Complex Analysis and Riemann Surfaces with Prof. Phong

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1 Complex Analysis and Riemann Surfaces with Prof. Phong Leonardo Abbrescia May 8, 2014 Overview of the Course In this class we will find L 2 estimates of Hörmander and their applications. These involve the Kodaira embedding formula, lower bounds for the Bergan Method, and canonical metrics and stability. 1 Bochner-Kodaira Formula 1.1 First Class We will begin by reviewing things. Let = µ µ be a complex n-manifold, where µ is a coordinate chart. We require that that ϕ µ ϕ 1 ν be holomorphic with invertible differentials. Definition 1.1 (Vector Bundles. We say that E is a holomorphic vector bundle of rank r if and only if it can be characterized by the transition functions t µν β (z. These must be holomorphic on µ ν, 1, β r. We can think of t µν as r r invertible matrices. Also, t ν ν β = δ β. Finally, we have t µν = t µk = t kν on µ k ν. Definition 1.2 (Smooth Sections of Vector Bundles. We say that ϕ Γ(, E if and only if ϕ µ(z µ, 1 r is smooth on µ and satisfies ϕ µ(z µ = t µν β (zϕβ ν (z ν on µ ν. Definition 1.3 (Covariant Derivatives on Holomorphic Vector Bundles. Let E be a holomorphic vector bundle, and let ϕ Γ(, E. Then we define ϕ Γ(, E Λ 0,1 on µ by ϕ := z µ j ϕ µ d z µ. j Now howe do we glue these together on different coordinate charts? Recall that ϕ µ(z µ = t µν β (zϕβ ν (z ν. Taking the and using the fact that t µν is holomorphic gives us z j µ ϕ µ(z µ = t µν β (z z µ j = t µν β (z ( z k ν z j µ ( ϕ β ν (z ν ϕ β ν z ν k (z ν This confirms the fact that ϕ Γ(, E Λ 0,1 (it is a section ( of E Λ 0,1, where we define Λ 0,1 to be the zν anti-holomorphic vector bundle with transition functions k. z j µ 1

2 In order to differentiate in the z j direction, we need a connection. An important connection is a unitary connection: Let H = Hᾱβ (z be a Hetmitian metric on the vector bundle E. We require (H µ ᾱβ to be a positive definite metric satisfying for ϕ Γ(, E Hence Now we can define a metric: ϕ 2 H = (H µ ᾱβ ϕ µϕ β µ = (H ν γδ ϕ γ νϕ δ ν on µ ν = (H µ ᾱβ t µν γ ϕγ νt µν β δ ϕδ ν = (H ν γδ ϕ γ νϕ δ ν. (H µ ᾱβ t µν γ t µν β δ = (H ν γδ. (1 Definition 1.4. A metric H = Hᾱβ on E is an assignment of (H µ ᾱβ on µ such that (1 holds, ϕ 2 H 0, and equality holding ϕ 0. Now that we have defined a metric H on E, we can use it to define the corresponding covariant derivative on Γ(, E. We do so as follows: Definition 1.5 (Unitary Connection. Let ϕ Γ(, E. Then we define ϕ := j ϕ dz j and j ϕ := H γ ( j H γβ ϕ β = H γ ( H γβ z j φ β Where H γ H γβ = δ β. : Γ(, E Γ(, E Λ1,0. This makes sense because H γβ ϕ β is a section of E, which is anti-holomorphic and so j H γβ ϕ β Γ(, E Λ 1,0. Tensoring with E now gives us a section of E Λ 1,0. Another way we like to write this is We directly compute this to get ( j ϕ = H γ j ( H γβ ϕ β. ( j ϕ = H γ j ( H γβ ϕ β = H γ ( H γβ j ϕ β + ( j H γβ ϕ β = δβ j ϕ β + H γ j H γβ }{{} A j β β ϕ = j ϕ + A j β ϕβ. (2 Definition 1.6 (A general connection. A connection on E is an assignment A j β such that (2 defines section in Γ(, E Λ 1,0 γ. This means that a unitary connection is a connection such that A j β = H γ j H γβ. We can view this in matrix notation as A j = H 1 j H. From here we move on to commutation relations. Note that jϕ = jφ. We leave it as an exercise to show that [ j, k] = 0 and [ j, k ] = 0. We will now show however that [ j, k ]φ = F jk β ϕβ : [ j, k ]φ = j( k ϕ k ( jϕ ( k ϕ + A kβ ϕ β ( k ( jϕ ( + A kβ jϕ β = j = ( ja k β ϕ β Then we have F jk = ja β kβ. In matrix notation, we are able to write this as follows: F jk = ja k = ( H 1 k H. We will also need the corresponding form: j F = i 2π F jk β dzk d zj Γ(, E E Λ 1,1 = Γ(, End(E Λ 1,1. (3 2

3 Now we start going back to the original problem: the Bochner-Kodaira Formulas. Let be a complex manifold which for simplicity we assume to be compact and without boundary. Let E be a holomorphic vector bundle. Now we consider the complex: Γ(, E Λ p,q Γ(, E Λ p,q+1 Now we give a more precise definition. We say φ Γ(, E Λ p,q when it has the form Now we apply the definition of : ϕ = 1 p!q! ϕ j 1 j qi 1 i p dz ip dz i1 d z jq d z j1. ϕ = 1 ( p!q! ϕ j 1 j qi 1 i p dz ip dz i1 d z jq d z j1 = 1 ( kϕ j p!q! 1 j qi 1 i p d z k dz ip dz i1 d z jq d z j1. Before class ends we give some examples: Example 1.7. We apply on Γ(, E Λ 0,0 = ϕ = kϕ d z k. Example 1.8. We apply on Γ(, E Λ 0,1. We know that ϕ in this space is given by ϕ = ϕ ᾱ j d zj. Now we compute: ϕ = ( ϕ ᾱ j d z j = ( kϕ ᾱ j d zk d z j = 1 ( kϕ ᾱ j 2 jϕ ᾱ k d z k d z j. This tells us that ( ϕ j k = kϕ j jϕ k. 1.2 Second Class We go back to our problem of the Bochner-Kodaira Formula. Let E be a holomorphic vector bundle of rank r. Recall the complex: Recall that if ϕ Γ(, E Λ p,q, then Γ(, E Λ p,q Γ(, E Λ p,q+1 ϕ = 1 p!q! ϕ j 1 j qi 1 i p dz ip dz i1 d z jq d z j1 and ϕ = 1 ( kϕ j p!q! 1 j qi 1 i p d z k dz ip dz i1 d z jq d z j1 1 [ ( ] anti-sym = ϕ dz ip dz i1 d z jq d z j1 p!(q + 1! j 1 j qi 1 i p k 3

