Yau's Proof of the Calabi Conjecture

Yau’s proof of the Calabi Conjecture

Mathew George

1 Introduction

Calabi conjecture can be stated as follows.

α β α β Let M be a compact Kahler¨ manifold with Kahler¨ metric gαβdz ⊗ dz¯ . Let R˜ αβdz ⊗ dz¯ be a tensor whose i ˜ α β associated (1, 1)-form 2π Rαβdz ∧ dz¯ represents the ﬁrst Chern class of M. Then we can ﬁnd a Kahler¨ metric α β α β g˜αβdz ⊗ dz¯ which is cohomologous to the original metric and whose Ricci tensor is given by R˜ αβdz ⊗ dz¯ .

This was proposed by Eugenio Calabi in 1954 and a proof was published in 1978 by S.T. Yau. This accomplishment by Yau lead him to win the Fields medal later on. One direct consequence of this theorem is the existence of Ricci ﬂat Kahler¨ manifolds, now also called as Calabi-Yau manifolds. These special surfaces ﬁnd applications in String theory as well. In these notes we go through Yau’s proof in detail.

2 Proof of the Calabi conjecture

The proof can be divided into four steps:

1. Reformulating the statement in terms of a PDE called the complex Monge-Ampere equation.

2. Finding apriori estimates for all derivatives of the solution upto the second order.

3. Finding apriori estimates for the third order derivatives of the solution.

4. Using the method of continuity to solve the PDE.

1 The complex Monge-Ampere equation can be written as

∂2φ det g + = det(g ) exp(F) ij¯ ∂zi∂z¯j ij¯ k where the function F ∈ C (M) with k > 3 is given and satisﬁes

exp(F) = Vol(M) ZM It can be seen by integrating both sides of the previous equation that this condition on F is necessary. We will be showing that this indeed is sufﬁcient for the existence of a solution φ. Since the proof as a whole can be rather disorienting to read all at once, let’s break down the main ideas of each section.

1. Reformulating the statement in terms of a PDE

Observe that the first Chern class of M, by definition, represents the real cohomology class i containing times the Ricci form. So the conjecture now implies that the given tensor differs 2π from the Ricci tensor of the given metric by an exact form (say ∂∂F¯ ). For Kahler¨ manifolds there ∂2 is a simple identity for Ricci tensor given by − log(det(g )). Since the Kahler¨ metrics ∂zi∂z¯j ij¯ mentioned in the theorem are cohomologous to each other, we can relate them by an exact form ∂∂φ¯ , for some scalar-valued function φ. Note that any exact form can be globally written in this form as a consequence of the ∂∂¯-lemma. So if we were to have the above tensor R˜ as the Ricci tensor of a the new Kahler¨ metric (say g0), then it is clear that the following has to be satisfied.

∂2 ∂2φ log(det g + (det(g ))−1 exp(−F)) = 0 ∂zi∂z¯j ij¯ ∂zi∂z¯j ij¯

This implies that

∂2φ log(det g + (det(g ))−1 exp(−F)) = C ij¯ ∂zi∂z¯j ij¯

by maximum principle on a compact manifold. Now the above condition M exp(F) = Vol(M) would imply that C = 0, and the resulting expression is the complex Monge-Ampere equation. R We can now see that if there is a function φ satisfying this equation, then the metric g0 deﬁned ∂2φ as g + has all the required properties. ij¯ ∂zi∂z¯j

The difficult part of the proof is finding apriori estimates for the derivatives. So we focus on that first.

2.1 Estimates upto the second order We will be studying the equation

∂2φ det g + (det(g ))−1 = exp(F) (1) ij¯ ∂zi∂z¯j ij¯ 2 3 ∂ φ where F is assumed to be C (M). We want solutions φ of (1) such that the matrix gij¯ + i j is ∂z ∂z¯ ij positive deﬁnite Hermitian. Then this would deﬁne another Kahler¨ metric on M. Denote

∂2φ g0 = g + ij¯ ij¯ ∂zi∂z¯j

2 We assume that φ ∈ C5(M). The aim of this section is to obtain estimates on all derivatives of φ upto the second order. To guarantee that the solutions are unique, we impose the normalization

φ = 0 (2) ZM Differentiating (1) wrt zk we have

3 ∂gij¯ ∂ φ ∂gij¯ ∂F g0ij + − gij¯ = (3) ∂zk ∂zi∂z¯j∂zk ∂zk ∂zk This differentiation is done by ﬁrst taking log on both sides and then using the formula for the derivative of log of determinant of a matrix ?. Differentiating (3) again wrtz ¯l

2 3 3 2 4 ! ∂ F ∂g ¯ ∂ φ ∂g ¯ ∂ φ ∂ g ¯ ∂ φ = − g0tj¯g0in¯ tj + ij + + g0ij¯ ij + ∂zk∂z¯l ∂z¯l ∂zt∂z¯j∂z¯l ∂zk ∂zi∂z¯n∂zk ∂zk∂z¯l ∂zi∂z¯j∂zk∂z¯l (4) 2 ∂g ¯ ∂g ¯ ∂ g ¯ + gtj¯gij¯ tj ij − gij¯ ij ∂z¯l ∂zk ∂zk∂z¯l

0ij¯ 0 ∂g ∂gtj¯ Here we used the formula = −g0tj¯g0ij¯ for the derivative of the inverse of a matrix. ∂z¯l ∂z¯l Let ∆0 be the Laplacian associated with the metric g0. Then

