On J-Equation

Home , Song Sun, Xiuxiong Chen

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n 1 n ω − ω χ 0 c 0 =0. ∧ (n 1)! − 0 n! ZM − When χ is a Kähler form, it is well known that the critical point of the χ functional is exactly the solution to the J-equation. It was the second appearanceJ of the J-equation in the literature. Following this formula, using the interpola- tion of the K-energy and the χ functional, Chen-Cheng [6, 7, 8] proved that the existence of cscK metric isJ equivalent to the geodesic stability of K-energy. However, the relationship between the existence of cscK metrics and the uniform K-stability is still open. When we replace the K-energy by the χ functional for a Kähler form χ, the analogy of the K-stability and the slopeJ stability conditions were proposed by Lejmi and Székelyhidi [28]. See also Section 6 of [19] for the extension to non-projective case. The main theorem of this paper proves the equivalence between the existence of the critical point of functional, the solvability of Jχ J-equation, the coerciveness of χ functional, and the uniform J-stability as well as the uniform slope J-stability.J Theorem 1.1. (Main Theorem) Fix a Kähler manifold M n with Kähler metrics χ and ω0. Let c0 > 0 be the constant such that

n 1 n ω − ω χ 0 = c 0 , ∧ (n 1)! 0 n! ZM − ZM then the followings are equivalent: (1) There exists a unique smooth function ϕ up to a constant such that ω = ω + √ 1∂∂ϕ¯ > 0 satisﬁes the J-equation ϕ 0 −

trωϕ χ = c0; (2) There exists a unique smooth function ϕ up to a constant such that ω = ω + √ 1∂∂ϕ¯ > 0 satisﬁes the J-equation ϕ 0 − n 1 n ω − ω χ ϕ = c ϕ ; ∧ (n 1)! 0 n! −

2 (3) There exists a unique smooth function ϕ up to a constant such that ϕ is the critical point of the functional; Jχ (4) The χ functional is coercive, in other words, there exist ǫ1.1 > 0 and another constantJ C such that (ϕ) ǫ (ϕ) C ; 1.2 Jχ ≥ 1.1Jω0 − 1.2 (5) (M, [ω0], [χ]) is uniformly J-stable, in other words, there exists ǫ1.1 > 0 such that for all Kähler test configurations ( , Ω) defined as Definition 2.10 of X [19], the numerical invariant J[χ]( , Ω) defined as Definition 6.3 of [19] satisfies J ( , Ω) ǫ J ( , Ω); X [χ] X ≥ 1.1 [ω0] X (6) (M, [ω0], [χ]) is uniformly slope J-stable, in other words, there exists ǫ1.1 > 0 such that for any subvariety V of M, the degeneration to normal cone ( , Ω) defined as Example 2.11 (ii) of [19] satisfies J ( , Ω) ǫ J ( , Ω); X [χ] X ≥ 1.1 [ω0] X (7) There exists ǫ1.1 > 0 such that

p p 1 (c (n p)ǫ )ω pχ ω − 0 0 − − 1.1 0 − ∧ 0 ≥ ZV for all p-dimensional subvarieties V with p =1, 2, ..., n. Remark 1.2. Lejmi and Sz´ekelyhidi’s original conjecture is that the solvability of

trωϕ χ = c0 is equivalent to p p 1 c ω pχ ω − > 0 0 0 − ∧ 0 ZV for all p-dimensional subvarieties V with p = 1, 2, ..., n 1 [28]. However, it seems that our uniform version is more natural from geometri− c point of view.

Remark 1.3. It is well known that there exists a constant C1.3 depending on n such that the functional Jω0 1 n 1 n 1 n 1 ω − ω tω − ( ϕ(ω tϕ n tϕ ))dt = (√ 1 ∂ϕ ∂ϕ¯ tϕ )dt 0 ∧ (n 1)! − n! − ∧ ∧ (n 1)! Z0 ZM − Z0 ZM − and Aubin’s I-functional

n n n ¯ k n k ϕ(ω0 ωϕ)= √ 1 ∂ϕ ∂ϕ ω0 ωϕ− M − − M ∧ ∧ ∧ Z Z kX=0 satisfy 1 n n n n C− ϕ(ω ω ) (ϕ) C ϕ(ω ω ). 1.3 0 − ϕ ≤ Jω0 ≤ 1.3 0 − ϕ ZM ZM For example, Collins and Székelyhidi used this fact and their Definition 20 in n n [13] replaced ω0 (ϕ) by M ϕ(ω0 ωϕ) in the definition of the coerciveness which was called “properness”J in [13].− By (3) of [2], Aubin’s I-functional can also be replaced by Aubin’s J-functionalR in the definition of coerciveness. Accordingly, in the definition of uniform stability, the numerical invariant J ( , Ω) can be [ω0] X

3 replaced by the minimum norm of ( , Ω) defined as Definition 2.18 of [19]. By X (62) of [16], Aubin’s J-functional can be further replaced by the d1 distance in the definition of the coerciveness when ϕ is normalized such that the Aubin-Mabuchi energy of ϕ is 0. Remark 1.4. By Proposition 2 of [5], if the solution to the J-equation exists, it is unique up to a constant. It is easy to see that (1) and (2) are equivalent. The equivalence between (2) and (3) follows from the formula n 1 n d ∂ϕ ω − ω Jχ = (χ ϕ c ϕ ). dt ∂t ∧ (n 1)! − 0 n! ZM − By Proposition 21 and Proposition 22 of [13] and Remark 1.3, (1) and (4) are equivalent. By Corollary 6.5 of [19], (4) implies (5). It is trivial that (5) implies (6). By [28], (6) implies (7) in the projective case if ǫ1.1 is replaced by 0. However, it is easy to see that it is also true in non-projective case and for positive ǫ1.1. Thus, we only need to prove that (7) implies (1) in Theorem 1.1. Remark that there is a simpler proof that (1) implies (7). Let χ = δij and ωϕ = λiδij , then for any c> 0, the condition p p 1 cω pχ ω − 0 ϕ − ∧ ϕ ≥ is equivalent to p 1 c λ ≤ j=1 ij X for all distinct p numbers i ,i , ..., i 1, 2, ..., n . So tr χ = c as well as { 1 2 p}⊂{ } ωϕ 0 the upper bounds of λi imply that for small enough ǫ1.1 > 0, p p 1 (c (n p)ǫ )ω pχ ω − 0 0 − − 1.1 ϕ − ∧ ϕ ≥ for all p =1, 2, ..., n. (4) follows from the fact that

p p 1 p p 1 (c (n p)ǫ )ω pχ ω − = (c (n p)ǫ )ω pχ ω − . 0 − − 1.1 0 − ∧ 0 0 − − 1.1 ϕ − ∧ ϕ ZV ZV

When c1(M) < 0, we can choose χ as a K¨ahler form in c1(M). Since x log x −n n R ωϕ ωϕ is bounded from below for any x , the entropy M log( ωn ) n! is also bounded ∈ 0 from below. So the coerciveness of functional implies the coerciveness of K- χ R energy. This observation appearedJ as Remark 2 of [5]. Using this observation, as a corollary of Theorem 1.3 of [7] and Theorem 1.1, we can ﬁnd many cscK metrics with c1(M) < 0.

Corollary 1.5. If c1(M) < 0, and ǫ1.1 > 0, then for any K¨ahler class [ω0] such that n 1 n[c1(M)] [ω0] − p p 1 ((− · (n p)ǫ )ω pω − ( c (M)) 0 [ω ]n − − 1.1 0 − 0 ∧ − 1 ≥ ZV 0 for all p-dimensional subvarieties V with p = 1, 2, ..., n, there exists a cscK metric in [ω0].

4 Remark 1.6. If there exists ωϕ [ω0] such that Ric(ωϕ) < 0 and ωϕ has constant scalar curvature, then the∈ condition above is also necessary. Besides the appearances in the moment map picture and the study of the cscK problem, J-equation also arises from the study of mirror symmetry. In fact, using the following observation of Collins-Jacob-Yau [12]

n n π 1 lim k( arctan(kλi)) = , k 2 − λ →∞ i=1 i=1 i X X the J-equation is exactly the limit of the deformed Hermitian-Yang-Mills equation n arctan λi = constant, i=1 X where λi are the eigenvalues of ωϕ with respect to χ. It plays an important role in the study of mirror symmetry [35, 29]. Motivated by the J-equation, Collins-Jacob-Yau [12] conjectured that the solvability of the deformed Hermitian-Yang-Mills equation is also equivalent to a notion of stability. In this paper, we prove the uniform version of their conjecture when the angle is in ( nπ π , nπ ): 2 − 4 2 n Theorem 1.7. Fix a K¨ahler manifold M with K¨ahler metrics χ and ω0. Let ˆ nπ π nπ θ ( 2 4 , 2 ) be a constant. Then there exists a unique smooth function ϕ up∈ to a constant− such that n arctan λi = θˆ i=1 X for eigenvalues λ of ω = ω + √ 1∂∂ϕ¯ > 0 with respect to χ if and only if i ϕ 0 − there exists a constant ǫ1.1 > 0 and for all p-dimensional subvarieties V with p =1, 2, ..., n (V can be chosen as M), there exist smooth functions θV (t) from (n p)π pπ [1, ) to [θˆ − + (n p)ǫ , ) such that for all t [1, ), ∞ − 2 − 1.1 2 ∈ ∞

p p pπ (χ + √ 1tω0) =0,θV (t) = arg( (χ + √ 1tω0) ), lim θV (t)= . − 6 − t 2 ZV ZV →∞

Moreover, when V = M, it is required that θM (1) = θˆ.

Remark 1.8. We only study the case when θˆ ( nπ π , nπ ) in this paper. So it ∈ 2 − 4 2 is natural to assume that [ω0] is a Kähler class. However, usually we need extra conditions in addition to [ω ] being Kähler to make sure (χ + √ 1tω )p is 0 V − 0 not 0 so that θV (t) is well defined. When p =1, 2, θV (t) is always well defined and increasing for t ( , ) without any extra assumption.R In addition, ∈ −∞ ∞ θV (0) = 0. When p =3,

(χ + √ 1tω )3 = χ3 3t2 χ ω2 + √ 1t(3 χ2 ω t2 ω3). − 0 − ∧ 0 − ∧ 0 − 0 ZV ZV ZV ZV ZV

5 So if the inequality

( χ3)( ω3) < 9( χ ω2)( χ2 ω ) 0 ∧ 0 ∧ 0 ZV ZV ZV ZV in Proposition 3.3 of [14] holds, then θV (t) is well deﬁned for t ( , ). Moreover, if θ (1) > π, then θ (t) is increasing for t [1, ).∈ In−∞ addition,∞ V V ∈ ∞ θV (0) = 0. So the choice of θV (1) in this paper is the same as the choice of θV (1) in Proposition 8.4 of [15]. In higher dimensions, more inequalities are involved. ˆ (n p)π Remark 1.9. Collins-Jacob-Yau conjectured that θV (1) > θ −2 for all p = 1, 2, ..., n 1 is equivalent to the solvability of the deformed− Hermitian- Yang-Mills equation− [12]. However, it seems that our uniform version is more natural because Deﬁnition 8.10 (2) of [15] also assumed the uniform positive lower bound. Remark 1.10. By Theorem 1.1 of [26], the solution to the deformed Hermitian- Yang-Mills equation is unique up to a constant if it exists. The “only if” part of Theorem 1.7 is a combination of Proposition 3.1 of [12] and Remark 1.4. So we only need to prove the “if” part of Theorem 1.7. Theorem 1.7 will be proved in Section 5 using the same method of the proof of Theorem 1.1. Instead of Theorem 1.1, we will prove the following stronger statement by induction:

n Theorem 1.11. Fix a K¨ahler manifold M with K¨ahler metrics χ and ω0. Let 1 1 n 1 c> 0 be a constant and f > ( ) − be a smooth function satisfying − 2n c n n n 1 χ ω ω − f = c 0 χ 0 0, n! n! − ∧ (n 1)! ≥ ZM ZM ZM − then there exists ω = ω + √ 1∂∂ϕ¯ > 0 satisfying the equation ϕ 0 − χn χ f c trωϕ + n = ωϕ and the inequality n 1 n 2 cω − (n 1)χ ω − > 0 ϕ − − ∧ ϕ if there exists ǫ1.1 > 0 such that

p p 1 (c (n p)ǫ )ω pω − χ 0 − − 1.1 0 − 0 ∧ ≥ ZV for all p-dimensional subvarieties V with p =1, 2, ..., n. Remark 1.12. By Remark 1.4, Theorem 1.1 is a corollary of Theorem 1.11 by choosing f =0.

