abstract. arxiv:2105.10725v3 [math.dg] 10 jun 2021

A NAKAI–MOISHEZON TYPE CRITERION FOR SUPERCRITICALDEFORMED HERMITIAN–YANG–MILLS EQUATION

JIANCHUN CHU, MAN-CHUN LEE, AND RYOSUKE TAKAHASHI

Abstract. The deformed Hermitian–Yang–Mills equation is a complex Hessian equa-tion on compact Kahler manifolds that corresponds to the special Lagrangian equationin the context of the Strominger–Yau–Zaslow mirror symmetry [SYZ96]. Recently,Chen [Che21] proved that the existence of the solution is equivalent to a uniform sta-bility condition in terms of holomorphic intersection numbers along test families. Inthis paper, we establish an analogous stability result not involving a uniform constantin accordance with a recent work on the J-equation by Song [Son20], which makesfurther progress toward Collins–Jacob–Yau’s original conjecture [CJY15] in the super-critical phase case. In particular, we confirm this conjecture for projective manifoldsin the supercritical phase case.

Contents

1. Introduction 12. Preliminaries 63. The twisted dHYM equation on the resolution Y 144. The twisted dHYM equation on the product space Y 205. The mass concentration and local smoothing 236. Gluing construction 347. Completion of the proof of Theorem 1.3 40References 42

1. Introduction

Let X be an n-dimensional compact complex manifold with a real (1, 1)-cohomologyclass α and a Kahler class β. For a given Kahler form χ ∈ β, the deformed Hermitian–Yang–Mills (dHYM) equation is defined by

Im(e−√−1ϑ0(χ+

√−1ω)n

)= 0, (1.1)

where ω ∈ α is the desired real (1, 1)-form and ϑ0 is a constant. Integrating (1.1),one can easily observe that the constant ϑ0 is uniquely determined (mod. 2π) by a

Date: June 11, 2021.2020 Mathematics Subject Classification. Primary 53C55; Secondary 35A01.Key words and phrases. deformed Hermitian–Yang–Mills equation, Thomas–Yau conjecture, Nakai–

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2 J. CHU, M.-C. LEE, AND R. TAKAHASHI

cohomological condition (∫X

(χ+√−1ω)n

)e−√−1ϑ0 ∈ R>0.

The dHYM equation first appeared in the physics literature [MMMS00] and [LYZ01]in the mathematical side. The dHYM equation corresponds to the special Lagrangianequation in the setting of the Strominger–Yau–Zaslow mirror symmetry [SYZ96], andnow is studied extensively (e.g. [CCL20, Che21, CJY15, CXY17, CY18, HJ20, HY19,JY17, Pin19, SS19, Tak20, Tak21]).

It is also useful to consider another expression for (1.1). Let λi be the eigenvalues of ωwith respect to χ. Then by the computation in [JY17], the equation (1.1) is equivalentto

n∑i=1

arccot(λi) = θ0, (1.2)

where θ0 and ϑ0 are related as θ0 := nπ2− ϑ0.

One of the main topics in the study of special Lagrangians is Thomas–Yau conjec-ture [TY02], which asserts that a given Lagrangian embedded in a Calabi–Yau manifoldcould be deformed to a special one by Hamiltonian deformations if and only if it is “sta-ble”. Thereafter Joyce [Joy15] proposed a broad update to the Thomas-Yau conjecturein terms of Bridgeland stability and the Lagrangian mean curvature flow, and also therehave been some significant progress on the symplectic side (e.g. [IJS16, Nev13]). How-ever, this conjecture is still widely open. So it is natural to consider the mirror versionof Thomas–Yau conjecture in the complex geometric side. Collins–Jacob–Yau [CJY15,Conjecture 1.4] predicted that the existence of the solution to (1.1) is equivalent to acertain stability condition in terms of holomorphic intersection numbers for any analyticsubvarieties Y ⊂ X, modeled on the Nakai–Moishezon criterion, and confirmed the con-jecture for complex surfaces. Recently, Chen [Che21] proved a Nakai–Moishezon typecriterion for the supercritical dHYM equation (and for the J-equation as well) undera slightly stronger condition that these intersection numbers have a uniform positivelower bound independent of Y . In a recent work of Datar-Pingali [DP20], the authorsextended the technique in [Che21] to the case of generalized Monge-Ampere and as aresult, the Collins–Jacob–Yau’s conjecture is confirmed when X is a three dimensionalprojective manifold and tan(ϑ0) > 0 (i.e. cot(θ0) 6 0) (cf. [DP20, Remark 1.6]). Jacob-Sheu [JS20] showed that the conjecture holds on the blowup of complex projectivespace.

Meanwhile, when θ0 is very close to zero, there are similarities between the theory ofthe dHYM equation and the J-equation (cf. [Che00, Don99]) for a Kahler form ω ∈ αdefined by

n∑i=1

1

λi=nωn−1 ∧ χ

ωn.

The both are complex Hessian equations on Kahler manifolds and related to each othervia the “small radius limit” as observed in [CXY17, CY18], i.e. we replace ω with sω

SUPERCRITICAL DEFORMED HERMITIAN–YANG–MILLS EQUATION 3

(s ∈ (0,∞)) and consider the limit s→∞. Then we have

lims→∞

sn∑i=1

arccot(sλi) =n∑i=1

1

λi, θ0 → 0.

Very recently, a remarkable progress for the study of the J-equation was made by Song[Son20]. He further extended the method of [Che21], and showed a Nakai–Moishezontype criterion for the J-equation without assuming a uniform lower bound for theintersection numbers, which confirms Lejmi–Szekelyhidi’s conjecture [LS15].

From now on, we focus on the supercritical dHYM equation i.e. θ0 ∈ (0, π). In thiscase, Collins–Jacob–Yau’s conjecture [CJY15, Conjecture 1.4] can be stated as follows.Actually, the direction (2) ⇒ (1) in Conjecture 1.1 only holds in the supercritical case(see [Che21, Remark 1.10]).

Conjecture 1.1 (Conjecture 1.4 of [CJY15]). Let X be an n-dimensional compactcomplex manifold, α a real (1, 1)-cohomology class, β a Kahler class on X and θ0 ∈(0, π) a constant. Assume that∫

X

(Re(α +

√−1β)n − cot(θ0)Im(α +

√−1β)n

)= 0.

Then the following two conditions are equivalent:

(1) for any m-dimensional proper analytic subvariety Y of X (m = 1, . . . , n − 1),we have ∫

Y

(Re(α +

√−1β)m − cot(θ0)Im(α +

√−1β)m

)> 0.

(2) for any (or some) Kahler form χ ∈ β, there exists a unique solution to thedeformed Hermitian–Yang–Mills equation (1.2).

In this paper, we strengthen Chen’s result [Che21] by using the same method as[Son20]. For the reader’s convenience, let us recall some terminologies and notationsfirst. Following [Che21], we say that a smooth family ωt,0 is a test family (emanatingfrom a real (1, 1)-form ω0 ∈ α) if the following conditions hold:

(A) ω0,0 = ω0 ∈ α.

(B) For any t1 < t2, we have ωt1,0 < ωt2,0.

(C) There exists a number T > 0 such that ωt,0 − cot(θ0n

)χ > 0 holds for all

t ∈ [T,∞).

For instance, we have a test family by setting ωt.0 := ω0 + tχ for any real (1, 1)-formω0 ∈ α. For any constant θ0 ∈ (0, π), we say that the triple (X,α, β) is stable along atest family ωt,0 if for any m-dimensional analytic subvariety1 Y of X (m = 1, . . . , n),we have

F Stabθ0

(Y, ωt,0, t) := (Re(ωt,0 +√−1χ)m − cot(θ0)Im(ωt,0 +

√−1χ)m) · Y

:=

∫Y

(Re(ωt,0 +√−1χ)m − cot(θ0)Im(ωt,0 +

√−1χ)m) > 0

(1.3)

1In this paper, the terminology “m-dimensional (sub)variety” means that it is reduced, but doesnot have to be pure m-dimensional or irreducible.


for all t ∈ [0,∞), and the strict inequality holds if m < n for all t ∈ [0,∞). If thereexists ε > 0 such that for any m-dimensional analytic subvariety Y of X (m = 1, . . . , n),we have

F Stabθ0

(Y, ωt,0, t) > (n−m)ε

∫Y

χm

for all t ∈ [0,∞), then we say that the triple (X,α, β) is uniform stable along a testfamily ωt,0.

In [Che21], Chen proved that the existence of the solution to the dHYM equation(1.2) is equivalent to the uniform stability condition.

Theorem 1.2. [Theorem 1.7 of [Che21]] Let X be an n-dimensional compact complexmanifold, α a real (1, 1)-cohomology class, β a Kahler class on X and θ0 ∈ (0, π) aconstant. Assume that∫

X

(Re(α +


√−1β)n

)= 0. (1.4)

Then the following three conditions are equivalent:

(1) the triple (X,α, β) is uniform stable along any test family ωt,0 ∈ αt.(2) the triple (X,α, β) is uniform stable along some test family ωt,0 ∈ αt.(3) for any (or some) Kahler form χ ∈ β, there exists a unique solution to the

deformed Hermitian–Yang–Mills equation (1.2).

Motivated by [Son20], it is very natural to ask if the uniform stability condition inTheorem 1.2 can be replaced by the stability condition alone. We give an affirmativeanswer to this question. The following theorem is the main theorem.

Theorem 1.3. Let X be an n-dimensional compact complex manifold, α a real (1, 1)-cohomology class, β a Kahler class on X and θ0 ∈ (0, π) a constant. Assume that∫

X

(Re(α +


√−1β)n

)= 0. (1.5)

Then the following three conditions are equivalent:

(1) the triple (X,α, β) is stable along any test family ωt,0 ∈ αt.(2) the triple (X,α, β) is stable along some test family ωt,0 ∈ αt.(3) for any (or some) Kahler form χ ∈ β, there exists a unique solution to the

deformed Hermitian–Yang–Mills equation (1.2).

Theorem 1.3 is not exactly same as Conjecture 1.1 in that it involves conditions fortest families. However, we can derive the following Corollary 1.4 from Theorem 1.3,which does not involve the assumptions of test families.

Corollary 1.4. Let X be an n-dimensional compact complex manifold, α a real (1, 1)-cohomology class, β a Kahler class on X and θ0 ∈ (0, π) a constant. Assume that∫

X

(Re(α +


√−1β)n

)= 0.

Then the following two conditions are equivalent:


(1) there exists a Kahler class γ on X such that for any 1 6 k 6 n, we have∫X

(Re(α +

√−1β)k − cot(θ0)Im(α +

√−1β)k

)∧ γn−k > 0,

and for any proper m-dimensional analytic subvariety Y of X (m = 1, . . . , n−1)and 1 6 k 6 m, we have∫

Y

(Re(α +


√−1β)k

)∧ γm−k > 0.

(2) for any (or some) Kahler form χ ∈ β, there exists a unique solution to thedeformed Hermitian–Yang–Mills equation (1.2).

Let us discuss the relationship between Conjecture 1.1 and Corollary 1.4. Roughlyspeaking, Conjecture 1.1 predicts a kind of numerical criterion for the “positivity” of(m,m)-class

Re(α +√−1β)m − cot(θ0)Im(α +

√−1β)m

for all 1 6 m 6 n. This is very similar to the numerical criterion of the Kahlerclass proved by Demailly-Paun [DP04]. Indeed, the assumptions of Corollary 1.4 aremotivated by [DP04, Theorem 4.2]. In this sense, Corollary 1.4 can be regarded asa counterpart of [DP04, Theorem 4.2] in the supercritical dHYM setting, and also aweaker version of Conjecture 1.1. In particular, when X is projective, Corollary 1.4implies Conjecture 1.1.

Corollary 1.5. If X is projective, then Conjecture 1.1 is true.

We should mention that there are some significant differences from the argumentsin [Son20]. First, we consider the stability condition not only for a single class α butalso for a family of cohomology classes αt as in [Che21]. Indeed, we need the notion oftest families to choose the correct brunch cut of the arccot function. Also, since thereis no canonical choice of test families, we check that every test family has a “nice” lifton the resolution of singularities of Y (cf. Lemma 3.3 and Lemma 3.5). Second, in ourinduction argument, we have to assume that the ambient space X is always smooth,whereas [Son20, Theorem 1.1] can deal with the case when X is possibly singular andembedded in a Kahler manifold. The reason is that we take different kinds of approachesaccording to the case m < n and m = n. The former case corresponds to Theorem3.1. A key observation for Theorem 3.1 is that we can always take a uniformly positivetwisting function in the twisted dHYM equation (4.4) on the product space (cf. Lemma4.1), and this property holds only for the case m < n (cf. Remark 4.2). On the otherhand, we can treat the later case essentially by the same argument as [Che21, Section5] where the smoothness of X is used (see Section 7 for more details).

The organization of the paper is as follows. First of all, we should mention that ourarguments are mostly based on the seminal works [Che21, Son20]. So we recommendreaders to refer the both papers appropriately. Specifically, Section 3, 6 in our papercorresponds to [Son20, Section 4, 6], and Section 4, 5 corresponds to [Son20, Section 5]respectively. Our Section 3, 4, 5, 6, 7 are also based on [Che21, Section 5].