4 Lets see why this operator is well defined. In the simple case, consider ϕ Γ(, Λ 1. We know that ϕ = ϕ i dx i = ϕ k dy k. From the chain rule we know that dy k = yk x j dx j and so we have the glueing relation Taking the differential gives us ϕ j = ϕ k y k x j. dϕ = dϕ j dx j = ϕ j dx l dx j x l = 1 ( ϕ j 2 x l ϕ l x j dx l dx j. Differentiating the glueing relation gives us ϕ j x l = ( y k x l ϕ k x j + ϕ y k k x l x j = ϕ k y m y k y m x l x j + ϕ 2 y k k x l x j. Notice that the second term is symmetric with respect to the super indices i and j, and so they cancel out in the differential. This shows how this form is in fact well defined when anti-symmetrizing. Now we move on a bit and introduce a metric Hᾱβ on E and a metric g kj on T 1,0 (. We know this to be the well defined vector fields T 1,0 ( V = V j z i.e., ϕ 2 j H = H ᾱβϕ ϕ β is a scalar and g kj V j V k is a scalar. Now we will attempt to introduce an L 2 metric on Γ(, E Λ p,q ϕ, ψ: ϕ, ψ := 1 ϕ ᾱ JI p!q! ψβ KL K H β g Jg I L ωn where of course by definition g K J = g k1 j1 g kq j q, ω = i 2 g kjdz j d z k. Now we define the formal adjoint of as satisfying ϕ, ψ = ϕ, ψ for any ϕ C (, E Λ p,q and ψ C (, E Λ p,q+1. Hence we can draw Γ(, E Λ p,q Γ(, E Λ p,q+1 Now I define a certain type of laplacian : Γ(, E Λ p,q Γ(, E Λ p,q by := +. This raises the question that will give us a Vanishing Theorem : when do we have dim ker Γ(, Λ p,q = 0? Note that this is analogous to finding harmonic functions (because they are kernel s of. In order to find a question we will need to deduce the Bochner-Kodaira formula. I claim that ( ϕ ᾱ JĪ = gk l k lϕ ᾱ JI + Torsion terms + (Curvature termsϕ. (4 4

5 We will hence be working with Käbler metrics, which are metrics in which the Torsion terms vanish. We will see that after some integration by parts, we will have ϕ, ϕ = lϕ ᾱ JĪ 2 + Curvature ϕ, ϕ. This boils down our problem to showing that the curvature > 0 because this would imply that ker = 0. In order to do this we must explicitly compute. Lets start out with in the second part of the following: Γ(, E Λ 0,0 Γ(, E Λ 0,1 Γ(, E Λ 0,2. Lets begin by defining our functions: Γ(, E Λ 0,1 ϕ = ϕ ᾱ j d zj ϕ = kϕ ᾱ j d z Γ(, E Λ 0,2 ψ = 1 2 ψ j kd z k d z j. Recall that we can anti-symmetrize ϕ by ( ϕ ᾱ j k = kϕ ᾱ j jϕ ᾱ k and we impose ϕ, ψ = ϕ, ψ. The LHS of this will become LHS = 1 ( 2 ϕ m p ψ β j k H βg j m k p ωn g = 1 ( pϕ m m ϕ ᾱ p ψ β 2 j k H βg j m k p ωn g We will now write this in terms of covariant derivatives. Recall that the covariant derivatives in the bar direction on anti-holomorphic sections act exactly how covariant derivatives in the unbar direction act on holomorphic sections: where Γ l p m = g k l( p g mk. Finally we can write p ϕ ᾱ m = m ϕ ᾱ m + g k l( p g mk = m ϕ ᾱ m + Γ l p m ϕ ᾱ l p ϕ ᾱ m m ϕ ᾱ p = p ϕ ᾱ m m ϕ ᾱ p + (Γ l p m Γ l m p ϕ ᾱ l. We will denote the last term as the torsion tensor and Γ l p m is the torsion of the covariant derivative p. Now we can formally define Kähler metric: a metric g kj is said to be Kähler if Γ l m p = Γ l p m p g mk = m g pk p g km = m g kp. It is easy to see that this condition would make our calculations much easier because there would be no torsion terms in (4. More importantly for our calculations, this means that p ϕ ᾱ m m ϕ ᾱ p = p ϕ ᾱ m m ϕ ᾱ p. We will see next time how this applies to our calculations (that involve integration by parts on our inner product. 1.3 Third Class Lets review the context that we are in. We have E a holomorphic vector bundle, Hᾱβ as metric on E, g kj a metric on T 1,0 (. We introduce the complex. Γ(, E Λ p,q Γ(, E Λ p,q+1 5

6 where is defined to satisfy ϕ, ψ = ϕ, ψ for ϕ C (, E Λ p,q and ψ C (, E Λ p,q+1. We also defined the laplacian := + : Γ(, E Λ p,q Γ(, E Λ p,q. For simplicity in the computations we consider the case where p = 0 and q = 1. Our main goal is to find the kernel of. Our complex is as follows: Γ(, E Λ 0,0 Γ(, E Λ 0,1 Γ(, E Λ 0,2. It is obvious that our goal is to explicitly find. First we find it acting on Γ(, E Λ 0,1. We have ϕ, ψ = ϕ, ψ for ϕ C (, E and ψ C (, E Λ 0,1. Suppose we have Γ(, E ϕ = ϕ (z. Then applying gives us ϕ = jϕ d z j. Take ψ = ψ ᾱ k d zk. We pair up the two with our inner product to get jϕ ψ β k k j ωn H β g = ϕ ( ψ β ω n H β (5 where ω = i 2 g kjdzj d z k. In order to reach an expression for we need to study how to integrate by parts on manifolds. Our first step is to write the LHS of (5 as follows LHS = jϕ ψ β k ωn Hᾱβ gj k = jϕ W j. The first step is justified because we have conjugated the subindices β ᾱβ and similarly with the j k superindeces. Finally, notice that in the ψ β k Hᾱβ g j k term, the β and k indices are both run over completely and we only have the and j indices. This justifies the second step. Since differentiating on holomorphic sections in the bar direction is equivalent to taking the covariant derivative in the bar direction we have the following identities ( jϕ W j ( jϕ W j = j (ϕ ( W j ϕ jw j. This is essentially the integration by parts formula on manifolds. Then we plug this into the LHS of (5 to get ( jϕ W j det g qp = (ϕ ( W j det g qp ϕ jw j det g qp. (6 The reason why there is that det g qp term is because from the definition of ω, we have ω n j = det g kj n j=1 i 2 dzj d z k. Claim. If the metric g kj is Kähler, then j (ϕ W j det g qp = 0. Proof. We write V j = ϕ W j. We want to show ( jv j det g qp = 0. 6

7 Indeed ( ( jv j det g qp = = j jv j + Γ j j k V k deg g qp (V j det g qp V j ( j det g qp + Γ j j k V k deg g qp. From a formula that will not be proved, the derivative in the bar direction of the determinant function is j det g qp = (det g qp g l m jg ml = Γ m j m det g qp. Then we have that the latter two terms will vanish when g kj is Kähler. Then we have that our form is exact and so its integral is zero. So now we go back to our integration by parts. Plugging in what we know from our Claim into (6, we have ( jϕ W j = ϕ jw j det g qp. Opening up the W j term yields jϕ ψ β k ωn Hᾱβ gj k = ϕ j ( ψ β k ω n Hᾱβ g j k. Now we rewrite the RHS to get an expression in the form of (5: ( ϕ j ψ β k ω n ( Hᾱβ g j k = ϕ ω gj k jψ H β k n β. Going back to our imposition of ϕ, ψ = ϕ, ψ gives us ( ψ β = g j k j ψ β k. (7 We are done with half of the battle. Now we need to find the formal adjoint acting on Γ(, E Λ 0,2. We again impose ϕ, ψ = ϕ, ψ where ϕ C (, E Λ 0,1 and ψ C (, E Λ 0,2. We write ψ = 1 2 ψ ᾱ l m d zm d z l ϕ = ϕ ᾱ j d zj ϕ = kϕ ᾱ j d zk d z j = 1 2 = 1 2 ( kϕ ᾱ j jϕ ᾱ k ( kϕ ᾱ j jϕ ᾱ k d z k d z j d z k d z j where the last equality comes from imposing g kj to be Kähler. Now we do the inner product! The LHS of the inner product becomes LHS = 1 ( kϕ ᾱ j 2 jϕ ᾱ k ψ β l m H β g l j m k ωn g ( = kϕ ᾱ j ψ β l m H β g l j m k ωn g = ϕ ᾱ j ( g k m k ψ β l l j ωn m H β g 7

