Kodaira embedding theorem

Remarks

Proof idea: choose a basis $s_0, \ldots, s_N$ of the space of holomorphic sections of $L^{\otimes k}$ and consider $x \mapsto [s_0(x) : \cdots : s_N(x)]$. This is an embedding as soon as the sections separate points and tangent vectors. To get sections with prescribed values at one or two points, one blows up these points. Then the Kodaira vanishing theorem on the blow-up kills the obstruction in $H^1$; for large $k$ the needed positivity holds.

Equivalent form: a compact complex manifold is projective if and only if it has a Kähler form $\omega$ with a rational cohomology class. Rational means that $[\omega]$ lies in the image of $H^2(X; \Q)$; after scaling, one may take it integral. Such $X$ are called Hodge manifolds. Then $L$ is the line bundle with first Chern class $[\omega]$. By Chow's theorem, $X$ is then the zero set of homogeneous polynomials in $\mathbb{CP}^N$.

Example: on $\mathbb{CP}^n$ the hyperplane bundle $\mathcal{O}(1)$ is positive, with the Fubini–Study metric, and already $k = 1$ gives the identity embedding.

The positivity hypothesis cannot be dropped. A generic complex torus $\C^2 / \Lambda$ is Kähler, but its Kähler classes are not rational. It has no positive line bundle and it is not projective; it even has no nonconstant meromorphic functions.

Going deeper

For a student

The theorem is the bridge between complex differential geometry and algebraic geometry: a curvature condition implies an algebraic conclusion. Combined with the theory of Kähler manifolds it says that projective manifolds are exactly the compact Kähler manifolds with an integral Kähler class. A positive line bundle is thus ample, and the converse also holds. It shows, for instance, that every compact Riemann surface is a projective algebraic curve. Via the Riemann bilinear relations, it also tells which complex tori are abelian varieties.