Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 22 (Complex Manifolds)

The holomorphic normal bundle is the quotient $N_V=T^{1,0}M{|}_V/T^{1,0}V$, of rank $r$. The tangent sequence is a short exact sequence of holomorphic vector bundles:Taking determinant line bundles gives $\det (T^{1,0}M){|}_V\cong \det (T^{1,0}V)\otimes \det N_V$. Dualizing and rearranging proves adjunction for a smooth submanifold:The determinant identity follows locally by adjoining lifts of a quotient frame to a subbundle frame; changing the lifts adds only off-diagonal blocks, so does not change the determinant.

For a smooth divisor, choose reduced local defining functions $f_i$, with $f_i=u_{ij}{f}_j$ and $u_{ij}$ holomorphic and nowhere zero. The holomorphic line bundle associated to a divisor $[V]={\mathcal{O}}_M(V)$ has local frames $e_i=f_i^{-1}$, with $e_j=u_{ij}{e}_i$. The products $f_i{e}_i$ define its canonical section vanishing on $V$. Along $V$,Thus $v\mapsto d{f}_i(v)e_i$ patches to a map $T^{1,0}M{|}_V\to [V]{|}_V$, whose kernel is $T^{1,0}V$ and which is surjective because each reduced defining function has nonzero differential there. Hence $N_V\cong [V]{|}_V$. This is the normal bundle of a smooth analytic hypersurface. The divisor construction uses a closed hypersurface: an arbitrary nonclosed embedded hypersurface need not be a divisor. For example, a punctured line in ${\mathbb{C}}^2$ cannot be the support of a divisor, since its missing limit point would also lie in every local holomorphic zero set containing that line.

Let $s$ be the section of $E$ defining $V$. The intended hypothesis is that it is a regular zero locus: $ds$ has rank $r$ on $V$. In a local trivialization, differentiation of the component functions definesThis is intrinsic: differentiating a change-of-frame matrix introduces terms multiplied by $s$, which vanish on $V$. Its kernel is $T^{1,0}V$, so it induces an isomorphism of holomorphic normal bundles. ConsequentlyThis is the normal bundle of a regular zero locus. Here $\det E=L_1\otimes \cdots \otimes L_r$.

Regularity is essential if “vanishing” is read only as equality of underlying sets. On ${\mathbb{P}}^2$, the section $Z_0^2$ of $\mathcal{O}(2)$ vanishes set-theoretically on a smooth line $V$. Its derivative vanishes on $V$, andThus smoothness of the underlying zero set alone does not establish the claimed normal-bundle isomorphism. The preceding proof applies when the equations define the reduced smooth submanifold, equivalently are transverse to zero.

For Complex projective space, put $\mathcal{O}(a)=[aH]$. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Quotienting $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ by its scalar maps into $\ell$ yields the Euler sequenceIts determinant gives $K_{{\mathbb{P}}^n}\cong \mathcal{O}(-n-1)=[-(n+1)H]$. For a smooth complete intersection defined regularly by the homogeneous equations, apply the normal-bundle result to $E={⨁}_i\mathcal{O}(d_i)$:This is the canonical bundle of a regular projective complete intersection. The same regularity qualification is needed here: the squared-line example would otherwise predict ${\mathcal{O}}_{{\mathbb{P}}^1}(-1)$ for its canonical bundle, whereas the line has $K_V={\mathcal{O}}_{{\mathbb{P}}^1}(-2)$.

A connection on a vector bundle is a $\mathbb{C}$-linear map $D:A^0(E)\to A^1(E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth functions. Its extension is the covariant exterior derivativeSquaring gives a $C^{\infty }$-linear operator $D^{2}s=\Theta s$, defining the vector-bundle curvature $\Theta \in A^2(\mathrm{End}E)$.

For a line bundle, $\mathrm{End}E$ is canonically the trivial line bundle, so this vector-bundle curvature is an ordinary global complex two-form. In a local frame with $D=d+A$, the scalar identity $A\wedge A=0$ gives $\Theta =dA$. Therefore $d\Theta =0$. The calculations in the two subparts also explain the gluing directly.

Use a row of local frame sections $e=(e_1,\dots ,e_m)$ and columns of coefficients: $D(eu)=e(du+Au)$. If $e^{\prime }=eg$ for a smooth invertible matrix-valued function $g$, thenHence the change of frame of a vector-bundle connection isThis convention specifies both the placement of the frame matrix and the sign of the inhomogeneous term. The vector-bundle curvature transforms by ${\Theta }^{\prime }=g^{-1}\Theta g$, either by applying $D^2$ to the new frame or by substituting in the formula of the next subpart.

For a section represented by a column $u$, expand the covariant exterior derivative using the graded Leibniz rule:Therefore the Cartan curvature matrix equation isThe product includes endomorphism composition; it need not vanish for a matrix-valued one-form. For a scalar frame change $g$, $A^{\prime }=A+g^{-1}dg$ and $d(g^{-1}dg)=0$, confirming that the line-bundle two-forms $dA$ patch globally.

The smooth exponential sequence isLocal logarithms make it exact as a sequence of sheaves. A line bundle has a transition class in $H^1(M,{\mathcal{C}}^{\infty }({\mathbb{C}}^{\times }))$; its image under the connecting map defines the First Chern class. For a holomorphic line bundle the holomorphic exponential sequence gives the same class by naturality.

