Notes on Kähler-Ricci ow

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1 Notes on Kähler-Ricci ow immediate December 8, 2016 Contents 1 Kähler metrics on complex manifolds 1 2 Kähler-Ricci ow: maximal existence time 6 3 Finite time singularity C 0 -estimate Higher regularity of the limit Normalized Kähler-Ricci ow on Fano manifolds 16 5 Long time behavior KRF 24 1 Kähler metrics on complex manifolds Denition 1. Almost complex structure: J : T R M T R M saties J 2 = Id. Denition 2. complex manifolds) An holomorphic atlas: {U α, z α = {zα} i : U α C n )} such that the transition function z α z 1 β : U α U β C n is holomorphic. If M is a complex manifold, then there is a natural almost complex structure on T R M induced by the standard complex structure on C n = R n : J x i = y i, J y i = x i. 1

2 The complexied tangent bundle T C M = T C M C splitts into ±i-eigenspace of J: { T 1,0) M = span C z = 1 i 2 x i )}, T 0,1) M = T i y 1,0) M. i J is integrable in the sense that [T 1,0), T 1,0) ] T 1,0) M. Reversely we have the Newlander-Nirenberg theorem: Theorem 1 Newlander-Nirenberg). If J is a C 2n+α almost complex structure on M that is integrable, then there exists holomorphic atlas that makes M a complex manifold. Remark 1. This should be compared with Frobenius theorem for integrable distributions and can be seen as a generalization of the isothermal coordinate theorem in real dimensional 2 to higher dimensions. From now on, assume M, J) is a complex manifold. g is a Riemannian metric that is compatible with J in the sense that: gjx, JY ) = gx, Y ). In this case, we can dene a 2-form by: ωx, Y ) = gjx, Y ), In this case we say that g is an Hermitian metric and call ω g to be the Kähler form associated to g. Notice that gt 1,0) M, T 1,0) M) = 0. So locally under holomorphic chart U, {z i }), we can write: g = i,j g i j dz i d z j + d z j dz i). 1) where g i j forms a positive denite Hermitian matrix: g i j = g jī and g i jξ i ξj > 0 if ξ = ξ 1,..., ξ n ) 0. The associated Kähler form is a 1,1) form that has local expression as: ω = 1 g i jdz i d z j. 2) i,j Denition 3. A Riemannian metric g on a complex manifold M, J) is called a Kähler metric if g is an Hermitian metric and dω g = 0. Proposition 1. Assume g is an Hermitian metric on M, J). Then the following condition is equivalent: 2

3 1. LC J = 0, or equivalently LC ω g = dω g = 0, equivalently under local holomorphic coordinate chart we have g i j z k = g k j, i, j, k. 3) zi 3. At any point p M, there exists a holomorphic chart U, {z i }) centered at p i.e. z i p) = 0) such that g i j = δ ij + O z 2 ). Proof. Because d = e i LC e i, 1 implies 2 is clear. To see 2 implies 1, we rst derive the following basic consequence of dω g = 0 or equivalently of 3). Under holomorphic coordinate chart {z i }, the following components of the Levi-Civita connection form vanish: In other words, we have: and Γ i j = 0, or k; Γ k ij = 0 = Γ k ī j, i, j, k. z i z j = 0, z j zi = 0, i, j. 4) z i z j T 1,0) M, and z i z j T 0,1) M. From the above equality, we see that T 1,0) and T 0,1) are preserved under parallel transport. Now we can calculate to see that J = 0. i J) j ) = i J j ) J i j ) and i J) j) = 0. 3 implies 2 is clear. Assume 2 and write: = 1Γ k ij k J Γ k ij k ) = 0, ω g = 1g i jdz i z j with g i j = δ ij + a i jkz k + a i j k z k + O z 2 ). Because g i j = g jī, we have a i jk = a jī k. Consider the coordinate change z i = w i ci jk wj w k + O w 2 ). Then we have: ω g = 1 δ ij + a i jkw k + a i j k w k) dw i + c i pqw p dw q ) d w j + c j rs w r d w s ) = ) 1 dw i d w i + a i jk + c jki )wk dw i d w j + a i j k + c ikj ) wk dw i d w j. So we need to choose c j ki = a i jk, so that c i kj = a jīk = a i j k. On the other hand, we have a i jk = c j ki = cj ik = a k ji which is guarantted exactly because a i jk = k g i j z=0 = i g k j z=0 = a k ji. 3

4 The curvature tensor of a Kähler metric is easier to compute using the complex coordinates. zi zj = Γ k ij zk, with Γ k ij = g k l zj g i l. Similarly, Γ kī j = gl k zi g l j. zi zj = 0 = zi zj. So the curvature is use i = zi and j = zj ) R zi, zj ) zk = R l i jk z l, with R l i jk = z j Γ l ik = g l s jg r s )g r q i g k q ) g l q i jg k q. So we get: The Ricci curvature R i jk l = 2 g k l z i z j + g r q g k q z i g r l z j. R i j = R ji = R k k ji = j g k l ) i g k l = j i log detg k l). So we get the simple expression for the Ricci curvature of the Ricω) = 1 log ω n = 1 i,j 2 z i z j log detg k l)dz i d z j. We can give the Chern-Weil explanation of this formula. Any volume form Ω induces an Hermitian metric on K 1 M by z1 zn 2 Ω = 2 n Ω 1 n dz1 d z 1 dz n d z n. By abuse of notation, we denote the Chern curvature of the K 1 M, Ω) to be RicΩ) then we see that Ricω) = Ricω n ). So we see that the Ricci form is a closed 1,1) form representing the class 2πc 1 M). Sω φ ) = g i j φ Ricω φ) i j, S = n2πc 1M))[ω] n 1, [M] [ω] n, [M] are the scalar curvature of ω φ and average of scalar curvature. topological constant. Note that S is a Lemma 1 -Lemma). If [ω 1 ] = [ω 2 ], then there exists φ C M) such that ω 2 ω 1 = 1 φ. 4

