Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 13 (Complex Manifolds)

On the polydisc, put $b_i(z)=f_i(z)-f_i(a)$ and defineEach integral is along the straight coordinate segment, which remains in the polydisc. Holomorphic dependence on parameters and differentiation under the integral show that $F$ is a holomorphic function. For a fixed $j$, terms with $k<j$ do not depend on $z_j$, and the $k=j$ term contributes $b_j(z_1,\dots ,z_j,a_{j+1},\dots ,a_n)$. If $k>j$, use the assumed symmetry ${\partial }_j{b}_k={\partial }_k{b}_j$ to obtainThese differences telescope. ThereforeThis is the coordinate-integral primitive on a polydisc.

For a closed holomorphic one-form $\beta ={\sum }_i{c}_{i}d{z}_i$, closedness is exactly ${\partial }_i{c}_j={\partial }_j{c}_i$. The preceding construction plus ${\sum }_i{c}_i(a)(z_i-a_i)$ gives a local holomorphic function $G$ with $\partial G=\beta$. The kernel of $\partial$ on the structure sheaf of a complex manifold consists of locally constant functions: holomorphicity already gives $\bar{\partial }G=0$, and $\partial G=0$ then implies $dG=0$. Thus the holomorphic Poincaré lemma gives the short exact sequence of sheavesThe first object is the complex constant sheaf, and surjectivity is local, on stalks. It does not say that every global closed holomorphic one-form has a global primitive.

Now assume compactness. The constancy assertion is for a connected complex manifold; otherwise it holds on each connected component. A smooth complex-valued $u$ with $\partial \bar{\partial }u=0$ is a pluriharmonic function. Restricting to each complex coordinate line makes it a harmonic function of one complex variable. Choose $x_0$ where $|u|$ has its global maximum and a coordinate polydisc about $x_0$. The allowed harmonic maximum principle makes the first coordinate-line restriction constant. Every point on that line still has the same maximal modulus and the same value, so the second coordinate may then vary without changing $u$. Continue through all coordinates to conclude that $u$ equals $u(x_0)$ on the whole polydisc. The set where it has this value is closed and, by the same argument at each of its points, open. Connectedness therefore gives $u$ constant. In particular global holomorphic functions on a compact connected complex manifold are constant.

For the top-degree pairing, write locally $\theta =gd{z}_1\wedge \cdots \wedge d{z}_n$. With the complex orientation,It is everywhere nonnegative and is positive on an open set if $\theta$ is not identically zero. HenceThe phase factor is necessary: the unnormalized integral need not be a positive real number. This is the integral pairing of complex top forms.

If $\psi$ is a holomorphic differential form of degree $n-1$, then $d\psi =\partial \psi$ has type $(n,0)$ and $d\overline{d\psi }=0$. The graded Leibniz rule and Stokes theorem giveThe positive pairing just proved forces $d\psi =0$. Thus every holomorphic $(n-1)$-form on a compact complex $n$-manifold is closed, without a Kähler metric assumption.

On a compact complex surface, this proves closedness of all holomorphic one-forms. Sending such a form to its de Rham cohomology class defines the required inclusion. Indeed, if $\alpha =df$, its type $(1,0)$ gives $\bar{\partial }f=0$, so $f$ is a global holomorphic function and is constant. Therefore $\alpha =0$.

To prove the zero intersection, let $\alpha ,\beta$ be holomorphic one-forms and suppose $[\alpha ]=[\overline{\beta }]$. There is a smooth complex-valued $f$ with $\alpha -\overline{\beta }=df$. Its type components are $\partial f=\alpha$ and $\bar{\partial }f=-\overline{\beta }$. Closedness of $\beta$ gives $\partial \overline{\beta }=0$, hence $\partial \bar{\partial }f=0$. Compact pluriharmonic rigidity now makes $f$ constant and $\alpha =\beta =0$. The two spaces are therefore linearly independent in de Rham cohomology, and conjugation preserves their dimensions:For the other bound, take the long exact sequence in sheaf cohomology of the short exact sequence above. The map $H^0(M,\underline{\mathbb{C}})\to H^0(M,{\mathcal{O}}_M)$ is an isomorphism, componentwise. Since every global holomorphic one-form is closed, the relevant portion isIdentify the middle term with first de Rham cohomology, and the last with $H_{\bar{\partial }}^{0,1}(M)$ by the Dolbeault theorem. Taking dimensions givesThese are the Betti and Hodge bounds for compact complex surfaces.