8 Comparing this to the RHS with ϕ, ψ = ϕ, ψ gives us ( ψ β l = g k m k ψ β l m. (8 Now we need to compute what ( ϕ and ( ϕ is. Again we define ϕ = ϕ ᾱ j d zj Γ(, E Λ 0,1 and we compute by plugging in (7 ( ϕ ( λ = g j k ( j ϕ ᾱ k = l g j k j ϕ ᾱ k d z l = g j k l j ϕ ᾱ k. The other piece of the laplacian will come from computing ϕ. Recall that if g kj is Kähler then we are allowed to say ϕ = 1 ( kϕ ᾱ j 2 jϕ ᾱ k d z k d z j. Picking out the indices we have again gives us ( ϕ ᾱ j k = kϕ ᾱ j jϕ ᾱ k. Plugging this into (8 gives us ( ϕ β l = g k m k ( ϕ β l m = g k m k ( m ϕ β l lϕ β m = g k m k m ϕ β l + g k m k lϕ β m. We put everything together to finally get the explicit form of := + = g k m k m ϕ ᾱ l + g k m k lϕ ᾱ m g k m l k ϕ m (9 This is the Bochner-Kodaira-Formula. Notice that can be though of as the canonical laplacian + curvature terms. ( ϕ ᾱ l = g k m k m ϕ ᾱ l + g k m [ k, l]ϕ ᾱ m. We will write the curvature term as [ k, l]ϕ ᾱ m = F lk β ϕβ m }{{} curvature of Hᾱβ + p R lk m ϕᾱ p }{{} curvature of g kj. Now recall that and by analogy we have F lk β = l ( H γ k H γβ. ( p R lk m = g pq r R lk q g mr = g pq ( l ( g r s k g sq g mr. Now we have the following version of the Bochner-Kodaira-Formula: ϕ ᾱ l = g k m k m ϕ ᾱ l + F m l β ϕβ m + R p l ϕ ᾱ p (10 Where g R p l k m p R lk m is the so called Ricci Curvature. 8

9 This allows us to prove a simple but basic corollary. Let L be a positive holomorphic line bundle over a compact manifold i.e., h metric on L with j k log h > 0. Set ω = i 2 log h. In components this would be i 2 ( j k log h dz j d z k. If L is positive then ω is a metric and is Kähler. The reason is simple: g kj = j k log h l g kj = l j k log h = j g kl. Then for m >> 1 we will show next time that ker Γ(,Lm Λ 0,1 = Fourth Class We return to our discussion of the Bochner-Kodaira Formula. Let E be a holomorphic vector bundle, and Hᾱβ a metric on E and g kj a metric on T 1,0 (. We were trying to find an accurate description of the laplacian = + + on Γ(, E Λ p,q. In our case, to lighten the notation, we let p = 0 and q = 1. Let Γ(, E Λ 0,1 ϕ = ϕ ᾱ j d zj. Thn our Bochner-Kodaira Formula is the following ( ϕ ᾱ l = g j k j kϕ ᾱ l + g m p [ m, l]ϕ ᾱ p. (11 Notice that the subscript l on the LHS reminds us that it is a 0, 1-form and the superscript reminds us that it has coefficients in E. We can also rewrite it in the following way ( ϕ ᾱ l = g j k { } j kϕ ᾱ l + g m p F lm γ ϕᾱ q p + R lm p ϕᾱ q = g j k j kϕ ᾱ l + F p l γ ϕγ p + R p l ϕ ᾱ p. As a consequence, now we take the inner product with ϕ: ( ϕ ᾱ l ϕβ mh m l ωn β g = g j k j kϕ ᾱ l ϕβ mh m l ωn ( β g + F p l γ ϕγ p + ϕ R p l ᾱ p ϕ β mh m l ωn β g We can then do our integration by parts to get ϕ, ϕ = kϕ ᾱ l β m g j kh m l ωn β g }{{} + E Λ 0,1 Λ 0,1 jϕ ( = ϕ 2 + = ϕ 2 + We can get some corollaries out of this. F m p βγ ϕγ pϕ β m + R m p H β ϕ ᾱ p ϕ β m ( F β m p + Rm p H β ϕ ᾱ p ϕ β m ω n Corollary 1.9. If F m p βγ + Rm p H β > 0, then ker E 0,1 = 0. ( F p l γ ϕγ p + ϕ R p l ᾱ p ϕ β mh m l ωn β g Corollary Let E be a positive line bundle L. Recall that positivity mans that there exists a metric Hᾱβ = H 11 = h with ω = i 2 log h > 0. We equipt our manifold with a Kähler metric ω and consider on L M 0,1. Then if M >> 1, ker = 0. ω n Proof. When E = L, our inner product is much simpler: (F ϕ, ϕ = ϕ 2 + m p + R m p ϕ pϕ mh ωn 9

10 However, I haven t explained why we factored out the metric h. The reason is this: F kj β = k (H γ j H γβ F m p β = g m kg j p F kj h = F m p h. Now if we replace L with L M, R m p is going to be unchanged because it deals with the matrix on the base manifold. However, F m p becomes MF m p because h becomes h p. In the case with positive line bundles, we will use a metric on the line bundle h whose curvature makes a metric on the base : F }{{} m p = g m p }{{}. curvature on L metric on Then finally on L M for a positive line bundle L, we have (Mg ϕ, ϕ = ϕ 2 + m p + R m p ϕ pϕ mh ωn. Since g m p > 0, we can chose M large enough so that Mg m p + R m p > 0. With this I hope I have inspired the reader to think about more Bochner-Kodaira formulas. For example, we can consider = + on Γ(, E Λ p,q. This can give us a new theorem: Theorem 1.11 (Kodaira-Akizuki-Nakao Theorem. Let E = L be a line bundle. Then where Λ is the Hodge operator: Γ(, E Λ p+1,q+1 Φ = + [F, Λ] As an exercise one can show [, Λ] = and [, Λ] =. Proof of K.A.N. Λ (ΛΦ KJ = g l p Φ p l KJ Γ(, E Λ p,q. = [, Λ] + [, Λ] + [, Λ] + [, Λ] = ( + Λ Λ ( + = [ +, Λ] = [F, Λ]. In practice, we will apply the following lemma when we apply the K.A.N. 2 Kodaira Imbedding Theorem and Cohomolgy of Sheafs 2.1 Fifth Class Let E be a holomorphic vector bundle (we ll take E = L be a line bundle and let Hᾱβ = h be a metric on E and g kj be a Kähler metric on. Recall the complex: Γ(, E Λ p,q Γ(, E Λ p,q+1 10

