---- > [!theorem] Theorem. ([[Hodge decomposition]]) > Let $(M,g)$ be a [[compact]] [[orientation of a Euclidean submanifold|oriented]] [[Riemannian manifold|Riemannian]] [[smooth manifold|manifold]] of dimension $n$. Let $0 \leq p \leq n$, and write $\mathcal{H}^{p}(M)$ for the [[Laplace-Beltrami operator|space of harmonic]] $p$-[[differential form|forms]] on $M$. Write $d:\Omega^{p}(M) \to\Omega^{p+1}(M)$ for the $p$th[[exterior derivative|exterior differential]], and $\delta=d^{*}:\Omega^{p-1}(M) \to \Omega^{p}(M)$ for its [[adjoint]], the $p$th [[Hodge star|Hodge codifferential]]. > > Then: > 1. $\text{dim }\mathcal{\mathcal{H}^{p}}(M) < \infty$ > 2. One has decompositions $\begin{align} > \Omega^{p}(M) &= \mathcal{H}^{p}(M) \oplus \operatorname{im }(\Delta: \Omega^{p}(M) \to \Omega^{p}(M)) \\ > &= \mathcal{H}^{p}(M) \oplus \operatorname{im } d\delta \oplus \operatorname{im } \delta d \\ > &= \underbrace{ \mathcal{H}^{p}(M) }_{ \text{harmonic} } \oplus \underbrace{ \operatorname{im }d }_{ \text{exact} } \oplus \underbrace{ \operatorname{im } \delta }_{ \text{coexact} } > \end{align}$ > with the direct summands [[L2 inner product of differential forms|L2]]-[[orthogonal direct sum of modules|orthogonal]]. > [!proposition] Corollary. > > Write $H_{\text{dR}}^{*}(M)$ for the [[de Rham cohomology]] of $M$. > > Each $a \in H^{p}_{\text{dR}}=\frac{\operatorname{ker }d^{p}}{\operatorname{im }d^{p-1}=}$ has a unique harmonic representative $\alpha \in \mathcal{H}^{p}(M)$, $[\alpha]=a$, making the map $\begin{align} > \mathcal{H}^{p}(M) & \xrightarrow{\cong} H^{p}_{\text{dR}}(M) \\ > \alpha &\mapsto [\alpha] \\ > \alpha& \leftarrow_|a > \end{align}$ > an [[isomorphism]]. > [!proof] Proof of corollary. > **Uniqueness.** Suppose $\alpha_{1},\alpha_{2}$ are both harmonic representatives for the same [[(co)homology of a complex|cohomology class]], $\alpha_{1},\alpha_{2} \in\mathcal{H}^{p}(M)$, $[\alpha_{2}]=[\alpha_{1}]$. Then $\alpha_{1}-\alpha_{2} \in \mathcal{H}^{p}(M)$ is harmonic also. But since $[\alpha_{2}-\alpha_{1}]=0$, we have $\alpha_{2}-\alpha_{1} \in \operatorname{im }d^{p-1}$. By the Hodge decomposition, $\operatorname{im }d^{p-1}$ belongs to the [[orthogonal complement]] of $\mathcal{H}^{p}(M)$, so this means $\alpha_{2}-\alpha_{1}=0$, i.e., $\alpha_{1}=\alpha_{2}$. We can also prove this easily without invoking the Hodge decomposition: write $\alpha_{2}-\alpha_{1}=d\beta$. Then using adjointness of $d$ and $\delta$: $\|d \beta\|^{2}=\langle \langle d \beta, d \beta \rangle \rangle = \langle \langle \beta, \delta d \beta \rangle \rangle =\langle \langle \beta, \delta \alpha_{2} \rangle \rangle - \langle \langle \beta, \delta \alpha_{1} \rangle \rangle $ > but $\delta \alpha_{1}$ and $\delta \alpha_{2}$ are zero because [[Laplace-Beltrami operator|harmonic forms are coclosed]]. So $d\beta=\alpha_{2}-\alpha_{1}=0$, i.e., $\alpha_{2}=\alpha_{1}$. > > **Existence.** Here there is no getting around using the decomposition. Suppose we are given a [[de Rham cohomology]] class $a \in H^{p}_{\text{dR}}(M)$, say, $a=[\widetilde{\alpha}]$ with $\widetilde{\alpha} \in \Omega^{p}(M)$ and $d\widetilde{\alpha}=0$, and we want to represent $a$ with a harmonic form. Using the Hodge decomposition, write $\widetilde{\alpha}=\alpha+d \beta+\delta \gamma$ with $\alpha \in \mathcal{H}^{p}(M)$ harmonic (closed + coclosed). We claim that $\delta \gamma=0$, meaning that $\widetilde{\alpha}$ and $\alpha$ are represented by the same cohomology class because their difference is a boundary. Indeed, $0=d \widetilde{\alpha}= \cancel{ d \alpha } ^{\alpha \text{ closed}}+ \cancel{ d ^{2} \beta }^{0}+ d \delta \gamma,$ > so $\delta\gamma$ is [[closed form|closed]]. Now, $0=\langle \langle d \delta \gamma, \gamma \rangle \rangle = \langle \langle \delta \gamma, \delta \gamma \rangle \rangle =\|\delta \gamma\|^{2} ,$ > so $\delta \gamma=0$, i.e., $\gamma$ is coclosed. > [!proof]- Proof. ([[Hodge decomposition]]) > Consider A. Kovalev's essay: it will include this proof and more! ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```