Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 17 (Complex Manifolds)

Let ${\mathcal{A}}^{p,q}(X)$ be the space of smooth complex differential forms of type $(p,q)$. The Dolbeault operator satisfies ${\bar{\partial }}^2=0$, and the Dolbeault cohomology isFor $q=0$ the denominator is zero; forms outside the dimension range are zero.

A sequence of sheaves is an exact sequence of sheaves precisely when its sequence of stalks at every point is exact. Here this means the first map is injective on stalks, the last is surjective on stalks, and the image equals the kernel at the middle stalk. In particular, a section of the last sheaf need only have local lifts; surjectivity on all global sections is not required.

The associated long exact sequence in sheaf cohomology of Čech cohomology, understood in the direct limit over open covers, isThe degree-zero groups are global sections. The connecting map is obtained by locally lifting a Čech cocycle to the middle sheaf and taking its Čech coboundary, which takes values in the first sheaf. Different lifts change it by a Čech coboundary.

The Dolbeault theorem identifieswhere ${\Omega }_X^p$ is the sheaf of holomorphic differential forms of degree $p$.

The local analytic ingredient is the Dolbeault-Poincaré lemma: a $\bar{\partial }$-closed smooth $(p,q)$-form with $q>0$ is locally $\bar{\partial }$-exact. Here is a local proof. On a polydisc, the one-variable Cauchy-Green operator in coordinate $z_j$ iswith a smooth cutoff supported in the coordinate disc and equal to one on a smaller disc. The fundamental-solution identity ${\partial }_{\bar{z}}(1/(\pi z))={\delta }_0$ gives ${\partial }_{{\bar{z}}_j}{T}_{j}g=g$ on that smaller disc. The operator is smooth in the parameters and commutes with derivatives in the other coordinates.

For a $(0,q)$-form, write $\alpha =d{\bar{z}}_m\wedge \beta +\gamma$, with neither $\beta$ nor $\gamma$ containing $d{\bar{z}}_m$. Subtract $\bar{\partial }(T_m\beta )$. The remainder contains no $d{\bar{z}}_m$, remains closed, and its coefficients are holomorphic in $z_m$. Repeat with $z_{m-1}$ and then the other coordinates, shrinking discs as needed. Each new integral preserves the holomorphic dependence already achieved. At the end the closed remainder has positive antiholomorphic degree but contains no antiholomorphic differential, so it is zero. Factoring out each holomorphic basis form $d{z}_I$ gives the same assertion for $(p,q)$-forms, with the fixed degree sign included in the primitive. At degree zero, the kernel of $\bar{\partial }$ consists precisely of holomorphic coefficients.

Thus the following is an exact Dolbeault resolution of holomorphic differential forms of sheaves:Each sheaf of smooth forms is a fine sheaf: multiplication by a smooth partition of unity supplies endomorphisms supported in its open cover. Fine sheaves on a paracompact manifold have zero positive sheaf cohomology; the Čech contraction sums a cochain against the partition of unity, giving $\delta h+h\delta =I$ in positive degrees. The general acyclic resolution theorem therefore computes ${\check{H}}^q(X,{\Omega }_X^p)$ as the cohomology of the global-section complex above. That complex is exactly the Dolbeault resolution of holomorphic differential forms, proving the claimed isomorphism. The sheaf-cohomology/Čech comparison and the acyclic-resolution principle are the stated general Čech properties used here.

Cover the product by $U_0={\mathbb{C}}_z\times {\mathbb{C}}_w$ and $U_{\infty }={\mathbb{C}}_{\zeta }\times {\mathbb{C}}_w$, with $\zeta =1/z$ on their intersection ${\mathbb{C}}^{\ast }\times \mathbb{C}$. The permitted vanishing and the Dolbeault theorem make this an acyclic cover for both $\mathcal{O}$ and ${\Omega }^2$. The acyclic cover theorem lets its two-term Čech complex compute the cohomology. In particular, all groups with $q\ge 2$ vanish.

For $p=0$, a global holomorphic function is constant on each compact complex projective line fibre, by the maximum modulus principle. Its remaining dependence on $w$ is entire. Hence $H_{\bar{\partial }}^{0,0}=\mathcal{O}(\mathbb{C})$. For the first Čech group, a function on the intersection has a Laurent seriesEach coefficient is entire in $w$ by its Cauchy integral formula. The nonnegative powers extend to $U_0$, and the negative powers extend to $U_{\infty }$ in coordinate $\zeta$. These two series converge locally uniformly with the parameter $w$, by the usual Laurent estimates on compact parameter sets. Thus every intersection function is a Čech coboundary and $H_{\bar{\partial }}^{0,1}=0$.

For $p=2$, write a two-form on the intersection as $F(z,w)dz\wedge dw$. The other chart hasForms extending from $U_0$ have coefficient powers $j\ge 0$; those extending from $U_{\infty }$ have powers $j\le -2$. A globally defined two-form would need to have both types of expansion, so it is zero. In the first Čech quotient, precisely the $z^{-1}$ term remains, and its coefficient is an arbitrary entire function of $w$. ConsequentlyThe second nonzero group is represented in Čech cohomology by $a(w)z^{-1}dz\wedge dw$. This explicit residue description is the Dolbeault cohomology of the projective line times the affine line; it also identifies that group naturally with the holomorphic one-forms on the affine factor.