11 The key property that makes this a complex is that 2 = 0. Then this implies that Range ker Γ(,E Λ p,q Γ(,E Λ. p,q+1 This makes sense because for Γ(, E Λ p,q ϕ, ϕ Range by definition, but since 2 = 0, Γ(,E Λ p,q this is also in ker Now we will define the cohomoly group of the Γ(,E Λ p,q+1. complex as H p,q (, E Λp,q := ker Γ(,E Λ p,q+1 / Range (12 Γ(,E Λ. p,q Before we analyze this lets look at some motivation. This is the analogue of the de Rham cohomolgy group. Let be a smooth manifolds. Then examine the d complex: d Γ(, Λ p d Γ(, Λ p+1 where d 2 = 0. Then of course this implies Range d ker d Γ(,E Λ p,q d Γ(,E Λ p,q+1 by the same reasoning of the complex. Then the de Rham cohomology group is gong to be defined as / H p dr ( := ker d Range d Γ(,Λp Γ(,Λ p 1. Now we will attempt to see what these groups measure. One can see that these groups are actually topologically invariant from. However in the case for the cohomology of the complex, it depends very heavily on the complex structure of both and E. Now one can raise the question of where the group vanish, and similarly, how does one compute this group? There are two universal approaches to computing H p ( (or at least when this group is = 0. One is Hodge Theory and the other is by the L 2 estimates of Hörmander. The main idea of Hodge Theory is as follows: define the Laplacian as = dd + d d and define a metric g ij. Then one can show that there is a metric dependent isomorphism H p dr ( = ker. Our vanishing theorems then tell us when ker = 0 = H p dr ( = 0. The main idea of the L 2 estimate approach is as follows: Consider the equation du = v for v L 2 (, Λ p+1, dv = 0. Then we attempt to find conditions under which this equation admits solutions. Notice that this would tell us something about the de Rham cohomology group because we will have ker d = Range d for solutions. One advantage of the L 2 estimate approach is that this equation deals with one derivative, while deals with two. Here are some key steps: 1. The key step in the Hodge theory approach: construction of a certain operator G such that G = G = I Π where Π : L 2 (, Λ p ker is the orthogonal projections. Furthermore assume dg = Gd and d G = Gd and assume existence of G. Then from G = I Π we have u = Gu + Πu = d(d u + d (du + Πu for any smooth u. Then if u ker d = du = 0 and so we have u = Gu + Πu = d(d u + Πu 11

12 and so [u] = [Πu] in terms of equivalence classes because d(d u is in the range of d Γ(,Λ p. As a sanity check, we can see that since Π is the orthogonal projection, this isomorphism of equivalence classes is metric dependent, as desired! Additionally now we have a candidate that solves du = v for v L 2 (, Λ p+1, dv = 0. Just take d Gv = u. Then we can check that du = d(d Gv = ( d dgv = Gv = v Πv. The proof of the existence of G comes from the following regularity theorem: u 2 W k+2,2 ( ( u C 2 W k,2 ( + u 2 W k,2 ( for all u W k,2 ( and elliptic. 2. The key step in the L 2 estimate approach: Let L be a holomorphic line bundle, h a metric on L, and g kj a Kähler metric on. Recall the complex: Γ(, E Λ p,q Γ(, E Λ p,q+1 We will define derivatives in the sense of distributions (i.e. the weak sense for f L 1 loc (Rn. Let f L 1 loc (Rn. Then the derivative f x 1 in those sense is the following functional: take C0 (R n ϕ f ϕ. x 1 In the case where f Cloc 1 (Rn, we can do integration by parts and have the functional equal f ϕ. x 1 In this case the derivative in the sense of distributions is equivalent to the usual derivative. For the most general functions, we have this as ϕ gϕ. Now lets go over some definitions Definition 2.1 (Domain of. We define Dom p,q := {ϕ L 2 (, L Λ p,q ϕ L 2 (, L Λ p,q+1 such that ϕ = ψ in the weak sense}. Definition 2.2 (Domain of. We define Dom := {ψ L 2 (, L Λ p,q+1 ϕ L 2 (, L Λ p,q s.t. ψ = ϕ in the weak sense AND λ, ψ = λ, ϕ λ Dom }. We see that restricting the latter to λ C0 where we obtain the weaker condition that ψ = ϕ in the sense of distributions. This will be our main Lemma for next time: Lemma 2.3. Let g kj be a Kähler metric on, h a metric on L. Assume the following holds: 1 u u 2 2 Au, u u Dom 1 Dom 0. (13 Then for any f L 2 (, L Λ 0,1 with f = 0 in the sense of distributions, v = f and v 2 A 1 f, f. 12

13 Lets discuss some of our notation that we introduced in the Lemma. Our operators are the ones in the following sense 0 1 L 2 (, EL Λ 0,0 L 2 (, L Λ 0,1 L 2 (, E Λ 0, More explicitly u = u jd z j with u j a section of L and and so Also v L 2 (, L Λ 0,1 and v 2 = So the final result is Au, u = A l j u ju lh Au, u = v 2 h ωn, v 2 h ωn A l j u ju lh ωn. A 1 f, f = (A 1 l j f jf lh ωn. (A 1 l j f jf lh ωn. The beautiful thing about this is that there is no constant on the right hand side of the equatio 2.2 Sixth Class Recall that we were looking at the L 2 Estimate approach to see when the cohomology group of the complex vanishes. Let L be a holomorphic line bundle over a compact complex manifold. Let h be a metric on L and g kj a Kähler metric on. Consider L 2 (, L Λ 0,0 0 L 2 (, L Λ 0,1 1 L 2 (, L Λ 0,2 And let ϕ L 2 (, L Λ 0,0,. Since this ϕ is not necessarily smooth we introduce and similarly Dom := {ϕ L 2 (, L Λ 0,q ψ L 2 (, L Λ 0,q+1 with ϕ = ψ in the weak sense}, Dom := {ϕ L 2 (, L Λ 0,q ψ L 2 (, L Λ 0,q 1 with ϕ = ψ in the weak sense This will be the main lemma: and f, ϕ = f, ψ f Dom }. Lemma 2.4. Assume ϕ 2 + ϕ 2 Aϕ, ϕ ϕ Dom 0 Dom 1 (14 where A is a positive definite matrix. Then f L 2 (, L Λ 0,1 with f = 0, there exists u L 2 (, L with u = f and u 2 A 1 f, f. 13

14 Notation. We have u 2 = Aϕ, ϕ = A 1 f, f = u ūh ωn A k j ϕ jϕ kh ωn (A 1 k j f jf kh ωn. Proof of Lemma 2.4. Lets keep note of our goal: we want to find u L 2 (, L such that u = f for f L 2 (, L Λ 0,1. In order to keep going we must prove the following lemma: Lemma 2.5. Let ϕ Dom 0 and write ϕ = ϕ 1 + ϕ 2 where ϕ 1 ker 1 and ϕ 2 (ker 1. Then we will show that ϕ 1, ϕ 2 Dom 0. Proof of Lemma 2.5. This makes sense from studies of linear algebra and that ker 1 is a closed subspace. Since we are working with a complex and 1 0 = 0 we have Range 0 ker 1 and so if ϕ 2 is perpendicular to ker 1, then ϕ 2 (Range 0 and so we can write by definition of our inner product as 0 = 0 ψ, ϕ 2 for ψ Dom 0. This is still true and we can write 0, ψ, ϕ 2 = ψ, 0 ψ Dom 1 and so ϕ 2 satisfies exactly the conditions of being in Dom 0 and so ϕ 2 Dom 0. By a very similar argument ϕ 1 ker 1 implies that ϕ 1 Dom 0 and so we are done with the lemma. Now we can go back to Lemma Consider the following functional T : 0 ϕ ϕ, f for ϕ Dom 0. Then we decompose ϕ = ϕ 1 + ϕ 2 where ϕ 1 ker 1 and ϕ 2 (ker 1 and by our previous lemma and construction and the fact that f = 0 and ϕ 2 is orthogonal to ker we get ϕ, f = ϕ 1, f + ϕ 2, f = ϕ 1, f. Since A is positive definite, we can view this as a norm and see that by the Cauchy-Schwarz inequality [ ] [ ] ϕ, f 2 Aϕ 1, ϕ 1 A 1 f, f ( ϕ ϕ 1 2 [ ] A 1 f, f ϕ 1 2 A 1 f, f and so this functional T is well defined. Now recall the Hahn-Banach Theorem: Let V B be a subspace of a Banach space and T a linear functional V v T v with T (v A v. Then there exists an extension T of T to the whole of B satisfying T v A v v B. We apply this theorem to get T : L 2 (, L C extenuating T. Then there exists a u L 2 (, L such that T (ψ = ψ, u and u = T for any ϕ L 2. Then in particular we can take ψ = 0 ϕ for a ϕ C 0. Then we have that 0 ϕ, u = T ( 0ϕ = ϕ, f which is exactly the prerequisite to imply u = f in the weak sense. Now that we have this done, we need to determine when (14 holds. Recall the Bochner-Kodaira Formula: for C (, L Λ 0,1 : ϕ, ϕ = ϕ 2 + ϕ 2 for = + and for smooth forms we were able to integrate by parts to get ϕ, ϕ = kϕ j 2 L 2 + (F kj + R kj ϕ j ϕ k h ωn 14