∂2 ∂2φ ∆0(∆φ) = g0kl¯ gij¯ ∂zk∂z¯l ∂zi∂z¯j ∂4φ ∂2gij¯ ∂2φ = g0kl¯gij¯ + g0kl¯ (5) ∂zi∂z¯j∂zk∂z¯l ∂zk∂z¯l ∂zi∂z¯j ij¯ 3 3 ∂g ∂ φ ∂g ¯ ∂ φ + g0kl¯ + g0kl¯ ij ∂zk ∂zi∂z¯j∂z¯l ∂z¯l ∂zi∂z¯j∂zk Since LHS is deﬁned independent of the co-ordinate system, we compute the RHS by picking normal ∂gij¯ ∂gij¯ co-ordinates at a point so that g = δ , = = 0. ij¯ ij ∂zk ∂z¯l Multiplying (4) by gkl¯ and adding over k, l gives

2 2 ∂ g ¯j ∂ gij¯ ∆F + g0tj¯g0in¯ gkl¯φ φ = gkl¯gij¯φ + gkl¯g0ij¯ i − gij¯gkl¯ (6) tn¯ l¯ ijk¯ ijk¯ l¯ ∂zk∂z¯l ∂zk∂z¯l Plugging this in (5) gives

2 ij¯ 2 2 2 ∂ g ∂ φ ∂ g ¯ ∂ g ¯ ∆0(∆φ) = ∆F + g0kj¯g0in¯ φ φ + g0kl¯ + gkl¯g0ij¯ ij − gij¯gkl¯ ij kn¯ l¯ ijl¯ ∂zk∂z¯l ∂zi∂z¯j ∂zk∂z¯l ∂zk∂z¯l 0kj¯ 0in¯ 0kl¯ 0ij¯ = ∆F + g g φkn¯ l¯φijl¯ + g Rijk¯ l¯φij¯ + g Rijl¯ l¯ − Riil¯ l¯ (7) i j l i l X, X X, ∂2g ij¯ ij¯ In the last step we used the fact that g = δij and Rijk¯ l¯ = . ∂zk∂z¯l

Now choose a co-ordinate system so that in addition to gij¯ = δij, we also have φij¯ = δijφii¯. This is possible because these are symmetric matrices and hence are simultaneously diagonalizable. Then we have

0ij¯ 1 g = δij (8) (1 + φii¯) and

3 0ij¯ 0kl¯ g Rijl¯ l¯ − Riil¯ l¯ + g Rijk¯ l¯φij¯ l i l i j X X, X, φ φ = − R ii¯ + R ii¯ iil¯ l¯ 1 + φ iil¯ l¯ 1 + φ i l ii¯ i l ll¯ X, X, φ (φ − φ ) = − R ii¯ ll¯ ii¯ (9) iil¯ l¯ (1 + φ )(1 + φ ) i l ii¯ ll¯ X, Now using the symmetry of this expression wrt i and l,

  1 φ (φ − φ ) φ (φ − φ ) = − R ii¯ ll¯ ii¯ − R ll¯ ii¯ ll¯ 2  iil¯ l¯ (1 + φ )(1 + φ ) iil¯ l¯ (1 + φ )(1 + φ ) i l ii¯ ll¯ i l ii¯ ll¯ X, X, 1 (φ − φ )2 = R ll¯ ii¯ 2 iil¯ l¯ (1 + φ )(1 + φ ) i l ii¯ ll¯ X, From (9) and (7) we see that   0 0kj¯ 0in¯ 1 + φii¯ 2 ∆ (∆φ) > ∆F + g g φknl¯φijl¯ + (inf Riil¯ l¯)  − m  (10) ¯ i6=l 1 + φ l i l ll¯ X X, This is true because 2 φ (φ − φ ) 1 + φ + φ ¯ + φ ¯ − ii¯ ll¯ ii¯ = ii¯ ll ii − 1 (1 + φii¯)(1 + φll¯) (1 + φii¯)(1 + φll¯) When is taken, we get terms of the form i,l P 2 2 φii¯ + φll¯ + 1 + 1 + φll¯ + φii¯ + φii¯ + φll¯ (1 + φii¯)(1 + φll¯) Then we rearrange and factor this as needed. On the other hand

φ ∆0φ = ii¯ 1 + φ i ii¯ X 1 = m − (11) 1 + φ i ii¯ X Let C be a positive constant. Then we have

0 2 0ij¯ 0ij¯ ∆ (exp(−Cφ)(m + ∆φ)) = C exp(−Cφ)(g φiφj¯)(m + ∆φ) − C exp(−Cφ)g φij¯(m + ∆φ) 0ij¯ − C exp(−Cφ)g φj¯(∆φ)i (12) 0ij¯ − C exp(−Cφ)g φi(∆φ)j¯ (13) + exp(−Cφ)∆0(∆φ) q q 0ij¯ 0ij¯ 0ij¯ Now using g φi(∆φ)j¯ 6 g φiφj¯ g (∆φ)i(∆φ)j¯ on (12) and (13), followed by a2 b2 ab + with = C(m + ∆φ), we get a cancellation with the ﬁrst term. We are left with 6 2 2

0ij¯ exp(−Cφ)g (∆φ)i(∆φ)j¯ ∆0(exp(−Cφ)(m + ∆φ)) − − C exp(−Cφ)∆0φ(m + ∆φ) > m + ∆φ + exp(−Cφ)∆0(∆φ) (14)

