Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 118 (Complex Manifolds)

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

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The restriction of $J$ to the $J$-invariant bundle $TY$ is an almost complex structure. Its Nijenhuis tensor is the restriction of the ambient Nijenhuis tensor because vector fields tangent to an embedded submanifold have tangent Lie bracket. The ambient tensor vanishes since $X$ is a complex manifold, so the Newlander-Nirenberg theorem makes the induced structure on $Y$ integrable. The inclusion has complex-linear differential and is therefore holomorphic; hence $Y$ is a complex submanifold.

For a complex submanifold, the holomorphic normal bundle isIf $Y$ is a smooth hypersurface, taking top exterior powers in the holomorphic conormal sequencegivesThe normal bundle of a hypersurface is $N_{Y/X}\cong {\mathcal{O}}_X(Y){|}_Y$, so the Adjunction formula is

On $X={\mathbb{CP}}^n\times {\mathbb{CP}}^m$, a bihomogeneous polynomial of bidegree $(d_1,d_2)$ is a section of the holomorphic line bundle $\mathcal{O}(d_1,d_2)$. Its zero locus is smooth precisely when the section is transverse to the zero section, equivalently when $F$ and all of its homogeneous first partial derivatives have no common projective zero. Sincethe Adjunction formula yieldsThus a smooth $F$ with $d_1=n+1$ and $d_2=m+1$ defines a complex submanifold with trivial canonical bundle.

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Choose holomorphic coordinates centered at $p$. Locally, the blowup of a complex manifold at a point iswith $\sigma (z,[\ell ])=z$; away from $p$ this is an isomorphism, so it glues to $X\setminus \{p\}$. The exceptional divisor is $E={\sigma }^{-1}(p)\cong {\mathbb{CP}}^{n-1}$.

The proper transform is the closure of ${\sigma }^{-1}(Y\setminus \{p\})$. If $p\notin Y$, it is isomorphic to $Y$. If $p\in Y$, the holomorphic implicit function theorem supplies coordinates in which $Y=\{z_1=0\}$. In the blowup chart with $z_i=t$ and $z_j=t{u}_j$ for $j\ne i$, every chart with $i\ne 1$ describes the proper transform by $u_1=0$, while the chart $i=1$ does not meet it. These are smooth coordinate hypersurfaces, so $\tilde{Y}$ is smooth.

For a divisor $D$, the line bundle associated to a divisor $\mathcal{O}(D)$ consists locally of meromorphic functions $f$ such that $(f)+D\ge 0$. Pulling back a local defining function for $Y$ shows that its divisor iswhere $m$ is the order of vanishing at $p$ of a local defining function for $Y$. ThereforeBecause $Y$ is smooth, $m=1$ when $p\in Y$ and $m=0$ when $p\notin Y$.

Applying the definition with $D=E$ gives directlyThe map sends a section to its local meromorphic coefficient relative to the canonical meromorphic section of $\mathcal{O}(E)$; the divisor inequality is exactly the condition that these coefficients define a holomorphic section, and the inverse construction is local multiplication by that canonical section.

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A Riemannian metric $g$ on a complex manifold is a Kähler metric when its complex-linear extension is a Hermitian form on each tangent space and its fundamental two-formis closed. Equivalently, $\omega$ is a positive real closed $(1,1)$-form, making $X$ a Kähler manifold.

The Lefschetz operator of a Kähler manifold and its adjoint areBecause $d\omega =0$ and $\omega$ has type $(1,1)$, both $\partial \omega$ and $\bar{\partial }\omega$ vanish. The graded Leibniz rule therefore gives $[L,\partial ]=[L,\bar{\partial }]=0$.

Writing formal adjoints with stars, define the three Laplacians byThe supplied Kähler identities identity $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$ gives, by complex conjugation and taking adjoints,Expanding these commutators and using $\partial \bar{\partial }+\bar{\partial }\partial =0$ shows that the mixed terms in $\Delta$ vanish and that ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Since $d=\partial +\bar{\partial }$, it follows that

The adjoint of $[L,\partial ]=0$ gives $[\Lambda ,{\partial }^{\ast }]=0$. Hencewhere the last equality is the adjoint of $\partial \bar{\partial }+\bar{\partial }\partial =0$. Thus $\Lambda$, and therefore every power ${\Lambda }^k$, commutes with ${\Delta }_{\partial }$. It follows that ${\Lambda }^k$ sends every $\partial$-harmonic $(p,q)$-form to a $\partial$-harmonic $(p-k,q-k)$-form whenever $0\le k\le \min (p,q)$.

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