Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 18 (Complex Manifolds)

A holomorphic atlas on a real smooth manifold of dimension $2n$ consists of compatible smooth charts ${\phi }_a:U_a\to V_a\subseteq {\mathbb{C}}^n$ whose transition maps ${\phi }_b{\phi }_a^{-1}$ are biholomorphic on their domains. An almost complex structure is a smooth real vector bundle endomorphism $J:TX\to TX$ with $J^2=-\mathrm{id}$. In each chart defineThe differential of a holomorphic map is a complex-linear map, so the transition differentials commute with multiplication by $i$. Consequently the local definitions agree on overlaps. They are smooth and square to $-\mathrm{id}$, yielding the natural almost complex structure induced by a complex atlas. This construction uses the atlas, rather than an arbitrary choice of coordinates on the underlying real smooth manifold.

Extend $J$ complex-linearly to the complexification of a real vector bundle $T_{\mathbb{C}}X=TX{\otimes }_{\mathbb{R}}\mathbb{C}$. Since $J^2=-\mathrm{id}$, its $+i$ and $-i$ eigenbundles have smooth projectionsEach has complex rank $n$: complex conjugation interchanges them, and together they have rank $2n$. These are the type decomposition of the complexified tangent bundle. For an arbitrary almost complex structure, they are smooth complex vector bundles; a holomorphic vector bundle structure requires integrability.

When $J$ comes from a holomorphic atlas, write a holomorphic coordinate as $z_j=x_j+i{y}_j$. Its Wirtinger derivatives areThe induced almost complex structure has $J{\partial }_{x_j}={\partial }_{y_j}$ and $J{\partial }_{y_j}=-{\partial }_{x_j}$. Hence the displayed vector fields are respectively $+i$ and $-i$ eigenvectors. They are linearly independent, and each collection has $n$ elements. Thus they give local frames for $T^{1,0}X$ and $T^{0,1}X$ respectively, with $T^{1,0}X$ the holomorphic tangent bundle. The repeated $T^{0,1}X$ in the first sentence of the printed item must be read as the two complementary eigenbundles.

With the normalization used here, $N=2N_J$, where $N_J$ is the standard Nijenhuis tensor. First check that it is a tensor: the Lie bracket identity $[fU,V]=f[U,V]-V(f)U$ givesSkew symmetry gives the same $C^{\infty }$-linearity in the second input. In coordinates from a holomorphic atlas, the real coordinate vector fields ${\partial }_{x_j},{\partial }_{y_j}$ commute, and $J$ sends each to another such field with a constant sign. Every Lie bracket in $N_J$ therefore vanishes on these coordinate fields. Tensoriality then gives $\boxed{N=0}$ on all smooth vector fields. The factor two in this paper does not change this vanishing conclusion.

Use the complex-bilinear extension of the Nijenhuis tensor expression to complex vector fields. If $U,V\in T^{1,0}X$, then $JU=iU$ and $JV=iV$, soFor $U,V\in T^{0,1}X$, the same substitution with $-i$ gives $N(U,V)=-8[U,V{]}^{1,0}$. For inputs of opposite types, the two leading Lie brackets cancel, as do the two terms involving $J$, so $N(U,V)=0$. Applying the type decomposition of the complexified tangent bundle to both inputs consequently givesThus $N=0$ precisely when the Lie bracket of any two sections of either eigenbundle remains in that same eigenbundle. In other words, $N=0$ if and only if both tangent-type distributions are involutive distributions. This proves the equivalence directly and does not assume that an arbitrary almost complex structure already has holomorphic coordinates.

We use complex-valued smooth differential forms. The sheaf of smooth differential forms is ${\mathcal{A}}^k(U)=\Gamma (U,{\bigwedge }^k(T^{\ast }X\otimes \mathbb{C}))$, with the usual restriction maps. Dualizing the type decomposition of the complexified tangent bundle and taking exterior powers decomposes this bundle into the summandsTheir smooth sections form the sheaf of differential forms of type (p, q) ${\mathcal{A}}^{p,q}$. Locally a section is a sum of $f_{I,J}d{z}_I\wedge d{\bar{z}}_J$ with $|I|=p$, $|J|=q$ and smooth coefficients. Holomorphic transition maps preserve types, so the local decompositions agree globally. HenceThe exterior derivative splits as $d=\partial +\bar{\partial }$, with bidegrees $(1,0)$ and $(0,1)$. Its square being zero gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault cohomology is thereforeForms in negative or out-of-range bidegrees are understood to be zero.

