) H_{1}(Y,\mathbb{Q}) H 1 ( Y , Q ) Λ ⊂ X = Y × C ∗ \Lambda\subset X = Y \times\mathbb{C}^{*} Λ ⊂ X = Y × C ∗ Γ \Gamma Γ C ∗ \mathbb{C}^{*} C ∗ Γ \Gamma Γ Γ ⊂ R 5 \Gamma\subset\mathbb{R}^{5} Γ ⊂ R 5 Λ \Lambda Λ
The same construction can be carried out in Hulek’s domain Y H Y_H Y H X = Y H × C ∗ X = Y_H \times\mathbb{C}^{*} X = Y H × C ∗ dim C X = 3 \dim_{\mathbb{C}} X = 3 dim C X = 3 Y H ⊂ C 2 Y_H \subset\mathbb{C}^{2} Y H ⊂ C 2 Γ ⊂ Y H \Gamma\subset Y_H Γ ⊂ Y H O ( Y H ) \mathcal{O}(Y_H) O ( Y H ) Y H Y_H Y H Γ \Gamma Γ Γ \Gamma Γ O ( Y H ) \mathcal{O}(Y_H) O ( Y H ) Γ \Gamma Γ Γ \Gamma Γ
Acknowledgments. The author obtained the proof of Theorem 1 in the fall of 2025, but all attempts to remove assumption (1) were unsuccessful. The counterexample of Theorem 2 was eventually found with the help of AI, specifically Claude Fable 5.1, which proposed the original construction; the construction presented here has been substantially simplified. AI was also used to assist with editing the exposition. The author bears full responsibility for the correctness of all proofs and for the content of this paper. The author thanks Purvi Gupta for many valuable discussions related to the problem, and in particular for identifying H 2 ( X , Z ) ≅ Q H_2(X,\mathbb{Z}) \cong\mathbb{Q} H 2 ( X , Z ) ≅ Q
Background: periods, Hodge classes, Levi form
Sections 2.1 and 2.2 collect in one place the facts relating homology and cohomology that are used throughout the paper: the universal coefficient theorem, periods of closed forms, integral (Hodge) classes, and the structure of H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z ) [10 Section 3.1] for the universal coefficient theorem and the properties of Ext, and in [13 Chapter 18] for the de Rham theorem in terms of integration over smooth singular chains. We include short proofs to fix the notation. Sec. 2.3 discusses the Levi form with some estimates that will be used in the proof of the main results.
In the next two subsections X X X
Periods and the universal coefficient theorem
If β \beta β k k k X X X C C C k k k ∫ C β \int_C \beta ∫ C β [ β ] [\beta] [ β ] [ C ] ∈ H k ( X , Z ) [C] \in H_k(X,\mathbb{Z}) [ C ] ∈ H k ( X , Z ) c ∈ H k ( X , R ) c \in H^k(X,\mathbb{R}) c ∈ H k ( X , R ) H k ( X , C ) H^k(X,\mathbb{C}) H k ( X , C ) ⟨ c , ⋅ ⟩ : H k ( X , Z ) → R \langle c,\cdot\rangle: H_k(X,\mathbb{Z}) \to\mathbb{R} ⟨ c , ⋅ ⟩ : H k ( X , Z ) → R C \mathbb{C} C period homomorphism of $c. It vanishes on torsion classes, because R \mathbb{R} R C \mathbb{C} C
For every abelian group A A A
0 → Ext ( H k − 1 ( X , Z ) , A ) → H k ( X , A ) → Hom ( H k ( X , Z ) , A ) → 0 , (2) 0 \to\operatorname{Ext}(H_{k-1}(X,\mathbb{Z}), A) \to H^k(X,A) \to\operatorname{Hom}(H_k(X,\mathbb{Z}), A) \to0, \tag*{(2)} 0 → Ext ( H k − 1 ( X , Z ) , A ) → H k ( X , A ) → Hom ( H k ( X , Z ) , A ) → 0 , (2) which is natural in A A A X X X
Proposition 1. We have the following.
(a) The period map H 2 ( X , R ) → Hom ( H 2 ( X , Z ) , R ) H^2(X,\mathbb{R}) \to\operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{R}) H 2 ( X , R ) → Hom ( H 2 ( X , Z ) , R ) X X X H 2 ( X , Z ) → R H_2(X,\mathbb{Z}) \to\mathbb{R} H 2 ( X , Z ) → R H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R ) H 2 ( X , Q ) → H 2 ( X , R ) H^2(X,\mathbb{Q}) \to H^2(X,\mathbb{R}) H 2 ( X , Q ) → H 2 ( X , R )
(b) Let H X \mathcal{H}_X H X H 2 ( X , Z ) → H 2 ( X , R ) H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{R}) H 2 ( X , Z ) → H 2 ( X , R ) c ∈ H 2 ( X , R ) c \in H^2(X,\mathbb{R}) c ∈ H 2 ( X , R ) H X \mathcal{H}_X H X ⟨ c , x ⟩ ∈ Z \langle c,x\rangle\in\mathbb{Z} ⟨ c , x ⟩ ∈ Z x ∈ H 2 ( X , Z ) x \in H_2(X,\mathbb{Z}) x ∈ H 2 ( X , Z ) H X \mathcal{H}_X H X Hom ( H 2 ( X , Z ) , Z ) \operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{Z}) Hom ( H 2 ( X , Z ) , Z )
(c) If Hom ( H 2 ( X , Z ) , Z ) = 0 \operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{Z}) = 0 Hom ( H 2 ( X , Z ) , Z ) = 0 H X = 0 \mathcal{H}_X = 0 H X = 0
Proof. (a) The groups R \mathbb{R} R Q \mathbb{Q} Q Z \mathbb{Z} Z Ext ( H 1 ( X , Z ) , A ) = 0 \operatorname{Ext}(H_1(X,\mathbb{Z}),A) = 0 Ext ( H 1 ( X , Z ) , A ) = 0 A = R , Q A = \mathbb{R},\mathbb{Q} A = R , Q k = 2 k = 2 k = 2 H 2 ( X , A ) ≅ Hom ( H 2 ( X , Z ) , A ) H^2(X,A) \cong\operatorname{Hom}(H_2(X,\mathbb{Z}),A) H 2 ( X , A ) ≅ Hom ( H 2 ( X , Z ) , A ) A = R A = \mathbb{R} A = R A A A H 2 ( X , Q ) → H 2 ( X , R ) H^2(X,\mathbb{Q}) \to H^2(X,\mathbb{R}) H 2 ( X , Q ) → H 2 ( X , R ) Q ⊂ R \mathbb{Q} \subset\mathbb{R} Q ⊂ R Hom ( H 2 ( X , Z ) , Q ) → Hom ( H 2 ( X , Z ) , R ) \operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{Q}) \to\operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{R}) Hom ( H 2 ( X , Z ) , Q ) → Hom ( H 2 ( X , Z ) , R ) Q \mathbb{Q} Q
(b) If c c c c ′ ∈ H 2 ( X , Z ) c' \in H^2(X,\mathbb{Z}) c ′ ∈ H 2 ( X , Z ) ⟨ c , x ⟩ = ⟨ c ′ , x ⟩ ∈ Z \langle c,x\rangle= \langle c',x\rangle\in\mathbb{Z} ⟨ c , x ⟩ = ⟨ c ′ , x ⟩ ∈ Z A A A c c c χ = ⟨ c , ⋅ ⟩ \chi= \langle c,\mathord{\cdot}\rangle χ = ⟨ c , ⋅ ⟩ Hom ( H 2 ( X , Z ) , Z ) \operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{Z}) Hom ( H 2 ( X , Z ) , Z ) A = Z A = \mathbb{Z} A = Z c ′ ∈ H 2 ( X , Z ) c' \in H^2(X,\mathbb{Z}) c ′ ∈ H 2 ( X , Z ) χ \chi χ H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R ) χ \chi χ c c c
(c) follows from (b). □
A Hodge form on a complex manifold X X X Ω \Omega Ω [ Ω ] ∈ H X [\Omega] \in\mathcal{H}_X [ Ω ] ∈ H X Ω \Omega Ω Hom ( H 2 ( X , Z ) , Z ) = 0 \operatorname{Hom}(H_2(X,\mathbb{Z}),\mathbb{Z}) = 0 Hom ( H 2 ( X , Z ) , Z ) = 0 d d c φ dd^c\varphi d d c φ φ \varphi φ
The decomposition of H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z )
Suppose that H 2 ( X , Z ) = F ⊕ T H_2(X,\mathbb{Z}) = \mathcal{F} \oplus\mathcal{T} H 2 ( X , Z ) = F ⊕ T F \mathcal{F} F T \mathcal{T} T H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z ) F \mathcal{F} F T \mathcal{T} T H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z ) F \mathcal{F} F pr : H 2 ( X , Z ) → F \operatorname{pr}: H_2(X,\mathbb{Z}) \to\mathcal{F} pr : H 2 ( X , Z ) → F T \mathcal{T} T H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z ) Q , R \mathbb{Q},\mathbb{R} Q , R T \mathcal{T} T F \mathcal{F} F ⟨ c , x ⟩ = ⟨ c , pr ( x ) ⟩ \langle c,x\rangle= \langle c,\operatorname{pr}(x)\rangle ⟨ c , x ⟩ = ⟨ c , pr ( x )⟩
Corollary 1. If H 2 ( X , Z ) = F ⊕ T H_2(X,\mathbb{Z}) = \mathcal{F} \oplus\mathcal{T} H 2 ( X , Z ) = F ⊕ T F \mathcal{F} F H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R ) Hom ( F , R ) \operatorname{Hom}(\mathcal{F},\mathbb{R}) Hom ( F , R ) H X \mathcal{H}_X H X Hom ( F , Z ) \operatorname{Hom}(\mathcal{F},\mathbb{Z}) Hom ( F , Z ) H 2 ( X , Q ) H^2(X,\mathbb{Q}) H 2 ( X , Q ) Hom ( F , Q ) \operatorname{Hom}(\mathcal{F},\mathbb{Q}) Hom ( F , Q ) F \mathcal{F} F b b b C 1 , … , C b C_1,\ldots,C_b C 1 , … , C b F \mathcal{F} F
[ β ] ⟼ ( ∫ C 1 β , … , ∫ C b β ) [\beta] \longmapsto\left(\int_{C_1}\beta,\ldots,\int_{C_b}\beta\right) [ β ] ⟼ ( ∫ C 1 β , … , ∫ C b β ) is an isomorphism H 2 ( X , R ) ≅ R b H^2(X,\mathbb{R}) \cong\mathbb{R}^b H 2 ( X , R ) ≅ R b H X \mathcal{H}_X H X Z b \mathbb{Z}^b Z b Q b \mathbb{Q}^b Q b C 1 , … , C b C_1,\ldots,C_b C 1 , … , C b
We shall also need two algebraic lemmas. Both are used for infinitely generated F \mathcal{F} F
Lemma 1. Let F \mathcal{F} F I ⊂ F I \subset\mathcal{F} I ⊂ F S = { x ∈ F : n x ∈ I for some n ≥ 1 } S = \{x \in\mathcal{F}: nx \in I\text{ for some }n \ge1\} S = { x ∈ F : n x ∈ I for some n ≥ 1 } I I I S S S F \mathcal{F} F F / S \mathcal{F}/S F / S
Proof. Fix a basis of F \mathcal{F} F I I I F M F_M F M F = F M ⊕ F M ′ \mathcal{F}=F_M\oplus F'_M F = F M ⊕ F M ′ F M ′ F'_M F M ′ n x ∈ F M nx\in F_M n x ∈ F M n ≠ 0 n\ne0 n = 0 F M ′ F'_M F M ′ x x x S ⊂ F M S\subset F_M S ⊂ F M F M / S F_M/S F M / S S S S F M F_M F M S S S F \mathcal{F} F F / S ≅ F M / S ⊕ F M ′ \mathcal{F}/S\cong F_M/S\oplus F'_M F / S ≅ F M / S ⊕ F M ′
Lemma 2. Let H 2 ( X , Z ) = F ⊕ T H_2(X,\mathbb{Z})=F\oplus T H 2 ( X , Z ) = F ⊕ T F = S ⊕ B F=S\oplus B F = S ⊕ B F F F ψ : S → R \psi:S\to\mathbb{R} ψ : S → R χ ψ : H 2 ( X , Z ) → R \chi_\psi:H_2(X,\mathbb{Z})\to\mathbb{R} χ ψ : H 2 ( X , Z ) → R
χ ψ ( s + b + t ) = ψ ( s ) , s ∈ S , b ∈ B , t ∈ T . \chi_\psi(s+b+t)=\psi(s),\qquad s\in S,\ b\in B,\ t\in T. χ ψ ( s + b + t ) = ψ ( s ) , s ∈ S , b ∈ B , t ∈ T . Then there is a unique class c ψ ∈ H 2 ( X , R ) c_\psi\in H^2(X,\mathbb{R}) c ψ ∈ H 2 ( X , R ) χ ψ \chi_\psi χ ψ ψ \psi ψ S S S 0 0 0 B ⊕ T B\oplus T B ⊕ T N ≥ 1 N\ge1 N ≥ 1 N ψ ( S ) ⊂ Z N\psi(S)\subset\mathbb{Z} N ψ ( S ) ⊂ Z N c ψ ∈ H X Nc_\psi\in H_X N c ψ ∈ H X
Proof. Since H 2 ( X , Z ) = S ⊕ B ⊕ T H_2(X,\mathbb{Z})=S\oplus B\oplus T H 2 ( X , Z ) = S ⊕ B ⊕ T χ ψ \chi_\psi χ ψ c ψ c_\psi c ψ N ψ ( S ) ⊂ Z N\psi(S)\subset\mathbb{Z} N ψ ( S ) ⊂ Z N χ ψ N\chi_\psi N χ ψ N c ψ ∈ H X Nc_\psi\in H_X N c ψ ∈ H X
The Levi form
For a real C 2 C^2 C 2 f f f R m \mathbb{R}^m R m Hess x f \operatorname{Hess}_x f Hess x f x x x Hess x f ( v , w ) = ∑ j , k ∂ 2 f ∂ x j ∂ x k ( x ) v j w k \operatorname{Hess}_x f(v,w)=\sum_{j,k}\frac{\partial^2 f}{\partial x_j\partial x_k}(x)v_jw_k Hess x f ( v , w ) = ∑ j , k ∂ x j ∂ x k ∂ 2 f ( x ) v j w k ∥ Hess x f ∥ \|\operatorname{Hess}_x f\| ∥ Hess x f ∥ Hess x f \operatorname{Hess}_x f Hess x f f f f C 2 C^2 C 2 X X X d f ( q ) = 0 df(q)=0 df ( q ) = 0 Hess q f \operatorname{Hess}_q f Hess q f T q X T_qX T q X
The rest of this subsection is local, so we work in C n \mathbb{C}^n C n z = x + i y z=x+iy z = x + i y x , y ∈ R n x,y\in\mathbb{R}^n x , y ∈ R n C n \mathbb{C}^n C n R 2 n \mathbb{R}^{2n} R 2 n ξ = a + i b ∈ C n \xi=a+ib\in\mathbb{C}^n ξ = a + ib ∈ C n v = ( a , b ) ∈ R 2 n v=(a,b)\in\mathbb{R}^{2n} v = ( a , b ) ∈ R 2 n J v = ( − b , a ) Jv=(-b,a) J v = ( − b , a ) i ξ i\xi i ξ C 2 C^2 C 2 u u u z z z
L u ( z ; ξ ) = ∑ j , k = 1 n ∂ 2 u ∂ z j ∂ z ‾ k ( z ) ξ j ξ ‾ k . L_u(z;\xi)=\sum_{j,k=1}^{n}\frac{\partial^2u}{\partial z_j\partial\overline{z}_k}(z)\xi_j\overline{\xi}_k. L u ( z ; ξ ) = j , k = 1 ∑ n ∂ z j ∂ z k ∂ 2 u ( z ) ξ j ξ k . It is related to the real Hessian and to d d c dd^c d d c
L u ( z ; ξ ) = 1 4 ( Hess z u ( v , v ) + Hess z u ( J v , J v ) ) = d d c u ( v , J v ) . (3) L_u(z;\xi)=\frac{1}{4}\left(\operatorname{Hess}_z u(v,v)+\operatorname{Hess}_z u(Jv,Jv)\right)=dd^c u(v,Jv). \tag*{(3)} L u ( z ; ξ ) = 4 1 ( Hess z u ( v , v ) + Hess z u ( J v , J v ) ) = d d c u ( v , J v ) . (3) where d c = i 4 ( ∂ ‾ − ∂ ) d^c=\frac{i}{4}(\overline{\partial}-\partial) d c = 4 i ( ∂ − ∂ ) d d c = d ∘ d c = i 2 ∂ ∂ ‾ dd^c=d\circ d^c=\frac{i}{2}\partial\overline{\partial} d d c = d ∘ d c = 2 i ∂ ∂ g ( λ ) = u ( z + λ ξ ) g(\lambda)=u(z+\lambda\xi) g ( λ ) = u ( z + λ ξ ) λ = t + i s ∈ C \lambda=t+is\in\mathbb{C} λ = t + i s ∈ C 0 0 0 ∂ λ ∂ λ ‾ g ( 0 ) = L u ( z ; ξ ) \partial_\lambda\partial_{\overline{\lambda}}g(0)=L_u(z;\xi) ∂ λ ∂ λ g ( 0 ) = L u ( z ; ξ ) ∂ λ ∂ λ ‾ = 1 4 ( ∂ t 2 + ∂ s 2 ) \partial_\lambda\partial_{\overline{\lambda}}=\frac{1}{4}(\partial_t^2+\partial_s^2) ∂ λ ∂ λ = 4 1 ( ∂ t 2 + ∂ s 2 ) ∂ t 2 g ( 0 ) = Hess z u ( v , v ) \partial_t^2g(0)=\operatorname{Hess}_z u(v,v) ∂ t 2 g ( 0 ) = Hess z u ( v , v ) ∂ s 2 g ( 0 ) = Hess z u ( J v , J v ) \partial_s^2g(0)=\operatorname{Hess}_z u(Jv,Jv) ∂ s 2 g ( 0 ) = Hess z u ( J v , J v ) ∂ t \partial_t ∂ t ∂ s \partial_s ∂ s v v v J v Jv J v d d c u = i 2 ∑ j , k ∂ 2 u ∂ z j ∂ z ‾ k d z j ∧ d z ‾ k dd^c u=\frac{i}{2}\sum_{j,k}\frac{\partial^2u}{\partial z_j\partial\overline{z}_k}\,dz_j\wedge d\overline{z}_k d d c u = 2 i ∑ j , k ∂ z j ∂ z k ∂ 2 u d z j ∧ d z k d z j ∧ d z ‾ k ( v , J v ) = ξ j ⋅ i ξ ‾ k − i ξ j ⋅ ξ ‾ k = − 2 i ξ j ξ ‾ k dz_j\wedge d\overline{z}_k(v,Jv)=\xi_j\cdot i\overline{\xi}_k-i\xi_j\cdot\overline{\xi}_k=-2i\xi_j\overline{\xi}_k d z j ∧ d z k ( v , J v ) = ξ j ⋅ i ξ k − i ξ j ⋅ ξ k = − 2 i ξ j ξ k u u u L u ( z ; ξ ) > 0 L_u(z;\xi)>0 L u ( z ; ξ ) > 0 z z z ξ ≠ 0 \xi\ne0 ξ = 0
For example, if u ( x + i y ) = F ( x ) + σ ∣ y ∣ 2 u(x+iy)=F(x)+\sigma|y|^2 u ( x + i y ) = F ( x ) + σ ∣ y ∣ 2 F ∈ C 2 ( R n ) F\in C^2(\mathbb{R}^n) F ∈ C 2 ( R n ) σ ∈ R \sigma\in\mathbb{R} σ ∈ R Hess z u \operatorname{Hess}_z u Hess z u Hess x F \operatorname{Hess}_x F Hess x F 2 σ I d 2\sigma\,\mathrm{Id} 2 σ Id
L u ( z ; ξ ) = 1 4 ( Hess x F ( a , a ) + Hess x F ( b , b ) + 2 σ ∣ ξ ∣ 2 ) ≥ 1 4 ( 2 σ − ∥ Hess x F ∥ ) ∣ ξ ∣ 2 . (4) L_u(z;\xi)=\frac{1}{4}\left(\operatorname{Hess}_x F(a,a)+\operatorname{Hess}_x F(b,b)+2\sigma|\xi|^2\right)\ge\frac{1}{4}\left(2\sigma-\|\operatorname{Hess}_x F\|\right)|\xi|^2. \tag*{(4)} L u ( z ; ξ ) = 4 1 ( Hess x F ( a , a ) + Hess x F ( b , b ) + 2 σ ∣ ξ ∣ 2 ) ≥ 4 1 ( 2 σ − ∥ Hess x F ∥ ) ∣ ξ ∣ 2 . (4) Finally, we record two product formulas, which hold on any complex manifold. For real C 2 C^2 C 2 f f f g g g d c ( f g ) = f d c g + g d c f d^c(fg)=f\,d^cg+g\,d^cf d c ( f g ) = f d c g + g d c f d d d
d d c ( f g ) = f d d c g + g d d c f + d f ∧ d c g + d g ∧ d c f . (5) dd^c(fg)=f\,dd^cg+g\,dd^cf+df\wedge d^cg+dg\wedge d^cf. \tag*{(5)} d d c ( f g ) = f d d c g + g d d c f + df ∧ d c g + d g ∧ d c f . (5) For f = g = v f=g=v f = g = v d v ∧ d c v = i 2 ( ∂ v + ∂ ˉ v ) ∧ ( ∂ ˉ v − ∂ v ) = i ∂ v ∧ ∂ ˉ v dv\wedge d^cv=\frac{i}{2}(\partial v+\bar{\partial}v)\wedge(\bar{\partial}v-\partial v)=i\,\partial v\wedge\bar{\partial}v d v ∧ d c v = 2 i ( ∂ v + ∂ ˉ v ) ∧ ( ∂ ˉ v − ∂ v ) = i ∂ v ∧ ∂ ˉ v
d d c ( v 2 ) = i ∂ v ∧ ∂ ˉ v + 2 v d d c v . (6) dd^c(v^2)=i\,\partial v\wedge\bar{\partial}v+2v\,dd^cv. \tag*{(6)} d d c ( v 2 ) = i ∂ v ∧ ∂ ˉ v + 2 v d d c v . (6) Preliminary Results on Kähler Forms
In this section, we prove some preliminary results concerning Kähler forms and their cohomology classes. Lemmas 3 and 4 are well-known, but proofs are hard to find. We provide them below for the sake of completeness.