In Section 2, we define notations and prove basic properties for functions on thespace of Hermitian matrices. Then we recall some results for subsolutions proved in


[Che21, CJY15]. In the end of the section, we will give the proof for the easier part (1)⇒ (2), (3) ⇒ (1) of Theorem 1.3. We also show how to apply Theorem 1.3 to proveCorollary 1.4, and apply Corollary 1.4 to prove Corollary 1.5. In the remaining partof the paper, we will consider the harder part (2) ⇒ (3) of Theorem 1.3, in particular,Theorem 3.1. In Section 3, we consider the twisted dHYM equation on the resolution

of singularities Y for any proper analytic subvariety Y ⊂ X. Then we also study the

twisted dHYM equation on the product space Y = Y × Y in Section 4. In Section 5,we establish some estimates for the fiber integration by using the mass concentrationargument, and construct a subsolution Ω as a current. We show that on each coordinate

ball of a sufficiently finer finite covering of Y , the smoothing of Ω for sufficiently small

scale satisfies the condition for subsolutions away from the exceptional set E0 ⊂ Y onwhich χ degenerates. In Section 6, we glue these local subsolutions together by takingthe regularized maximum. In Section 7, we give a proof of the harder part (2) ⇒ (3)of Theorem 1.3 based on the argument in [Che21, Section 5], and accomplish the proofof Theorem 1.3.

Acknowledgment. M.-C. Lee was partially supported by NSF grant DMS-1709894and EPSRC grant number P/T019824/1. R. Takahashi was supported by Grant-in-Aidfor Early-Career Scientists (20K14308) from JSPS.

2. Preliminaries

By following [Che21, Section 5], for any positive integers k 6 n, we define functionsPk,n, Qk,n on Rn by

Pk,n(λ1, . . . , λn) := maxI

∑i∈I

arccot(λi),

Qk,n(λ1, . . . , λn) := maxJ

∑j∈J

arccot(λj),

where P1,n := 0 and the maximum is taken over all subsets I ⊂ 1, . . . , n (resp.J ⊂ 1, . . . , n) of cardinality k − 1 (resp. k).2 Let Γ be the set which consists of allλ ∈ Rn satisfying Qn,n(λ) < π.

Proposition 2.1. We have the following:

(1) The functions Pk,n and Qk,n are monotone on Rn, i.e. if (λ1, . . . , λn), (λ′1, . . . , λ′n) ∈

Rn satisfies λi 6 λ′i for all i = 1, . . . , n, then we have

Pk,n(λ′1, . . . , λ′n) 6 Pk,n(λ1, . . . , λn), Qk,n(λ′1, . . . , λ

′n) 6 Qk,n(λ1, . . . , λn).

(2) If k 6 k′, thenPk,n 6 Pk′,n, Qk,n 6 Qk′,n

hold on Rn.

(3) Γ is a convex subset of Rn whose boundary is a smooth hypersurface.

(4) The functions cot(Pk,n(·)) and cot(Qk,n(·)) are concave on Γ.

2Although we get immidiately Pk+1,n = Qk,n by the definition, we will use the separate notationsto avoid confusion.


Proof. (1) and (2) are obvious, and (3) follows from [Yua06, Lemma 2.1]. For (4), theconcavity of the function Qn,n on Γ was proved in the course of the proof of [Tak20,Theorem 1.1]. By using this fact, one can observe that

cot(Pk,n(λ1, . . . , λn)) = minI

cot(∑i∈I

arccot(λi))

and each cot(∑

i∈I arccot(λi))

is concave on Γ. This shows that cot(Pk,n(·)) is concave.The proof for cot(Qk,n(·)) is similar.

Let Herm(n) be the space of all n× n Hermitian matrices and A,B ∈ Herm(n). Weregard A,B as Hermitian forms on Cn. Assume that A is positive definite and let λibe the eigenvalues of the matrix A−1B. Then PA,k,n(B) (resp. QA,k,n(B)) is definedas Pk,n(λ1, . . . , λn) (resp. Qk,n(λ1, . . . , λn)). Since the eigenvalues of A−1B is invariantunder linear transformations, we may always assume that A = In and B is diagonal. LetΓA ⊂ Herm(n) be the subset of n× n Hermitian matrices B satisfying QA,n,n(B) < π.Then by Proposition 2.1, we know that:

Proposition 2.2. We have the following:

(1) The functions PA,k,n(·) and QA,k,n(·) are monotone on Herm(n).

(2) If k, k′ ∈ N satisfy k 6 k′, then

PA,k,n 6 PA,k′,n, QA,k,n 6 QA,k′,nhold on Herm(n).

(3) ΓA is a convex subset of Herm(n) whose boundary is a smooth hypersurface.

(4) cot(PA,k,n(·)) and cot(QA,k,n(·)) are concave on ΓA.

Proof. First we assume A = In and let λ1 6 . . . 6 λn (resp. λ′1 6 . . . 6 λ′n) be theeigenvalues of an n× n Hermitian matrix B (resp. B′). Then it is well known that thecondition B 6 B′ implies λi 6 λ′i for all i = 1, . . . , n. Thus the statement (1) followsfrom the monotonicity of the functions Pk,n, Qk,n. (2) is clear, and (3) follows from[Yua06, Lemma 2.1]. As for (4), it is well-known (cf. [And94, Ger96, Spr05]) that if thefunction f(λ1, . . . , λn) is symmetric and concave, then the associated function FA(B) :=f(λ1, . . . , λn) in terms of a symmetric function of the eigenvalues λi is concave. Thusthe function QA,n,n is concave on ΓA. In general case, the function f(·) := cot(Pk,n(·))(or f(·) := cot(Qk,n(·))) may not be differentiable at points (λ1, . . . , λn) where λi arenot distinct. However, we can approximate f with a sequence of smooth symmetricconcave functions. Indeed, for any B,B′ ∈ ΓA, we take a small constant ε > 0 such thatB,B′ ∈ ΓεA, where the convex subset ΓεA b ΓA is given by ΓεA := B′′ ∈ ΓA|QA,n,n(B′′) <π − ε. Also let Γε := λ ∈ Γε|Qn,n(λ) < π − ε be the associated convex subset ofΓε. Since the derivative of Qn,n is uniformly bounded on Γ, we know that there existsr0 > 0 such that for any r ∈ (0, r0) and λ ∈ Γε, we have Br(λ) b Γ, where Br(λ)denotes a Euclidean ball of radius r centered at λ. Then for any r ∈ (0, r0), we definethe smoothing of f as in Definition 5.3 and denote it by f (r)(λ) (λ ∈ Γε). We note thatf (r) converges uniformly to f as r → 0 on any compact subset of Γε. Also one can easilysee that from the definition of the smoothing and concavity, f (r) is concave on Γε sincef is so by Proposition 2.1 (4). Let Sn denote the symmetric group which acts linearly


on Γε ⊂ Rn by λ := (λ1, . . . , λn) 7→ λσ := (λσ(1), . . . , λσ(n)). Then the symmetrization

fr(λ) := 1n!

∑σ∈Sn f

(r)(λσ) is still concave and converges to f on Γε as r → 0 since

f (r) → f and f is symmetric. Let FA,r denote the associated function with respect tofr on ΓεA. Then we have FA,r → FA as r → 0. We can apply the result [Spr05] to eachfr to know that

FA,r(sB + (1− s)B′) > sFA,r(B) + (1− s)FA,r(B′)

for any s ∈ [0, 1]. By taking the limit r → 0, we have

FA(sB + (1− s)B′) > sFA(B) + (1− s)FA(B′)

as desired.

It is also sometimes useful to consider the variational characterization of PA,k,n andQA,k,n. Let A = In be the identity matrix for simplicity and B ∈ Herm(n). For any(k − 1)-dimensional subspace V of Cn, there exists n× (k − 1) matrix U ∈ Mat(n, k −1;C) such that U

TU = Ik−1 and a basis of V consists of the columns of U . We

choose U such that UTBU = diag(λ′1, . . . , λ

′k−1) and S ∈ U(n) such that Sij = Uij

(i = 1, . . . , n; j = 1, . . . , k − 1). Then the Schur-Horn theorem says that the diagonal

(λ′1, . . . , λ′k−1, (S

TBS)kk, . . . , (S

TBS)nn) of the matrix S

TBS lies in the convex full of

the vectors obtained by permuting the entries of (λ1, . . . , λn), where λi are eigenvaluesof B. Thus by the concavity of the function cot(Pk,n(·)) on Γ, we know that

PIn,k,n(B) = maxI

∑i∈I

arccot(λi) = maxU∈Mat(n,k−1;C), U

TU=Ik−1, U

TBU=diag(λ′1,...,λ

′k−1)

k−1∑i=1

arccot(λ′i)

(2.1)for all B ∈ ΓIn . A similar observation shows that

QIn,k,n(B) = maxJ

∑i∈J

arccot(λi) = maxK∈Mat(n,k;C), K

TK=Ik, K

TBK=diag(λ′′1 ,...,λ

′′k)

k∑i=1

arccot(λ′′i )

(2.2)for all B ∈ ΓIn . These are simple extensions of formulas obtained in [Che21, Section 5].

Let X be an n-dimensional Kahler manifold X, α a real (1, 1)-cohomology classand β a Kahler class on X. For any closed real (1, 1)-form ω ∈ α and Kahler formχ ∈ β, we can also define functions Pχ,k,n(ω), Qχ,k,n(ω) in a standard manner. Let0 < θ0 6 Θ0 6 π be constants and Y an analytic subvariety of X. Let ω0 ∈ α be afixed real (1, 1)-form on X. We define the set Γχ,α,θ0,Θ0(Y,X) which consists of germsof real closed (1, 1)-forms ω ∈ α|Y,X satisfying Pχ,n,n(ω) < θ0 and Qχ,n,n(ω) < Θ0 onY ⊂ X, where the class α|Y,X consists of all germs of real closed (1, 1)-forms ω at Y ⊂ Xsuch that ω = ω0 +

√−1∂∂φω for some smooth function φω defined on a neighborhood

of Y in X. Then one can prove the following lemma exactly in the same way as [CJY15,Lemma 8.2]:

Lemma 2.3. For any ω ∈ Γχ,α,θ0,π(Y,X), on a neighborhood of Y in X, we have

Im(e−√−1θ0(ω +

√−1χ)p) < 0


for all p = 1, . . . , n− 1, where negativity is to be understood in the sense of (p, p)-forms(cf. [Dem12, Chapter III, Section 1.A]).

Proof. Actually, this lemma is equivalent to [CJY15, Lemma 8.2]. Since we use differentnotations, for the reader’s convenience, we include a brief proof here. It suffices to showthat for any simple positive (n− k, n− k)-form Ω,

Im(e−√−1θ0(ω +

√−1χ)k ∧ Ω

)< 0.

We choose a coordinates such that χij = δij and ωij = λiδij. Then

(ω +√−1χ)k = (

√−1)kk!

∑J

∏j∈J

(λi +√−1)dzJ ∧ dzJ

= (√−1)kk!

∑J

e√−1

∑j∈J arccot(λi)

∏j∈J

(√λ2i + 1

)dzJ ∧ dzJ .

where the sum is over all subsets J ⊂ 1, . . . , n such that |J | = k. The form Ω can bewritten as

Ω = (√−1)n−k

∑J

cJdzJc ∧ dzJc + Ω′,

where J c denotes the complement of J and Ω′ is smooth form such that Ω′∧dzJ∧dzJ = 0for all J . Since Ω is a simple positive form, then cJ > 0 for all J and at least one cJ ispositive. Since Pχ(ω) < θ and |J | = k 6 n− 1, then

∑j∈J arccot(λi)− θ < 0 for all J .

Therefore,

Im(e−√−1θ(α +

√−1χ)k ∧ Ω

)χn

=k!

n!

∑J

cJIm(e√−1(

∑j∈J arccot(λi)−θ)

)∏j∈J

(√λ2i + 1

)< 0.

If Y is smooth and Y = X, we often write Pχ,n,n(ω), Qχ,n,n(ω), Γχ,α,θ0,Θ0(Y, Y ) asPχ(ω), Qχ(ω), Γχ,α,θ0,Θ0(Y ) for simplicity. We remark that if χ 6 χ and ω is Kahler,then we have

Pχ,k,n(ω) 6 Pχ,k,n(ω), Qχ,k,n(ω) 6 Qχ,k,n(ω). (2.3)

This property does not hold for general (1, 1)-form ω, especially when ω is not Kahler.However, we can show the following uniform semi-continuity when k = n, which issimilar to [Che21, Definition 5.10].

Proposition 2.4. For 0 < θ < π, there is c0(n, θ) > 0 such that the following holds.Suppose ω, χ are closed real (1, 1)-forms where χ is Kahler, then for all ε < θ,

(a) Qχ(ω + εχ) < θ − c0ε if Qχ(ω) < θ;

(b) Pχ(ω + εχ) < θ − c0ε if Pχ(ω) < θ.


Proof. We prove (a) first. We let λ1 6 · · · 6 λn be the eigenvalues of ω with respect toχ. If Qχ(ω) 6 1

2θ, then the conclusion holds directly. So it suffices to consider the case

12θ < Qχ(ω) < θ. In this case, we have |λ1| 6 C(n, θ) and hence,

Qχ(ω + εχ) =n∑i=1

arccot(λi + ε)

6n∑i=2

arccot(λi) + arccot(λ1 + ε)

6 Qχ(ω)− c0(n, θ)ε.

Here we have applied mean value theorem on the function arccot(x) and used the factthat λ1 is uniformly bounded. The conclusion (b) can be proved analogously.

Proposition 2.5. For θ < π, there is σ0(n, θ) > 0 such that the following holds: letχ1, χ2 and χ3 are Kahler and ω be a closed real (1, 1) form which satisfies χ1, χ2 6 4χ3

and |χ1 − χ2| 6 σ5χ3 for some σ < σ0. Then

(a) Qχ1(ω + σχ3) < θ if Qχ2(ω) < θ;

(b) Pχ1(ω + σχ3) < θ if Pχ2(ω) < θ.