For a differential form of type (p, q), $d\alpha =0$ is equivalent to both $\partial \alpha =0$ and $\bar{\partial }\alpha =0$, because the two resulting types are distinct. Moreover $\partial \bar{\partial }\beta$ has type $(p,q)$ when $\beta$ has type $(p-1,q-1)$, andby the square-zero and anticommutation identities. Thus the denominator defining Bott-Chern cohomology is a vector subspace of its numerator, making the quotient well defined.

Complex conjugation of differential-form type sends a $d$-closed $(p,q)$-form to a $d$-closed $(q,p)$-form. For a representative in the denominator,The minus sign does not change the denominator vector subspace. Conjugation consequently induces a conjugate-linear bijection $[\alpha ]\mapsto [\bar{\alpha }]$ between the two Bott-Chern cohomology spaces; applying it twice is the identity. Equivalently, $\boxed{H_{BC}^{p,q}(X)\cong \overline{H_{BC}^{q,p}(X)}}$ as complex vector spaces.

We prove the Bott-Chern Poincaré lemma on a polydisc $B$. Let $p,q\ge 1$ and $d\alpha =0$, with $\alpha$ of type $(p,q)$. The allowed Dolbeault-Poincaré lemma first gives $\alpha =\bar{\partial }{b}_0$, where $b_0$ has type $(p,q-1)$. Since $\bar{\partial }\partial b_0=-\partial \alpha =0$, repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form $\partial b_{q-1}$ has type $(p+q,0)$, is holomorphic because its Dolbeault operator vanishes, and is $\partial$-closed. The holomorphic Poincaré lemma on a polydisc gives a holomorphic form $h$ of type $(p+q-1,0)$ with $\partial h=\partial b_{q-1}$. This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincaré lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take $h=0$.

Now $b_{q-1}-h$ is $\partial$-closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives $b_{q-1}-h=\partial c_{q-1}$, since its holomorphic degree is positive. Moving backwards, if the last modified form is $\partial c_j$, thenApply the conjugate Dolbeault-Poincaré lemma again to obtain $b_{j-1}+\bar{\partial }{c}_j=\partial c_{j-1}$. The modification does not change its Dolbeault operator, so this process continues down to $b_0$. In the case $q=1$, subtract $h$ at this last step instead. Finally,Thus $\boxed{H_{BC}^{p,q}(B)=0(p,q\ge 1)}$.

This vanishing fails on general complex manifolds. On the complex projective line, let $\omega$ be its Fubini-Study form. It is a $d$-closed $(1,1)$-form with ${\int }_{{\mathbb{P}}^1}\omega >0$. If it were $\partial \bar{\partial }f$, it would be the exact differential form $d(\bar{\partial }f)$, whose integral is zero by Stokes theorem. Therefore $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

A $d$-closed differential form of type (p, q) is $\bar{\partial }$-closed, and a Bott-Chern cohomology coboundary is a Dolbeault cohomology coboundary becauseHence taking the same representative defines a complex-linear map $\varphi :H_{BC}^{p,q}(X)\to H_{\bar{\partial }}^{p,q}(X)$.

If $X$ is compact and Kähler, the Dolbeault Hodge decomposition supplies a $\bar{\partial }$-harmonic representative $\alpha$ of every Dolbeault cohomology class. The scalar Kähler Laplacian identity is ${\Delta }_d=2{\Delta }_{\bar{\partial }}$, so $\alpha$ is also $d$-harmonic. Consequentlyand $d\alpha =0$. It therefore defines a Bott-Chern cohomology class mapping to the original class. Thus $\boxed{\varphi \text{ is surjective on a compact Ka¨hler manifold}.}$

For clarity, the scalar Kähler Laplacian identity follows from the Kähler identities: they make the mixed anticommutators $\{\partial ,{\bar{\partial }}^{\ast }\}$ and $\{\bar{\partial },{\partial }^{\ast }\}$ zero, while the curvature-zero instance of the identity proved in 4(b) gives ${\Delta }^{\prime }={\Delta }^{\prime \prime }$. Expanding ${\Delta }_d$ yields ${\Delta }^{\prime }+{\Delta }^{\prime \prime }=2{\Delta }^{\prime \prime }$.

Order the index set of the open cover $\mathcal{U}=(U_i)$. For the sheaf of abelian groups $\mathcal{F}$, the Čech cochain group isIt is a product, not a finite-support sum. The Čech coboundary isEvery term in ${\delta }^{2}c$ occurs twice with opposite signs, so ${\delta }^2=0$. The Čech cohomology of the cover is ${\check{H}}^p(\mathcal{U},\mathcal{F})=\ker \delta /\mathrm{im}\delta$. Refinements induce maps on these groups; different choices of refinement indices give maps related by a cochain homotopy, hence the same map on cohomology. Taking the direct limit over covers givesOne may use locally finite covers, since a complex manifold is a paracompact space and these covers are cofinal. In degree zero this recovers the group of global sections by the sheaf gluing axiom.