A rank-$r$ holomorphic vector bundle is a complex manifold $E$ with a holomorphic projection to $X$ and local holomorphic trivializations $E{|}_{U_i}\cong U_i\times {\mathbb{C}}^r$ that are complex-linear on each fibre. Its transition maps have the form $(x,v)\mapsto (x,g_{ij}(x)v)$ with holomorphic $g_{ij}:U_i\cap U_j\to G{L}_r(\mathbb{C})$ satisfying the cocycle identities.

The holomorphic Picard group consists of isomorphism classes of holomorphic line bundles, with tensor product as multiplication, the trivial line bundle as identity and the dual bundle as inverse. The classification on the complex projective line givesThe inverse is degree; $\mathcal{O}(1)$ is the positive generator.

In a local trivialization, identify the lines in each fibre with $U_i\times {\mathbb{P}}^1$. On overlaps glue byThis action is well defined because $g_{ij}(x)$ is invertible and scaling $v$ does not change the resulting line. It is holomorphic: in affine projective coordinates it is a ratio of holomorphic functions on the open set where its denominator is nonzero. The transition maps satisfy the cocycle condition and have holomorphic inverses. They give the holomorphic projectivization by lines its complex-manifold atlas, of dimension ${\dim }_{\mathbb{C}}X+1$, on the usual projective-bundle topology. Points over different base points are separated by base neighborhoods; points over the same base point are separated inside a common local product chart. A countable trivializing cover and the standard charts of the complex projective line give second countability.

The local projections agree, defining a holomorphic map $p:\mathbb{P}(E)\to X$ whose fibre is $\mathbb{P}(E_x)\cong {\mathbb{P}}^1$.

Define the relative tautological line bundle $S\subset p^{\ast }E$ by $S_l=l$. Its local construction is holomorphic, and its restriction to a fibre is ${\mathcal{O}}_{{\mathbb{P}}^1}(-1)$. The required relative hyperplane line bundle is thereforeThis fixes the lines convention for projectivization and the sign of the fibre degree.

Pullback and tensor product preserve holomorphic line bundles and their isomorphisms. Interpret negative tensor powers of $L$ as powers of $L^{\ast }$. Thus the formula defines a group homomorphism on isomorphism classes.

To prove injectivity, suppose $p^{\ast }M\otimes L^{\otimes n}$ is trivial. Restrict it to any fibre. A line pulled back from its base point is trivial there, so its fibre restriction is ${\mathcal{O}}_{{\mathbb{P}}^1}(n)$. By the Picard group classification, $n=0$. It remains to show that $p^{\ast }M$ trivial implies $M$ trivial.

Let $\sigma$ be a nowhere-zero holomorphic section of $p^{\ast }M$. On a sufficiently small open set where both $E$ and $M$ are trivial, write $\sigma =a_i(x,[v])p^{\ast }{e}_i$ for a frame $e_i$ of $M$. For each fixed $x$, $a_i(x,\cdot )$ is a holomorphic function on the compact complex projective line, hence is constant. It is therefore $b_i(x)$; evaluating at a constant projective point in the local trivialization shows $b_i$ is holomorphic in $x$. The section is nowhere zero, so each $b_i$ is nowhere zero. Its overlap laws are exactly those of a section of $M$, and $b_i{e}_i$ glue to a global holomorphic frame. Thus $M$ is trivial.

The kernel consists only of $({\mathcal{O}}_X,0)$, provingThis is the Picard injection for holomorphic projective bundles. The argument uses local fibrewise constancy, so no global section of the projective bundle is required.

Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.

The unheaded definitions use the complex structure $J$ on the real tangent bundle. Compatibility means $g(Ju,Jv)=g(u,v)$. Its fundamental Hermitian form is $\omega (u,v)=g(Ju,v)$; the compatibility identity makes this real, alternating and of type $(1,1)$. The metric is a Kähler metric precisely when $d\omega =0$. In complex dimension one a real three-form is zero, so every compatible metric is a Kähler metric.

Now assume closedness and prove the Kähler normal holomorphic coordinates condition. Start with holomorphic coordinates $w$ centered at the point and make a complex-linear change so that the positive Hermitian coefficient matrix satisfies $h(0)=I$. Closedness of the $(1,1)$ form givesDefine $a_{i\bar{j},k}={\partial }_{w_k}{h}_{i\bar{j}}(0)$; it is symmetric in $i,k$. Choose new coordinates implicitly byThe derivative of this holomorphic map at zero is the identity, so the holomorphic inverse function theorem makes $z$ valid local coordinates. In the new coordinates,At zero, $h^{\prime }=I$, and differentiating givesHermitian symmetry makes all antiholomorphic first derivatives zero as well. Taylor's theorem for a smooth function now yieldsThus condition (a) implies condition (b), with the exact factor $i/2$ in the fundamental-form convention retained.