Proof. We will focus on proving (a) while (b) can be proved using similar argument. Itwill be sufficient to prove Qχ1(ω+σχ3) 6 θ+σ under Qχ2(ω) < θ and |χ1−χ2| 6 σ4χ3

since we can use the fact that χ1 6 4χ3 and apply Proposition 2.4 to show that for σ0

sufficiently small,

Qχ1(ω + 5σχ3 + 5c−10 σχ3) < Qχ1(ω + σχ3 + c−1

0 σχ1) 6 (θ + σ)− c0

(c−1

0 σ)

= θ.

The result will follow by shrinking σ0 further.To show Qχ1(ω + σχ3) 6 θ + σ, it is equivalent to show that

Im(e−√−1(θ+σ)(ω + σχ3 +

√−1χ1)n

)6 0.

We will assume σ0 < 1. For notational convenience, we let θ′ = θ+σ and χ1 = χ2 +χd.Then

Im(e−√−1(θ+σ)(ω + σχ3 +

√−1χ1)n

)= Im

(e−√−1θ′

[(ω +

√−1χ2) + (σχ3 +

√−1χd)

]n)= Im

(e−√−1θ′(ω +

√−1χ2)n

)+

n∑k=1

(n

k

)Im(e−√−1θ′(ω +

√−1χ2)n−k ∧ (σχ3 +

√−1χd)

k).


By using Qχ2(ω) < θ < θ′, the first term is non-positive. We will show that each termin the partial sum is non-positive. For 1 6 k 6 n,

Im(e−√−1θ′(ω +

√−1χ2)n−k ∧ (σχ3 +

√−1χd)

k)

=k−1∑l=0

(k

l

)Im(e−√−1θ′(ω +

√−1χ2)n−k ∧

[(σχ3)l ∧ (

√−1χd)

k−l])+ Im

(e−√−1θ′(ω +

√−1χ2)n−k

)∧ (σχ3)k

= S + G.

We now show that if σ0 is sufficiently small, then S will be dominated by the negativeterm G. We first show that e−

√−1θ′(ω +

√−1χ2)k is controlled by its imaginary part

due to the squeezed angle.

Claim 2.6. We have

Re(e−√−1θ′(ω +

√−1χ2)k

)6 − 2

σIm(e−√−1θ′(ω +

√−1χ2)k

).

Proof of Claim 2.6. To show this, we choose a coordinate such that (χ2)ij = δij andωij = λiδij. Then

Re(e−√−1θ′(ω +

√−1χ2)k

)= k!

∑J

[cos

(θ′ −

∑j∈J

arccot(λi)

)∏j∈J

√λ2i + 1

](√−1)kdzJ ∧ dzJ

where the sum is taken over all subset J ⊂ 1, ..., n so that |J | = k. Similarly,

− Im(e−√−1θ′(ω +

√−1χ2)k

)= k!

∑J

[sin

(θ′ −

∑j∈J

arccot(λi)

)∏j∈J

√λ2i + 1

](√−1)kdzJ ∧ dzJ .

Recalling that Qχ2(ω) < θ and θ′ = θ + σ, we have

θ′ −∑j∈J

arccot(λi) = θ −∑j∈J

arccot(λi) + σ > σ.

Hence, the elementary inequality sinx > 12x for x < σ implies

−Im(e−√−1θ′(ω +

√−1χ2)k

)>

1

2σ · k!

∑J

[∏j∈J

√λ2i + 1

](√−1)kdzJ ∧ dzJ

>1

2σRe

(e−√−1θ′(ω +

√−1χ2)k

).

This proves Claim 2.6.

Now we compare S and G.


Claim 2.7. For all 0 6 l 6 k − 1, we have

Im(e−√−1θ′(ω +

√−1χ2)n−k ∧

[(σχ3)l ∧ (

√−1χd)

k−l])6 −2σk+2Im

(e−√−1θ′(ω +

√−1χ2)n−k

)∧ χk3.

Proof of Claim 2.7. We first consider the case when k−l is even. In this case,(√−1χd

)k−lis real and hence,

Im(e−√−1θ′(ω +

√−1χ2)n−k ∧

[(σχ3)l ∧ (

√−1χd)

k−l])= ±Im

(e−√−1θ′(ω +

√−1χ2)n−k

)∧ (σχ3)l ∧ χk−ld

6 −σ4k−3l · Im(e−√−1θ′(ω +

√−1χ2)n−k

)∧ χk3

6 −σk+3 · Im(e−√−1θ′(ω +

√−1χ2)n−k

)∧ χk3,

where we have used −σ4χ3 6 χd 6 σ4χ3 and Im(e−√−1θ′(ω +

√−1χ2)n−k

)< 0 (cf.

Lemma 2.3).

When k − l is odd, we have (√−1χd)

k−l is imaginary and hence, by Claim 2.6,

Im(e−√−1θ′(ω +

√−1χ2)n−k ∧

[(σχ3)l ∧ (

√−1χd)

k−l])= ±Re

(e−√−1θ′(ω +

√−1χ2)n−k

)∧ (σχ3)l ∧ χk−ld

6 −2σ3(k−l)+k−1 · Im(e−√−1θ′(ω +

√−1χ2)n−k

)∧ χk3

6 −2σ2+k · Im(e−√−1θ′(ω +

√−1χ2)n−k

)∧ χk3,

where we have used −σ4χ3 6 χd 6 σ4χ3 and Im(e−√−1θ′(ω +

√−1χ2)n−k

)< 0 (cf.

Lemma 2.3) again.

By using Claim 2.7, we arrive at the following:

G + S 6 σk(1− Cnσ2

)· Im

(e−√−1θ′(ω +

√−1χ2)n−k

)∧ χk3.

Therefore, if σ0 is small enough, we have G + S < 0 for all 1 6 k 6 n. This completesthe proof.

Following [CJY15], we would also like to call the element of Γχ,α,θ0,Θ0(X) as subsolu-tions. Indeed, it was proved in [CJY15] that the existence of subsolutions leads to theexistence of genuine solutions to the dHYM equation (1.2). We will use the followingf -twisted version obtained by Chen [Che21, Proposition 5.5]:

Proposition 2.8. Let X be an n-dimensional compact complex manifold, χ a Kahlerform, α a real (1, 1)-cohomology class and 0 < θ0 < Θ0 < π constants. Then thereexists a constant c > 0 depending only on n, θ0,Θ0 such that the following statementholds. Assume the following:


(1) When n = 1, 2, 3, f > 0 is a constant satisfying∫X

fχn =

∫X

(Re(ω0 +√−1χ)n − cot(θ0)Im(ω0 +

√−1χ)n) > 0.

(2) When n > 4, f > −c is a smooth function satisfying∫X

fχn =

∫X

(Re(ω0 +√−1χ)n − cot(θ0)Im(ω0 +

√−1χ)n) > 0.

(3) Γχ,α,θ0,Θ0(X) 6= ∅.Then there exists a solution ω to the twisted dHYM equation

Re(ω +√−1χ)n − cot(θ0)Im(ω +

√−1χ)n − fχn = 0, (2.4)

where ω ∈ Γχ,α,θ0,Θ0(X).

Remark 2.9. In the above proposition, a smooth function f is allowed to be slightlynegative, which is crucial in the argument of [Che21], but is not used in the proofof Theorem 3.1. Also it is important to note that we cannot take θ0 and Θ0 as thesame constant. Indeed, if f is negative at some point z ∈ X and there is a solutionω ∈ Γχ,α,θ0,θ0(X), then the equation (2.4) automatically yields that Qχ(ω(z)) > θ0.

Now we are ready to prove the easy part of Theorem 1.3.

Proof of (1) ⇒ (2), (3) ⇒ (1) in Theorem 1.3. (1) ⇒ (2) is trivial. So we will show(3) ⇒ (1). Assume that there exists the solution ω0 to (1.2). Then we have ω0 ∈Γχ,α,θ0,π(X). By Lemma 2.3 and the assumption of Theorem 1.3, for any test familyωt,0 ∈ αt (t ∈ [0,∞)) emanating from ω0 and m-dimensional analytic subvariety Y ofX, we have

F Stabθ0

(Y, ωt,0, 0) > 0,

and the strict inequality holds if m < n. Since ωt,0 > ω0 for all t ∈ [0,∞), themonotonicity of Pχ, Qχ implies that ωt,0 ∈ Γχ,αt,θ0,π(X) for all t ∈ [0,∞). Thus byusing Lemma 2.3 again, we observe that

d

dtF Stabθ0

(Y, ωt,0, t) = m

∫Y

d

dtωt,0∧

(Re(ωt,0+

√−1χ)m−1−cot(θ0)Im(ωt,0+

√−1χ)m−1

)> 0

for all t ∈ [0,∞). So the triple (X,α, β) is stable along any test family ωt,0 emanatingfrom ω0. Finally, we remark that for any ω′0 ∈ α and test family ω′t,0 ∈ αt (t ∈ [0,∞))emanating from ω′0, we can always obtain a test family ωt,0 (t ∈ [0,∞)) emanating fromω0 by setting

ωt,0 := ω0 − ω′0 + ω′t,0 ∈ αt,and we have

F Stabθ0

(Y, ω′t,0, t) = F Stabθ0

(Y, ωt,0, t)for all t ∈ [0,∞) by the cohomological invariance. This completes the proof.

Now we are in a position to prove Corollary 1.4 by Theorem 1.3.


Proof of Corollary 1.4. (2)⇒ (1) follows from Lemma 2.3. Next we will apply Theorem1.3 to show (1) ⇒ (2). Let σ ∈ γ be a Kahler form on X. It is clear that

ωt,0 := ω0 + tσ, t ∈ [0,∞)

is a test family emanating from ω0. By Theorem 1.3, it suffices to show that the triple(X,α, β) is stable along this test family. For any m-dimensional analytic subvariety Yof X (m = 1, . . . , n) and t ∈ [0,∞), we compute∫

Y

(Re(ωt,0 +

√−1χ)m − cot(θ0)Im(ωt,0 +

√−1χ)m

)= − 1

sin(θ0)

∫Y

Im(e−√−1θ0(ωt,0 +

√−1χ)m

)= − 1

sin(θ0)

∫Y

Im(e−√−1θ0(ω0 +

√−1χ+ tσ)m

)= − 1

sin(θ0)

m∑k=0

tm−k(m

k

)∫Y

Im(e−√−1θ0(ω0 +

√−1χ)k

)∧ σm−k

=m∑k=0

tm−k(m

k

)∫Y

(Re(ω0 +

√−1χ)k − cot(θ0)Im(ω0 +

√−1χ)k

)∧ σm−k.

The assumption shows each coefficient in this polynomial (of t) is positive when m < nand non-negative when m = n. Then the triple (X,α, β) is stable along the test familyωt,0. This completes the proof of Corollary 1.4.

Next we show that Corollary 1.4 implies Corollary 1.5.

Proof of Corollary 1.5. We follow the argument of [DP04, Theorem 4.5] (see also [CT16,Theorem 1.1]). Since X is projective, then we choose γ to be the first Chern class of avery ample line bundle L over X. For any m-dimensional analytic subvariety Y of X(m = 1, . . . , n) and 1 6 k 6 m, there exist generic members H1, . . . , Hm−k of the linearsystem |L| such that Y ∩H1 ∩ . . . ∩Hm−k is a k-dimensional analytic subvariety of Xand ∫

Y

(Re(α +


√−1β)k

)∧ γm−k

=

∫Y ∩H1∩...∩Hm−k

(Re(α +


√−1β)k

).

Then Corollary 1.5 follows from Corollary 1.4.

3. The twisted dHYM equation on the resolution Y

We will prove the remaining part (2) ⇒ (3) of Theorem 1.3 by induction argumentfor m := dimY . We will prove the following:

Theorem 3.1. Let X be an n-dimensional compact complex manifold, α a real (1, 1)-cohomology class and β a Kahler class on X. Let θ0 ∈ (0, π) be a constant satisfying

(Re(α +√−1β)n − cot(θ0)Im(α +

√−1β)n) ·X > 0, (3.1)


and assume the triple (X,α, β) is stable along some test family ωt,0 ∈ αt (t ∈ [0,∞))emanating from ω0 ∈ α. Then for any Kahler form χ ∈ β, we have

Γχ,α,θ0,θ0(Y,X) 6= ∅ (3.2)

for all proper m-dimensional subvarieties Y of X (m = 1, . . . , n− 1).

In the above Theorem, we need the stronger result (3.2) instead of Γχ,α,θ0,π(Y,X) 6=∅ since unlike the function Qχ, the upper bound for Pχ is not preserved under theextension argument (see the proof of Theorem 3.1 in Section 7). Also we remark thatTheorem 3.1 is not true when m = n.3 However, we can prove Theorem 1.3 by thesame argument as [Che21, Section 5] as long as Theorem 3.1 holds (see Section 7 formore details).

So in the remaining part of the paper, we mainly focus on Theorem 3.1. We will proveTheorem 3.1 by induction on dimension m := dimY for proper analytic subvarietiesY ⊂ X. The following lemma demonstrates the case when m = 1 which initiates theinduction argument.

Lemma 3.2. Under the assumption of Theorem 3.1, for any subvariety Y of X withdimY 6 1, we have

Γχ,α,θ0,θ0(Y,X) 6= ∅.Proof. According to the value of dimY , there are two cases:

Case 1 (dimY = 0): Lemma 3.2 is obvious since every cohomology class in a suffi-ciently small neighborhood of a point is trivial.