Lemma 3. Let X X X
(a) Every class in H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R ) ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X
(b) Let D ⊂ X D\subset X D ⊂ X O ( X ) O(X) O ( X ) w w w ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X D D D ω \omega ω X X X [ ω ] = [ w ] [\omega]=[w] [ ω ] = [ w ] H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R ) ω = w \omega=w ω = w D D D
In particular, taking D = ∅ D=\varnothing D = ∅ H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R )
Proof. (a) Let u u u u = u 2 , 0 + u 1 , 1 + u 0 , 2 u=u^{2,0}+u^{1,1}+u^{0,2} u = u 2 , 0 + u 1 , 1 + u 0 , 2 ( 0 , 3 ) (0,3) ( 0 , 3 ) d u = 0 du=0 d u = 0 ∂ ˉ u 0 , 2 \bar{\partial}u^{0,2} ∂ ˉ u 0 , 2 u 0 , 2 u^{0,2} u 0 , 2 ∂ ˉ \bar{\partial} ∂ ˉ X X X H 0 , 2 ( X ) = 0 H^{0,2}(X)=0 H 0 , 2 ( X ) = 0 ( 0 , 1 ) (0,1) ( 0 , 1 ) p p p ∂ ˉ p = u 0 , 2 \bar{\partial}p=u^{0,2} ∂ ˉ p = u 0 , 2 γ = p + p ˉ \gamma=p+\bar{p} γ = p + p ˉ w = u − d γ w=u-d\gamma w = u − d γ ( 0 , 2 ) (0,2) ( 0 , 2 ) d γ d\gamma d γ ∂ ˉ p = u 0 , 2 \bar{\partial}p=u^{0,2} ∂ ˉ p = u 0 , 2 w 0 , 2 = 0 w^{0,2}=0 w 0 , 2 = 0 w w w w 2 , 0 = w 0 , 2 ‾ = 0 w^{2,0}=\overline{w^{0,2}}=0 w 2 , 0 = w 0 , 2 = 0 w w w ( 1 , 1 ) (1,1) ( 1 , 1 ) u u u
(b) Step 1: an exhaustion function. Since w w w D D D O O O D D D D D D O ( X ) O(X) O ( X ) Ψ \Psi Ψ X X X Ψ < 0 \Psi<0 Ψ < 0 D D D Ψ > 0 \Psi>0 Ψ > 0 X ∖ O X\setminus O X ∖ O [11 Theorem 5.1.6] . (If D = ∅ D=\varnothing D = ∅ O = ∅ O=\varnothing O = ∅ Ψ > 0 \Psi>0 Ψ > 0 { Ψ ≤ 1 / λ } \{\Psi\leq1/\lambda\} { Ψ ≤ 1/ λ } λ → ∞ \lambda\to\infty λ → ∞ { Ψ ≤ 0 } ⊂ O \{\Psi\leq0\}\subset O { Ψ ≤ 0 } ⊂ O Ψ \Psi Ψ λ Ψ \lambda\Psi λ Ψ λ > 0 \lambda>0 λ > 0
{ Ψ ≤ 1 } ⊂ O . \{\Psi\leq1\}\subset O. { Ψ ≤ 1 } ⊂ O . Step 2: a convex function G G G For j ≥ 1 j\geq1 j ≥ 1
ν j : = max { − w ( v , J v ) d d c Ψ ( v , J v ) : x ∈ { Ψ ≤ j + 1 } , 0 ≠ v ∈ T x X } . \nu_j:=\max\left\{\frac{-w(v,Jv)}{dd^c\Psi(v,Jv)}:x\in\{\Psi\leq j+1\},\ 0\ne v\in T_xX\right\}. ν j := max { d d c Ψ ( v , J v ) − w ( v , J v ) : x ∈ { Ψ ≤ j + 1 } , 0 = v ∈ T x X } . This is finite, because { Ψ ≤ j + 1 } \{\Psi\leq j+1\} { Ψ ≤ j + 1 } d d c Ψ ( v , J v ) > 0 dd^c\Psi(v,Jv)>0 d d c Ψ ( v , J v ) > 0 v ≠ 0 v\ne0 v = 0 x ∈ { Ψ ≤ j + 1 } x\in\{\Psi\leq j+1\} x ∈ { Ψ ≤ j + 1 } s > ν j s>\nu_j s > ν j w + s d d c Ψ w+s\,dd^c\Psi w + s d d c Ψ x x x s j > max { ν j , 0 } s_j>\max\{\nu_j,0\} s j > max { ν j , 0 } G G G ( − ∞ , 0 ] (-\infty,0] ( − ∞ , 0 ] s 1 s_1 s 1 [ 0 , 2 ] [0,2] [ 0 , 2 ] s j s_j s j [ j , j + 1 ] [j,j+1] [ j , j + 1 ] j ≥ 2 j\geq2 j ≥ 2 G ′ G' G ′ j ≥ 2 j\geq2 j ≥ 2 G : R → R G:\mathbb{R}\to\mathbb{R} G : R → R
G = 0 on ( − ∞ , 0 ] , G ′ ≥ 0 , G ′ > ν j on [ j , j + 1 ] ( j ≥ 1 ) . G=0\ \text{on }(-\infty,0],\qquad G'\geq0,\qquad G'>\nu_j\ \text{on }[j,j+1]\ (j\geq1). G = 0 on ( − ∞ , 0 ] , G ′ ≥ 0 , G ′ > ν j on [ j , j + 1 ] ( j ≥ 1 ) . Step 3: the Kähler form. Put ω = w + d d c ( G ∘ Ψ ) \omega=w+dd^c(G\circ\Psi) ω = w + d d c ( G ∘ Ψ ) ( 1 , 1 ) (1,1) ( 1 , 1 )
d d c ( G ∘ Ψ ) = G ′ ( Ψ ) d d c Ψ + G ′ ′ ( Ψ ) d Ψ ∧ d c Ψ ≥ G ′ ( Ψ ) d d c Ψ ≥ 0 , dd^c(G\circ\Psi)=G'(\Psi)\,dd^c\Psi+G''(\Psi)\,d\Psi\wedge d^c\Psi\geq G'(\Psi)\,dd^c\Psi\geq0, d d c ( G ∘ Ψ ) = G ′ ( Ψ ) d d c Ψ + G ′′ ( Ψ ) d Ψ ∧ d c Ψ ≥ G ′ ( Ψ ) d d c Ψ ≥ 0 , since G ′ ≥ 0 G' \ge0 G ′ ≥ 0 G ′ ′ ≥ 0 G'' \ge0 G ′′ ≥ 0 ω \omega ω x ∈ X x \in X x ∈ X Ψ ( x ) ≤ 1 \Psi(x) \le1 Ψ ( x ) ≤ 1 x ∈ O x \in O x ∈ O w > 0 w > 0 w > 0 ω ≥ w > 0 \omega\ge w > 0 ω ≥ w > 0 x x x Ψ ( x ) ∈ [ j , j + 1 ] \Psi(x) \in[j,j+1] Ψ ( x ) ∈ [ j , j + 1 ] j ≥ 1 j \ge1 j ≥ 1 x ∈ { Ψ ≤ j + 1 } x \in\{\Psi\le j+1\} x ∈ { Ψ ≤ j + 1 } G ′ ( Ψ ( x ) ) > ν j G'(\Psi(x)) > \nu_j G ′ ( Ψ ( x )) > ν j x x x
ω ≥ w + G ′ ( Ψ ( x ) ) d d c Ψ > 0. \omega\ge w+G'(\Psi(x))\mathrm{d}\mathrm{d}^c\Psi> 0. ω ≥ w + G ′ ( Ψ ( x )) d d c Ψ > 0. Hence, ω \omega ω w w w G ∘ Ψ = 0 G \circ\Psi= 0 G ∘ Ψ = 0 { Ψ < 0 } ⊃ D \{\Psi< 0\} \supset D { Ψ < 0 } ⊃ D ω = w \omega= w ω = w D D D
Lemma 4 (d d c \mathrm{d}\mathrm{d}^c d d c Let X X X
(a) Let η \eta η ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X η \eta η d d d h ∈ C ∞ ( X ) h \in C^\infty(X) h ∈ C ∞ ( X ) d d c h = η \mathrm{d}\mathrm{d}^c h = \eta d d c h = η
(b) Fix a Hermitian metric on X X X n = dim C X n = \dim_{\mathbb{C}} X n = dim C X U ⋐ X U \Subset X U ⋐ X η \eta η ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X d d d U ‾ \overline{U} U h ∈ C ∞ ( U ‾ ) h \in C^\infty(\overline{U}) h ∈ C ∞ ( U ) d d c h = η \mathrm{d}\mathrm{d}^c h = \eta d d c h = η U U U
∥ h ∥ C 2 ( U ) ≤ C ∥ η ∥ C n + 2 ( U ) , (7) \lVert h\rVert_{C^2(U)} \le C\lVert\eta\rVert_{C^{n+2}(U)}, \tag*{(7)} ∥ h ∥ C 2 ( U ) ≤ C ∥ η ∥ C n + 2 ( U ) , (7) where C C C U U U
Proof. (a) Since η \eta η d d d 1 1 1 ξ = ξ 1 , 0 + ξ 0 , 1 \xi= \xi^{1,0}+\xi^{0,1} ξ = ξ 1 , 0 + ξ 0 , 1 d ξ = η d\xi= \eta d ξ = η ( 0 , 2 ) (0,2) ( 0 , 2 ) d ξ d\xi d ξ ∂ ˉ ξ 0 , 1 \bar{\partial}\xi^{0,1} ∂ ˉ ξ 0 , 1 η \eta η ( 1 , 1 ) (1,1) ( 1 , 1 ) ∂ ˉ ξ 0 , 1 = 0 \bar{\partial}\xi^{0,1}=0 ∂ ˉ ξ 0 , 1 = 0 X X X H 0 , 1 ( X ) = 0 H^{0,1}(X)=0 H 0 , 1 ( X ) = 0 f f f ∂ ˉ f = ξ 0 , 1 \bar{\partial}f=\xi^{0,1} ∂ ˉ f = ξ 0 , 1 η \eta η
η = d ξ = ∂ ξ 0 , 1 + ∂ ˉ ξ 1 , 0 = ∂ ( ∂ ˉ f ) + ∂ ˉ ( ∂ f ˉ ) = ∂ ∂ ˉ ( f − f ˉ ) . \eta=d\xi=\partial\xi^{0,1}+\bar{\partial}\xi^{1,0}
=\partial(\bar{\partial}f)+\bar{\partial}(\partial\bar{f})
=\partial\bar{\partial}(f-\bar{f}). η = d ξ = ∂ ξ 0 , 1 + ∂ ˉ ξ 1 , 0 = ∂ ( ∂ ˉ f ) + ∂ ˉ ( ∂ f ˉ ) = ∂ ∂ ˉ ( f − f ˉ ) . Let h = 4 Im ( f ) h=4\operatorname{Im}(f) h = 4 Im ( f ) f − f ˉ = 2 i Im ( f ) = i 2 h f-\bar{f}=2i\operatorname{Im}(f)=\frac{i}{2}h f − f ˉ = 2 i Im ( f ) = 2 i h η = ∂ ∂ ˉ ( i 2 h ) = i 2 ∂ ∂ ˉ h = d d c h \eta=\partial\bar{\partial}\left(\frac{i}{2}h\right)=\frac{i}{2}\partial\bar{\partial}h=\mathrm{d}\mathrm{d}^c h η = ∂ ∂ ˉ ( 2 i h ) = 2 i ∂ ∂ ˉ h = d d c h
(b) Write ∥ ⋅ ∥ s \lVert\cdot\rVert_s ∥ ⋅ ∥ s W s W^s W s U U U [16 Lemma 3.2.1] , applied on the compact manifold with boundary U ‾ \overline{U} U η \eta η ξ ∈ C ∞ ( U ‾ ) \xi\in C^\infty(\overline{U}) ξ ∈ C ∞ ( U ) d ξ = η d\xi=\eta d ξ = η
∥ ξ ∥ s ≤ C 1 ∥ η ∥ s − 1 ( s ≥ 1 ) . \lVert\xi\rVert_s \le C_1\lVert\eta\rVert_{s-1}\qquad(s\ge1). ∥ ξ ∥ s ≤ C 1 ∥ η ∥ s − 1 ( s ≥ 1 ) . Replacing ξ \xi ξ ξ \xi ξ ∂ ˉ ξ 0 , 1 = 0 \bar{\partial}\xi^{0,1}=0 ∂ ˉ ξ 0 , 1 = 0 U U U L 2 L^2 L 2 ∂ ˉ \bar{\partial} ∂ ˉ ( 0 , 1 ) (0,1) ( 0 , 1 ) L 2 ( U ) L^2(U) L 2 ( U ) ∂ ˉ \bar{\partial} ∂ ˉ L 2 ( U ) L^2(U) L 2 ( U ) ∂ ˉ \bar{\partial} ∂ ˉ ( 0 , 1 ) (0,1) ( 0 , 1 ) U U U H ( 0 , 1 ) ξ 0 , 1 H_{(0,1)}\xi^{0,1} H ( 0 , 1 ) ξ 0 , 1 [4 Corollary 5.3.11] , there exists f ∈ C ∞ ( U ‾ ) f \in C^\infty(\overline{U}) f ∈ C ∞ ( U ) ∂ ˉ f = ξ 0 , 1 \bar{\partial}f=\xi^{0,1} ∂ ˉ f = ξ 0 , 1
∥ f ∥ s + 1 2 ≤ C 2 ∥ ξ 0 , 1 ∥ s . \lVert f\rVert_{s+\frac{1}{2}} \le C_2\lVert\xi^{0,1}\rVert_s. ∥ f ∥ s + 2 1 ≤ C 2 ∥ ξ 0 , 1 ∥ s . As in (a), h : = 4 Im ( f ) ∈ C ∞ ( U ‾ ) h:=4\operatorname{Im}(f)\in C^\infty(\overline{U}) h := 4 Im ( f ) ∈ C ∞ ( U ) d d c h = η \mathrm{d}\mathrm{d}^c h=\eta d d c h = η U U U
Now take s = n + 3 s=n+3 s = n + 3 U U U 2 n 2n 2 n W s ( U ) ↪ C 2 ( U ‾ ) W^s(U)\hookrightarrow C^2(\overline{U}) W s ( U ) ↪ C 2 ( U ) s > 2 + n s>2+n s > 2 + n
∥ h ∥ C 2 ( U ) ≤ 4 ∥ f ∥ C 2 ( U ) ≤ C 3 ∥ f ∥ s ≤ C 3 C 2 ∥ ξ 0 , 1 ∥ s ≤ C 4 ∥ ξ ∥ s ≤ C 5 ∥ η ∥ s − 1 ≤ C ∥ η ∥ C n + 2 ( U ) , \lVert h\rVert_{C^2(U)}\le4\lVert f\rVert_{C^2(U)}
\le C_3\lVert f\rVert_s
\le C_3C_2\lVert\xi^{0,1}\rVert_s
\le C_4\lVert\xi\rVert_s
\le C_5\lVert\eta\rVert_{s-1}
\le C\lVert\eta\rVert_{C^{n+2}(U)}, ∥ h ∥ C 2 ( U ) ≤ 4 ∥ f ∥ C 2 ( U ) ≤ C 3 ∥ f ∥ s ≤ C 3 C 2 ∥ ξ 0 , 1 ∥ s ≤ C 4 ∥ ξ ∥ s ≤ C 5 ∥ η ∥ s − 1 ≤ C ∥ η ∥ C n + 2 ( U ) , where the last step uses that U U U
Remark 1. By Lemma 4(a), an exact Kähler form ω \omega ω X X X ω = d d c φ \omega=\mathrm{d}\mathrm{d}^c\varphi ω = d d c φ φ ∈ C ∞ ( X ) \varphi\in C^\infty(X) φ ∈ C ∞ ( X ) d d c φ = d ( d c φ ) \mathrm{d}\mathrm{d}^c\varphi=d(\mathrm{d}^c\varphi) d d c φ = d ( d c φ ) d d c \mathrm{d}\mathrm{d}^c d d c
Next, we prove the following.
Lemma 5. Let X X X U ⊂ X U \subset X U ⊂ X K ⊂ U K \subset U K ⊂ U φ : H 2 ( X , Z ) → R \varphi: H_{2}(X,\mathbb{Z}) \to\mathbb{R} φ : H 2 ( X , Z ) → R H 2 ( U , Z ) → H 2 ( X , Z ) H_{2}(U,\mathbb{Z}) \to H_{2}(X,\mathbb{Z}) H 2 ( U , Z ) → H 2 ( X , Z ) ( 1 , 1 ) (1,1) ( 1 , 1 ) β \beta β X X X
∫ S β = φ ( S ) \int_{S}\beta=\varphi(S) ∫ S β = φ ( S ) S ∈ H 2 ( X , Z ) S\in H_{2}(X,\mathbb{Z}) S ∈ H 2 ( X , Z )
β = 0 \beta=0 β = 0 K K K
Proof. By Prop. 1(a) there is a class in H 2 ( X , R ) H^{2}(X,\mathbb{R}) H 2 ( X , R ) φ \varphi φ Lemma 3 (a), it is represented by a smooth, closed, real ( 1 , 1 ) (1,1) ( 1 , 1 ) α \alpha α X X X S ′ S' S ′ U U U ∫ S ′ α = φ ( [ S ′ ] ) = 0 \int_{S'}\alpha=\varphi([S'])=0 ∫ S ′ α = φ ([ S ′ ]) = 0 [ S ′ ] [S'] [ S ′ ] H 2 ( U , Z ) H_{2}(U,\mathbb{Z}) H 2 ( U , Z ) α ∣ U \alpha|_{U} α ∣ U U U U U U U Lemma 4 (a) gives f ∈ C ∞ ( U ) f\in C^{\infty}(U) f ∈ C ∞ ( U ) d d c f = α dd^{c}f=\alpha d d c f = α U U U
Choose a neighbourhood W W W K K K ρ : X → [ 0 , 1 ] \rho:X\to[0,1] ρ : X → [ 0 , 1 ] ρ = 1 \rho=1 ρ = 1 W W W supp ρ ⋐ U \operatorname{supp}\rho\Subset U supp ρ ⋐ U ρ f \rho f ρ f X X X
β = α − d d c ( ρ f ) \beta=\alpha-dd^{c}(\rho f) β = α − d d c ( ρ f ) is a smooth, closed, real ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X d d c ( ρ f ) dd^{c}(\rho f) d d c ( ρ f ) β \beta β α \alpha α W W W β = α − d d c f = 0 \beta=\alpha-dd^{c}f=0 β = α − d d c f = 0
Finally, we need the following approximation result.
