Hodge decomposition

Remarks

The idea of the proof: by the Hodge theorem, each class in de Rham cohomology has a unique harmonic representative for the Laplacian $\Delta_d$. The Kähler identities give $\Delta_d = 2\Delta_{\bar\partial}$, so $\Delta_d$ preserves the type $(p, q)$ of forms, and the harmonic part of a form splits by type. The $\Delta_{\bar\partial}$-harmonic $(p,q)$-forms represent Dolbeault cohomology, which gives part 3; part 2 holds because $\Delta_d$ is a real operator.

Example: for a compact Riemann surface of genus $g$, $H^1(X, \C) = H^{1,0} \oplus H^{0,1}$ with $H^{1,0}$ the $g$-dimensional space of holomorphic $1$-forms, so $b_1 = 2g$.

Consequence: the Hodge numbers $h^{p,q} = \dim H^{p,q}$ satisfy $h^{p,q} = h^{q,p}$, so every odd Betti number $b_{2m+1} = \sum_{p+q = 2m+1} h^{p,q}$ is even. The Kähler hypothesis is needed: the Hopf surface, diffeomorphic to $S^1 \times S^3$, is a compact complex manifold with $b_1 = 1$, so no such decomposition with symmetry exists on it.

Going deeper

For a student

The decomposition makes $H^k(X, \Z)$ modulo torsion into a Hodge structure of weight $k$, which is the starting point of period maps and of the Torelli problem. Combined with the hard Lefschetz theorem and the Hodge–Riemann bilinear relations, it gives polarized Hodge structures. It also links topology to holomorphic invariants: for example $h^{1,0}$, the number of independent holomorphic $1$-forms, equals half the first Betti number.