Introduction to Hodge Theory

1 Introduction

In this notes we are going to introduce some of the concepts concerning with Hodge Theory and give some interesting applications of the main theorems. We will follow both real and complex case where we can see how the theory provides beautiful connections between geometry and topology of manifolds. In the ﬁnal section we will discuss how the Hodge theorem can provide some topological restrictions for a complex manifold to be K¨ahler.

2 Background 2.1 Hodge star operator and Laplacian Let (M n,g) be a closed manifold, i.e., compact without boundary, Riemannian and oriented. We denote Λk(M) the space of sections of the bundle of diﬀerential k-forms on the manifold, i.e., Λk(M) := Γ(Ωk(M)). First we deﬁne the Hodge star operator:

? :Λk(M) → Λn−k(M) by requiring α ∧ ?β =< α, β > dV olM , where <, > is the induced inner product deﬁned on the bundles of k forms. If we choose a positive orthonormal basis {e1, . . . , en} for the tangent space TxM at x ∈ M, then

?(e1 ∧ ... ∧ ek) = ek+1 ∧ ... ∧ en. p In particular, in local coordinates around x we have dvolM = det(gij)dx1 ∧ ... ∧ dxn.

1 i Example 1. Let ω ∈ Λ (M), ω = ω dxi, then

j−1 ij i ˆ ?ω = (−1) g ω dx1 ∧ · · · ∧ dxj ∧ · · · ∧ dxn

. Exercise 1. Check that ?? = (−1)k(n−k)Id :Λk(M) → Λk(M).

Our next step is to deﬁne the L2 inner product of α, β ∈ Λk(M) by: R R (α, β) := M < α, β > dvolM = M α ∧ ?β

1 Recall the exterior derivative d :Λk(M) → Λk+1(M), the fundamental property of d is that d ◦ d = 0. We say ω ∈ Λk(M) is closed if dω = 0 and when ω = dη then ω is called exact form. From d ◦ d = 0 we see that exact forms are also closed forms, therefore we can take the quotient of the subspace of closed forms over the exact forms. This quotient is a R vector space called the k − th de Rham cohomology group and denoted by

k HDR(M). In the presence of the L2 inner product (, ) we can ask for the adjoint of the operator d which we will denote by d?. In other words,

d? :Λk(M) → Λk−1(M), satisfying (dα, β) = (α, d?β). Lemma 1. d? :Λk(M) → Λk−1(M) satisﬁes

d? = (−1)n(k−1)−1 ? d ? .

Proof. Let α be a k − 1 form and β a k form. Recall the Leibniz rule for the d operator: d(α ∧ ?β) = dα ∧ β + (−1)k−1α ∧ d ? β. Using Stokes theorem we get: Z Z dα ∧ β = (−1)kα ∧ d ? β M M Now we observe that d?β is a n−k+1 form. Therefore ?? = (−1)(n−k+1)(k−1)Id. Z Z (dα, β) = dα ∧ ?β = α ∧ ?[(−1)(n−k+1)(k−1)(−1)k ? d ? β] = M M Z = α ∧ ?[(−1)n(k−1)−1 ? d ? β] = (α, (−1)n(k−1)−1 ? d ? β). M Therefore, d? = (−1)n(k−1)−1 ? d ? . Deﬁnition 1. The Laplace operator on Λk(M) is the linear operator

∆ = dd? + d?d :Λk(M) → Λk(M).

ω is called harmonic if ∆(ω) = 0.

Properties of the Laplacian

1. (∆α, β) = (α, ∆β), i.e, (∆ is self-adjoint). 2. ?∆ = ∆?. 3.∆ α = 0, if and only if dα = d?α = 0.

Proof. (∆α, α) = (dd?α, α) + (d?dα, α) = (d?α, d?α) + (dα, dα) = |dα|2 + |dα|2

2 Example 2. If k = 0 then ∆ = d?d and in local coordinates we have for any f ∈ C∞(M): 1 ∂ ijq ∂f ∆f = −p [g det(gij) ] det(gij) ∂xi ∂xj Example 3. In the Euclidean case there is a natural expression for the laplacian I of a k form. If ω = ω dxI then

n X ∂2ωI ∆ ω = − dx . e ∂x2 I m=1 m We have seen that any harmonic k−form is closed and therefore it represents k a nontrivial cohomological class in HDR(M). Indeed, if ω is harmonic and exact then |ω|2 = (ω, ω) = (ω, dη) = (d?ω, η) = 0 which implies that ω = 0. Let’s denote the space of k harmonic forms by Hk(M), hence we have constructed a injective map k k H (M) → HDR(M). We can try to ﬁnd a minimizing element in each cohomological class with respect the L2 inner product (·, ·). If ω is a closed smooth k form and has minimal norm in its cohomological class then d?ω = 0. Indeed, let η ∈ Λk−1(M)