Case 2 (dimY = 1): Since any Kahler classes on Y are proportional to each other, bythe condition (α− cot(θ0)β) ·Y > 0 and the extension theorem (cf. [DP04, Proposition3.3]), we know that there exists a real (1, 1)-form ωY ∈ α|Y,X on a neighborhood UY ofY such that

ωY − (cot(θ0) + 2ε)χ > 0

for some small ε > 0. Let Ysing be the singular set of Y . We take ϕ ∈ PSH(UY \Ysing, εχ)∩C∞(UY \Ysing) such that ϕ→ −∞ at Ysing. This implies that

ωY +√−1∂∂ϕ− (cot(θ0) + ε)χ > 0.

on UY \Ysing. For simplicity, we assume that Ysing is a point p. So by Case 1, there existsa neighborhood Up of p and ϕp ∈ C∞(Up) such that

Qχ,n,n(ωp) < θ0, ωp := ωY +√−1∂∂ϕp

on Up b UY . By subtracting a large constant from ϕp we have ϕp < ϕ − 2 on UY \Vpfor some neighborhood Vp b Up of p. Also since ϕ has positive Lelong number alongYsing, we know that ϕ(x)→ −∞ as x→ p. So there exists a neighborhood Wp b Vp ofp such that ϕp > ϕ+ 2 on Wp. Let Y1, . . . , Y` be the components of Y in UY \Wp whichare disjoint smooth open curves. Define a function

ϕ := ϕ+ Ad2

3We take θ0 that makes (3.1) an equality. If Theorem 3.1 is true for Y = X, then the condition(3.2) contradicts the choice of θ0.


for some constant A > 0, where d is the distance function to Y1, . . . , Y` with respectto any fixed Kahler metric on Y . So ωY := ωY +

√−1∂∂ϕ produces large positive

eigenvalues along normal direction of⋃`i=1 Yi as A increases. This implies that there

exists a large constant A such that

Qχ,n,n(ωY ) < θ0

on some sufficiently small open neighborhood U of Y \Wp in X. After shrinking U if

necessary, we may assume that ϕp < ϕ− 1 on U\Vp and ϕp > ϕ+ 1 on U ∩Wp. Then

we may take the regularized maximum ϕY of (U , ϕ) and (Vp, ϕp) (cf. [Dem12, Section5.E])). The form ωU := ωY +

√−1∂∂ϕY on a sufficiently small open neighborhood U

of Y satisfies the desired condition as in Lemma 3.2. Indeed, as pointed out in [Che21,Section 4], the regularized maximum preserves the upper bound Qχ,n,n(ωU) < θ0 dueto the monotonicity and concavity properties (cf. Proposition 2.2).

Using the induction argument, we assume that Theorem 3.1 is true for all quadruple(Y,X, α, β) such that (X,α, β) is stable along some test family and dimY 6 m− 1 6n−2. Now let X be an n-dimensional analytic variety satisfying the stability conditionas in Theorem 3.1 for some test family ωt,0 ∈ αt (t ∈ [0,∞)) emanating from ω0 ∈ α,and Y an m-dimensional analytic subvariety of X. So we have to show that

Γχ,α,θ0,θ0(Y,X) 6= ∅.By Lemma 3.2, we may assume that n > 3 and 2 6 m 6 n− 1. We take the resolution

of singularities Φ: X ′ → X for Y such that the strict transform Y of all componentsof Y by Φ is a disjoint union of smooth submanifolds of X ′. We assume that Y isirreducible for simplicity in later arguments (in general case, we apply Theorem 6.1 toeach component separately, and then prove Theorem 3.1 by the same argument). Let

Ysing be the singular set of Y and set E0 := Y ∩Φ−1(Ysing) so that χ is non-degenerate on

Y \E0. We can choose Φ to be successive blow-ups along smooth centers. This impliesthat there exist a metric hE0 on the line bundle associated to E0 and a small constant

κ0 > 0 such that χ−κFhE0is a Kahler form on Y for all κ ∈ (0, κ0], where FhE0

denotesthe curvature of hE0 . In particular, we set

ξ := χ− κ0FhE0,

andφ := κ0 log |σE0|2hE0

∈ PSH(Y , χ) ∩ C∞(Y \E0), (3.3)

where σE0 is the defining section of the line bundle associated to E0. Then the functionφ has positive Lelong number along E0 and satisfies

ξ = χ+√−1∂∂φ

on Y \E0. By the definition of test families, changing variables if necessary, we mayassume that

ω1,0 − cot(θ0

n

)χ > 0. (3.4)

For % > 0, set

ω0 = ω0(t, %) := ωt,0 + %tξ, χ = χ(t, %) := χ+ (%t)nξ.


Then the associated cohomology class is given by

α = α(t, %) := αt + %t[ξ], β = β(t, %) := β + (%t)n[ξ]

respectively.

Lemma 3.3. There exists a constant %0 ∈ (0,min(

tan(θ02n

))1/(n−1), 1/100) and cY >

0 such that for any t ∈ [0, 1], % ∈ [0, %0] and p-dimensional subvariety V of Y (p =1, . . . ,m), we have(

Re(α +√−1β)m − cot(θ0)Im(α +

√−1β)m

)· Y > cY [ξ]m · Y

and (Re(α +

√−1β)p − cot(θ0)Im(α +

√−1β)p

)· V > cY (%t)p[ξ]m · V.

Proof. We may assume that V is irreducible and set

A%,t := 1 +√−1(%t)n−1.

Case 1 (p = m): In this case, a direct computation shows that

Im(e−√−1θ0(α +

√−1β)m

)· Y

= Im(e−√−1θ0(αt +

√−1β)m

)· Y

+Im(e−√−1θ0

m−1∑i=0

(%t)m−i(m

i

)(αt +

√−1β)i · Am−i%,t [ξ]m−i

)· Y .

By the stability assumption and m 6 n − 1, we know that the first term is negativefor all t ∈ [0, 1]. We can observe that for all t ∈ [0, 1], there exists c > 0 such that

Im(e−√−1θ0(αt+

√−1β)m) · Y < −c[ξ]m · Y if % > 0 is sufficiently small since the second

term is of order O(%t) and αt is fixed.

Case 2 (p = 1, . . . ,m− 1): Assume t% > 0. Then a direct computation shows that

Im(e−√−1θ0(α +

√−1β)p

)· V

= Im(e−√−1θ0(αt +

√−1β)p

)· V

+

p−1∑i=1

(%t)p−i(p

i

)Im(e−√−1θ0(αt +

√−1β)i · Ap−i%,t [ξ]p−i

)· V

+(%t)pIm(e−√−1θ0Ap%,t[ξ]

p)· V

=5∑j=1

Ij,


where

I1 := Im(e−√−1θ0(αt +

√−1β)p

)· V,

I2 :=

p−1∑i=1

(%t)p−i(p

i

)Re(Ap−i%,t )Im

(e−√−1θ0(αt +

√−1β)i

)· [ξ]p−i · V,

I3 :=

p−1∑i=1

(%t)p−i(p

i

)Im(Ap−i%,t )Re

(e−√−1θ0(αt +

√−1β)i

)· [ξ]p−i · V,

I4 := −(%t)pRe(Ap%,t) sin(θ0)[ξ]p · V,

I5 := (%t)pIm(Ap%,t) cos(θ0)[ξ]p · V.

Let W := Φ(V ), so W is a subvariety of X of dimW 6 p 6 m − 1. By the induc-tion hypothesis, there exists a real (1, 1)-form ωW ∈ Γχ,α,θ0,θ0(W,X). We consider thesmooth family ωW,t emanating from ωW :

ωW,t := ωt,0 + ωW − ω0 ∈ αt.

Since ωW,t > ωW , we have ωW,t ∈ Γχ,αt,θ0,θ0(W,X) for all t ∈ (0, 1], which shows that

Im(e−√−1θ0(ωW,t +

√−1χ)k) < 0

on a neighborhood of W in X for all k = 1, . . . , n− 1 and t ∈ (0, 1] by Lemma 2.3. LetωV,t := Φ∗ωW,t. Then we have

Im(e−√−1θ0(ωV,t +

√−1χ)k) 6 0,

on a neighborhood of V in X ′ for all k = 1, . . . , n − 1 and t ∈ (0, 1]. In particular, wehave I1 6 0 for all t ∈ (0, 1]. To compute I2 and I4, we observe that

Re(Ai%,t) > 0, Im(Ai%,t) > 0

for all i = 1, . . . , p and t ∈ (0, 1] by taking % > 0 sufficiently small so that %n−1 <tan(π2n

). So we have I2 6 0 and I4 < 0.

To deal with bad terms I3, I5, we observe that exists a constant C > 0 (dependingonly on θ0, ωt,0 and χ) such that

Re(e−√−1θ0(ωt,0 +

√−1χ)i

)6 Cξi

on Y for all i = 1, . . . , p− 1 and t ∈ (0, 1]. This shows that

I3 < I ′3 := C

p−1∑i=1

(%t)p−i(p

i

)Im(Ap−i%,t )[ξ]p · V.

We observe that

Im(Ai%,t) = (%t)n−1 +O((%t)n).


Then the coefficients of [ξ]p · V in I ′3, I5 are polynomials of %t, which are of orderO((%t)n), O((%t)n+p−1) respectively. On the other hand, the coefficients of [ξ]p · V in I4

is a polynomials of %t of the form

− sin(θ0)(%t)p +O((%t)2n+p−2).

Then there exists c > 0 independent of V such that I ′3 + I4 + I5 < −c(%t)p[ξ]m · V forall t ∈ (0, 1] and sufficiently small % > 0. This completes the proof.

We will consider the following twisted dHYM equation for ω ∈ α on Y

Re(ω +√−1χ)m − cot(θ0)Im(ω +

√−1χ)m − ct,%χm = 0, (3.5)

where ct,% is a normalized constant defined by

ct,%βm · Y = (Re(α +

√−1β)m − cot(θ0)Im(α +

√−1β)m) · Y . (3.6)

The following lemma is crucial, and a direct consequence from Lemma 3.3.

Lemma 3.4. The constant ct,% depends smoothly on t, %, and is positive for all t ∈ [0, 1]and % ∈ [0, %0].

We take a constant Θ0 ∈ (θ0, π) (depending only on n and θ0) sufficiently close to θ0

so that

Θ0 − θ0 <π −Θ0

n.

For any fixed % ∈ (0, %0], we set

T% := t ∈ (0, 1]|(3.5) has a smooth solution ω(t, %) ∈ Γχ,α,θ0,Θ0(Y ).

We remark that for any t ∈ T%, the solution ω ∈ Γχ,α,θ0,Θ0(Y ) to (3.5) automaticallysatisfies

Pχ(ω) < Qχ(ω) < θ0 (3.7)

since Qχ(ω) < Θ0 < π and ct,% > 0.

Lemma 3.5. For any fixed % ∈ (0, %0], the set T% is open in (0, 1]. Moreover, we have1 ∈ T%.

Proof. Openness follows from Proposition 2.8 and the fact that ct,% > 0 for all t ∈ (0, 1].When t = 1, by (3.4), we know that

ω0 > cot(θ0

n

)χ+ %ξ, χ = χ+ %nξ,

and

Pχ(ω0) < Qχ(ω0) 6 Qχ

(cot(θ0

n

)χ+ %ξ

).

Let µi be the eigenvalues of χ with respect to ξ. Then

Qχ

(cot(θ0

n

)χ+ %ξ

)=

m∑i=1

arccotcot(θ0n

)µi + %

µi + %n<

m∑i=1

arccot

(cot(θ0

n

))< θ0

since %n−1 6 %n−10 < tan

(θ0n

). Thus we have 1 ∈ T% by Proposition 2.8.

Set t% := inf T%.


Lemma 3.6. For any fixed % ∈ (0, %0], we have t% = 0.

Proof. Assume t% > 0. Take any element η ∈ α(t%) and consider the test familyemanating from η

ζs,t% := η + sξ, s ∈ [0,∞)

with respect to the background metric χ(t%). There exists a constant C > 0 such that

χ(t%) 6 Cξ. For any p-dimensional analytic subvariety V of Y (p = 1, . . . ,m), let usconsider the function F Stab

θ0(V, ζs,t%, s). By Lemma 3.3, we have

F Stabθ0

(V, ζs,t%, 0) > cY (%t%)p[ξ]p · V > cY (C−1%t%)

p[χ(t%)]p · V

By the definition of T%, for any t ∈ (t%, 1] we have a solution ω(t) ∈ Γχ(t),α(t),θ0,Θ0(Y )to (3.5). In order to compute F Stab

θ0(V, ζs,t%, s), for any t ∈ (t%, 1], let us consider the

test family

ζs,t := ω(t) + sξ, s ∈ [0,∞)

emanating from ω(t) with the background metric χ(t). Since ω(t) ∈ Γχ(t),α(t),θ0,Θ0(Y ),

we know that ζs,t ∈ Γχ(t),α(t),θ0,Θ0(Y ) by the monotonicity of Pχ(t), Qχ(t), and henceddsF Stabθ0

(V, ζs,t, s) > 0 for all s ∈ [0,∞). Moreover, since [ζs,t]→ [ζs,t% ] as t→ t% anddds

[ζs,t] = dds

[ζs,t% ] = [ξ], we have

d

dsF Stabθ0

(V, ζs,t%, s) = limt→t%

d

dsF Stabθ0

(V, ζs,t, s) > 0

for all s ∈ [0,∞), where we used the fact that ddsF Stabθ0

(V, ζs,t, s) is a cohomologicalinvariant. This shows that

F Stabθ0

(V, ζs,t%, s) > F Stabθ0

(V, ζs,t%, 0) > cY (C−1%t%)p[χ(t%)]

p · V

for all s ∈ [0,∞), so the triple (Y , α(t%), β(t%)) together with the test family ζs,t% satisfythe uniform stable assumption of [Che21, Proposition 5.2]. Then there exists a solution

ω(t%) ∈ Γχ(t%),α(t%),θ0,Θ0(Y ) to (3.5). This is a contradiction.

