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 fe1; : : : ; eng for the tangent space TxM at x 2 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 ! 2 Λ (M), ! = ! dxi, then j−1 ij i ^ ?! = (−1) g ! 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 2 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 prop- erty of d is that d ◦ d = 0. We say ! 2 Λk(M) is closed if d! = 0 and when ! = dη then ! 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): ! is called harmonic if ∆(!) = 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) = jdαj2 + jdαj2 2 Example 2. If k = 0 then ∆ = d?d and in local coordinates we have for any f 2 C1(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 ! = ! dxI then n X @2!I ∆ ! = − 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 ! is harmonic and exact then j!j2 = (!; !) = (!; dη) = (d?!; η) = 0 which implies that ! = 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 ! is a closed smooth k form and has minimal norm in its cohomological class then d?! = 0. Indeed, let η 2 Λk−1(M) @ 0 = j! + tdηj2j @t t=0 @ 0 = (! + tdη; ! + tdη) = 2(!; dη) @t This is equivalent to (d?!; η) = 0 for all η 2 Λk−1(M) ie, d?! = 0. Conversely, if ! is closed and d?! = 0 (i.e., ! is harmonic) then ! has minimal norm. It is enough to prove that j! + dηj2 > j!j2: j! + dηj2 = (! + dη; ! + dη) = j!j2 + jdηj2 + 2(!; 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 cohomo- logical 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 (Poincar´eDuality). 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; r) a Riemannian manifold with the Levi Civita conection associated to its metric (rg = 0). Recall the Cuvarture Tensor R which is a tensor (C1(M) linear in each variable): R : X (M) × X (M) × X (M) !X (M) defined by: R(X; Y; Z) := rX rY Z − rY rX Z − r[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 2 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 fe1; : : : ; eng 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 = r?rα + Ric(α); where r? is the adjoint of r 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) = (rα; rα) + (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 jrαj + M Ric(α ; α ) = 0. An immediate consequence is that if Ric > 0 then b1 = 0. If Ric ≥ 0 then we get rα = 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.