∂ 0 = |ω + tdη|2| ∂t t=0 ∂ 0 = (ω + tdη, ω + tdη) = 2(ω, dη) ∂t This is equivalent to (d?ω, η) = 0 for all η ∈ Λk−1(M) ie, d?ω = 0. Conversely, if ω is closed and d?ω = 0 (i.e., ω is harmonic) then ω has minimal norm. It is enough to prove that |ω + dη|2 > |ω|2.

|ω + dη|2 = (ω + dη, ω + dη) = |ω|2 + |dη|2 + 2(ω, dη)

So we are done since the last term on the right hand side is zero. A natural question is if it possible to minimize the norm in each cohomological class such that the minimum is attained by a smooth diﬀerential form? The answer is positive thanks to the following theorem: Theorem 1 (Hodge). Let M n be an oriented closed Riemannian manifold of k dimension n.Then HDR(M) has ﬁnite dimension and in each cohomological class we have precisely one harmonic k form.

Remark In general, the above theorem can also be seen in the following form: M Λk(M) = Hk(M) Im∆,

where Hk(M) = Ker(∆) has ﬁnite dimension.

Exercise 2. Check that the above decomposition implies the Hodge Theorem. The ﬁrst application is the famous Poincar´eduality:

3 Corollary 1 (PoincaréDuality). Let M n a closed oriented differentiable man- k n−k ifold then we have an isomorphism between HDR(M) and HDR (M). Proof. First remember that any differentiable manifold can be equipped with a Riemannian metric. By Hodge Theorem it is enough to prove that we have a isomorphism between Hk(M) and Hn−k(M). For this we just use the property that the Hodge star operator sends harmonic forms into harmonic forms. Indeed, just recall ?∆ = ∆? Therefore, ? is the isomorphim we are looking for. Definition 2. For each k from 0 to n we define the k − th Betti Number as k bk(M) := dimHDR(M).

By Poincar´eDuality we have bk(M) = bn−k(M) We would like to give another important application of the Hodge Theo- rem in the scenary of Riemannian Geometry. Let’s consider a (M n, g, ∇) a Riemannian manifold with the Levi Civita conection associated to its metric (∇g = 0). Recall the Cuvarture Tensor R which is a tensor (C∞(M) linear in each variable): R : X (M) × X (M) × X (M) → X (M) deﬁned by: R(X,Y,Z) := ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ]Z

R(X,Y,Z,W ) = g(R(X,Y,Z),Z). Note that R(X,Y,Z,W ) = −R(Y,X,Z,W ) and R(X,Y,Z,W ) = R(Z, W, X, Y ).

If we fixe x, y ∈ TpM then Rx,y : TpM → TpM defined by Rx,y(z) = R(x, z, y) is a linear map on TpM, so it makes sense to take its trace which we will denote by Ric(x, y). If we choose a orthonormal basis {e1, . . . , en} at TpM then we can define the Ricci curvature as:

n X Ric(X,Y ) := Trace R(X, ·,Y ) = R(X, ei, Y, ei). i=1 Since the Ricci curvature is symmetric, it deﬁnes on each tangent space an endomorphism which is selfadjoint with respect to the metric g, that we still denote by Ric. In particular, we can extent its action to an endormophism on Γ(T ?(M)) = Λ1(M).

Ric(α)(v) := α(Ric(v)). Now we can state a simple case of a famous identy in Riemannian Geometry, known as the Weitzenb¨ock formula for 1-forms: ∆α = ∇?∇α + Ric(α), where ∇? is the adjoint of ∇ with respect the L2 inner product.

4 Theorem 2. Let (M n, g) is a closed oriented Riemannian manifold with non- negative Ricci curvature then the ﬁrst betti number b1(M) ≤ n. Besides, if Ric is positive then b1(M) = 0. Proof. If α is a harmonic 1-form then

0 = (∆α, α) = (∇α, ∇α) + (Ric(α), α).

Now, Z (Ric(α), α) = g(Ric(α), α)dvg. M But, X g(Ric(α), α) = α(ei)Ric(α)(ei) = i X = α(ei)α(Ric(ei)) = i X ] ] = g(Ric(ei), α )g(α , ei) = i = g(Ric(α]), α]) = Ric(α], α]).