Lemma 6 (Rational approximation). Let X X X H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) X X X ω \omega ω X X X Σ ⊂ H 2 ( X , Z ) \Sigma\subset H_{2}(X,\mathbb{Z}) Σ ⊂ H 2 ( X , Z ) ∫ S ω = 0 \int_{S}\omega=0 ∫ S ω = 0 S ∈ Σ S\in\Sigma S ∈ Σ U U U [ ω ] [\omega] [ ω ] H 2 ( X , R ) H^{2}(X,\mathbb{R}) H 2 ( X , R ) k ≥ 0 k\geq0 k ≥ 0 ϵ > 0 \epsilon>0 ϵ > 0 D ⊂ X D\subset X D ⊂ X ( 1 , 1 ) (1,1) ( 1 , 1 ) ω ~ D \widetilde{\omega}_{D} ω D X X X
(a) [ ω ~ D ] ∈ H 2 ( X , Q ) ∩ U [\widetilde{\omega}_{D}]\in H^{2}(X,\mathbb{Q})\cap U [ ω D ] ∈ H 2 ( X , Q ) ∩ U ∫ S ω ~ D = 0 \int_{S}\widetilde{\omega}_{D}=0 ∫ S ω D = 0 S ∈ Σ S\in\Sigma S ∈ Σ
(b) ω ~ D \widetilde{\omega}_{D} ω D D D D ∥ ω ~ D − ω ∥ C k ( D ) < ϵ \|\widetilde{\omega}_{D}-\omega\|_{C^{k}(D)}<\epsilon ∥ ω D − ω ∥ C k ( D ) < ϵ
Proof. Since H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) H 2 ( X , Z ) = F ⊕ T H_{2}(X,\mathbb{Z})=F\oplus T H 2 ( X , Z ) = F ⊕ T § 2.2 , with F ≅ Z b F\cong\mathbb{Z}^{b} F ≅ Z b C 1 , … , C b C_{1},\ldots,C_{b} C 1 , … , C b F F F Lemma 3 (a), there are real closed ( 1 , 1 ) (1,1) ( 1 , 1 ) e 1 , … , e b e_{1},\ldots,e_{b} e 1 , … , e b X X X
∫ C l e j = δ j l , 1 ≤ j , l ≤ b . \int_{C_l}e_j=\delta_{jl},\qquad1\leq j,l\leq b. ∫ C l e j = δ j l , 1 ≤ j , l ≤ b . We identify H 2 ( X , R ) H^{2}(X,\mathbb{R}) H 2 ( X , R ) R b \mathbb{R}^{b} R b [ β ] ↦ x = ( ∫ C 1 β , … , ∫ C b β ) [\beta]\mapsto x=\left(\int_{C_{1}}\beta,\ldots,\int_{C_{b}}\beta\right) [ β ] ↦ x = ( ∫ C 1 β , … , ∫ C b β ) H 2 ( X , Q ) = Q b H^{2}(X,\mathbb{Q})=\mathbb{Q}^{b} H 2 ( X , Q ) = Q b S ∈ Σ S\in\Sigma S ∈ Σ S = ∑ j a j ( S ) [ C j ] + t S=\sum_{j}a_{j}(S)[C_{j}]+t S = ∑ j a j ( S ) [ C j ] + t a j ( S ) ∈ Z a_{j}(S)\in\mathbb{Z} a j ( S ) ∈ Z t ∈ T t\in T t ∈ T ∫ S β = ∑ j a j ( S ) ∫ C j β \int_{S}\beta=\sum_{j}a_{j}(S)\int_{C_{j}}\beta ∫ S β = ∑ j a j ( S ) ∫ C j β a ( S ) = ( a 1 ( S ) , … , a b ( S ) ) ∈ Z b a(S)=(a_{1}(S),\ldots,a_{b}(S))\in\mathbb{Z}^{b} a ( S ) = ( a 1 ( S ) , … , a b ( S )) ∈ Z b
Density of rational points. Let a 1 , … , a r ∈ Z b a^{1},\ldots,a^{r}\in\mathbb{Z}^{b} a 1 , … , a r ∈ Z b Q \mathbb{Q} Q { a ( S ) : S ∈ Σ } \{a(S):S\in\Sigma\} { a ( S ) : S ∈ Σ } A A A r × b r\times b r × b a 1 , … , a r a^{1},\ldots,a^{r} a 1 , … , a r Σ \Sigma Σ L = ker R A ⊂ R b L=\ker_{\mathbb{R}}A\subset\mathbb{R}^{b} L = ker R A ⊂ R b A A A Q \mathbb{Q} Q R \mathbb{R} R ker Q A \ker_{\mathbb{Q}}A ker Q A Q \mathbb{Q} Q b − rank A = dim R L b-\operatorname{rank}A=\dim_{\mathbb{R}}L b − rank A = dim R L Q \mathbb{Q} Q ker Q A \ker_{\mathbb{Q}}A ker Q A R \mathbb{R} R L L L L ∩ Q b L\cap\mathbb{Q}^{b} L ∩ Q b L L L
Choice of the periods. Write λ j = ∫ C j ω \lambda_j=\int_{C_j}\omega λ j = ∫ C j ω λ = ( λ 1 , … , λ b ) ∈ L \lambda=(\lambda_1,\ldots,\lambda_b)\in L λ = ( λ 1 , … , λ b ) ∈ L U U U k k k ϵ \epsilon ϵ D D D ω \omega ω D D D c 0 > 0 c_0>0 c 0 > 0 ( 1 , 1 ) (1,1) ( 1 , 1 ) α \alpha α ∥ α − ω ∥ C 0 ( D ) < ϵ 0 \|\alpha-\omega\|_{C^{0}(D)}<\epsilon_0 ∥ α − ω ∥ C 0 ( D ) < ϵ 0 D D D M D = max { 1 , max j ∥ e j ∥ C k ( D ) } M_D=\max\{1,\max_{j}\|e_j\|_{C^{k}(D)}\} M D = max { 1 , max j ∥ e j ∥ C k ( D ) } δ ∈ ( 0 , min { ϵ , ϵ 0 } / ( b M D ) ] \delta\in(0,\min\{\epsilon,\epsilon_0\}/(bM_D)] δ ∈ ( 0 , min { ϵ , ϵ 0 } / ( b M D )] x ∈ R b x\in\mathbb{R}^{b} x ∈ R b ∣ x j − λ j ∣ < δ |x_j-\lambda_j|<\delta ∣ x j − λ j ∣ < δ j j j U U U q ∈ L ∩ Q b q\in L\cap\mathbb{Q}^{b} q ∈ L ∩ Q b ∣ λ j − q j ∣ < δ |\lambda_j-q_j|<\delta ∣ λ j − q j ∣ < δ j j j Choice of ω ~ D \widetilde{\omega}_D ω D Define
ω ~ D = ω − ∑ j = 1 b ( λ j − q j ) e j . \widetilde{\omega}_D=\omega-\sum_{j=1}^{b}(\lambda_j-q_j)e_j. ω D = ω − j = 1 ∑ b ( λ j − q j ) e j . This is a smooth, closed, real ( 1 , 1 ) (1,1) ( 1 , 1 ) ∫ C l ω ~ D = λ l − ( λ l − q l ) = q l \int_{C_l}\widetilde{\omega}_D=\lambda_l-(\lambda_l-q_l)=q_l ∫ C l ω D = λ l − ( λ l − q l ) = q l l l l [ ω ~ D ] [\widetilde{\omega}_D] [ ω D ] q ∈ L ∩ Q b q\in L\cap\mathbb{Q}^b q ∈ L ∩ Q b U \mathcal{U} U δ \delta δ ∫ S ω ~ D = a ( S ) ⋅ q = 0 \int_S\widetilde{\omega}_D=a(S)\cdot q=0 ∫ S ω D = a ( S ) ⋅ q = 0 S ∈ Σ S\in\Sigma S ∈ Σ
∥ ω ~ D − ω ∥ C k ( D ) ≤ ∑ j = 1 b ∣ λ j − q j ∣ ∥ e j ∥ C k ( D ) < b δ M D ≤ min { ϵ , ϵ 0 } . \left\|\widetilde{\omega}_D-\omega\right\|_{C^k(D)}
\leq\sum_{j=1}^{b}|\lambda_j-q_j|\left\|e_j\right\|_{C^k(D)}
<b\delta M_D\leq\min\{\epsilon,\epsilon_0\}. ∥ ω D − ω ∥ C k ( D ) ≤ j = 1 ∑ b ∣ λ j − q j ∣ ∥ e j ∥ C k ( D ) < b δ M D ≤ min { ϵ , ϵ 0 } . Since ∥ ⋅ ∥ C 0 ( D ) ≤ ∥ ⋅ ∥ C k ( D ) \|\cdot\|_{C^0(D)}\leq\|\cdot\|_{C^k(D)} ∥ ⋅ ∥ C 0 ( D ) ≤ ∥ ⋅ ∥ C k ( D ) ω ~ D \widetilde{\omega}_D ω D D D D
Proof of Theorem 1
We first give the proof for finite topology, then extend it to the general case.
Case 1: H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z )
Step 1: Rational approximation. Let n = dim C X n=\dim_{\mathbb{C}}X n = dim C X X X X C k C^k C k U U U K K K K = { h ≤ 0 } K=\{h\leq0\} K = { h ≤ 0 } ω = d d c h \omega=dd^ch ω = d d c h U U U
Choose an open set U 1 U_1 U 1 K ⊂ U 1 ⋐ U K\subset U_1\Subset U K ⊂ U 1 ⋐ U m 1 : = min ∂ U 1 h > 0 m_1:=\min_{\partial U_1}h>0 m 1 := min ∂ U 1 h > 0 h > 0 h>0 h > 0 U ∖ K U\setminus K U ∖ K 0 < c < c ′ < m 1 0<c<c'<m_1 0 < c < c ′ < m 1 h h h
V = { x ∈ U 1 : h ( x ) < c } , V ′ = { x ∈ U 1 : h ( x ) < c ′ } . V=\{x\in U_1:h(x)<c\},\qquad V'=\{x\in U_1:h(x)<c'\}. V = { x ∈ U 1 : h ( x ) < c } , V ′ = { x ∈ U 1 : h ( x ) < c ′ } . Then K ⊂ V ⋐ V ′ ⋐ U K\subset V\Subset V'\Subset U K ⊂ V ⋐ V ′ ⋐ U ω = d d c h \omega=dd^ch ω = d d c h V ′ V' V ′ V V V { h = c } \{h=c\} { h = c } Ψ \Psi Ψ X X X R ∈ R R\in\mathbb{R} R ∈ R V ′ ⊂ { Ψ < R } V'\subset\{\Psi<R\} V ′ ⊂ { Ψ < R } D = { Ψ ≤ R } D=\{\Psi\leq R\} D = { Ψ ≤ R } D D D
V ‾ ⊂ V ′ ‾ ⊂ int D ⊂ D . \overline{V}\subset\overline{V'}\subset\operatorname{int}D\subset D. V ⊂ V ′ ⊂ int D ⊂ D . Let ϵ > 0 \epsilon>0 ϵ > 0 V V V D D D Lemma 4 (b) for V V V ρ \rho ρ ϵ \epsilon ϵ
Let Σ ⊂ H 2 ( X , Z ) \Sigma\subset H_2(X,\mathbb{Z}) Σ ⊂ H 2 ( X , Z ) H 2 ( V ′ , Z ) → H 2 ( X , Z ) H_2(V',\mathbb{Z})\to H_2(X,\mathbb{Z}) H 2 ( V ′ , Z ) → H 2 ( X , Z ) ω = d d c h \omega=dd^ch ω = d d c h V ′ V' V ′ ∫ S ω = 0 \int_S\omega=0 ∫ S ω = 0 S ∈ Σ S\in\Sigma S ∈ Σ Lemma 6 with this Σ \Sigma Σ k = n + 2 k=n+2 k = n + 2 ϵ \epsilon ϵ D D D ( 1 , 1 ) (1,1) ( 1 , 1 ) ω ~ D \widetilde{\omega}_D ω D X X X
ω ~ D > 0 at every point of D , ∥ ω ~ D − ω ∥ C n + 2 ( D ) < ϵ , \widetilde{\omega}_D>0\quad\text{at every point of }D,\qquad
\left\|\widetilde{\omega}_D-\omega\right\|_{C^{n+2}(D)}<\epsilon, ω D > 0 at every point of D , ∥ ω D − ω ∥ C n + 2 ( D ) < ϵ , and whose class [ ω ~ D ] ∈ H 2 ( X , Q ) [\widetilde{\omega}_D]\in H^2(X,\mathbb{Q}) [ ω D ] ∈ H 2 ( X , Q )
∫ S ω ~ D = 0 for all S ∈ Σ . \int_S\widetilde{\omega}_D=0\quad\text{for all }S\in\Sigma. ∫ S ω D = 0 for all S ∈ Σ. Put Θ = ω ~ D − ω \Theta=\widetilde{\omega}_D-\omega Θ = ω D − ω 2 2 2 S ′ S' S ′ V ′ V' V ′ H 2 ( X , Z ) H_2(X,\mathbb{Z}) H 2 ( X , Z ) Σ \Sigma Σ
∫ S ′ Θ = ∫ S ′ ω ~ D − ∫ S ′ ω = 0. \int_{S'}\Theta=\int_{S'}\widetilde{\omega}_D-\int_{S'}\omega=0. ∫ S ′ Θ = ∫ S ′ ω D − ∫ S ′ ω = 0. By Prop. 1(a), applied to V ′ V' V ′ Θ \Theta Θ d d d V ′ V' V ′ V ‾ \overline{V} V V ⊂ D V\subset D V ⊂ D
∥ Θ ∥ C n + 2 ( V ) ≤ ∥ Θ ∥ C n + 2 ( D ) < ϵ . \|\Theta\|_{C^{n+2}(V)}\leq\|\Theta\|_{C^{n+2}(D)}<\epsilon. ∥Θ ∥ C n + 2 ( V ) ≤ ∥Θ ∥ C n + 2 ( D ) < ϵ . Step 2: Solving the d d c dd^{c} d d c By Step 1, Θ = ω ~ D − ω \Theta=\widetilde{\omega}_{D}-\omega Θ = ω D − ω ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X d d d V ‾ \overline{V} V ∥ Θ ∥ C n + 2 ( V ) < ϵ \lVert\Theta\rVert_{C^{n+2}(V)}<\epsilon ∥ Θ ∥ C n + 2 ( V ) < ϵ V V V u ∈ C ∞ ( V ‾ ) u\in C^{\infty}(\overline{V}) u ∈ C ∞ ( V ) d d c u = Θ dd^{c}u=\Theta d d c u = Θ V V V
∥ u ∥ C 2 ( V ) ≤ C ∥ Θ ∥ C n + 2 ( V ) < C ϵ . \lVert u\rVert_{C^{2}(V)}\leq C\lVert\Theta\rVert_{C^{n+2}(V)}<C\epsilon. ∥ u ∥ C 2 ( V ) ≤ C ∥ Θ ∥ C n + 2 ( V ) < C ϵ . where C C C V V V
Step 3: Gluing. Fix an open neighbourhood W W W K K K W ‾ ⊂ V \overline{W}\subset V W ⊂ V ρ : X → [ 0 , 1 ] \rho:X\to[0,1] ρ : X → [ 0 , 1 ] ρ ≡ 1 \rho\equiv1 ρ ≡ 1 W W W supp ρ ⋐ V \operatorname{supp}\rho\Subset V supp ρ ⋐ V ρ u \rho u ρ u X X X
ω ~ = ω ~ D − d d c ( ρ u ) . \widetilde{\omega}=\widetilde{\omega}_{D}-dd^{c}(\rho u). ω = ω D − d d c ( ρ u ) . This is a smooth, closed, real ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X [ ω ~ ] = [ ω ~ D ] [\widetilde{\omega}]=[\widetilde{\omega}_{D}] [ ω ] = [ ω D ] (5) to d d c ( ρ u ) dd^{c}(\rho u) d d c ( ρ u ) d d c u = ω ~ D − ω dd^{c}u=\widetilde{\omega}_{D}-\omega d d c u = ω D − ω V V V V V V
ω ~ = ( 1 − ρ ) ω ~ D + ρ ω − E , where E = u d d c ρ + d ρ ∧ d c u + d u ∧ d c ρ . \widetilde{\omega}=(1-\rho)\widetilde{\omega}_{D}+\rho\omega-E,\qquad\text{where }E=u\,dd^{c}\rho+d\rho\wedge d^{c}u+du\wedge d^{c}\rho. ω = ( 1 − ρ ) ω D + ρ ω − E , where E = u d d c ρ + d ρ ∧ d c u + d u ∧ d c ρ . Outside supp ρ \operatorname{supp}\rho supp ρ ω ~ = ω ~ D \widetilde{\omega}=\widetilde{\omega}_{D} ω = ω D W W W E = 0 E=0 E = 0 ω ~ = ω \widetilde{\omega}=\omega ω = ω C ρ C_{\rho} C ρ ∥ ρ ∥ C 2 \lVert\rho\rVert_{C^{2}} ∥ ρ ∥ C 2
∥ E ∥ C 0 ( V ) ≤ C ρ ∥ u ∥ C 2 ( V ) < C ρ C ϵ . \lVert E\rVert_{C^{0}(V)}\leq C_{\rho}\lVert u\rVert_{C^{2}(V)}<C_{\rho}C\epsilon. ∥ E ∥ C 0 ( V ) ≤ C ρ ∥ u ∥ C 2 ( V ) < C ρ C ϵ . Step 4: Global positivity and integrality. Let g g g α ≥ β \alpha\geq\beta α ≥ β α − β \alpha-\beta α − β c 0 > 0 c_{0}>0 c 0 > 0 α ≥ − c 0 ∥ α ∥ C 0 ( Q ) g \alpha\geq-c_{0}\lVert\alpha\rVert_{C^{0}(Q)}g α ≥ − c 0 ∥ α ∥ C 0 ( Q ) g Q Q Q ( 1 , 1 ) (1,1) ( 1 , 1 ) α \alpha α Q Q Q
Since ω \omega ω D D D m > 0 m>0 m > 0 ω ≥ 2 m g \omega\geq2mg ω ≥ 2 m g D D D ϵ > 0 \epsilon>0 ϵ > 0
c 0 ϵ ≤ m and c 0 C ρ C ϵ < m . c_{0}\epsilon\leq m\qquad\text{and}\qquad c_{0}C_{\rho}C\epsilon<m. c 0 ϵ ≤ m and c 0 C ρ C ϵ < m . This choice depends only on D D D V V V C C C ρ \rho ρ C ρ C_{\rho} C ρ
Positivity on D D D Since ∥ ω ~ D − ω ∥ C 0 ( D ) < ϵ \lVert\widetilde{\omega}_{D}-\omega\rVert_{C^{0}(D)}<\epsilon ∥ ω D − ω ∥ C 0 ( D ) < ϵ ω ~ D ≥ ω − c 0 ϵ g ≥ m g \widetilde{\omega}_{D}\geq\omega-c_{0}\epsilon g\geq mg ω D ≥ ω − c 0 ϵ g ≥ m g D D D V V V ( 1 − ρ ) ω ~ D + ρ ω (1-\rho)\widetilde{\omega}_{D}+\rho\omega ( 1 − ρ ) ω D + ρ ω ≥ m g \geq mg ≥ m g
ω ~ ≥ m g − c 0 ∥ E ∥ C 0 ( V ) g > ( m − c 0 C ρ C ϵ ) g ≥ 0. \widetilde{\omega}\geq mg-c_{0}\lVert E\rVert_{C^{0}(V)}g>(m-c_{0}C_{\rho}C\epsilon)g\geq0. ω ≥ m g − c 0 ∥ E ∥ C 0 ( V ) g > ( m − c 0 C ρ C ϵ ) g ≥ 0. On D ∖ supp ρ D\setminus\operatorname{supp}\rho D ∖ supp ρ ω ~ = ω ~ D ≥ m g \widetilde{\omega}=\widetilde{\omega}_{D}\geq mg ω = ω D ≥ m g ω ~ \widetilde{\omega} ω D = { Ψ ≤ R } D=\{\Psi\leq R\} D = { Ψ ≤ R }
Global positivity. The set D = { Ψ ≤ R } D=\{\Psi\leq R\} D = { Ψ ≤ R } O ( X ) O(X) O ( X ) Ψ \Psi Ψ [11 Sec. 5.1] . Since ω ~ \widetilde{\omega} ω D D D Ω \Omega Ω X X X [ Ω ] = [ ω ~ ] = [ ω ~ D ] [\Omega]=[\widetilde{\omega}]=[\widetilde{\omega}_{D}] [ Ω ] = [ ω ] = [ ω D ] Ω = ω ~ \Omega=\widetilde{\omega} Ω = ω D D D W ⊂ V ⊂ D W\subset V\subset D W ⊂ V ⊂ D Ω = ω ~ = ω = d d c h \Omega=\widetilde{\omega}=\omega=dd^{c}h Ω = ω = ω = d d c h W W W