15 where F kj = j k log h, R kj = Ricci Curvature. So then we can actually define F kj + R kj = A and note that the above lemma will hold if F kj + R kj > 0 from our vanishing theorems and (14 extends from C0 to Dom 0 Dom 1. Lets now examine the baby version of this extension. Note that on a compact manifold, C0 is dense in Dom 0 Dom 1 with respect to the norm u L 2 + u L 2 + u L 2. If we assume this, then we have the following theorem. Theorem 2.6. Let L be a holomorphic line bundle, g kj a Kähler metric. Let h have the form h = e ϕ and assume j kϕ + R kj ɛg kj for some ɛ > 0. Then for any f L 2 (, L Λ 0,1 satisfying f = 0, there exists u L 2 (, L solving u = f and u 2 ϕ ωn e 1 g k j ϕ ωn f jf ke ɛ. A useful variation of this set up is to solve u = f for f L 2 (L Λ n,1. L 2 (, L Λ n,0 u 2 = u ūe ϕ Observe that for u and we don t need the volume form! Observe that L Λ n,1 = (L Λ n,0 Λ 0,1. Then notice that we can let L Λ n,0 =: L as a new line bundle. Then we can apply the Theorem 1.17 by replacing L with L and the then imposing ɛg kj F kj + R kj = F kj R kj + R kj (15 = F kj Where step (15 is left as an exercise. Note that all of these theorems extend to not compact, but complete, so we can define topologies and our estimate (14 will still extend from C0 to Dom 0 Dom 1. If (, g kj is not necessarily complete but g kj which is Kähler and complete, then we can apply the theorem to (, g δ kj = g kj + δg kj. Then we will get a u δ that converges weakly to a solution u. 2.3 Seventh Class Lets recall that we just found out a basic condition for our vanishing theorem to apply. Let L be a holomorphic line bundle, h a smooth metric on L, and g kj a Kähler metric on. Then we found the following condition of on Λ 0,1 : if j k log h > ɛg kj (16 for some ɛ > 0 then we will have ker L Λ n,1 / Range L Λ n,0 = 0. Our goal is to extend the theory for h not smooth, but still satisfying an analogue of (16. For this we must begin to analyze the theory of subharmonic functions in R n. Lemma 2.7. Let u C 2 (Ω, Ω R n. Then u = 0 and u satisfying the MVP are equivalent statements. Proof. Ommited. 15

16 Lemma 2.8. Assume u C 2 (Ω. Then the following are equivalent: u 0 in Ω, and for B r (x Ω; u(x u(x u(ydσ y ; B r(x u(ydy. B r(x Functions satisfying these conditions are called subharmonic functions. Lemma 2.9. Let u L 1 loc (Ω. Then the following are equivalent: 1. u 0 in the weak sense 2.!u which is equivalent to u almost everywhere, is upper semi-continuous, and u (x u(ydσ y B r(x 3.!u which is equivalent to u almost everywhere, is upper semi-continuous, and u (x u(ydx. B r(x Note that u L 1 loc (Ω means that u is measurable and for any B Ω we have u <. Proof. Note that u = 0 in the weak sense means that for any ϕ C0 (Ω, 0 = u ϕdx B Ω which is done in the proof of the MVP. So then we have trivially 2 = 3 = 1. We need to prove 1 = 2, which will be done next class. 2.4 Eighth Class Recall that we have the following Lemma left to prove Lemma Let u L 1 loc (Ω. Then the following are equivalent: 1. u 0 in the weak sense 2.!u which is equivalent to u almost everywhere, is upper semi-continuous, and u (x u(ydσ y B r(x 3.!u which is equivalent to u almost everywhere, is upper semi-continuous, and u (x u(ydx. B r(x 16

17 Proof. We will use mollifiers. Pick χ C0 ({ x < 1} such that its integral over R n is normalized to one with χ(x = χ( x. Then for δ > 0 define u δ = u χ δ where χ δ (z = 1 δ χ( z n δ and one can easily see that supp χ δ { x < δ}. Lets recall some basic properties of the mollification of u. One can easily see that u δ C and if u L p for 1 p < then u δ u in L p. Now assume u 0 in the weak sense. Then u δ 0 in the usual sense and u δ u (i.e. u δ u δ if δ δ. Define u (x := lim δ 0 u δ (x and note that u (x might be. Now recall that, by definition, f is upper semi-continuous when {f < c} is open c. Now let u j u and each u j be upper semi-continuous. Then we claim that u is upper semi-continuous. The proof is one line: {u < c} = {u j < c} j but we need to understand why it is true. Let x {u < c}. Then j such that u j (x < c. Convercely, if x j {u j < c} Then there exists j 0 such that u j0 < c < c and u j u implies that for any j j 0 we have u j u j0 < c < c and so taking the limit gives us lim u j c < c. j This finishes the claim and tells us that u (x is upper semi-continuous. Now lets go back to the task at hand. Since u δ (x is smooth and subharmonic, then from Lemma 1.19 we have u δ u δ (xdσ y. (17 B r(x Fix some δ 0 > 0 and since u δ is continuous, for any x B r (x we have u δ0 c. Since u δ is a decreasing sequence, δ δ 0 we have u δ (x u δ0 c 0 C u δ (x. But then u δ u implies that C u δ is an monotonically increasing sequence and since C u δ C u we have from the monotone convergence theorem that C u δ = C u lim u δ = u lim δ 0 and so from (17 we have our required claim. Now lets prove uniqueness. Let v be another such function with the same properties of u. Then v(x v(ydσ y = B δr (c v(x + rδωdσ ω. B 1(x We then multiply both sides by χ(rr n 1 and integrate to get [ ] v(x χ(rr n 1 dr = v(x + rδωχ(rr n 1 drdσ ω 0 B 1(x 0 = 1 v(x + δzχ(zdz ω n R n = 1 v δ ω n = 1 ω n u δ 17

18 where we have used the face that χ(x = χ( x and that v = u almost everywhere, so equality is reached under the integral. This then implies [ ] 1 v(x u δ (x χ(rr 0 n 1 dr ω n = u δ (x δ v(x u where we have taken the limit in the last line. Now to prove uniqueness, I claim that we can t have a strict inequality. Assume you can and that for some x 0 we have v(x 0 < c < u (x 0. Since v is upper semi-continuous we will have that v(x < c for all x B k (x 0 where k 1. Then by the sub-mean value property and the fact that v u almost everywhere we have u(x < c almost everywhere. Then this implies that and we have a contradiction. u(x 0 u(x c < u (x 0 B k2 (x Now we quote some properties of subharmonic properties without proof. A function u is subharmonic if and only if K Ω, ϕ continuous on K and ϕ = 0 in K, u ϕ on K, then u ϕ on all of K. Another property is if u j u and u j is subharmonic, then u is subharmonic. Lets move on to something else. We say that u is plurisubharmonic if u L is subharmonic where L is a complex line. Example Let f k (z be holomorphic in Ω C n for 1 k N. Then ( N ϕ(z = log f k (z 2 is plurisubharmonic. Lets see why. Define ( N ϕ δ (z = log f k (z 2 + δ 2 k=1 k=1 where δ > 0 and notice that ϕ C and ϕ δ ϕ as δ 0. For functions that are C plurisubharmonicity is equivalent to j kϕ δ 0 as a matrix. Define f N+1 = ɛ and lets compute ( N+1 j kϕ δ = j k log f l (z 2 jk l=1 ( N+1 l=1 = f l(z k f l (z j N+1 l=1 f l(z 2 j f l (z k f l (z fl (z = k f l (z j f l f l (z fl 2 ( f l ( ( = ( f l 2 2 j f l (z k f l (z fl 2 f l (z k f l (z j f l f l (z Acting this on V j V k we get 2 ( ϕ 1 = z j z k ( f l 2 2 j fv j 2 fl 2 ( j f l V j fl k f l V k f l 0 by Cauchy-Schwarz. 18

