Kodaira–Nakano vanishing theorem

Remarks

The theorem is also called the Akizuki–Nakano vanishing theorem. A compact manifold with a positive line bundle is Kähler, with Kähler form $\omega = i\Theta_h$ given by the curvature of a positive metric $h$. By Dolbeault's theorem, a class in $H^q(X, \Omega^p_X \otimes L)$ is represented by an $L$-valued harmonic $(p,q)$-form $u$. The Bochner–Kodaira–Nakano identity gives $\langle [i\Theta_h, \Lambda] u, u \rangle \le 0$ for such $u$, and with $\omega = i\Theta_h$ the operator $[i\Theta_h, \Lambda]$ is multiplication by $p + q - n$ on $(p,q)$-forms. So $(p + q - n)\norm{u}^2 \le 0$, which forces $u = 0$ when $p + q > n$.

The case $p = n$ is the Kodaira vanishing theorem: $\Omega^n_X = K_X$ and $H^q(X, K_X \otimes L) = 0$ for $q \ge 1$. By Serre duality the theorem is equivalent to $H^q(X, \Omega^p_X \otimes L^{-1}) = 0$ for $p + q < n$.

Positivity of $L$ cannot be dropped. For $X = \mathbb{P}^1$, $n = 1$, $p = q = 1$ and $L$ trivial, one has $H^1(\mathbb{P}^1, \Omega^1) \cong \C \ne 0$, though $p + q = 2 > 1$.

Going deeper

For a student

The proof is a model of the Bochner technique: a curvature identity shows that harmonic representatives must vanish. It extends Kodaira vanishing, which is the case $p = n$, to all bundles of holomorphic forms. It is used to prove the Lefschetz hyperplane theorem for cohomology and to control cohomology of $\Omega^p$ on projective manifolds. For vector bundles the analogous result needs Nakano positivity, not just Griffiths positivity.