5 Proof. ω 2 ω 1 = dη = η 1,0) + η 0,1). Consider the Hodge decomposition of η 1,0) with respect to the operator, we get η 1,0) = α + β + γ where γ Ker ) = Ker ). So η 1,0) = α + β. On the other hand, because 0 = ω 2 ω 1 ) = η 1,0) = β and =, we have 0 = β, β = β, β = β 2 L 2. So β = 0. The term η 0,1) is dealt with similarly. Exercise 1. Use the -Poincaré lemma and holomorphic Poincaré lemma to prove the local -lemma on polydisks. This Lemma is very useful because it reduces equations on Kähler metriccs to equations involving Kähler potentials. Denition 4. Fix a reference metric ω and dene the space of smooth Kähler potentials as H := H ω = {φ C M) ω φ := ω + 1 φ > 0} 5) Remark 2. The set H depends on reference Kähler metric ω. However in the following, we will omit writing down this dependence, because it's clear that H is also the set of Hermitian metrics h on L whose curvature form ω h := 1 log h is a positive 1,1) form on X. Since ω φ determines φ up to the addition of a constant, H/C is the space of smooth Kähler metric in the Kähler class [ω]. By abuse of language, sometimes we will not distinguish H and H/C. Calabi problem: Consider the Ricci curvature mapping: [ ω] 2πc 1 M) ω Ricω) Question: Sujectivity Yau's theorem on prescribing Ricci curvature) and injectivity by maximum principle). Transform to the PDE for the potential function: for any η 2πc 1 M) and ω = ω + 1 ϕ, we have: Ricω) η = Ricω) Ric ω) + Ric ω) η = 1 log ωn ω n + 1 f. 5

6 So the equation Ricω) = η is equivalent to the equation involving only the potential function ϕ: ω + 1 ϕ) n = e f c ω n. This is a complex Monge-Ampére type equation, since in local holomorphic coordinates, we have: ) det ĝ i j + 2 ϕ = e f c det g z i z i j). j Basic examples: Kähler manifolds with constant holomorphic sectional curvatures: R i jk l = µg i jg k l + g i lg k j), µ = 1, 0, 1. Notation: B n = {z C n ; z < 1}. P n = C n+1 {0})/C = C n P n 1. B n ω B n = 1 log1 z 2 ) B n /Γ; Γ < PSUn, 1) C n ω C n = 1 z 2 C n /Λ; Λ = Z 2n P n ω FS = 1 log1 + z 2 ) P n 2 Kähler-Ricci ow: maximal existence time g ω = Ricg) = Ricω). 6) t t The Ricci ow preserves the Kähler condition of the metric: dω = dω t t = d Ricω)) = 0. ω t = Ricω), ω0) = ω 0. 7) Evolution of cohomology class: d dt [ω t] = 2πc 1 X) = 2πc 1 K X ) [ω t ] = [ω 0 ] + 2πtc 1 K X ). From now on, we x the initial Kähler class α = [ω 0 ] H 1,1 M, R). The Kähler class is positive denitite if and only if t [0, T max ) where T max := T max α) = sup {t; α + t2πc 1 K M ) > 0}. 6

7 Theorem 2 Tian-Zhang). Fix a Kähler class α > 0 H 1,1 M, R). For any Kähler metric ω 0 α, the Kähler-Ricci ow exists if and only if t [0, T max ). Example 1. Let M = CP 2 CP 2 = Bl p CP 2. Let π : M CP 2 be the blow down map. Denote by H the pull back of the hyperplane class of CP 2. The Kähler cone is given by: KC = {ah be; a > 0, b > 0, a > b}. c 1 K X ) = 3H +E. Fix a Kähler class [ω 0 ] = ah be KC. So [ω t ] = a 3t)H b t)e. There are three cases: 1. b < 3a. T max = b. M CP b = 3a. T max = b = 3a. M pt. 3. b > 3a. T max = a b 2. M P1. Proof. Fix any T [0, T max ), there exists ɛ > 0 such that because [ω 0 ] + T + ɛ)2πc 1 K M ) > 0, there exists a Kähler metric ˆω T +ɛ [ω 0 ] + T + ɛ)2πc 1 K M ) so that ˆω T +ɛ ω 0 T + ɛ)2πc 1 K M ). By the -lemma, there exists a smooth volume form Ω = Ω T +ɛ such that ψ := RicΩ) = 1 log Ω = ˆω T +ɛ ω 0 )/T + ɛ). Dene the changing reference metric ˆω t = ω 0 + t RicΩ)) = 1 t T + ɛ ) ω 0 + t T + ɛ ˆω T +ɛ [ω 0 ] + t2πc 1 K M ). Then by the -lemma, we can write ω t = ˆω t + 1 ϕ with ϕ C M). Taking derivative with respect to t on both sides, we get: ω t t On the other hand, we have: = ˆω t t + ϕ 1 t = RicΩ) + ϕ 1 t. Ricω) = Ricω) + RicΩ) RicΩ) = 1 log ωn Ω RicΩ). So the Kähler-Ricci ow is equivalent to the ow: ϕ 1 t = 1 log ωn Ω 7