19 2.5 Ninth Class Recall that last time we discussed plurisubharmonic functions. We say that φ is psh in Ω C n if φ is upper semi-continuous, and φ L is subharmonic for every complex line L C n. A basic example is the function φ = log ( N f j (z 2 For f j hol c. Today we use psh functions to prove L 2 estimates with singular weights. j=1 Definition We say that f vanishes of order k at z 0 if f(z = k a (z z 0 in a Taylor expansion around z 0. Lemma Assume that f is hol c in a neighborhood of 0 in C n, such that f 2 e φ < B ρ(0 For φ = γ log z 2. Then k, there exists a γ depending only on k and n, such that f vanishes of order k at 0. Proof. For convenience, we rewrite the hypothesis as B ρ f 2 z 2γ < On B ρ, we can express f(z = γ a z + E(z, with E(z C z γ+1. By the triangle inequality, we obtain: a z ( f 2 z 2γ 2 + E(z 2 z 2γ < γ We compute the integral on the LHS in polar coordinates: z = rω, with ω S 2n 1. ρ a r ω 2 r 2γ r 2n 1 drdσ(ω < 0 S 2n 1 ρ 0 γ γ } a 2 r 2 r {S 2n 1 2γ ω 2 dσ(ω dr < 2n 1 The LHS diverges unless a = 0 whenever 2 + 2n 1 2γ 1, which is to say γ n. Therefore all we need to do is take γ = k + n. We want to extend this result to complex manifolds (, ω, with ω Kahler. Lemma If ω = i 2 g kjdz j d z k is a Kahler metric, then locally ω can be written as Where K C ( is called the Kahler potential. ω = i 2 K Definition φ is ω-psh if φ + K is psh in C n. Formally this means j k(φ + K 0, i.e. i 2 φ ω. (Note that we could not expect strict positivity for φ. That s because φ has a maximum on a compact manifold, and the Hessian is negative at the maximum. Therefore we need to shift positivity by K. Now we are in a position to state and prove: 19

20 Theorem 2.16 (L 2 estimates with singular weights. Let L be a hol c line bundle over a compact Kahler manifold (, ω. Assume that there is a metric h = e φ, with φ upper semi-continuous (in general, not smooth satisfying: i φ ɛω For some ɛ > 0. Then for all f L 2 (, L Λ n,1 such that f = 0, there exists u L 2 (, L Λ n,0 with u = f in the sense of distributions, and u 2 e φ 1 f 2 e φ ɛ Remark. i φ ɛω is equivalent to φ (ɛ + 1K is psh in C n. Remark. The theorem was already proved for the case of φ smooth. Proof. The idea is to use φ j φ, φ j C, such that i φ j ɛω. A set of theorems by D ly, as well as a theorem by Blocki and Kolodziej, guarantee the existence of these functions. We apply Hormander s theorem with smooth weights, and obtain u j such that u j = f, and u j 2 e φj 1 f 2 e φj ɛ Define h j = e φj, then obviously h j h = e φ. Moreover, h 1 h j h. Then we have: u j 2 h 1 u j 2 h j 1 f 2 h j 1 ɛ ɛ f 2 h We see that, in the space L 2 h 1 (, L Λ n,0, u j L 2 C, for a constant C independent of j. This means we are within the assumptions of the Banach-Alaoglu theorem, stated below. Applying it, we obtain a subsequence {u jl } of {u j }, such that u jl u, for some u L 2 h 1. The key is to note that this implies u jl u in L 2 h k, for all k. Indeed, h 1 /h k is a strictly positive and C function on a compact manifold. We can then write: ( u jl vh 1 = u jl v h 1 h k h k i.e. we replace the C function v by another C function to get the desired result. Then, since u jl u in L 2 h k and since weak convergence implies semi-continuity of norms (proved below, we obtain u 2 h k lim inf u jl 2 h k 1 f 2 h ɛ Finally, we apply the Lebesgue monotone convergence theorem, and get that u 2 h 1 f 2 h ɛ For completeness, we give below the two results that have been used in the proof. Theorem 2.17 (Banach-Alaoglu. Let u j L 2 (, Euclidean space, and u j L2 C. Then there exists a subsequence u jl u, for some u L 2 (. This weak convergence means, by definition, that whenever v L 2 ( we have u jl v u v This property is also known as weak compactness. Proof. Omitted. 20

21 Remark. u j u does not imply u j u 0. To see an example of this, consider {u j } an orthonormal basis of L 2. By Pythagora, u j u l = 2, so {u j } doesn t converge. However, for any v L 2, we have: v, u j 2 = v 2 < Therefore v, u j 0 as j, which means u j 0. Lemma If u j u, then u 2 lim inf u j 2. j Proof. uū = lim u j ū, so by Cauchy-Schwartz u 2 lim inf u j u. Therefore u 2 lim inf u j Tenth Class Lets state the theorem for the first time. Theorem Let L be a positive Line bundle over a compact Kähler complex manifold. Let H 0 (C, L k = {space of holomorphic sections of L k } with N k +1 = dim H 0 (, L k. Let {s (z} N =0 be a basis for H 0 (, L k. The Kodaira map is defined by z z k [s 0 (z;... ; S Nk (z] CP N k. Then k 0 such that for k k 0, the map z k is an imbedding of inside CP N k. Note that in order for this map to be well defined we need to require that at least one coordinate be non-zero. The key technical tool for this are Hörmander s estimates (which we have already done and the tying formalism is Cech cohomology. The pieces are as follows: 1. We will need to establish Cech Cohomology. Need to define sheafs, look at Ȟq (, F and look at computational tools that allow us to look at short exact sequences of sheaves as long exact sequences of cohomology groups. 2. We will make the Kodaira Imbedding Theorem have a Cech cohomology formulation. Given N points z 1,... z n and N integers k 1,..., k n, we can define a sheaf I k1z 1,...,k nz n consisting of germs of holomorphic functions that vanish of order k i at z i. Kodaira s theorem will follow if k 1,... k n, k 0 such that Ȟ 0 (, L k Ȟ0 (, L k O/I k1z 1,...,k nz n (18 is surjective, where O is defined as the sheaf of holomorphic germs. Example I z1 will be the sheaf of holomorphic functions vanishing at z 1. Then O/I z1 C because it is essentially values of the germs of O at z 1. Then O O/I z1 is obviously surjective. 3. The map (18 fits inside the following exact sequence and by 1. this sequence implies 0 I k1z 1,...,k nz n O O/I 0 Ȟ 0 (, L k I k1z 1,...,k nz n Ȟ0 (, L k O Ȟ0 (, L k O/I k1z 1,...,k nz n Ȟ1 (, L k I k1z 1,...,k nz n and so we are reduced to showing that Ȟ1 (, L k I k1z 1,...,k nz n = 0 to show surjectivity. This links to the following point. 21