2 R ] ] Therefore, we have |∇α| + M Ric(α , α ) = 0. An immediate consequence is that if Ric > 0 then b1 = 0. If Ric ≥ 0 then we get ∇α = 0, in other words, α is parallel and so its norm is constant and consequently if we have b1 linearly indepedent harmonic 1 forms then in ? each tangent space we will have b1 linearly indepedent vectors on Tp (M) and therefore b1 ≤ n. The above theorem gives a topological obstruction for the existence of met- rics with positive Ricci curvature. For example, T n = Rn/Zn and S1 × S2 do not have a such metric since b1 = n and b1 = 1 respectively.

3 Applications of Hodge Theory on Complex Geometry

For this section we are going to consider a compact complex manifold (M 2n, g, J), where g is a Riemannian metric compatible with the complex structure J. The 1,0 complexiﬁcation of tangent space TxM ⊗ C can be decomposed as Tx (M) ⊕ T 0,1(M), where T 1,0(M) is the vector space generated by { ∂ ,..., ∂ } with x x ∂z1 ∂zn 1,0 Tx0, 1(M) = Tx (M). In the same way, we can express the complexiﬁcation of the space of k forms as k k p 1,0 q 0,1 (A (M))C = A (M) ⊗ C = ⊕[Λ T (M) ⊗ Λ T (M)]. From now on, we use the following notation for the decompsition above Ak = ⊕Ap,q. Recall the natural operators in this setting:

∂ : Ap,q(M) → Ap+1,q(M) ∂ : Ap,q(M) → Ap,q+1(M)

5 The action of ∂ on a particular (p, q)-form ω = ωIJ dzI ∧ dzJ is given by:

∂ IJ I J ∂ω = ω dzk ∧ dz ∧ dz ∂zk

The action of ∂ is analogous. Recal that d = ∂ + ∂. Since d ◦ d = 0, we have ∂ ◦ ∂ = 0, ∂ ◦ ∂ = 0.

Deﬁnition 3. Denoting the space of (p, q)−closed forms wrt ∂ as Zp,q(M) then ∂ the groups Zp,q(M) Hp,q(M) = ∂ ∂ ∂Ap,q−1(M) are called the Dolbeault cohomology groups.

From our Riemannian metric g we can obtain a hermitian metric on (TxM)C by extending this product linearly with respec to C. We will denote this hermitian product by h.

Excercise Check the following properties of h :(TxM)C × (TxM)C → C: 1. h(Z, W ) = h(Z,W ); 2. h(Z, Z) > 0; 3. h(Z,W ) = 0, if Z,W ∈ T 1,0(M).

1,0 There is a natural isomorphism between TxM and Tx (M) given by v 7→ v − iJ(v). Therefore, using this isomorphism we can get the following formula:

h(v − iJv, w + iJw) = 2g(v, w) − 2ig(Jv, w).

That expression just means that we can regain the riemannian metric from the hermitian metric by taking the real part of it. Conversely, for any hermitian product (satisfying the properties 1, 2 and 3) we can produce a riemannian metric on the manifold by taking the real part. In the same way, by taking the imaginary part of a hermitian product we get a 2-form ω ∈ Λ2(M), in our case ω = g(J·, ·). In local coordinates, hij = h(∂zi, ∂zj) = hji. The real 2-form ω(v, w) = g(Jv, w)is called the fundamental form of the complex manifold.

Exercise 3. In local coordinates ω = −ihijdzi ∧ dzj.

n Exercise 4. ω = n!dvolM . Definition 4. We say that a complex manifold (M 2n, g, J) is a Kählermanifold i if the fundamental form ω = − 2 hijdzi ∧ dzj is closed. Proposition 1. If (M, g, J, ω) is a Kählermanifold then [ω] represents a non- 2 i 2i trivial cohomological class in HDR(M). The same is true for [ω ] ∈ HDR(M), i ∈ {1, . . . , n}. Proof. If ωi = dη then ωn = dη ∧ ωn−i = d(η ∧ ωn−i) − (−1)2i−1η ∧ dωn−i = d(η ∧ ωn−i) because ω is closed. Therefore, applying Stokes Theorem we have n!vol(M) = 0 which is a contradiction.

6 The Hodge star operator ? can be naturally extend to a map ? : Ak(M) → A2n−k. It is possible to see that in this way the Hodge star satisﬁes:

α ∧ ?β = h(α, β)dV olM .

We can observe that the hermitian product h is non-zero only if we apply it to forms of type (p, q) and (q, p) respectively. It is an easy exercise to see that

? : Ap,q(M) → An−q,n−p(M).