4 Using gij¯ = δij and φij¯ = φii¯δij at a point, (14) can be simpliﬁed using (10)

−1 0ij¯ 0 −1 −1 2 −(m + ∆φ) g (∆φ)i(∆φ)j¯ + ∆ (∆φ) > −(m + ∆φ) (1 + φii¯) | φkki¯ | i k X X −1 −1 + ∆F + (1 + φkk¯ ) (1 + φii¯) φki¯j¯φikj¯ i j k X, ,   1 + φll¯ 2 + (inf Riil¯ l¯)  − m  (15) i6=l 1 + φ i l ii¯ X, Here we used

kl¯ klφ¯ kl¯ (∆φ)i(∆φ)j¯ = (g φkl¯)i(g )j kl¯ kl¯ = (g φkli¯ )(g φklj¯ ) = φkki¯ φkkj¯ i j k X, , ∂gkl¯ since = 0. ∂zi But

φ (m + ∆φ)−1 (1 + φ )−1| φ |2 = (m + ∆φ)−1 (1 + φ )−1 kki¯ (1 + φ )1/2 ii¯ kki¯ ii¯ (1 + φ )1/2 kk¯ i k i k kk¯ X X X X   ! −1 −1 −1 6 (m + ∆φ)  (1 + φii¯) (1 + φkk¯ ) φkki¯ φkk¯ i¯ (1 + φkk¯ ) i k k X, X −1 −1 = (1 + φii¯) (1 + φkk¯ ) φkki¯ φkk¯ i¯ i k X, −1 −1 = (1 + φii¯) (1 + φkk¯ ) φikk¯ φki¯k¯ (16) i k X, The second line is using the Cauchy-Schwarz inequality and in the third line we just wrote (1 + k φkk¯ ) = m + ∆φ. The last two lines follow from P

φkki¯ = φkki¯ = φkik¯ = φikk¯ and

φkk¯ i¯ = φkk¯ i¯ = φki¯k¯ As a side note, the third covariant derivatives of a function commute with the second one only if they are of the same type. When these derivatives are of the opposite type an extra curvature term will appear. This does not happen with ﬁrst and second covariant derivatives though. Continuing the computation above

−1 −1 6 (1 + φkk¯ ) (1 + φii¯) φki¯j¯φikj¯ (17) i j k X, , −1 0ii¯ since we are just adding extra positive stuff with (1 + φii¯) = g > 0 and φki¯j¯ = φikj¯ . So plugging this into (15) gives (after cancellation)

  −1 0ij¯ 0 1 + φll¯ 2 −(m + ∆φ) g (∆φ)i(∆φ)j¯ + ∆ (∆φ) > ∆F + inf Riil¯ l¯  − m  (18) i6=l 1 + φ i l ii¯ X,

5 Inserting this into (14) we get

   0 1 + φll¯ 2 ∆ (exp(−Cφ)(m + ∆φ)) > exp(−Cφ) ∆F + inf Riil¯ l¯  − m  i6=l 1 + φ i l ii¯ X, − C exp(−Cφ)(∆0φ)(m + ∆φ) (19)

Inserting (11) into this gives

  0 2 1 + φii¯ ∆ (exp(−Cφ)(m + ∆φ)) > exp(−Cφ) ∆F − m inf Riil¯ l¯ + exp(−Cφ) inf Riil¯ l¯   i6=l i6=l 1 + φ i l ll¯ X, ! 1 − C exp(−Cφ)m(m + ∆φ) + C exp(−Cφ)(m + ∆φ) 1 + φ i ii¯ X 2 = exp(−Cφ) ∆F − m inf Riil¯ l¯ − C exp(−Cφ)m(m + ∆φ) i6=l ! 1 + (C + inf Riil¯ l¯) exp(−Cφ)(m + ∆φ) (20) i6=l 1 + φ i ii¯ X where we used 1 + φii¯ = m + ∆φ. i P (1 + φ ) 1 m−1 ii¯ Notice that the expansion of would contain i and everything else is posi- i 1 + φii¯ P(1 + φii¯) P i tive. This gives the following inequality Q

1   m−1 (1 + φii¯) 1 i 1 + φ > P(1 + φ ) i ii¯ ii¯ X i Q Therefore

1 1 −F (m + ∆φ) m−1 exp (21) 1 + φ > m − 1 i ii¯ X where we have used

0 det(gij¯) = exp(F) det(gij¯)

(1 + φii¯) = exp(F) i Y !− 1 m−1 F (1 + φ ) = exp − ii¯ m − 1 i Y Choose C so that

C + inf Riil¯ l¯ > 1 i6=l Then it follows from the previous computation that

0 2 ∆ (exp(−Cφ)(m + ∆φ)) > exp(−Cφ)(∆F − m inf Riil¯ l¯) + (C + inf Riil¯ l¯) exp(−Cφ) i6=l i6=l F 1+ 1 × exp − (m + ∆φ) m−1 − C exp(−Cφ)m(m + ∆φ) (22) m − 1

6 Now we use this to estimate exp(−Cφ)(m + ∆φ). In fact it must achieve its maximum at some point p so that the RHS of (22) is non-positive (since ∆0(...) has to be non-positive. This does no depend on the metric used to compute the Laplacian). At this point

F 1 2 1+ m−1 0 > ∆F − m inf Riil¯ l¯ − Cm(m + ∆φ) + (C + inf Riil¯ l¯) exp − (m + ∆φ) i6=l i6=l m − 1

This implies that (m + ∆φ)(p) has an upper bound C1 depending only on sup{−∆F}, sup | inf Riil¯ l¯|, Cm M M i6=l 1+ 1 1+ 1 1+ 1 and sup F. This is because y m−1 6 ay + b would imply that either y m−1 6 2ay or y m−1 6 2b. M Since exp(−Cφ)(m + ∆φ) achieves its maximum at p, we have the following inequality