22 4. We shall show that ϕ that is plurisubharmonic such that k 1,..., k n, there exists a k depending only on k 1,..., k n such that { } f holomorphic near z f 2 e ϕ < J ϕ I k1z 1,...,k nz n U z and e ϕ is a singular metric on L k with i ϕ ɛω where ω is Kähler. Then ker Lk Λ n,0 J ϕ / Im Lk Λ n,0 J ϕ = H 1(, L k Λ n,0 J ϕ = 0 by our vanishing theorems. 5. Our last piece is H 1(, L k Λ n,0 J ϕ = Ȟ1 (, L k Λ n,0 J ϕ. Now lets review some of the basics of Cech cohomology. Definition Let be a topological space. A presheaf F of abelian groups on consists of the data (a for every open subset U, an abelian group F (U, and (b for every inclusion V U of open subsets of, a morphism of abelian groups ρ UV : F (U F (V, subject to the conditions (i F ( = 0 (ii ρ UU is the identity map F(U F(U, and (iii if W V U are three open subsets, then ρ UW = ρ V W ρ UV. Definition A presheaf F on a topological space is a sheaf if, on top of (a and (b, it satisfies (a If U is an open set, if {V i } is an open covering of U, and if s F (U is an element such that s Vi = 0 for all i, then s = 0. (b if U is an open set, if {V i } is an open covering of U, and if we have elements s i F (V for each i, with the property that for each i and j, s i Vi V j = s j Vi V j, then there is an element s F (U such that s Vi = s i for each i. Note that the previous (a makes s unique. Definition For a given z, a stalk E z is defined as E z = F (U/ where for s F (U and t F (V we define s t if W z, W U V such that s W = t W. z U 22

23 2.7 Eleventh Class Let be a topological space, = U and define the covering {U } = U. Given a sheaf E, we can identify sections of E over U U Γ(U, E. Example Γ(U, E = {space of p-forms over U }, where E = Λ p Example Γ(U, E = {f holomorphic on U, with f V = 0} Example Γ(U, J = {f holomorphic on U, with U f 2 e ϕ < }. Now lets talk about the Cech Cohomology of E. We will define C p = {p-cycles σ; U 0,..., U p σ 0 p Γ(U 0 U 1, E}. We define the operator δ : C p C p+1 by (δσ 0 1 = ( 1 j σ 0 j p+1 Γ(U 0 U p+1, E and one can tell that δ 2 = 0. With this we define Ȟ p (U, E = ker δ C p / Im δ C p 1. We want to stretch this definition to the full covering. Let U = {U } and U = {U β } be two coverings of. We say that U is a refinement of U if ϕ : {} {β} with U U β. Then we have a map of cycles (ϕσ 0 p = σ ϕ(0 ϕ( p Γ(U, E. This induces a map on the Cech cohomology Ȟp (U, E Ȟp (U, E. Now consider a sequence of refinements U j+1 U j and define Ȟ p (, E = lim Ȟ p (U j, E j j Ȟ p (U j, E/ where σ Ȟp (U j, E and τ Ȟp (U k, E are equivalent if m > j, k with ϕ jm σ = ϕ km τ. From this we have the following Lemma, quoted without proof. Lemma If {U} is a covering with Ȟp (U 0 U p, E = 0 p 1, then From this lets go through a couple of examples. Ȟ p (, E = Ȟp (U, E. Example From definition Ȟ0 (, E = {σ C 0 δσ = 0}, where σ : U σ Γ(U, E. Then we have that C 1 (δσ β = σ β σ be definition of the delta operator. Then δσ = 0 implies that σ = σ β on U U β, and by the axioms of sheaf theory, {σ } makes a global section of E. This implies that Ȟ 0 (, E = Γ(, E. Example 2.29 (Cousin Problem 1. Let Ω C n. Given {p j } Ω, does there exists a meromorphic function f with a pole of order n j at p j? Locally, this is trivial because you can just look at a tiny neighborhood of each p j. What about globally? Consider σ C 1 (, O where O is the sheaf of holomorphic functions. Now look at the cycle σ jk = f j f k on U j U k, which is now holomorphic because the poles have cancelled. Now the punchline is that if Ȟ1 (, O = 0, then a global function with this property exists, because if the Cech Cohomology group is tribal then this implies that τ C 0 (, O with σ = (δτ and σ jk = τ j τ k. Then this implies that f j f k = τ j τ k = f j τ j = f k τ k which gives us a global meromorphic function because the indices match. 23

24 Example 2.30 (Cousin Problem 2. Given {p j } Ω, does there exists a holomorphic function with a given order of vanishing at p j? The answer is yes, if H 1 (, O = 0 where Γ(U, O = {non-vanishing holomorphic functions on U}. Now we will show that an element σ C 1 (, O gives rise to a line bundle L. First note that C 1 (, O σ : U U β σ β Γ(U U β, O reminds us of transition functions of line bundles. Since these sections are multiplicative, we have that δσ = 0 σ β = σ γ σ β. This means that Ȟ 1 (, O = {line bundles with transition functions σ β }/ where σ β σ β if σ β = σ β f f 1 β. Now given one line bundle L with transition function σ β, we have Γ(, L ϕ = {ϕ }, ϕ = σ ϕ β ( and similarly with a line bundle L, we ( have Γ(, L ϕ = {ϕ }, ϕ = σ β ϕ β. Then ϕ = σ β ϕ β = σ β f f 1 β ϕ β f 1 ϕ = σ β f 1 β ϕ β. This gives us a mapping Γ(, L ϕ ϕ = f 1 ϕ Γ(, L and this one-to-one correspondence tells us that if we have two equivalent cycles, we have two equivalent line bundles. Now lets talk about short exact sequences of sheaves. Assume that we have sheaves E, F, G and maps Φ Ψ 0 E F G 0 which are exact as maps of sheaves i.e. z, the sequence of maps between stalks Φ Ψ 0 E z F z G z 0 is exact as a sequence between groups. Then we will show that we have the following long exact sequence of Čech cohomology groups 0 Ȟ 0 (, E Ȟ 0 (, F Ȟ 0 (, G δ Ȟ 1 (, E It turns out that the δ operator is in fact the Chern operator. Lets see how to construct δ, which we call the coboundary operator. Let σ C p (, G such that δσ = 0. Let τ C p (, F with Ψ(τ = σ (which makes sense because Ψ is surjective. Then I claim that δτ = Φ(λ for λ C p+1 (, E. This is because Ψ(δτ = δψ(τ = δσ = 0. Furthermore, Φ(δλ = δφ(λ = δ 2 τ = 0 and so this implies that δτ = 0 because Φ is injective, which yields a map C p (, G σ δ λ C p+1 (, E called the coboundary operator. With this I want to have a tool that equates de Rham cohomology groups with Čech cohomology groups. We say that E is a fine sheaf if χ = 1, supp χ U, then there is a well defined map Γ(U, E s χ s Γ(U, E. Example Suppose Γ(U, E is the sheaf of smooth p-forms on U. Then this is a fine sheaf. However the sheaf of holomorphic functions is not a fine sheaf because multiplying by a partition of unity destroys holomorphicity. 24