8 So up to a constant we get the following ow equation: ϕ t = log ˆω t + 1 n ϕ), ϕ0) = 0. 8) Ω Notice that this is equivalent to the Monge-Ampere type equation: ˆω t + 1 ϕ) n = e ϕ t Ω, ϕ0) = 0. We want to show that this ow has a solution for t [0, T + ɛ). There are several steps: C 0 -estimate of ϕ and ϕ/t) Assume min x X ϕx, t) = ϕx t ) for x t X. Then at x t we have: The last inequality is because: ˆω t = 1 t T + ɛ ϕ t x t) log ˆωn t Ω x t) n log ɛ + A. 9) ) ω 0 + t T + ɛ ˆω T +ɛ ɛ T + ɛ ω 0. By the minimum principle, we get min ϕ is non-decreasing so that For the upper bound, we get: u max t ϕx, t) n log ɛ + A)t. 10) log ˆωn t Ω max log ω 0 + ˆω T +ɛ ) n x X Ω = B. The last inequality is because ˆω t ω 0 + ˆω T +ɛ. This implies that: u u max Bt. for t [0, T ]. Next we consider the bound for u/t. Taking derivative of 8), we get: t ϕ = nˆω t + 1 ϕ) n 1 RicΩ) + 1 ϕ t ) = tr ωt RicΩ)) + t ϕ. Using the equality ˆω t = ω 0 tricω), it follows: ) t t t ϕ ϕ) = t t ϕ + t tr ωt RicΩ) + t ϕ) = t t ϕ ϕ) + tr ωt ω t ˆω t ) t tr ωt RicΩ) = t t ϕ ϕ) + n tr ωt ω 0. 8

9 We rewrite it as: t t ϕ ϕ nt) = tt ϕ ϕ nt) tr ωt ω 0 ). 11) Because tr ωt ω 0 > 0, by maximal principle we know that the maximum t ϕ u nt is nondecreasing, so we get: t ϕ t ϕ nt 0. By combining the local existence for small time and the uniform upper bound of ϕ, we get: ϕ t n + t 1 sup ϕ n + C. M To get lower bound of ϕ, we calculates: t t T + ɛ t) ϕ + ϕ) = T + ɛ t) t ϕ = T + ɛ t) tr ω t ψ + t ϕ) = t T + ɛ t) ϕ + ϕ) tr ωt ω t ˆω t ) T + ɛ t)tr ωt ψ = t T + ɛ t) ϕ + ϕ) n + tr ωt ω 0 T + ɛ)ψ). We rewrite it as: ) t t T + ɛ t) ϕ + ϕ + nt) = tr ωt ω 0 T + ɛ)ψ). 12) Because tr ωt ω 0 T + ɛ)ψ) = tr ωt ˆω T +ɛ > 0, so by the maximal principle, we get: T + ɛ t) ϕ ϕ + ϕ + nt T + ɛ) min t t=0 t > C. Exercise 2. ϕ t = Rt) where Rt) is the scalar curvature of ω t. So the bound of ϕ/t is equivalent to the bound on the scalar curvature of ω t. There is always a uniform lower bound of Rt) for general Ricci ow. So there is a uniform upper bound of ϕ = ϕ/t. We also have: ϕ = R + tr ωt ψ. 9

10 1. There is a uniform upper bound of ϕ depending only on T max. One can actually directly use Exercise 2 to get this. 2. Integrate twice the upper bound of ϕ to get upper bound of ϕ. ) 3. Let ˆω t = 1 t T max ω 0 + t T max ˆω Tmax. Then ˆω t is still smooth Kähler metric for t [0, T max ). Previous calculation shows that: ) t t T max t) ϕ + ϕ + nt) = t T max t) ϕ + ϕ) n + tr ω ˆω Tmax. So by the maximal principle, we get: T max t) ϕ t + ϕ + nt T ϕ max min t=0 t C. By the upper bound of u/t, we get uniform lower bound of ϕ. C 2 -estimate of u) Lemma 2. For any family of Kähler metrics ˆω t : t t )tr ωt ˆω t = tr ω tˆω) + ˆR i jk lg i j g k l g i j g k lĝ p q k ĝ i q lĝ p j. 13) Proof. Choose local coordinate {z i } so that k g i jx) = 0 for any i, j, k. Then we an calculate: g i jĝ ij ) = g i jĝ ) i j g k l = g i q k g p q g p jĝ ) i j + g i j k ĝ i j g k l k l l = g i q k lg p q )g p jĝ ) i j + g i j k lĝ i j g k l = g i q R p qk lg p jĝ i jg k l g i j ˆRi jk lg k l + g i j g k l k ĝ i q )ĝ p q lĝ p j) t g i jĝ i j) = g i q ġ p q g p jĝ i j + g i j ĝi j. Combining the above two identities, we get the identity. Recall that ĝ = g 0 tψ. So we get t t ) log tr ωt ˆω t = t t )tr ωt ˆω t + tr ω t ˆω t ) 2 tr ωt ˆω t tr ωt ˆω t ) 2 = tr ω t ψ) + ˆR i jk lg i j g k l g i jĝ i j We deal with each term on the right-hand-side: gi j g k lĝ i q,k ĝ p j, lĝ p q g i jĝ i j + gi jĝ i j,pg k lĝ k l, q g p q g i jĝ i j) 2 10

