Application of Hodge decomposition Hodge decomposition states any $p$ form can be decomposed into three orthogonal $L^2$ components: exact form, co-exact form and hamonic form. But actually we don't know how to decompose a general one. So how to apply the decomposition to other fields? Even if we can decompose one, what does it imply?
 A: I'm not sure this is the sort of thing you're looking for but a nice and quick application is a proof of Poincare duality: On an $n$-dimensional manifold $M$, the Hodge star operator, *, gives an isomorphism between harmonic $k$ forms and $n-k$ forms.  This is because a form $\alpha$ is harmonic if and only if $d \alpha = 0$ and $*d*\alpha = 0$.  Thus $H^k(M;\mathbb R) \simeq H^{n-k}(M;\mathbb R)$.
A: For the interplay of Hodge theory and physics, see various papers of Cantarella, deTurck, and Gluck.
A: If $X$ is a smooth projective variety over $\mathbb C$ (equivalently,
a closed complex submanifold of $\mathbb C P^n$), then the cohomology
of $X$ (with $\mathbb C$ coeficients) has its Hodge decomposition: $H^n(X,
\mathbb C) = \oplus_{p+ q = n} H^{p,q}$; here $H^{p,q}$ consists of class that
can be represented by harmonic $(p,q)$-forms.  
(The decomposition arises by combining Hodge theory as described in the OP
with extra stucture induced by the fact that $X$ is a Kahler manifold.)
One has Hodge symmetry: complex conjugation interchanges $H^{p,q}$
and $H^{q,p}$, and this implies that they have the same dimension.
The Hodge decomposition and Hodge symmery together imply, for example, that if $n$ is odd then the dimension of $H^n(X,\mathbb C)$ is even.  This is a major
topological constraint on the topology of complex projective varieties.
E.g it implies that the Hopf surface $(\mathbb C^2 \setminus \{0\})/ 2^{\mathbb Z}$ (here $2^{\mathbb Z}$ acts by scalar multiplication), which is a compact complex manifold, can't be embedded into projective space.  (Its $H^1$ is one-dimensional.)

Some other example applications:


*

*Lefschetz decomposition

*Torelli theorems

*Many more, but I'm going to post this now
