Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 15 (Complex Manifolds)

At a point $[X]\in {\mathbf{P}}^n$, the complex tautological line bundle has fiber $\mathbb{C}X\subseteq {\mathbb{C}}^{n+1}$. The hyperplane line bundle is its dual, $[H]=\mathcal{O}(1)$, and $\mathcal{O}(m)=\mathcal{O}(1{)}^{\otimes m}$. On $U_i=\{X_i\ne 0\}$ let $t_i=X/X_i$ be the tautological frame and $e_i$ its dual. On overlaps $e_j=(X_j/X_i)e_i$. The sheaf of holomorphic sections of a vector bundle of this dual bundle is the sheaf ${\mathcal{O}}_{{\mathbf{P}}^n}(1)$.

If a nonzero linear form $\ell$ cuts out the projective hyperplane $H$, it gives a holomorphic section $s_H$ of $\mathcal{O}(1)$ with a simple zero along $H$. Dividing a local section $s$ by $s_H$ gives a meromorphic function with no poles away from $H$ and at most a simple pole on $H$. Conversely multiplication by $s_H$ turns such a function into a holomorphic section. These mutually inverse local constructions agree on overlaps and prove the holomorphic line bundle associated to a divisor identification

The nonnegative hyperplane twist cohomology by restriction calculation givesTo derive it from the permitted vanishing, use the hyperplane exact sequence for twisting sheaveswhere $i:H\hookrightarrow {\mathbf{P}}^n$ and $H\cong {\mathbf{P}}^{n-1}$. Local multiplication by a defining equation is injective, and restriction gives exactly its cokernel. Induct on $n$, starting from a point, and then on $m$. For $m=0$, higher sheaf cohomology vanishes by assumption and global holomorphic functions are constant by compactness and the maximum principle. For $m>0$, the long exact sequence in sheaf cohomology, the inductive vanishing for $\mathcal{O}(m-1)$ and the dimension induction on $H$ give vanishing in every positive degree andSumming this recurrence gives $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. Homogeneous degree-$m$ monomials give that many independent sections, so they form a basis. The printed extra equality to zero cannot hold in degree zero, already because $h^0({\mathbf{P}}^n,\mathcal{O})=1$; the boxed calculation gives the intended dimensions.

For the differential of the projective quotient map, the differential of a smooth map is computed by applying a tangent vector to coordinate functions. Since $x_j=X_j/X_0$,ThusUnder replacing $X$ by $\lambda X$, $d\pi$ on a constant ambient tangent vector scales by ${\lambda }^{-1}$, while a homogeneous linear form $L$ scales by $\lambda$. Therefore $d{\pi }_X(L(X){\partial }_{X_j})$ is independent of the representative. Its chart coefficients are holomorphic, so it is a global holomorphic vector field. The radial Euler vector field lies in the kernel of $d\pi$, either from the same formulas or because it differentiates a scaling orbit. Hence

Describe the Euler sequence on complex projective space fiberwise as well as in coordinates. At a line $\ell =\mathbb{C}X$, the middle bundle has fiber $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ and the holomorphic tangent bundle has fiber $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Send a homomorphism to its composition with the quotient. The map from $\mathcal{O}$ sends a scalar to that scalar times the inclusion $\ell \hookrightarrow {\mathbb{C}}^{n+1}$. In homogeneous notation these maps areA local section of $\mathcal{O}(1)$ is a local degree-one homogeneous function on the punctured cone, so the second expression is representative-independent. Quotienting is surjective, and its kernel consists exactly of scalar inclusion maps. This proves the short exact sequence of locally free sheavesSince it is locally split, dualizing and tensoring by $\mathcal{O}(1)$ preserves exactness and gives the cotangent Euler sequence in homogeneous coordinatesOn global sections the last map identifies ${\mathbb{C}}^{n+1}$ with the basis of linear forms in $H^0(\mathcal{O}(1))$. Its kernel is zero. This proves the vanishing of global sections of the once-twisted projective cotangent bundle; left exactness gives $\boxed{h^0({\mathbf{P}}^n,{\Omega }^1(1))=0}$.