6 Remark 1.13. When n = 1, Theorem 1.11 is trivial. When n = 2, Theorem 1.11 is the statement that the Demailly-Paun’s characterization [18] for [cω0 χ] being K¨ahler implies the solvability of the Calabi conjecture −

(cω χ)2 = (cf + 1)χ2 ϕ − by Yau [40]. In the toric case when f is a non-negative constant, Theorem 1.11 was proved by Collins and Sz´ekelyhidi [13]. There are several steps to prove Theorem 1.11. Step 1: Prove the following:

n Theorem 1.14. Fix a K¨ahler manifold M with K¨ahler metrics χ and ω0. Let 1 1 n 1 c> 0 be a constant and f > ( ) − be a smooth function satisfying − 2n c n n n 1 χ ω ω − f = c 0 χ 0 0, n! n! − ∧ (n 1)! ≥ ZM ZM ZM − then there exists ω = ω + √ 1∂∂ϕ¯ satisfying the equation ϕ 0 − χn χ f c trωϕ + n = ωϕ and the inequality n 1 n 2 cω − (n 1)χ ω − > 0 ϕ − − ∧ ϕ if n 1 n 2 cω − (n 1)χ ω − > 0. 0 − − ∧ 0

We will use the continuity method to prove Theorem 1.14. The details will be provided in Section 2.

Remark 1.15. Let χ = δij and ωϕ = λiδij , then the equation χn χ f c trωϕ + n = ωϕ is equivalent to n 1 f + = c. λ n λ i=1 i i=1 i X 1 Q Remark 1.16. Suppose i=k λ c for all k =1, 2, ..., n and 6 i ≤ P n 1 f + = c, λ n λ i=1 i i=1 i X 1 1 n 1 Q 1 then as long as f > 2n ( c ) − , it is easy to see that i=k λ

n Theorem 1.18. Fix a Kähler manifold M with Kähler metrics χ and ω0. Suppose that for all t > 0, there exist a constant ct > 0 and a smooth Kähler form ω [(1 + t)ω ] satisfying t ∈ 0 n 1 n 2 cω − (n 1)χ ω − > 0 t − − ∧ t and χn trωt χ + ct n = c. ωt Then there exist a constant ǫ > 0 and a current ω [ω ǫ χ] such that 1.4 1.5 ∈ 0 − 1.4 n 1 n 2 cω − (n 1)χ ω − 0 1.5 − − ∧ 1.5 ≥ in the sense of Definition 3.3.

Remark 1.19. In general we can only take the wedge product of ωϕ when ϕ is 2 in C . Bedford-Taylor [3] proved that it can also be defined when ϕ is in L∞. In our case, ϕ is unbounded, so we have to figure out the correct definition of

p p 1 cω pχ ω − 0 ϕ − ∧ ϕ ≥ for unbounded ϕ and p =1, 2, ..., n. This will be done in Definition 3.3. Remark 1.20. When n =2, it is same as Theorem 2.12 of [18]. Now let us sketch the proof here. It is analogous to the proof of Theorem 2.12 of Demailly-Paun’s paper [18]. Consider the diagonal ∆ inside the product manifold M M. Cover it by finitely many open coordinate balls B . Since ∆ × j is non-singular, we can assume that on Bj , gj,k, k =1, 2, ..., 2n are coordinates and ∆ = gj,k =0, 1 k n . Assume that θj are smooth functions supported { 2≤ ≤ } in Bj such that θj = 1 in a neighborhood of ∆. For ǫ1.6 > 0, define P n ψ = log( θ2 g 2 + ǫ2 ). ǫ1.6 j | j,k| 1.6 j X Xk=1 Define χM M = π1∗χ + π2∗χ, × and ¯ χM M,ǫ . ,ǫ . = χM M + ǫ1.7√ 1∂∂ψǫ . . × 1 6 1 7 × − 1 6 Let χ2n χ2n M M,ǫ . ,ǫ . ct M M,ǫ . ,ǫ . f × 1 6 1 7 > × 1 6 1 7 , t,ǫ1.6,ǫ1.7 = 2n 1+ n 2n 1 χM M − c χM M − × ×

8 then by Lemma 2.1 (ii) of [18], there exists ǫ1.7 > 0 such that for ǫ1.6 small enough, χ2n M M,ǫ1.6,ǫ1.7 1 1 2n 1 × 2n 1 > ( ) − . χM M − −4n (n + 1)c × 1 Now we consider ω0,M M,t = π1∗ωt + π2∗χ. By Theorem 1.14, there exists × c ωt,ǫ . ,ǫ . [ω0,M M,t] such that 1 6 1 7 ∈ × 2n χM M trωt,ǫ ,ǫ χM M + ft,ǫ . ,ǫ . × = (n + 1)c. 1.6 1.7 × 1 6 1 7 ω2n t,ǫ1.6,ǫ1.7

Now deﬁne ω1,t,ǫ1.6,ǫ1.7 by

n 1 c − n ω1,t,ǫ1.6,ǫ1.7 = (π1) (ωt,ǫ ,ǫ π2∗χ). n ∗ 1.6 1.7 M nχ ∧

Fix ǫ1.7 and let t and ǫ1.6 convergeR to 0. For small enough ǫ1.4, let ω1.5 be the weak limit of ω ǫ χ. Then we shall expect 1,t,ǫ1.6,ǫ1.7 − 1.4 n 1 n 2 cω − (n 1)χ ω − 0 1.5 − − ∧ 1.5 ≥ in the sense of Definition 3.3. The details will be provided in Section 3. Step 3: Consider the set I of t 0 such that there exist a constant ct 0 and a smooth Kähler form ω [(1≥ + t)ω ] satisfying ≥ t ∈ 0 n 2 (cω (n 1)χ) ω − > 0 t − − ∧ t and χn trωt χ + ct n = c. ωt By Theorem 1.14, it suffices to show that 0 I. When t is large enough, the condition of Theorem 1.14 is satisfied. So t ∈I. It is easy to see that if t I, then for nearby t, the condition of Theorem∈ 1.14 is also satisfied. So I is open.∈ Still by Theorem 1.14, as long as t I, then for all t′ t, t′ I. Thus, in ∈ ≥ ∈ order to prove the closedness of I, it suffices to show that if t I for all t>t0, then t I. After replacing (1 + t )ω by ω , we can without∈ loss of generality 0 ∈ 0 0 0 assume that t0 = 0. In particular we can apply Theorem 1.18 to get ω1.5. Let ν(x) be the Lelong number of ω1.5 at x. For ǫ1.8 > 0 to be determined, let Y be the set Y = x : ν(x) ǫ . { ≥ 1.8} By the result of Siu [32], Y is a subvariety with dimension p

2 ω = ω + √ 1∂∂¯(Proj∗ ϕ + C d (., Y ) ) 1.11 0 − Y 1.9 1.10 χ

9 satisﬁes n p n 1 n 2 (c − ǫ )ω − (n 1)χ ω − > 0 − 2 1.1 1.11 − − ∧ 1.11 on a tubular neighborhood of Y , where ProjY means the projection to Y . Bya generalization of the result of Blocki and Kolodziej [4], we can glue the smoothing of ω outside Y and ω near Y into ω = ω + √ 1∂∂ϕ¯ satisfying 1.5 1.11 1.12 0 − 1.12 n 1 n 2 cω − (n 1)χ ω − > 0 1.12 − − ∧ 1.12 on M. Then we are done by Theorem 1.14. In general, Y is singular and we need to use Hironaka’s desingularization theorem to resolve it. The details will be provided in Section 4.

Acknowledgement

The author wishes to thank Xiuxiong Chen for suggesting him this problem and providing valuable comments. The author is also grateful to Jingrui Cheng for pointing out a gap in the ﬁrst version of this paper, Helmut Hofer for a discussion about symplectic geometry as well as Simone Calamai and Long Li for minor suggestions. This material is based upon work supported by the National Science Foundation under Grant No. 1638352, as well as support from the S. S. Chern Foundation for Mathematics Research Fund.

2 The analysis part

In this section, we use the continuity method twice to prove Theorem 1.14. First of all, for t [0, 1], deﬁne χ by ∈ t c χ = tχ + (1 t) ω t − n 0 and deﬁne f 0 as the constant such that t ≥ n n n 1 n n 1 χ ω ω − ω ω − f t = c 0 χ 0 = t(c 0 χ 0 ) 0. t n! n! − t ∧ (n 1)! n! − ∧ (n 1)! ≥ ZM ZM ZM − ZM ZM − Now we consider the set I consisting of all t [0, 1] such that there exists ω = ω + √ 1∂∂ϕ¯ > 0 for smooth ϕ satisfying∈ t 0 − t t n χt trωt χt + ft n = c ωt and n 1 n 2 cω − (n 1)χ ω − > 0. t − − t ∧ t Then it is easy to see that 0 I. Remark that the equation is the same as ∈ n n 1 n ω ω − χ c t χ t = f t . n! − t ∧ (n 1)! t n! −

10 The linearization is

n n 1 1 n 1 n 2 ∂ϕt ∂ χt ∂χt ωt − (cω − (n 1)χ ω − ) √ 1∂∂¯ = (f )+ . (n 1)! t − − t ∧ t ∧ − ∂t ∂t t n! ∂t ∧ (n 1)! − − Assume that t I, then the left hand side is a second order elliptic equation on ∂ϕt ∈ ∂t . On the other hands, the integrability condition implies that the integral of the right hand side is 0. By standard elliptic theory and the implicit function theorem, I is open when we replace the smoothness assumption of ϕ by C100,α. However, standard elliptic regularity theory implies that any C100,α solution is automatically smooth. So I is in fact open. Assume that we are able to show the closedness of I, then we have proved Theorem 1.14 for f replaced by f1. We can use another continuity path by ﬁxing χ and ω0 but choosing fˆt = tf1 + (1 t)f. However, it is the same as before ˆ 1 1 n 1 − except that ft > 2n ( c ) − is a function instead of a constant. Thus, we only − 1 1 n 1 need to prove the a priori estimate of ωt by assuming that ft > 2n ( c ) − is a function. We start from the following proposition which is analogous− to Lemma 3.1 of Song-Weinkove’s paper [34]: Proposition 2.1. Assume that t I and ω = ω + √ 1∂∂ϕ¯ is the corre- ∈ t 0 − t sponding solution, then there exist constants C2.1 and C2.2 depending only on c, 2 ω , the C∞-norm of χ with respect to ω , the C -norm of f with respect to 0 t 0 || t|| ω0 such that C . (ϕt inf ϕt) tr ω C e 2 1 − . χt t ≤ 2.2

i ¯j i ¯j Proof. In local coordinates, χ = √ 1χ ¯dz dz and ω = √ 1g ¯dz dz . t − ij ∧ t − ij ∧ Fix any point x, choose a χt-normal coordinate such that χi¯j = δi¯j , χi¯j,k = χi¯j,k¯ = 0 and gi¯j = λiδi¯j at x, where the derivatives are all ordinary derivatives. Then the equation is i¯j det χαβ¯ g χi¯j + ft = c. det g ¯ i,j αβ X Deﬁne an operator ∆˜ by

i¯l g det χ ¯ ˜ i¯j kl¯ αβ ∆u = ( g g χk¯j + ft )u,i¯l, det gαβ¯ Xi,l Xj,k then it is easy to see that ∆˜ is independent of the choice of local coordinates. At x, 1 1 1 ∆˜ u = ( + f )u ¯. λ2 t λ n λ ,ii i i i α=1 α X 1 1 1 n 1 Since ( ) − , it is easyQ to see that λα − 2n c 1 1 1 2 + ft n > 0 λi λi α=1 λα Q 11 for all i. So ∆˜ is a second order elliptic operator. ˜ ˜ i¯j Now we compute ∆(log trχt ωt)= ∆(log( i,j χ gi¯j )). It equals to

i¯i P2 (g ¯ ¯ + (χ ) ¯λi) g ¯ 1 1 1 i ii,kk ,kk | i ii,k| f ( 2 )( 2 + t n )) λi − ( λi) λ λk λα k P i P i k α=1 X at x. P P Q Now we diﬀerentiate the equation

i¯j det χαβ¯ g χi¯j + ft = c, det g ¯ i,j αβ X then

i¯j i¯b a¯j det χαβ¯ g χi¯j,k g ga¯b,kg χi¯j + (ft,k − det g ¯ i,j αβ X i,j,a,bX i¯j i¯j +f (χ χ ¯ g g ¯ ))=0. t ij,k − ij,k i,j X So

1 1 1 1 1 2 2 ( χ ¯ ¯ g ¯ ¯ + ( g ¯ + g ¯ ¯ ) λ λ ii,kk − λ2 ii,kk λ2 λ | ij,k| | ij,k| i i i i i i i,j i j Xk X X X 1P 1 2 1 2 1 + (f ( ( g ¯ ) + χ ¯ ¯ + g ¯ g ¯ ¯) λ t | λ ii,k | ii,kk λ λ | ij,k| − λ ii,kk α α i i i i,j i j i i X X X X Q 1 1 +f ¯ f ¯ ( g ¯ ) f ( g ¯ ¯))) = 0 t,kk − t,k λ ii,k − t,k λ ii,k i i i i X X at x. By K¨ahler condition, gi¯i,kk¯ = gkk,i¯ ¯i, gi¯j,k = gk¯j,i and gi¯j,k¯ = gik,¯ ¯j . Using the i¯i 1 bounds that χ ¯ ¯ + (χ ) ¯ + f + f ¯ + f ¯ + f + < C for all ii,kk ,kk t,k t,k t,kk t λi 2.3 i, k, it is easy| to see| that| by taking| | the| sum| of| the| previous| | | two equations,