A strict example is the scalar Hopf surface $X=({\mathbb{C}}^2\setminus \{0\})/⟨z\mapsto 2z⟩$. Polar coordinates identify it smoothly with $S^3\times S^1$: the dilation fixes the angular coordinate and shifts $\log |z|$ by $\log 2$. Its first Betti number is one, since its fundamental group is $\mathbb{Z}$. The proved lower bound then forces $h^{1,0}=0$, and the upper bound forces $h^{0,1}\ge 1$. Thus $h^{1,0}<h^{0,1}$ on this compact complex surface; no unsupported cohomology computation is needed.

Fix the input-index conventionExtend the connection on a vector bundle by the graded Leibniz rule, $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Applying it twice givessoThis is the Cartan curvature equation with input indices. In the more usual coefficient-column convention $A={\theta }^t$, and $(A\wedge A{)}^t=-\theta \wedge \theta$ because the entries have degree one. Thus $F=dA+A\wedge A={\Theta }^t$; the sign difference is entirely conventional.

The Chern connection of a Hermitian vector bundle with holomorphic structure is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. To construct it, use a holomorphic local frame and let $h_{ij}=⟨e_i,e_j⟩$, with the Hermitian inner product linear in the first variable. The second defining property makes $\theta$ type $(1,0)$. The first givesand therefore the only possible matrix isConversely, this formula has type $(1,0)$ and satisfies both type components of the metric identity, because $h={\overline{h}}^t$. Under a holomorphic frame change $e^{\prime }=ge$, $h^{\prime }=gh{\overline{g}}^t$ and direct differentiation givesThis is precisely the transformation obtained by differentiating $e^{\prime }=ge$. The locally constructed connections glue, proving existence, and the forced formula proves uniqueness of the Chern connection.

The derivative of $h^{-1}$ gives $\partial \theta =\theta \wedge \theta$. Consequently the Chern curvature isThe second assertion is ${\bar{\partial }}^2\theta =0$ in a holomorphic local frame; holomorphic transition functions make it intrinsic. Thus the curvature Dolbeault class is $\Psi =[\Theta ]\in H^1(M,{\Omega }_M^1\otimes \mathrm{End}E)$ by the Dolbeault theorem. Here ${\Omega }^1(\mathrm{End}E)$ denotes this tensor product. We use the identification with bundle-valued Dolbeault forms in which the antiholomorphic exterior factor is placed first, so $\bar{\partial }\theta$ has the stated positive descent sign.

For the stated transitions $e^{(\beta )}=g_{\alpha \beta }{e}^{(\alpha )}$, differentiating the frame relation yieldsTransport the alpha matrix to the beta frame before subtracting it. The difference is the holomorphic sectionOn a triple intersection, $g_{\alpha \gamma }=g_{\beta \gamma }{g}_{\alpha \beta }$, soThis is exactly the Čech cocycle condition after transporting all representatives to the gamma frame.

To identify its class, let $D_{\alpha }$ be the local holomorphic connection making the alpha frame parallel. Then $t_{\alpha }=D-D_{\alpha }$ is a genuine local section of $A^{1,0}(\mathrm{End}E)$, represented by ${\theta }_{\alpha }$, with $\bar{\partial }{t}_{\alpha }=\Theta$. Its Čech coboundary, using $t_{\beta }-t_{\alpha }$, is the intrinsic section represented by ${\sigma }_{\alpha \beta }$ in the beta frame. This local-primitive construction is the Čech–Dolbeault identification, and proves that $({\sigma }_{\alpha \beta })$ maps to $\Psi$. For an arbitrary trivializing cover use its natural map to sheaf cohomology; a Stein refinement gives the usual Čech computation. The resulting object is the Čech cocycle for the curvature Dolbeault class.