11 1. tr ωt ψ tr ωt C 1ˆω t ) = C 1 tr ωt ˆω t. 2. We can simultaneously diagonalize: ĝ i j = δ ij and g i j = µ i δ ij. ˆR i jk lg i j g k l g i jĝ i j = i,k ˆR iīk kµ 1 i µ 1 k i µ 1 i K k µ 1 k = K tr ωt ˆω t. 3. The last two terms combined is equal to: i,p µ 1 i µ 1 k T i pkt pī k i µ 1 i + µ 1 i T iīp µ 1 k i µ 1 i ) 2 T k k pµ 1 p The miracle is that this is non-positive: µ 1 p µ 1 i T iīp 2 = µ 1 p µ 1/2 i T iīp µ 1/2 i 2 p p Combining the above estimate, we get: p,j,l µ 1 p µ 1 p i i µ 1 i j µ 1 j T j jp 2 ) µ 1 j T j lp 2 i t t ) log tr ωt ˆω t C 1 + Ktr ωt ˆω t. Since t u = tr ωt ω t ˆω t ) = n tr ωt ˆω t, we get: µ 1 i t t ) logtr ωt ˆω t ) λϕ) C + K λ)tr ωt ˆω t = C + K λ) tr ωt ˆω t )e λϕ) e λϕ ). By choosing λ = K + 1 and ft) = max x X logtr ωt ˆω t ) λϕ), then f t) C 1 C 2 e f e f f t) C 1 e f C 2 So So d dt e C 1 t e f) C 2 e C 1t = e C 1t e ft) C 2 e C 1t 1)/C 1. ft) logc 1 /C 2 ) + C 1 t loge C 1t 1) = Ct). 11

12 So we get: logtr ωt ˆω t ) λϕ Ct) = tr ωt ˆω t Ct)e λϕ C t). Now because ω n t = e ϕ/t Ω, we get C ɛˆω t ω t C ɛˆω t. 3 Finite time singularity 3.1 C 0 -estimate Notice that the above discussion depends on choosing ˆω t for t [0, T + ɛ). Now that we know the maximal existence time, we can choose any smooth form ˆω Tmax = ω 0 tψ [ω 0 ] + T max 2πc 1 M) with ψ = RicΩ) without assuming its positivity. We can still dene: ˆω t = ω 0 tˆω Tmax. Again in general we don't have the positivity of ˆω t. However we still get the Monge- Ampére ow: t ϕ = log ˆω t + 1 n ϕ), ϕx, 0) = 0. Ω By exercise 2, one always has upper bound ϕ C for nite time singularity. By integrating we get uniform upper bound ϕ CT. If there exists a smooth semi-positive form ˆω Tmax [ω 0 ] + T max 2πc 1 K M ), then we can get uniform bound of u. To see this, we can choose ɛ = T max T. The same calculation in 12) shows that ) t t T max t) ϕ + ϕ + nt) = tr ωt ˆω Tmax ) 0. So by maximal principle, we get: So More generally, we have: T max t) ϕ + ϕ + nt T max min x M ϕx, 0) C. ϕ C nt T max t) ϕ C. 12

13 Denition 5. Let [α] be a nef 1,1) class. We say the closed positive current with minimal singularities in the class [α] has bounded potential, if the following condition is satised. There exists a constant C 0 > 0 such that for every ɛ > 0 there exists η ɛ C X, R) such that α + 1 η ɛ ɛω 0 and sup X η ɛ C 0. Theorem 3. If the closed positive current with minimal singularities in the class [ω Tmax ] has bounded potential, then the KRF solution has uniform lower bound ux, t) C. Proof. We compute as in 11) and 12) to get: ) t t T max t) ϕ + ϕ + nt) ɛt ϕ ϕ nt) η ɛ ) = tr ωt ω0 T max ψ + ɛω η ɛ ) 0. So by minimum principle, we get: T max t ɛt) ϕ ɛ)ϕ η ɛ C C. Because ϕ C we get ϕ C on X [0, T ). Exercise 3. Conversely if u C on X [0, T max ), then the positive closed current with minimal singularities in [ω Tmax ] has bounded potential. Question: Under what condition does [ω Tmax ] satisfy the condition: positive closed current with bounded potential? This question is open in general. Theorem 4. If X is projective and [ω 0 ] = 2πc 1 L), then [ω Tmax ] contains a smooth semi-positive form and hence satisfy the condition. This follows from Kawamata's base-point-free theorem. Theorem 5. Assume L is nef and L K X is nef and big. For m 1, ml is base point free. 3.2 Higher regularity of the limit Proposition 2. The following set is equivalent: {x X U x open, C > 0, s.t. Rmt) ωt) C on U [0, T )} = {x X U x open, C > 0, s.t. Rt) C on U [0, T )} = {x X U x open, ω U Kähler metric on U, s.t. ω t ω U in C U) as t T }. 13