Integrality. Since [ Ω ] = [ ω ~ D ] [\Omega]=[\widetilde{\omega}_{D}] [ Ω ] = [ ω D ] H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) N ≥ 1 N\geq1 N ≥ 1 N [ Ω ] N[\Omega] N [ Ω ] H X \mathcal{H}_{X} H X Ω H : = N Ω \Omega_{H}:=N\Omega Ω H := N Ω X X X W W W K K K
Ω H = N d d c h = d d c ( N h ) . \Omega_{H}=Ndd^{c}h=dd^{c}(Nh). Ω H = N d d c h = d d c ( N h ) . By [2 ] , K K K X X X
Case 2: the general case
We now only assume (1): H 2 ( X , Z ) = F ⊕ T H_{2}(X,\mathbb{Z})=F\oplus T H 2 ( X , Z ) = F ⊕ T F F F T T T pr : H 2 ( X , Z ) → F \operatorname{pr}:H_{2}(X,\mathbb{Z})\to F pr : H 2 ( X , Z ) → F T T T
Step 1: Setup. Let n n n V ⋐ V ′ ⋐ U V \Subset V' \Subset U V ⋐ V ′ ⋐ U Ψ \Psi Ψ R R R D = { Ψ ≤ R } D = \{\Psi\le R\} D = { Ψ ≤ R } W W W ρ \rho ρ ϵ > 0 \epsilon> 0 ϵ > 0 V V V D D D Lemma 4 (b) for V V V ρ \rho ρ
Choose a regular value R 1 > R R_1 > R R 1 > R Ψ \Psi Ψ X 1 = { Ψ < R 1 } X_1 = \{\Psi< R_1\} X 1 = { Ψ < R 1 } X 1 X_1 X 1 X ‾ 1 \overline{X}_1 X 1 H 2 ( X 1 , Z ) H_2(X_1,\mathbb{Z}) H 2 ( X 1 , Z )
Φ 1 : H 2 ( X 1 , Z ) ⟶ H 2 ( X , Z ) , K 1 = ker Φ 1 . \Phi_1:H_2(X_1,\mathbb{Z})\longrightarrow H_2(X,\mathbb{Z}),\qquad K_1=\ker\Phi_1. Φ 1 : H 2 ( X 1 , Z ) ⟶ H 2 ( X , Z ) , K 1 = ker Φ 1 . where Φ 1 \Phi_1 Φ 1
Step 2: Rational approximation on X 1 X_1 X 1 Σ ⊂ H 2 ( X 1 , Z ) \Sigma\subset H_2(X_1,\mathbb{Z}) Σ ⊂ H 2 ( X 1 , Z ) K 1 K_1 K 1 H 2 ( V ′ , Z ) → H 2 ( X 1 , Z ) H_2(V',\mathbb{Z}) \to H_2(X_1,\mathbb{Z}) H 2 ( V ′ , Z ) → H 2 ( X 1 , Z ) ω ∣ X 1 \omega|_{X_1} ω ∣ X 1 Σ \Sigma Σ H 2 ( V ′ , Z ) H_2(V',\mathbb{Z}) H 2 ( V ′ , Z ) ω = d d c h \omega= dd^c h ω = d d c h V ′ V' V ′ K 1 K_1 K 1 ∫ S ω = ⟨ [ ω ] , Φ 1 ( S ) ⟩ = 0 \int_S\omega= \langle[\omega],\Phi_1(S)\rangle= 0 ∫ S ω = ⟨[ ω ] , Φ 1 ( S )⟩ = 0 S ∈ K 1 S \in K_1 S ∈ K 1
Apply Lemma 6 to the Stein manifold X 1 X_1 X 1 ω ∣ X 1 \omega|_{X_1} ω ∣ X 1 Σ \Sigma Σ k = n + 2 k=n+2 k = n + 2 ϵ \epsilon ϵ D ⊂ X 1 D \subset X_1 D ⊂ X 1 ( 1 , 1 ) (1,1) ( 1 , 1 ) ω ~ D \widetilde{\omega}_D ω D X 1 X_1 X 1 D D D ∥ ω ~ D − ω ∥ C n + 2 ( D ) < ϵ \|\widetilde{\omega}_D-\omega\|_{C^{n+2}(D)} < \epsilon ∥ ω D − ω ∥ C n + 2 ( D ) < ϵ [ η ] : = [ ω ~ D ] ∈ H 2 ( X 1 , Q ) [\eta]:=[\widetilde{\omega}_D] \in H^2(X_1,\mathbb{Q}) [ η ] := [ ω D ] ∈ H 2 ( X 1 , Q ) Σ \Sigma Σ
Steps 1–3 of Case 1 and Positivity on D D D of Step 4 now apply verbatim, with X X X X 1 X_1 X 1 ω ~ D \widetilde{\omega}_D ω D ( 1 , 1 ) (1,1) ( 1 , 1 ) D D D H 2 ( V ′ , Z ) H_2(V',\mathbb{Z}) H 2 ( V ′ , Z ) ( 1 , 1 ) (1,1) ( 1 , 1 ) ω ~ \widetilde{\omega} ω X 1 X_1 X 1
[ ω ~ ] = [ η ] ∈ H 2 ( X 1 , R ) , ω ~ > 0 on D , ω ~ = ω on W . [\widetilde{\omega}]=[\eta]\in H^2(X_1,\mathbb{R}),\qquad\widetilde{\omega}>0\text{ on }D,\qquad\widetilde{\omega}=\omega\text{ on }W. [ ω ] = [ η ] ∈ H 2 ( X 1 , R ) , ω > 0 on D , ω = ω on W . Step 3: Extending the class to X X X I 1 = pr ( Φ 1 ( H 2 ( X 1 , Z ) ) ) ⊂ F I_1=\operatorname{pr}(\Phi_1(H_2(X_1,\mathbb{Z})))\subset\mathcal{F} I 1 = pr ( Φ 1 ( H 2 ( X 1 , Z ))) ⊂ F S 1 S_1 S 1 F \mathcal{F} F S 1 S_1 S 1 F \mathcal{F} F F = S 1 ⊕ B \mathcal{F}=S_1\oplus B F = S 1 ⊕ B
The prescribed periods. Define ϕ : Φ 1 ( H 2 ( X 1 , Z ) ) → Q \phi:\Phi_1(H_2(X_1,\mathbb{Z}))\to\mathbb{Q} ϕ : Φ 1 ( H 2 ( X 1 , Z )) → Q ϕ ( Φ 1 ( z ) ) = ⟨ [ η ] , z ⟩ \phi(\Phi_1(z))=\langle[\eta],z\rangle ϕ ( Φ 1 ( z )) = ⟨[ η ] , z ⟩ [ η ] [\eta] [ η ] K 1 K_1 K 1 Q \mathbb{Q} Q [ η ] [\eta] [ η ] Φ 1 ( z ) ∈ T \Phi_1(z)\in T Φ 1 ( z ) ∈ T n Φ 1 ( z ) = 0 n\Phi_1(z)=0 n Φ 1 ( z ) = 0 n ≥ 1 n\ge1 n ≥ 1 n z ∈ K 1 nz\in K_1 n z ∈ K 1 ⟨ [ η ] , z ⟩ = 0 \langle[\eta],z\rangle=0 ⟨[ η ] , z ⟩ = 0 ϕ \phi ϕ Φ 1 ( H 2 ( X 1 , Z ) ) ∩ T \Phi_1(H_2(X_1,\mathbb{Z}))\cap T Φ 1 ( H 2 ( X 1 , Z )) ∩ T ϕ = ϕ ‾ ∘ pr \phi=\overline{\phi}\circ\operatorname{pr} ϕ = ϕ ∘ pr ϕ ‾ : I 1 → Q \overline{\phi}:I_1\to\mathbb{Q} ϕ : I 1 → Q S 1 / I 1 S_1/I_1 S 1 / I 1 Q \mathbb{Q} Q ϕ ‾ \overline{\phi} ϕ ψ : S 1 → Q \psi:S_1\to\mathbb{Q} ψ : S 1 → Q ψ ( x ) = ϕ ‾ ( n x ) / n \psi(x)=\overline{\phi}(nx)/n ψ ( x ) = ϕ ( n x ) / n n x ∈ I 1 nx\in I_1 n x ∈ I 1 C 1 , … , C m C_1,\ldots,C_m C 1 , … , C m S 1 S_1 S 1 N ≥ 1 N\ge1 N ≥ 1 ψ ( C 1 ) , … , ψ ( C m ) \psi(C_1),\ldots,\psi(C_m) ψ ( C 1 ) , … , ψ ( C m ) N ψ ( S 1 ) ⊂ Z N\psi(S_1)\subset\mathbb{Z} N ψ ( S 1 ) ⊂ Z
Let c ψ ∈ H 2 ( X , R ) c_\psi\in H^2(X,\mathbb{R}) c ψ ∈ H 2 ( X , R ) S = S 1 S=S_1 S = S 1
(i) c ψ ∣ X 1 = [ η ] c_\psi|_{X_1}=[\eta] c ψ ∣ X 1 = [ η ] z ∈ H 2 ( X 1 , Z ) z\in H_2(X_1,\mathbb{Z}) z ∈ H 2 ( X 1 , Z ) pr ( Φ 1 ( z ) ) ∈ I 1 ⊂ S 1 \operatorname{pr}(\Phi_1(z))\in I_1\subset S_1 pr ( Φ 1 ( z )) ∈ I 1 ⊂ S 1
⟨ c ψ ∣ X 1 , z ⟩ = χ ψ ( Φ 1 ( z ) ) = ψ ( pr ( Φ 1 ( z ) ) ) = ϕ ‾ ( pr ( Φ 1 ( z ) ) ) = ϕ ( Φ 1 ( z ) ) = ⟨ [ η ] , z ⟩ . \langle c_\psi|_{X_1},z\rangle=\chi_\psi(\Phi_1(z))=\psi(\operatorname{pr}(\Phi_1(z)))=\overline{\phi}(\operatorname{pr}(\Phi_1(z)))=\phi(\Phi_1(z))=\langle[\eta],z\rangle. ⟨ c ψ ∣ X 1 , z ⟩ = χ ψ ( Φ 1 ( z )) = ψ ( pr ( Φ 1 ( z ))) = ϕ ( pr ( Φ 1 ( z ))) = ϕ ( Φ 1 ( z )) = ⟨[ η ] , z ⟩ . By Prop. 1(a) on X 1 X_1 X 1 H 2 ( X 1 , R ) H^2(X_1,\mathbb{R}) H 2 ( X 1 , R )
(ii) N c ψ ∈ H X Nc_\psi\in\mathcal{H}_X N c ψ ∈ H X
Step 4: Gluing and global positivity. By Lemma 3 (a), we may represent c ψ c_\psi c ψ ( 1 , 1 ) (1,1) ( 1 , 1 ) Ω ψ \Omega_\psi Ω ψ X X X [ Ω ψ ] [\Omega_\psi] [ Ω ψ ] [ ω ~ ] [\widetilde{\omega}] [ ω ] H 2 ( X 1 , R ) H^2(X_1,\mathbb{R}) H 2 ( X 1 , R ) ω ~ − Ω ψ \widetilde{\omega}-\Omega_\psi ω − Ω ψ X 1 X_1 X 1 Lemma 4 (a), there is a real function f ∈ C ∞ ( X 1 ) f\in C^\infty(X_1) f ∈ C ∞ ( X 1 ) ω ~ = Ω ψ + d d c f \widetilde{\omega}=\Omega_\psi+dd^c f ω = Ω ψ + d d c f X 1 X_1 X 1 ρ ~ : X → [ 0 , 1 ] \widetilde{\rho}:X\to[0,1] ρ : X → [ 0 , 1 ] ρ ~ = 1 \widetilde{\rho}=1 ρ = 1 D D D supp ρ ~ ⊂ X 1 \operatorname{supp}\widetilde{\rho}\subset X_1 supp ρ ⊂ X 1
ω ^ = Ω ψ + d d c ( ρ ~ f ) . \widehat{\omega}=\Omega_\psi+dd^c(\widetilde{\rho}f). ω = Ω ψ + d d c ( ρ f ) . This is a smooth, closed, real ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X [ ω ^ ] = c ψ [\widehat{\omega}]=c_\psi [ ω ] = c ψ ω ^ = ω ~ \widehat{\omega}=\widetilde{\omega} ω = ω D D D ω ^ \widehat{\omega} ω D D D ω ^ = ω \widehat{\omega}=\omega ω = ω W W W
The set D = { Ψ ≤ R } D=\{\Psi\le R\} D = { Ψ ≤ R } O ( X ) \mathcal{O}(X) O ( X ) ω ^ \hat{\omega} ω ^ D D D Lemma 3 (b) there is a Kähler form Ω \Omega Ω X X X [ Ω ] = [ ω ^ ] = c ψ [\Omega]=[\hat{\omega}]=c_{\psi} [ Ω ] = [ ω ^ ] = c ψ Ω = ω ^ \Omega=\hat{\omega} Ω = ω ^ D D D W ⊂ V ⊂ D W\subset V\subset D W ⊂ V ⊂ D ω ^ = ω = d d c h \hat{\omega}=\omega=dd^{c}h ω ^ = ω = d d c h W W W Ω = d d c h \Omega=dd^{c}h Ω = d d c h W W W
By (ii), Ω H : = N Ω \Omega_{H}:=N\Omega Ω H := N Ω X X X W W W K K K
Ω H = N d d c h = d d c ( N h ) . \Omega_{H}=N\,dd^{c}h=dd^{c}(Nh). Ω H = N d d c h = d d c ( N h ) . By [2 ] , K K K X X X
Trivial extension
In this subsection we prove a proposition showing that if the topology of X X X K K K H 2 ( X , Z ) H_{2}(X,\mathbb{Z}) H 2 ( X , Z ) d d c h dd^{c}h d d c h K K K K ^ O ( X ) = { x ∈ X : ∣ f ( x ) ∣ ≤ max K ∣ f ∣ for all f ∈ O ( X ) } \widehat{K}_{\mathcal{O}(X)}=\{x\in X:|f(x)|\le\max_{K}|f|\text{ for all }f\in\mathcal{O}(X)\} K O ( X ) = { x ∈ X : ∣ f ( x ) ∣ ≤ max K ∣ f ∣ for all f ∈ O ( X )}
Proposition 2. Let X X X U U U h h h K = { h ≤ 0 } K=\{h\le0\} K = { h ≤ 0 } d d c h dd^{c}h d d c h ω \omega ω X X X K ^ O ( X ) \widehat{K}_{\mathcal{O}(X)} K O ( X ) W W W H 2 ( W , Z ) → H 2 ( X , Z ) H_{2}(W,\mathbb{Z})\to H_{2}(X,\mathbb{Z}) H 2 ( W , Z ) → H 2 ( X , Z ) φ \varphi φ X X X d d c φ = d d c h dd^{c}\varphi=dd^{c}h d d c φ = d d c h K K K [2 ] , K K K
Proof. By [11 Theorem 5.1.6] there is a smooth strictly plurisubharmonic exhaustion function Ψ \Psi Ψ X X X Ψ < 0 \Psi<0 Ψ < 0 K ^ O ( X ) \widehat{K}_{\mathcal{O}(X)} K O ( X ) Ψ > 0 \Psi>0 Ψ > 0 X ∖ W X\setminus W X ∖ W R 0 R_{0} R 0 R 0 ′ R'_{0} R 0 ′ max K ^ O ( X ) Ψ < R 0 < R 0 ′ < 0 \max_{\widehat{K}_{\mathcal{O}(X)}}\Psi<R_{0}<R'_{0}<0 max K O ( X ) Ψ < R 0 < R 0 ′ < 0
X 0 = { Ψ ≤ R 0 } , U 0 = { Ψ < R 0 ′ } . X_{0}=\{\Psi\le R_{0}\},\qquad U_{0}=\{\Psi<R'_{0}\}. X 0 = { Ψ ≤ R 0 } , U 0 = { Ψ < R 0 ′ } . Then X 0 X_{0} X 0 K ⊂ K ^ O ( X ) ⊂ int X 0 K\subset\widehat{K}_{\mathcal{O}(X)}\subset\operatorname{int}X_{0} K ⊂ K O ( X ) ⊂ int X 0 U 0 ⊂ W U_{0}\subset W U 0 ⊂ W H 2 ( U 0 , Z ) → H 2 ( X , Z ) H_{2}(U_{0},\mathbb{Z})\to H_{2}(X,\mathbb{Z}) H 2 ( U 0 , Z ) → H 2 ( X , Z ) H 2 ( W , Z ) H_{2}(W,\mathbb{Z}) H 2 ( W , Z )
Hence, the homomorphism ⟨ [ ω ] , ⋅ ⟩ : H 2 ( X , Z ) → R \langle[\omega],\mathord{\cdot}\rangle:H_{2}(X,\mathbb{Z})\to\mathbb{R} ⟨[ ω ] , ⋅ ⟩ : H 2 ( X , Z ) → R H 2 ( U 0 , Z ) H_{2}(U_{0},\mathbb{Z}) H 2 ( U 0 , Z ) Lemma 5 , applied to U 0 U_{0} U 0 X 0 X_{0} X 0 ( 1 , 1 ) (1,1) ( 1 , 1 ) β \beta β X X X ⟨ [ ω ] , ⋅ ⟩ \langle[\omega],\mathord{\cdot}\rangle ⟨[ ω ] , ⋅ ⟩ X 0 X_{0} X 0 ω ^ = ω − β \hat{\omega}=\omega-\beta ω ^ = ω − β ω ^ \hat{\omega} ω ^ ω ^ \hat{\omega} ω ^ ω ^ = ω > 0 \hat{\omega}=\omega>0 ω ^ = ω > 0 X 0 X_{0} X 0
The set X 0 X_{0} X 0 O ( X ) \mathcal{O}(X) O ( X ) Ψ \Psi Ψ Lemma 3 (b) there is a Kähler form Ω \Omega Ω X X X [ Ω ] = [ ω ^ ] = 0 [\Omega]=[\hat{\omega}]=0 [ Ω ] = [ ω ^ ] = 0 Ω = ω ^ = ω \Omega=\hat{\omega}=\omega Ω = ω ^ = ω X 0 X_{0} X 0 Ω \Omega Ω Ω = d d c h \Omega=dd^{c}h Ω = d d c h K ⊂ int X 0 K\subset\operatorname{int}X_{0} K ⊂ int X 0 Lemma 4 (a) there is φ ∈ C ∞ ( X ) \varphi\in C^{\infty}(X) φ ∈ C ∞ ( X ) d d c φ = Ω dd^{c}\varphi=\Omega d d c φ = Ω φ \varphi φ Ω \Omega Ω
Proof of Theorem 2
We first outline the idea; t = log ∣ w ∣ t=\log|w| t = log ∣ w ∣ X = Y × C ∗ X=Y\times\mathbb{C}^{*} X = Y × C ∗ Λ \Lambda Λ Γ ⊂ Y \Gamma\subset Y Γ ⊂ Y 0 ∈ R 5 0\in\mathbb{R}^{5} 0 ∈ R 5 Γ \Gamma Γ 1 1 1 Λ \Lambda Λ 0 0 0 C 0 , C 1 C_{0},C_{1} C 0 , C 1 0 0 0 1 1 1 A = { 0 } × { 1 ≤ ∣ w ∣ ≤ e } A=\{0\}\times\{1\le|w|\le e\} A = { 0 } × { 1 ≤ ∣ w ∣ ≤ e } B ⊂ A B\subset A B ⊂ A Ω \Omega Ω Λ \Lambda Λ B B B A A A A − B A-B A − B Λ \Lambda Λ h h h Λ \Lambda Λ d d c h dd^{c}h d d c h C 0 ∪ C 1 C_{0}\cup C_{1} C 0 ∪ C 1 Γ \Gamma Γ C 0 ∪ C 1 C_{0}\cup C_{1} C 0 ∪ C 1 Lemma 13 ).
The manifold
Lemma 7. There exist an open set U M ⊂ R 5 U_M \subset\mathbb{R}^{5} U M ⊂ R 5 Y ⊂ C 5 Y \subset\mathbb{C}^{5} Y ⊂ C 5 Y ∩ R 5 = U M Y \cap\mathbb{R}^{5} = U_M Y ∩ R 5 = U M U M U_M U M Y Y Y
H 1 ( Y , Z ) ≅ Q , H 2 ( Y , Z ) = 0. H_{1}(Y,\mathbb{Z}) \cong\mathbb{Q}, \qquad H_{2}(Y,\mathbb{Z}) = 0. H 1 ( Y , Z ) ≅ Q , H 2 ( Y , Z ) = 0. Proof. Recall that the mapping cylinder of a continuous map f : A → B f\colon A \to B f : A → B
M f = ( ( A × [ 0 , 1 ] ) ⊔ B ) / ( ( a , 1 ) ∼ f ( a ) ) , M_f = ((A \times[0,1]) \sqcup B)/((a,1) \sim f(a)), M f = (( A × [ 0 , 1 ]) ⊔ B ) / (( a , 1 ) ∼ f ( a )) , obtained by attaching the end A × { 1 } A \times\{1\} A × { 1 } A × [ 0 , 1 ] A \times[0,1] A × [ 0 , 1 ] B B B f f f A ≅ A × { 0 } A \cong A \times\{0\} A ≅ A × { 0 } B B B B B B { a } × [ 0 , 1 ] \{a\} \times[0,1] { a } × [ 0 , 1 ] f ( a ) f(a) f ( a )
Given a sequence of maps A 1 → f 1 A 2 → f 2 ⋯ A_1 \xrightarrow{f_1} A_2 \xrightarrow{f_2} \cdots A 1 f 1 A 2 f 2 ⋯ mapping telescope is the union of the mapping cylinders M f k M_{f_k} M f k A k + 1 A_{k+1} A k + 1 M f k M_{f_k} M f k A k + 1 × { 0 } A_{k+1} \times\{0\} A k + 1 × { 0 } M f k + 1 M_{f_{k+1}} M f k + 1 [10 Ch. 0 and Sec. 3.F] .