If the metric changes, the difference $B=D^{\prime }-D$ is a global smooth $(1,0)$ form with values in $\mathrm{End}E$. The local curvature formulas give ${\Theta }^{\prime }-\Theta =\bar{\partial }B$, so $\Psi$ is metric independent. If holomorphic frames change by $e_{\alpha }^{\prime }=u_{\alpha }{e}_{\alpha }$, the corresponding local tensors $t_{\alpha }^{\prime }-t_{\alpha }$, expressed in the old frame, are $B_{\alpha }=u_{\alpha }^{-1}\partial u_{\alpha }$. They are holomorphic, and their differences change the cocycle by $\delta B$. Hence $\Psi$ depends only on the holomorphic bundle, not on its metric or trivializations.

Apply exterior product and endomorphism composition to curvature to defineThese are closed for the Dolbeault operator by the product rule. Their independence can also be verified without commuting matrices. If ${\Theta }_1-{\Theta }_0=\bar{\partial }B$, thenThis follows by the noncommutative telescoping identity and the even total degree of curvature. The matrix products are interpreted in the fixed frame convention; powers of the degree-two curvature transpose to the corresponding usual curvature powers. The trace contraction is invariant under frame changes, so its class is well defined. This proves metric independence of curvature Dolbeault powers; for $k>\dim M$ the forms and groups in question vanish.

Finally take a smooth unitary frame. Metric compatibility gives ${\theta }^{†}=-\theta$, and the Cartan curvature equation with input indices gives ${\Theta }^{†}=-\Theta$. ThusThe second expression is therefore a real differential form. It is also $d$-closed: differentiating Cartan's equation gives $d\Theta =\theta \wedge \Theta -\Theta \wedge \theta$, and cyclicity of trace cancels the resulting commutator in $d\mathrm{tr}({\Theta }^k)$. On a compact Kähler manifold, the Hodge decomposition theorem for compact Kähler manifolds identifies $H^k(M,{\Omega }_M^k)$ with the $(k,k)$ component of complex de Rham cohomology. HenceIt is $k!$ times the degree-$2k$ component of the Chern character. Reality does not assert integrality of these individual power-sum classes.

The specified bidegree is the conjugate-linear Hodge star, not the complex-linear extension of the real star. If the pointwise Hermitian inner product is linear in its first argument, it is defined byIt is the real Hodge star operator extended complex-linearly and composed with complex conjugation; ${\ast }^2=(-1{)}^{p+q}$. On a compact Hermitian manifold,is a positive definite Hermitian inner product on smooth forms. The formal adjoint is ${\bar{\partial }}^{\ast }=-\ast \bar{\partial }\ast$ in this convention; the final star in this identity is present in the PDF but is corrupted in the TeX aid.

State the relevant Hodge theorem as follows. On a compact manifold without boundary, the Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$ is elliptic, self-adjoint and nonnegative. Its harmonic kernel ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional, and smooth forms split orthogonally as this kernel plus the image of ${\Delta }_{\bar{\partial }}$. There are the orthogonal projection $H$ and a smooth Green operator $G$ withFor any form $u$, expand the Laplacian term to obtainHarmonic forms are both closed and coclosed, since $({\Delta }_{\bar{\partial }}v,v)=\Vert \bar{\partial }v{\Vert }^2+\Vert {\bar{\partial }}^{\ast }v{\Vert }^2$. They are orthogonal to both other terms; the two images are orthogonal because $(\bar{\partial }a,{\bar{\partial }}^{\ast }b)=({\bar{\partial }}^{2}a,b)=0$. This proves the Dolbeault Hodge decompositionThe closed forms are exactly the first two summands: if $\bar{\partial }u=0$, then $u$ is orthogonal to the adjoint image, and conversely the first two summands are closed. It follows that every Dolbeault cohomology class has a unique harmonic representative. Negative or out-of-range bidegrees mean zero spaces. Compactness and the absence of boundary are necessary for this stated global decomposition.

