LECTURE 25: THE HODGE LAPLACIAN 1. The Hodge star operator Let (M; g) be an oriented Riemannian manifold of dimension m. Then in lecture 3 we have seen that for any orientation-preserving chart, the Riemannian volume form (which is independent of the choice of coordinates) is given by p 1 m !g = Gdx ^ · · · ^ dx : Now let p 2 M. Then the Riemannian metric g induces a dual inner product ∗ structure on Tp M via i j ij h!idx ; ηjdx i = g !iηj: k ∗ More generally, one can define an inner product on Λ Tp M as follows: For any 1 m ∗ orthogonal basis θ ; ··· ; θ of Tp M, we require the set i1 ik fθ ^ · · · ^ θ j i1 < ··· < ikg k ∗ form an orthonormal basis of Λ Tp M. One can check that this definition is inde- pendent of the choice of θi's. Note that in particular we have h!g;!gi = 1 since in normal coordinates (gij) = I at p. As in the case of functions, the pointwise inner product induces an L2 inner k product structure on Ωc (M) via Z (!; η) := h!; ηi!g: M To define the Hodge-Laplacian of a differential form, one need to define the so-called Hodge star operator. We first use the pointwise inner product to get an k ∗ k ∗ ∗ k ∗ identification between Λ Tp M and (Λ Tp M) that sends β 2 Λ Tp M to k ∗ m ∗ Lβ :Λ Tp M ! R = Λ Tp M; α 7! hα; βi!g: On the other hand, the wedge product gives us a non-degenerate pairing k ∗ m−k ∗ m ∗ ^ :Λ Tp M × Λ Tp M ! R = Λ Tp M; (α; β) 7! α ^ β: k ∗ ∗ m−k ∗ which identifies any element in (Λ Tp M) as an element in Λ Tp M. In particular, k ∗ m−k ∗ for β 2 Λ Tp M one can get an element ?β 2 Λ Tp M that is identified with Lβ, i.e. α ^ ?β = hα; βi!g: 1 2 LECTURE 25: THE HODGE LAPLACIAN This construction gives us a linear isomorphism k ∗ m−k ∗ ? :Λ Tp M ! Λ Tp M at each point. Glue these constructions together, we are able to define Definition 1.1. The Hodge star operator ? :Ωk(M) ! Ωm−k(M) maps any k-form η 2 Ωk(M) to the (m − k)-form ?η 2 Ωm−k(M) so that for any ! 2 Ωk(M), ! ^ ?η = h!; ηi!g: Remark. Obviously ? is C1(M)-linear. 2 k Remark. The L inner product structure on Ωc (M) can be written as Z (!; η) = ! ^ ?η: M Note that by definition, ?1 = !g; ?!g = 1;! ^ ?η = η ^ ?!: More generally, if we let !1; ··· ;!m be a (local) basis of T ∗M so that 1 m ! ^ · · · ^ ! = f!g is a positive m-form, then for i1 < ··· < ik, if we let j1 < ··· < jm−k be the complement indeces of i's, i.e., such that fi1; ··· ; ik; j1; ··· ; jm−kg = f1; ··· ; mg; we have h!; !i ?(!i1 ^ · · · ^ !ik ) = ± !j1 ^ · · · ^ !jm−k ; f where we denoted ! = !i1 ^ · · · ^ !ik , and the sign ± is chosen so that ! ^ ?! = h!; !i!g; i.e. so that !i1 ^ · · · ^ !ik ^ !j1 ^ · · · ^ !jm−k = ±!1 ^ · · · ^ !m: One can check that in this case ± = (−1)i1+···+ik+1+···+k: Lemma 1.2. For ! 2 Ωk(M), ??! = (−1)k(m−k)!: Proof. By C1(M)-linearity, we may assume without loss of generality that ! = !i1 ^ · · · ^ !ik ; where !1; ··· ;!m is an orthonormal basis at one point p. Then the above compu- tations show ?! = (−1)i1+···+ik+1+···+k!j1 ^ · · · ^ !jm−k LECTURE 25: THE HODGE LAPLACIAN 3 and thus ??! = (−1)i1+···+ik+1+···+k(−1)j1+···+jm−k+1+···+m−k!: It remains to check the following elementary identity m(m + 1) k(k + 1) (m − k)(m − k + 1) + + ≡ k(m − k) (mod2): 2 2 2 As a consequence, we see ? is in fact a linear isometry: Corollary 1.3. For any !; η 2 Ωk(M), one has h?!; ?ηi = h!; ηi: Proof. We have k(m−k) h?!; ?ηi!g = (?!) ^ ?(?η) = (−1) (?!) ^ η = η ^ ?! = hη; !i!g: So h?!; ?ηi = hη; !i = h!; ηi. Remark. The Hodge star operator is of particular important in dimension 4. In fact, 2 ∗ 2 ∗ for m = 4 and k = 2, the linear map ? :Λ Tp M ! Λ Tp M satisfies ?2 = I: So one can decompose (according to eigenvalues of ?) 2 ∗ 2 ∗ 2 ∗ Λ Tp M = Λ+Tp M ⊕ Λ−Tp M: 2 ∗ 2 ∗ Sections of Λ+T M are called self-dual 2-forms, while sections of Λ−T M are called anti-self-dual 2-forms. 2. The Hodge-Laplace operator Using the Hodge star operator, one can define Definition 2.1. The co-differential of ! 