0 < m + ∆φ 6 C1 exp(C(φ − inf φ)) (23) M

The ﬁrst inequality is true because (1 + φii¯) > 0 for all i. ij¯ 0 We use this to estimate sup |φ|. Since ∆φ + m = 1 + φii¯ = g gij¯ > 0, we can estimate sup φ as M i M follows. P Let G(p, q) be the Green’s function of the operator ∆ on M. Let K be a constant such that

G(p, q) + K > 0 Then

φ(p) = − G(p, q)∆φ(q)dq ZM = − (G(p, q) + K)∆φ(q)dq (24) ZM since M ∆φ = 0. R Therefore

sup(φ) 6 m sup (G(p, q) + K)dq M p∈M ZM | sup φ| 6 m sup |G(p, q) + K|dq (25) M p∈M ZM since ∆φ > −m. So we get

|φ| 6 | sup(φ) − φ| + | sup(φ)| ZM ZM M ZM M 6 (sup φ)Vol(M) + | sup(φ)|Vol(M) − φ M M ZM 6 2mVol(M) sup (G(p, q) + K)dq (26) p∈M ZM Note that the right hand sides of the above estimate does not depend on φ, since the Green’s function is prescribed for the manifold independent of φ. So far we have estimated sup(φ) and ||φ||L1(M). To M estimate inf(φ) two different proofs are offered. The ﬁrst one works for the case when m = 2 and the M second one works in general. We show both the proofs here. ?

7 For m = 2, we renormalize (by adding a constant) φ so that sup(φ) 6 −1. This is possible because of M (25). Let p be any positive number greater than or equal to 1. Then

|φ |2 ∆0(−φ)p = p(p − 1) i − p(−φ)p−1∆0φ 1 + φ i ii¯ X ! |φ |2 1 = p(p − 1)(−φ)p−2 i − p(−φ)p−1 m − (27) 1 + φ 1 + φ i ii¯ i ii¯ X X When m = 2 we ﬁnd that

|φ |2 ∆0(−φ)p = p(p − 1)(−φ)p−2 i − 2p(−φ)p−1 + p(−φ)p−1(1 + φ )−1(1 + φ )−1(2 + ∆φ) 1 + φ 11¯ 22¯ i ii¯ X |φ |2 = p(p − 1)(−φ)p−2 i − 2p(−φ)p−1 + 2p(−φ)p−1 + 2p(φ)p−1 exp(−F) 1 + φ i ii¯ X + exp(−F)(−∆(−φ)p + p(p − 1)(−φ)p−2|∇φ|2) (28) since

p(−φ)p−1∆φ = −∆(−φ)p + p(p − 1)(−φ)p−2|∇φ|2 0 (1 + φ11¯ )(1 + φ22¯ ) = det(g ) = exp(F)

Multiplying (28) by exp(F) and integrating wrt the volume form of the original metric g we have

|φ |2 2p (−φ)p−1(1 − exp(F)) = p(p − 1) (−φ)p−2 i exp(F) − ∆(−φ)p + p(p − 1) (−φ)p−2|∇φ|2 1 + φ ZM ZM ii¯ ZM ZM 2 p−2 |φi| 4(p − 1) p 2 = p(p − 1) (−φ) exp(F) exp(F) + |∇(−φ) 2 | 1 + φ p ZM ii¯ ZM Since φ is negative this implies that for p > 2

2 p 2 p p−1 |∇(−φ) 2 | (1 − exp(F))(−φ) 6 2(p − 1) ZM ZM p−1 6 pC2 |φ| (29) ZM where C2 depends only on sup(F). M Now we use the Sobolev inequality. It states that for any g ∈ W1,r with r < dim(M) = n, we have the following inequality

||g||Lq(M) 6 C(||g||Lr(M) + ||∇g||Lr (M)) nr for some constant C that depend only on M. Here q can be any number in [p, n−r ]. Observing that p the real dimension of M is 2m = 4 and then applying Sobolev inequality to the function g = |φ| 2 with r = 2 and q = 4, we can ﬁnd a constant C3 depending only on M such that

1 2 2p p p 2 |φ| 6 C3 |φ| + C3 |∇(−φ) 2 | ZM ZM ZM p p−1 6 C3 |φ| + pC2C3 |φ| (30) ZM ZM The last line follows from

8 p 2 p p |∇(−φ) 2 | = − ∆(−φ) 2 (−φ) 2 (Green’s identity) ZM ZM and p p p p p−2 2 p p−1 ∆(−φ) 2 .(−φ) 2 = − 1 (−φ) |∇φ| − (−φ) ∆φ 2 2 2 Since φ 6 −1, we derive another constant C4 depending only on M such that

2 2p 2 p |φ| 6 C4p |φ| (31) ZM ZM 2 In order to use (31), we need another estimate of M |φ| . But this follows from the estimate of M |φ| combined with the Poincare inequality and (29) as follows R R

|φ|2 = (−φ)2 ZM ZM 2 6 C |∇(−φ)| ZM 6 2CC2 |φ| (32) ZM

We now claim that we can ﬁnd a constant C5 depending only on M such that

p p 2 |φ|p Cp for all integers p 1 (33) 6 5 2 > ZM In fact, let p0 be the ﬁrst integer such that for p > p0

p+1 p−3 p + 2 2 p + 1 2 C (1 + Vol(M)) (34) 4 4 6 2

p−3 p+1 2 2 This is possible because lim = . p→ p+1 p+2 2 4 ∞ ∞ Notice there are only ﬁnitely many values that p cannot assume. So we can choose a constant C5 such that (33) is valid for all 1 6 p 6 2p0. This can be done because ||φ||L1(M) and ||φ||L2(M) are estimated above and ||φ||Lp(M) can be bounded by using (31) repeatedly with Holder¨ inequality. We prove (33) for all integers p > 1 by induction. There are two cases. If p + 1 is divisible by two, then from (31) we have