Now assume a Kähler metric. Set $L=\omega \wedge -$ and $\Lambda =L^{\ast }$. The permitted Kähler identities, stated with signs, arewith ordinary commutators because $\Lambda$ has even degree. Thus ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$, andby ${\partial }^2=0$. The companion identity gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ in the same way.

Define ${\Delta }_d=d{d}^{\ast }+d^{\ast }d$ and ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$. Expanding $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$ makes the two mixed anticommutators vanish, so ${\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}$. To prove equality of the latter two, substitute the same commutator formulas and use $\partial \bar{\partial }=-\bar{\partial }\partial$:The displayed operators are identical. Therefore the Kähler Laplacian identity is

Let $\eta$ be a $\bar{\partial }$-exact form. It is closed and orthogonal to harmonic forms. The Dolbeault Hodge decomposition, or the Green identity with $\alpha =G\eta$, therefore givesFor the Green construction, $\bar{\partial }\alpha =0$ follows since $G$ commutes with $\bar{\partial }$; equivalently the coexact component of a closed form vanishes. This yields exactly the stated factorization.

If also $\partial \eta =0$, put $\beta ={\bar{\partial }}^{\ast }\alpha$ and $\xi =\partial \beta$. The identities already proved giveThus $\xi$ is harmonic for ${\Delta }_{\bar{\partial }}$ and hence for ${\Delta }_{\partial }$. But it is $\partial$-exact, so it is orthogonal to the harmonic summand in the Hodge decomposition theorem. Consequently $\xi =0$, proving that $\beta ={\bar{\partial }}^{\ast }\alpha$ is $\partial$-closed.

Furthermore $\beta$ is orthogonal to the common harmonic space, since $(\beta ,h)=(\alpha ,\bar{\partial }h)=0$. Its $\partial$-Hodge decomposition then has neither a harmonic term nor a coexact term: closedness kills the coexact term. Therefore $\beta =\partial \chi$ for a form $\chi \in A^{p-1,q-1}$. It follows thatThis proves the requested ddbar lemma, including the intermediate harmonicity and orthogonality steps. The hypotheses $p,q\ge 1$ ensure that the potential has a valid bidegree.

Finally $\eta ={\omega }_2-{\omega }_1$ is a real closed $(1,1)$ form and is $d$-exact. Integration by parts makes it orthogonal to every $d$-harmonic form. By the Kähler Laplacian identity it is also orthogonal to the Dolbeault harmonic space; because it is $\bar{\partial }$-closed, the Dolbeault Hodge decomposition makes it $\bar{\partial }$-exact. Apply the proved ddbar lemma to get $\eta =\partial \bar{\partial }\varphi$ for a smooth complex-valued function. Reality gives $\eta =-\partial \bar{\partial }\overline{\varphi }$, so averaging yieldsWith $f=\mathrm{Im}\varphi$,The potential is defined up to a real constant on each connected component. This is a statement about cohomologous Kähler forms, not about an arbitrary pair of such forms.

The canonical bundle is $K_M={\bigwedge }^m(T^{1,0}M{)}^{\ast }$, the holomorphic line bundle of holomorphic top forms on a complex $m$-manifold. For an embedded complex submanifold $V$ of dimension $n$, the holomorphic normal bundle is the quotientTaking determinants and duals gives adjunction for a smooth submanifold:For a closed smooth hypersurface, $[V]={\mathcal{O}}_M(V)$ is its divisor line bundle, equipped with a holomorphic section whose zero divisor is $V$ with multiplicity one. If $f_{\alpha }$ locally defines $V$, local meromorphic generators $1/f_{\alpha }$ transform by the holomorphic units $f_{\alpha }/f_{\beta }$. The stated normal-bundle relation isEquivalently the conormal is ${\mathcal{O}}_M(-V){|}_V$. The closed-divisor interpretation is needed to define $[V]$; an arbitrary nonclosed local submanifold does not define that global divisor bundle.

