Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 118 / 4

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 $H_{\bar{\partial }}^{r,r}(X)$.

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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 $H_{\bar{\partial }}^{1,1}(X)$.

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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.

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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

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