An almost complex structure is a smooth endomorphism $J:TM\to TM$ with $J^2=-I$; in particular the real dimension is even. Complexifying the tangent bundle splits it into the $i$ and $-i$ eigensubbundlesinterchanged by conjugation. On a complex manifold, these are locally spanned by ${\partial }_{z^j}$ and ${\partial }_{{\bar{z}}^j}$. An integrable almost complex structure is one arising from a holomorphic coordinate atlas. Its $T^{0,1}$ sections are closed under the Lie bracket. The obstruction is the Nijenhuis tensorExpanding this expression in the two eigentypes shows that $N_J=0$ is equivalent to involutivity of $T^{0,1}$. Holomorphic coordinates make this condition necessary; the Newlander-Nirenberg theorem gives its sufficiency for smooth $J$. Integrability also gives $d=\partial +\bar{\partial }$, with ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault operator on $T^{1,0}M=T_M^{\prime }$ defines its holomorphic vector bundle structure.

A real connection on a vector bundle $\nabla$ on $TM$ preserves $J$ when $\nabla J=0$. Its complexification then preserves both eigensubbundles. Restricting to $T_M^{\prime }$ gives a complex connection $D$. Conversely any complex connection on $T_M^{\prime }$ gives such a real connection by adjoining its conjugate on $T^{0,1}$ and restricting to the conjugation-fixed real tangent bundle. Equivalently, transfer $D$ through the real isomorphism $v\mapsto (v-iJv)/2$. These operations are inverse. Preserving $J$ alone does not impose the holomorphic compatibility condition $D^{0,1}={\bar{\partial }}_{T^{\prime }}$.

For integrable $J$, the latter condition has a precise real-connection interpretation: the mixed-type torsion tensor vanishes. In holomorphic coordinates it says $D_{{\partial }_{{\bar{z}}^i}}{\partial }_{z^j}=0$. Reality also gives ${\nabla }_{{\partial }_{z^j}}{\partial }_{{\bar{z}}^i}=0$. Since these coordinate fields commute, their torsion is zero. Conversely a $J$-preserving real connection with zero mixed torsion has these two derivatives equal; they belong to opposite eigentypes, so both vanish, proving $D^{0,1}={\bar{\partial }}_{T^{\prime }}$. This is the holomorphic tangent connection and mixed torsion criterion.

Full torsion-freeness imposes the additional symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in $D_{{\partial }_{z^i}}{\partial }_{z^j}={\Gamma }_{ij}^k{\partial }_{z^k}$, and its conjugate. Such a $J$-preserving torsion-free connection exists on any complex manifold: local holomorphic-coordinate flat connections have these properties, and a real partition of unity combines them globally. Preservation of $J$ and zero torsion persist because both conditions are affine in the connection. Conversely if $J$ is merely almost complex, a torsion-free connection preserving it forces $N_J=0$: replace brackets by ${\nabla }_{X}Y-{\nabla }_{Y}X$ in the displayed tensor and use $\nabla J=0$; all terms cancel. Thus this connection condition detects integrability, without imposing any metric condition.

A Hermitian metric is a real Riemannian metric $g$ with $g(JX,JY)=g(X,Y)$. It induces a Hermitian metric $h$ on $T_M^{\prime }$ and the fundamental two-form $\omega (X,Y)=g(JX,Y)$. The unique Chern connection satisfies both $D^{0,1}={\bar{\partial }}_{T^{\prime }}$ and metric compatibility. In a holomorphic frame with metric matrix $H$ and coefficient-column convention, its matrix is $H^{-1}\partial H$. It need not be torsion-free when transferred to $TM$.

In holomorphic coordinates its coefficients are ${\Gamma }_{ij}^k=h^{k\bar{l}}{\partial }_i{h}_{j\bar{l}}$. Their symmetry in $i,j$ is equivalent to ${\partial }_i{h}_{j\bar{l}}={\partial }_j{h}_{i\bar{l}}$, which says $\partial \omega =0$. The conjugate gives $\bar{\partial }\omega =0$. The result that a torsion-free Chern tangent connection characterizes a Kähler metric follows: the transferred Chern connection is torsion-free exactly when $\boxed{d\omega =0}$, the Kähler metric condition. In that case it equals the Levi-Civita connection, by uniqueness of the metric-compatible torsion-free connection. Equivalently a Hermitian metric is Kähler exactly when its Levi-Civita connection preserves $J$. In the reverse direction $\nabla J=0$ implies $\nabla \omega =0$, and zero torsion gives $d\omega =0$.