The printed assertion for every cohomological degree is false. The valid conclusion supplied by the stated Hartogs extension theorem is the degree-zero isomorphism. Indeed, around each removed point choose a small coordinate ball on which the holomorphic vector bundle $E$ is trivial. A holomorphic section on the punctured ball has finitely many holomorphic coefficient functions, each of which extends by Hartogs extension theorem when the complex dimension is at least two. Uniqueness of holomorphic extension makes these local extensions agree with the original section on overlaps, givingThis also identifies the direct image sheaf $j_{\ast }\mathcal{O}(E{|}_{X_0})$ with $\mathcal{O}(E)$ for the inclusion $j:X_0\hookrightarrow X$. It does not identify higher sheaf cohomology.

Here is an explicit higher-degree obstruction. Take $X={\mathbb{P}}^2$, remove $[0:0:1]$, and take the trivial line bundle. Cover $X_0$ by $U=\{x\ne 0\}$ and $V=\{y\ne 0\}$. On $U$ use coordinates $t=y/x$, $s=z/x$; on $V$ use $u=x/y=1/t$, $v=z/y=s/t$. Both charts are ${\mathbb{C}}^2$, and the overlap is ${\mathbb{C}}^{\ast }\times \mathbb{C}$. For every integer $m\ge 2$, the holomorphic functiondefines a Čech cocycle. It cannot be a Čech coboundary $b(1/t,s/t)-a(t,s)$ with $a,b$ entire on their charts. Taking the coefficient of $s^m$, the term from $a$ has only nonnegative powers of $t$, whereas the term from $b$ has the form $t^{-m}{b}_m(1/t)$ and only powers at most $-m$. Neither can provide the coefficient $t^{-1}$. Uniqueness of Laurent series proves the contradiction. The same argument applied to each fibre degree proves that all the classes $[g_m]$, $m\ge 2$, are linearly independent.

These classes remain nonzero in Čech cohomology of $X_0$, rather than merely of this cover. The low-degree Mayer-Vietoris sequence for sheaf cohomology injects the quotient of overlap sections by the two chart-section groups into $H^1(X_0,\mathcal{O})$. Concretely, a partition of unity gives smooth $f_U,f_V$ with $f_V-f_U=g$; their common Dolbeault operator defines a global $(0,1)$-form. If that form were $\bar{\partial }$-exact, subtracting its global smooth primitive would make $f_U,f_V$ holomorphic and split $g$. The nonsplitting just proved therefore gives the same obstruction in Dolbeault cohomology, and the Dolbeault theorem identifies it with sheaf cohomology. Thus $H^1(X_0,\mathcal{O})$ is infinite dimensional. In contrast, $H^1({\mathbb{P}}^2,\mathcal{O})$ is finite dimensional by compact Dolbeault Hodge decomposition for the Fubini-Study form. Hence the two degree-one groups cannot be isomorphic, disproving the printed all-degree request already in complex dimension two.

In complex dimension one, even the degree-zero conclusion fails. Take $X={\mathbb{P}}^1$, remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected ${\mathbb{P}}^1$ are constant by the maximum modulus principle, whereas $X_0\cong \mathbb{C}$ has nonconstant entire functions such as $z$. Therefore the restriction map is not onto even in degree zero.

The complex tautological line bundle on ${\mathbb{P}}^1$ has fibre at a line $\ell \subseteq {\mathbb{C}}^2$ equal to $\ell$ itself:Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$, $\mathcal{O}(k)=\mathcal{O}(1{)}^{\otimes k}$ for $k>0$, $\mathcal{O}(0)$ to be trivial, and $\mathcal{O}(k)=\mathcal{O}(-1{)}^{\otimes (-k)}$ for $k<0$. These are the tensor powers of the hyperplane line bundle.

On the two given charts, use the holomorphic local frames $e_0(w)=(1,w)$ and $e_1(z)=(z,1)$ of $\mathcal{O}(-1)$. The maps $([1:w],a)\mapsto ([1:w],a(1,w))$ and $([z:1],b)\mapsto ([z:1],b(z,1))$ are its trivializations. On the overlap $z=1/w$,Thus the frame-transition factor from $e_0$ to $e_1$ is $w^{-1}$, while the fibre-coordinate transition from chart 0 to chart 1 is $w$. Specifying both avoids an inverse-convention ambiguity.