4. The twisted dHYM equation on the product space Y

In this section, we consider the twisted dHYM equation on the product space Y :=

Y × Y as in [Che21, DP04, Son20]. Set ∆ := (p, p) ∈ Y|p ∈ Y and suppose that∆ is locally defined by holomorphic functions fj,kj,k on finitely many domains Ujcovering Y . We define

ψs := log

(∑j

ρj∑k

|fj,k|2 + s2

)for s ∈ (0, 1] where ρj is a partition of unity for Uj. Let π1, π2 be the projection

maps from Y → Y and set

χY := π∗1χ+ π∗2χ, (4.1)

χY,s := χY − `κ0(π∗1FhE0+ π∗2FhE0

) + `2√−1∂∂ψs, (4.2)


where the function φ ∈ PSH(Y , χ) ∩ C∞(Y \E0) is defined in Section 3, and the smallconstant ` > 0 is determined later. If we set∫

Yχ2mY,s = (1 + ct,%,`)

∫Yχ2mY , (4.3)

then we have lim`→0 ct,%,` = 0 uniformly for t ∈ [0, 1] and % ∈ [0, %0] (we note that theconstant ct,%,` is independent of s ∈ (0, 1]). We fix a constant K > 0 (depending onlyon m and θ0) such that

K > cot(π − θ0

n

),

and define constants ζK , θ0 by

ζK := m arccot(K), 0 < θ0 := θ0 + ζK < π.

Now let us consider the twisted dHYM equation for ωY,s ∈ π∗1α +Kπ∗2β on Y

Re(ωY,s +√−1χY)2m − cot(θ0)Im(ωY,s +

√−1χY)2m − FY,sχ2m

Y = 0, (4.4)

where the smooth function FY,s is given by

FY,s :=χ2mY,s

χ2mY

+ ct,%sin(θ0)

sin(θ0)(K2 + 1)m/2 − (1 + ct,%,`). (4.5)

Also we set

RK := Re(K +√−1)m, IK := Im(K +

√−1)m > 0.

Then we have IK > 0 since 0 < ζK < π. Also we note that RK/IK = cot(ζK).

Lemma 4.1. There exists ` > 0 such that the equation (4.4) has a solution ωY,s ∈ΓχY ,π∗1 α+Kπ∗2 β,θ0,θ0

(Y) for all t ∈ (0, 1], % ∈ (0, %0] and s ∈ (0, 1].

Proof. In order to prove the existence of the solution, we may check the conditionsimposed in Proposition 2.8. First, we show the existence of subsolution. Set ω†Y :=π∗1ω+Kπ∗2χ, where ω is a solution to (3.5). Let λi be the eigenvalues of ω with respect

to χ. Then the eigenvalues of ω†Y with respect to χY are given by

(λ1, . . . , λm︸︷︷︸m

, K, . . . ,K︸︷︷︸m

).

Combining with (3.7), we have

QχY (ω†Y) = Qχ(ω) +m arccot(K) < θ0 +m arccot(K) = θ0.

Next, we check the integrals of both sides of (4.4) on Y are equal:∫YFY,sχ

2mY =

(Re(ωY,s +

√−1χY)2m − cot(θ0)Im(ωY,s +

√−1χY)2m

)· Y .


Using (4.3) and (3.6), we compute∫YFY,sχ

2mY

= ct,%sin(θ0)

sin(θ0)(K2 + 1)m/2 ·

(2m

m

)·(βm · Y

)2

= − (K2 + 1)m/2

sin(θ0)·(

2m

m

)·(βm · Y

)∫Y

Im(e−√−1θ0(α +

√−1β)m

)= − (K2 + 1)m/2

sin(θ0)·(

2m

m

)·

Im

∫Ye−√−1θ0π∗2(e

√−1 arccot(K)β)m ∧ π∗1(α +

√−1β)m

= − 1

sin(θ0)

Im

∫Ye−√−1θ0

[(π∗1α +Kπ∗2β) +

√−1(π∗1β + π∗2β)

]2m

=(


√−1χY)2m

)· Y .

Last, we check infY FY,s > 0. From the argument in [DP04, Lemma 2.1 (ii)], thereexists A > 1 such that for any s ∈ (0, 1], we have

A(π∗1ξ + π∗2ξ) +√−1∂∂ψs > 0.

This shows that for any ε > 0,

χY,s − (1− ε)χY > π∗1γ` + π∗2γ`,

where

γ` := εχ− `κ0FhE0− A`2ξ

= (ε− A`2)

(χ− `κ0(1− A`)

ε− A`2FhE0

),

which shows that if ` < min1/2A,√ε/2A, ε, we have γ` > 0, and hence

χY,s > (1− ε)χY > 0.

This implies that

FY,s =χ2mY,s

χ2mY

+ ct,%sin(θ0)

sin(θ0)(K2 + 1)m/2 − (1 + ct,%,`)

> εχ2mY,s

χ2mY

+ (1− ε)2m+1 + ct,%sin(θ0)

sin(θ0)(K2 + 1)m/2 − (1 + ct,%,`).

(4.6)

We find that the third term of the RHS has uniform positive lower bound for t ∈ (0, 1]and % ∈ (0, %0] by Lemma 3.4. Combining with the fact that lim`→0 ct,%,` = 0, we cantake ε > 0 sufficiently small so that for all s ∈ (0, 1], t ∈ (0, 1] and % ∈ (0, %0],

FY,s > εχ2mY,s

χ2mY

> 0. (4.7)


So we can apply Proposition 2.8 to get the solution ωY,s, which automatically satisfies

PχY (ωY,s) < QχY (ωY,s) < θ0

since infY FY,s > 0. This completes the proof.

Remark 4.2. In the above proof, the crucial point is that the third term of the RHS of(4.6) has a uniform positive lower bound, which is not true for the case m = n sincethe constant ct,% may be equal to zero when t = % = 0.

In later arguments, we fix ε > 0 and ` > 0 in the proof of Lemma 4.1. In the courseof the proof, we also find that

χY,s > π∗1γ` + π∗2γ` > 0,

so χY,s is Kahler for any s ∈ (0, 1], t ∈ [0, 1] and % ∈ [0, %0].

5. The mass concentration and local smoothing

In this section, we will construct a subsolution as a current, and establish someestimates for its local smoothing. Compared to [Son20, Section 5], we need to makesome modifications to deal with the problem that the solution ωY,s to (4.4) is no longerKahler. As pointed out in [Che21, Section 5], we can overcome this by using a uniformlower bound for ωY,s which we will discuss at first. Since ωY,s ∈ ΓχY ,π∗1 α+Kπ∗2 β,θ0,θ0

(Y),

we get a lower bound of a real (1, 1)-form ΨY,s defined by

ΨY,s := ωY,s + Cθ0χY > cθ0χY (5.1)

for some fixed constants Cθ0 , cθ0 > 0 depending only on θ0. Thus ΨY,s is Kahler andafter passing to a sequence si, we know that for all k = 1, . . . ,m, the (k, k)-formΨkY,si converges to some closed positive (k, k)-current Ξk = Ξk(t, %) weakly when si → 0

and t, % are fixed. For any η > 0, let ∆η be the η-neighborhood of the diagonal set ∆with respect to any fixed Kahler metric on Y . We set

V := Im(K +√−1)mβm · Y > Im(K +

√−1)mβm · Y > 0.

Lemma 5.1. We have the following:

(1) For k = 1, . . . ,m−1 and any positive (m−k,m−k)-form γ, the (m,m)-currentΞk ∧ γ has no concentration of mass on ∆, i.e. for any ε > 0, we have∫

∆η

Ξk ∧ γ < ε

for sufficiently small η > 0.(2) There exists δ0 ∈ (0, 1) (independent of t, % and the choice of si) such that

for any t ∈ (0, 1] and % ∈ (0, %0], we have Ξm > 100Vδ0[∆].

Proof. By [Dem12, Corollary III. 2.11], we have 1∆Ξk = 0 and so 1∆(Ξk∧γ) = 0, whichimplies (1).

As for (2), let us recall the twisted dHYM equation (4.4):


√−1χY)2m − FY,sχ2m

Y = 0.


Combining this with (5.1) and (4.7), we obtain

Ψ2mY,s > c0FY,sχ

2mY > c0ε

(χY,s(t, %)

)2m> c0ε

(χY,s(0, 0)

)2m, (5.2)

where c0 > 0 is a constant depending only on θ0 and ε is the fixed constant in the endof Section 4. On the other hand, we know that 1∆Ξm is a closed positive current withsupport in ∆ by Skoda’s extension theorem. Hence 1∆Ξm = δ′[∆] for some δ′ > 0 bythe support theorem. In the same way as [DP04, Proposition 2.6] (or the argumentsin [Che21, Section 3] for the J-equation case), one can see that there exists δ0 ∈ (0, 1)such that Ξm > 1∆Ξm = 100Vδ0[∆]. This completes the proof.

For k = 1, . . . ,m, we set

ω+Y,k,s :=

∑i:even

(k

i

)Ciθ0

Ψk−iY,s ∧ χ

iY , ω−Y,k,s := −

∑i:odd

(k

i

)Ciθ0

Ψk−iY,s ∧ χ

iY ,

Then we know that ±ω±Y,k,s defines a closed positive (k, k)-form with a decomposition

ωkY,s = (ΨY,s − Cθ0χY)k = ω+Y,k,s + ω−Y,k,s, ω+

Y,k,s > ΨkY,s.

The form ω+Y,k,si , ω

−Y,k,si converges to

Υ+k = Υ+

k (t, %) :=∑i:even

(k

i

)Ciθ0

Ξk−i ∧ χiY , Υ−k = Υ−k (t, %) := −∑i:odd

(k

i

)Ciθ0

Ξk−i ∧ χiY .

as si → 0 respectively. Again ±Υ±k are closed positive (k, k)-currents. We remark thatby Lemma 5.1, for any fixed positive (m− k,m− k)-form γ, Υ+

k ∧ γ (k = 1, . . . ,m− 1)and −Υ−k ∧ γ (k = 1, . . . ,m) have no concentration of mass on ∆, whereas the currentΥ+m satisfies Υ+

m > Ξm > 100Vδ0[∆].Now we recall the argument in [Che21, Section 5]. We define a real closed (1, 1)-form

ωs by the fiber integration:

ωs := ωs(t, %) = V−1

bm−12c∑

k=0

(−1)km!

(m− 2k)!(2k + 1)!(π1)∗(ω

m−2kY,s ∧ π∗2χ2k+1).

Then one can check that ωs ∈ α as follows:

[ωs] = V−1

bm−12c∑

k=0

(−1)km!

(m− 2k)!(2k + 1)!(π1)∗

([ωY,s]

m−2k ∧ [π∗2χ]2k+1)

= V−1

bm−12c∑

k=0

(−1)k1

m− 2k

(m

2k + 1

)(π1)∗

([π∗1ω +Kπ∗2χ]m−2k ∧ [π∗2χ]2k+1

)= V−1

bm−12c∑

k=0

(−1)k(

m

2k + 1

)(π1)∗

([π∗1ω] ∧ [Kπ∗2χ]m−2k−1 ∧ [π∗2χ]2k+1

)= V−1(π1)∗

[π∗1ω] ∧ [π∗2Im(Kχ+

√−1χ)m]

= [ω].


Now we will compute Qχ(ωs). We write

ωY,s = ωH + ωM + ωMT + ωV ,

where ωH is the horizontal component, ωV is the vertical component, ωM and ωMT are

the off-diagonal or the mixed components of ωY,s respectively. At any point p ∈ π−11 (z)

for z ∈ Y , we can choose a coordinate so that

χH = π∗1χ =√−1

m∑i=1

dzi ∧ dzi, χV = π∗2χ =√−1

m∑i=1

dwi ∧ dwi,

ωH =√−1

m∑i=1

µidzi ∧ dzi, ωV =√−1

m∑i=1

λidwi ∧ dwi,

ωM =m∑

i,j=1

√−1ωMij dzi ∧ dwj, ω

MT = ωM .

Then the calculation of [Che21, Proof of Proposition 5.13] shows that ωs can be ex-pressed as

ωs = V−1(π1)∗

(ωY,s ∧ Im(ωV +

√−1χV )m

), (5.3)

where in this coordinate, the form ωY,s is given by

ωY,s :=m∑

i,j,l=1

(µiδij − ωMil

Im∏m

k=1,k 6=l(λk +√−1)

Im∏m

k=1(λk +√−1)

ωMjl

)dzi ∧ dzj.

By Lemma 4.1 and [Che21, Lemma 5.12], we know that for all t ∈ (0, 1], % ∈ (0, %0] ands ∈ (0, 1], we have

QχH (ωY,s) +QχV (ωV ) 6 θ0, (5.4)

where QχH (ωY,s) is taken for the horizontal component of ωY,s. In particular, this showsthat the (m,m)-form Im(ωV +

√−1χV )m defines a positive measure when restricted to

each fiber.

Lemma 5.2. For any t ∈ (0, 1], % ∈ (0, %0] and s ∈ (0, 1], we have

Qχ(ωs) 6 θ0.


Proof. The proof is identical to [Che21, Proposition 5.13]. By using the monotonicity

and concavity property for cot(Qχ(·)) with (5.4), for any z ∈ Y we observe that

1

cot(Qχ(ωs(z)))− cot(θ0)6 V−1

∫π−11 (z)

Im(ωV +√−1χV )m

cot(QχH (ωY,s))− cot(θ0)

6 V−1

∫π−11 (z)


cot(θ0 −QχV (ωV ))− cot(θ0)

= V−1

∫π−11 (z)

cot(QχV (ωV ))− cot(θ0)

1 + cot2(θ0)Im(ωV +

√−1χV )m

= V−1

∫π−11 (z)

Re(ωV +√−1χV )m − cot(θ0)Im(ωV +

√−1χV )m

1 + cot2(θ0)

= V−1

∫Y

Re(Kχ+√−1χ)m − cot(θ0)Im(Kχ+

√−1χ)m

1 + cot2(θ0)

=RK − cot(θ0)IK

(1 + cot2(θ0))IK=

1

cot(θ0)− cot(θ0),

where we used RK/IK = cot(ζK) and θ0 = θ0 + ζK in the last equality.

