Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 118 (Complex Manifolds)

An almost complex structure on a smooth manifold $M$ is a smooth bundle endomorphism $J:TM\to TM$ satisfying $J^2=-1$. On the complexified tangent bundle, $J$ has the eigenbundle decompositionwith eigenvalues $i$ and $-i$. A differential form of type (p, q) is a section of ${\bigwedge }^p(T^{1,0}M{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}M{)}^{\ast }$. The operators $\partial$ and $\bar{\partial }$ are the type-$(1,0)$ and type-$(0,1)$ components of the exterior derivative.

We prove the three stated conditions are equivalent. If $X,Y$ have type $(1,0)$ and $\alpha$ has type $(0,1)$, the formula for the exterior derivative givesThus closure of $T^{1,0}M$ under the Lie bracket is equivalent to the vanishing of the $(2,0)$ component of $d\alpha$ for every $(0,1)$-form. Complex conjugation of differential-form type gives the corresponding vanishing of the $(0,2)$ component on $(1,0)$-forms. Since every complex one-form is a sum of these two types, this is preciselyThis proves (i)$⟺$(ii).

Under (ii), the $(2,0)$ component of $d^{2}f=0$ is ${\partial }^{2}f$, proving (iii). Conversely, for $X,Y$ of type $(1,0)$,If this vanishes for every smooth complex-valued $f$, then $[X,Y{]}^{0,1}=0$, so (iii) implies (i). These conditions define an integrable almost complex structure.

For a complex manifold, a holomorphic chart $z^j=x^j+i{y}^j$ defines $J_M({\partial }_{x^j})={\partial }_{y^j}$ and $J_M({\partial }_{y^j})=-{\partial }_{x^j}$. The derivative of a holomorphic coordinate change is complex linear, hence commutes with multiplication by $i$; the definitions therefore glue and are independent of coordinates. Locally $T^{1,0}M$ is spanned by the commuting fields $\partial /\partial z^j$, so it is closed under brackets and $J_M$ is integrable. This is the almost complex structure induced by a complex atlas.

We next prove the local Dolbeault-Poincaré lemma. Write a $\bar{\partial }$-closed $(0,q)$-form on a slightly larger polydisc aswhere neither $\alpha$ nor $\beta$ contains $d{\bar{z}}_n$. Apply the supplied one-variable Cauchy-Green operator coefficientwise to $\beta$, obtaining $\gamma$ with $\partial \gamma /\partial {\bar{z}}_n=\beta$. Then $\phi -\bar{\partial }\gamma$ contains no $d{\bar{z}}_n$, and $\bar{\partial }\phi =0$ says that its coefficients are holomorphic in $z_n$ and $\bar{\partial }$-closed in the first $n-1$ variables. Induction on $n$, with the one-variable formula as the base case, makes this remainder $\bar{\partial }$-exact. Hence every $\bar{\partial }$-closed $(0,q)$-form with $q>0$ is $\bar{\partial }$-exact on each bounded polydisc.

Finally, translation by $i$ leaves $d\bar{z}$ unchanged, so this form descends to $\mathbb{C}/(z\sim z+i)$. It is in fact exact on this noncompact complex cylinder, because the invariant function $z+\bar{z}=2\mathrm{Re}z$ descends and satisfies

A holomorphic line bundle is a complex line bundle with holomorphic transition functions, and a holomorphic section is one whose coefficient in every holomorphic local frame is holomorphic. Given a Hermitian metric on a holomorphic vector bundle $h$, its Chern connection is the connection $\nabla$ satisfying

Let $e$ be a nonvanishing holomorphic local frame, put $H=h(e,e)>0$, and write $\nabla e=Ae$. The first condition forces $A^{0,1}=0$, while metric compatibility forcesThis determines $\nabla$ uniquely and also constructs it. If $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $A^{\prime }=A+g^{-1}\partial g$, exactly the connection one-form transformation law, so the local constructions glue.

For a line bundle, $A\wedge A=0$, and the curvature form of a connection isIt has type $(1,1)$. Under $e^{\prime }=ge$, the extra term $g^{-1}\partial g$ is closed, so the curvature is unchanged and therefore global. This is the local formula for the Chern connection on a line bundle.

Any other Hermitian metric has the form $\hat{h}=e^{u}h$ for a global smooth real function $u$. Its local squared norm is $\hat{H}=e^{u}H$, whence

Connections ${\nabla }_1$ and ${\nabla }_2$ induce the tensor product connectionIts connection form in a product frame is $A_1+A_2$, so the curvature of a tensor product connection is $F_1+F_2$. For Chern connections, equip $L_1\otimes L_2$ with the product metricThe tensor product connection has the correct $(0,1)$ part and preserves this metric, so uniqueness identifies it with the Chern connection of $h$.