For $\mathcal{O}(k)$, let $e_i^{(k)}$ be the induced frames; they satisfy $e_1^{(k)}=w^k{e}_0^{(k)}$. Identifying overlap sections using $e_0^{(k)}$, the Čech cochain groups and Čech coboundary for $k\ge 0$ areHere $\mathcal{O}(\mathbb{C})$ means entire functions in the coordinate of the corresponding chart. Global sections are $\ker \delta$, so $a(w)=w^{k}b(1/w)$, or $b(z)=z^{k}a(1/z)$. Expanding the entire function $a(w)={\sum }_{j\ge 0}{a}_j{w}^j$ shows that $b$ is holomorphic at zero exactly when $a_j=0$ for $j>k$. Thus a global section is determined by a polynomial of degree at most $k$, with basis $1,w,\dots ,w^k$ in the chart-0 frame. Therefore $\boxed{\dim H^0({\mathbb{P}}^1,\mathcal{O}(k))=k+1(k\ge 0)}$.

The Chern connection of a Hermitian metric on a holomorphic vector bundle is the unique connection on a vector bundle compatible with that metric and whose $(0,1)$ part is the bundle's Dolbeault operator ${\bar{\partial }}_E$. Fix a holomorphic local frame and column coefficients for sections. Write the metric as $h(s,t)={\bar{s}}^{T}Ht$, conjugate-linear in the first argument, and write the connection as $Ds=ds+As$. The condition $D^{0,1}={\bar{\partial }}_E$ forces $A$ to have type $(1,0)$. Metric compatibility requireswhere the dagger conjugates the differential-form coefficients as well as transposing the matrix. Taking the $(1,0)$ part gives $\boxed{A=H^{-1}\partial H}$. Its conjugate-transpose supplies the $(0,1)$ metric equation because $H$ is Hermitian. This proves uniqueness and local existence.

Under a holomorphic change of frame $e^{\prime }=eg$, the metric matrix becomes $H^{\prime }=g^{†}Hg$. The local formula for the Chern connection on a vector bundle then givesThis is precisely the transformation rule for a connection on a vector bundle, so the local connections glue and establish global existence. No Kähler hypothesis is needed for this part.

Extend $D$ to vector-bundle-valued differential forms by the graded Leibniz rule. The curvature form of a connection is the tensorial square $F_h=D^2$, acting by exterior multiplication. In the chosen frame,Indeed $\partial A+A\wedge A=0$ by differentiating $H^{-1}H=I$. It follows that $F_h$ has type $(1,1)$; the gauge change is $F_h^{\prime }=g^{-1}{F}_{h}g$. Thus it is a global smooth two-form with values in $\mathrm{End}(E)$, namely an element of ${\mathcal{A}}^2(\mathrm{End}(E))$.

The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

A positive holomorphic line bundle $F$ admits a Hermitian metric whose Chern connection curvature satisfies that $i{F}_F$ is a positive real (1, 1)-form. In a local holomorphic local frame with squared length $e^{-\phi }$, the local formula for the Chern connection on a line bundle gives $F_F=\partial \bar{\partial }\phi$, so positivity means $i\partial \bar{\partial }\phi$ is positive definite. Its closedness makes $\omega =i{F}_F$ a Kähler form; use this form to define the operators below.

Let $n={\dim }_{\mathbb{C}}X>0$, choose a Hermitian metric on $E$, and equip $G_m=E\otimes F^{-m}$ with the tensor-product metric. The curvature of a tensor product connection givesOn $G_m$-valued zero-forms, the Lefschetz commutator is $[L,\Lambda ]=-n\mathrm{id}$. Thus the Bochner-Kodaira-Nakano identity givesThis last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite $C$ such that $⟨R_{E}v,v⟩\ge -C\Vert v{\Vert }^2$ at every point. For a holomorphic section $s$ of $G_m$, $D^{\prime \prime }s=0$ and $D^{\prime \prime \ast }s=0$ by degree, so integrating the identity givesChoose an integer $m_0$ with $m_{0}n>C$. Then $\boxed{H^0(X,E\otimes F^{-m})=0(m\ge m_0)}$. The threshold depends on the fixed bundle $E$, as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.

No threshold can work uniformly over all holomorphic vector bundles $E$. Given a proposed threshold $M$, choose an integer $m\ge M$ and let $E=F^m$. The canonical trivialization of a line bundle tensored with its dual givesThis contradicts vanishing at that value of $m$. In the proof above, the curvature of $E=F^m$ exactly offsets the curvature contribution from $F^{-m}$, explaining why the bound cannot be independent of $E$.