Now let us introduce the local smoothing on the manifold Y .

Definition 5.3. Let T be a locally L1-integrable function or closed (1, 1)-current on Y ,

and R > 0 a constant. Then the local smoothing T (r) = T (r)j j∈J of T with respect to

a finte covering Bj,3Rj∈J of Y and a scale r ∈ (0, R) is defined as follows:

(1) For each j ∈ J , Bj,4R is a coordinate chart of Y which is biholomorphic to aEuclidean ball B4R(0) of radius 4R centered at the origin in Cm with respect tothe standard Euclidean metric gj. We assume that Bj,R ' BR(0) ⊂ Cm (j ∈ J )

is also a covering of Y .

(2) T(r)j (z) is the smoothing on Bj,3R defined by the following convolution

T(r)j (z) :=

∫Br(0)

r−2mρ

(|y|r

)T (z − y)dVolgj(y).

for r ∈ (0, R), where ρ(t) is a smooth non-negative function with support in [0, 1],and is constant on [0, 1/2]. Moreover, the function ρ satisfies the normalizationcondition ∫

B1(0)

ρ(|y|)dVolgj(y) = 1.

By taking R > 0 sufficiently small, we can always assume that

1

2gj 6 ξ 6 2gj (5.5)

on each Bj,4R. Also for any small ε ∈ (0, 1], we can choose a sufficiently small R > 0and a finite covering Bj,3Rj∈J so that there exists a smooth family of Kahler metrics


χj = χj(t, %) with constant coefficients such that

χj 6 χ 6 χj + ε ξ (5.6)

on each Bj,4R for all t ∈ (0, 1] and % ∈ (0, %0]. We may further assume that for any fixedR > 0 and a finite covering Bj,3Rj∈J , there exists r0 ∈ (0, R) such that for all r < r0,

χ− 1

10ξ 6 χ(r) 6 χ+

1

10ξ,

9

10ξ 6 ξ(r) 6

11

10ξ (5.7)

on each Bj,3R since χ and ξ are smooth and convergence of the smoothing is uniformon Bj,3R. Also we may assume that

χ 6 2ξ (5.8)

for all t ∈ (0, 1] and % ∈ (0, %0] since %0 < 1/100. In this section, we will show thefollowing:

Theorem 5.4. Let δ0 > 0 be a constant determined in Lemma 5.1. Then there existconstants ε = ε(ω0, χ, δ0, θ0) ∈ (0, 1], ε0 = ε0(ω0, χ, δ0, θ0) > 0 such that for any finitecovering Bj,3R and r0 ∈ (0, R) satisfying (5.5), (5.6) and (5.7), there exists a closed(1, 1)-current Ω ∈ α− δ0β satisfying the following properties:

(1) Ω + Aχ is a Kahler current for some constant A > 0, and has positive Lelongnumber along E0.

(2) For any r ∈ (0, r0), we have

Qχ(Ω(r)) 6 θ0 − ε0 (5.9)

on each Bj,3R\E0.

In order to prove Theorem 5.4, we perform the truncation as in [Che21, Son20]. Forany z ∈ Y and η > 0, we set

Ωs,η(z) := V−1

(∫π−11 (z)\∆η

ωY,s∧Im(ωV +√−1χV )m+

∫π−11 (z)∩∆η

KχY∧Im(ωV +√−1χV )m

).

Lemma 5.5. For any t ∈ (0, 1], % ∈ (0, %0], s ∈ (0, 1] and η > 0, we have

1

cot(Qχ(Ωs,η(z)))− cot(θ0)

61

cot(θ0)− cot(θ0)+

cot(θ0)− cot(θ0)

1 + cot2(θ0)V−1

∫π−11 (z)∩∆η

Im(ωV +√−1χV )m.


Proof. By using the monotonicity and concavity of cot(Qχ(·)) with (5.4), we compute

1


6 V−1

∫π−11 (z)\∆η


cot(QχH (ωY,s))− cot(θ0)+ V−1

∫π−11 (z)∩∆η


cot(QχH (KχH))− cot(θ0)

6 V−1

∫π−11 (z)\∆η


cot(θ0 −QχV (ωV ))− cot(θ0)+ V−1

∫π−11 (z)∩∆η


cot(QχH (KχH))− cot(θ0)

= V−1

∫π−11 (z)\∆η



√−1χV )m

+V−1

∫π−11 (z)∩∆η



√−1χV )m.

We apply the computation of Lemma 5.2 (from third line to last line) to the first term,and obtain

1


61

cot(θ0)− cot(θ0)− V−1

∫π−11 (z)∩∆η



√−1χV )m

+V−1

∫π−11 (z)∩∆η



√−1χV )m

=1

cot(θ0)− cot(θ0)+ V−1

∫π−11 (z)∩∆η

cot(θ0)− cot(QχV (ωV ))


√−1χV )m

61



1 + cot2(θ0)V−1

∫π−11 (z)∩∆η

Im(ωV +√−1χV )m,

where we used QχV (ωV ) 6 θ0 (cf. (5.4)) in the last inequality.

Since ∫π−11 (z)∩∆η

Im(ωV +√−1χV )m 6

∫π−11 (z)

Im(ωV +√−1χV )m = V ,

from the above lemma, we immediately obtain the following:

Corollary 5.6. There exist constants θ1 ∈ (θ0, θ0), (depending only on θ0) such thatfor any t ∈ (0, 1], % ∈ (0, %0], s ∈ (0, 1] and η > 0, we have

Qχ(Ωs,η) 6 θ1.

Lemma 5.7. There is σ0(θ0, n) ∈ (0, δ0) such that we can choose ε = ε(σ0, θ0) ∈ (0, 1]so that the following property holds: assume that χ satisfies (5.6) on each Bj,4R forthe constant ε. Then for any t ∈ (0, 1], % ∈ (0, %0] and r ∈ (0, R), there exist η0 =


η0(ε, t, %, r) > 0, s0 = s0(ε, t, %, r) ∈ (0, 1] such that for any η ∈ (0, η0), s ∈ (0, s0), wehave

Qχ(Ω(r)s,η + σ0ξ) < θ0

on each Bj,3R.

Proof. We first let σ0 be the constant obtained from Proposition 2.5. Decreasing σ0 ifneeded, we may assume σ0 < δ0. By Proposition 2.5, choosing ε sufficiently small, forany z ∈ Bj,3R, r ∈ (0, R) and y ∈ Br(0), we have

Qχj(Ωs,η + σ0ξ)|(z+y) < θ0.

Since χj has constant coefficients, combining the above with the concavity of cot(Qχj(·)),Lemma 5.5 and Proposition 2.5, decreasing ε if necessary, for any z ∈ Bj,3R andr ∈ (0, R), we have

1

cot(Qχj(Ω(r)s,η(z) + σ0ξ(r)(z)))− cot(θ0)

6∫y∈Br(0)

r−2mρ(r−1|y|) 1

cot(Qχj(Ωs,η + σ0ξ)|(z+y))− cot(θ0)dVolgj(y)

6∫y∈Br(0)

r−2mρ(r−1|y|) 1

cot(Qχ(Ωs,η)(z + y))− cot(θ0)dVolgj(y)

61



1 + cot2(θ0)2mr−2mV−1

∫∆η

Im(ωV +√−1χV )m ∧ π∗1ξm.

(5.10)

We note that∫∆η

Im(ωV +√−1χV )m ∧ π∗1ξm =

∫∆η

Im(ωY,s +√−1χV )m ∧ π∗1ξm.

So this is a linear combination of∫∆η

ω±Y,m−2k−1,s ∧ χ2k+1V ∧ π∗1ξm, k = 0, . . . ,

⌊m− 1

2

⌋,

which has no concentration of mass on ∆ by Lemma 5.1. This yields that for any ε > 0,t ∈ (0, 1], % ∈ (0, %0] and r ∈ (0, R), there exist η0 = η0(ε, t, %, r) > 0, s0 = s0(ε, t, %, r) ∈(0, 1] such that for any η ∈ (0, η0), s ∈ (0, s0) we have


1 + cot2(θ0)2mr−2mV−1

∫∆η

Im(ωV +√−1χV )m ∧ π∗1ξm 6

ε

2.

Otherwise, there exists some k, a constant ε3 > 0 and sequences ηi → 0, si → 0 (wemay further assume that these sequences are monotone) such that

±∫

∆ηi

ω±Y,m−2k−1,si∧ χ2k+1

V ∧ π∗1ξm > ε3


holds for all i. In particular, for any i > i′ we have

±∫

∆ηi′

ω±Y,m−2k−1,si∧ χ2k+1

V ∧ π∗1ξm > ε3.

This yields a contradiction by letting i→∞ and followed by i′ →∞.

By (5.10), we may choose ε sufficiently small, so that Qχj(Ω(r)s,η+σ0ξ

(r)) < θ0 + 1100c0σ0

where c0 is the constant from Proposition 2.4. By Proposition 2.1, (5.8) and (5.7), weconclude that

Qχj(Ω(r)s,η + 10σ0ξ) 6 Qχj(Ω

(r)s,η + σ0ξ

(r))− 5c0σ0 < θ0.

Using Proposition 2.5 and (5.6) again, we obtain

Qχ(Ω(r)s,η + 100σ0ξ) < θ0.

This completed the proof by relabelling the constants.

As with (5.3), we observe that

bm−12c∑

k=0

(−1)km!

(m− 2k)!(2k + 1)!

∫π−11 (z)∩∆η

ωm−2kY,s ∧π

∗2χ

2k+1 =

∫π−11 (z)∩∆η

ωY,s∧Im(ωV +√−1χV )m.

Now we rewrite Ωs,η as

Ωs,η = ωs − ωs,η − ω′s,η + ω′′s,η,

where

ωs,η(z) := V−1

∫π−11 (z)∩∆η

ω+Y,m,s ∧ π

∗2χ,

ω′s,η(z) := V−1

∫π−11 (z)∩∆η

ω−Y,m,s ∧ π∗2χ

+ V−1

bm−12c∑

k=1

(−1)km!

(m− 2k)!(2k + 1)!

∫π−11 (z)∩∆η

ωm−2kY,s ∧ π∗2χ2k+1,

ω′′s,η(z) := V−1

∫π−11 (z)∩∆η

KχY ∧ Im(ωV +√−1χV )m.

and we will consider each term separately.

Lemma 5.8. For any t ∈ (0, 1], % ∈ (0, %0], r ∈ (0, r0), η > 0 and z ∈ Y , there exists1 = s1(t, %, r, η, z) > 0 such that for all s ∈ (0, s1) we have

(ωs,η + 100δ0

√−1∂∂φ)(r) > 80δ0ξ

at z ∈ Bj,3R.

Proof. Assume the statement does not hold at some point z ∈ Bj,3R. So there exists a

sequence si → 0 and vi ∈ T 1,0z Y with |vi|gj = 1 such that for all i we have

(ωsi,η + 100δ0

√−1∂∂φ)(r)(vi, vi) < 80δ0ξ(vi, vi)


at z ∈ Bj,3R. After passing to a further subsequence, we may assume that ΨmY,s converges

weakly to a closed positive current Ξm > 100Vδ0[∆]. Then we compute

limi→∞

ωsi,η(r)(z)

= limi→∞

∫y∈Br(0)

r−2mρ(r−1|y|)ωsi,η(z + y)dVolgj(y)

= V−1 limi→∞

∫y∈Br(0)

r−2mρ(r−1|y|)(∫

w∈π−11 (z+y)∩∆η

ω+Y,m,si(z + y, w) ∧ χV (w)

)dVolgj(y)

> V−1 limi→∞

∫y∈Br(0)

r−2mρ(r−1|y|)(∫

w∈π−11 (z+y)∩∆η

ΨmY,si(z + y, w) ∧ χV (w)

)dVolgj(y)

= V−1 limi→∞

∫(y,w)∈(Br(0)×Y )∩∆η

r−2mρ(r−1|y|)ΨmY,si(z + y, w) ∧ χV (w)dVolgj(y)

> 100δ0

∫(y,w)∈(Br(0)×Y )∩∆

r−2mρ(r−1|y|)χV (w)dVolgj(y)

= 100δ0

∫y∈Br(0)

r−2mρ(r−1|y|)χ(z + y)dVolgj(y),

where we used w = z + y on Bj,4R in the last equality. So by definition of φ, we have

limi→∞

(ωsi,η + 100δ0

√−1∂∂φ)(r)(z)

> 100δ0

∫y∈Br(0)

r−2mρ(r−1|y|)(χ+√−1∂∂φ)(z + y)dVolgj(y)

> 100δ0ξ(r)(z)

> 90δ0ξ(z)

by (5.7). In particular, for sufficiently large i, we have

(ωsi,η + 100δ0

√−1∂∂φ)(r)(z)(vi, vi) > 85δ0ξ(z)(vi, vi).

This is a contradiction.

Lemma 5.9. For any t ∈ (0, 1], % ∈ (0, %0], r ∈ (0, R) and z ∈ Y , there existss2 = s2(t, %, r, z) > 0, η1 = η1(t, %, r, z) > 0 such that for any s ∈ (0, s2), η ∈ (0, η1), wehave

ω′′s,η(r) 6 δ0ξ

at z ∈ Bj,3R.