This explains the role of Kähler metrics: they make the real Riemannian and holomorphic connection theories coincide. A general complex manifold has torsion-free $J$-preserving connections and has Hermitian metrics, but need not have a connection satisfying both requirements simultaneously. On a Kähler manifold, the Kähler identities then link the Dolbeault Laplacian and Hodge Laplacian, giving ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$ and the resulting harmonic Hodge decomposition theorem for compact Kähler manifolds.

For a smooth complex vector bundle $E$, a connection is a complex-linear map $D:\Gamma (E)\to A^1(M;E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth complex functions $f$. Local trivializations supply flat local connections. A locally finite smooth partition of unity $({\rho }_i)$ subordinate to them gives $D={\sum }_i{\rho }_i{D}_i$; local finiteness makes the sum smooth and ${\sum }_i{\rho }_i=1$ gives the Leibniz rule. Thus connections on a vector bundle always exist on the usual paracompact smooth manifold.

Extend $D$ to bundle-valued forms by $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Squaring shows that $D^2(fs)=f{D}^{2}s$, so the curvature form is the endomorphism-valued two-form $\Theta =D^2$. With a row frame $e=(e_1,\dots ,e_r)$, write $De=eA$; for coefficient columns $s=ev$, $Ds=e(dv+Av)$. Applying $D$ again and using the graded Leibniz rule givesThe cross terms $-A\wedge dv$ and $A\wedge dv$ cancel. Hence the Cartan curvature matrix equation in this convention isFor a frame change $e^{\prime }=eg$, the connection matrix and curvature transform as $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Trace is invariant under conjugation, so the local forms $\mathrm{Tr}\Theta$ patch into the global trace of vector-bundle curvature.

The induced determinant connection on $\det E={\Lambda }^{r}E$ iswhere the one-form factor of each $D{s}_j$ is placed in front of the section factors. In the local determinant frame $e_1\wedge \cdots \wedge e_r$, only the diagonal terms survive, so its connection form is $\mathrm{Tr}A$. Its curvature is $d\mathrm{Tr}A$. Sinceby pairing off-diagonal terms and using $A_{ii}\wedge A_{ii}=0$, we obtainIt is closed locally by $d^2=0$, and thus closed globally.

For two connections, $B=D_1-D_0$ is a global endomorphism-valued one-form: the derivative terms cancel in their Leibniz rules. In a local frame $A_1=A_0+B$, soTaking traces cancels the mixed terms, because both factors are one-forms, and cancels $\mathrm{Tr}(B\wedge B)$. Therefore the trace curvature transgression formula isThe right-hand side is globally exact, proving independence of the de Rham cohomology class without needing a sheaf-cohomology identification.

Finally let $e$ be a holomorphic frame for the holomorphic line bundle, with $h=\Vert e{\Vert }^2>0$. Define $De=e\partial \log h$, extending by the Leibniz rule. If $e^{\prime }=eg$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andThis is exactly the line-bundle frame-change law for a connection, so the local definitions glue globally. Its $(0,1)$ part is the Dolbeault operator. Metric compatibility follows from $d\log h=\partial \log h+\bar{\partial }\log h$, or $dh=h(A+\overline{A})$ for $A=\partial \log h$. These conditions uniquely force this $A$, identifying the Chern connection. Since a scalar one-form wedges with itself to zero,This agrees with the local formula for the Chern connection on a line bundle; keeping the order of the two Dolbeault differentials fixes the sign.