Let S 1 , S 2 , … S_1,S_2,\ldots S 1 , S 2 , … f k : S k → S k + 1 f_k\colon S_k \to S_{k+1} f k : S k → S k + 1 z ↦ z k + 1 z \mapsto z^{k+1} z ↦ z k + 1 M k M_k M k M M M
S 1 → f 1 S 2 → f 2 S 3 → f 3 ⋯ . S_1 \xrightarrow{f_1} S_2 \xrightarrow{f_2} S_3 \xrightarrow{f_3} \cdots. S 1 f 1 S 2 f 2 S 3 f 3 ⋯ . This is obtained as the union of the M k M_k M k M k ∩ M k + 1 = S k + 1 M_k \cap M_{k+1} = S_{k+1} M k ∩ M k + 1 = S k + 1 S 1 S^1 S 1 f k f_k f k S 1 S^1 S 1 f k + 1 f_{k+1} f k + 1
Homology of M M M Put T 0 = S 1 T_0 = S_1 T 0 = S 1 T k = M 1 ∪ ⋯ ∪ M k T_k = M_1 \cup\cdots\cup M_k T k = M 1 ∪ ⋯ ∪ M k k ≥ 1 k \ge1 k ≥ 1 T k T_k T k S k + 1 S_{k+1} S k + 1 H 1 ( T k , Z ) ≅ Z H_1(T_k,\mathbb{Z}) \cong\mathbb{Z} H 1 ( T k , Z ) ≅ Z [ S k + 1 ] [S_{k+1}] [ S k + 1 ] H 2 ( T k , Z ) = 0 H_2(T_k,\mathbb{Z}) = 0 H 2 ( T k , Z ) = 0 T k ⊂ T k + 1 T_k \subset T_{k+1} T k ⊂ T k + 1 S k + 2 S_{k+2} S k + 2 S k + 1 S_{k+1} S k + 1 f k + 1 f_{k+1} f k + 1 k + 2 k+2 k + 2 H 1 ( S k + 1 , Z ) → H 1 ( S k + 2 , Z ) H_1(S_{k+1},\mathbb{Z}) \to H_1(S_{k+2},\mathbb{Z}) H 1 ( S k + 1 , Z ) → H 1 ( S k + 2 , Z ) k + 2 k+2 k + 2
Every compact subset of M M M T k T_k T k [10 Prop. 3.33]
H 1 ( M , Z ) ≅ l i m → ( Z → ⋅ 2 Z → ⋅ 3 Z → ⋅ 4 ⋯ ) ≅ Q , H 2 ( M , Z ) = 0. H_1(M,\mathbb{Z}) \cong\varinjlim(\mathbb{Z} \xrightarrow{\cdot2} \mathbb{Z} \xrightarrow{\cdot3} \mathbb{Z} \xrightarrow{\cdot4} \cdots) \cong\mathbb{Q}, \qquad H_2(M,\mathbb{Z}) = 0. H 1 ( M , Z ) ≅ lim ( Z ⋅ 2 Z ⋅ 3 Z ⋅ 4 ⋯ ) ≅ Q , H 2 ( M , Z ) = 0. To evaluate this direct limit, map the k k k Z \mathbb{Z} Z Q \mathbb{Q} Q 1 1 1 1 k ! \frac{1}{k!} k ! 1 k k k ( k + 1 ) (k+1) ( k + 1 ) k + 1 k+1 k + 1 ( k + 1 ) ⋅ 1 ( k + 1 ) ! = 1 k ! (k+1) \cdot\frac{1}{(k+1)!} = \frac{1}{k!} ( k + 1 ) ⋅ ( k + 1 )! 1 = k ! 1 Q \mathbb{Q} Q p q \frac{p}{q} q p p ( q − 1 ) ! q ! \frac{p(q-1)!}{q!} q ! p ( q − 1 )! q q q
Embedding M M M R 5 \mathbb{R}^{5} R 5 Triangulate M M M M k M_k M k M k + 1 M_{k+1} M k + 1 S k + 1 S_{k+1} S k + 1 f k f_k f k M k + 1 M_{k+1} M k + 1 S k + 1 S_{k+1} S k + 1 M M M 2 2 2 v 1 , v 2 , … v_1,v_2,\ldots v 1 , v 2 , … p j = ( j , j 2 , j 3 , j 4 , j 5 ) ∈ R 5 p_j=(j,j^2,j^3,j^4,j^5)\in\mathbb{R}^{5} p j = ( j , j 2 , j 3 , j 4 , j 5 ) ∈ R 5 p j p_j p j ( 1 , t , t 2 , t 3 , t 4 , t 5 ) (1,t,t^2,t^3,t^4,t^5) ( 1 , t , t 2 , t 3 , t 4 , t 5 ) Φ : M → R 5 \Phi\colon M \to\mathbb{R}^{5} Φ : M → R 5 v j v_j v j p j p_j p j j j j M M M v j 0 , … , v j k v_{j_0},\ldots,v_{j_k} v j 0 , … , v j k p j 0 , … , p j k p_{j_0},\ldots,p_{j_k} p j 0 , … , p j k M M M 6 6 6 Φ \Phi Φ v j v_j v j j ≤ J j \le J j ≤ J { x 1 > J } \{x_1 > J\} { x 1 > J } M M M R 5 \mathbb{R}^{5} R 5 M M M
Thickening. Since M M M R 5 \mathbb{R}^{5} R 5 U M ⊂ R 5 U_M \subset\mathbb{R}^{5} U M ⊂ R 5 [15 Ch. 3] . By [14 Lemma 6] , there exists a Stein domain Y ⊂ C 5 Y \subset\mathbb{C}^{5} Y ⊂ C 5 Y ∩ R 5 = U M Y \cap\mathbb{R}^{5}=U_M Y ∩ R 5 = U M U M U_M U M Y Y Y Y Y Y M M M
Remark 2. For k = 1 k=1 k = 1 G = Q G=\mathbb{Q} G = Q [14 Theorem 5] produces a Runge domain D ⊂ C 5 D \subset\mathbb{C}^{5} D ⊂ C 5 H 1 ( D , Z ) ≅ Q H_{1}(D,\mathbb{Z})\cong\mathbb{Q} H 1 ( D , Z ) ≅ Q H 2 ( D , Z ) H_{2}(D,\mathbb{Z}) H 2 ( D , Z ) M M M H 2 ( M , Z ) = 0 H_{2}(M,\mathbb{Z})=0 H 2 ( M , Z ) = 0
From now on Y Y Y U M U_M U M z ↦ z − q z\mapsto z-q z ↦ z − q C 5 \mathbb{C}^{5} C 5 q ∈ R 5 q\in\mathbb{R}^{5} q ∈ R 5 R 5 \mathbb{R}^{5} R 5 Y Y Y U M U_M U M U M U_M U M 0 ∈ U M 0\in U_M 0 ∈ U M e 1 , … , e 5 e_1,\ldots,e_5 e 1 , … , e 5 R 5 \mathbb{R}^{5} R 5 R x j = { x ∈ R 5 : x k = 0 for k ≠ j } \mathbb{R}_{x_j}=\{x\in\mathbb{R}^{5}:x_k=0\text{ for }k\ne j\} R x j = { x ∈ R 5 : x k = 0 for k = j } x j x_j x j z = x + i y z=x+iy z = x + i y x = Re z x=\operatorname{Re}z x = Re z y = Im z y=\operatorname{Im}z y = Im z R 5 \mathbb{R}^{5} R 5 C 5 \mathbb{C}^{5} C 5 r > 0 r>0 r > 0 B r = { z ∈ C 5 : ∣ z ∣ < r } B_r=\{z\in\mathbb{C}^{5}:|z|<r\} B r = { z ∈ C 5 : ∣ z ∣ < r } B r ∩ R 5 B_r\cap\mathbb{R}^{5} B r ∩ R 5 r r r R 5 \mathbb{R}^{5} R 5 X = Y × C ∗ X=Y\times\mathbb{C}^{*} X = Y × C ∗ ( z , w ) (z,w) ( z , w ) π : X → Y \pi:X\to Y π : X → Y t = log ∣ w ∣ t=\log|w| t = log ∣ w ∣
Lemma 8. X X X 6 6 6
H 2 ( X , Z ) ≅ Q , Hom ( H 2 ( X , Z ) , Z ) = 0 , H 2 ( X , R ) ≅ R , Im ( H 2 ( X , Z ) → H 2 ( X , R ) ) = 0. (8) H_{2}(X,\mathbb{Z})\cong\mathbb{Q},\qquad\operatorname{Hom}(H_{2}(X,\mathbb{Z}),\mathbb{Z})=0,\qquad H^{2}(X,\mathbb{R})\cong\mathbb{R},\qquad\operatorname{Im}\bigl(H^{2}(X,\mathbb{Z})\to H^{2}(X,\mathbb{R})\bigr)=0. \tag*{(8)} H 2 ( X , Z ) ≅ Q , Hom ( H 2 ( X , Z ) , Z ) = 0 , H 2 ( X , R ) ≅ R , Im ( H 2 ( X , Z ) → H 2 ( X , R ) ) = 0. (8) Moreover, $H^{2}(X,\mathbb{Z})\cong\operatorname{Ext}(\mathbb{Q},\mathbb{Z})\ne0.
Proof. X X X H ∗ ( C ∗ , Z ) ≅ H ∗ ( S 1 , Z ) H_{*}(\mathbb{C}^{*},\mathbb{Z})\cong H_{*}(S^{1},\mathbb{Z}) H ∗ ( C ∗ , Z ) ≅ H ∗ ( S 1 , Z )
H 1 ( X , Z ) ≅ ( H 1 ( Y ) ⊗ H 0 ( S 1 ) ) ⊕ ( H 0 ( Y ) ⊗ H 1 ( S 1 ) ) ≅ ( Q ⊗ Z ) ⊕ ( Z ⊗ Z ) ≅ Q ⊕ Z , H_{1}(X,\mathbb{Z})\cong\bigl(H_{1}(Y)\otimes H_{0}(S^{1})\bigr)\oplus\bigl(H_{0}(Y)\otimes H_{1}(S^{1})\bigr)\cong(\mathbb{Q}\otimes\mathbb{Z})\oplus(\mathbb{Z}\otimes\mathbb{Z})\cong\mathbb{Q}\oplus\mathbb{Z}, H 1 ( X , Z ) ≅ ( H 1 ( Y ) ⊗ H 0 ( S 1 ) ) ⊕ ( H 0 ( Y ) ⊗ H 1 ( S 1 ) ) ≅ ( Q ⊗ Z ) ⊕ ( Z ⊗ Z ) ≅ Q ⊕ Z , H 2 ( X , Z ) ≅ ( H 2 ( Y ) ⊗ H 0 ( S 1 ) ) ⊕ ( H 1 ( Y ) ⊗ H 1 ( S 1 ) ) ⊕ ( H 0 ( Y ) ⊗ H 2 ( S 1 ) ) ≅ 0 ⊕ ( Q ⊗ Z ) ⊕ 0 ≅ Q , H_{2}(X,\mathbb{Z})\cong\bigl(H_{2}(Y)\otimes H_{0}(S^{1})\bigr)\oplus\bigl(H_{1}(Y)\otimes H_{1}(S^{1})\bigr)\oplus\bigl(H_{0}(Y)\otimes H_{2}(S^{1})\bigr)\cong0\oplus(\mathbb{Q}\otimes\mathbb{Z})\oplus0\cong\mathbb{Q}, H 2 ( X , Z ) ≅ ( H 2 ( Y ) ⊗ H 0 ( S 1 ) ) ⊕ ( H 1 ( Y ) ⊗ H 1 ( S 1 ) ) ⊕ ( H 0 ( Y ) ⊗ H 2 ( S 1 ) ) ≅ 0 ⊕ ( Q ⊗ Z ) ⊕ 0 ≅ Q , where all homology groups are taken with coefficients in Z \mathbb{Z} Z
The remaining statements follow from Prop. 1. Indeed, Hom ( Q , Z ) = 0 \operatorname{Hom}(\mathbb{Q},\mathbb{Z})=0 Hom ( Q , Z ) = 0 f f f x ∈ Q x\in\mathbb{Q} x ∈ Q f ( x ) f(x) f ( x ) n ≥ 1 n\ge1 n ≥ 1 f ( x ) = 0 f(x)=0 f ( x ) = 0 Hom ( Q , R ) ≅ R \operatorname{Hom}(\mathbb{Q},\mathbb{R})\cong\mathbb{R} Hom ( Q , R ) ≅ R H 2 ( X , R ) ≅ R H^{2}(X,\mathbb{R})\cong\mathbb{R} H 2 ( X , R ) ≅ R H X = 0 H_X=0 H X = 0
H 2 ( X , Z ) ≅ Ext ( H 1 ( X , Z ) , Z ) = Ext ( Q ⊕ Z , Z ) = Ext ( Q , Z ) , H^{2}(X,\mathbb{Z})\cong\operatorname{Ext}(H_{1}(X,\mathbb{Z}),\mathbb{Z})=\operatorname{Ext}(\mathbb{Q}\oplus\mathbb{Z},\mathbb{Z})=\operatorname{Ext}(\mathbb{Q},\mathbb{Z}), H 2 ( X , Z ) ≅ Ext ( H 1 ( X , Z ) , Z ) = Ext ( Q ⊕ Z , Z ) = Ext ( Q , Z ) , which is nonzero (it is a Q \mathbb{Q} Q [6 Example 1] ). □
So X X X H 1 ( X , O ∗ ) ≅ H 2 ( X , Z ) ≠ 0 H^{1}(X,\mathcal{O}^{*})\cong H^{2}(X,\mathbb{Z})\ne0 H 1 ( X , O ∗ ) ≅ H 2 ( X , Z ) = 0 X X X
Let ℓ : H 1 ( Y , Z ) ≅ Q ↪ R \ell:H_{1}(Y,\mathbb{Z})\cong\mathbb{Q}\hookrightarrow\mathbb{R} ℓ : H 1 ( Y , Z ) ≅ Q ↪ R a ∈ H 1 ( Y , Z ) a\in H_{1}(Y,\mathbb{Z}) a ∈ H 1 ( Y , Z ) ℓ ( a ) = 1 \ell(a)=1 ℓ ( a ) = 1 H 0 ( Y , Z ) H_{0}(Y,\mathbb{Z}) H 0 ( Y , Z ) k = 1 k=1 k = 1 H 1 ( Y , C ) → Hom ( H 1 ( Y , Z ) , C ) H^{1}(Y,\mathbb{C})\to\operatorname{Hom}(H_{1}(Y,\mathbb{Z}),\mathbb{C}) H 1 ( Y , C ) → Hom ( H 1 ( Y , Z ) , C ) ℓ \ell ℓ C \mathbb{C} C H 1 ( Y , C ) H^{1}(Y,\mathbb{C}) H 1 ( Y , C ) Y Y Y H 1 ( Y , C ) H^{1}(Y,\mathbb{C}) H 1 ( Y , C ) [17 ] . Thus, there is a closed holomorphic 1-form β \beta β Y Y Y ℓ \ell ℓ
∫ c β = ℓ ( c ) for every c ∈ H 1 ( Y , Z ) . \int_c\beta=\ell(c)\qquad\text{for every }c\in H_{1}(Y,\mathbb{Z}). ∫ c β = ℓ ( c ) for every c ∈ H 1 ( Y , Z ) . Put α = Re β \alpha=\operatorname{Re}\beta α = Re β ℓ \ell ℓ ∫ c α = ℓ ( c ) \int_c\alpha=\ell(c) ∫ c α = ℓ ( c ) β = d G \beta=dG β = d G G G G α = d Re G \alpha=d\operatorname{Re}G α = d Re G X X X
μ = 1 2 ( d w w − π ∗ β ) . (9) \mu=\frac{1}{2}\left(\frac{dw}{w}-\pi^{*}\beta\right). \tag*{(9)} μ = 2 1 ( w d w − π ∗ β ) . (9) Locally μ = ∂ v \mu= \partial v μ = ∂ v v = log ∣ w ∣ − Re G ∘ π v = \log|w| - \operatorname{Re} G \circ\pi v = log ∣ w ∣ − Re G ∘ π 2 Re μ = d t − π ∗ α 2\operatorname{Re}\mu= \mathrm{d}t - \pi^{*}\alpha 2 Re μ = d t − π ∗ α ℓ : H 1 ( Y , Z ) → R \ell:H_{1}(Y,\mathbb{Z}) \to\mathbb{R} ℓ : H 1 ( Y , Z ) → R a ∈ H 1 ( Y , Z ) a \in H_{1}(Y,\mathbb{Z}) a ∈ H 1 ( Y , Z ) α \alpha α Y Y Y
ℓ ( a ) = 1 , ∫ c α = ℓ ( c ) for all c ∈ H 1 ( Y , Z ) . (10) \ell(a)=1,\qquad\int_{c}\alpha=\ell(c)\quad\text{for all }c\in H_{1}(Y,\mathbb{Z}). \tag*{(10)} ℓ ( a ) = 1 , ∫ c α = ℓ ( c ) for all c ∈ H 1 ( Y , Z ) . (10) The curve and the torus
Lemma 9. There exist a smooth immersion γ : R / 2 Z → U M ⊂ Y \gamma:\mathbb{R}/2\mathbb{Z}\to U_{M}\subset Y γ : R /2 Z → U M ⊂ Y
(a) γ ( s ) = γ ( s ′ ) \gamma(s)=\gamma(s') γ ( s ) = γ ( s ′ ) s ≠ s ′ s\ne s' s = s ′ { s , s ′ } = { 0 , 1 } \{s,s'\}=\{0,1\} { s , s ′ } = { 0 , 1 } γ ( 0 ) = γ ( 1 ) = 0 \gamma(0)=\gamma(1)=0 γ ( 0 ) = γ ( 1 ) = 0
(b) the loops γ ∣ [ 0 , 1 ] \gamma|_{[0,1]} γ ∣ [ 0 , 1 ] γ ∣ [ 1 , 2 ] \gamma|_{[1,2]} γ ∣ [ 1 , 2 ] a a a − a -a − a H 1 ( Y , Z ) H_{1}(Y,\mathbb{Z}) H 1 ( Y , Z )
(c) there is r Γ ∈ ( 0 , 1 2 ) r_{\Gamma}\in(0,\frac{1}{2}) r Γ ∈ ( 0 , 2 1 ) γ ( s ) = s e 1 \gamma(s)=se_{1} γ ( s ) = s e 1 ∣ s ∣ < r Γ |s|<r_{\Gamma} ∣ s ∣ < r Γ γ ( s ) = ( s − 1 ) e 2 \gamma(s)=(s-1)e_{2} γ ( s ) = ( s − 1 ) e 2 ∣ s − 1 ∣ < r Γ |s-1|<r_{\Gamma} ∣ s − 1∣ < r Γ ∣ γ ( s ) ∣ ≥ r Γ |\gamma(s)|\ge r_{\Gamma} ∣ γ ( s ) ∣ ≥ r Γ s ∈ R / 2 Z s\in\mathbb{R}/2\mathbb{Z} s ∈ R /2 Z
In particular, for every r ∈ ( 0 , r Γ ] r\in(0,r_{\Gamma}] r ∈ ( 0 , r Γ ] Γ = γ ( R / 2 Z ) \Gamma=\gamma(\mathbb{R}/2\mathbb{Z}) Γ = γ ( R /2 Z ) Γ ∩ B r = ( R x 1 ∪ R x 2 ) ∩ B r \Gamma\cap B_{r}=(\mathbb{R}x_{1}\cup\mathbb{R}x_{2})\cap B_{r} Γ ∩ B r = ( R x 1 ∪ R x 2 ) ∩ B r
Proof. Fix r Γ ∈ ( 0 , 1 2 ) r_{\Gamma}\in(0,\frac{1}{2}) r Γ ∈ ( 0 , 2 1 ) { x ∈ R 5 : ∣ x ∣ ≤ r Γ } ⊂ U M \{x\in\mathbb{R}^{5}:|x|\le r_{\Gamma}\}\subset U_{M} { x ∈ R 5 : ∣ x ∣ ≤ r Γ } ⊂ U M M M M U M U_{M} U M U M ↪ Y U_{M}\hookrightarrow Y U M ↪ Y U M U_{M} U M 0 0 0 a a a − a -a − a 0 0 0 U M U_{M} U M R 5 \mathbb{R}^{5} R 5 b 1 , b 2 : [ 0 , 1 ] → U M b_{1},b_{2}:[0,1]\to U_{M} b 1 , b 2 : [ 0 , 1 ] → U M 0 0 0 a a a − a -a − a