Proof. Suppose that the statement does not hold. So there exist sequences si → 0,

ηi → 0 and vi ∈ T 1,0z Y with |vi|gj = 1 such that

ω′′si,ηi(r)(vi, vi) > δ0ξ(vi, vi)

at some point z ∈ Bj,3R. Let γ be any positive (m − 1,m − 1)-form on Bj,4R withconstant coefficients. We regard ω′′si,ηi

(r) ∧ γ as a real number by using the Euclidean


volume form Volgj at z. Then after passing to a further subsequence, we have

limi→∞

(ω′′si,ηi(r) ∧ γ)(z)

= limi→∞

∫y∈Br(0)

r−2mρ(r−1|y|)ω′′si,ηi(z + y) ∧ γ(z + y)

= V−1 limi→∞

∫y∈Br(0)

r−2mρ(r−1|y|)(∫

w∈π−11 (zi+y)∩∆ηi

(KχH ∧ Im(ωY,si +

√−1χV )m

)(z + y, w)

)∧γ(z + y)

6 V−1r−2m limi→∞

∫∆ηi

KχH ∧ Im(ωY,si +√−1χV )m ∧ ρ(r−1|y|)π∗1γ

= 0.

since the last integral has no concentration of mass on ∆ by the similar observation asin the proof of Lemma 5.7. Thus for sufficiently large i, we have

ω′′si,ηi(r)(z)(vi, vi) 6

δ0

2ξ(zi)(vi, vi).

This is a contradiction.

Recall that the form ω′s,η(z) can be written as the linear combination of the fiberintegral of

ω+Y,m−2k,s ∧ π

∗2χ

2k+1, k = 1, . . . ,

⌊n− 1

2

⌋and

ω−Y,m−2k,s ∧ π∗2χ

2k+1, k = 0, . . . ,

⌊n− 1

2

⌋on π−1

1 (z) ∩ ∆η, which have no concentration of mass on ∆. Thus we can show thefollowing lemma in the same way as Lemma 5.9:

Lemma 5.10. For any t ∈ (0, 1], % ∈ (0, %0], r ∈ (0, R) and z ∈ Y , there existss3 = s3(t, %, r, z) > 0, η2 = η2(t, %, r, z) > 0 such that for any s ∈ (0, s3), η ∈ (0, η2), wehave

ω′s,η(r) > −δ0ξ

at z ∈ Bj,3R.

Lemma 5.11. There exists a constant ε = ε(σ0, θ0) ∈ (0, 1] satisfying the followingproperties: assume that for ε, the form χ satisfies (5.6) on each Bj,4R. Then for any

t ∈ (0, 1], % ∈ (0, %0], r ∈ (0, r0) and z ∈ Y , there exists s4 = s4(ε, t, %, r, z) > 0 suchthat for any s ∈ (0, s4), we have

Qχ

((ωs − 25δ0χ+ 100δ0

√−1∂∂φ

)(r))< θ0

at z ∈ Bj,3R.


Proof. For any given t ∈ (0, 1], % ∈ (0, %0], r ∈ (0, r0) and z ∈ Y , constants s0, s2, s3, η0,η1, η2 are determined by the previous lemmas. We fix η > 0 so that η < minη0, η1, η2.Then for this η, we have s1 = s1(t, %, r, η, z) as in Lemma 5.8. We take s4 > 0 so thats4 < mins0, s1, s2, s3. Then for any s ∈ (0, s4), we have(

ωs − 25δ0χ+ 100δ0

√−1∂∂φ

)(r)

=(Ωs,η + (ωs,η + 100δ0

√−1∂∂φ)− 25δ0χ+ ω′s,η − ω′′s,η

)(r)

> Ω(r)s,η + 80δ0ξ − 55δ0ξ − δ0ξ − δ0ξ

> Ω(r)s,η + σ0ξ

at z ∈ Bj,3R, where we used (5.7), (5.8) in the third line, and σ0 < δ0 in the last line.Then Lemma 5.7 shows

Qχ

((ωs − 25δ0χ+ 100δ0

√−1∂∂φ

)(r))6 Qχ

(Ω(r)s,η + σ0ξ

)< θ0

at z ∈ Bj,3R as desired.

Now we are in a position to prove Theorem 5.4.

Proof of Theorem 5.4. By Lemma 5.2, we have ωs − cot(θ0)χ > 0 and so

ωs − 25δ0χ+ 100δ0

√−1∂∂φ+ Aχ > 100δ0ξ

for A := | cot(θ0)| + 200δ0 > 0. Thus after passing to a subsequence, we have a closedcurrent

ω? = ω?(t, %) := limsi→0

(ωsi − 25δ0χ+ 100δ0

√−1∂∂φ) ∈ α− 25δ0β.

On each Bj,4R, we take a quasi-PSH function hsi such that

ωsi − 25δ0χ+ 100δ0

√−1∂∂φ =

√−1∂∂hsi .

Then after passing to a subsequence, we may further assume that hsi converges to

some quasi-PSH function h? in L1 with ω? =√−1∂∂h?. Thus for any r ∈ (0, r0), h

(r)si

converges to h?(r) uniformly on Bj,3R. On the other hand, Lemma 5.11 implies that for

any t ∈ (0, 1], % ∈ (0, %0], r ∈ (0, r0) and z ∈ Y , there exists s4 > 0 such that

Qχ(√−1∂∂hsi

(r)) < θ0

hold at z ∈ Bj,3R for all i satisfying si < s4. Thus for any t ∈ (0, 1], % ∈ (0, %0] andr ∈ (0, r0), we have

Qχ(√−1∂∂h?(r)) 6 θ0

on Bj,3R. The current ω? inherits a lower bound

ω? + Aχ > 100δ0ξ.

So we can take a subsequence tk, %k → 0 so that ω?(tk, %k) converges to a closed currentω?0 ∈ α− 25δ0β with

ω?0 + Aχ > 100δ0ξ. (5.11)


By the similar argument as above, we know that the current ω?0 ∈ α− 25δ0β satisfies

Qχ(ω?0(r)) 6 θ0

for all r ∈ (0, r0) and Bj,3R\E0. Now we define a current Ω ∈ α− δ0β as

Ω := ω?0 + 24δ0χ+ 24δ0

√−1∂∂φ.

This implies thatΩ > ω?0 + 24δ0ξ. (5.12)

Combining with (5.7), we have

Ω(r) > ω?0(r) + 24δ0ξ

(r) > ω?0(r) + δ0χ.

By Proposition 2.4, we set ε0 = c0δ0 and obtain

Qχ(Ω(r)) 6 Qχ

(ω?0

(r) + δ0χ)6 θ0 − c0δ0 = θ0 − ε0

on Bj,3R\E0. On the other hand, by (5.11) and (5.12), we observe that

Ω + Aχ > 100δ0ξ.

We note that Ω+Aχ has positive Lelong number along E0 since φ is so. This completesthe proof.

6. Gluing construction

In this section, we perform a gluing construction to the local subsolutions obtainedin Theorem 5.4, and show the following:

Theorem 6.1. There exists ϕY ∈ C∞(Y \Ysing) such that

(1) The (1, 1)-form ωY := ω0 +√−1∂∂ϕY on Y \Ysing satisfies

Qχ(ωY ) < θ0.

(2) The function ϕY tends to −∞ uniformly at Ysing.

Let Ω be a closed current constructed in Theorem 5.4. Continuing from the previoussection, we have to deal with the problem that Ω is no longer Kahler. So we take aconstant A > 0 such that Ω+Aχ is a Kahler current and ω0−δ0χ+Aχ is semipositive.We take a potential ϕ so that

Ω = ω0 − δ0χ+√−1∂∂ϕ,

and hence ϕ ∈ PSH(Y , ω0 − δ0χ + Aχ). For a constant τ ∈ (0, 110000

), by takingsufficiently small R > 0 and finer covering Bj,3R if necessary, we may assume that

(1− τ)gj 6 ξ 6 (1 + τ)gj,

ξ =√−1∂∂φj,ξ,

∣∣φj,ξ − |z|2∣∣ 6 τR2, φj,ξ(0) = 0

on Bj,4R, where φj,ξ is a local potential for ξ, and |z|2 denotes the square of the Euclideandistance to z from the origin. In other words, each Bj,4R is very close to the normalcoordinates with respect to ξ. On each Bj,4R, we choose a local bounded potentials ofω0 and χ so that

ω0 =√−1∂∂φj,ω0 , φj,ω0(0) = 0,


χ =√−1∂∂φj,χ, φj,χ(0) = 0,

and assume that

|∇(φj,ω0 − δ0φj,χ + Aφj,χ)| 6 KR

for some fixed large constant K > 0. We choose a constant µ0 so that

µ0 <δ3

0R2

8A. (6.1)

By replacing r0 with a smaller number if necessary, we may assume that

φj,χ − µ0 6 φ(r)j,χ 6 φj,χ + µ0 (6.2)

on each Bj,3R for all r ∈ (0, r0). We set

Ω =√−1∂∂ϕj, ϕj := φj,ω0 − δ0φj,χ + ϕ

on Bj,4R and

ϕj,r := ϕ(r)j − φj,ω0 + δ0φj,χ − δ3

0φj,ξ + δ20φ

for r ∈ (0, R) on Bj,3R, where the function φ is defined in (3.3). Since δ0 ∈ (0, κ0), wehave

χ+ δ0

√−1∂∂φ = (1− δ0)χ+ δ0ξ > δ0ξ > δ2

0ξ.

Thus

ω0 +√−1∂∂ϕj,r = Ω(r) − δ3

0ξ + δ0(χ+ δ0

√−1∂∂φ) > Ω(r),

which yields that

Qχ(ω0 +√−1∂∂ϕj,r) 6 θ0 − ε0 (6.3)

on each Bi,3R\E0 for all r ∈ (0, r0). For any z ∈ Bj,3R and r ∈ (0, R), we set a PSHfunction ψj and its approximation ψj,r as

ψj := ϕj + Aφj,χ, ψj,r(z) := supBj,r(z)

ψj,

where Bj,r(z) denotes a Euclidean ball of radius r centered at z in Bj,4R. For anyr ∈ (0, R/2), we define

νj,ψj(z, r) :=ψj, 3

4R(z)− ψj,r(z)

log(

34R)− log r

, z ∈ Bj,3R.

As r → 0, the quantity νj,ψj(z, r) converges decreasingly to the Lelong number of ϕ atz (with respect to a reference form ω0 − δ0χ+ Aχ):

limr→0

νj,ψj(z, r) = νϕ(z),

where we note that the Lelong number νϕ(z) is independent of j ∈ J . We will use thefollowing comparison formula for these functions which was proved in [Che21, Lemma4.2] based on the computation of [BK07]:

Lemma 6.2. For any r ∈ (0, R/2) and z ∈ Bj,3R, the following estimates hold:

(1) 0 6 ψj,r(z)− ψj, r2(z) 6 (log 2)νj,ψj(z, r),


(2) 0 6 ψj,r(z)−ψ(r)j (z) 6 ηνj,ψj(z, r), where the function ρ is defined in Definition

5.3, and a constant η > 0 is defined by

η :=32m−1

22m−3log 2 + Vol(∂B1(0))

∫ 1

0

log

(1

t

)t2m−1ρ(t)dt.

We let Eε be the analytic subvariety of Y defined by

Eε := z ∈ Y |νϕ(z) > ε,where a constant ε > 0 is taken so that

ε < min

δ3

0R2

6 + 4η,1

4infz∈E0

νϕ(z)

. (6.4)

We note that Eε is an analytic subvariety of Y by Siu’s result [Siu74], and contains E0

by (6.4). Since dim Φ(Eε) 6 dimEε 6 m − 1, we apply the induction assumption toΦ(Eε). So there exist a real (1, 1)-form ωU ∈ α|Φ(Eε),X such that

Qχ,n,n(ωU) < θ0

on some open neighbourhood U of Φ(Eε) in X, and hence

Qχ,m,n(ωU) 6 θ0 − ε0on U by replacing ε0 > 0 and shrinking U if necessary. This implies that

Qχ,m,n(Φ∗ωU) 6 θ0 − ε0on Φ−1(U)\Φ−1(Ysing). We set U := Φ−1(U) ∩ Y and ωU := (Φ∗ωU)|U . By (2.2), we

observe that U is a open neighbourhood of Eε in Y and satisfies

Qχ(ωU) 6 θ0 − ε0 (6.5)

on U\E0. We write ωU as

ωU := ω0 +√−1∂∂ϕU

for some uniformly bounded, smooth function ϕU on U .

Lemma 6.3. There exist r1 ∈ (0,minr0, R/2) and an open neighbourhood V b U of

Eε in Y such that the following estimates hold for all r ∈ (0, r1):

(1) If z ∈ Y \V , thenmax

j∈J |z∈Bj,3Rνj,ψj(z, r) 6 2ε (6.6)

andmax

j∈J |z∈Bj,3Rϕj,r(z) > sup

U

ϕU + 3ε log r + 1.

(2) Ifmax

j∈J |z∈Bj,3Rνj,ψj(z, r) > 4ε,

thenmax

j∈J |z∈Bj,3Rϕj,r(z) 6 inf

UϕU + 3ε log r − 1.


(3) If

maxj∈J |z∈Bj,3R

νj,ψj(z, r) 6 4ε,

then

maxj∈J |z∈Bj,3R\Bj,2R

ϕj,r(z) 6 maxj∈J |z∈Bj,R

ϕi,r(z)− 2ε.

Proof. The main difference from [Son20, Lemma 6.3] is that we need the additional term

Aφ(r)j,χ to deal with νj,ψj(z, r). However, one can prove in the same way as in [Son20,

Lemma 6.3] by using Lemma 6.2 and the two sided bound (6.2). We give the full prooffor the sake of completeness.