Conversely, choose the stated normal holomorphic coordinates at an arbitrary point. The coefficient matrix equals $I+O(|z{|}^2)$, so all of its first derivatives at the center vanish. Since the coordinate differentials themselves are closed,Every point can serve as the center, hence $d\omega =0$ on all of $X$. Combined with the preceding construction, this proves the equivalence of (a) and (b); it is a first-jet characterization rather than a claim that the metric is flat on a neighborhood.

To prove (a) implies (c), shrink to a contractible coordinate neighborhood. The real Poincare lemma gives a real one-form $\eta$ with $\omega =d\eta$. Write $\eta ={\eta }^{1,0}+{\eta }^{0,1}$. Because $\omega$ has type $(1,1)$, $\bar{\partial }{\eta }^{0,1}=0$ and $\partial {\eta }^{1,0}=0$. The Dolbeault-Poincaré lemma supplies a smooth function $\psi$ with ${\eta }^{0,1}=\bar{\partial }\psi$. Reality of $\eta$ gives ${\eta }^{1,0}=\partial \overline{\psi }$. ThusTaking the real function $f=2\mathrm{Im}\psi$ gives the local real potential for a closed (1,1)-form. Conversely, $d(i\partial \bar{\partial }f)=0$, by ${\partial }^2={\bar{\partial }}^2=0$ and anticommutation. Hence (a) and (c) are equivalent, completing all three conditions.

For the unheaded radial continuation, take the real potential $f$ as in (c). Rotation invariance makes $f$ constant on every circle of radius $r>0$, sois well defined and smooth by composition. This avoids treating a smooth function as if it had a convergent Taylor series. At nonzero $z$, put $t=\log |z{|}^2$. The identities ${\partial }_{z}t=1/z$ and ${\partial }_{\bar{z}}t=1/\bar{z}$ giveLet $q(z)=f_{z\bar{z}}(z)$, which is smooth on the whole plane. ThenAt the origin, rotation invariance gives the finite Taylor expansion $f(z)=f(0)+b|z{|}^2+O(|z{|}^4)$, with $b=q(0)$. Alternatively continuity of $q$ directly givesThe metric is positive away from the origin exactly when $u^{\prime \prime }(t)>0$ for all finite $t$, and it is positive at the origin exactly when this limit is positive. Smoothness at the origin is already supplied by the original smooth potential. Consequently the positivity criterion for a radial Kähler potential isClosedness is automatic from the potential, and positivity completes the Kähler condition. No completeness of this metric is asserted. The rotation-invariant Kähler potential on the complex plane is understood as real; if a complex radial potential is initially allowed with real $\omega$, its smooth radial imaginary part is harmonic and hence constant, so that constant can be removed.

Use the convention that a Hermitian metric $h$ is complex-linear in its first argument and conjugate-linear in its second. It is a smoothly varying positive-definite Hermitian form on each fibre. A connection on a vector bundle is a complex-linear operator $D:\Gamma (E)\to {\Omega }^1(X,E)$ satisfying $D(as)=da\otimes s+aDs$ for smooth complex functions $a$. Metric compatibility means, for every real vector field $V$,This is the metric-compatible connection condition with the sesquilinear convention fixed.

Apply the smooth Gram-Schmidt process to a local frame to obtain an $h$-orthonormal smooth frame $s_1,\dots ,s_r$. Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write $D{s}_j={\sum }_i{A}_{ij}{s}_i$. Differentiating $h(s_i,s_j)={\delta }_{ij}$ and using compatibility givesfor every real $V$. Thus $\boxed{A(V{)}^{†}=-A(V)}$: the connection matrix is skew-Hermitian. This smooth unitary frame for a Hermitian connection is generally not holomorphic; the requested local-frame assertion requires only a smooth frame.

The holomorphic dual vector bundle $E^{\ast }$ is obtained by dualizing fibres and using transition matrices $g_{ij}^{-T}$ when $E$ has transitions $g_{ij}$. Their entries are holomorphic because matrix inversion is holomorphic on $G{L}_r(\mathbb{C})$. Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre $(E_x{)}^{\ast }$.

The conjugate vector bundle $\overline{E}$ has the same underlying real fibres but opposite scalar action: $\lambda \cdot \overline{v}=\overline{\overline{\lambda }v}$. Its smooth transition matrices are $\overline{g_{ij}}$; they need not be holomorphic on $X$. The conjugate bundle is used here as a smooth complex bundle, whereas $E^{\ast }$ is holomorphic.

Define the smooth dual tensor $H$ byIt is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of $h$ into linearity. Thus $H\in \Gamma ((E\otimes \overline{E}{)}^{\ast })$.

The conjugate connection is defined on real vector fields by ${\overline{D}}_V(\overline{s})=\overline{D_{V}s}$ and extended complex-linearly on the conjugate bundle. The tensor product connection isIts dual connection $D_0$ is uniquely characterized byThe Leibniz rule makes this a genuine connection on $(E\otimes \overline{E}{)}^{\ast }$. Evaluate it on the decomposable tensor $u=s\otimes \overline{t}$:Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every $V,s,t$:This is metric compatibility as parallelism of a Hermitian tensor.