(1) b 1 ( s ) = s e 1 b_{1}(s)=se_{1} b 1 ( s ) = s e 1 b 1 ( 1 − s ) = − s e 2 b_{1}(1-s)=-se_{2} b 1 ( 1 − s ) = − s e 2 b 2 ( s ) = s e 2 b_{2}(s)=se_{2} b 2 ( s ) = s e 2 b 2 ( 1 − s ) = − s e 1 b_{2}(1-s)=-se_{1} b 2 ( 1 − s ) = − s e 1 0 ≤ s < r Γ 0\le s<r_{\Gamma} 0 ≤ s < r Γ ∣ b j ( s ) ∣ ≥ r Γ |b_{j}(s)|\ge r_{\Gamma} ∣ b j ( s ) ∣ ≥ r Γ s ∈ [ r Γ , 1 − r Γ ] s\in[r_{\Gamma},1-r_{\Gamma}] s ∈ [ r Γ , 1 − r Γ ]
(2) each b j b_{j} b j [ 0 , 1 ) [0,1) [ 0 , 1 ) b 1 ( s ) ≠ b 2 ( s ′ ) b_{1}(s)\ne b_{2}(s') b 1 ( s ) = b 2 ( s ′ ) s , s ′ ∈ { 0 , 1 } s,s'\in\{0,1\} s , s ′ ∈ { 0 , 1 }
Thus b 1 b_{1} b 1 0 0 0 + e 1 +e_{1} + e 1 e 2 e_{2} e 2 − e 2 -e_{2} − e 2 b 2 b_{2} b 2 + e 2 +e_{2} + e 2 e 1 e_{1} e 1 − e 1 -e_{1} − e 1 γ ( s ) = b 1 ( s ) \gamma(s)=b_{1}(s) γ ( s ) = b 1 ( s ) s ∈ [ 0 , 1 ] s\in[0,1] s ∈ [ 0 , 1 ] γ ( s ) = b 2 ( s − 1 ) \gamma(s)=b_{2}(s-1) γ ( s ) = b 2 ( s − 1 ) s ∈ [ 1 , 2 ] s\in[1,2] s ∈ [ 1 , 2 ] γ ( s ) = s e 1 \gamma(s)=se_{1} γ ( s ) = s e 1 ∣ s ∣ < r Γ |s|<r_{\Gamma} ∣ s ∣ < r Γ R / 2 Z \mathbb{R}/2\mathbb{Z} R /2 Z γ ( s ) = ( s − 1 ) e 2 \gamma(s)=(s-1)e_{2} γ ( s ) = ( s − 1 ) e 2 ∣ s − 1 ∣ < r Γ |s-1|<r_{\Gamma} ∣ s − 1∣ < r Γ ∣ γ ( s ) ∣ ≥ r Γ |\gamma(s)|\ge r_{\Gamma} ∣ γ ( s ) ∣ ≥ r Γ γ \gamma γ 0 < r ≤ r Γ 0<r\le r_{\Gamma} 0 < r ≤ r Γ γ ( s ) \gamma(s) γ ( s ) ∣ s ∣ < r Γ |s|<r_{\Gamma} ∣ s ∣ < r Γ ∣ s − 1 ∣ < r Γ |s-1|<r_{\Gamma} ∣ s − 1∣ < r Γ R x 1 ∪ R x 2 \mathbb{R}x_{1}\cup\mathbb{R}x_{2} R x 1 ∪ R x 2 Γ \Gamma Γ B r B_{r} B r □ \square □
By Lemma 9 (b) and (10) , ∫ R / 2 Z γ ∗ α = ℓ ( a ) + ℓ ( − a ) = 0 \int_{\mathbb{R}/2\mathbb{Z}}\gamma^{*}\alpha=\ell(a)+\ell(-a)=0 ∫ R /2 Z γ ∗ α = ℓ ( a ) + ℓ ( − a ) = 0 τ : R / 2 Z → R \tau:\mathbb{R}/2\mathbb{Z}\to\mathbb{R} τ : R /2 Z → R d τ = γ ∗ α \mathrm{d}\tau=\gamma^{*}\alpha d τ = γ ∗ α τ ( 0 ) = 0 \tau(0)=0 τ ( 0 ) = 0 Lemma 9 (b) and (10) , τ ( 1 ) = ∫ γ ∣ [ 0 , 1 ] α = ℓ ( a ) = 1 \tau(1)=\int_{\gamma|_{[0,1]}}\alpha=\ell(a)=1 τ ( 1 ) = ∫ γ ∣ [ 0 , 1 ] α = ℓ ( a ) = 1
Λ = { ( γ ( s ) , e τ ( s ) + i θ ) ∈ Y × C ∗ : s ∈ R / 2 Z , θ ∈ R / 2 π Z } . \Lambda=\left\{(\gamma(s),e^{\tau(s)+i\theta})\in Y\times\mathbb{C}^{*}:s\in\mathbb{R}/2\mathbb{Z},\ \theta\in\mathbb{R}/2\pi\mathbb{Z}\right\}. Λ = { ( γ ( s ) , e τ ( s ) + i θ ) ∈ Y × C ∗ : s ∈ R /2 Z , θ ∈ R /2 π Z } . By Lemma 9 (a) and τ ( 1 ) ≠ τ ( 0 ) \tau(1)\ne\tau(0) τ ( 1 ) = τ ( 0 ) ( s , θ ) ↦ ( γ ( s ) , e τ ( s ) + i θ ) (s,\theta)\mapsto(\gamma(s),e^{\tau(s)+i\theta}) ( s , θ ) ↦ ( γ ( s ) , e τ ( s ) + i θ ) Λ ⊂ X \Lambda\subset X Λ ⊂ X
Λ ∩ π − 1 ( 0 ) = C 0 ∪ C 1 , C 0 = { 0 } × { ∣ w ∣ = 1 } , C 1 = { 0 } × { ∣ w ∣ = e } . \Lambda\cap\pi^{-1}(0)=C_{0}\cup C_{1},\qquad C_{0}=\{0\}\times\{|w|=1\},\qquad C_{1}=\{0\}\times\{|w|=e\}. Λ ∩ π − 1 ( 0 ) = C 0 ∪ C 1 , C 0 = { 0 } × { ∣ w ∣ = 1 } , C 1 = { 0 } × { ∣ w ∣ = e } . A global Kähler form and a first local potential
Let r Γ r_{\Gamma} r Γ Lemma 9 , and fix r B ∈ ( 0 , r Γ ] r_{B}\in(0,r_{\Gamma}] r B ∈ ( 0 , r Γ ] B r B ⊂ Y B_{r_{B}}\subset Y B r B ⊂ Y Lemma 9 ,
Γ ∩ B r B = ( R x 1 ∪ R x 2 ) ∩ B r B . (11) \Gamma\cap B_{r_{B}}=(\mathbb{R}x_{1}\cup\mathbb{R}x_{2})\cap B_{r_{B}}. \tag*{(11)} Γ ∩ B r B = ( R x 1 ∪ R x 2 ) ∩ B r B . (11) Since B r B B_{r_{B}} B r B G G G B r B B_{r_{B}} B r B d G = β \mathrm{d}G=\beta d G = β G ( 0 ) = 0 G(0)=0 G ( 0 ) = 0 γ ( s ) ∈ B r B \gamma(s)\in B_{r_{B}} γ ( s ) ∈ B r B ∣ s ∣ < r B |s|<r_{B} ∣ s ∣ < r B ∣ s − 1 ∣ < r B |s-1|<r_{B} ∣ s − 1∣ < r B Lemma 9 (c), we have d ( Re G ∘ γ ) = γ ∗ α = d τ \mathrm{d}(\operatorname{Re}G\circ\gamma)=\gamma^{*}\alpha=\mathrm{d}\tau d ( Re G ∘ γ ) = γ ∗ α = d τ γ ( 0 ) = γ ( 1 ) = 0 \gamma(0)=\gamma(1)=0 γ ( 0 ) = γ ( 1 ) = 0 G ( 0 ) = τ ( 0 ) = 0 G(0)=\tau(0)=0 G ( 0 ) = τ ( 0 ) = 0 τ ( 1 ) = 1 \tau(1)=1 τ ( 1 ) = 1
Re G ( γ ( s ) ) = τ ( s ) for ∣ s ∣ < r B , Re G ( γ ( s ) ) = τ ( s ) − 1 for ∣ s − 1 ∣ < r B . (12) \operatorname{Re}G(\gamma(s))=\tau(s)\quad\text{for }|s|<r_{B},\qquad\operatorname{Re}G(\gamma(s))=\tau(s)-1\quad\text{for }|s-1|<r_{B}. \tag*{(12)} Re G ( γ ( s )) = τ ( s ) for ∣ s ∣ < r B , Re G ( γ ( s )) = τ ( s ) − 1 for ∣ s − 1∣ < r B . (12) For x = ( x 1 , … , x 5 ) ∈ R 5 x=(x_1,\ldots,x_5)\in\mathbb{R}^5 x = ( x 1 , … , x 5 ) ∈ R 5
F 0 ( x ) = ( x 1 x 2 ) 2 + x 3 2 + x 4 2 + x 5 2 . (13) F_0(x)=(x_1x_2)^2+x_3^2+x_4^2+x_5^2. \tag*{(13)} F 0 ( x ) = ( x 1 x 2 ) 2 + x 3 2 + x 4 2 + x 5 2 . (13) Then F 0 ≥ 0 F_0\geq0 F 0 ≥ 0 F 0 − 1 ( 0 ) = R x 1 ∪ R x 2 F_0^{-1}(0)=\mathbb{R}_{x_1}\cup\mathbb{R}_{x_2} F 0 − 1 ( 0 ) = R x 1 ∪ R x 2 (11) F 0 − 1 ( 0 ) ∩ B r B = Γ ∩ B r B F_0^{-1}(0)\cap B_{r_B}=\Gamma\cap B_{r_B} F 0 − 1 ( 0 ) ∩ B r B = Γ ∩ B r B ( x 1 , 0 , 0 , 0 , 0 ) (x_1,0,0,0,0) ( x 1 , 0 , 0 , 0 , 0 ) x 1 ≠ 0 x_1\neq0 x 1 = 0 F 0 F_0 F 0 diag ( 0 , 2 x 1 2 , 2 , 2 , 2 ) \operatorname{diag}(0,2x_1^2,2,2,2) diag ( 0 , 2 x 1 2 , 2 , 2 , 2 ) R x 1 = T Γ \mathbb{R}_{x_1}=T\Gamma R x 1 = T Γ R x 2 \mathbb{R}_{x_2} R x 2 x 1 x_1 x 1 x 2 x_2 x 2
Lemma 10. There are a function F ∈ C c ∞ ( R 5 ) F\in C^\infty_c(\mathbb{R}^5) F ∈ C c ∞ ( R 5 ) F ≥ 0 F\geq0 F ≥ 0 N N N Γ \Gamma Γ R 5 \mathbb{R}^5 R 5 r 3 > 0 r_3>0 r 3 > 0
(a) F − 1 ( 0 ) ∩ N = Γ F^{-1}(0)\cap N=\Gamma F − 1 ( 0 ) ∩ N = Γ
(b) ker Hess z F = T z Γ \ker\operatorname{Hess}_z F=T_z\Gamma ker Hess z F = T z Γ z ∈ Γ ∖ { 0 } z\in\Gamma\setminus\{0\} z ∈ Γ ∖ { 0 }
(c) F = F 0 F=F_0 F = F 0 B r 3 ∩ R 5 B_{r_3}\cap\mathbb{R}^5 B r 3 ∩ R 5
Proof. We first construct F F F 0 < r 1 < r 2 < r B 0<r_1<r_2<r_B 0 < r 1 < r 2 < r B χ 0 ∈ C ∞ ( B r B ∩ R 5 , [ 0 , 1 ] ) \chi_0\in C^\infty(B_{r_B}\cap\mathbb{R}^5,[0,1]) χ 0 ∈ C ∞ ( B r B ∩ R 5 , [ 0 , 1 ]) χ 0 = 1 \chi_0=1 χ 0 = 1 B r 2 ∩ R 5 B_{r_2}\cap\mathbb{R}^5 B r 2 ∩ R 5 Γ ′ = Γ ∖ B r 1 \Gamma'=\Gamma\setminus B_{r_1} Γ ′ = Γ ∖ B r 1 δ = dist ( ⋅ , Γ ′ ) \delta=\operatorname{dist}(\mathord\cdot,\Gamma') δ = dist ( ⋅ , Γ ′ ) 0 0 0 Γ \Gamma Γ Γ ∖ B r 2 \Gamma\setminus B_{r_2} Γ ∖ B r 2 Γ ′ \Gamma' Γ ′ δ 2 \delta^2 δ 2 O O O Γ ∖ B r 2 \Gamma\setminus B_{r_2} Γ ∖ B r 2 Γ ′ \Gamma' Γ ′ Hess z δ 2 \operatorname{Hess}_z\delta^2 Hess z δ 2 Γ \Gamma Γ z z z T z Γ T_z\Gamma T z Γ N 1 = O ∪ ( B r 2 ∩ R 5 ) N_1=O\cup(B_{r_2}\cap\mathbb{R}^5) N 1 = O ∪ ( B r 2 ∩ R 5 ) Γ \Gamma Γ
F l o c = χ 0 F 0 + ( 1 − χ 0 ) δ 2 . F_{\mathrm{loc}}=\chi_0F_0+(1-\chi_0)\delta^2. F loc = χ 0 F 0 + ( 1 − χ 0 ) δ 2 . This is clearly smooth and nonnegative. Further, F l o c − 1 ( 0 ) = Γ F_{\mathrm{loc}}^{-1}(0)=\Gamma F loc − 1 ( 0 ) = Γ N 1 N_1 N 1
Let z ∈ Γ ∖ { 0 } z\in\Gamma\setminus\{0\} z ∈ Γ ∖ { 0 } z ∈ B r 2 z\in B_{r_2} z ∈ B r 2 F l o c = F 0 F_{\mathrm{loc}}=F_0 F loc = F 0 z z z ker Hess z F l o c = T z Γ \ker\operatorname{Hess}_zF_{\mathrm{loc}}=T_z\Gamma ker Hess z F loc = T z Γ F 0 F_0 F 0 z ∉ B r B z\notin B_{r_B} z ∈ / B r B z ∉ supp χ 0 z\notin\operatorname{supp}\chi_0 z ∈ / supp χ 0 F l o c = δ 2 F_{\mathrm{loc}}=\delta^2 F loc = δ 2 z z z δ 2 \delta^2 δ 2 z ∈ B r B ∖ B r 2 z\in B_{r_B}\setminus B_{r_2} z ∈ B r B ∖ B r 2 F 0 F_0 F 0 δ 2 \delta^2 δ 2 z z z z z z (11) and z ∈ Γ ′ z\in\Gamma' z ∈ Γ ′ z z z g = F 0 − δ 2 g=F_0-\delta^2 g = F 0 − δ 2 g ( z ) = 0 g(z)=0 g ( z ) = 0 d g ( z ) = 0 dg(z)=0 d g ( z ) = 0 χ 0 \chi_0 χ 0 Hess z ( χ 0 g ) \operatorname{Hess}_z(\chi_0 g) Hess z ( χ 0 g ) F l o c = δ 2 + χ 0 ( F 0 − δ 2 ) F_{\mathrm{loc}}=\delta^2+\chi_0(F_0-\delta^2) F loc = δ 2 + χ 0 ( F 0 − δ 2 ) d F l o c ( z ) = 0 dF_{\mathrm{loc}}(z)=0 d F loc ( z ) = 0
Hess z F l o c = χ 0 ( z ) Hess z F 0 + ( 1 − χ 0 ( z ) ) Hess z δ 2 . \operatorname{Hess}_z F_{\mathrm{loc}}=\chi_0(z)\operatorname{Hess}_z F_0+(1-\chi_0(z))\operatorname{Hess}_z\delta^2. Hess z F loc = χ 0 ( z ) Hess z F 0 + ( 1 − χ 0 ( z )) Hess z δ 2 . Both Hessians are positive semidefinite with kernel T z Γ T_z\Gamma T z Γ 1 1 1 ker Hess z F l o c = T z Γ \ker\operatorname{Hess}_z F_{\mathrm{loc}}=T_z\Gamma ker Hess z F loc = T z Γ
Finally, choose an open neighbourhood N ⋐ N 1 N\Subset N_1 N ⋐ N 1 Γ \Gamma Γ θ ∈ C ∞ ( N 1 , [ 0 , 1 ] ) \theta\in C^\infty(N_1,[0,1]) θ ∈ C ∞ ( N 1 , [ 0 , 1 ]) θ = 1 \theta=1 θ = 1 N N N F = θ F l o c F=\theta F_{\mathrm{loc}} F = θ F loc F = F l o c F=F_{\mathrm{loc}} F = F loc N N N r 3 ≤ r 2 r_3\le r_2 r 3 ≤ r 2 B r 3 ∩ R 5 ⊂ N B_{r_3}\cap\mathbb{R}^5\subset N B r 3 ∩ R 5 ⊂ N B r 3 ∩ R 5 B_{r_3}\cap\mathbb{R}^5 B r 3 ∩ R 5 θ = χ 0 = 1 \theta=\chi_0=1 θ = χ 0 = 1 F = F 0 F=F_0 F = F 0
From now on F F F N N N r 3 r_3 r 3 σ ≥ 1 \sigma\ge1 σ ≥ 1 2 σ > sup ∥ Hess F ∥ 2\sigma>\sup\lVert\operatorname{Hess}F\rVert 2 σ > sup ∥ Hess F ∥ π : X → Y \pi:X\to Y π : X → Y z = x + i y z=x+iy z = x + i y
Lemma 11. Define
ρ ( z , w ) = σ ∣ Im z ∣ 2 + F ( Re z ) , ( z , w ) ∈ X = Y × C ∗ . (14) \rho(z,w)=\sigma\lvert\operatorname{Im}z\rvert^2+F(\operatorname{Re}z),\qquad(z,w)\in X=Y\times\mathbb{C}^*. \tag*{(14)} ρ ( z , w ) = σ ∣ Im z ∣ 2 + F ( Re z ) , ( z , w ) ∈ X = Y × C ∗ . (14) and let V = { z ∈ Y : Re z ∈ N } V=\{z\in Y:\operatorname{Re}z\in N\} V = { z ∈ Y : Re z ∈ N } Γ \Gamma Γ Y Y Y ρ \rho ρ X X X π \pi π
(a) For q ∈ X q\in X q ∈ X ξ ∈ T q X \xi\in T_qX ξ ∈ T q X d d c ρ ( ξ , J ξ ) > 0 dd^c\rho(\xi,J\xi)>0 d d c ρ ( ξ , J ξ ) > 0 π ∗ ξ ≠ 0 \pi_*\xi\ne0 π ∗ ξ = 0 d d c ρ ( ξ , ⋅ ) = 0 dd^c\rho(\xi,\mathord\cdot)=0 d d c ρ ( ξ , ⋅ ) = 0 π ∗ ξ = 0 \pi_*\xi=0 π ∗ ξ = 0 ρ \rho ρ d d c ρ dd^c\rho d d c ρ
(b) { ρ = 0 } ∩ π − 1 ( V ) = π − 1 ( Γ ) \{\rho=0\}\cap\pi^{-1}(V)=\pi^{-1}(\Gamma) { ρ = 0 } ∩ π − 1 ( V ) = π − 1 ( Γ ) d ρ = 0 d\rho=0 d ρ = 0 π − 1 ( Γ ) \pi^{-1}(\Gamma) π − 1 ( Γ )
(c) For q ∈ π − 1 ( Γ ∖ { 0 } ) q\in\pi^{-1}(\Gamma\setminus\{0\}) q ∈ π − 1 ( Γ ∖ { 0 }) Hess q ρ \operatorname{Hess}_q\rho Hess q ρ { ξ ∈ T q X : π ∗ ξ ∈ T π ( q ) Γ } \{\xi\in T_qX:\pi_*\xi\in T_{\pi(q)}\Gamma\} { ξ ∈ T q X : π ∗ ξ ∈ T π ( q ) Γ }
(d) ρ ( z , w ) = σ ∣ y ∣ 2 + F 0 ( x ) \rho(z,w)=\sigma\lvert y\rvert^2+F_0(x) ρ ( z , w ) = σ ∣ y ∣ 2 + F 0 ( x ) ( z , w ) ∈ X (z,w)\in X ( z , w ) ∈ X z = x + i y z=x+iy z = x + i y ∣ x ∣ < r 3 \lvert x\rvert<r_3 ∣ x ∣ < r 3
Proof. Put ρ Y ( z ) = σ ∣ Im z ∣ 2 + F ( Re z ) \rho_Y(z)=\sigma\lvert\operatorname{Im}z\rvert^2+F(\operatorname{Re}z) ρ Y ( z ) = σ ∣ Im z ∣ 2 + F ( Re z ) z ∈ C 5 z\in\mathbb{C}^5 z ∈ C 5 ρ = ρ Y ∘ π \rho=\rho_Y\circ\pi ρ = ρ Y ∘ π ρ ≥ 0 \rho\ge0 ρ ≥ 0 π \pi π ρ Y \rho_Y ρ Y 1 4 ( 2 σ − sup ∥ Hess F ∥ ) ∣ ξ ∣ 2 > 0 \frac{1}{4}(2\sigma-\sup\lVert\operatorname{Hess}F\rVert)\lvert\xi\rvert^2>0 4 1 ( 2 σ − sup ∥ Hess F ∥) ∣ ξ ∣ 2 > 0 ρ Y \rho_Y ρ Y C 5 \mathbb{C}^5 C 5