(1) We fix any open V b U . Then we can choose r1 ∈ (0,minr0, R/2) sufficiently

small so that (6.6) holds since the Lelong number of ϕ at any point in Y \V is less thanε, and νj,ψj(·, r) is increasing in r > 0 and upper semi-continuous for any fixed r > 0.

So for any z ∈ Y \V and j ∈ J such that z ∈ Bj,3R, we have νj,ψj(z, r) 6 2ε. By thedefinition of νj,ψj(z, r), for all r ∈ (0, r1), we observe that

ψj,r(z) > 2ε log r − C1

for some constant C1 > 0 depending only on R and ψj, 34R. This shows that

ϕj,r(z)− supU

ϕU − 3ε log r − 1

= ψ(r)j (z)− Aφ(r)

j,χ(z)− φj,ω0(z) + δ0φj,χ(z)− δ30φj,ξ(z) + δ2

0φ(z)− supU

ϕU − 3ε log r − 1

> −ε log r − ηνj,ψj(z, r)− Aφ(r)j,χ(z)− φj,ω0(z) + δ0φj,χ(z)− δ3

0φj,ξ(z) + δ20φ(z)− sup

U

ϕU − 1− C1

> −ε log r − 2εη − A supBj,3R

φj,χ − Aµ0 − supBj,3R

φj,ω0 + δ0 supBj,3R

φj,χ − δ30 supBj,3R

φj,ξ + δ20 infY \V

φ

− supU

ϕU − 1− C1

> −ε log r − C2,

where the constant C2 > 0 depends only on R, φj,ω0 , φj,χ, φj,ξ, φ|Y \V and ϕU . Thus we

may choose r1 < e−C2/ε.(2) We will prove by contradiction. So assume that there exists j ∈ J such thatz ∈ Bj,3R and

ϕj,r(z) > infUϕU + 3ε log r − 1.

Then for any r ∈ (0,minr0, R/2), we compute

ψj,r(z) > ψ(r)j (z)

= ϕj,r(z) + φj,ω0(z)− δ0φj,χ(z) + δ30φj,ξ(z)− δ2

0φ(z) + Aφ(r)j,χ(z)

> 3ε log r + infBj,3R

φj,ω0 − δ0 supBj,3R

φj,χ + δ30 infBj,3R

φj,ξ − δ20 supBj,3R

φ+ A infBj,3R

φj,χ − Aµ0 + infUϕU − 1

> 3ε log r − C3,


where the constant C3 depends only on φj,ω0 , φj,χ, φj,ξ, φ and ϕU . Now let i ∈ J beany index such that z ∈ Bi,3R. From our choice of the finite covering Bi′,3Ri′∈J , onecan easily see that Bj, r

2(z) ⊂ Bi,r(z) for all r ∈ (0,minr0, R/2). Thus we have

ψj, r2(z) 6 sup

Bj, r2(z)

ϕ+ supBj, r2

(z)

(φj,ω0 − δ0φj,χ + Aφj,χ)

6 supBi,r(z)

ϕ+ 4KR2

6 ψi,r(z) + supBi,r(z)

(− (φi,ω0 − δ0φi,χ + Aφi,χ)

)+ 4KR2

6 ψi,r(z) + 8KR2,

and hence

ψi,r(z) > 3ε log

(r

2

)− 8KR2 − C3.

This shows that

4ε

(log

(3

4R

)− log r

)−(ψi, 3

4R(z)− ψi,r(z)

)> −ε log r − C4

for some constant C4 > 0 depending only on K, R, C3 and ψi, 34R. So if we take

r1 < e−C4/ε, we have

νi,ψi(z, r) < 4ε

for all r ∈ (0, r1).

(3) We assume that a point z ∈ Y satisfies

z ∈ (Bj,3R\Bj,2R) ∩Bi,R, maxj′∈J |z∈Bj′,3R

νj′,ψj′ (z, r) 6 4ε,

where we note that the set i ∈ J |z ∈ Bi,R is not empty since we assume that Bi,Ris also a covering of Y . Then for any r < min(r0, R/2), we compute

ϕj,r(z)

= ψ(r)j (z)− Aφ(r)

j,χ(z)− φj,ω0(z) + δ0φj,χ(z)− δ30φj,ξ(z) + δ2

0φ(z)

6 ψj, r2(z) + (log 2)νj,ψj(z, r)− Aφ

(r)j,χ(z)− φj,ω0(z) + δ0φj,χ(z)− δ3

0φj,ξ(z) + δ20φ(z)

6 ψj, r2(z) + (log 2)νj,ψj(z, r)− Aφj,χ(z) + Aµ0 − φj,ω0(z) + δ0φj,χ(z)− δ3

0φj,ξ(z) + δ20φ(z)

6 supBj, r2

(z)

ϕ+ supBj, r2

(z)

(φj,ω0 − δ0φj,χ + Aφj,χ − (φj,ω0 − δ0φj,χ + Aφj,χ)(z)

)+ 4ε+ Aµ0

−δ30φi,ξ(z)− δ3

0(φj,ξ(z)− φi,ξ(z)) + δ20φ(z).


Since Bj, r2(z) ⊂ Bi,r(z) for all r < min(r0, R/2), we have

ϕj,r(z)

6 supBi,r(z)

ϕ+KRr + 4ε+ Aµ0 − δ30φi,ξ(z)− δ3

0

((7

4R

)2

−(

5

4R

)2)+ δ2

0φ(z)

6 ψi,r(z) + supBi,r(z)


)+KRr + 4ε+ Aµ0 − δ3

0φi,ξ(z)− 3

2δ3

0R2 + δ2

0φ(z)

6 ψ(r)i (z) + sup

Bi,r(z)


)+ (4 + 4η)ε+KRr + Aµ0 − δ3

0φi,ξ(z)

−3

2δ3

0R2 + δ2

0φ(z)

6 ϕi,r(z) + supBi,r(z)

((φi,ω0 − δ0φi,χ + Aφi,χ)(z)− (φi,ω0 − δ0φi,χ + Aφi,χ)

)+ (4 + 4η)ε

+KRr + 2Aµ0 −3

2δ3

0R2

6 ϕi,r(z) + (4 + 4η)ε+ 2KRr + 2Aµ0 −3

2δ3

0R2

6 ϕi,r(z)− 2ε

from the choice of ε (6.4) and µ0 (6.1) as long as r 6 δ30R

8K.

Thus we may take the regularized maximum of (Bj,3R, ϕj,r) and (U , ϕU). By Lemma6.3, we know that the resulting function ϕ is smooth. If we set

Ω′ := ω0 +√−1∂∂ϕ,

then by (6.3), (6.5), we know that Ω′ is smooth on Y and satisfies

Qχ(Ω′) 6 θ0 − ε0on Y \E0. Thus there exists a sufficiently small constant s > 0 such that

Qχ(Ω′ − sχ) < θ0

on Y \E0 by using the fact that arccot(λ− s) 6 arccot(λ) + s for all λ ∈ R and s > 0.We set

Ω := Ω′ + s√−1∂∂φ = Ω′ − sχ+ sξ > Ω′ − sχ.

Then the monotonicity of Qχ shows that

Qχ(Ω) < θ0

on Y \E0. Moreover, if we write Ω as

Ω = ω0 +√−1∂∂ϕY , ϕY := ϕ+ sφ,

then ϕY → −∞ uniformly at E0.

Proof of Theorem 6.1. Theorem 6.1 follows immediately from the above observations

and the fact that Y \Ysing = Φ(Y \E0

).


7. Completion of the proof of Theorem 1.3

First, we will give the proof of Theorem 3.1.

Proof of Theorem 3.1. The argument is similar as the J-equation case [Son20, Theorem3.1]. By the induction assumption, there exists an open neighbourhood U of Ysing in Xand ϕU ∈ C∞(U) such that

Qχ,n,n(ωU) < θ0, ωU := ω0 +√−1∂∂ϕU .

We take ϕY as in Theorem 6.1, and open neighbourhoods W b V b U of Ysing in Xsuch that

(1) both Y \W and Y \V are a union of finitely many open smooth analytic subva-rieties of Y ;

(2) ϕY < ϕU − 2 in Y ∩W ;

(3) ϕY > ϕU + 2 in Y ∩ (U\V )

by subtracting a large constant from ϕU if necessary since ϕY tends to −∞ uniformlyalong Ysing. We take a smooth extension ϕ′Y of ϕY in Y \W and set

ϕ′′Y := ϕ′Y + Ad2

for a constant A > 1, where d denotes the distance function to Y \W with respect toany fixed Kahler metric on X. By taking a sufficiently large A, we observe that

Qχ,n,n(ω′′Y ) < θ0, ω′′Y := ω0 +√−1∂∂ϕ′′Y

in an open neighbourhood of Y \W in X. Since ϕ′′Y is continuous, there exists a suffi-

ciently small open neighbourhood U of Y \W in X such that

ϕ′′Y < ϕU − 1

in U ∩W andϕ′′Y > ϕU + 1

in U\V . So if we let ϕ be the regularized maximum of (U , ϕ′′Y ) and (V, ϕU), then wehave

Qχ,n,n(ω) < θ0, ω := ω0 +√−1∂∂ϕ

on a neighbourhood of Y in X as desired.

Now we will prove Theorem 1.3. Again we stress that once we obtain Theorem 3.1,we can prove Theorem 1.3 by using exactly the same argument as [Che21, Section 5].However, we will give the outline of the proof for reader’s convenience.

Outline of the proof of Theorem 1.3. Assume that the triple (X,α, β) is stable along atest family ωt,0 ∈ αt (t ∈ [0,∞)). By Theorem 3.1, the definition (B) of the test familyand monotonicity of Qχ, we know that

Γχ,αt,θ0,θ0(Y,X) 6= ∅ (7.1)

for all proper analytic subvariety Y ⊂ X and t ∈ [0,∞). Let us consider the followingtwisted dHYM equation for ωt ∈ αt

Re(ωt +√−1χ)n − cot(θ0)Im(ωt +

√−1χ)n − ctχn = 0, (7.2)


where the normalized constant ct is non-negative for all t ∈ [0,∞) and c0 = 0 by thestability assumption and choice of θ0. We take a constant Θ0 ∈ (θ0, π) (depending onlyon n and θ0) sufficiently close to θ0 so that

Θ0 − θ0 <π −Θ0

n,

and define

T := t ∈ [0,∞)|(7.2) has a solution ωt ∈ Γχ,αt,θ0,Θ0(X) for t.By (C) of the definition of the test family, monotonicity of Pχ, Qχ and Proposition 2.8,we know that there exists a large enough T > 0 such that [T,∞) ⊂ T . Also T is openby Proposition 2.8. To show the closeness, we have to prove that t0 ∈ T for t0 := inf T .Now we apply the same argument in Section 4, 5 (essentially based on [Che21, Section5]) with suitable small changes: first, we take a constant K > 0 (depending only on nand θ0) so that

cot(π −Θ0

n

)< K < cot(Θ0 − θ0), (7.3)

and set0 < θ0 := θ0 + n arccot(K) < Θ0 := Θ0 + n arccot(K) < π.

Set X := X ×X, and define the quantities χX , χX ,s, FX ,s corresponding to (4.1), (4.2),(4.5) respectively (but we set FhE0

= 0 in this case). We consider the twisted dHYMequation for ωX ,s ∈ π∗1αt +Kπ∗2β (t ∈ (t0,∞))

Re(ωX ,s +√−1χX )2n − cot(θ0)Im(ωX ,s +

√−1χX )2n − FX ,sχ2n

X = 0. (7.4)

Then by using the condition (7.3), one can easily see that ω†X := π∗1ωt +Kπ∗2χ satisfies

PχX (ω†X ) < θ0 and QχX (ω†X ) < Θ0, where ωt denotes the solution to (7.2) for t. Also bychoosing a sufficiently small constant ` > 0 in the definition of χX ,s, one can observethat inf FX ,s > −c where the constant c > 0 is determined in Proposition 2.8. So weobtain a solution ωX ,s ∈ ΓχX ,π∗1αt+Kπ∗2β,θ0,Θ0

(X ) for all t ∈ (t0,∞). Then performing the

truncation and using the mass concentration argument, one can proceed the computa-tions for Pχ, Qχ almost in parallel, by using the monotonicity and concavity of thesefunctions. As a consequence, we obtain a constant δ0 > 0 as in Lemma 5.1 and ananalogous result of Theorem 5.4, where the condition (5.9) for a current Ω ∈ αt0 − δ0βshould be replaced by

Pχ(Ω(r)) 6 θ0 − ε0, Qχ(Ω(r)) 6 Θ0 − ε0.Finally, we consider the gluing argument as in Section 6 (essentially based on [Che21,Section 5]). If Ω + Aχ has no positive Lelong number, we are done. If not, we apply(7.1) to the ε-sublevel set Eε of Lelong number of Ω + Aχ for some suitable smallconstant ε > 0 and then glue them together. As a result, we obtain a subsolutionωX ∈ Γχ,αt0 ,θ0,Θ0(X). By Proposition 2.8, this implies that t0 ∈ T . Thus we have asolution of (7.2) at t = 0. This completes the proof.


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(Jianchun Chu) Department of Mathematics, Northwestern University, 2033 SheridanRoad, Evanston, IL 60208

Email address: [email protected]

(Man-Chun Lee) Mathematics Institute, Zeeman Building, University of Warwick,Coventry CV4 7AL; Department of Mathematics, Northwestern University, 2033 Sheri-dan Road, Evanston, IL 60208

Email address: [email protected], [email protected]

(Ryosuke Takahashi) Faculty of Mathematics, Kyushu University, 744, Motooka, Nishi-ku, Fukuoka, 819-0395, JAPAN

Email address: [email protected]

abstract. arxiv:2105.10725v3 [math.dg] 10 jun 2021

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