A <Hermitian metric on a holomorphic vector bundle> $E\to X$ is a smoothly varying family of positive-definite <Hermitian form>[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 is
$$
\theta=h^{-1}\partial h=\partial\log h,
\qquad \nabla e=\theta\otimes e.
$$
If $e'=ge$ for a nowhere-zero <holomorphic function> $g$, then $h'=|g|^2h$ and
$$
\theta'=\partial\log h'=\theta+g^{-1}\partial g,
$$
which is exactly the <connection on a vector bundle>[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 $U_i=\{Z_i\ne0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frame
$$
s_i(w)=(w_0,\ldots,w_{i-1},1,w_{i+1},\ldots,w_n).
$$
Then
$$
h_i(s_i,s_i)=1+\sum_{j\ne i}|w_j|^2,
\qquad
\theta_i=\frac{\sum_{j\ne i}\overline w_j\,dw_j}{1+\sum_{j\ne i}|w_j|^2}.
$$
The <curvature form of a connection> is
$$
F_\nabla=\bar\partial\theta_i=-\partial\bar\partial\log(1+|w|^2).
$$
Consequently <First Chern class>[Chern-Weil theory] gives the closed representative
$$
c_1(\mathcal O(-1))=\left[\frac{iF_\nabla}{2\pi}\right]
=\left[-\frac{i}{2\pi}\partial\bar\partial\log(1+|w|^2)\right]
=-\left[\frac{\omega_{FS}}{2\pi}\right],
$$
where $\omega_{FS}$ is the <Fubini-Study form>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 118 / 1 / Solution

The curvature of the <Chern connection> on the complex line bundle $TX$ has type $(1,1)$, and $iF_\nabla$ is real. Since $X$ has complex dimension one, every real two-form is a unique smooth multiple of its nonvanishing area form, so
$$
iF_\nabla=\lambda\omega_X
$$
for some real smooth function $\lambda$. By <First Chern class>[Chern-Weil theory],
$$
\frac1{2\pi}\int_X\lambda,d\operatorname{vol}_g
=\int_Xc_1(TX)
=(3-d)\int_Xh.
$$
The hyperplane class has degree $d$ on a degree-$d$ plane curve, so
$$
\frac1{2\pi}\int_X\lambda,d\operatorname{vol}_g=d(3-d).
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 118 / 1 / d / Solution

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

Solved by gpt-5.6-sol high.

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

{c}
{title2=$c_1(E)$}
{wiki=Chern_class}

The first Chern class classifies complex line bundles topologically. For a Hermitian holomorphic line bundle with Chern curvature $F_\nabla$, Chern-Weil theory gives $c_1(E)=[iF_\nabla/(2\pi)]$.


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