14 Proof. Set 3 Set 1 Set 2. So just need to show Set 2 Set 3. On the open set U, ϕ = R C = ϕ C = u CT. t Letting ˆω t ω 0 in 13), we get ) t t log tr ωt ω 0 Ctr ωt ω 0. So we get ) t t logtr ωt ω 0 ) + Ct ϕ ϕ nt)) Ctr ωt ω 0 Ctr ωt ω ) By maximal principle, tr ωt ω 0 e Ct ϕ ϕ nt) C on U. The higher order estimate can be derived using parabolic version of Krylov-Evan's theory. So we get ω t C k C for any k 1. Denote by S any of the above sets and call Σ := X \ S the singularity formation set. Corollary 1. For every nite time singularity of the KRF the singularity formation set Σ is nonempty and we have lim sup t T Rt) = +. For a nef class [α] H 1,1 X, R), dene Nullα) = { Z Z subvariety of X, Z Z } α dimz) = 0. Theorem 6 Collins-Tosatti). Let [α] = [ω Tmax ]. Then Σ = Nullα). Proof. Consider w = T max + δ t) ϕ + ϕ nt φ where φ is from the next Theorem 7. Then as in 12), we get: ) t t w = tr ωt ω 0 T max + δ)ψ + 1 φ) So by minimum principle, min w is increasing. So = tr ωt ω 0 T max ψ + 1 φ) δψ) 0. T max + δ t) ϕ + ϕ nt φ C. 14

15 So we get lower estimate of ϕ: 1 ϕ T max + δ t u + nt + φ C) 1 δ φ C δ. 15) The partial C 2 -estimate can be done as in 14). Higher order estimate follows from Krylov-Evans. The above proof depends on the following construction of barrier functions. Theorem 7. Let [α] be a closed real 1,1) calss such that α is nef and X αn > 0. Then there exists an upper semi-continuous L 1 function φ : X R { }, which equals on Nullα), which is nite and smooth on X \ Nullα) and such that on X \ Nullα) for some ɛ > 0. α + 1 φ ɛω 0. Theorem 7 generalizes following theorem is known due to Kodaira's lemma. Theorem 8. Let D be a nef and big divisor on a projective manifold X. Then thre is an eective R-divisor E such that D E is ample. We can choose e φ = e ϕ = e ϕ e φ 1 where e ϕ is a positively curved metric on D E σ 2 σ 2 h and 1 is a singular metric on E where E = {σ = 0}. σ 2 1 φ = 1 ϕ + φ1 ) + log σ 2 h 1 ϕ ɛω 0. If we dene augmented base locus no-ample locus): B + D) = E = m N {E; E eective and D E ample } BmD A) = m N Bs kmd E). k N Then actually B + D) = NullD) Nakamaye's theorem). Theorem 9. If [ω 0 ] = 2πc 1 L) > 0. Then ω t converges smoothly to a Kähler metric ω Tmax on X \ B + D) where D = L + T max K X. Remark 3. There is an analytic characterization of the augmented base locus. One can dene the non-kähler locus of a 1,1)-class [α] as: E nk α) = SingT ) T [α] where the intersection ranges over all Kähler currents T = α + 1 ϕ in the calss [α]. Then Boucksom showed that B + D) = E nk D), and moreover there is a Kähler current with analytic singularities along E nk D). 15

16 4 Normalized Kähler-Ricci ow on Fano manifolds A special nite time singularity happens when X is a Fano manifold and [ω 0 ] = 2πc 1 X) > 0. In this case, the maximal existence time T max = 1 and the X converges in the Gromov-Hausdor sense to a point as t 1. To get more information, we consider the normalized Kähler-Ricci ow. Assume ω t is a solution to the unnormalized KR ow. Dene ω = ω/1 t) so that Vol ω t ) is xed. Then Notice that Ricω t ) = Ric ω t ). normalized Kähler-Ricci ow t ω = ω 1 t) = ω t t 1 t 1 1 t ω Ricω t)). If we dene t = log1 t), then we get the t ω = Ric ω) + ω, ω0) = ω 0. 16) This ow exists for any time t [0, ). Let ω t be the solution to the normalized KRF 16) and g t be the associated Kähler metric. If Rm gt)) < and the diameter of gt) is uniformly bounded, then g t converges in the Cheeger-Gromov sense to a metric g on X Proposition 3. If Rm gt)) and diam gt)) is uniformly bounded, then the Cheeger- Gromov limit g is a Kähler-Ricci soliton. Proof. 1. Recall Perelman's W-entropy and µ-energy: W g, f, τ) = τr + f 2 ) + f n ) 4πτ) n/2 e f dv g. Under the evolution equation: We have X s g = 2Ricg), s f = f + f 2 R + n 2τ, sτ = 1. d W gs), fs), τs)) = ds The µ-energy: µg, τ) = inf X 2τ R ij + f ij 1 2τ g ij 2 4πτ) n/2 e f dv g. { } W g, f, τ); 4πτ) n/2 e f dv g = 1. 17) X 16

17 Then µgs), τs)) is increasing. It's strictly increasing unless gs) is a Kähler- Ricci soliton. Notice that we have the rescaling invariance: W g, f, τ) = W τ 1 g, f, 1) and µg, τ) = µτ 1 g, 1). To adapt to the our setting, we let t = 2s, τs) = 1 s = t), gt) = 1 2 t) 1 g = 2τ) 1 g. Then d dt W gt), f, 1 2 ) = 1 d W gs), fs), τs)) 2 ds = 1 2s 2 R 2 ij + f ij g ij 2π) n/2 e f dv g So if t = log1 t) = log1 2s), then we get: d d t W g t), f t), 1 ) = 2 R ij + f ij g ij 2π) n/2 e f dv g. under the evolution equation: X X g t = Ric g) + g, f t = f + f 2 g R + n. 2. For simplicity, we will deal with the normalized functional and remove the tilde symbol. Let f ti +A be the minimizer of W gt i + A), f, 1/2). Then 0 W gt i + A), f ti +A, 1/2) W gt i ), f ti +At i ), 1/2) A R j k + j k f ti +At i + s) g j k 2 dv ti +sds + 0 X A j f ti +At i + s) 2 + j k f ti +At i + s) 2) dv ti +sds 0 X µgt i + A), 1/2) µgt i ), 1/2). µg ti, 1/2) µg, 1/2) is uniformly bounded. So one can show that as i +, f ti +As) converges to f s) that is a weak solution to the equation: R j kg s)) + f s)) j k g ) j k = 0, f s)) jk = f s)) j k = 0. By elliptic regularity, f C X) so g is a Kähler-Ricci soliton. 17