As we did in the Riemannian case, let’s deﬁne the L2 inner product (·, ·) in the space Ak(M) as Z (α, β) = α ∧ ?β M k p,q Lemma 2. The decomposition A (M) = ⊕{p+q=k}A is orthogonal with respect to (·, ·)L2 . Let’s ﬁrst recall that d? = − ? d? because the dim M = 2n. The next proposition provides the ajoint of the operators ∂ and ∂. ? Proposition 2. Consider ∂? := − ? ∂? and ∂ := − ? ∂?, then

(∂α, β) = (α, ∂?β)

? (∂α, β) = (α, ∂ β) Proof. The idea is again to use the Leibniz rule together with Stokes Theorem again, just remember that d = ∂ when applied to form of type (n, n − 1). The details are left to the reader. The Laplace operators are deﬁning by

? ? ∆d = dd + d d (1) ? ? ∆∂ = ∂ ∂ + ∂∂ (2) ? ? ∆∂ = ∂∂ + ∂ ∂. (3)

p,q Definition 5. Let ω ∈ A (M), we say ω is ∂−harmonic if ∆∂ ω = 0. The spaces of ∂−harmonic (p, q)-form is denoted by Hp,q(M). ∂ Remark I should mention here that I am making an abuse of notation by using Hk(M) for both the space of real k-harmonic forms and the complex k- harmonic forms.To be precise, the second space is just the complexification of the first one. The same remark should be applied to the De Rham cohomology groups in the real and complex case. ? Lemma 3. ω ∈ Ap,q(M) is ∂−harmonic if and only if ∂ω = 0 and ∂ ω = 0. For the sake of clarity, let’s state again the Hodge theorem, Theorem 3 (Hodge). We have the following orthogonal decomposition:

p,q k k−1 ? k+1 A (M) = Hd(M) ⊕ dA (M) ⊕ d A (M) .

7 Corollary 2 (Hodge). In each cohomological class there is only one harmonic k k form. Moreover, there is a natural isomorphism between Hd (M) = Hd(M). Each ∂−harmonic form associates a cohomological cass in the Dolbeault cohomology groups. The technique used to prove Hodge theorem can be used again to prove the following: Theorem 4 (Dolbeault Theorem). We have the following orthogonal decomposition: ? Ap,q(M) = Hp,q(M) ⊕ ∂Ap,q−1(M) ⊕ ∂ Ap,q+1(M). ∂ Corollary 3. In each Dolbault cohomological class there is only one ∂−harmonic form. Moreover, there is a natural isomorphism between Hp,q(M) and Hp,q(M). ∂ ∂ Exercise 5. Let C be a smooth projective curve or equivalently a compact Rie- mann surface of genus g > 0. Prove using Hodge theory that any map P1 → C is constant. Proposition 3 (K¨ahlerIdentity). If (M 2n, g, J, ω) is a K¨ahlermanifold then we have the following identity:

∆d = 2∆∂ = 2∆∂ . The L2 product gives a orthogonal decomoposition for Ak(M)

k p,q A (M) = ⊕p+q=kA (M).

It is clear from our deﬁnitions that the ∆∂ preserves such decomposition. The same is in general not true for the ∆d. However, the above propostion says that this remarkable property holds when we considering a K¨ahlermanifold. In other words,

Hk(M) = ⊕Hp,q(M) = ⊕Hp,q(M) ∂ is a orthogonal decomposition of the space of k-harmonic forms in harmonic forms of type (p, q), p+q = k. By Hp,q(M) we mean Hk(M)∩Ap,q(M). Observe that from the above proposition Hp,q(M) = Hp,q(M). ∂ k p,q The Betti numbers bk(M) = dim HDR(M, R) is equal dimC H (M). Hence, X bk(M) = bp,q(M). p+q=k Because d is a real operator; thus α is harmonic iﬀ α is harmonic. As a consequence, in the K¨ahlercase,

Hp,q(M) = Hq,p.

Therefore, bp,q(M) = bq,p(M). An immediate consequence is the following: Corollary 4. If (M 2n, J, ω) is a compact K¨ahlermanifold, X bk(M) = bp,q(M) (4) p+q=k

bp,q(M) = bq,p(M) (5)

bk(M) is even for odd k. (6)

8 The above corollary is a strong topological obstruction for a complex manifold be K¨ahler.See the next example:

Example 4 (Hopf Manifold). Consider in Cn \{0} the free and discrete Z- action by: n n n · (z1, . . . , zn) = (λ z1, . . . , λ zn), where λ is a complex number of |λ| < 1. The quotient of Cn \{0} by this action is a complex manifold X called the Hopf manifold. One can easily show that X 1 2n−1 1 2n−1 is diffeomorphic to S ×S . In particular, we get b1(X) = b1(S ×S ) = 1. Since the first Betti number of a Kählermanifold is even, a Hopf manidold does not admit a Kählerstructure.