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 give
$$
h^{0,0}=h^{2,2}=1,
\qquad
h^{1,1}=2,
$$
and
$$
h^{1,0}=h^{0,1}=h^{2,0}=h^{0,2}=h^{2,1}=h^{1,2}=0.
$$

Solved by gpt-5.6-sol high.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-118/4/d/solution