Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 118 / 3

Use the integral normalization of the Chern connection curvature. For a holomorphic local frame $e$, put $h_e=h(e,e)>0$. The fundamental form of a Hermitian holomorphic line bundle isThis follows from the local formula for the Chern connection on a line bundle, $F_h=\bar{\partial }\partial \log h_e$. It is a closed real (1, 1)-form representing the First Chern class. If $h_e=e^{-\phi }$, the formula becomes $(i/(2\pi ))\partial \bar{\partial }\phi$.

The Hermitian metric is positive when this is a positive real (1, 1)-form: for every nonzero $v\in T^{1,0}M$,Equivalently, the matrix $({\partial }^2\phi /\partial z_j\partial {\bar{z}}_k)$ is a positive-definite matrix in every frame. Such a metric makes $\mathcal{L}$ a positive holomorphic line bundle. Omitting the normalization factor $2\pi$ changes none of these positivity conditions.

Apply Serre duality for compact complex manifolds to $\mathcal{L}\otimes K_M$, where $K_M$ is the canonical bundle:We can show the group on the right vanishes directly from the embedding. Since ${\mathcal{L}}^{-1}=i^{\ast }\mathcal{O}(-1)$, the tautological bundle inclusion gives an inclusion of holomorphic vector bundlesA global holomorphic section of ${\mathcal{L}}^{-1}$ consequently determines $r+1$ global holomorphic functions on $M$. Compactness and the maximum modulus principle make these functions constant on every connected component. On such a component the section is therefore a constant vector $v$ lying in the tautological line $i(x)$ for every $x$.

If $v\ne 0$, all the points $i(x)$ equal $[v]$. This contradicts the fact that $i$ is an embedding of a component of positive dimension $n\ge 1$. Hence $v=0$ on every component, and $H^0(M,{\mathcal{L}}^{-1})=0$. Duality proves

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore

For the Calabi-Yau threefold in the stated convention, the Hodge decomposition theorem for compact Kähler manifolds givesIn particular, the Dolbeault theorem yields $H^1(M,{\mathcal{O}}_M)=0$. By Serre duality for compact complex manifolds in dimension three,The canonical bundle is trivial, so the group on the right equals $H^1(M,{\mathcal{O}}_M)=0$. Thus $H^2(M,{\mathcal{O}}_M)=0$, and part (c) gives