2 g ¯ 1 1 1 ˜ ω C | i ii,k| f ∆(log trχ t) 2.4 ( 2 )( 2 + t n ) ≥− − ( λi) λ λk λα k P i k α=1 X 1 1 1 2 P2 1 Q1 2 + ( ( g ¯ + g ¯ ¯ )+ (f ( ( g ¯ ) λ λ2 λ | ij,k| | ij,k| λ t | λ ii,k | i i i,j i j α α i i Xk X X P 1 2 1 Q 1 + g ¯ ) f ¯ ( g ¯ ) f ( g ¯ ¯))). λ λ | ij,k| − t,k λ ii,k − t,k λ ii,k i,j i j i i i i X X X

12 Remark that 1 1 1 1 f ¯ ( g ¯ ) = f ( g ¯ ¯) | λ t,k λ ii,k | | λ t,k λ ii,k | α α i i α α i i X X Q 1 1 Q 1 = f ¯ ( g ¯ ) C g ¯ | t,k|| λ λ ii,k |≤ 2.5 |λ2 ii,k| i α α i i i X X 1 1 2 Q 1 1 1 2 g ¯ + C g ¯ ¯ + C ≤ 4 λ3 | ii,k| 2.6 λ ≤ 4 λ3 | ii,k| 2.7 i i i i i i X X X and

ft 1 2 nft 1 2 1 1 2 ( g ¯ ) g ¯ g ¯ ¯ . − λ | λ ii,k | ≤− λ λ2 | ii,k| ≤ 2 λ3 | ii,k| α α i i α α i i i i X X X So Q Q 2 g ¯ 1 1 1 ˜ ω C | i ii,k| f ∆(log trχt t) 2.8 ( 2 )( 2 + t n ) ≥− − ( λi) λ λk λα k P i k α=1 X 1 1 1 P 2 ft 1Q 2 + ( g ¯ + g ¯ ). λ λ2 λ | ij,k| λ λ λ | ij,k| i i i,j i j α α i,j i j Xk X X We have used P Q 1 1 2 1 2 g ¯ ¯ g ¯ ¯ λ2 λ | ij,k| ≥ λ3 | ii,k| i,j i j i i X X here. By Cauchy-Schwarz inequality and the fact that gi¯i,k = gk¯i,i,

2 1 1 1 ( g ¯ )( + f ) | ii,k| λ2 t λ n λ i k k α=1 α Xk X 1 1Q 1 gi¯i,k gj¯j,k ( 2 + ft n ) ≤ | || | λk λk α=1 λα i,j,kX Q 2 1 1 1 2 1 1 1 gi¯i,k ( 2 + ft n ) gj¯j,k ( 2 + ft n ) ≤ | | λ λk λα | | λ λk λα i,j s k k α=1 s k k α=1 X X X Q Q 2 1 1 1 2 = ( gi¯i,k ( 2 + ft n )) | | λ λk λα i s k k α=1 X X 2 gi¯i,k 1 Q 1 1 ( λi) | | ( 2 + ft n ) ≤ λi λ λk λα i i k k α=1 X X X 2 g ¯ 1 1 Q1 = ( λ ) | ik,k| ( + f ) i λ λ2 t λ n λ i k i i α=1 α X Xi,k 2 g ¯ 1 1 Q 1 ( λ ) | ij,k| ( + f ), ≤ i λ λ2 t λ n λ i j i i α=1 α X i,j,kX Q 13 ˜ ˜ so ∆(log trχt ωt) C2.9 at x. However, since x is arbitrary and ∆ is indepen- ≥− ˜ dent of the local coordinates, we see that ∆(log trχt ωt) C2.9 on M. c ≥− Choose ǫ2.10 < 2n as a small constant such that

n 1 n 2 n 1 cω − (n 1)χ ω − > 2ǫ ω − , 0 − − ∧ 0 2.10 0 then n 1 n 2 n 1 cω − (n 1)χ ω − > 2ǫ ω − 0 − − t ∧ 0 2.10 0 2C2.9 by the deﬁnition of χt. Choose C2.1 as , then at the maximal point of ǫ2.10 log tr ω C ϕ , χt t − 2.1 t i¯l g det χ ¯ ˜ i¯j kl¯ αβ 0 ǫ2.10 ∆ϕt = ( g g χk¯j + ft )(gi¯l gi¯l) < . − − det gαβ¯ − 2 Xi,l Xj,k If i¯j kl¯ 0 c g g χ ¯(2g ¯ g ) <ǫ , − kj il − i¯l 2.10 Xi,l Xj,k by the proof of Lemma 3.1 of [34], tr ω C . If χt t ≤ 2.11 i¯j kl¯ 0 c g g χ ¯(2g ¯ g ) ǫ , − kj il − i¯l ≥ 2.10 Xi,l Xj,k then

i¯l g det χαβ¯ 0 ǫ2.10 i¯j ǫ2.10 det χαβ¯ ft (gi¯l gi¯l) < + c g χi¯j = + ft , − det g ¯ − − 2 − − 2 det g ¯ αβ i,j αβ Xi,l X

det gαβ¯ 1 so i λi = < C2.12. Using the fact that λi > , trχt ωt = i λi C2.13 det χαβ¯ c ≤ is also true. This completes the proof of the proposition. Q P By adding a constant if necessary, we can without loss of generality assume 0 that supM ϕt = 0. Then we have the following C estimate: Proposition 2.2.

ϕ 0 C . || t||C ≤ 2.14 1 Moreover, C− χ ω C χ . 2.15 t ≤ t ≤ 2.15 t Proof. Lemma 3.3 and Lemma 3.4 and Proposition 3.5 of [34] only used the inequality in Proposition 2.1. So they are still true in our case. Proposition 2.3. I is closed. Proof. First of all, we want to check the uniform ellipticity and concavity for the Evans-Krylov estimate. The equation is

i¯j det χαβ¯ g χi¯j ft = c. − − det gαβ¯ −

14 View it as a function in terms of gi¯j , χi¯j and ft, then the partial derivative in the ga¯b direction is i¯b a¯j det χαβ¯ a¯b g g χi¯j + ft g . det gαβ¯ At x, it equals to 1 1 f t δ . ( 2 + ) a¯b λa λa i λi It has uniform upper bound and lower bound.Q The second order derivative in ga¯b and gcd¯ direction is

id¯ c¯b a¯j i¯b ad¯ c¯j det χαβ¯ a¯b cd¯ det χαβ¯ ad¯ c¯b g g g χi¯j g g g χi¯j ft g g ft g g . − − − det gαβ¯ − det gαβ¯

At x, when taking the product with wa¯bwcd¯ and summing a,b,c,d for any matrix wi¯j , we get

1 2 1 2 ft waa¯ 2 ft 1 2 w ¯ w ¯ ( ) w ¯ . − λ2 λ | ab| − λ2λ | ab| − λ λ − λ λ λ | ab| a b b a i i a a i i a b Xa,b Xa,b X Xa,b Q Q It is easy to see that it is non-positive. Thus, if we replace the complex second derivatives by real second derivatives, the uniform ellipticity and concavity for the Evans-Krylov estimate [23, 24, 27, 38] are satisﬁed. By checking Evans-Krylov’s estimate carefully, it is easy to see that in our complex case, the estimate

[(ϕ ) ¯] α C t ij C ≤ 2.16 is still true. By standard elliptic estimate, ϕ 101,α is bounded. By Arzela-Ascoli the- || t||C orem, if ti t and ti I, then a subsequence of ϕt converges to ϕt∞ in C100,α-norm.→ By∞ Remark∈ 1.16,

n 1 n 2 cω − (n 1)χ ω − > 0. t∞ − − t ∧ t∞

So by standard elliptic regularity, ϕt∞ is smooth. In other words, t I. ∞ ∈ 3 Concentration of mass and its application

In this section, we prove Theorem 1.18. However, before that, we need to ﬁgure out the correct deﬁnition of

p p 1 cω pχ ω − 0 − ∧ ≥ when ω is only a current. Recall the following deﬁnition of the smoothing:

15 Definition 3.1. Fix a smooth non-negative function ρ supported in [0,1] such that 1 2n 1 ρ(t)t − Vol(∂B1(0))dt =1 Z0 and ρ is a positive constant near 0. For any δ > 0, the smoothing ϕδ is defined by 2n y ϕδ(x)= ϕ(x y)δ− ρ( )dVoly. Cn − | δ | Z We can define the smoothing of a current using similar formula. It is easy to see that the smoothing commutes with derivatives. So (√ 1∂∂ϕ¯ ) = √ 1∂∂¯(ϕ ). − δ − δ

Recall that √ 1∂∂ϕ¯ 0 if and only if √ 1∂∂ϕ¯ δ 0 for all δ > 0. As an analogy, we can define− ≥ − ≥ p p 1 cω pχ ω − 0 − ∧ ≥ for a closed positive (1,1) current ω using smoothing. Remark that any closed positive (1,1) current can be written as √ 1∂∂¯ acting on a real function locally. − Definition 3.2. Suppose that χ is a Kähler form with constant coefficients on an open set O Cn. Then we say that ⊂ p p 1 c(√ 1∂∂ϕ¯ ) pχ (√ 1∂∂ϕ¯ ) − 0 − − ∧ − ≥ on O if for any δ > 0, the smoothing ϕδ satisfies

p p 1 c(√ 1∂∂ϕ¯ ) pχ (√ 1∂∂ϕ¯ ) − 0 − δ − ∧ − δ ≥ on the set O = x : B (x) O . δ { δ ⊂ } We can also define it without the constant coefficients assumption. Definition 3.3. We say that

p p 1 cω pχ ω − 0 − ∧ ≥ if on any coordinate chart, for any open subset O, as long as χ χ on O for a ≥ 0 K¨ahler form χ0 with constant coeﬃcients, then

p p 1 cω pχ ω − 0. − 0 ∧ ≥

Remark 3.4. p p 1 c(√ 1∂∂ϕ¯ ) pχ (√ 1∂∂ϕ¯ ) − 0 − − 0 ∧ − ≥ is a convex property for ϕ. So if ω is smooth, then

p p 1 cω pχ ω − 0 − ∧ ≥ on O pointwise if and only if it is true on O in the sense of Deﬁnition 3.3.

16 For simplicity, for any positive deﬁnite n n matrix A, we deﬁne P (A) by × I 1 1 P (A) = max( )= max (tr(A )− ), I − V k λj V n 1 Cn | j=k ⊂ X6 where λj are the eigenvalues of A. Then

n 1 n 2 cω − (n 1)χ ω − 0 − − ∧ ≥ is equivalent to Pχ(ω) c. Now we need a lemma:≤ Lemma 3.5.