2 2 p+1 p + 1 p+1 |φ| C |φ| 2 6 4 2 ZM ZM 2 p+1 p + 1 p + 1 2 C Cp+1 by induction hypothesis 6 4 2 5 4

Now apply (34) to this to get

p+1 p + 1 2 |φ|p+1 Cp+1 6 5 2 ZM which is (33) for p + 1. On the other hand, if p + 1 is not divisible by two, then by (31)

9 2 2 p+1 p + 1 p+1 |φ| C |φ| 2 since (31) is true for non-integers as well 6 4 2 ZM ZM 2 2(p+1) p + 1 p+2 p+2 2 C |φ| 2 (Vol(M)) p+2 6 4 2 ZM 2 p+1 p + 1 p+1 p + 2 2 2 C C (Vol(M)) p+2 6 4 2 5 4 p+1 p + 2 2 Cp+1 6 5 4

p + 2 The second inequality is using Holder¨ inequality with conjugate exponents and p + 2. The second p + 1 2 to last line follows by induction hypothesis since (Vol(M)) p+2 6 Vol(M) if Vol(M) > 1. So we get (33) for all p. Using this we have the following

p 2 k 2p exp(kφ ) 6 |φ| ∞ p! M p=0 M Z X Z p k 2 p p 6 (C5) p (35) ∞ p! p=0 X By Stirling’s formula √ p+ 1 −p p! > 2πp 2 e for all positive integers p. This implies that

2 2 p − 1 exp(kφ ) 6 (kC5e) p 2 (36) ∞ M p=1 Z X When

−2 −1 k < C5 e (37) 2 the RHS of (36) is ﬁnite and we have thus obtained an estimate of M exp(kφ ) with any k satisfying (37). R We can now estimate sup |φ|. Rewrite M

∆φ = f (38) where

−m 6 f 6 C1 exp(sup(φ)) exp(−C inf(φ)) (39) M M We will now use Schauder estimates which essentially gives gradient estimates for the solution of (38) in terms of C(M)-norm of f and ||φ||L1(M). So there is a constant C6 depending only on M and C1 exp(C sup(φ)) such that M sup |∇φ| 6 C6 exp(−C inf φ) + |φ| (40) M M ZM

10 Since we have already proved estimates for sup(φ) and M |φ|, we have a constant C7 depending only M on M and C such that R

sup |∇φ| 6 C7(exp(−C inf(φ)) + 1) (41) M M

Let q be a point in M where φ(q) = inf(φ). Then in the geodesic ball, with center q and radius M 1 1 − inf(φ)C−1(exp(−C inf(φ)) + 1)−1, φ is not greater that inf(φ), because of the mean value theorem 2 M 7 M 2 M applied with (41). Since we may assume that − inf(φ) to be large, we can take the radius of the geodesic ball to be smaller M than the injectivity radius of M. If − inf(φ) was too small for this, then we have an estimate in terms M of the injectivity radius anyway. Therefore the integral of exp(kφ2) in this ball is not less than

2m 1 2 1 −2m −2m C8 exp( k inf(φ) ) − inf(φ) C (inf(φ) + 1) 4 M 2 M 7 M 2 k 2 where C8 is a positive constant depending only on M. Here we have used exp(kφ ) 6 exp (inf(φ)) 4 M 2 and comparing to the volume of the corresponding ball in TqM. Since we have estimated M exp(kφ ), the last quantity is estimated. This, of course gives an estimate for | inf φ|. M R ? Now we give an estimate of | inf(φ)| without assuming m = 2. Let N be any positive number. Then M (20) shows that

0 2 ∆ (exp(−Nφ)(m + ∆φ)) > exp(−Nφ)(∆F − m inf)Riil¯ l¯ − N exp(−Nφ)m(m + ∆φ) i6=l ! 1 + (N + inf Riil¯ l¯) exp(−Nφ)(m + ∆φ) (42) i6=l 1 + φ i ii¯ X Choose N so that

1 N + inf Riil¯ l¯ > N (43) i6=l 2

Then by (21)

1 N F m m−1 (N + inf Riil¯ l¯)(m + ∆φ) > exp − (m + ∆φ) (44) i6=l 1 + φii¯ 2 m − 1

There is a constant C9 depending only on sup(F) and m such that M

1 F m N exp − (m + ∆φ) m−1 2Nm(m + ∆φ) − NC (45) 2 m − 1 > 9

n since the function (any − by n−1 ) attains a minimum that depends only on n, a and b. Inserting (43), (44) and (45) into (42) we ﬁnd

0 2 ∆ (exp(−Nφ)(m + ∆φ)) > exp(−Nφ)(∆F − m inf Riil¯ l¯ − NC9) + N exp(−Nφ)m(m + ∆φ) (46) i6=l