For a holomorphic subbundle $F\subset E$, define $D_{F}s=\pi D_{E}s$ on sections of $F$, with $\pi$ acting on the bundle factor of a bundle-valued form. It obeys the Leibniz rule. Since $F$ is holomorphic, $D_E^{0,1}s={\bar{\partial }}_{E}s$ lies in $F$, so $D_F^{0,1}={\bar{\partial }}_F$. For $s,t\in F$, their inner products with the normal components of $D_{E}s,D_{E}t$ vanish; hence projecting preserves metric compatibility. The uniqueness proved for the Chern connection identifiesThis is the projected Chern connection.

Choose the adapted smooth unitary frame and retain Question 2's input-index convention. The connection matrix is skew-Hermitian. Its upper-left block is ${\theta }_1$ for $D_F$; writing the lower-left block as $-A$ forces the upper-right block to be ${\overline{A}}^t$:Holomorphicity of $F$ makes the normal part of $D_E^{0,1}s$ vanish, so $B={\overline{A}}^t$ has type $(1,0)$, and $A$ has type $(0,1)$. The matrix $B$ represents the second fundamental form of a holomorphic subbundle in this input-index convention. These types would appear in the opposite blocks with coefficient-column indices; they must not be swapped while retaining the printed minus-sign Cartan equation.

Take the upper-left block of the Cartan curvature equation with input indices. It isTherefore the curvature formula for a holomorphic subbundle isThe sign and the order of the factors both follow from the displayed block multiplication.

Now let $M$ be a complex torus and $V$ a closed smooth hypersurface. The torus has a translation-invariant flat Kähler metric, and $T^{1,0}M$ and $K_M$ are holomorphically trivial. Give $E=T^{1,0}M{|}_V$ its flat metric and apply the preceding formula to $F=T^{1,0}V$. Since ${\Theta }_E=0$,For a $(1,0)$ tangent vector $X$, the latter coefficient is $\mathrm{tr}(B(X)B(X{)}^{†})\ge 0$, so $i{\Theta }_{K_V}$ is a semipositive real $(1,1)$ form. Thus $K_V$ is a semipositive holomorphic line bundle. This proves the semipositive canonical bundle of a submanifold of a complex torus assertion directly from curvature. By adjunction for a smooth submanifold, $K_V\cong [V]{|}_V$.

Restrict a positive metric on $L$ to $V$. The tensor-product metric onhas curvature $i{\Theta }_{L{|}_V}+ai{\Theta }_{K_V}$, which is positive for every nonnegative integer $a$. HenceThe integer condition is implicit in a tensor power. No assertion of strict positivity of $[V]{|}_V$ is required.

For vanishing, write $E_a=L\otimes [V{]}^{\otimes a}$. The allowed Kodaira vanishing theorem says that on a compact Kähler manifold $X$, $H^i(X,K_X\otimes P)=0$ for $i>0$ when $P$ is a positive holomorphic line bundle. Since $K_M$ is trivial, its application on the torus gives $H^i(M,E_0)=H^i(M,L)=0$ for $i>0$.

For every $a\ge 1$, the divisor section gives the exact sequencewhere $i:V\hookrightarrow M$. Crucially,and $P_{a-1}$ is positive by the proved restriction statement. The induced Kähler metric makes $V$ compact Kähler, so Kodaira vanishing theorem on $V$ gives $H^i(V,E_a{|}_V)=0$ for all $i>0$. In the long exact sequence in sheaf cohomology, the segmentthen has zero outside terms by induction. ThereforeThis proves vanishing for divisor twists on a complex torus using restriction positivity and the divisor exact sequence, rather than presuming ambient positivity of every divisor twist.