1 T 1 A C P (A CB− C¯ ) + tr(B− ) P ( ) I − ≤ I C¯T B

Proof. By restricting on the codimension 1 subspaces, it suﬃces to prove that

1 1 T 1 1 A C − tr((A CB− C¯ )− ) + tr(B− ) tr( ). − ≤ C¯T B It is easy to see that

1 1 T I CB− A C I O A CB− C¯ O − T 1 T = − . O I C¯ B B− C¯ I O B − So

1 T 1 1 1 I O A CB− C¯ O − I CB− A C − 1 T − − = T . B− C¯ I O B O I C¯ B − After taking traces, the left hand side equals to

1 T 1 1 1 T 1 T 1 1 tr((A CB− C¯ )− ) + tr(B− ) + tr(B− C¯ (A CB− C¯ )− CB− ). − −

Now we start the proof of Theorem 1.18. By assumption, for any t > 0, there exist c > 0 and ω [(1 + t)ω ] satisfying t t ∈ 0 n 1 n 2 cω − (n 1)χ ω − > 0 t − − ∧ t and χn trωt χ + ct n = c. ωt 1 Consider ω0,M M,t = π1∗ωt + π2∗χ and χM M = π1∗χ + π2∗χ. At each point, × c × diagonalizing them so that χi¯j = δi¯j and (ωt)i¯j = λiδi¯j . Then the eigenvalues on the product manifold are λ , ...λ , 1 , ..., 1 . Their inverses are 1 , ..., 1 , c, ..., c. 1 n c c λ1 λj

17 So the sum of them is at most (n + 1)c because ct > 0. In particular, the sum of (2n-1) distinct elements among them is also at most (n + 1)c. Deﬁne ft,ǫ1.6,ǫ1.7 as in Section 1, then there exists ǫ1.7 > 0 such that for ǫ1.6 small 1 1 2n 1 enough, f > ( ) − . So we can apply Theorem 1.14 to get t,ǫ1.6,ǫ1.7 − 4n (n+1)c ωt,ǫ . ,ǫ . [ω0,M M,t] such that PχM×M (ωt,ǫ . ,ǫ . ) < (n + 1)c and 1 6 1 7 ∈ × 1 6 1 7 2n χM M trωt,ǫ ,ǫ χM M + ft,ǫ . ,ǫ . × = (n + 1)c. 1.6 1.7 × 1 6 1 7 2n ωt,ǫ1.6,ǫ1.7

(1) (1) For each point (x1, x2), we assume that z1 , ..., zn are the local coordinates (2) (2) on M x , and z , ..., zn are the local coordinates on x M. Then we ×{ 2} 1 { 1}× can express ωt,ǫ1.6,ǫ1.7 as

(1) (2) (1,2) (2,1) ωt,ǫ1.6,ǫ1.7 = ωt,ǫ1.6,ǫ1.7 + ωt,ǫ1.6,ǫ1.7 + ωt,ǫ1.6,ǫ1.7 + ωt,ǫ1.6,ǫ1.7 , where n (1) (1) (1) (1) ωt,ǫ ,ǫ = √ 1ω ¯dz dz¯ , 1.6 1.7 − t,ǫ1.6,ǫ1.7,ij i ∧ j i,j=1 X n (2) (2) (2) (2) ωt,ǫ ,ǫ = √ 1ω ¯dzi dz¯j , 1.6 1.7 − t,ǫ1.6,ǫ1.7,ij ∧ i,j=1 X n (1,2) (1,2) (1) (2) ωt,ǫ ,ǫ = √ 1ω ¯dz dz¯ , 1.6 1.7 − t,ǫ1.6,ǫ1.7,ij i ∧ j i,j=1 X ω(2,1) ω(1,2) . t,ǫ1.6,ǫ1.7 = t,ǫ1.6,ǫ1.7 (2) After changing the deﬁnition of zi if necessary, we can assume that n (2) (2) π∗χ = √ 1 dz dz¯ 2 − i ∧ i i=1 X and n (2) (2) (2) ω = √ 1 λidz dz¯ t,ǫ1.6,ǫ1.7 − i ∧ i i=1 X at (x1, x2).

Now consider ω1,t,ǫ1.6,ǫ1.7 deﬁned as

ω1,t,ǫ1.6,ǫ1.7 n 1 c − n = (π1) (ωt,ǫ ,ǫ π2∗χ) n ∗ 1.6 1.7 M nχ ∧ n 1 R c − (1) (2) n 1 = (π1) (nωt,ǫ ,ǫ (ωt,ǫ ,ǫ ) − π2∗χ) n ∗ 1.6 1.7 1.6 1.7 M nχ ∧ ∧ n 1 Rc − (1,2) (2,1) (2) n 2 + (π1) (n(n 1)ωt,ǫ ,ǫ ωt,ǫ ,ǫ (ωt,ǫ ,ǫ ) − π2∗χ). n ∗ 1.6 1.7 1.6 1.7 1.6 1.7 M nχ − ∧ ∧ ∧ R 18 At (x1, x2),

(1) (2) n 1 nω (ω ) − π∗χ t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ 2 n n (1) (1) (1) 1 (2) n = √ 1ω ¯dz dz¯ ( )(ωt,ǫ ,ǫ ) , − t,ǫ1.6,ǫ1.7,ij i ∧ j ∧ λ 1.6 1.7 i,j=1 α=1 α X X and

(1,2) (2,1) (2) n 2 n(n 1)ω ω (ω ) − π∗χ − t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ 2 n 1 1 = √ 1ω(1,2) ω(1,2) dz(1) dz¯(1) ( )(ω(2) )n t,ǫ ,ǫ ,ik¯ t,ǫ ,ǫ ,jk¯ i j t,ǫ1.6,ǫ1.7 − − 1.6 1.7 1.6 1.7 ∧ ∧ λk λα i,j,k=1 α=k X X6 n n (1,2) (1,2) (1) (1) 1 1 (2) n √ 1ω ¯ω ¯dzi dz¯j ( )(ωt,ǫ ,ǫ ) . ≥− − t,ǫ1.6,ǫ1.7,ik t,ǫ1.6,ǫ1.7,jk ∧ ∧ λ λ 1.6 1.7 k α=1 α i,j,kX=1 X By Lemma 3.5,

n n (1) 1 (1,2) (1,2) (1) (1) ∗ Pπ χ(√ 1 ((ω ¯ ω ¯ω ¯)dzi dz¯j )) 1 − t,ǫ1.6,ǫ1.7,ij − λ t,ǫ1.6,ǫ1.7,ik t,ǫ1.6,ǫ1.7,jk ∧ i,j=1 k X Xk=1 P ω (2) π χ χM×M ( t,ǫ1.6,ǫ1.7 ) trω ( 2∗ ) ≤ − t,ǫ1.6,ǫ1.7

n c (2) π χ . ( + 1) trω ( 2∗ ) ≤ − t,ǫ1.6,ǫ1.7 Now we view n 1 c − (2) n (2) (tr π2∗χ)(ωt,ǫ . ,ǫ . ) n ωt,ǫ ,ǫ 1 6 1 7 M nχ 1.6 1.7 as a measure on x1 R M, then it is easy to see that { }× n 1 c − (2) n (tr (2) π∗χ)(ω ) n ω 2 t,ǫ1.6,ǫ1.7 nχ x M t,ǫ1.6 ,ǫ1.7 M Z{ 1}× n 1 R c − (2) n 1 = (π∗χ) (ω ) − χn 2 t,ǫ1.6,ǫ1.7 M x1 M ∧ Z{ }× n 1 Rc − χ n 1 = χ ( ) − χn ∧ c M ZM =1R.

19 By the monotonicity and convexity of Pχ,

n 1 c − 2 (2) n P (ω ) (n + 1)c (tr (2) π∗χ) (ω ) χ 1,t,ǫ1.6,ǫ1.7 n ω 2 t,ǫ1.6,ǫ1.7 nχ t,ǫ1.6 ,ǫ1.7 ≤ − M x1 M Z{ }× (2) n 2 R n 1 ( (tr (2) π∗χ)(ω ) ) c x1 M ω 2 t,ǫ1.6,ǫ1.7 (n + 1)c − { }× t,ǫ1.6,ǫ1.7 ≤ − nχn R (ω(2) )n M x M t,ǫ1.6,ǫ1.7 { 1}× n 1 χ n 1 2 Rnc − ( ( ) − R χ) = (n + 1)c M c ∧ − χn ( χ )n M R M c = c. R R Up to here, we have not used the equation

2n χM M trωt,ǫ ,ǫ χM M + ft,ǫ . ,ǫ . × = (n + 1)c. 1.6 1.7 × 1 6 1 7 ω2n t,ǫ1.6,ǫ1.7

f ω2n t,ǫ1.6 ,ǫ1.7 χ2n By the equation, t,ǫ1.6,ǫ1.7 (n+1)c M M . So as in Proposition 2.6 of [18], ≥ × ωn t ǫ it is easy to see that for any weak limit Θ of t,ǫ1.6,ǫ1.7 when and 1.6 converging to 0, 1∆Θ= ǫ3.1[∆] for a constant ǫ3.1 > 0. Let ∆ǫ3.2 be the ǫ3.2 neighborhood of ∆ with respect to χM M . Then for any δ > 0, for any small enough ǫ3.2 and × ǫ3.3, the smoothing of

n 1 n 1 c − n c − ω π∗χ ǫ χ n t,ǫ1.6,ǫ1.7 2 n 3.1 nχ ( x M) ∆ǫ ∧ − nχ M Z { 1}× ∩ 3.2 M R R is at least ǫ3.3χ for small enough t and ǫ1.6. Similarly,− locally for any n + 1 dimensional subvariety V containing ∆, for n 1 any weak limit Θ′ of ω − ,1 Θ′ = ǫ [V ] for ǫ 0. Since the dimension t,ǫ1.6,ǫ1.7 V 3.4 3.4 ≥ of ∆ is strictly smaller then V , for any ﬁxed smoothing function, for any ǫ3.3 > 0, there exists ǫ3.2 > 0 such that the smoothing of

n 1 c − (2) n 1 π∗χ((ω ) − π∗χ) n 1 t,ǫ1.6,ǫ1.7 2 nχ ( x M) ∆ǫ ∧ M Z { 1}× ∩ 3.2 R is at most ǫ3.3χ for small enough t and ǫ1.6. Now let ǫ3.5 be an arbitrary small positive number. Then we can choose ǫ3.3 n−1 ǫ3.3 1 c such that + ǫ3.3 < n ǫ3.1. Then we choose the number ǫ3.2 depending ǫ3.5 2 RM nχ on ǫ . For any Kähler form ω restricted to the first n coordinates of M M, 3.3 × after choosing a good coordinate, assume that π1∗χ = δi¯j and ωi¯j = λiδi¯j . We ∗ ∗ 1 define the truncation T π χ (ω) by (T π χ (ω)) ¯ = min λi, δ ¯. Now consider 1 1 ij ǫ3.5 ij ǫ . ǫ . { } ∗ 3 5 3 5 π χ ( 1 ) ω ǫ3.5 the truncation 1,t,ǫ1.6,ǫ1.7 defined as

n 1 c − (1) (2) ∗ n 1 (π1) (T π χ (˜ωt,ǫ ,ǫ ) (ωt,ǫ ,ǫ ) − π2∗χ), n ∗ 1 1.6 1.7 1.6 1.7 M nχ ǫ3.5 ∧ ∧ R 20 (1) where the (1,1)-formω ˜t,ǫ1.6,ǫ1.7 is deﬁned by

(1) (2) n 1 nω (ω ) − π∗χ t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ 2 (1,2) (2,1) (2) n 2 + n(n 1)ω ω (ω ) − π∗χ − t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ 2 (1) (2) n 1 =ω ˜ (ω ) − π∗χ. t,ǫ1.6,ǫ1.7 ∧ t,ǫ1.6,ǫ1.7 ∧ 2 The smoothing of

∗ π1 χ n 1 ( ǫ ) c − ω ω 3.5 ǫ χ 1,t,ǫ1.6,ǫ1.7 1,t,ǫ1.6,ǫ1.7 n 3.1 − − M nχ

ǫ3.3 is at least χ ǫ3.3χ for small enough t andR ǫ1.6. In fact, the sum of the − ǫ3.5 − ﬁrst two terms is non-negative outside ∆3.2, the sum of the ﬁrst and third term ǫ3.3 inside ∆3.2 is at least ǫ3.3χ and the second term inside ∆3.2 is at least χ. − − ǫ3.5 By the choice of ǫ3.3, the smoothing of ∗ π χ ( 1 ) 1 cn 1 ω ω ǫ3.5 − ǫ χ 1,t,ǫ1.6,ǫ1.7 1,t,ǫ1.6,ǫ1.7 n 3.1 − − 2 M nχ is nonnegative for small enough t and ǫ1.6. On theR other hands,

∗ P ∗ T π χ ω P ∗ ω n ǫ π1 χ( 1 ( )) π1 χ( ) ( 1) 3.5 ǫ3.5 − ≤ − for any (1,1)-form ω on the ﬁrst n coordinates of M M. So using the estimate × of Pχ(ω1,t,ǫ1.6,ǫ1.7 ), it is easy to see that

∗ π χ ( 1 ) ǫ3.5 Pχ(ω ) c + (n 1)ǫ3.5. 1,t,ǫ1.6,ǫ1.7 ≤ − So if χ χ on the support of the smoothing function for a K¨ahler form χ with 0 ∗ 0 ≥ π χ ( 1 ) P ω ǫ3.5 constant coeﬃcients, then χ0 acting on the smoothing of 1,t,ǫ1.6,ǫ1.7 is at most n−1 c n ǫ P ω 1 c ǫ χ + ( 1) 3.5. So χ0 acting on the smoothing of 1,t,ǫ1.6,ǫ1.7 2 R nχn 3.1 is − − M 1 cn−1 also at most c+(n 1)ǫ3.5. Let ω1.5 be a weak limit of ω1,t,ǫ1.6,ǫ1.7 n ǫ3.1χ, 2 RM nχ − n−1 − ω ω ǫ χ ǫ 1 c ǫ P then 1.5 [ 0 1.4 ], where 1.4 = 2 R nχn 3.1. Moreover, χ0 acting on the ∈ − M smoothing of ω1.5 is at most c + (n 1)ǫ3.5. Since ǫ3.5 is arbitrary, it is at most c. This completes the proof of Theorem− 1.18.

