If $E$ has local frame transition matrices $g_{ab}$ and the cotangent bundle has transitions $J_{ab}^{-T}$, then $E\otimes\Lambda^rT^*B$ has local trivializations with transitions
$$
g_{ab}\otimes\Lambda^r(J_{ab}^{-T}),
$$
which satisfy the cocycle condition. Thus it is a well-defined <tensor product of vector bundles>.

A <connection on a vector bundle> is a linear map
$$
\nabla:\Gamma(E)\to\Omega^1(B;E)
$$
satisfying $\nabla(fs)=df\otimes s+f\nabla s$. Contracting with a vector field gives the covariant derivative $\nabla_Xs$. In a local frame, $\nabla=d+A$ for a matrix-valued one-form $A$; under a frame change $g$ the matrix transforms as
$$
A'=g^{-1}Ag+g^{-1}dg.
$$
Its <covariant exterior derivative> is defined by
$$
d_A(s\otimes\alpha)=\nabla s\wedge\alpha+s\otimes d\alpha,
$$
and locally
$$
d_A\eta=d\eta+A\wedge\eta.
$$
This formula and the graded Leibniz rule show that definitions in different frames agree.

The <curvature form of a connection> is $F(A)=d_A^2$. Locally,
$$
F(A)=dA+A\wedge A.
$$
Its covariant derivative satisfies the <Bianchi identity>
$$
d_AF=dF+A\wedge F-F\wedge A=0.
$$
Indeed, substituting $F=dA+A\wedge A$, using $d^2=0$, and applying the graded Leibniz rule leaves equal and opposite terms.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 / 3 / a / Solution

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.

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

{title2=$\nabla$}
{wiki=Connection_(vector_bundle)}

A connection is a linear map $\nabla:\Gamma(E)\to\Omega^1(B;E)$ satisfying $\nabla(fs)=df\otimes s+f\nabla s$. In a local frame it has the form $d+A$ for a matrix-valued one-form $A$.


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