The Hermitian metric induces pointwise Hermitian inner products on the exterior powers of the complex cotangent bundle, with different bidegrees orthogonal, and a positive volume form $d{V}_g$. Thus the global $L^2$ inner product on smooth differential forms of type $(p,q)$ isLet ${\bar{\partial }}^{\ast }$ be the formal adjoint of the Dolbeault operator, defined by integration by parts, and let ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. This Dolbeault Laplacian preserves bidegree, is elliptic and nonnegative, and hasOn a compact Hermitian manifold without boundary, the Dolbeault Hodge decomposition on a compact Hermitian manifold says that the smooth harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$ is finite-dimensional andMore precisely, there are the orthogonal harmonic projection $H$ and a Dolbeault Green operator $G$ taking smooth forms to smooth forms, with $G=0$ on harmonics and ${\Delta }_{\bar{\partial }}G=G{\Delta }_{\bar{\partial }}=I-H$. Elliptic regularity ensures these are statements about smooth forms, not only $L^2$ completions. The Green operator commutes with $\bar{\partial }$ and ${\bar{\partial }}^{\ast }$, since the Laplacian does and its inverse is unique on the orthogonal complement of the kernel.

Expanding $\alpha =H\alpha +{\Delta }_{\bar{\partial }}G\alpha$ givesThe three summands are orthogonal: harmonic forms are killed by both adjoints, and $⟨\bar{\partial }u,{\bar{\partial }}^{\ast }v⟩=⟨{\bar{\partial }}^{2}u,v⟩=0$. If $\alpha$ is $\bar{\partial }$-closed, its coexact component is orthogonal to every closed form; taking its inner product with $\alpha$ shows that component has norm zero. Conversely the first two components are closed. Thus $\boxed{\ker \bar{\partial }={\mathcal{H}}_{\bar{\partial }}^{p,q}\oplus \bar{\partial }{A}^{p,q-1}}$, with out-of-range form spaces taken to be zero.

On the Kähler manifold define $L\alpha =\omega \wedge \alpha$ using the Kähler form; the operator $\Lambda =L^{\ast }$ is the adjoint Lefschetz operator, the pointwise contraction adjoint to wedging with $\omega$. It lowers bidegree by $(1,1)$. Setand keep ${\Delta }_{\bar{\partial }}$ as above. These are the Hodge Laplacian and the holomorphic and antiholomorphic Laplacians respectively. The permitted identity among the Kähler identities $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$ gives ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. Expanding rather than assuming the desired anticommutation,because ${\partial }^2=0$. Hence $\boxed{\partial {\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\partial =0}$.

Complex conjugation preserves the real Kähler form and interchanges $\partial ,\bar{\partial }$ and their formal adjoints. Conjugating the permitted identity among the Kähler identities therefore yields $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$. The same nilpotence calculation as in part (a) gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$. Now expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$:Both mixed anticommutators vanish, proving $\boxed{{\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}}$.

Use both Kähler identities from the preceding parts, so that ${\partial }^{\ast }=i[\Lambda ,\bar{\partial }]$ and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion givesSince $\partial \bar{\partial }+\bar{\partial }\partial =0$, we obtainTogether with part (b), this proves the full Kähler Laplacian identity

Now take the exact form $\eta$. Being $\bar{\partial }$-exact, it is orthogonal to the harmonic space, so $H\eta =0$, and it is also $\bar{\partial }$-closed. Choose $\alpha =G\eta$ for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives $\bar{\partial }\alpha =G\bar{\partial }\eta =0$. ThereforeSuppose also $\partial \eta =0$, and put $v={\bar{\partial }}^{\ast }\alpha$, $u=\partial v$. The mixed identities and $({\bar{\partial }}^{\ast }{)}^2=0$ giveThus $u=\partial {\bar{\partial }}^{\ast }\alpha$ is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also $\partial$-harmonic and hence ${\partial }^{\ast }u=0$. But it is $\partial$-exact, soConsequently $\boxed{\partial {\bar{\partial }}^{\ast }\alpha =0}$: the form $v$ is $\partial$-closed.

For every harmonic form $h$ of the same bidegree as $v$, $⟨v,h⟩=⟨\alpha ,\bar{\partial }h⟩=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the Dolbeault Hodge decomposition. Since $v$ is $\partial$-closed, it has no ${\partial }^{\ast }$-coexact component either. Thus $v=\partial \varphi$, with $\varphi \in A^{p-1,q-1}$. We conclude the requested ddbar lemma:For example $\varphi ={\partial }^{\ast }Gv$ is a choice, since the same Green operator serves ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta =0$. Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.