4 Regularization

In this section, we prove Theorem 1.11. By Remark 1.13, the n = 1 and n =2 cases have been proved. By induction, we can assume that Theorem 1.11 has been proved in dimension 1, 2, ..., n 1. By Section 1, we can in addition assume that the condition of Theorem− 1.18 are satisﬁed. So by Theorem 1.18, there exist a constant ǫ > 0 and a current ω [ω ǫ χ] such that 1.4 1.5 ∈ 0 − 1.4 n 1 n 2 cω − (n 1)χ ω − 0 1.5 − − ∧ 1.5 ≥

21 in the sense of Definition 3.3. 1 Pick small enough ǫ4.1 < 10000 such that ω 100ǫ ω (1 + ǫ )2(ω ǫ χ). 0 − 4.1 0 ≥ 4.1 0 − 1.4 Then there exists a current ω = ω 100ǫ ω + √ 1∂∂ϕ¯ such that 4.2 0 − 4.1 0 − 4.2 c ωn 1 n χ ωn 2 2 4.−2 ( 1) 4.−2 0 (1 + ǫ4.1) − − ∧ ≥ in the sense of Definition 3.3. Now we pick a finite number of coordinate balls B2r(xi) such that Br(xi) is a cover of M. Moreover, we require that

i i χ0 <χ< (1 + ǫ4.1)χ0

i on B2r(xi) for Kähler forms χ0 with constant coefficients. We also assume that (1 ǫ )√ 1∂∂¯ z 2 ω (1 + ǫ )√ 1∂∂¯ z 2 − 4.1 − | | ≤ 0 ≤ 4.1 − | | B x ϕi √ ∂∂ϕ¯ i ω B x on 2r( i). Let ω0 be potential such that 1 ω0 = 0 on 2r( i). Then we also assume that − i 2 2 ϕ z ǫ4.1r . | ω0 − | | |≤ Let ϕi be the smoothing of ϕ + (1 100ǫ )ϕi . When δ < r , it is well δ 4.2 − 4.1 ω0 5 defined on B 9 (xi). By assumption, it is easy to see that 5 r c √ ∂∂ϕ¯ i n 1 n χi √ ∂∂ϕ¯ i n 2 . 2 ( 1 δ) − ( 1) 0 ( 1 δ) − 0 (1 + ǫ4.1) − − − ∧ − ≥ So c ¯ i n 1 ¯ i n 2 (√ 1∂∂ϕδ) − (n 1)χ (√ 1∂∂ϕδ) − > 0. 1+ ǫ4.1 − − − ∧ − i i i Now define the function ϕ from B 9 (xi) to R as ϕ ϕ , our goal is to show 4.3 5 r δ ω0 that for any x M, − ∈ 2 i i ǫ4.1r + max ϕ4.3(x) < max ϕ4.3(x). i:x B 9 r(xi) B 8 r (xi) i:x Br(xi) { ∈ 5 \ 5 } { ∈ }

i If this is true, then for the smooth function ϕ4.4 = ˜maxϕ4.3 on M, where ˜max means the regularized maximum by choosing the parameters “ηi” in Lemma 2 ǫ4.1r ¯ I.5.18 of [17] to be smaller than 3 , the K¨ahler form ω4.4 = ω0 + √ 1∂∂ϕ4.4 n 1 n 2 − will satisfy cω4.−4 (n 1)χ ω4.−4 > 0. In general, this is not true. However, using the proof of− the results− ∧ of Blocki and Kolodziej [4], it is in fact true if the Lelong number is small enough. The details consist the rest of this section. r i j It is easy to see that if x B 9 r(xi) B 9 r(xj ) and δ < 10 , B δ (x) Bδ (x). ∈ 5 ∩ 5 2 ⊂ r i For any δ < and x B 9 r(xi), we deﬁneϕ ˆδ by 20 ∈ 5 i i ϕˆδ(x) = sup (ϕ4.2 + (1 100ǫ4.1)ϕω0 ) i − Bδ (x)

22 and define νi(x, δ) by i i ϕˆ r (x) ϕˆδ(x) νi(x, δ)= 16 − . log( r ) log δ 16 − Then νi(x, δ) is monotonically non-decreasing in δ. Recall that the Lelong number is defined by νi(x) = lim νi(x, δ). δ 0 → It is independent of i and can be denoted as ν(x) instead. Recall the definition of ρ in Definition 3.1. Let

2 ǫ4.1r ǫ4.5 = , 1 1 2n 1 32n−1 5( 0 log( t )Vol(∂B1(0))t − ρ(t)dt + log 2 + 22n−3 log 2) R then by the result of Siu [32], the set Y = x : ν(x) ǫ4.5 is a subvariety. For simplicity, we assume that Y is smooth.{ The≥ singular} case will be done at the end of this section. Since Y is smooth by our assumption, as in Section 1, there exists a smooth function ϕ1.11 in a neighborhood O of Y such that

n p n 1 n 2 (c − ǫ )ω − (n 1)χ ω − > 0 − 2 1.1 1.11 − − ∧ 1.11 on O. Now we pick smaller neighborhoods O′ and O′′ such that O O and ′ ⊂ O O′. We need to prove the following proposition: ′′ ⊂ r Proposition 4.1. (1) For small enough δ < 20 , if

max νi(x, δ) 2ǫ , ≤ 4.5 i:x B 9 r(xi) { ∈ 5 } then

ϕ ǫ δ ǫ r2 < ϕi x ϕi x . sup 1.11 +3 4.5 log + 4.1 max ( δ( ) ω0 ( )) ′ − O i:x B 9 r(xi) { ∈ 5 }

r (2) For small enough δ < 20 , if

ϕ ǫ δ ǫ r2 ϕi x ϕi x , inf 1.11 +3 4.5 log 4.1 max ( δ( ) ω0 ( )) O′ − ≤ − i:x B 9 r(xi) { ∈ 5 } then i max ν (x, δ) < 4ǫ4.5. i:x B 9 r(xi) { ∈ 5 } r (3) For small enough δ < 20 , if

max νi(x, δ) 4ǫ . ≤ 4.5 i:x B 9 r(xi) { ∈ 5 }

23 then

i i 2 i i max (ϕ (x) ϕ (x)) + ǫ4.1r < max (ϕ (x) ϕ (x)). δ − ω0 δ − ω0 i:x B 9 r (xi) B 8 r(xi) i:x Br(xi) { ∈ 5 \ 5 } { ∈ }

If Proposition 4.1 is true, for small enough δ, we can deﬁne ϕ1.12 as the i i regularized maximum of ϕ1.11(x)+3ǫ4.5 log δ on O and ϕ ϕ on B 9 (xi). ′ δ ω0 5 r − i Since ν(x) <ǫ4.5 for x Y , for small enough δ, max ν (x, δ) 2ǫ4.5 i:x B 9 r(xi) 6∈ { ∈ 5 } ≤ for all x O′′. So by Proposition 4.1 (1), we do not need to worry about the 6∈ discontiuty near the boundary of O′. By Proposition 4.1 (2) and (3), there is also no need to worry about the discontinuity near the boundary of B 9 (xi). In 5 r conclusion, ϕ1.12 will be smooth and satisfy

n 1 n 2 cω − (n 1)χ ω − > 0 1.12 − − ∧ 1.12 on M as long as Y is smooth and Proposition 4.1 is true. In order to prove Proposition 4.1, we need the following lemma of Bloc ki and Kolodziej [4].

r Lemma 4.2. For any δ < and x B 9 (xi), the following estimates hold: 20 5 r i i i ∈ (1) 0 ϕˆδ ϕˆ δ ν (x, δ) log a for all a 1, a ≤ − ≤ ≥ − i i i 1 1 2n 1 32n 1 (2) 0 ϕˆ ϕ ν (x, δ)( log( )Vol(∂B (0))t − ρ(t)dt + 2n−3 log 2). ≤ δ − δ ≤ 0 t 1 2 Proof. For readers’ convenience,R we almost line by line copy the paper [4] here: i (1) It follows from the logarithmical convexity ofϕ ˆδ and the deﬁnition of νi(x, δ). i (2) Deﬁne another regularizationϕ ˜δ by

i 1 i ϕ˜ (x)= (ϕ4.2 + (1 100ǫ4.1)ϕ )dVol. δ Vol(∂B (x)) − ω0 δ Z∂Bδ(x) Then by the Poisson kernel for subharmonic functions [4] and the estimate in (1), 32n 1 32n 1 ϕi x ϕi x − ϕi ϕi − νi x,tδ ˆtδ( ) ˜tδ( ) 2n 2 (ˆtδ ˆtδ/2) ( 2n 2 log 2) ( ) − ≤ 2 − − ≤ 2 − for all t (0, 1]. By monotonicity, ∈ 32n 1 32n 1 ϕi x ϕi x − νi x,tδ − νi x, δ . ˆtδ( ) ˜tδ( ) ( 2n 2 log 2) ( ) ( 2n 2 log 2) ( ) − ≤ 2 − ≤ 2 − Deﬁne 2n 1 ρ˜(t) = Vol(∂B1(0))t − ρ(t), 1 then 0 ρ˜(t)=1. So R 1 1 32n 1 ϕ˜i ϕi = (˜ϕi ϕ˜i )˜ρ(t)dt (ˆϕi ϕˆi )˜ρ(t)dt + ( − log 2)νi(x, δ). δ − δ δ − tδ ≤ δ − tδ 22n 2 Z0 Z0 −

24 By the estimate in (1) again, 1 ϕî ϕî νi(x, δ) log( ). δ − tδ ≤ t The other side of inequality 0 ϕî ϕi is trivial. ≤ δ − δ r It is easy to see that there exists a constant C4.6 such that for any δ < 20 i and x B 9 r(xi), ν (x, δ) < C4.6. Now we are ready to prove Proposition 4.1. ∈ 5 r (1) Suppose δ < , x B 9 r(xi) and 20 ∈ 5 i i ϕˆ r (x) ϕˆδ(x) νi(x, δ)= 16 − 2ǫ , log( r ) log δ ≤ 4.5 16 − then

i i r ϕˆδ(x) ϕˆ r (x)+2ǫ4.5(log δ log( )) C4.7 +2ǫ4.5 log δ. ≥ 16 − 16 ≥− By Lemma 4.2 (2), ϕi (x) C +2ǫ log δ. δ ≥− 4.8 4.5 It is easy to see that for δ small enough,

2 i i sup ϕ1.11 +3ǫ4.5 log δ + ǫ4.1r < ϕδ(x) ϕω0 (x) O′ −

i because ϕ is uniformly bounded on B 9 (xi). ω0 5 r r (2) Suppose δ < , x B 9 r(xi) and 20 ∈ 5 ϕ ǫ δ ǫ r2 ϕi x ϕi x , inf 1.11 +3 4.5 log 4.1 δ( ) ω0 ( ) O′ − ≤ − then as before ϕˆi (x) ϕi (x) C +3ǫ log δ. δ ≥ δ ≥− 4.9 4.5 i By Lemma 4.2 (1) and the deﬁnition ofϕ ˆ δ (x), 2

sup ϕ4.2 C4.10 +3ǫ4.5 log δ. i ≥− B δ (x) 2

i j If x B 9 r(xj ), then B δ (x) Bδ (x) and therefore ∈ 5 2 ⊂

sup ϕ4.2 sup ϕ4.2 C4.10 +3ǫ4.5 log δ. j ≥ Bi (x) ≥− Bδ (x) δ 2

j j j By the deﬁnition ofϕ ˆδ(x) and ν (x, δ), it is easy to see that ν (x, δ) < 4ǫ4.5 if δ is small enough.

25 r (3) Suppose δ < , x (B 9 r(xi) B 8 r(xi)) Br(xj ) and 20 ∈ 5 \ 5 ∩ max νi(x, δ) 4ǫ , ≤ 4.5 i:x B 9 r(xi) { ∈ 5 } then

i i 2 2 2 7 2 ϕ δ x ϕ x ϕ ǫ r r δ r ǫ r , ˆ ( ) ω0 ( ) sup 4.2 +2 4.1 + (2 + ) (2 ) 100 4.1( ) 2 − ≤ i − − 5 B δ (x) 2 and 6 ϕ ϕj x ϕj x ǫ r2 r δ 2 r 2 ǫ r 2. sup 4.2 ˆδ( ) ω0 ( )+2 4.1 + (2 + ) (2 ) + 100 4.1( ) j ≤ − − 5 Bδ (x)

By Lemma 4.2 (1),

i i i ϕˆδ ϕˆ δ ν (x, δ) log 2 4ǫ4.5 log 2. − 2 ≤ ≤ By Lemma 4.2 (2), ϕi ϕˆi and δ ≤ δ 1 2n 1 j j 1 2n 1 3 − ϕˆ ϕ 4ǫ ( log( )Vol(∂B (0))t − ρ(t)dt + log 2). δ − δ ≤ 4.5 t 1 22n 3 Z0 −

Since supBi (x) ϕ4.2 sup j ϕ4.2, by summing everything together, for δ small δ Bδ (x) 2 ≤ ϕi x ϕi x ǫ r2 < ϕj x ϕj x Y enough, δ( ) ω0 ( )+ 4.1 δ( ) ω0 ( ). We are done if is smooth. In general −Y is singular. By Hironaka’s− desingularization theorem, there exists a blow-up M˜ of M obtained by a sequence of blow-ups with smooth centers such that the proper transform Y˜ of Y is smooth. Without loss of generality, assume that we only need to blow up once. Let π be the projection of M˜ to M. Let E be the exceptional divisor. Let s be the defining section of E. Let h √ 1 ¯ 2 be any smooth metric on the line bundle [E], then 2−π ∂∂ log s h = [E]+ ω4.11 by the Poincaré-Lelong equation. Then it is well known that the| | smooth (1,1)- form ω [E] on M˜ and ω > C π∗ω . For example, see Lemma 3.5 4.11 ∈− 4.11 − 4.12 0 of [18] for the explanation. Define ω4.13 = C4.12π∗ω0 + ω4.11. Then ω4.13 is a Kähler form on M˜ .