For a holomorphic vector bundle $E$, the sheaf of vector-bundle-valued differential forms iswhere sections are smooth and restrictions are the usual ones. In a holomorphic local frame of $E$, the Dolbeault operator ${\bar{\partial }}_E$ acts coefficientwise and satisfies ${\bar{\partial }}_E^2=0$. Its Dolbeault cohomology with values in a holomorphic vector bundle is the cohomology of the global section complex ${\mathcal{A}}^{p,\bullet }(X,E)$.

The bundle-valued Dolbeault theorem identifies this with sheaf cohomology:Here ${\Omega }_X^p$ is the sheaf of holomorphic differential forms, and $\mathcal{O}(E)$ is the sheaf of holomorphic sections of a vector bundle. Equivalently, the complexis a fine sheaf resolution, by the local Dolbeault-Poincaré lemma and smooth partitions of unity. In particular $p=0$ computes $H^q(X,\mathcal{O}(E))$; for $p>0$ the holomorphic differential-form factor must be retained.

The Kähler form and the Hermitian metric give the $L^2$ inner product on vector-bundle-valued differential forms, using volume ${\omega }^n/n!$. Define the formal adjoint ${\bar{\partial }}_E^{\ast }$ and the elliptic, self-adjoint, nonnegative Dolbeault LaplacianIts harmonic space isThe equality follows from $⟨{\Delta }_E^{\prime \prime }\alpha ,\alpha ⟩=\Vert {\bar{\partial }}_E\alpha {\Vert }^2+\Vert {\bar{\partial }}_E^{\ast }\alpha {\Vert }^2$. On compact $X$, the bundle-valued Dolbeault Hodge decomposition states that this space is finite dimensional and thatThe sum is orthogonal for the $L^2$ inner product and all summands here consist of smooth forms. Every Dolbeault cohomology class has a unique harmonic representative, giving $H_{\bar{\partial }}^{p,q}(X,E)\cong {\mathcal{H}}^{p,q}(X,E)$ and, by the Dolbeault theorem, the corresponding sheaf cohomology isomorphism. This is a decomposition for ${\bar{\partial }}_E$, whose square is zero; it does not require the full Chern connection to be flat.

Take the threshold $k_0\ge 1$ in the stated spectral gap for powers of a positive line bundle. For $q\ge 1$ and $k\ge k_0$, it givesA harmonic form must therefore be zero. The Dolbeault Hodge decomposition and the Dolbeault theorem identify its zero harmonic space with $H^q(X,\mathcal{O}(L^k))$. Hence $\boxed{H^q(X,L^k)=0(q\ge 1,k\ge k_0)}$. For $q>{\dim }_{\mathbb{C}}X$ the form spaces already vanish. The gap controls the lowest eigenvalue including zero, so it rules out harmonic forms rather than just controlling the positive spectrum.

Choose a positive Hermitian metric on $L$. Its curvature form $\omega =i{F}_h$ is a Kähler form on the compact complex manifold of dimension one. Part (c) gives a threshold $k_0$ with $H^1(X,\mathcal{O}(L^k))=0$ for every $k\ge k_0$. We use this to prescribe a finite principal part of a meromorphic section.

Fix a smooth cutoff $\chi$ supported inside the given coordinate chart and equal to one near $x_0$. Let $P(z)={\sum }_{j=-r}^{-1}{a}_j{z}^j$. On the punctured manifold define $t_k=\chi P(z){\zeta }^k$ inside the chart, extended by zero outside. Its Dolbeault operatoris smooth globally: it vanishes near the pole and near the boundary of the chart. It is ${\bar{\partial }}_{L^k}$-closed, either by the square-zero identity or because there are no $(0,2)$-forms in complex dimension one. The vanishing of $H^1(X,\mathcal{O}(L^k))$, together with the Dolbeault theorem, therefore gives a global smooth section $u_k$ with ${\bar{\partial }}_{L^k}{u}_k={\eta }_k$.

The section $s_k=t_k-u_k$ is holomorphic on $X\setminus \{x_0\}$. Near $x_0$, ${\eta }_k=0$, so $u_k=f_k(z){\zeta }^k$ with $f_k$ holomorphic through zero. Its Taylor series then giveswhere the second series converges near zero and its coefficients are those of $-f_k$. The threshold is determined by the fixed bundle $L$ and $X$, and is independent of the prescribed coefficients. In fact the same threshold works for every finite pole order $r$, since the Dolbeault cohomology obstruction is the same $H^1(X,L^k)$ for every such cutoff construction.