(a) Since π \pi π d d c ρ = π ∗ d d c ρ Y dd^c\rho=\pi^*dd^c\rho_Y d d c ρ = π ∗ d d c ρ Y π ∗ J = J π ∗ \pi_*J=J\pi_* π ∗ J = J π ∗
d d c ρ ( ξ , J ξ ) = d d c ρ Y ( π ∗ ξ , J π ∗ ξ ) , dd^c\rho(\xi,J\xi)=dd^c\rho_Y(\pi_*\xi,J\pi_*\xi), d d c ρ ( ξ , J ξ ) = d d c ρ Y ( π ∗ ξ , J π ∗ ξ ) , which is positive if π ∗ ξ ≠ 0 \pi_*\xi\ne0 π ∗ ξ = 0 d d c ρ ( ξ , ⋅ ) = 0 dd^c\rho(\xi,\mathord\cdot)=0 d d c ρ ( ξ , ⋅ ) = 0 π ∗ ξ = 0 \pi_*\xi=0 π ∗ ξ = 0 ρ \rho ρ
(b) We have ρ ( z , w ) = 0 \rho(z,w)=0 ρ ( z , w ) = 0 Im z = 0 \operatorname{Im}z=0 Im z = 0 F ( Re z ) = 0 F(\operatorname{Re}z)=0 F ( Re z ) = 0 z ∈ V z\in V z ∈ V z ∈ R 5 z\in\mathbb{R}^5 z ∈ R 5 z ∈ F − 1 ( 0 ) ∩ N = Γ z\in F^{-1}(0)\cap N=\Gamma z ∈ F − 1 ( 0 ) ∩ N = Γ ρ ≥ 0 \rho\ge0 ρ ≥ 0 π − 1 ( Γ ) \pi^{-1}(\Gamma) π − 1 ( Γ ) d ρ = 0 d\rho=0 d ρ = 0
(c) Let q = ( z , w ) q=(z,w) q = ( z , w ) z ∈ Γ ∖ { 0 } z\in\Gamma\setminus\{0\} z ∈ Γ ∖ { 0 } ( z , w ) (z,w) ( z , w ) π \pi π Hess q ρ ( ξ , ξ ) = Hess z ρ Y ( π ∗ ξ , π ∗ ξ ) \operatorname{Hess}_q\rho(\xi,\xi)=\operatorname{Hess}_z\rho_Y(\pi_*\xi,\pi_*\xi) Hess q ρ ( ξ , ξ ) = Hess z ρ Y ( π ∗ ξ , π ∗ ξ ) z ∈ R 5 z\in\mathbb{R}^5 z ∈ R 5 ρ Y \rho_Y ρ Y z z z 2 σ ∣ ⋅ ∣ 2 2\sigma\lvert\mathord\cdot\rvert^2 2 σ ∣ ⋅ ∣ 2 i R 5 i\mathbb{R}^5 i R 5 Hess z F \operatorname{Hess}_zF Hess z F R 5 \mathbb{R}^5 R 5 T z Γ T_z\Gamma T z Γ
(d) follows from Lemma 10(c). □
Define
ω ^ = d d c ρ + i μ ∧ μ ‾ . \widehat{\omega}=dd^c\rho+i\,\mu\wedge\overline{\mu}. ω = d d c ρ + i μ ∧ μ . where μ \mu μ ω ^ \widehat{\omega} ω μ \mu μ X X X i μ ∧ μ ‾ i\,\mu\wedge\overline{\mu} i μ ∧ μ μ = d w 2 w \mu=\frac{dw}{2w} μ = 2 w d w d d c ρ dd^c\rho d d c ρ
Let W ⊂ X W\subset X W ⊂ X Λ \Lambda Λ Λ \Lambda Λ 2 Re μ = d t − π ∗ α 2\operatorname{Re}\mu=dt-\pi^*\alpha 2 Re μ = d t − π ∗ α { θ ↦ ( γ ( 0 ) , e i θ ) } \{\theta\mapsto(\gamma(0),e^{i\theta})\} { θ ↦ ( γ ( 0 ) , e i θ )} d τ = γ ∗ α d\tau=\gamma^*\alpha d τ = γ ∗ α s ↦ ( γ ( s ) , e τ ( s ) ) s\mapsto(\gamma(s),e^{\tau(s)}) s ↦ ( γ ( s ) , e τ ( s ) ) W W W v v v W W W d v = μ dv=\mu d v = μ d v = d t − π ∗ α dv=dt-\pi^*\alpha d v = d t − π ∗ α v ∣ Λ = 0 v|_\Lambda=0 v ∣ Λ = 0 ∂ ∂ ‾ v = ∂ ‾ ∂ v = 0 \partial\overline{\partial}v=\overline{\partial}\partial v=0 ∂ ∂ v = ∂ ∂ v = 0 v v v Λ \Lambda Λ ( s , θ ) ↦ ( γ ( s ) , e τ ( s ) + i θ ) (s,\theta)\mapsto(\gamma(s),e^{\tau(s)+i\theta}) ( s , θ ) ↦ ( γ ( s ) , e τ ( s ) + i θ )
form d v = d t − π ∗ α dv = dt - \pi^*\alpha d v = d t − π ∗ α d τ − γ ∗ α = 0 d\tau- \gamma^*\alpha= 0 d τ − γ ∗ α = 0 v ∣ Λ v|_\Lambda v ∣ Λ Λ \Lambda Λ v v v v ∣ Λ ≡ 0 v|_\Lambda\equiv0 v ∣ Λ ≡ 0
Put
h 0 = ρ + v 2 on W , d d c h 0 = d d c ρ + i ∂ v ∧ ∂ ‾ v = ω ^ , h_0 = \rho+ v^2 \quad\text{on } W,\qquad dd^c h_0 = dd^c \rho+ i\,\partial v \wedge\overline{\partial}v = \widehat{\omega}, h 0 = ρ + v 2 on W , d d c h 0 = d d c ρ + i ∂ v ∧ ∂ v = ω , where the second identity follows from (6) .
Lemma 12. On a neighbourhood of Λ \Lambda Λ h 0 h_0 h 0
(a) h 0 ≥ 0 h_0 \ge0 h 0 ≥ 0 h 0 ∣ Λ = 0 h_0|_\Lambda= 0 h 0 ∣ Λ = 0 d h 0 ∣ Λ = 0 dh_0|_\Lambda= 0 d h 0 ∣ Λ = 0
(b) For q ∈ Λ ∖ ( C 0 ∪ C 1 ) q \in\Lambda\setminus(C_0 \cup C_1) q ∈ Λ ∖ ( C 0 ∪ C 1 ) Hess q h 0 \operatorname{Hess}_q h_0 Hess q h 0 T q X / T q Λ T_qX/T_q\Lambda T q X / T q Λ
(c) On π − 1 ( B r B ) \pi^{-1}(\mathbb{B}_{r_B}) π − 1 ( B r B ) z = x + i y z = x + iy z = x + i y
r = t − Re G ( z ) = log ∣ w ∣ − Re G ( z ) , r = t - \operatorname{Re}G(z) = \log|w| - \operatorname{Re}G(z), r = t − Re G ( z ) = log ∣ w ∣ − Re G ( z ) , where G G G (12) . Near C 0 C_0 C 0 ρ = σ ∣ y ∣ 2 + F 0 ( x ) \rho= \sigma|y|^2 + F_0(x) ρ = σ ∣ y ∣ 2 + F 0 ( x ) Lemma 11 (d), and
Λ = { x 2 = x 3 = x 4 = x 5 = 0 , y = 0 , r = 0 } , h 0 = ρ + r 2 . \Lambda= \{x_2 = x_3 = x_4 = x_5 = 0,\ y = 0,\ r = 0\},\qquad h_0 = \rho+ r^2. Λ = { x 2 = x 3 = x 4 = x 5 = 0 , y = 0 , r = 0 } , h 0 = ρ + r 2 . At a point q ∈ Λ q \in\Lambda q ∈ Λ C 0 C_0 C 0 x 1 = x 1 ( q ) x_1 = x_1(q) x 1 = x 1 ( q )
Hess q h 0 = 2 σ ∑ j = 1 5 d y j 2 + 2 x 1 2 d x 2 2 + 2 ∑ j = 3 5 d x j 2 + 2 d r 2 , (15) \operatorname{Hess}_q h_0 = 2\sigma\sum_{j=1}^{5}dy_j^2 + 2x_1^2dx_2^2 + 2\sum_{j=3}^{5}dx_j^2 + 2dr^2, \tag*{(15)} Hess q h 0 = 2 σ j = 1 ∑ 5 d y j 2 + 2 x 1 2 d x 2 2 + 2 j = 3 ∑ 5 d x j 2 + 2 d r 2 , (15) where d y j 2 = d y j ⊗ d y j dy_j^2 = dy_j \otimes dy_j d y j 2 = d y j ⊗ d y j q q q d y 1 , … , d y 5 , d x 2 , … , d x 5 , d r dy_1,\ldots,dy_5,dx_2,\ldots,dx_5,dr d y 1 , … , d y 5 , d x 2 , … , d x 5 , d r q q q T q Λ T_q\Lambda T q Λ C 1 C_1 C 1 z 1 ↔ z 2 z_1 \leftrightarrow z_2 z 1 ↔ z 2 G G G G + 1 G+1 G + 1 r r r r − 1 r-1 r − 1
Proof. (a) holds because both summands of h 0 h_0 h 0 Λ \Lambda Λ q ∈ Λ q \in\Lambda q ∈ Λ Hess q h 0 = Hess q ρ + 2 d v ⊗ d v \operatorname{Hess}_q h_0 = \operatorname{Hess}_q \rho+ 2\,dv \otimes dv Hess q h 0 = Hess q ρ + 2 d v ⊗ d v ξ \xi ξ Hess q h 0 \operatorname{Hess}_q h_0 Hess q h 0 q ∉ C 0 ∪ C 1 q \notin C_0 \cup C_1 q ∈ / C 0 ∪ C 1 π ( q ) ∈ Γ ∖ { 0 } \pi(q) \in\Gamma\setminus\{0\} π ( q ) ∈ Γ ∖ { 0 } Lemma 11 (c) this means that π ∗ ξ ∈ T π ( q ) Γ \pi_*\xi\in T_{\pi(q)}\Gamma π ∗ ξ ∈ T π ( q ) Γ d v ( ξ ) = 0 dv(\xi) = 0 d v ( ξ ) = 0 ξ \xi ξ π ∗ ξ ∈ T π ( q ) Γ \pi_*\xi\in T_{\pi(q)}\Gamma π ∗ ξ ∈ T π ( q ) Γ E E E ker π ∗ \ker\pi_* ker π ∗ π ∗ α \pi^*\alpha π ∗ α d v = d t dv = dt d v = d t ker π ∗ \ker\pi_* ker π ∗ d v ∣ E ≠ 0 dv|_E \ne0 d v ∣ E = 0 Hess q h 0 \operatorname{Hess}_q h_0 Hess q h 0 E ∩ ker d v E \cap\ker dv E ∩ ker d v T q Λ T_q\Lambda T q Λ v ∣ Λ = 0 v|_\Lambda= 0 v ∣ Λ = 0 T q Λ T_q\Lambda T q Λ
For (c), the points of Λ \Lambda Λ s ≈ 0 s \approx0 s ≈ 0 ( γ ( s ) , e τ ( s ) + i θ ) (\gamma(s),e^{\tau(s)+i\theta}) ( γ ( s ) , e τ ( s ) + i θ ) γ ( s ) = ( s , 0 , 0 , 0 , 0 ) \gamma(s) = (s,0,0,0,0) γ ( s ) = ( s , 0 , 0 , 0 , 0 ) (12) , r = τ ( s ) − Re G ( γ ( s ) ) = 0 r = \tau(s) - \operatorname{Re}G(\gamma(s)) = 0 r = τ ( s ) − Re G ( γ ( s )) = 0 s ≈ 1 s \approx1 s ≈ 1 r = 1 r = 1 r = 1 Λ \Lambda Λ C 0 C_0 C 0 v = r v = r v = r d t − π ∗ α dt - \pi^*\alpha d t − π ∗ α d Re G = α d\operatorname{Re}G = \alpha d Re G = α B r B \mathbb{B}_{r_B} B r B Λ \Lambda Λ h 0 = ρ + r 2 h_0 = \rho+ r^2 h 0 = ρ + r 2 ρ = σ ∣ y ∣ 2 + F 0 ( x ) \rho= \sigma|y|^2 + F_0(x) ρ = σ ∣ y ∣ 2 + F 0 ( x ) Lemma 11 (d). Since r ( q ) = 0 r(q) = 0 r ( q ) = 0 F 0 F_0 F 0 2 x 1 2 d x 2 2 + 2 ∑ j ≥ 3 d x j 2 2x_1^2dx_2^2 + 2\sum_{j\ge3}dx_j^2 2 x 1 2 d x 2 2 + 2 ∑ j ≥ 3 d x j 2 ( x 1 , 0 , 0 , 0 , 0 ) (x_1,0,0,0,0) ( x 1 , 0 , 0 , 0 , 0 ) (15) . The covectors d y j dy_j d y j d x 2 , … , d x 5 dx_2,\ldots,dx_5 d x 2 , … , d x 5 d r = d t − d Re G dr = dt - d\operatorname{Re}G d r = d t − d Re G d t dt d t d x , d y dx,dy d x , d y T q Λ T_q\Lambda T q Λ ∂ x 1 + ∂ x 1 Re G ∂ t \partial_{x_1} + \partial_{x_1}\operatorname{Re}G\,\partial_t ∂ x 1 + ∂ x 1 Re G ∂ t C 1 C_1 C 1 (12) for s s s 1 1 1 □ \square □
Thus the only degeneracy of h 0 h_0 h 0 ∂ x 2 \partial_{x_2} ∂ x 2 C 0 C_0 C 0 ∂ x 1 \partial_{x_1} ∂ x 1 C 1 C_1 C 1 { h 0 = 0 } \{h_0 = 0\} { h 0 = 0 } Λ \Lambda Λ { z 1 = z 3 = z 4 = z 5 = 0 , z 2 ∈ R , r = 0 } \{z_1 = z_3 = z_4 = z_5 = 0,\ z_2 \in\mathbb{R},\ r = 0\} { z 1 = z 3 = z 4 = z 5 = 0 , z 2 ∈ R , r = 0 } C 0 C_0 C 0 h 0 h_0 h 0 Λ \Lambda Λ
Correcting the Hessian
We remove this degeneracy by adding to h 0 h_0 h 0 Re z 2 2 = x 2 2 − y 2 2 \operatorname{Re}z_2^2 = x_2^2 - y_2^2 Re z 2 2 = x 2 2 − y 2 2 C 0 C_0 C 0 Re z 1 2 \operatorname{Re}z_1^2 Re z 1 2 C 1 C_1 C 1 r 0 > 0 r_0 > 0 r 0 > 0
N 0 = { ∣ x 1 ∣ < 3 r 0 , ∣ y 1 ∣ < 3 r 0 , ∣ z j ∣ < 3 r 0 ( j = 2 , … , 5 ) , ∣ r ∣ < 3 r 0 } . N_0 = \left\{|x_1| < 3r_0,\ |y_1| < 3r_0,\ |z_j| < 3r_0\ (j=2,\ldots,5),\ |r| < 3r_0\right\}. N 0 = { ∣ x 1 ∣ < 3 r 0 , ∣ y 1 ∣ < 3 r 0 , ∣ z j ∣ < 3 r 0 ( j = 2 , … , 5 ) , ∣ r ∣ < 3 r 0 } . where z = x + i y z=x+iy z = x + i y r = log ∣ w ∣ − Re G ( z ) r=\log|w|-\operatorname{Re}G(z) r = log ∣ w ∣ − Re G ( z ) r 0 r_0 r 0 N 0 N_0 N 0 C 0 C_0 C 0 W ∩ π − 1 ( B r 3 ) W\cap\pi^{-1}(B_{r_3}) W ∩ π − 1 ( B r 3 ) r r r N 0 N_0 N 0 ρ = σ ∣ y ∣ 2 + F 0 ( x ) \rho=\sigma|y|^2+F_0(x) ρ = σ ∣ y ∣ 2 + F 0 ( x ) F 0 F_0 F 0 v = r v=r v = r N 0 N_0 N 0
Λ ∩ N 0 = { ∣ x 1 ∣ < 3 r 0 , y 1 = 0 , z 2 = z 3 = z 4 = z 5 = 0 , r = 0 } . \Lambda\cap N_0=\{|x_1|<3r_0,\ y_1=0,\ z_2=z_3=z_4=z_5=0,\ r=0\}. Λ ∩ N 0 = { ∣ x 1 ∣ < 3 r 0 , y 1 = 0 , z 2 = z 3 = z 4 = z 5 = 0 , r = 0 } . Define N 1 N_1 N 1 C 1 C_1 C 1 z 1 z_1 z 1 z 2 z_2 z 2 r r r r − 1 r-1 r − 1 t = r + Re G ( z ) t=r+\operatorname{Re}G(z) t = r + Re G ( z ) N 0 N_0 N 0 N 1 N_1 N 1 N 0 ∩ N 1 = ∅ N_0\cap N_1=\varnothing N 0 ∩ N 1 = ∅ r 0 r_0 r 0 χ ∈ C ∞ ( R , [ 0 , 1 ] ) \chi\in C^\infty(\mathbb{R},[0,1]) χ ∈ C ∞ ( R , [ 0 , 1 ]) χ = 1 \chi=1 χ = 1 [ − r 0 , r 0 ] [-r_0,r_0] [ − r 0 , r 0 ] supp χ ⊂ ( − 2 r 0 , 2 r 0 ) \operatorname{supp}\chi\subset(-2r_0,2r_0) supp χ ⊂ ( − 2 r 0 , 2 r 0 )
ψ 0 = χ ( x 1 ) ϑ ( y 1 , r , z 2 , z 3 , z 4 , z 5 ) . \psi_0=\chi(x_1)\vartheta(y_1,r,z_2,z_3,z_4,z_5). ψ 0 = χ ( x 1 ) ϑ ( y 1 , r , z 2 , z 3 , z 4 , z 5 ) . where ϑ \vartheta ϑ ( y 1 , r , z 2 , … , z 5 ) (y_1,r,z_2,\ldots,z_5) ( y 1 , r , z 2 , … , z 5 ) [ 0 , 1 ] [0,1] [ 0 , 1 ] { ∣ y 1 ∣ , ∣ r ∣ , ∣ z j ∣ < 2 r 0 } \{|y_1|,|r|,|z_j|<2r_0\} { ∣ y 1 ∣ , ∣ r ∣ , ∣ z j ∣ < 2 r 0 } ψ 0 ∈ C c ∞ ( N 0 ) \psi_0\in C^\infty_c(N_0) ψ 0 ∈ C c ∞ ( N 0 ) ψ 0 = χ ( x 1 ) \psi_0=\chi(x_1) ψ 0 = χ ( x 1 ) Λ ∩ N 0 \Lambda\cap N_0 Λ ∩ N 0 ψ 1 ∈ C c ∞ ( N 1 ) \psi_1\in C^\infty_c(N_1) ψ 1 ∈ C c ∞ ( N 1 ) z 1 z_1 z 1 z 2 z_2 z 2 ψ 0 \psi_0 ψ 0 ψ 1 \psi_1 ψ 1 X X X κ > 0 \kappa>0 κ > 0
g κ = κ 2 ( ψ 0 Re z 2 2 + ψ 1 Re z 1 2 ) ∈ C c ∞ ( X ) , h = h 0 + g κ on W , ω κ = ω ^ + d d c g κ . g_\kappa=\frac{\kappa}{2}\left(\psi_0\operatorname{Re}z_2^2+\psi_1\operatorname{Re}z_1^2\right)\in C^\infty_c(X),\qquad h=h_0+g_\kappa\ \text{on }W,\qquad\omega_\kappa=\widehat{\omega}+dd^cg_\kappa. g κ = 2 κ ( ψ 0 Re z 2 2 + ψ 1 Re z 1 2 ) ∈ C c ∞ ( X ) , h = h 0 + g κ on W , ω κ = ω + d d c g κ . Then ω κ \omega_\kappa ω κ ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X ω ^ \widehat{\omega} ω ω κ = d d c h \omega_\kappa=dd^ch ω κ = d d c h W W W
Lemma 13. For κ > 0 \kappa>0 κ > 0
(a) ω κ \omega_\kappa ω κ X X X