Therefore

11 0 2 exp(F)∆ (exp(−Nφ)(m + ∆φ)) > exp(−Nφ) exp(F)(∆F − m inf Riil¯ l¯ − NC9) + N exp(−Nφ) i6=l × exp(inf F)m(m + ∆φ) M 2 2 = exp(−Nφ) exp(F)(∆F − m inf Riil¯ l¯ − NC9) + m N exp(inf F) i6=l M − mN exp(inf F) exp(−Nφ)∆φ M (47) 2 2 = exp(−Nφ) exp(F)(∆F − m inf Riil¯ l¯ − NC9) + m N exp(inf F) i6=l M + m exp(inf F)(−∆ exp(−Nφ) + N2 exp(−Nφ)|∇φ|2) M > −C10 exp(−Nφ) + m exp(inf F)(−∆ exp(−Nφ) M + N2 exp(−Nφ)|∇φ|2) where C10 depends only on N, F and M. In the second to last line we used

∆ exp(−Nφ) = exp(−Nφ)N2|∇φ|2 + N exp(−Nφ)∆φ Integrating we obtain 1 1 |∇ exp − Nφ |2 = N2 exp(−Nφ)|∇φ|2 2 4 ZM ZM 1 −1 6 C10m exp(− inf F) exp(−Nφ) (48) 4 M ZM 0 since ∆ = ∆ = 0 and exp(F − inf F) > 1. M M M We claimR that forR each N satisfying (43), the inequalities (48) and (26) furnish an estimate of

exp(−Nφ) ZM that depends only on N, F and M. We are going to prove this statement by contradiction. Suppose that there exists a sequence {φi} satisfying (26) and (48) such that lim exp(−Nφi) = . i→ M Then we deﬁne R ∞ ∞ −1 exp(−Nφ˜ i) = exp(−Nφi) exp(−Nφi) (49) ZM It follows from (48) that the sequence

2 1 ∇ exp − Nφ˜i 2 ZM ˜ is uniformly bounded from above by a constant depending only on N, F and M. Since M exp(−Nφi) = 1 for all i, this last fact implies that a subsequence of {exp − 1 Nφ˜ } converges in L2(M) to some function 2 R f ∈ L2(M). This is a consequence of W1,2 ⊂⊂ L2(M). We assume that this subsequence is denoted using the same indices. On the other hand, we know that, for any λ > 0

1 2 1 Vol x λ exp − Nφ˜i = Vol x log(λ) + log exp(−Nφi) −φi (50) 6 2 N N 6 ZM 1 Since lim exp − Nφ˜i = we conclude that for i large enough i→ M 2 R ∞ 1 ∞ 2 1 Vol x λ exp − Nφ˜i Vol x 0 < log(λ) + log exp(−Nφi) |φi| (51) 6 2 6 N N 6 ZM

12 By (26) M |φi| is uniformly bounded and the inequality (51) implies that

R 1 lim Vol x λ 6 exp − Nφ˜i = 0 (52) i→ 2 for all λ > 0. Clearly ∞

1 1 1 1 Vol x λ f Vol x f − exp − Nφ˜i + Vol x exp − Nφ˜i 6 6 2 6 2 2 6 2

2 4 1 1 1 f − exp − Nφ˜i + Vol x exp − Nφ˜i (53) 6 λ2 2 2 6 2 ZM 1 1 ˜ First inequality is true because if λ 6 f, then 2 λ can lie either between f and exp − 2 Nφi or be less 1 ˜ than exp − 2 Nφi . The last line is just the Chebyshev’s inequality. 1 ˜ 2 Since exp − 2 Nφi converges to f in L (M),(52) and (53) shows that

Vol x λ 6 f = 0 (54) 2 1 ˜ Since f is the L -limit of exp − 2 Nφi , this implies that f = 0 almost everywhere. This is a contradiction 2 because M f = 1. So we get the conclusion that whenever N satisfies (43), exp(−Nφ) has an estimate above that R M depends only on N, F and M. R We can now repeat the previous argument to find an estimate for | inf(φ)|.(41) is valid for any dimen- M 1 sion. As before, we find a geodesic ball with radius − inf(φ)C−1(exp(−C inf(φ)) + 1)−1 (which is not 2 M 7 M greater than the injectivity radius), such that φ is not greater than 1 inf(φ) in this ball. Then we choose 2 M N so large that (43) is satisfied. Now the integral of exp(−Nφ) in the above geodesic ball is not less than

2m 1 1 −1 −2m C12 exp − N inf(φ) − inf(φ) (C7) (exp(−C inf(φ)) + 1) 2 M 2 M M 1 since φ 6 −1 and φ 6 inf(φ) in this ball. Here C12 is a positive constant depending only on M. Now 2 M if we make N larger than 4mC, this gives an estimate for − inf(φ). Since we had already found an M estimate for sup(φ), we combine this to get an estimate for sup |φ|.(41) and (23) then gives estimates M M for sup |∇φ| and sup(m + ∆φ). on the other hand, since (δij + φij¯) is a positive-deﬁnite Hermitian M M m matrix, we can ﬁnd upper estimates for 1 + φii¯ for each i. The equation (1 + φii¯) = exp(F) then i=1 gives a positive lower estimate for 1 + φii¯ for each i. In conclusion, we haveQ proved the following:

i j Proposition 2.1. Let M be a compact Kahler¨ manifold with the metric gij¯dz ⊗ dz¯ . Let φ be a real-valued ∂2φ function in C4(M) such that φ = 0 and g + (det(g ))−1dzi ⊗ dz¯j deﬁnes another metric on M ij¯ ∂zi∂z¯j ij¯ M. Suppose that φ satisﬁes R ∂2φ g + (det(g ))−1 = exp(F) ij¯ ∂zi∂z¯j ij¯ Then there are positive constants C1, C2, C3 and C4 depending only on inf F, sup F, inf ∆F and M such that M M M

sup |φ| 6 C1 M sup |∇φ| 6 C2 M

0 < C3 6 1 + φii¯ 6 C4 for all i.