6n Lemma 4.3. Let C4.14 = . Then for all small enough t and q-dimensional ǫ1.1 subvarieties V of M˜ , as long as q

n q 2 q (c − ǫ )((1 + C t)π∗ω + C t ω ) − 3n 1.1 4.14 0 4.14 4.13 ZV 2 q 1 2 q((1 + C t)π∗ω + C t ω ) − (π∗χ + t ω ). ≥ 4.14 0 4.14 4.13 ∧ 4.13 ZV

26 Proof. By assumption,

ǫ1.1 q q 1 ǫ1.1 q q 1 (c )π∗ω qπ∗ω − π∗χ = (c )ω qω − χ 0. − 3 0 − 0 ∧ − 3 0 − 0 ∧ ≥ ZV Zπ(V ) So

ǫ1.1 q q 1 (c )((1+C t)π∗ω ) q((1+C t)π∗ω ) − ((1+C t)π∗χ) 0. − 3 4.14 0 − 4.14 0 ∧ 4.14 ≥ ZV It suﬃces to show that

ǫ1.1 2 q (c )((1 + C4.14t)π∗ω0 + C4.14t ω4.13) V − 3 Z 2 q 1 2 q((1 + C t)π∗ω + C t ω ) − (π∗χ + t ω ) − 4.14 0 4.14 4.13 ∧ 4.13 ǫ1.1 q (c )((1 + C4.14t)π∗ω0) ≥ V − 3 Z q 1 q((1 + C t)π∗ω ) − ((1 + C t)π∗χ). − 4.14 0 ∧ 4.14

Since it only depends on the cohomology classes, we want to replace ω0 by a better representative in its cohomology class. Remark that π(E) is smooth by assumption. So we can apply Theorem 1.11 to π(E). As in Section 1, there exists a smooth function ϕ4.15 on a neighborhood O4.16 of π(E) in M such that ω = ω + √ 1∂∂ϕ¯ satisﬁes 4.15 0 − 4.15 ǫ1.1 n 1 n 2 (c )ω − (n 1)χ ω − > 0 − 2 4.15 − − ∧ 4.15

2 log s h on O4.16. Define ϕ4.17 = π | | on M π(E). Recall the definition of the ∗ 4πC4.12 \ regularized maximum in Lemma I.5.18 of [17]. For large enough C4.18, let ϕ4.19 be the regularized maximum of ϕ4.17 + C4.18 and ϕ4.15. Then ϕ4.19 is smooth on M and ω4.19 = ω0 +√ 1∂∂ϕ¯ 4.19 > 0 on M. Moreover, there exists a smaller neighborhood O of π(−E) such that ϕ = ϕ on O O . 4.20 4.19 4.15 4.20 ⊂ 4.16 After replacing ω0 by ω4.19, it suffices to show that

q ǫ1.1 q! q i 2 i (c ) ((1 + C t)π∗ω ) − (C t ω ) − 3 i!(q i)! 4.14 4.19 ∧ 4.14 4.13 i=1 − q X (q 1)! q i i 1 2 i q − ((1 + C t)π∗ω ) − C − (t ω ) − (q i)!(i 1)! 4.14 4.19 ∧ 4.14 4.13 i=1 X − − q 1 − (q 1)! q 1 i q − ((1 + C t)π∗ω ) − − − i!(q 1 i)! 4.14 4.19 i=1 − − X 2 i (C t ω ) (1 + C t)π∗χ ∧ 4.14 4.13 ∧ 4.14 2 q 1 +q((1 + C t)π∗ω + C t ω ) − C tπ∗χ 4.14 4.19 4.14 4.13 ∧ 4.14 0. ≥

27 By definition of C4.14, (q 1)! ǫ q! q − < 1.1 C (q i)!(i 1)! 6 i!(q i)! 4.14 − − − for all i = 1, 2, ..., q. So we can combine the first term and the second term. If 1 the point is inside π− (O ), then for all i =1, 2, ..., q 1, 4.20 − ǫ1.1 q i q 1 i (c )(π∗ω ) − (q i)(π∗ω ) − − π∗χ − 2 4.19 ≥ − 4.19 ∧ because ǫ1.1 n 1 n 2 (c )ω − (n 1)ω − χ − 2 4.19 ≥ − 4.19 ∧ on O4.20. So the sum of the first three terms is non-negative if i =1, 2, ..., q 1. So we are done because the i = q term and the fourth term are non-negative.− If 1 the point is outside π− (O4.20), then there exists C4.21 such that

1 C4.21π∗χ>ω4.13 > C4−.21π∗χ and 1 C4.21π∗χ > π∗ω4.19 > C4−.21π∗χ on M˜ π 1(O ). The only ﬁrst order term in t is \ − 4.20 q 1 qπ∗ω − C tπ∗χ. 4.19 ∧ 4.14 Since it is positive, for small enough t, we also get the required inequality. Now we pick t> 0 such that t satisﬁes Lemma 4.3 and c ǫ c > c 1.1 , . 2 max 1+ C4.14t + C4.12C4.14t { − 4n 1+ ǫ4.1 }

We apply Theorem 1.11 to the lower dimensional smooth manifold Y˜ with the 2 2 K¨ahler forms (1 + C4.14t)π∗ω0 + C4.14t ω4.13 and π∗χ + t ω4.13. As in Section 1, there exists a smooth function ϕ4.22 on a neighborhood of Y˜ in M˜ such that

2 ω = (1+ C t)π∗ω + C t ω + √ 1∂∂ϕ¯ 4.22 4.14 0 4.14 4.13 − 4.22 satisﬁes ǫ1.1 n 1 2 n 2 (c )ω − (n 1)(π∗χ + t ω ) ω − > 0 − 4n 4.22 − − 4.13 ∧ 4.22 near Y˜ . Similarly, let ϕ4.23 be the potential near E. For large enough constant C4.24, deﬁne

1 ϕ = ˜max ϕ , ϕ + C− π∗ϕ + C 4.25 { 4.23 4.22 4.24 4.17 4.24} and 2 ω = (1+ C t)π∗ω + C t ω + √ 1∂∂ϕ¯ . 4.25 4.14 0 4.14 4.13 − 4.25

28 Then ǫ1.1 n 1 2 n 2 (c )ω − (n 1)(π∗χ + t ω ) ω − > 0 − 4n 4.25 − − 4.13 ∧ 4.25 on a neighborhood O of Y˜ E in M˜ . Since t2ω > 0, it is easy to see that ∪ 4.13 ǫ1.1 n 1 n 2 (c )(π ω4.25) − (n 1)χ (π ω4.25) − > 0 − 4n ∗ − − ∧ ∗ on π(O E). Now we choose neighborhoods O′ and O′′ of Y π(E) in M such \ ∪ that O π(O) and O O′. Then as before, for small enough δ, we can ′ ⊂ ′′ ⊂ deﬁne ϕ4.26 as the regularized maximum of π ϕ4.25 +3ǫ4.5 log δ on O′ π(E) i i ∗ \ and ϕ ϕ on B 9 (xi). Then ϕ4.26 is smooth and bounded on M π(E). δ ω0 5 r Moreover,− for \

2 ω4.26 =(1+ C4.14t)ω0 + C4.14t π ω4.13 + √ 1∂∂ϕ¯ 4.26 ∗ − 2 2 =(1+ C4.14t + C4.12C4.14t )ω0 + C4.14t π ω4.11 + √ 1∂∂ϕ¯ 4.26, ∗ − it is easy to see that

ǫ1.1 c n 1 n 2 (max c , )ω4.−26 (n 1)χ ω4.−26 > 0 { − 4n 1+ ǫ4.1 } − − ∧

2 on M π(E) because C4.14tω0 + C4.14t π ω4.13 > 0. Now we deﬁne \ ∗ 2 ω4.26 C4.14t π ω4.11 + √ 1∂∂ϕ¯ 4.26 ω ω ∗ , 4.27 = 2 = 0 + − 2 1+ C4.14t + C4.12C4.14t 1+ C4.14t + C4.12C4.14t then by the choice of t,

n 1 n 2 cω − (n 1)χ ω − > 0 4.27 − − ∧ 4.27 on M π(E). For large enough constant C , deﬁne \ 4.28 2 C4.14t 2 π log s + ϕ4.26 ϕ 2π ∗ h C , ϕ , 4.29 = ˜max | | 2 + 4.28 4.15 {1+ C4.14t + C4.12C4.14t } then ϕ is smooth on M and ω = ω + √ 1∂∂ϕ¯ satisﬁes 4.29 4.29 0 − 4.29 n 1 n 2 cω − (n 1)χ ω − > 0 4.29 − − ∧ 4.29 on M. We are done.

5 Deformed Hermitian-Yang-Mills Equation

In this section, we prove Theorem 1.7. The equation

n arctan λi = θˆ i=1 X

29 for eigenvalues λi of ωϕ = ω0 + √ 1∂∂ϕ¯ > 0 with respect to χ is the same as the equation − n 1 nπ arctan( )= θ.ˆ λ 2 − i=1 i X nπ ˆ To simplify the notations, deﬁne θ0 = 2 θ. Then the equation is equivalent to the inequality − 1 arctan( ) <θ0 λi i=j X6 for j =1, 2, ..., n and the equation

Im(ω + √ 1χ)n = tan(θ )Re(ω + √ 1χ)n. ϕ − 0 ϕ − Inspired by the work of Pingali in the toric case [30], the analogy of Theorem 1.11 is the following:

n Theorem 5.1. Fix a K¨ahler manifold M with K¨ahler metrics χ and ω0. Let θ (0, π ) be a constant and f > 1 be a smooth function satisfying 0 ∈ 4 − 100n χn tan(θ )Re(ω + √ 1χ)n Im(ω + √ 1χ)n f = ( 0 ϕ − ϕ − ) 0, n! n! − n! ≥ ZM ZM then there exists a smooth function ϕ satisfying the equation

Im(ω + √ 1χ)n + fχn = tan(θ )Re(ω + √ 1χ)n ϕ − 0 ϕ − and the inequality 1 arctan( ) <θ0 λi i=j X6 for j = 1, 2, ..., n and eigenvalues λ of ω = ω + √ 1∂∂ϕ¯ > 0 with respect i ϕ 0 − to χ if there exists a constant ǫ1.1 > 0 and for all p-dimensional subvarieties V with p =1, 2, ..., n (V can be chosen as M), there exist smooth functions θV (t) from [1, ) to [ pπ θ + (n p)ǫ , pπ ) such that for all t [1, ), ∞ 2 − 0 − 1.1 2 ∈ ∞

p p pπ (χ + √ 1tω0) =0,θV (t) = arg( (χ + √ 1tω0) ), lim θV (t)= . − 6 − t 2 ZV ZV →∞

When n = 1, it is trivial. In higher dimensions, we need to prove it by induction. Inspired by the work of Collins-Jacob-Yau [12], the analogy of Theorem 1.14 is the following:

n Theorem 5.2. Fix a K¨ahler manifold M with K¨ahler metrics χ and ω0. Let θ (0, π ) be a constant and f > 1 be a smooth function satisfying 0 ∈ 4 − 100n χn tan(θ )Re(ω + √ 1χ)n Im(ω + √ 1χ)n f = ( 0 ϕ − ϕ − ) 0, n! n! − n! ≥ ZM ZM

30 then there exists a smooth function ϕ satisfying the equation

Im(ω + √ 1χ)n + fχn = tan(θ )Re(ω + √ 1χ)n ϕ − 0 ϕ − and the inequality 1 arctan( ) <θ0 λi i=j X6 for j =1, 2, ..., n and eigenvalues λi of ωϕ = ω0 + √ 1∂∂ϕ¯ > 0 with respect to χ if − 1 arctan( ) <θ0 λi,0 i=j X6 for j =1, 2, ..., n and eigenvalues λi,0 of ω0 > 0 with respect to χ.