2 Ωk(M) is δ! 2 Ωk−1(M) defined by δ! = (−1)km+m+1 ? d ? : k 2 The next lemma states that when we endow all Ωc (M)'s with this L structure, k k−1 the co-differential operator δ :Ωc (M) ! Ωc (M) is the adjoint of the differential k−1 k operator d :Ωc (M) ! Ωc (M). k k−1 Lemma 2.2. For any ! 2 Ωc (M) and η 2 Ωc (M), (!; dη) = (δ!; η): Proof. By Stokes' theorem, we have Z Z Z (!; dη) = (dη; !) = dη^?! = d(η^?!)−(−1)k−1η^d?! = (−1)k η^d?! M M M 4 LECTURE 25: THE HODGE LAPLACIAN On the other hand, by lemma 1.2, Z Z ((−1)km+m+1?d?!; η) = η^(−1)m(k+1)+1??d?! = (−1)km+m+1(−1)(m−k+1)(k−1) η^d?!; M M km+m+1 (m−k+1)(k−1) k so the conclusion follows from the fact (−1) (−1) = (−1) . The following formula will be useful. i Proposition 2.3. Let feig be an orthonormal frame and f! g the dual frame. Let r be the Levi-Civita connection. Then i (1) d = ! ^ re . P i (2) δ = − j ιej rej . Proof. (1) One can check that the right hand side is independent of choice of basis. So at each point p that is fixed, with out loss of generality one may take ei = @i to be the coordinate vector field for a normal coordinate system centered at p. The dual basis is then dxi. Recall that by definition, at the point p one has, for any i; j; k, j j j (r@i dx )(@k) = r@i (dx (@k)) − dx (r@i @j) = 0: j So at p one has r@i dx = 0 for any i; j. Now we denote ¯ i i d = ! ^ rei = dx ^ r@i : Consider η = fdxi1 ^ · · · ^ dxik . Then at p, @f dη¯ = dxi ^ dxi1 ^ · · · ^ dxik = dη: @xi This implies d = d¯. (2) The proof is similar. We denote ¯ X X δ = − ιej rej = − ι@j r@j : j j Then at p, X ¯ j−1 i1 ij ik δη = − (−1) (@ij f)dx ^ · · · ^ dxd ^ · · · ^ dx : j On the other hand, by the definition of δ one can calculate δη and prove that at p, X j i1 ij ik δη = (−1) (@ij f)dx ^ · · · ^ dxd ^ · · · ^ dx : j This completes the proof. Definition 2.4. The Hodge-Laplace operator on k-forms is ∆ = dδ + δd :Ωk(M) ! Ωk(M): LECTURE 25: THE HODGE LAPLACIAN 5 Remark. Since d2 = 0;?2 = ±1, we immediately get δ2 = 0: As a consequence, ∆ = (d + δ)2: Example. One can check that when k = 0, the operator ∆ = δd equals with the Laplace-Beltrami operator ∆ that we defined in lecture 3. To see this, again we do computation in normal coordinates. Then for any f 2 C1(M) we have j df = (@jf)dx and thus X j−1 1 j m ?df = (@jf)(−1) dx ^ · · · ^ dxc ^ · · · ^ dx ; which implies X 1 m d ? df = @j(@jf)dx ^ · · · ^ dx : It follows X ∆f = δdf = − ? d ? df = − @j(@jf); which is exactly the Laplace-Beltrami operator we defined in lecture 3 (but now calculated in normal coordinates). One can also see this by applying proposition 2.3: 2 ∆f = δdf = −ιej rej df = −tr(r f): Like the Beltrami-Laplacian, the Hodge-Laplacian also have very nice proposi- tions: Proposition 2.5. We have (1)( !; ∆η) = (∆!; η), i.e. ∆ is symmetric. (2) (∆!; !) = jδ!j2 + jd!j2 ≥ 0, i.e. ∆ is non-negative. (3) ?∆ = ∆?. k Proof. By lemma 2.2, for any !; η 2 Ωc (M), (!; ∆η) = (!; dδη) + (!; δdη) = (δ!; δη) + (d!; dη): Both (1) and (2) follows. To prove (3), we let ! be any k-form. Then ?δ! = (−1)km+m+1 ? ?d ? ! = (−1)km+m+1(−1)(m−k+1)(k−1)d ? ! = (−1)kd ? !: Similarly δ ? ! = (−1)(m−k)m+m+1 ? d ? ?! = (−1)(m−k)m+m+1(−1)k(m−k) ? d! = (−1)k+1 ? d!: So we get ?dδ! = (−1)kδ ? δ! = δd ? ! 6 LECTURE 25: THE HODGE LAPLACIAN and ?δd! = (−1)k+1d ? d! = dδ ? !: It follows ?∆ = ?dδ + ?δd = δd ? ! + dδ? = ∆ ?: In problem set 1 we have seen that if M is connected, then ∆f = 0 if and only if f is a constant function. Corollary 2.6. ∆(f!g) = 0 if and only if f is a constant function. Proof. This follows from ∆(f!g) = ∆ ? f = ?∆f = (∆f)!g: Definition 2.7. A k-form ! is called harmonic if ∆! = 0. We will denote the set of all harmonic k-forms on (M; g) by Hk(M). It is obviously a vector space. Obviously if M is connected, 0 m H (M) ' R; H (M) ' R: According to proposition 2.3, if ! 2 Ωk(M) is parallel, i.e.