Positive holomorphic line bundle

A holomorphic line bundle is positive when it admits a Hermitian metric whose Chern connection has positive curvature form $i{F}_{\nabla }$, equivalently when $i{F}_{\nabla }$ is a positive real (1, 1)-form.

The positive-curvature definition uses a positive real (1, 1)-form.

On a compact Kähler manifold $X$, if $P$ is a positive holomorphic line bundle, then $H^i(X,K_X\otimes P)=0$ for $i>0$. The canonical-bundle twist is part of the hypothesis and conclusion; positivity alone does not remove it on a general manifold. For a complex torus, the canonical bundle is trivial, so higher cohomology of a positive line bundle vanishes without an additional twist.

Let $L$ be positive and $V$ a closed smooth hypersurface in a complex torus. The semipositive canonical bundle of a submanifold of a complex torus makes $P_a=(L\otimes \mathcal{O}(aV)){|}_V$ positive. Since $K_V=\mathcal{O}(V){|}_V$, the restriction of the $a$-th ambient twist is $K_V\otimes P_{a-1}$ when $a\ge 1$. Apply the Kodaira vanishing theorem on $V$, and use $0\to L((a-1)V)\to L(aV)\to L(aV){|}_V\to 0$ and its long exact sequence to induct from the positive bundle $L$ on the torus. No unproved ambient positivity of $\mathcal{O}(V)$ is needed.

Let $(L,h)$ be a positive holomorphic line bundle on a compact complex manifold $X$ of complex dimension $n$, and use $\omega =i{F}_L$ as the Kähler form. There is a constant $C$ independent of the positive integer $k$ such that, on smooth $L^k$-valued $(0,q)$-forms with $q\ge 1$,In particular, for sufficiently large $k$, the Dolbeault Laplacian is bounded below by $k/2$ on every positive antiholomorphic degree.

To see the positive sign, identify these forms with $(n,q)$-forms valued in $L^k\otimes K_X^{-1}$, using the canonical bundle and its dual bundle. The Bochner-Kodaira-Nakano identity writes the quadratic form as a nonnegative derivative term plus the curvature commutator. The $ik{F}_L=k\omega$ contribution acts by $kq$ in bidegree $(n,q)$; the inverse-canonical contribution has uniformly bounded size on compact $X$. This gives $kq-C$. Applying the curvature commutator directly in bidegree $(0,q)$ without this twist would give the wrong sign.^[ref]

The estimate rules out nonzero harmonic forms in these degrees. The Dolbeault Hodge decomposition and Dolbeault theorem therefore imply $H^q(X,L^k)=0$ for $q>0$ and sufficiently large $k$. The additive constant matters for small powers; the linear spectral gap is an eventual statement.

The Kodaira embedding theorem says that a compact complex manifold carrying a positive holomorphic line bundle is projective: sufficiently large tensor powers of the bundle have global holomorphic sections defining a holomorphic embedding into Complex projective space. Conversely, the restriction of $\mathcal{O}(1)$ gives a positive line bundle on every projective complex manifold.