13 2.2 Third order estimates

In this section we ﬁnd estimates for third order derivatives, φijk¯ of the solution φ. We assume that F ∈ C3(M). Consider the following function

0ir¯ 0sj¯ 0kt¯ S = g g g φijk¯ φrs¯ t¯ (55)

We are going to compute the Laplacian of S.√ For convenience, we shall introduce the following convention. We say that A ' B if |A − B| 6 C1 S + C√2 where C1 and C2 are constants that can be ∼ estimated. We also say that A = B if |A − B| 6 C3S + C4 S + C5 where C3, C4 and C5 are constants that can be estimated. As before, we diagonalize the metric and the Hessian (φij¯) at a point under consideration. Then by computation ? we have

0 ∼ −1 −1 −1 −1 ∆ S = (1 + φii¯) (1 + φjj¯) (1 + φkk¯ ) (1 + φαα¯ )  2 2 X −1 −1 ×  φij¯ kα¯ − φip¯ k¯ φpjα¯ (1 + φpp¯ ) + φijkα¯ − (φpiα¯ φpjk¯ + φpik¯ φpjα¯ )(1 + φpp¯ )  (56) p p X X where the ﬁrst summation is over all the indices. On the other hand by (10)

0 −1 −1 2 ∆ (∆φ) > (1 + φkk¯ ) (1 + φii¯) |φki¯j¯| − C6 (57) X where C6 is a constant that can be estimated. Therefore, by letting C7 be a large positive constant we ﬁnd

0 ∆ (S + C7∆φ) > C8S − C9 (58) where C7, C8 and C9 are positive constants√ that could be estimated. This is true because√ the ﬁrst term 2 in the RHS of (57) is greater than S and S can be estimated by S itself (S + b > 2b S for any positive constant b). Observe that at the point where S + C7∆φ achieves its maximum, (58) shows that C8S 6 C9 and hence

C8(S + C7∆φ) 6 C9 + C8C7∆φ (59)

Since we already have estimates for ∆φ, this gives an estimate for the quantity sup(S + C7∆φ) and M hence of sup(S). This in turn gives estimates of φijk¯ for all i, j and k. So we have proved the following M proposition.

i j Proposition 2.2. Let M be a compact Kahler¨ manifold with metric gij¯dz ⊗ dz¯ . Let φ be a real-valued funtion ∂2φ in C5(M) such that φ = 0 and g + dzi ⊗ dz¯j deﬁnes another metric on M. Suppose that M ij¯ ∂zi∂z¯j R ∂2φ g + (det(g ))−1 = exp(F) ij¯ ∂zi∂z¯j ij¯

i j Then there are estimates for the derivatives φijk¯ in terms of gij¯dz ⊗ dz¯ , sup |F|, sup |∇M|, sup sup |Fii¯| and M M M i sup sup |Fijk¯ |. M i,j,k Note that the last two factors in the proposition comes from the computation of ∆0S.

14 2.3 Solution of the equation With the estimates so far we can now solve the equation

∂2φ g + (det(g ))−1 = exp(F) ij¯ ∂zi∂z¯j ij¯ where F ∈ C3(M). If Ω is the Kahler¨ form of M, then the above equation is equivalent to

(Ω + ∂∂φ¯ )m = (exp(F))Ωm (60)

Hence integrating (60) we immediately see that

exp(F) = Vol(M) (61) ZM k Conversely, we shall now prove that if F ∈ C (M) with k > 3 satisﬁes (61), then we can ﬁnd a solution k+1,α φ of (60) with φ ∈ C (M) for any 0 6 α < 1. For this we use the continuity method. Let

2 −1 ∂ φ −1 S = t ∈ [0, 1] det g + (det(g )) = Vol(M) exp(tF) exp(tF) ij¯ ∂zi∂z¯j ij¯ ZM k+1,α has a solution in C (M) and (1 + φii¯) > 0 ∀i (62)

Note that 0 ∈ S, since φ ≡ 0 is a solution in this case. Hence we need to only show that S is both closed and open in [0, 1]. This will imply that 1 ∈ S and that our original equation has a solution in Ck+1,α. To see that S is open, we use the inverse function theorem. Let

k+1,α A = φ ∈ C (M) (1 + φii¯) > 0 ∀i and φ = 0 (63) M Z k−1 α B = f ∈ C , (M) f = Vol(M) (64)

ZM Then A is open in the Banach space Ck+1,α(M) and B is an afﬁne subspace of the Banach space Ck−1,α(M). We have a map G : A → B given by

∂2φ G(φ) = det g + (det(g ))−1 (65) ij¯ ∂zi∂z¯j ij¯ Its clear that G maps to B since all the metrics in the same cohomology class has the same volume.The differential of G at a point φ0 is given by

∂2φ det g + 0 (det(g ))−1∆ (66) ij¯ ∂zi∂z¯j ij¯ φ0

This can be seen as follows. Let φt = φ0 + tη. Then

2 d d ∂ φ −1 G(φt) = det gij¯ + i j (det(gij¯)) dt t=0 dt ∂z ∂z¯ ∂2φ ∂2φ˙ = g + 0 ( (g ))−1 t g0 ij¯ det ij¯ i j det ij¯ i j φ0 ∂z ∂z¯ ∂z ∂z¯ t=0

