I'm aware to varying extents of the existence of certain decompositions of the space of $k$forms on a compact complex or compact Riemannian manifold that split into closed, coclosed, and harmonic forms, and that the space of harmonic forms becomes isomorphic to the de Rham cohomology groups. However, these are defined differently, but there seems to be analogy here in the theorems and to some extent in the proofs; this is also suggested by the common name. Is there one? Or is there a more general way to encapsulate all this?
For any Riemannian (in particular, for any Hermitean) manifold there is the decomposition: all forms are the direct sum of the harmonic ones, exact ones and those in the image of the conjugate operator of the de Rham differential. Every de Rham cohomology class is represented by a unique $d$harmonic form.
For any complex (Hermitean) manifold there is a similar theory for the $\bar\partial$ operator, and we get a similar decomposition for each complex $({\cal E}^{p,\bullet},\bar\partial)$ where ${\cal E}^{p,q}$ stands for complex valued smooth $(p,q)$forms. See e.g. Chern, Complex manifolds. Every Dolbeault cohomology class is represented by a unique $\bar\partial$harmonic form.
For general complex hermitean manifolds the above decompositions have nothing to do with one another. However, if the metric is Kaehler, some miracles happen:
the Laplacians of $d$ and of $\bar\partial$ coincide (more precisely, one is twice the other), so the spaces of harmonic forms coincide as well.
the $(p,q)$projection of a $d$harmonic form is again $d$harmonic. Indeed, a local check shows that the $(p,q)$ projection operator commutes with the $d$Laplacian. So each of the $(p,q)$ components of a $d$harmonic form is closed and none is exact (if nonzero). So the $(p,q)$decomposition of forms gives a $(p,q)$decomposition of the cohomology classes (this does not exist for general complex manifolds). This is the Hodge decomposition.
since the conjugate of a $d$harmonic form is again $d$harmonic, the $(p,q)$ and the $(q,p)$ parts of the Hodge decomposition are conjugate to one another.
As a consequence of the above, the Hodgetode Rham spectral sequence degenerates in the first term. This implies that the $dd^c$lemma holds for Kaehler manifolds, and hence, they are formal. See Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kaehler manifolds.
The cohomological Hodge decomposition holds also for arbitrary bimeromorphically Kaehler compact complex manifolds (in particular, for smooth, complete but not necessarily projective complex algebraic varieties). See e.g. Peters, Steenbrink, Mixed Hodge structures, p. 49.

$\begingroup$ Thanks! Is it always possible to choose a Kahler metric on a complex manifold though in such a way that these two end up working together? $\endgroup$ – Akhil Mathew Jan 11 '10 at 11:49

2$\begingroup$ Not every compact complex manifold admits a Kahler metric. Perhaps the simplest examples are the Hopf surfaces obtained as the quotient of $\mathbb C^2 \setminus \{ (0,0) \}$ by a linear map of the form $(z,w) \mapsto (\alpha z, \beta w)$ where $\alpha$ and $\beta$ are nonzero complex numbers with modulus smaller than one. They are all diffeomorphic to $S^1 \times S^3$ and therefore have zero $H^2$. $\endgroup$ – Jorge Vitório Pereira Jan 11 '10 at 12:52
Hodge theory for Riemannian (real) and Hermitian (complex) compact manifolds are indeed completely analogous. The key tool in both cases is the fundamental theorem for elliptic operators as stated in Voisin's Hodge theory and complex algebraic geometry applied to the respective laplacian. The linked reference builds in parallel the Riemannian and the Hermitian Hodge theory.
Although the theories can be developed in parallel, I don't think there are strong relations between the Riemannian and Hermitian laplacians in general. But in compact Kahler manifolds the three distinct Laplacians satisfy the identities $$ \Delta = 2 \Delta_{\partial} = 2\Delta_{\overline \partial} , $$ which imply the Hodge decomposition $$ H^k(X,\mathbb C) = \bigoplus_{p+q=k} H^{p,q}(X) . $$ See Theorem 6.1 in the reference cited above.

$\begingroup$ Thanks for the response! Do you need a Hermitian metric on a complex manifold to be able to get Hodge theory? I was under the (vague) impression that it was only the complex structure needed to get the decomposition and the Laplacian, and that any metrics were solely an auxiliary tool. $\endgroup$ – Akhil Mathew Jan 11 '10 at 11:52

4$\begingroup$ The Laplacian and the decomposition do depend on the metric. Neverthless some of its consequences, like Serre duality for arbitrary compact complex manifolds and Hodge decomposition for compact Kahler manifolds, are metric independent. $\endgroup$ – Jorge Vitório Pereira Jan 11 '10 at 12:41