18 The assumption on the Riemannian curvature and diameter are used to make sure there is a convergence sequence in the Cheeger-Gromov sense. It turns out that the diameter is always bounded by Perelman's work. Theorem 10 Perelman). Let gt) be a normalized Kähler-Ricci ow on a Fano manifold X. There exists a uniform constant C so that Rgt)) C. diamx, gt)) < Get a uniform lower bound of the Ricci potential u). Let ωt) = ω φ and Ricωt)) ωt) = 1 ϕ. Then the normalized KRF is equivalent to: φt) = ut), t φ0) = 0. 18) Notice that we can write: Ricωt)) ωt) = Ricωt)) Ricω 0 ) + Ricω 0 ) ω 0 + ω 0 ωt) So the NKRF is equivalent to = 1 log ωt)n ω n 0 1 ϕ 0 1 φ. ωt)n φt) = log + φt) + u0) + c t ω0 n t, φ0) = 0. 19) Since u is dened up to a constant, we can normalize ut) so that: e u ωt) n = 2π) n. 20) X Notice that n R = u with n = 2m. W gt), u, 1 ) [ = R + u 2 ) + u m ] 2π) m/2 e u ωt) n 2 X [ = n u + u 2 ) + u 2n ] 2π) n e u ω n X = 2π) n ue u ω n n. X Because µg, 1/2) = inf{w g, f, 1/2); X 2π) n e f ωt) n } is increasing, we get W gt), u, 1/2) C. In other words, we have: 18

19 Lemma 3. There is a uniform constant C 1 = C 1 µg0), 1/2)) > 0 so that X ue u ω n C 1. Proof. Write u = u + + u with u + = max{0, u} and u = min{0, u}. The lemma follows from the fact that xe x is a bounded function for x 0. Lemma 4. The scalar curvature Rt) is uniformly bounded from below along the ow. Proof. Recall that along the un-normalized ow, we have: s ˆRs) = ˆ real ˆRs) + 2 Ricĝs)) 2. 21) Then by minimum principle, ˆRs) is uniformly bounded from below. For normalized ow, gs) = 1 2s) 1 ĝs). Rgs)) = 1 2s)Rĝs)) is also uniformly bounded from below. Rgs)) satises: s Rs) = ) 2Rĝs)) + 1 2s) ˆ ˆRs) + 2 Ricĝs)) 2ĝs) = 21 2s) 1 Rgt)) + 1 2s) 1 Rs) + 2 Ricgs)) 2 ). This implies for t = log1 2s), we have: t Rt) = 1 2 realrt) Rt) + Ricgt)) 2 real 22) Notice that real f = 2 f = g i j f i j + f ji). The Ricci potential ut) = φt) satises: t 1 ϕ = 1 ϕ = 1 t t t φ = 1 tr ωt) Ricωt)) + ωt)) + 1 φ = 1 u + u). Dene a = 2π) n X ue u dv. This implies u = u + u + a 23) t where a = X ue u 2π) n C by the previous lemma. 19

20 Lemma 5. The function ut) is uniformly bounded from below. Proof. Prove by contradiction. If the Ricci potential ut) is very negative for some time t 0 say ut) 2n + C 1 ). By C 1 very big, then t u = n R + u + a n + C 1) R < 0. 24) This implies ut) stays very negative for t t 0. If at t 0, uy 0 ) 0, then uy) 0 for all y U a neighborhood of y 0. So all t t 0, uy) 0 for y U. Then u C + u implies t Then φ = u yields ut)y) e t t 0 C + ut 0 )) Ce t for t t 0. 25) Because ωt) = ω φ, we have φt)y) φt 0 )y) Ce t t 0 C 1 e t. 26) 0 φt) = tr ω0 ωt) + n n. Assume φx t, t) = max M φy, t) and G 0 is the Green's function associated to the metric g0). By Green's formula, we have: 1 1 φx t, t) = φy, t)dv 0 0 φy, t)g 0 x t, y)dv 0 V ol 0 M) M V ol 0 M) M V ol 0M \ U) sup φ, t) + V ol 0U) φy, t)dv 0 + C V ol 0 M) M V ol 0 M) V ol 0M \ U) V ol 0 M) sup φ, t) Ce t + C. M for t t 0. Since V ol 0 M \ U)/V ol 0 M) < 1, we get: U max M φ, t) Cet + C. 27) On the other hand we can estimate φ, t) from below. Let uz t, t) = max M ut). Because M e u dv t = 2π) n, uz t, t) C. We also have: d ut) φt)) = u + a = n R + a. 28) dt 20