We will use the continuity method three times to prove Theorem 5.2. Letω ˜0 be the form ω0 and f˜ be the function f in Theorem 5.2. There exists a constant θ0 C5.1 such thatω ˜0 C5.1χ. We start from f = 0 and ω0 = cot( n )χ. In this case, ≥ θ0 1 θ0 ϕ = 0 is the solution. Then we let ω0 = t cot( n )χ + (1 t)(C5−.1 cot( n )+1)˜ω0 and f be the non-negative constant satisfying the integrability− condition as the 1 θ0 ﬁrst path. It will imply the result for ω0 = (C5−.1 cot( n )+1)˜ω0. Then we let ω0 = tω˜0 and f be the constant satisfying the integrability condition as the second path. f must be non-negative because d tan(θ )Re(tω + √ 1χ)n Im(tω + √ 1χ)n ( 0 0 − 0 − ) dt n! − n! ZM n 1 n 1 tan(θ )Re(tω + √ 1χ) − Im(tω + √ 1χ) − = ( 0 0 − 0 − ) ω > 0 (n 1)! − (n 1)! ∧ 0 ZM − − by the assumption on ω0 and Lemma 8.2 of [12]. The continuity method will imply the result forω ˜0 when f is the non-negative constant f0 satisfying the integrability condition. Finally, we let ω0 =ω ˜0 and f = tf˜+ (1 t)f0 be the third continuity path. It will imply Theorem 5.2. − It is easy to see the openness along the paths. Thus, we only need to prove the a priori estimate along the paths. It will be achieved by Sz´ekelyhidi’s estimates in [36]. First of all, we need to rewrite the equation. The equation

Im(ω + √ 1χ)n + fχn = tan(θ )Re(ω + √ 1χ)n ϕ − 0 ϕ − can be written as n n Im (λ + √ 1) + f = tan(θ )Re (λ + √ 1). i − 0 i − i=1 i=1 Y Y It is equivalent to

n 1 f n 1 sin( arctan( )) + = tan(θ ) cos( arctan( )), λ n 2 0 λ i=1 i i=1 λi +1 i=1 i X X Q p 31 which is the same as

n 1 f cos(θ ) sin(θ arctan( )) = 0 . 0 − λ n 2 i=1 i i=1 λi +1 X Let Γ be the region consisting of (λ , ..., λQ) Rpn such that λ > 0 and 1 n ∈ i 1 arctan( ) <θ0 λi i=j X6 for all j =1, 2, ..., n, then we want to study the function

n 1 f cos(θ ) F (f, λ , ..., λ ) = sin(θ arctan( )) 0 1 n 0 − λ − n 2 i=1 i i=1 λi +1 X 1 ¯ ¯ Q p on ( 100n , ) Γ, where Γ is the closure of Γ. For any Kähler form ω, we say ω −Γ if the∞ eigenvalues× of ω with respect to χ is in Γ. Similarly, we define ∈ χ Fχ(f,ω) as F (f, λi) for eigenvalues λi of ω with respect to χ. In order to apply Székelyhidi’s estimates in [36], we claim the following: Proposition 5.3. Assume that n 2. If f > 1 , then ≥ − 100n (1) ∂ F (f, λ) > 0 if λ Γ; ∂λi 2 ∈ ∂ λiδij (2) ∂λ ∂λ F (f, λ) cos(θ0) 2(λ2+1)2 if λ Γ and F (f, λ)=0; i j ≤− i ∈ ∂ ∂ (3) F (f, λ) F (f, λ) if λ Γ and λi λj ; ∂λi ≤ ∂λj ∈ ≥ (4) If λ ∂Γ, then F (f, λ) < 0; (5) For any∈ λ Γ, the set ∈

λ′ Γ: F (f, λ′)=0, λ′ λ { ∈ i ≥ i} is bounded. Proof. (1)

n 1 ∂ cos(θ0 arctan( )) f cos(θ ) λ F (f, λ)= − k=1 λk + 0 i . ∂λ λ2 +1 n 2 λ2 +1 i P i k=1 λk +1 i Using the deﬁnition of Γ, it is easy to see that Q p

n 1 nθ 0 < arctan( ) < 0 λk n 1 Xk=1 − on Γ. So n 1 1 cos(θ0 arctan( )) > cos(θ0) > . − λk √2 kX=1 It is easy to see that ∂ F (f, λ) > 0 if λ Γ. ∂λi ∈

32 (2)

n ∂2 1 2λ δ F f, λ θ i ij ( )= cos( 0 arctan( )) 2 2 ∂λi∂λj − − λk (λ + 1) k=1 i n X 1 1 sin(θ0 arctan( )) 2 2 − − λk (λi + 1)(λj + 1) kX=1 f cos(θ ) λ λ f cos(θ ) 1 λ2 0 i j + 0 − i δ . n 2 λ2 +1 λ2 +1 n 2 (λ2 + 1)2 ij − k=1 λk +1 i j k=1 λk +1 i

Q p 2λiδij Q p The ﬁrst term is at most cos(θ0) (λ2+1)2 . When F (f, λ) = 0, the second term − i f cos(θ0) 1 f equals to n 2 (λ2+1)(λ2+1) . When 0, it is a non-negative deﬁnite − Qk=1 √λk+1 i j ≥ matrix. When 0 >f > 1 , for any ξ , ...ξ R, − 100n 1 n ∈ n f cos(θ0) ξiξj − n 2 (λ2 + 1)(λ2 + 1) i,j=1 k=1 λk +1 i j X n Q 1 p ξiξj cos(θ0) | 5 | 5 ≤ 100n 2 4 2 4 i,j=1 (λi + 1) (λj + 1) X n 1 ξi 2 = cos(θ0)( | | 5 ) 100n 2 4 i=1 (λi + 1) X n 2 1 ξi cos(θ0) 5 ≤ 100 2 2 i=1 (λi + 1) X 1 n n ξ ξ δ cos(θ ) i j ij . ≤ 100 0 λ (λ2 + 1)2 i=1 j=1 i i X X

Since λi > cot(θ0) > 1 on Γ, it is easy to see that the second term is at most 1 4 times the first term. Similarly the third term and the fourth term are also − 1 at most 4 times the first term. (3) It− suffices to show that

d 1 f d x 2x f 1 x2 + = − + − dx x2 +1 n 2 dx x2 +1 (x2 + 1)2 n 2 (x2 + 1)2 k=1 λk +1 k=1 λk +1 p p is non-positiveQ for all x [λj , λi]. When f 0, it isQ trivial. When 0 >f > 1 , it is at most ∈ ≥ − 100n 2x 1 x2 1 2x 1 x2 1 . 2− 2 + 2 − 2 2− 2 + 2 − 2 (x + 1) 100nλi (x + 1) ≤ (x + 1) x (x + 1) This is indeed non-positive because x λ > 1. ≥ j

33 θ0 (4)The point λi = cot( n 1 ) belongs to ∂Γ and F at this point is negative because −

n 2 θ0 2 1 (cot ( n 1 )+1) θ0 1 − < sin( )= n 1 θ . 100n cos(θ ) n 1 2 0 0 cos(θ0) sin − ( n 1 ) − − Thus, it suffices to prove that if F (f, λ) = 0, then λ ∂Γ. If f 0, it is obvious. n 6∈ ≥ If 0 >f > 1 , then 0 θ arctan( 1 ) θ0 π . So λ ∂Γ 100n 0 i=1 λi n 1 4 i because − ≥ − ≥ − − ≥ − 6∈ 1 P 1 sin x < ( inf ( x2 + 1 arctan( )))( inf ). π 100n x (1, ) x x [ 4 ,0] x ∈ ∞ p ∈ − (5) is obvious. Compared to Székelyhidi’s conditions in [36], there are three major differ- ences. First of all, F also depends on f. Second of all, Γ does not contain the positive orthant. Finally, even if we fix the f variable, F is only concave when F = 0. However, we will show that his works still survive without much changes. Proposition 5 of [36] only requires the concavity of F when F = 0. So it still holds. Székelyhidi’s C0 estimate relies on the variant of Alexandroff-Bakelman- Pucci maximum principle similar to Lemma 9.2 of [25]. Clearly it does not take derivatives of f. So Székelyhidi’s C0 estimate is still true. The next step is to prove that

¯ 2 √ 1∂∂ϕ χ C5.2(1 + sup ϕ χ). | − | ≤ M |∇ |

We will use the same notations as in [36] except that letter f in [36] is replaced by F , the letter F is replaced by Fχ and the letter u is replaced by ϕ. It is easy to see that (78) of [36] still holds. Now we diﬀerentiate the equation Fχ(f,ωϕ) = 0. We see that

ij f Fχ gi¯j1 + Fχ f1 =0, and

pq,rs kk kk,f kk,f f Fχ gpq¯1grs¯1¯ + Fχ gkk¯11¯ + Fχ gkk¯1f1¯ + Fχ gkk¯1¯f1 + Fχ f11¯ =0

ff f cos(θ0) f because Fχ = 0. Since Fχ = n 2 1, the term Fχ f11¯ is bounded. | | |− Qi=1 √λi +1 |≤ 1 kk,f So the only additional term in (85) of [36] is C0λ1− Fχ gkk¯1 on the right hand side. Instead of (94) of [36], we get − | |

kk f Fχ gkkp¯ + Fχ fp =0.

f Since Fχ fp is bounded, the estimate in (95) still holds. So the only additional | | 1 kk,f term in (99) and (104) of [36] is C λ− F g ¯ on the right hand side. − 0 1 | χ kk1|

34 The case 1 in [36] will not happen. The additional term in (120) of [36] is also 1 kk,f C λ− F g ¯ . However, recall that (67) of [36] is that − 0 1 | χ kk1|

ij,rs F1 Fi F g ¯ g ¯ F g ¯ g ¯¯ − g ¯ . − χ ij1 rs¯1 ≥− ij ii1 jj1 − λ λ | i11| i>1 1 i X − (Remark that the letter f in [36] is replaced by F and the letter F is replaced by Fχ.) The term Fij gi¯i1gj¯j1¯ was thrown away. However, this term is at least λi 2 − 2(λ2+1)2 gi¯i1 . The term i | |

kk,f f cos(θ0) λi λi C0 Fχ gkk¯1 = C0 n 2 gi¯i1 C0 f 3 gi¯i1 2 λ +1 2 2 − | | − | k=1 λk +1 i |≥− | |(λi + 1) | |

λi 2Q p is at least 2(λ2+1)2 gi¯i1 C5.3. So Sz´ekelyhidi’s estimate − i | | − ¯ 2 √ 1∂∂ϕ χ C5.2(1 + sup ϕ χ) | − | ≤ M |∇ | still holds. Sz´ekelyhidi used the property that Γ contains the positive orthant to prove the C2 estimate [36]. We do not have this property. However, we can use Proposition 5.1 of [12] to achieve this. The Evans-Krylov estimate requires the uniform ellipticity and concavity of Fχ(f, .). Its relationship with the function F was cited as (63) and (64) of [36]. By Proposition 5.3, the conditions for the Evans-Krylov estimate are indeed true. The higher order estimate follows from standard elliptic theories. Finally, ωϕ will stay in the region Γχ along the continuity paths by Proposition 5.3 (4). The analogy of Theorem 1.18 is the following:

n Theorem 5.4. Fix a Kähler manifold M with Kähler metrics χ and ω0. Sup- pose that for all t> 0, there exist a constant ct > 0 and a smooth Kähler form ω [(1 + t)ω ] satisfying ω Γ , and t ∈ 0 t ∈ χ Im(ω + √ 1χ)n + c χn = tan(θ )Re(ω + √ 1χ)n. ϕ − t 0 ϕ −

Then there exist a constant ǫ5.4 > 0 and a current ω5.5 [ω0 ǫ5.4χ] such that ω Γ¯ in the sense of current. ∈ − 5.5 ∈ χ The deﬁnition of a current being in Γ¯χ is similar to Deﬁnition 3.3 except that we replace the condition

n 1 n 2 cω − (n 1)χ ω − 0 − − ∧ ≥ by ω Γ¯ for Kähler form ω. To simplify notations, for a Kähler form ω, we ∈ χ define Pχ(ω) and Qχ(ω) by 1 Pχ(ω) = max( arctan( )), j λi i=j X6

35 and 1 Q (ω)= arctan( ), χ λ i i X where λ are the eigenvalues of ω with respect to χ. Then ω Γ¯ is equivalent i ∈ χ to Pχ(ω) θ0. The analogy≤ of Lemma 3.5 is the following: Lemma 5.5. Suppose that A C I O > . C¯T B O I Then 1 T A C P (A CB− C¯ )+ Q (B) P ( ). I − I ≤ I C¯T B

Proof. It is easy to see that B > I. Moreover, for any ξ =0 Cn, 6 ∈ ¯T 1 ¯T ¯T ¯T 1 A C ξ ξ (A CB− C )ξ = ξ ξ CB− T 1 T − − C¯ B B− C¯ ξ − 2 1 T 2 > ξ + B− C¯ ξ , | | | | 1 T so A CB− C¯ > I. Therefore, both hand sides of the inequality are well deﬁned.− By restricting on the codimension 1 subspaces, it suﬃces to prove that

