Regarding Hodge's theorem

Question

We have $*$ the Hodge operator, and $d $ the exterior derivative. We define $\delta=\pm *d*$ and $\triangle=d\delta+\delta d $. Warner (pp. 223) says that we have $$ \triangle (E^p (M))=d\delta (E^p (M))\oplus \delta d (E^p (M))=d (E^{p-1}(M))\oplus\delta (E^{p+1}(M)) $$ I understand why the first space is a subspace of the second, and why the second is of the third. My question is why are there inverse inclusions?

Answer 1

Warner actually says more than that, and the part you omitted is the whole point.

If I remember correctly, this is what Warner calls the Hodge decomposition theorem:$$\begin{align}E^p(M)&=\Delta(E^p(M))\oplus\mathcal{H}^p(M)\\&=d\delta(E^p(M))\oplus\delta d(E^p(M))\oplus\mathcal{H}^p(M)\\&=d(E^{p-1}(M))\oplus \delta(E^{p+1}(M))\oplus\mathcal{H}^p(M).\end{align}$$

Now here is the explanation. Let us start with the last expression. The fact that the different summands are orthogonal to one another follows from the formal adjointness of $d$ and $\delta$, combined with the fact that a $p$-form is harmonic if and only if it is closed and coclosed. No analysis is needed here. However, this only gives the inclusion $$\delta(E^{p+1}(M))\oplus d(E^{p-1}(M))\subset\left(\mathcal{H}^p(M)\right)^\perp.$$ But the first line of the theorem (which is, more or less, the crucial one) reads $$\Delta(E^p(M))=\left(\mathcal{H}^p(M)\right)^\perp,$$ which provides the inclusion you want.

Answer 2

Original answer:

Since you know that "the first space is a subspace of the second, and why the second is of the third", it's sufficient to show that the first space is equal to the third. As in Hodge theory, we assume manifold M is compact and oriented.

Set $\mathbb{H}^k(M)=\{\omega \in E^k(M): \Delta \omega =0\}$, which is the space of all harmonic k-forms on M. First thing you should have known is that $\Delta \omega=0$ if and only if $d \omega = \delta \omega =0$, since $(\Delta \omega,\omega)=((d\delta+\delta d)\omega,\omega)=(d\omega,d\omega)+(\delta\omega,\delta\omega)=0$, and both the latter are non-negative.

Now for any $\omega \in \mathbb{H}^k,\ \eta \in E^{k-1}, \ \theta \in E^{k+1}$, we have$$(\omega,d\eta)=(\delta\omega,\eta)=0=(d\omega,\theta)=(\omega,\delta\theta),\ since\ d \omega = \delta \omega =0,\ and\ (d\eta,\delta\theta)=(d^2\eta,\theta)=0.$$ So these three space are tangent to each other. Take a $\omega$ orthogonal to $dE^{k-1} \oplus\delta E^{k+1}$, then $(\delta\omega,\eta)=(\omega,d\eta)=0,\ which\ implys\ \delta\omega=0$, and for the same reason $d\omega=0$, so its in $\mathbb{H}^k$.

It's sufficient to show $E^k=\mathbb{H}^k \oplus dE^{k-1} \oplus\delta E^{k+1}$, which is well-known as Hodge decomposition. Briefly, since $\mathbb{H}^k$ is a linear subspace of the infinite-dimension space $E^k$, there's a projection $\pi :E^k \to \mathbb{H}^k$. Theorem from PDE tells us the equation $\Delta \eta =\omega_0$, which is a ellipitic PDE of 2th order, must have a solution. (I read the details of this claim in a Chinese textbook so I cannot give you a reference, but I think it's easy to find in any PDE textbooks.)

We assume $\eta_0$ is the solution to $\Delta\eta= \omega -\pi(\omega)$, for an arbitary $\omega \in E^k$, and we can modify it, by taking $\eta_1=\eta_0-\pi(\eta_0)$. It's easy to show $\eta_1$ is another solution to the equation, as $\Delta\pi=\pi\Delta=0$. From disscusion above we find the Green's operator $$G:E^k \to (\mathbb{H}^k)^{\bot}=dE^{k-1} \oplus\delta E^{k+1}, \omega \mapsto \eta_1.$$ Then Hodge decomposition is given by $$\omega=\pi(\omega)+d\delta(G\omega)+\delta d(G\omega).$$