Here is the explicit Čech-de Rham curvature descent. On a good cover, write $e_j=e_i{g}_{ij}$ and choose logarithms $g_{ij}=\exp (2\pi i{f}_{ij})$. The integer cocyclerepresents $c_1(E)$. By the proved frame formula,Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta /(2\pi )=d{B}_i$. Then $\delta B=-df$ and $\delta f=c$. In the Čech-de Rham double complex, with total differential $D_{\mathrm{tot}}=\delta +(-1{)}^{p}d$ on Čech degree $p$,Thus the global closed differential form and the integer cocycle represent the same class under de Rham theorem:This proves integrality, including the sign and normalization. It identifies the image of the integral class; vector-bundle curvature alone cannot recover torsion classes lost in passage to complex coefficients.

Two connections on the same line bundle differ by a global scalar one-form $a$. Their vector-bundle curvatures satisfy ${\Theta }_1-\Theta =da$, so the normalized representatives differ by an exact differential form. Hence the class is independent of the connection.

Use the integral normalization of the Fubini-Study form. On the chart $Z_j\ne 0$, define the integrally normalized Fubini-Study form byThe potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial \bar{\partial }$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal{O}(1)$ has local squared norm $h=(1+\sum |z_k{|}^2{)}^{-1}$. Its Chern connection has $A=\partial \log h$ and vector-bundle curvature $\Theta =\bar{\partial }\partial \log h=\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$. Therefore $[{\omega }_{\mathrm{int}}]=c_1(\mathcal{O}(1))$, proving the requested integrality. With the unnormalized convention ${\omega }_{FS}=i\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$, this is ${\omega }_{\mathrm{int}}={\omega }_{FS}/(2\pi )$.

The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

The Hermitian metric on cotangent vectors is the dual metric, and its exterior-power metric is defined on decomposable vectors by the determinant of their pairwise inner products. Tensor this with the metric on $E$. In a unitary coframe and unitary frame of $E$, the forms obtained by wedging distinct holomorphic and antiholomorphic coframe elements and tensoring with a frame vector form an orthonormal basis. Let $dV={\omega }^n/n!$, and use inner products linear in the first variable.

The bundle-valued conjugate-linear Hodge star is characterized by contraction of the $E,E^{\ast }$ coefficients inIt is conjugate-linear, maps type $(p,q)$ to type $(n-p,n-q)$, and uses the metric identification of $E$ with its dual. For total degree $k=p+q$, the ordinary star-square rule giveswhere $E^{\ast \ast }$ is identified with $E$. The conjugate-linear convention is essential for the bidegrees specified here.

If $\psi$ has degree $k-2$ and $\eta$ degree $k$, the even degree of the real form $\omega$ givesSince ${\ast }_E\Lambda \eta =L{\ast }_E\eta$ for $\Lambda \eta =(-1{)}^k{\ast }_{E^{\ast }}L{\ast }_E\eta$, it follows that $(L\psi ,\eta )=(\psi ,\Lambda \eta )$ pointwise. Thus $\Lambda$ is the adjoint Lefschetz operator.

For holomorphic $E$, ${\bar{\partial }}_E$ differentiates coefficients in holomorphic local frames and uses the graded Leibniz rule on forms. Its bundle-valued Dolbeault adjoint is the formal differential operatorIt lowers the antiholomorphic degree by one. To verify this on compact $M$, define $⟨\psi ,\eta {⟩}_{L^2}={\int }_M(\psi ,\eta )dV$. For $\psi \in A^{p,q-1}(E)$ and $\eta \in A^{p,q}(E)$, Stokes applied to the scalar $(n,n-1)$-form $\psi \wedge {\ast }_E\eta$ yieldsThe star-square identity converts the right side into $⟨\psi ,{\bar{\partial }}_E^{\ast }\eta {⟩}_{L^2}$. Compactness and absence of boundary remove boundary terms. This establishes the asserted global adjoint relation on smooth sections.

The Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=B{B}^{\ast }+B^{\ast }B$, with $B={\bar{\partial }}_E$, is formally self-adjoint. Consequently the adjoint of a commutator satisfiesFor the $(1,0)$ part $D^{\prime }$ of the Chern connection, the explicit corresponding adjoint iswhere $D_{E^{\ast }}^{\prime }$ is the $(1,0)$ part of the dual Chern connection. This is the same integration-by-parts formula in holomorphic degree.

Now assume the metric is Kähler. Since $\bar{\partial }\omega =0$, $B$ commutes with $L$. Taking adjoints provesThe scalar Kähler identities needed areThey extend to the bundle-valued Kähler identitiesHere is a local justification of the extension. At any chosen point, take holomorphic normal coordinates for the Kähler metric and a holomorphic local frame of $E$ with $H=I$ and $\partial H=0$ at that point. This is a Chern connection in a normal holomorphic frame. Such a frame is obtained first by normalizing the metric matrix at the point and then prescribing the holomorphic local frame change's first derivatives to cancel $A=H^{-1}\partial H$. The first-order operators and their formal adjoints at the point are then the scalar operators acting componentwise. The scalar identities therefore give the two displayed bundle identities there. Since the point was arbitrary, they hold globally.

Let $A=D^{\prime }$. Its square and $B^2$ vanish, whilebecause the Chern curvature has pure type $(1,1)$. Use the two commutators already proved to calculateTaking adjoints, including conjugation of the scalar $-i$, gives the Lefschetz-Dolbeault curvature commutatorIf $\Theta =0$, commutation follows. Conversely, vanishing of the commutator on all $E$-valued forms implies $\Theta \wedge s=0$ for every smooth section of a vector bundle $s$. At any point, a bump-supported local section can have any prescribed fiber value. Thus $\Theta$ annihilates every vector in every fiber, forcing $\Theta =0$ identically. This proves the flatness criterion from Lefschetz commutation; the test on degree-zero forms is why the assertion is made on the entire form space.