1 T A C Q (A CB− C¯ )+ Q (B) Q ( ). I − I ≤ I C¯T B For any s [0, 1], it is easy to see that ∈ A + √ 1I sC det − sC¯ T B + √ 1I − 2 1 T A s C(B + √ 1I)− C¯ + √ 1I O = det − − − . O B + √ 1I − 2 1 T So we need to compute det(A s C(B + √ 1I)− C¯ + √ 1I). We already know that B >− I, so − −

1 1 1 1 (B + √ 1I)− = B− (I + √ 1B− )− − − ∞ k 2k 1 ∞ k 2k 2 = ( 1) B− − √ 1 ( 1) B− − . − − − − kX=0 Xk=0 Therefore 2 1 T A s C(B + √ 1I)− C¯ + √ 1I − − − 2 ∞ k 2k 1 T 2 ∞ k 2k 2 T = A s ( ( 1) CB− − C¯ )+ √ 1(I + s ( 1) CB− − C¯ ). − − − − Xk=0 Xk=0

36 2 1 T The real part is at least A s CB− C¯ > I and the imaginary part is also at least I. So − A + √ 1I sC arg det − sC¯ T B + √ 1I − 2 1 T = arg det(A s C(B + √ 1I)− C¯ + √ 1I) + arg det(B + √ 1I). − − − − if we deﬁne arg det(X + √ 1Y ) as QY (X) for X,Y > 0. In fact, this is true up to an integer times 2π. However,− both hand sides are continuous with respect to s and this equation holds for s = 0. So it holds for all s [0, 1]. Now it suﬃces to show that ∈

1 T 1 T arg det(A C(B + √ 1I)− C¯ + √ 1I) arg det(A CB− C¯ + √ 1I). − − − ≥ − − It follows from the facts that

∞ k 2k 2 T ∞ k+1 2k 1 T ( 1) CB− − C¯ ( 1) CB− − C¯ 0 − ≥ − ≥ kX=0 Xk=1 and 1 T A CB− C¯ > I. −

Choose C large enough such that θ + n arctan( 1 ) < π . The deﬁnitions 5.6 0 C5.6 4 of χM M , χM M,ǫ5.7,ǫ5.8 and ft,ǫ5.7,ǫ5.8 are still the same as in Section 1. As before,× there exists× ǫ > 0 such that for ǫ small enough, f > 1 . 5.8 5.7 t,ǫ5.7,ǫ5.8 − 200n Now we consider ω0,M M,t = π1∗ωt + C5.6π2∗χ. By Theorem 5.2, there exists × 1 ωt,ǫ . ,ǫ . [ω0,M M,t] such that PχM×M (ωt,ǫ . ,ǫ . ) <θ0 + n arctan( ) and 5 7 5 8 ∈ × 5 7 5 8 C5.6 √ 2n 2n Im(ωt,ǫ5.7,ǫ5.8 + 1χM M ) + ft,ǫ5.7,ǫ5.8 χM M − × × 1 2n = tan(θ0 + n arctan( ))Re(ωt,ǫ5.7,ǫ5.8 + √ 1χM M ) . C5.6 − ×

Deﬁne ω1,t,ǫ5.7,ǫ5.8 by

n−1 ⌊ 2 ⌋ k n! n 2k 2k+1 ( 1) (π1) (ωt,ǫ− ,ǫ π2∗χ ) k=0 (n 2k)!(2k+1)! ∗ 5.7 5.8 ω1,t,ǫ ,ǫ = − − ∧ . 5.7 5.8 Im(C χ + √ 1χ)n P M 5.6 −

Fix ǫ5.8 and let t and ǫ5.7 convergeR to 0. For small enough ǫ5.4, we shall expect ω to be the weak limit of ω ǫ χ. 5.5 1,t,ǫ5.7,ǫ5.8 − 5.4 As before, we write ωt,ǫ5.7,ǫ5.8 as

ω ω(1) ω(2) ω(1,2) ω(2,1) , t,ǫ5.7,ǫ5.8 = t,ǫ5.7,ǫ5.8 + t,ǫ5.7,ǫ5.8 + t,ǫ5.7,ǫ5.8 + t,ǫ5.7,ǫ5.8 and assume that n (2) (2) π∗χ = √ 1 dz dz¯ 2 − i ∧ i i=1 X

37 and n (2) (2) (2) ω = √ 1 λidz dz¯ t,ǫ5.7,ǫ5.8 − i ∧ i i=1 X at (x1, x2). To simplify notations, we omit t,ǫ5.7,ǫ5.8, then

n−1 ⌊ 2 ⌋ k k=0 ( 1) (π1) ωˆk ω1 = − ∗ , Im(C χ + √ 1χ)n MP 5.6 − whereω ˆk equals to R

n! (1) (2) n 2k 1 2k+1 ω (ω ) − − π∗χ (n 2k 1)!(2k + 1)! ∧ ∧ 2 − − n! (1,2) (2,1) (2) n 2k 2 2k+1 + ω ω (ω ) − − π∗χ (n 2k 2)!(2k + 1)! ∧ ∧ ∧ 2 − − n (1) (1) (1) √ 1ω ¯ dzi dz¯j 1 = − ij ∧ ( )(ω(2))n (2k + 1)! ∧ λ ...λ i,j=1 α1 α2k+1 X α1,...α2Xk+1distinct n (1,2) (1,2) (1) (1) √ 1ω ¯ ω ¯ dzi dz¯j 1 − il jl ∧ (ω(2))n. − (2k + 1)! ∧ λlλα1 ...λα2k+1 i,j,lX=1 α1,...α2kX+1,ldistinct Remark that

n−1 ⌊ 2 ⌋ ( 1)k 1 1 − ( ) (2k + 1)! λα1 ...λα2k+1 − λα1 ...λα2k+1 Xk=0 α1,...α2kX+1,ldistinct α1,...α2Xk+1distinct 1 1 1 1 1 = Re(1 + √ 1 )...(1 + √ 1 )(1 + √ 1 )...(1 + √ 1 ) −λl − λ1 − λl 1 − λl+1 − λn − 0, ≤ and

n−1 ⌊ 2 ⌋ ( 1)k 1 − (ω(2))n (2k + 1)! λα1 ...λα2k+1 Xk=0 α1,...α2Xk+1distinct 1 1 = Im(1 + √ 1 )...(1 + √ 1 )(ω(2))n − λ1 − λn (2) n = Im(ω + √ 1π∗χ) . − 2 By Lemma 5.5, n n (1) 1 (1,2) (1,2) (1) (1) Pπ∗χ(√ 1 ((ω ω ω )dz dz¯ )) 1 − i¯j − λ ik¯ jk¯ i ∧ j i,j=1 k X kX=1 ∗ (2) Pχ × (ω) Qπ χ(ω ) ≤ M M − 2 1 (2) <θ n Q ∗ ω . 0 + arctan( ) π2 χ( ) C5.6 −

38 So by the monotonicity and convexity of Pχ,

Pχ(ω1) 1 (2) (2) n (θ + n arctan( ) Q ∗ (ω ))Im(ω + √ 1π∗χ) x M 0 C5.6 π2 χ 2 < { 1}× − − Im(C χ + √ 1χ)n R M 5.6 − (Q ∗ (ω(2)))Im(ω(2) + √ 1π χ)n 1 R x M π2 χ 2∗ = θ + n arctan( ) { 1}× − . 0 n C5.6 − Im(C χ + √ 1χ) R M 5.6 − Using the convexity of x arctan x and theR fact that

(2) n (2) Im(ω + √ 1π2∗χ) Qπ∗χ(ω ) = arctan( − ), 2 Re(ω(2) + √ 1π χ)n − 2∗ it is easy to see that the minimum of

(2) (2) n Q ∗ ω ω √ π χ ( π2 χ( ))Im( + 1 2∗ ) x M − Z{ 1}× (2) 1 is achieved if Qπ∗χ(ω ) is a constant, which must be n arctan( ). Thus 2 C5.6 Pχ(ω1) <θ0. Back to our original notations, it means that Pχ(ω1,t,ǫ5.7,ǫ5.8 ) <θ0. ωn k The rest part of Section 3 still holds because t,ǫ−5.7,ǫ5.8 has no concentration of mass on the diagonal if k 1. Most part of Section 4 also holds because Γ¯χ is convex. We only need to prove≥ the following analogy of Lemma 4.3:

ǫ1.1 Lemma 5.6. Let C5.9 = cot( 6n ). Then for all small enough t and all q- dimensional subvarieties V of M˜ , as long as q

√ 1( ǫ1.1 θ ) 2 2 q Im e − 3 − 0 ((1 + C t)π∗ω + C t ω + √ 1(π∗χ + t ω )) 0. 5.9 0 5.9 4.13 − 4.13 ≤ ZV

Proof. We already know that

√ 1( ǫ1.1 θ ) q Im e − 3 − 0 ((1 + C t)π∗ω + √ 1(1 + C t)π∗χ) 0. 5.9 0 − 5.9 ≤ ZV So it suﬃces to show that

√ 1( ǫ1.1 θ ) 2 2 q Im e − 3 − 0 ((1 + C t)π∗ω + C t ω + √ 1(π∗χ + t ω )) 5.9 0 5.9 4.13 − 4.13 ZV ǫ √ 1( 1.1 θ ) q Im e − 3 − 0 ((1 + C t)π∗ω + √ 1(1 + C t)π∗χ) . ≤ 5.9 0 − 5.9 ZV As before, there exists a smooth function ϕ5.10 on a neighborhood O5.11 of π(E) in M such that ω = ω +√ 1∂∂ϕ¯ > 0 satisﬁes P (ω ) <θ ǫ1.1 5.10 0 − 5.10 χ 5.10 0 − 2 on O5.11. We deﬁne ϕ4.17 as before on M π(E). For any s > 0, by our assumption, there exists ϕ such that ω \ = (1+ s)ω + √ 1∂∂ϕ¯ > 0 5.12 5.12 0 − 5.12

39 π ω5.12 π satisﬁes Pχ(ω5.12) < θ0 < 4 . Choose s small enough such that Pχ( 1+s ) < 4 . Choose a large enough constant C5.13. Let ϕ5.14 be the regularized maximum ω5.12 1 of ϕ5.10 and 1+s + C5−.13ϕ4.17 + C5.13. Then it is easy to see that ϕ5.14 is smooth and P (ω ) < π on M for ω = ω + √ 1∂∂ϕ¯ > 0. Moreover, χ 5.14 4 5.14 0 − 5.14 there exists a smaller neighborhood O5.15 of π(E) such that ϕ5.14 = ϕ5.10 on O O . 5.15 ⊂ 5.11 After replacing ω0 by ω5.14, it suﬃces to show that

√ 1( ǫ1.1 θ ) 2 2 q Im(e − 3 − 0 ((1 + C t)π∗ω + C t ω + √ 1(π∗χ + t ω )) ) 5.9 5.14 5.9 4.13 − 4.13 √ 1( ǫ1.1 θ ) q Im(e − 3 − 0 ((1 + C t)π∗ω + √ 1(1 + C t)π∗χ) ). ≤ 5.9 5.14 − 5.9 First of all,

ǫ √ 1( 1.1 θ ) q Im(e − 3 − 0 ((1 + C t)π∗ω + √ 1(1 + C t)π∗χ) ) 5.9 5.14 − 5.9 √ 1( ǫ1.1 θ ) q Im(e − 3 − 0 ((1 + C t)π∗ω + √ 1π∗χ) ) − 5.9 5.14 − 1+C t 5.9 ǫ √ 1( 1.1 θ ) q 1 = Re(qe − 3 − 0 ((1 + C t)π∗ω + √ 1τπ∗χ) − π∗χ)dτ 5.9 5.14 − ∧ Z1 0 ≥ π using a calculation similar to Lemma 8.2 of [12] because Pχ(ω5.14) < 4 on M. 1 If the point is outside π− (O5.15), then as before, we get the required inequality 1 if t is small enough. If the point is inside π− (O5.15), then

ǫ √ 1( 1.1 θ ) 2 2 q Im(e − 3 − 0 ((1 + C t)π∗ω + C t ω + √ 1(π∗χ + t ω )) ) 5.9 5.14 5.9 4.13 − 4.13 √ 1( ǫ1.1 θ ) q Im(e − 3 − 0 ((1 + C t)π∗ω + √ 1π∗χ) ) − 5.9 5.14 − q 1 ǫ − √ 1( 1.1 θ ) q! k =Im(e − 3 − 0 ((1 + C t)π∗ω + √ 1π∗χ) k!(q k)! 5.9 5.14 − kX=0 − 2 2 q k (C t ω + √ 1t ω ) − ) ∧ 5.9 4.13 − 4.13 0 ≤ using a calculation similar to Lemma 8.2 of [12] because P (ω ) < θ ǫ1.1 χ 5.14 0 − 2 on O5.15. We are done.

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