The tautological bundle over Complex projective space isOn $U_i=\{z_i\ne 0\}$, the vector $e_i([z])=z/z_i$ is a holomorphic frame. On $U_i\cap U_j$,so the transition functions are holomorphic. Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$ and, for $k\in \mathbb{Z}$, define $\mathcal{O}(k)=\mathcal{O}(1{)}^{\otimes k}$ when $k\ge 0$ and $\mathcal{O}(k)=\mathcal{O}(-1{)}^{\otimes (-k)}$ when $k<0$. Its transition functions are the corresponding $k$th powers of those of $\mathcal{O}(1)$, equivalently the $(-k)$th powers of those of $\mathcal{O}(-1)$.

A nonzero linear functional $\ell \in ({\mathbb{C}}^{n+1}{)}^{\ast }$ restricts on each projective line to define a global holomorphic section $s_{\ell }$ of $\mathcal{O}(1)$. Its zero divisor is the projective hyperplane $H_{\ell }=\{[z]:\ell (z)=0\}$, so every hyperplane is an effective divisor of $\mathcal{O}(1)$.

Conversely, let $D$ be the divisor of a nonzero section of $\mathcal{O}(1)$. In the affine chart $z_0=1$, the section is represented by an entire function $f$ on ${\mathbb{C}}^n$. Compatibility with the other projective charts gives the linear growth bound in the question along every complex affine line. The supplied Liouville-type result makes $f$ affine-linear, and homogenizing it gives a linear functional $\ell$ on ${\mathbb{C}}^{n+1}$. Thus $D=H_{\ell }$.

In local coordinates centred at $x$, the blowup of a complex manifold at a point is modeled byOn the chart $v_i\ne 0$, write $u=z_i$ and $t_j=v_j/v_i$; then $z_j=u{t}_j$, giving holomorphic coordinates $(u,(t_j{)}_{j\ne i})$. These charts show that the blowup is a complex manifold, that $\sigma$ is holomorphic, and that its exceptional fiber is ${\mathbb{CP}}^{n-1}$. A biholomorphic coordinate change at the centre lifts by sending a punctured point together with its limiting tangent direction to its image; the chart formulas extend across the exceptional divisor. Hence the construction is independent of coordinates up to biholomorphism.

For $a(z)=-z$ on ${\mathbb{C}}^2$, defineThis is holomorphic and satisfies $\sigma \circ \tilde{a}=a\circ \sigma$. The blowup $\widetilde{mathbb{C}^2}$ is the total space of $\mathcal{O}(-1)\to {\mathbb{CP}}^1$, and $\tilde{a}$ acts as multiplication by $-1$ in every fiber. The invariant fiber coordinate is $w=u^2$. If $u_j=g_{ji}{u}_i$ are the transition laws for $\mathcal{O}(-1)$, then $w_j=g_{ji}^2{w}_i$, which are the transition laws for $\mathcal{O}(-2)$. Thus the local quotient maps $(t,u)\mapsto (t,u^2)$ glue, give the quotient a complex-manifold atlas even along the fixed zero section, and yield

Write $g=\mathrm{Re}h$ for the underlying Riemannian metric of the Hermitian manifold. Its fundamental form of a Hermitian manifold isIt is real and skew-symmetric. In a unitary coframe it is $\omega ={\sum }_{j=1}^n{e}^j\wedge J{e}^j$, which also shows that it has type $(1,1)$ and thatThe Hodge star operator is characterized, after complex-linear extension, byExpanding in the same unitary coframe gives

The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold says that every Dolbeault class has a unique ${\Delta }_{\bar{\partial }}$-harmonic representative andIf ${\Delta }_{\bar{\partial }}\eta =0$, thenso $\eta$ is $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed. If also $\eta =\bar{\partial }\xi$, then $\Vert \eta {\Vert }_2^2=⟨{\bar{\partial }}^{\ast }\eta ,\xi ⟩=0$, hence $\eta =0$.

Now suppose $X$ is compact and Kähler. With $d^c=i(\bar{\partial }-\partial )$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$, the Kähler identities make the mixed anticommutators vanish and give ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Consequently

Let $\alpha =d^c\gamma$ and $d\alpha =0$. The Kähler Laplacian identity implies that the $d$-Laplacian commutes with $d^c$. Since a harmonic form is $d^{c\ast }$-closed, $\alpha$ is orthogonal to every harmonic form. If $G$ is the Green operator of the Hodge Laplacian, thenwhere the $d^{\ast }dG\alpha$ term vanishes because $dG\alpha =Gd\alpha =0$. The Green operator commutes with $d^c$, and the anticommutation identity just proved givesTherefore, for the $(k-2)$-form $\beta =-d^{\ast }G\gamma$,This is the d d c lemma in the form required here.