Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 118 (Complex Manifolds)

On the affine chart $Z_i\ne 0$, divide the homogeneous polynomial $F$ by $Z_i^d$ to obtain a holomorphic function $f_i$ of two affine coordinates. The hypothesis $\mathrm{grad}F(p)\ne 0$ says that at every zero at least one affine partial derivative of $f_i$ is nonzero; the radial derivative contributes nothing on $F=0$ by Euler's homogeneous identity. The holomorphic implicit-function theorem therefore makes $f_i^{-1}(0)$ a complex submanifold of complex codimension one. These local loci agree on chart overlaps, so $X$ is a complex one-dimensional submanifold of ${\mathbb{CP}}^2$.

Solved by gpt-5.6-sol high.

The Adjunction formula, $K_X\cong (K_{{\mathbb{CP}}^2}\otimes \mathcal{O}(d)){|}_X$, and $K_{{\mathbb{CP}}^2}\cong \mathcal{O}(-3)$ giveIf $h=c_1(\mathcal{O}(1))$, thenWith the standard integral normalization of the Fubini-Study form, this becomes $c_1(TX)=(3-d)[{\omega }_{FS}{|}_X/(2\pi )]$ in real cohomology.

Solved by gpt-5.6-sol high.

A Kähler manifold has a real closed $(1,1)$-form $\omega$ for which $g(u,v)=\omega (u,Jv)$ is positive definite. The Fubini-Study form ${\omega }_{FS}$ has these properties on ${\mathbb{CP}}^2$. Pullback by the holomorphic inclusion $i:X\hookrightarrow {\mathbb{CP}}^2$ preserves reality, type, and closedness, while positivity restricts to every nonzero vector in $TX$. Hence ${\omega }_X=i^{\ast }{\omega }_{FS}$ is a Kähler form on $X$.

Solved by gpt-5.6-sol high.

The curvature of the Chern connection on the complex line bundle $TX$ has type $(1,1)$, and $i{F}_{\nabla }$ is real. Since $X$ has complex dimension one, every real two-form is a unique smooth multiple of its nonvanishing area form, sofor some real smooth function $\lambda$. By Chern-Weil theory,The hyperplane class has degree $d$ on a degree-$d$ plane curve, so

Solved by gpt-5.6-sol high.

The form $i{F}_{\nabla }$ is the Ricci form of a Kähler manifold. In complex dimension one, $i{F}_{\nabla }=K_g{\omega }_X=\frac{1}{2}{\mathrm{Scal}}_g{\omega }_X$. Thuswhere $K_g$ is the Gaussian curvature.

Solved by gpt-5.6-sol high.

Choose a small ball $B$ whose translates by distinct lattice elements are disjoint. The restrictions of the quotient map $\pi :{\mathbb{C}}^n\to {\mathbb{C}}^n/\Lambda$ to translates of $B$ supply holomorphic charts, because every transition map is a complex translation. A closed fundamental parallelepiped is compact and surjects onto the quotient, so the resulting complex torus $X$ is compact.

The standard formis translation invariant and therefore descends uniquely to a form ${\omega }_X$ with ${\pi }^{\ast }{\omega }_X={\omega }_0$. It remains closed, of type $(1,1)$, and positive, so it is a Kähler form.

Solved by gpt-5.6-sol high.

The forms $d{z}_I\wedge d{\overline{z}}_J$ are parallel for the flat metric. Consequently the Hodge Laplacian acts coefficientwise:If every coefficient is constant, this vanishes. Conversely, if $\Delta \alpha =0$, orthogonality of the constant frame gives $\Delta {\alpha }_{I,J}=0$ for every $I,J$. Each coefficient is a harmonic function on a compact connected manifold and hence is constant by the Strong maximum principle for harmonic functions.

Solved by gpt-5.6-sol high.

The Hodge decomposition theorem for compact Kähler manifolds stateswith each summand represented uniquely by harmonic forms of type $(p,q)$. Part b shows that these are precisely the constant-coefficient forms $d{z}_I\wedge d{\overline{z}}_J$. Therefore the Hodge numbers of a complex $n$-torus are

Solved by gpt-5.6-sol high.

All quotients ${\mathbb{C}}^n/\Lambda$ are diffeomorphic to the real torus ${\mathbb{R}}^{2n}/{\mathbb{Z}}^{2n}$ after choosing a real basis of the lattice. The complex structure is nevertheless visible in cohomology: under the cap-product period pairing, $H^{1,0}(X)$ is an $n$-dimensional subspace ofand its elements are exactly the period homomorphisms of holomorphic one-forms.

A biholomorphism pulls $H^{1,0}$ onto $H^{1,0}$ and induces an element of $GL(2n,\mathbb{Z})$ on integral first homology. The group $GL(2n,\mathbb{Z})$ is countable, so the orbit of any one period subspace is countable. On the other hand, varying a period parameter $\tau$ in the upper half-plane in lattices generated by $e_1,\dots ,e_n$ and $\tau e_1,i{e}_2,\dots ,i{e}_n$ produces uncountably many such subspaces. Two choices lying in distinct $GL(2n,\mathbb{Z})$-orbits therefore give diffeomorphic but nonbiholomorphic complex $n$-tori.