25 2.8 Twelfth Class Lets go over an observation. Let E be a fine sheaf, which means that if χ is a partition of unity subordinate to {U }, which also happens to be a covering of our manifold, then Γ(U, E σ χσ Γ(U, E is well defined. Then the observation is that Ȟq (, E = 0 for q 1. Proof. Let σ C q, (δσ = 0. We want σ = δτ. In fact, take τ 0 q 1 = χ γ σ γ0 q 1. We will verify the claim for q = 1 because for q 2 it is a trivial application of q = 1. For q = 1 let C 1 σ = σ β. Then τ = χ γ σ γ and from the definition of the δ operator we have (δτ β = τ β τ = χ γ σ γβ σχ σ γ = χ (σ β σ γ ( = χγ σ β = σ β. Now we go on to prove a very important theorem from cohomology. Theorem 2.32 (The de Rham Theorem. Let be a smooth connected manifold. Then / Ȟ q (, R = H q dr ( = ker d Λ q Im d Λ q 1. Proof. Consider the following sequence of sheaves d Λ q 1 d Λ q d Λ q+1 d (19 Note that we have to evaluate all of these differentials locally so we are actually doing this on stalks of the sheaves Λ q z = {germs of q-forms at z}. Note that Λ 0 Λ 1 is still well defined by giving a map from germs of functions to 1-forms. We put 0 R Λ 0 to make this an exact sequence. From exactness, if we have a with σ = dτ is well defined because we are dealing with germs. Now note that all of our sheaves except R are fine, and we need short exact sequences. Introduce Zz q which is the sheaf of germs of q-forms that are closed. Observe that Zz q is not fine because d(χσ 0. Then ( is equivalent to the following series of exact sequences σ Λ q z with dσ = 0 then τ Λ q 1 z 0 R Λ 0 Z 1 0 d (20 0 Z q Λ q Z q+1 0. Then from our observation from earlier today, we can apply our theorem to get the following exact sequence We can apply the de Rham theorem to get the following exact sequence: 0 Ȟ 0 (, R Ȟ 0 (, Λ 0 Ȟ 0 (, Z 1 d δ Ȟ 1 (, R Ȟ 1 (, Λ 0 25

26 Then from our observation that the Čech cohomology group of fine sheaves is zero, Ȟ0 (, Λ 0 = Ȟ1 (, Λ 1 = 0 and so from exactness / Ȟ(, R = Ȟ0 (, Z 1 dȟ0 (, Λ 0 (21 and we have proved the theorem for q = 1. Now consider a later piece in this sequence Ȟ q (, Λ 0 Ȟ q (, Z 1 δ Ȟ q+1 (, R Ȟ q+1 (, Λ 0. By the same reasoning that the Čech cohomology group of fine sheaves are trivial, Ȟ q (, Z 1 = Ȟq+1 (, R. (22 Now examine the second exact sequence from (20 to get 0 Ȟ 0 (, Z q Ȟ 0 (, Λ q Ȟ 0 (, Z q+1 d δ Ȟ 1 (, Z q Ȟ 1 (, Λ q and again our observation tells us that / Ȟ q (, Z = Ȟ0 (, Z q+1 dȟ0 (, Λ q = H q+1 dr (. (23 Now notice that we will be done if we show that Ȟ1 (, Z q = Ȟq (, Z by (22 and (23. Take a later piece in this sequence to see that Ȟ p (, Λ q Ȟ p (, Z q+1 δ Ȟ p+1 (, Z q Ȟ p+1 (, Λ q implies that Ȟp (, Z q+1 = Ȟp+1 (, Z q. Now just keep moving the indices to finish the proof. Theorem 2.33 (Theorem of Dolbeauilt. Let be complex and connected. Then Ȟ q (, Ω p = H p,q ( = ker / Λ p,q Im Λ p,q 1 where Ω p is the sheaf with stalks Ω p z of germs of holomorphic (p, 0-forms in a neighborhood of z. Proof. This proof is exactly as the de Rham theorem. We initially have 0 Ω p z Λ p,0 Λ p,1 Λ p,q 1 Λ p,q Λ p,q+1 and by the -Poincaré lemma, this is exact. We immediately get that Ȟq (, Ω = cohomology of global sections = H p,q (. 26

27 Now we will extend these ideas to multiplier ideal sheaves. Let L be a holomorphic line bundle, h = e ϕ be a metric on L (not smooth, (, ω a Kähler manifold, and define J ϕ (z = {germs of holomorphic functions near z, satisfying f 2 e ϕ < 0}. Now consider the following sequence of sheaves L q (Z = {germs of sections of L Λ n,q satisfying With theism we have a sequence of maps between sheaves and we mandate that it starts with L q 1 L q L q+1 U 2 e ϕ < and (weak with 0 L Λ n,0 J ϕ L 0 L 1 U U 2 e ϕ < }. We apply the previous machinery to get Ȟ q (, L Λ n,0 J ϕ = ker L q With this we are able to prove the following important theorem. / Im L q 1 (24 Theorem 2.34 (Nadel, Let L, (, ω Kähler and compact, and assume h satisfies i ϕ ɛω for ɛ > 0. Then Ȟ 1 (, L Λ n,0 J ϕ = 0. Proof. We apply Hörmander s theorem twice, once with being small balls to show that our sequence of L q is exact, and then again with being our whole manifold to show that the RHS of (24 is zero. 2.9 Thirteenth Class Through all of our hard work, we now have enough to prove the following main theorem. Theorem 2.35 (Kodaira Imbedding Theorem. Let L be a positive line bundle over a complex compact manifold. Then k 0 such that for k k 0 the map ι k : z [s 0 (z : : s Nk (z] CP N k is well defined and is an imbedding where {s (z} N k =0 is a basis of H0 (, L k, whose dimension is 1 + N k. Proof. I claim that it suffices to show that N, k and constants a, b > 0 such that the map H 0 (, L H 0 (, L k O/I k1z 1 k nz n is onto for any z 1,..., z n as long as N k a k i + b. i=1 27

28 Here, I z1 is the sheaf with stalks I z1 (w = {f holomorphic in a neighborhood of w; f(z 1 = 0}. Similarly, O is the sheaf with stalks O(w = {f holomorphic in a neighborhood of w}. Then just by observation we can see that { C w = z1 O/I z1 = 0 w z 1. For this reason we call it the sky-scraper sheaf. Lets recall exactly what it means for H 0 (, L k H 0 (, L k O/I z1 to be surjective: given any τ L k z 1, there exists s H 0 (, L k with s(z 1 = τ. More generally, H 0 (, L k H 0 (, L k O/I z1 z n being surjective means that τ 1,..., τ N, τ j L k z j, then s H 0 (, L k such that s(z j = τ j. Now we actually prove the statement. Recall Nadal s Vanishing theorem: Let L (, ω be a line bundle over a Kähler manifold and assume L admits a (possibly singular metric h such that i k log h ɛg kj. Then Ȟ1 (, L J ϕ = 0 where J ϕ (w = {f holomorphic in a neighborhood of w with U f 2 e ϕ < }. We now consider 0 J ϕ O O/J ϕ 0 which implies the long exact sequence 0 L k J ϕ L k O L k O/J ϕ 0 Ȟ 0 (, L k O Ȟ 0 (, L k O/J ϕ Ȟ 1 (, L k J ϕ We want to apply Nadal s theorem with K L = L k i.e. L = K 1 Lk. We shall now construct a metric h on L such that i log h ɛω (25 and J ϕ is a sub sheaf of I k1z 1 k nz n. If this indeed the case then Ȟ0 (, L k Ȟ0 (, L k O/J ϕ is surjective and O/J ϕ O/I k1z 1 k nz n is surjective. By hypothesis, L is a positive line bundle i.e. h 0 smooth on L such that w 0 = i 2 log h 0 > 0. Equip with the Kähler metric ω 0 (which is a metric on T 1,0 (. Such a metric induces a metric on K 1, which is just the maximal wedge product of T 1,0 (. Introduce a metric h on L k by h = h k 0e ϕ. For simplicity, set N = 1. Let = and take z 1 V W, χ C0 (W, and χ = 1 in a neighborhood of V. Define ϕ = γχ(z log z z 1 2 where γ is a big positive constant. We can calculate the curvature h as follows i h = k( i log h 0 + i (γχ(z log z z 1 2 = k( i log h 0 + γχ(z(i log z z 1 2 k 2 ω 0 28