21 Because R is uniformly bounded from below by Lemma 4, we get: This implies uz t, t) φz t, t) maxu, 0) φ, 0)) + Ct. M This contradicts 27). max M φt) φz t, t) uz t, t) Ct C Ct. 2. Bound u and Rt) = n u by C 0 -norm of Ricci potential ut). This can be achieved by considering evolution equation for u 2 u and, where B is u+2b u+2b a uniform constant such that u + B > 0. Proposition 4. There is a uniform constant C so that: u 2 Cu + C), R Cu + C). 29) Proof. We rst calculate: ) u 2 = t u 2 = u i ju īj u ij u ī j + u i u ī = u 2 u 2 + u 2. Let H = u 2, then we can calculate: u+2b Write: H = u 2 u 2 u + 2B + u 2 2B a) u + 2B) 2 + u 2 u + u 2 u u + 2B) 2 2 u 4 u + 2B) 3. u 2 u + u 2 u 2 u 4 u + 2B) 2 u + 2B) 3 = 1 ɛ) u H + u u u + 2B We estimate: + ɛ [ u 2 u + u 2 u u + 2B) 2 2 u 4 u + 2B) 3 u ī u j u j) i = u ī u ji u j + u ī u j u ji u 2 u 2 + u 2 ). 21 ].

22 So ɛ u u 2 ɛ u 2 u + u ) u + 2B) 2 u + 2B) 3/2 u + 2B) 1/2 Choose ɛ suciently small, we get: = ɛ u 4 2 u + 2B) + + u 2 3 4ɛ u 2 u + 2B H u 2 2B a) + 1 ɛ) u H + u H u + 2B) 2 u + 2B By maximal principle, we get: d dt H max u 2 u + 2B) 2 2B a ɛ u 2 2 u + 2B ɛ u 4 2 u + 2B). 3 ). 30) Corollary 2. This is a uniform constant C such that uy, t) Cdist 2 t x, y)+c, Ry, t) Cdist 2 t x, y)+c, u Cdist 2 t x, y)+c. 31) Proof. By Lemma 5, we can assume u δ > 0. From 29), it follows that u is Lipschitz function. Therefore uy, t) uz, t) Cdist t y, z). Choose z such that uz, t) = min M u, t). Then uz, t) K for a constant that does not depend on t. So we get uy, t) Cdist 2 t y, z) + C. The other two estimates follow from 29). 3. bound the diameter) Lemma 6. For every ɛ > 0, we can nd B2 k 1, 2 k 2 ), such that if diamm, gt)) is large enough, then V olb2 k 1, 2 k 2 )) < ɛ and V olb2 k 1, 2 k 2 )) 2 10n V olbk 1 + 2, k 2 2)). 32) 22

23 Proof. The rst estimate is easy. To get the second estimate, we rst need to show that: V olb2 k, 2 k+1 )) 2 2k 2 kn C. 33) To see this, rst we have R C2 2k on B2 k, 2 k+1 ) due to 31). Then 33) follows from the local noncollapsing property along the Ricci ow: V olbx i, 2 k ) 2 2kn C for any Bx i, 2 k ) B2 k 1, 2 k 2 ) \ B2 k 1+2, 2 k 2 2 ). Lemma 7. There exists r 1, r 2 and a uniform constant C such that 2 k 1 r 1 2 k1+1, 2 k 2 r 2 2 k2+1 such that R C V olb2 k 1, 2 k 2 )). Br 1,r 2 ) Proof. By mean value theorem, we know that there exists r 1 [2 k 1, 2 k 1+1 ] and r 2 [2 k 2 1, 2 k 2 ] such that V olsr 1 )) V olb2k 1, 2 k 2 )) 2 k 1+1, V olsr 2 )) V olb2k1, 2 k2 )) 2 k 2+1. So we can estimate V = V ol2 k 1, 2 k 2 )): R = R n) + nv olbr 1, r 2 )) Br 1,r 2 ) Br 1,r 2 ) = u + nv olbr 1, r 2 )) Br 1,r 2 ) u + u + nv olbr 1, r 2 )) Sr 1 ) Sr 2 ) V 2 k 1 2k V 2 k 2 C2k 2+1 = CV Proposition 5. There is a uniform constant C such that diamm, gt)) C. Proof. Proof by contradiction. Assume diamm, gt)) is not uniformly bounded in t. Then there exists a sequence t i + such that diamm, gt i )) +. Let ɛ i 0. By Lemma 6, there exist sequences k i 1 and k i 2 such that V ol ti B ti 2 ki 1, 2 k i 2 )) < ɛi, V olb ti 2 ki 1, 2 k i 2 )) 2 10n V olb ti 2 ki 1 +2, 2 ki 1 2 )). 34) 23