(b) h ∣ Λ = 0 h|_\Lambda=0 h ∣ Λ = 0 d h ∣ Λ = 0 dh|_\Lambda=0 d h ∣ Λ = 0 Hess q h \operatorname{Hess}_q h Hess q h T q X / T q Λ T_qX/T_q\Lambda T q X / T q Λ q ∈ Λ q\in\Lambda q ∈ Λ
(c) There are a neighbourhood W ′ ′ W'' W ′′ Λ \Lambda Λ c > 0 c>0 c > 0 h ≥ c dist ( ⋅ , Λ ) 2 h\ge c\operatorname{dist}(\cdot,\Lambda)^2 h ≥ c dist ( ⋅ , Λ ) 2 W ′ ′ W'' W ′′
Proof. (a) Put T = d d c ( 1 2 ψ 0 Re z 2 2 + 1 2 ψ 1 Re z 1 2 ) T=dd^c\left(\frac{1}{2}\psi_0\operatorname{Re}z_2^2+\frac{1}{2}\psi_1\operatorname{Re}z_1^2\right) T = d d c ( 2 1 ψ 0 Re z 2 2 + 2 1 ψ 1 Re z 1 2 ) d d c g κ = κ T dd^cg_\kappa=\kappa T d d c g κ = κ T ( 1 , 1 ) (1,1) ( 1 , 1 ) X X X S = supp ψ 0 ∪ supp ψ 1 S=\operatorname{supp}\psi_0\cup\operatorname{supp}\psi_1 S = supp ψ 0 ∪ supp ψ 1 g g g X X X ω ^ \widehat{\omega} ω S S S m > 0 m>0 m > 0 ω ^ ≥ 2 m g \widehat{\omega}\ge2mg ω ≥ 2 m g S S S C > 0 C>0 C > 0 T ≥ − C g T\ge-Cg T ≥ − C g S S S κ < m / C \kappa<m/C κ < m / C ω κ ≥ m g \omega_\kappa\ge mg ω κ ≥ m g S S S ω κ = ω ^ > 0 \omega_\kappa=\widehat{\omega}>0 ω κ = ω > 0 X ∖ S X\setminus S X ∖ S ω κ \omega_\kappa ω κ
(b) On N 0 N_0 N 0 Re z 2 2 \operatorname{Re}z_2^2 Re z 2 2 { z 2 = 0 } ⊃ Λ ∩ N 0 \{z_2=0\}\supset\Lambda\cap N_0 { z 2 = 0 } ⊃ Λ ∩ N 0 ψ 0 Re z 2 2 \psi_0\operatorname{Re}z_2^2 ψ 0 Re z 2 2 Λ \Lambda Λ q ∈ Λ ∩ N 0 q\in\Lambda\cap N_0 q ∈ Λ ∩ N 0
Hess q ( ψ 0 Re z 2 2 ) = ψ 0 ( q ) Hess q Re z 2 2 = χ ( x 1 ( q ) ) Hess q ( x 2 2 − y 2 2 ) . \operatorname{Hess}_q(\psi_0\operatorname{Re}z_2^2)=\psi_0(q)\operatorname{Hess}_q\operatorname{Re}z_2^2=\chi(x_1(q))\operatorname{Hess}_q(x_2^2-y_2^2). Hess q ( ψ 0 Re z 2 2 ) = ψ 0 ( q ) Hess q Re z 2 2 = χ ( x 1 ( q )) Hess q ( x 2 2 − y 2 2 ) . The same holds near C 1 C_1 C 1 h ∣ Λ = 0 h|_\Lambda=0 h ∣ Λ = 0 d h ∣ Λ = 0 dh|_\Lambda=0 d h ∣ Λ = 0 h h h d h dh d h Λ \Lambda Λ Hess q h ( ξ , η ) = 0 \operatorname{Hess}_q h(\xi,\eta)=0 Hess q h ( ξ , η ) = 0 ξ ∈ T q Λ \xi\in T_q\Lambda ξ ∈ T q Λ Hess q h \operatorname{Hess}_q h Hess q h T q X / T q Λ T_qX/T_q\Lambda T q X / T q Λ
Let q ∈ Λ ∩ N 0 q\in\Lambda\cap N_0 q ∈ Λ ∩ N 0 x 1 = x 1 ( q ) x_1=x_1(q) x 1 = x 1 ( q ) ψ 1 = 0 \psi_1=0 ψ 1 = 0 q q q
Hess q h = 2 σ d y 1 2 + ( 2 σ − κ χ ( x 1 ) ) d y 2 2 + 2 σ ∑ j = 3 5 d y j 2 + ( 2 x 1 2 + κ χ ( x 1 ) ) d x 2 2 + 2 ∑ j = 3 5 d x j 2 + 2 d r 2 . \operatorname{Hess}_q h=2\sigma\,dy_1^2+(2\sigma-\kappa\chi(x_1))\,dy_2^2+2\sigma\sum_{j=3}^{5}dy_j^2+(2x_1^2+\kappa\chi(x_1))\,dx_2^2+2\sum_{j=3}^{5}dx_j^2+2\,dr^2. Hess q h = 2 σ d y 1 2 + ( 2 σ − κ χ ( x 1 )) d y 2 2 + 2 σ j = 3 ∑ 5 d y j 2 + ( 2 x 1 2 + κ χ ( x 1 )) d x 2 2 + 2 j = 3 ∑ 5 d x j 2 + 2 d r 2 . By Lemma 12(c), the ten covectors appearing here form a basis of the dual of T q X / T q Λ T_qX/T_q\Lambda T q X / T q Λ 0 < κ ≤ 1 0<\kappa\le1 0 < κ ≤ 1 min ( κ , 2 r 0 2 ) \min(\kappa,2r_0^2) min ( κ , 2 r 0 2 ) 2 x 1 2 + κ χ ( x 1 ) ≥ κ 2x_1^2+\kappa\chi(x_1)\ge\kappa 2 x 1 2 + κ χ ( x 1 ) ≥ κ ∣ x 1 ∣ ≤ r 0 |x_1|\le r_0 ∣ x 1 ∣ ≤ r 0 ≥ 2 r 0 2 \ge2r_0^2 ≥ 2 r 0 2 ∣ x 1 ∣ ≥ r 0 |x_1|\ge r_0 ∣ x 1 ∣ ≥ r 0 2 σ − κ χ ( x 1 ) ≥ 1 2\sigma-\kappa\chi(x_1)\ge1 2 σ − κ χ ( x 1 ) ≥ 1 σ ≥ 1 \sigma\ge1 σ ≥ 1 κ ≤ 1 \kappa\le1 κ ≤ 1 C 1 C_1 C 1 q ∈ Λ ∖ ( N 0 ∪ N 1 ) q\in\Lambda\setminus(N_0\cup N_1) q ∈ Λ ∖ ( N 0 ∪ N 1 ) g κ g_\kappa g κ q q q q ∉ C 0 ∪ C 1 q\notin C_0\cup C_1 q ∈ / C 0 ∪ C 1 Hess q h = Hess q h 0 \operatorname{Hess}_q h=\operatorname{Hess}_q h_0 Hess q h = Hess q h 0 T q X / T q Λ T_qX/T_q\Lambda T q X / T q Λ
(c) By (b) and the compactness of Λ \Lambda Λ c > 0 c>0 c > 0 Hess q h ≥ 2 c ∣ ⋅ ∣ 2 \operatorname{Hess}_q h\ge2c|\cdot|^2 Hess q h ≥ 2 c ∣ ⋅ ∣ 2 T q Λ T_q\Lambda T q Λ q ∈ Λ q\in\Lambda q ∈ Λ Λ \Lambda Λ h ≥ c dist ( ⋅ , Λ ) 2 h\ge c\operatorname{dist}(\cdot,\Lambda)^2 h ≥ c dist ( ⋅ , Λ ) 2 W ′ ′ W'' W ′′ Λ \Lambda Λ
From now on κ \kappa κ ω = ω κ \omega=\omega_\kappa ω = ω κ ω \omega ω X X X ω = d d c h \omega=dd^ch ω = d d c h W W W
Proof of Theorem 2
Let c > 0 c>0 c > 0 Λ \Lambda Λ W ′ ′ W'' W ′′ W ′ ′ = { dist ( ⋅ , Λ ) < r 1 } W''=\{\operatorname{dist}(\cdot,\Lambda)<r_1\} W ′′ = { dist ( ⋅ , Λ ) < r 1 } r 1 > 0 r_1>0 r 1 > 0 W ′ ′ ⊂ W W''\subset W W ′′ ⊂ W h h h W ′ ′ W'' W ′′ ω = d d c h \omega=dd^ch ω = d d c h 0 < ε < c r 1 2 / 4 0<\varepsilon<cr_1^2/4 0 < ε < c r 1 2 /4
U = W ′ ′ , h ε = h − ε : U → R , K = { x ∈ U : h ε ( x ) ≤ 0 } . U=W'',\qquad h_\varepsilon=h-\varepsilon:U\to\mathbb{R},\qquad K=\{x\in U:h_\varepsilon(x)\leq0\}. U = W ′′ , h ε = h − ε : U → R , K = { x ∈ U : h ε ( x ) ≤ 0 } . Then K ⊂ { dist ( ⋅ , Λ ) ≤ ε / c } ⋐ U K\subset\{\operatorname{dist}(\cdot,\Lambda)\leq\sqrt{\varepsilon/c}\}\Subset U K ⊂ { dist ( ⋅ , Λ ) ≤ ε / c } ⋐ U Λ = { h = 0 } ⊂ int K \Lambda=\{h=0\}\subset\operatorname{int}K Λ = { h = 0 } ⊂ int K h ε h_\varepsilon h ε d d c h ε = ω ∣ U dd^ch_\varepsilon=\omega|_U d d c h ε = ω ∣ U ω \omega ω X X X h ε h_\varepsilon h ε h h h ε \varepsilon ε ε \varepsilon ε h h h K K K { h = ε } \{h=\varepsilon\} { h = ε } h h h Λ \Lambda Λ ∣ d h ∣ ≥ c ′ dist ( ⋅ , Λ ) |dh|\geq c'\operatorname{dist}(\cdot,\Lambda) ∣ d h ∣ ≥ c ′ dist ( ⋅ , Λ ) Λ \Lambda Λ c ′ > 0 c'>0 c ′ > 0 h h h Λ \Lambda Λ Λ \Lambda Λ { h ≤ ε } \{h\leq\varepsilon\} { h ≤ ε } Λ \Lambda Λ ε → 0 \varepsilon\to0 ε → 0 Λ ⊂ int K \Lambda\subset\operatorname{int}K Λ ⊂ int K
Isotropy and proof of (ii). By Lemma 13(b), d h = 0 dh=0 d h = 0 Λ \Lambda Λ T q X T_qX T q X ε \varepsilon ε d c h = − 1 4 d h ∘ J d^ch=-\frac{1}{4}dh\circ J d c h = − 4 1 d h ∘ J d c h d^ch d c h T q X T_qX T q X q ∈ Λ q\in\Lambda q ∈ Λ T q Λ T_q\Lambda T q Λ ι : Λ → X \iota:\Lambda\to X ι : Λ → X ι ∗ d c h = 0 \iota^*d^ch=0 ι ∗ d c h = 0 ι ∗ ( d d c h ) = d ( ι ∗ d c h ) = 0 \iota^*(dd^ch)=d(\iota^*d^ch)=0 ι ∗ ( d d c h ) = d ( ι ∗ d c h ) = 0 Ω \Omega Ω X X X λ d d c h \lambda\,dd^ch λ d d c h Λ \Lambda Λ λ ≥ 0 \lambda\geq0 λ ≥ 0 Ω ∣ Λ = 0 \Omega|_\Lambda=0 Ω ∣ Λ = 0 ι ∗ Ω = 0 \iota^*\Omega=0 ι ∗ Ω = 0 ω ∣ Λ = 0 \omega|_\Lambda=0 ω ∣ Λ = 0
For (iii) and (iv) we use the holomorphic annulus
A = { 0 } × { 1 ≤ ∣ w ∣ ≤ e } , A=\{0\}\times\{1\leq|w|\leq e\}, A = { 0 } × { 1 ≤ ∣ w ∣ ≤ e } , oriented as a complex curve, and the annulus B = { ( γ ( s ) , e τ ( s ) + i θ ) : 0 ≤ s ≤ 1 } ⊂ Λ B=\{(\gamma(s),e^{\tau(s)+i\theta}):0\leq s\leq1\}\subset\Lambda B = {( γ ( s ) , e τ ( s ) + i θ ) : 0 ≤ s ≤ 1 } ⊂ Λ d s ∧ d θ ds\wedge d\theta d s ∧ d θ C 0 C_0 C 0 C 1 C_1 C 1 θ \theta θ ∂ A = C 1 − C 0 = ∂ B \partial A=C_1-C_0=\partial B ∂ A = C 1 − C 0 = ∂ B Σ = A − B \Sigma=A-B Σ = A − B X X X
Lemma 14. Suppose that Λ \Lambda Λ Ω \Omega Ω X X X Ω ∣ Λ = 0 \Omega|_\Lambda=0 Ω ∣ Λ = 0 ∫ Σ Ω = ∫ A Ω > 0 \int_\Sigma\Omega=\int_A\Omega>0 ∫ Σ Ω = ∫ A Ω > 0 [ Ω ] ≠ 0 [\Omega]\ne0 [ Ω ] = 0 H 2 ( X , R ) H^2(X,\mathbb{R}) H 2 ( X , R )
Proof. Since B ⊂ Λ B\subset\Lambda B ⊂ Λ ∫ B Ω = 0 \int_B\Omega=0 ∫ B Ω = 0 ∫ A Ω > 0 \int_A\Omega>0 ∫ A Ω > 0 A A A Ω \Omega Ω ⟨ [ Ω ] , [ Σ ] ⟩ = ∫ A Ω − ∫ B Ω > 0 \langle[\Omega],[\Sigma]\rangle=\int_A\Omega-\int_B\Omega>0 ⟨[ Ω ] , [ Σ ]⟩ = ∫ A Ω − ∫ B Ω > 0
Proof of (iii). Let λ > 0 \lambda>0 λ > 0 Ω \Omega Ω Ω = d d c ( λ h ) \Omega=dd^c(\lambda h) Ω = d d c ( λh ) K ⊃ Λ K\supset\Lambda K ⊃ Λ Ω ∣ Λ = 0 \Omega|_\Lambda=0 Ω ∣ Λ = 0 [ Ω ] ≠ 0 [\Omega]\ne0 [ Ω ] = 0 H X = 0 \mathcal{H}_X=0 H X = 0 Ω \Omega Ω λ = N \lambda=N λ = N d d c ( N h ) dd^c(Nh) d d c ( N h )
Proof of (iv). We first show that A ⊂ Λ ^ M ( X ) A\subset\widehat{\Lambda}_{M(X)} A ⊂ Λ M ( X ) ∂ A = C 0 ∪ C 1 ⊂ Λ \partial A=C_0\cup C_1\subset\Lambda ∂ A = C 0 ∪ C 1 ⊂ Λ Z Z Z X X X A A A Λ \Lambda Λ Z ∩ A ≠ ∅ Z\cap A\ne\varnothing Z ∩ A = ∅ Z ∩ Λ = ∅ Z\cap\Lambda=\varnothing Z ∩ Λ = ∅ Z Z Z Σ \Sigma Σ H X = 0 \mathcal{H}_X=0 H X = 0
Since X X X Z Z Z s s s L = [ Z ] L=[Z] L = [ Z ] L L L c 1 ( L ) c_1(L) c 1 ( L ) ( 1 , 1 ) (1,1) ( 1 , 1 ) L L L H 2 ( X , Z ) → H 2 ( X , R ) H^2(X,\mathbb{Z})\to H^2(X,\mathbb{R}) H 2 ( X , Z ) → H 2 ( X , R ) [8 Ch. 1, §1] }. By Lemma 8 this image is 0, so
∫ Σ c 1 ( L ) = 0. \int_\Sigma c_1(L)=0. ∫ Σ c 1 ( L ) = 0. On the other hand, if e e e L L L s = f e s=fe s = f e c 1 ( L ) = − 1 π d d c log ∣ e ∣ 2 c_1(L)=-\frac{1}{\pi}dd^c\log|e|^2 c 1 ( L ) = − π 1 d d c log ∣ e ∣ 2 log ∣ s ∣ 2 = log ∣ f ∣ 2 + log ∣ e ∣ 2 \log|s|^2=\log|f|^2+\log|e|^2 log ∣ s ∣ 2 = log ∣ f ∣ 2 + log ∣ e ∣ 2 log ∣ f ∣ 2 \log|f|^2 log ∣ f ∣ 2 Z Z Z
c 1 ( L ) = − 1 π d d c log ∣ s ∣ 2 on X ∖ Z . c_1(L)=-\frac{1}{\pi}dd^c\log|s|^2\quad\text{on }X\setminus Z. c 1 ( L ) = − π 1 d d c log ∣ s ∣ 2 on X ∖ Z . Since B ⊂ Λ ⊂ X ∖ Z B\subset\Lambda\subset X\setminus Z B ⊂ Λ ⊂ X ∖ Z ∫ B c 1 ( L ) = − 1 π ∫ ∂ B d c log ∣ s ∣ 2 \int_B c_1(L)=-\frac{1}{\pi}\int_{\partial B}d^c\log|s|^2 ∫ B c 1 ( L ) = − π 1 ∫ ∂ B d c log ∣ s ∣ 2 A A A Z Z Z ∂ A ⊂ Λ \partial A\subset\Lambda ∂ A ⊂ Λ A ∩ Z = { p 1 , … , p r } A\cap Z=\{p_1,\ldots,p_r\} A ∩ Z = { p 1 , … , p r } A A A m j ≥ 1 m_j\ge1 m j ≥ 1 s ∣ A s|_A s ∣ A p j p_j p j A A A [8 Ch. 3, §2] , which here reduces to the identity 1 π d d c log ∣ w − w 0 ∣ 2 = δ w 0 \frac{1}{\pi}dd^c\log|w-w_0|^2=\delta_{w_0} π 1 d d c log ∣ w − w 0 ∣ 2 = δ w 0 1 π d d c log ∣ s ∣ 2 = ∑ j m j δ p j − c 1 ( L ) \frac{1}{\pi}dd^c\log|s|^2=\sum_jm_j\delta_{p_j}-c_1(L) π 1 d d c log ∣ s ∣ 2 = ∑ j m j δ p j − c 1 ( L ) A A A
∫ A c 1 ( L ) = ∑ j = 1 r m j − 1 π ∫ ∂ A d c log ∣ s ∣ 2 . \int_A c_1(L)=\sum_{j=1}^{r}m_j-\frac{1}{\pi}\int_{\partial A}d^c\log|s|^2. ∫ A c 1 ( L ) = j = 1 ∑ r m j − π 1 ∫ ∂ A d c log ∣ s ∣ 2 . Since ∂ A = ∂ B \partial A=\partial B ∂ A = ∂ B
∫ Σ c 1 ( L ) = ∫ A c 1 ( L ) − ∫ B c 1 ( L ) = ∑ j = 1 r m j ≥ 1 , \int_\Sigma c_1(L)=\int_A c_1(L)-\int_B c_1(L)=\sum_{j=1}^{r}m_j\ge1, ∫ Σ c 1 ( L ) = ∫ A c 1 ( L ) − ∫ B c 1 ( L ) = j = 1 ∑ r m j ≥ 1 , a contradiction. Hence, A ⊂ Λ ^ M ( X ) A\subset\widehat{\Lambda}_{M(X)} A ⊂ Λ M ( X )
Since A ∩ Λ = Λ ∩ π − 1 ( 0 ) = ∂ A A\cap\Lambda=\Lambda\cap\pi^{-1}(0)=\partial A A ∩ Λ = Λ ∩ π − 1 ( 0 ) = ∂ A K K K Λ \Lambda Λ ε → 0 \varepsilon\to0 ε → 0 ε \varepsilon ε A ⊄ K A\not\subset K A ⊂ K A ⊂ K A\subset K A ⊂ K Λ ^ M ( X ) ⊂ K ^ M ( X ) \widehat{\Lambda}_{M(X)}\subset\widehat{K}_{M(X)} Λ M ( X ) ⊂ K M ( X ) A ∖ K A\setminus K A ∖ K Λ \Lambda Λ K K K
Note that the fact that K K K [2 ] : a Hodge extension of d d c h dd^ch d d c h
Department of Mathematics, University of Western Ontario, London, Ontario N6A 5B7, Canada
Email address: shafikov@uwo.ca