= ∆φ0 η

15 ∂2φ where the subscript φ is used to denote quantities deﬁned using the metric det g + 0 . 0 ij¯ ∂zi∂z¯j

Note that B is the afﬁne space corresponding to the vector space f ∈ Ck−1,α(M) f = 0 . So the M

tangent space of B is same as this vector space. R

It is well-known (using Fredholm alternative for instance) that the condition for ∆φ0 φ = g to have a

weak solution on M is that M gdVφ0 = 0. Hence the condition for R ∂2φ det g + 0 (det(g ))−1∆ = f ij¯ ∂zi∂z¯j ij¯ φ0 −1 to have a weak solution is that M f = 0, since ∆φ0 φ = (det(gφ0 )) f det(gij¯) is solvable if and only if 0 = ( (g ))−1f (g )dV = fdV M det φ0 det ij¯ Rφ0 M g. R R Schauder theory makes sure that φ ∈ Ck+1,α(M) when f ∈ Ck−1,α(M). The solution is clearly unique if we require that M φ = 0 (by maximum principle two solutions will always differ by a constant). Hence the differential of G at φ is bijective and hence invertible. So G maps an open neighborhood R 0 of φ0 to an open neighborhood of G(φ0) in B. All solutions φ described in the set S that also satisfy −1 M φ = 0 are contained in the set A, since we can write Vol(M) M exp(tF) exp(tF) as ft ∈ B. Let −1 Rt0 ∈ S and φ0 be the corresponding solution with f0 = Vol(M) MR exp(t0F) exp(t0F). All functions ft with |t − t | < denote an open set in B around f and hence is the image of an open set in A 0 0 R around φ0 under G. This shows that |t − t0| < is contained in S and S is open. It remains to prove that the set S is closed. For this, we will set up a well-known bootstrapping technique in PDE, whereby we successively improve the regularity of the solution. If {tq} is a sequence k+1,α in S, then we have a sequence φq ∈ C (M) such that

2 −1 ∂ φq −1 det g + (det(g )) = Vol(M) exp(tqF) exp(tqqF) ij¯ ∂zi∂z¯j ij¯ ZM Normalizing we can assume that M φq = 0. Differentiating the above equation we have R 2 2 −1 ∂ φq 0 ij¯ ∂ ∂φq ∂ h i det g + g = Vol(M) exp(tqF) exp(tqF)(det(g )) (67) ij¯ ∂zi∂z¯j q ∂zi∂z¯j ∂zp ∂zp ij¯ ZM 2 0ij¯ ∂ φq where (gq ) is the inverse matrix of g + for each q. ij¯ ∂zi∂z¯j Proposition 2.1 shows that (φq)ij¯’s are uniformly bounded and hence the operator on the LHS of (67) is uniformly elliptic. Proposition 2.2 shows that the coefficients are Holder¨ continuous with exponent ∂φq 0 α 1. Then the Schauder estimate gives an estimate on the C2,α-norm of . Similarly we can 6 6 ∂zp ∂φq estimate the C2,α-norm of . From this information, we deduce that the coefficients of the LHS of ∂z¯p ∂φq ∂φq (67) are C1,α. The Schauder estimate again provides better differentiability for and . Iterating ∂zp ∂z¯p k+1,α this, one finds C -estimates of φq. Therefore, the sequence is uniformly bounded with a uniform bound on its derivatives. So it (a subsequence to be precise) converges in Ck+1,α-norm to a function φ by the Arzela-Ascoli theorem. Now taking limit as q → on the equation gives

∞ ∂2φ −1 det g + (det(g ))−1 = Vol(M) exp(t F) exp(t F) (68) ij¯ ∂zi∂z¯j ij¯ 0 0 ZM where t0 = lim tq. This proves that S is closed. So we have proved the following theorem. q→

i j k Theorem 2.3. Assume∞ that M is a compact Kahler¨ manifold with metric gij¯dz ⊗ dz¯ . Let F ∈ C (M) with k+1,α k > 3 and M exp(F) = Vol(M). Then there is a function φ ∈ C (M) for any 0 6 α < 1 such that ∂2φ g + R dzi ⊗ dz¯j deﬁnes a Kahler¨ metric and ij¯ ∂zi∂z¯j

16 ∂2φ det g + = exp(F) det(g ) ij¯ ∂zi∂z¯j ij¯ As a consequence of Theorem 2.3, we can prove the Calabi conjecture. We restate it here

To see this, notice the following well-known formula for Ricci curvature

∂2 R = − log(det(g )) (69) αβ¯ ∂zα∂z¯β ij¯ i Since the given tensor R˜ dzα ∧ dz¯β represents the ﬁrst Chern class of M, we can conclude that 2π αβ¯

∂2 R˜ = R − f (70) αβ¯ αβ¯ ∂zα∂z¯β for some smooth real-valued function f. This is because the first Chern class of M is the real cohomology class represented by the Ricci form of M. Now using Theorem 2.3, we can find a smooth function ∂2φ φ so that g + dzα ⊗ dz¯β defines a Kahler¨ metric with αβ¯ ∂zα∂z¯β ∂2φ det g + (det(g ))−1 = C exp(F) ij¯ ∂zi∂z¯j ij¯

where M C exp(f) = Vol(M). Now the Ricci tensor of this metric is given by R

∂2 R0 = − log(C exp(f) det(g )) αβ¯ ∂zα∂z¯β αβ¯ ∂2f = R − αβ¯ ∂zα∂z¯β = R˜ αβ¯

This proves the theorem.