TITLE: Regarding Hodge's theorem
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.