Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 118 (Complex Manifolds)

Let $(z^1,\dots ,z^n)$ be a complex manifold chart, with $z^j=x^j+i{y}^j$. The almost complex structure induced by a complex atlas isThus $J^2=-I$. On an overlap, the derivative of a holomorphic map is complex linear and hence commutes with multiplication by $i$. The two coordinate definitions of $J$ therefore agree, so $J$ is globally well-defined.

The complexified cotangent bundle splits into the $i$- and $-i$-eigenspaces of $J^{\ast }$, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a sumSplitting the exterior derivative according to type definesComplex conjugation sends $d{z}^j$ to $d{\bar{z}}^j$ and conjugates the coefficient derivatives. Term by term this gives the complex conjugation of differential-form type identity

Let $J$ act on a $k$-form byIt acts on a $(p,q)$-form by $i^{p-q}$. Therefore $J^{-1}dJ$ multiplies the $\partial$ component by $-i$ and the $\bar{\partial }$ component by $i$, proving the d c operator formula

A holomorphic vector field is a holomorphic section of $T^{1,0}X$. On the affine chart $U_0\subset {\mathbb{CP}}^n$, put $w_k=z_k/z_0$. The projection isDirect differentiation givesIf $\xi (z)$ is linear and homogeneous, the pushed-forward coefficients are consequently $\xi (1,w)$ when $j\ne 0$, and $-\xi (1,w)w_k$ when $j=0$. They are holomorphic on $U_0$.

More intrinsically, $\xi (z){\partial }_{z_j}$ is a linear vector field $Az$ on ${\mathbb{C}}^{n+1}$. Its flow $e^{tA}$ preserves complex lines and induces the projective transformationsDifferentiating gives a globally defined projectivization of a linear vector field, whose expression on $U_0$ is the field just computed. This proves extension across the other affine charts.

Finally choose distinct complex numbers ${\lambda }_0,\dots ,{\lambda }_n$ and the diagonal fieldIts projectivization vanishes at $[z]$ precisely when $({\lambda }_0{z}_0,\dots ,{\lambda }_n{z}_n)$ is proportional to $z$, so $[z]$ is an eigenline. The distinct eigenvalues leave exactly the $n+1$ coordinate points. Hence ${\mathbb{CP}}^n$ has a holomorphic vector field with finitely many zeroes.

An irreducible complex analytic hypersurface in a complex manifold $X$ is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface $Y$ at $x$ is a holomorphic function $f$ on a neighbourhood $U$ such thatThe necessary local algebra is that the stalk ${\mathcal{O}}_{X,x}$ is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives $f$, and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.

A divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{n}_{Y}Y$ of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small $U_{\alpha }$, local defining functions give a meromorphic equation $f_{\alpha }$ for $D$. The quotients $f_{\alpha }/f_{\beta }$ are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor $[D]$, and $f_{\alpha }{e}_{\alpha }$ gives its canonical meromorphic section with divisor $D$.

The Euler sequence on complex projective spaceimplies $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$. Taking the dual determinant yields the canonical bundle of complex projective spacewhere $H$ is a hyperplane divisor.

The hypotheses on the homogeneous polynomial $p$ say thatis a smooth projective hypersurface of degree $k$, so its divisor line bundle is $[V]\cong \mathcal{O}(kH)$. The Adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.

Now fix an isomorphism $\Phi :[P]\to [Q]$ and regard $s_P$ and ${\Phi }^{-1}{s}_Q$ as holomorphic sections of the same holomorphic line bundle. They have no common zero because $P\ne Q$. Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor $(f)=P-Q$. Thus $F^{-1}(0)=P$ and $F^{-1}(\infty )=Q$, both with multiplicity one. The degree of a holomorphic map $F$ is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.

A real (1, 1)-form is a form $\phi \in {\Omega }^{1,1}(X)$ satisfying $\bar{\phi }=\phi$. In holomorphic coordinates it has the formIt is a positive real (1, 1)-form whenfor every nonzero tangent vector $\xi$ of type $(1,0)$, equivalently when the Hermitian matrix $(h_{j\bar{k}})$ is positive definite.