24 By Lemma 7, there exists r i 1 [2 ki 1, 2 k i 1 +1 ] and r i 2 [2 ki 2 1, 2 ki 2 ] such that B ti r i 1,ri 2 ) R CV olbk i 1, k i 2)). Let η = η i be a sequence of cut-o functions such that ηz) = 1 for z [2 ki 1 +2, 2 ki 2 2 ] and equal to 0 for z, r1] [r i 2, i ). Let w i x) = e C i η i dist ti x, p i )) such that 2π) n M w2 i = 1. Let w 2 = 2π) n e f so that f = 2 log w n log 2π. We estimate each term of W g, f, 1/2) = R + f 2 + f 2n)2π) n e f dv g M = Rw w 2 w 2 log w 2 )dv g C. a) b) M M M R ti wi 2 e 2C i r e 2C i V olb2 k 1, 2 k 2 )) Br 1,r 2 ) 2 10n V olb2 k1+2, 2 k2 2 )) u 2 dv = 1. w 2 dv g = M e 2C η 2 dv g = e 2C V olb2 k 1, 2 k 2 )) 2 10n e 2C V olb2 k1+2, 2 k2 2 )) u 2 dv g = C. M M 5 Long time behavior KRF Example 2. Consider the product M = Σ 1 Σ 2 where Σ 1 is a torus, and Σ 2 is a Riemann surface of genus 2. Assume [ω 0 ] = η 1 + η 2 = a, b) where a = [η 1 ], [Σ 1 ] > 0 and b = [η 2 ], [Σ 2 ] > 0. Because 2πc 1 M) = 2πc 1 Σ 2 ) = 2π2 2g). So [ω t ] = [η 1 ] + [η 2 ] t2πc 1 Σ 2 ) = a, b 2πt2 2g)). So T max = and the volume V olm, ω t ) = a + b + 2πt2g 2) blows up. If we consider the normalized metric ω t = ω t /t = a/t, b/t 2π2 2g)) 0, 2π2g 2)). One can show that M, g t ) Σ 2, g ) in the Gromov-Hausdor sense, where g is the hyperbolic metric on Σ 2. 24

25 Let ˆω s, s [0, ) be a solution to the Kähler-Ricci ow. Then we know that K M is nef, i.e. c 1 K M ) C M. Denote ω s = ˆω s /1 + s). Then ω t satises the equation: Let t = log1 + s), we get: The Kähler class evolves like: ω s = sˆω s 1 + s ˆω s 1 + s) 2 = s Ricω s) ω s ). t ωt) = Ricωt)) ωt), ω0) = ω 0. 35) [ωt)] = [ω 0] 2πsc 1 M) 1 + s = e t [ω 0 ] 1 e t )2πc 1 M). So the Kähler class converges to the rst Chern class of the canonical line bundle: [ωt)] 2πc 1 M) = 2πc 1 K M ) as t c 1 M) = 0. Then ˆω s converges to a Ricci at metric on M. 2. c 1 M) is big: c M 1M) n > 0. Then ωt) converges to a singular Kähler- Einstein metric on M which is smooth on M \ NullK M ). 3. c 1 M) is not big. M c 1M) n = 0. Then conjecturally M, ωt)) converges to a lower dimensional space N of dimension given by: dim C N = max{k; c 1 M) k 0} =: νm). This is essentially the Abundance Conjecture: Conjecture 1. a) κm) = νm) where { } dim C H 0 M, mk M ) κm) = max k; lim sup > 0. m + m k b) K M is base-point-free. In other words, for l 1, lk M gives rise to a morphism M N with dim C N = κm). Example 3. The above example is a trivial elliptic bration. In general if we have an elliptic bration π : M Σ. Assume on Σ = Σ \ {p i }, the bration is smooth 25

26 and m i is the multiplicity of the possibly singular) bre π 1 p i ). Then we have the canonical bundle formula: ) k K M = π K Σ + f 1 O M ) )p i 36) m i Assume K M is nef. Then κm) = 1 if and only if ) k δπ) = χo X ) + 2g m 1 i ) > 0. In this case, one can show that there exists a Kähler metric ω on Σ satisfying the equation: k Ricω ) = ω + j ω W P + 2π 1 m 1 i )δ pi, where j : Σ M is the induced map from the Σ to the moduli space of Riemann surfaces of genus 1. Theorem 11. As t +, ωt) converges to π ω as currents. Furthermore, on π 1 Σ ), ωt) converges to π ω uniformly. In higher dimensions, assuming the Abundance Conjecture, if 0 < κm) < n, then there is a bration M N such that the generic bre is a Calabi-Yau manifold. We have a similar canonical bundle formula. One can dene a generalized Kähler- Einstein metric on the base N: i=1 i=1 i=1 Ricω ) = ω + ω W P on N. 37) To dene ω W P, we use the work by Tian-Todorov on smoothness of the deformation space of Calabi-Yau manifolds. ω W P = 1 log Ψ Ψ. M t where t are coordinates on N and Ψ is a holomorphic section of H 0 U, K M/N ). Choose a metric χ 2πc 1 M). Then 37) can be reduced a Monge-Ampére equation: χ + 1 φ) κ = F e φ χ κ. 38) 26

27 where F is dened in 40). On the other hand, the NKRF in 35) can be reduced to the equation: t ϕ = log en κ)t ˆω + 1 n ϕ) ϕ, ϕ0) = 0. 39) Ω where ˆωt) = e t ω 0 1 e t )χ and Ω is a volume form on M satisfying RicΩ) = χ. So the claim is that the solution to 39) converges to the solution to 38) as t +. We will only give a heuristic reason why this is true. First we can choose ω 0 such that ω 0 Mt Then is Ricci at and e n κ)t ˆω + 1 ϕ) n Ω π F = So the proof is reduced to show that: e n κ)t e t ω 0 1 e t )χ + 1 ϕ) n Ω ω n κ 0 χ κ. 40) = en κ)t ˆω + 1 ϕ) n n κ ω0 χ κ ω0 n κ χ κ Ω = en κ)t ˆω + 1 ϕ) n ω n κ 0 χ κ π F ) 1. ω n κ 0 χ κ χ + 1 φ) κ χ κ. 27