Solved by gpt-5.6-sol high.

Complexifying the cotangent bundle of an almost complex manifold gives the $\pm i$ eigenspace decomposition $T^{\ast }X\otimes \mathbb{C}=T^{1,0\ast }X\oplus T^{0,1\ast }X$. DefineFor a form $\alpha$ of type $(p,q)$, the Dolbeault operator is the type-$(p,q+1)$ projection of $d\alpha$:

Solved by gpt-5.6-sol high.

For $(0,1)$ vector fields $V,W$, direct evaluation givesup to the harmless overall sign fixed by the exterior-derivative convention. If $J$ is integrable, $(0,1)$ fields are closed under bracket, so the right side vanishes. Conversely, if it vanishes for every smooth complex function $f$, differentials separate tangent vectors and therefore $[V,W{]}^{1,0}=0$. Involutivity of $T^{0,1}X$, equivalently vanishing of the Nijenhuis tensor, is the Newlander-Nirenberg criterion for an integrable almost complex structure. Thus $J$ is integrable exactly when ${\bar{\partial }}^{2}f=0$ for every $f$.

Solved by gpt-5.6-sol high.

For a scalar form $\alpha \in {\Omega }^{p,q}(X)$ and a section $s$ of $E$, define the Dolbeault partial connection on decomposable forms byand extend linearly. This is independent of the chosen local expression precisely because the original operator obeys its Leibniz rule.

Applying the rule twice makes the two mixed terms cancel. Since $X$ is complex, ${\bar{\partial }}^{2}f=0$, and for every smooth function $f$ and $E$-valued form $\eta$ one obtainsThus ${\bar{\partial }}_E^2$ is linear over $C^{\infty }(X,\mathbb{C})$.

Solved by gpt-5.6-sol high.

Choose a local holomorphic frame $e$ of the holomorphic line bundle $E$ and defineIf $e^{\prime }=ge$ is another holomorphic frame, then $g$ is nowhere-zero and holomorphic, so $\bar{\partial }g=0$ and the two definitions agree. They therefore glue to a well-defined partial connection. In each holomorphic frame its square is ordinary ${\bar{\partial }}^2$, hence ${\bar{\partial }}_E^2=0$.

Solved by gpt-5.6-sol high.

The Dolbeault cohomology group isThe complex orientation of the codimension-$r$ submanifold $Z$ gives an integral Poincaré-dual class $[Z]\in H^{2r}(X;\mathbb{Z})$. Integration over $Z$ defines a closed current of type $(r,r)$, so under the Dolbeault identification its complexification belongs to $H_{\bar{\partial }}^{r,r}(X)$.

Solved by gpt-5.6-sol high.

The first assertion is false with the standard meaning of linear equivalence of divisors. For example, two distinct lines $D_1,D_2\subset {\mathbb{CP}}^2$ are smooth and linearly equivalent but intersect. Distinct fibers of a holomorphic map to ${\mathbb{CP}}^1$ are disjoint, so no map can have these lines as the fibers over $0$ and $\infty$. The ratio of defining sections gives only a meromorphic map, with an indeterminacy point at $D_1\cap D_2$. The assertion becomes true if the two sections have no common zero.

The requested cohomological conclusion is nevertheless valid. Linearly equivalent divisors define isomorphic holomorphic line bundles, and their Poincaré-dual Dolbeault classes both equal the First Chern class of that bundle. Hence $[D_1]=[D_2]$ in $H_{\bar{\partial }}^{1,1}(X)$.

Solved by gpt-5.6-sol high.

The projective linear group acts transitively on ${\mathbb{CP}}^2$. A projective automorphism carrying $p$ to another point $q$ lifts, by the universal construction of the blowup of a complex manifold at a point, to a biholomorphism between the two blowups. Thus the biholomorphism type is independent of the center.

If $D_1\sim D_2$, then $D_1-D_2$ is the divisor of a meromorphic function $f$. Pulling back givesso the total inverse-image divisors are linearly equivalent on $\tilde{X}$. Here ${\pi }^{-1}(D_i)$ must mean the total transform; strict transforms need not be linearly equivalent if their multiplicities at the blown-up point differ.

Solved by gpt-5.6-sol high.

The blowup is a compact Kähler surface. It is simply connected, so $b_1=b_3=0$, while $b_0=b_4=1$ and the supplied fact gives $b_2=2$. It has no holomorphic two-forms: blowing up a point does not change $H^{2,0}$, and $h^{2,0}({\mathbb{CP}}^2)=0$. Hodge decomposition and conjugation symmetry therefore giveand

Solved by gpt-5.6-sol high.