A holomorphic local trivialization of a holomorphic line bundle $L\to X$ is equivalently a nowhere-zero holomorphic local frame $e$. A connection $A$ is unitary when it preserves the fiberwise Hermitian inner product:The Chern connection is the unique unitary connection whose $(0,1)$ part is the bundle's Dolbeault partial connection ${\bar{\partial }}_L$.

In a holomorphic frame, put $h=h(e,e)$. The local formula for the Chern connection on a line bundle isThe curvature therefore has type $(1,1)$. Since a unitary connection has imaginary curvature, $\overline{F(A)}=-F(A)$, and henceThus $iF(A)$ is a real $(1,1)$-form.

For connections $A$ on $L$ and $\hat{A}$ on $\hat{L}$, the tensor product connection is defined on decomposable local sections byThe curvature of a tensor product connection on line bundles is additive:Consequentlywhich is positive whenever both summands are positive.

For the final assertion, simultaneously diagonalize the positive Hermitian matrices of $\phi$ and $\psi$ by congruence at the chosen point. In the resulting coframe,A direct wedge-product calculation givesIf $\xi$ and $\eta$ are linearly independent, at least one of these $2\times 2$ minors is nonzero. Every coefficient is positive, so the sum is strictly positive. This is wedge positivity for two positive (1, 1)-forms.

The Kähler manifold structure gives a Riemannian metric, its volume form, the complex orientation, and a Hermitian inner product on complex differential forms. The complex Hodge star operator is the complex-linear map characterized byOn $r$-forms in real dimension $m=2n$, the codifferential isequivalently the formal adjoint of $d$ for the $L^2$ inner product. Similarly ${\bar{\partial }}^{\ast }$ is the formal adjoint of $\bar{\partial }$. Define the Hodge Laplacian and Dolbeault Laplacian by

Expanding $d=\partial +\bar{\partial }$, the Kähler identities make the mixed anticommutators vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Hence the Kähler Laplacian identity is

Let $L\gamma =\omega \wedge \gamma$ be the Lefschetz operator of a Kähler manifold. The Kähler identities also implyThus, if ${\Delta }_{\bar{\partial }}\alpha =0$, thenThis is the fact that the Lefschetz operator preserves harmonic forms.

The Dolbeault Hodge decomposition on a compact Hermitian manifold states thatan orthogonal direct sum, where ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$.

Suppose $\eta =\bar{\partial }\gamma$ has type $(p,q)$. Apply this decomposition to $\gamma \in {\Omega }^{p,q-1}$. The harmonic and $\bar{\partial }$-exact pieces disappear after applying $\bar{\partial }$, so for some $\beta \in {\Omega }^{p,q}$,Put $\theta ={\bar{\partial }}^{\ast }\beta$. If also $\partial \eta =0$, thenThe Kähler anticommutation identity ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$ and $({\bar{\partial }}^{\ast }{)}^2=0$ giveTherefore $\partial \theta$ is ${\Delta }_{\bar{\partial }}$-harmonic. By ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$ it is also $\partial$-harmonic, but it is $\partial$-exact; orthogonality of harmonic and exact forms forcesThis proves both requested claims: $\partial {\bar{\partial }}^{\ast }\beta =0$ is harmonic, and ${\bar{\partial }}^{\ast }\beta =\theta$ is $\partial$-closed.

Finally, $\theta \in \mathrm{im}{\bar{\partial }}^{\ast }$ is orthogonal to $\ker \bar{\partial }$, and hence to every $\bar{\partial }$-harmonic form. Since the $\partial$- and $\bar{\partial }$-harmonic spaces agree on a compact Kähler manifold, the $\partial$-closed form $\theta$ has zero harmonic component in its $\partial$-Hodge decomposition. It follows that $\theta =\partial \varphi$ for some $\varphi \in {\Omega }^{p-1,q-1}(X)$. Hencewhich is the ddbar lemma in this case.