Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 18 / 4

The Chern connection of a Hermitian metric on a holomorphic vector bundle is the unique connection on a vector bundle compatible with that metric and whose $(0,1)$ part is the bundle's Dolbeault operator ${\bar{\partial }}_E$. Fix a holomorphic local frame and column coefficients for sections. Write the metric as $h(s,t)={\bar{s}}^{T}Ht$, conjugate-linear in the first argument, and write the connection as $Ds=ds+As$. The condition $D^{0,1}={\bar{\partial }}_E$ forces $A$ to have type $(1,0)$. Metric compatibility requireswhere the dagger conjugates the differential-form coefficients as well as transposing the matrix. Taking the $(1,0)$ part gives $\boxed{A=H^{-1}\partial H}$. Its conjugate-transpose supplies the $(0,1)$ metric equation because $H$ is Hermitian. This proves uniqueness and local existence.

Under a holomorphic change of frame $e^{\prime }=eg$, the metric matrix becomes $H^{\prime }=g^{†}Hg$. The local formula for the Chern connection on a vector bundle then givesThis is precisely the transformation rule for a connection on a vector bundle, so the local connections glue and establish global existence. No Kähler hypothesis is needed for this part.

Extend $D$ to vector-bundle-valued differential forms by the graded Leibniz rule. The curvature form of a connection is the tensorial square $F_h=D^2$, acting by exterior multiplication. In the chosen frame,Indeed $\partial A+A\wedge A=0$ by differentiating $H^{-1}H=I$. It follows that $F_h$ has type $(1,1)$; the gauge change is $F_h^{\prime }=g^{-1}{F}_{h}g$. Thus it is a global smooth two-form with values in $\mathrm{End}(E)$, namely an element of ${\mathcal{A}}^2(\mathrm{End}(E))$.

The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

A positive holomorphic line bundle $F$ admits a Hermitian metric whose Chern connection curvature satisfies that $i{F}_F$ is a positive real (1, 1)-form. In a local holomorphic local frame with squared length $e^{-\phi }$, the local formula for the Chern connection on a line bundle gives $F_F=\partial \bar{\partial }\phi$, so positivity means $i\partial \bar{\partial }\phi$ is positive definite. Its closedness makes $\omega =i{F}_F$ a Kähler form; use this form to define the operators below.

Let $n={\dim }_{\mathbb{C}}X>0$, choose a Hermitian metric on $E$, and equip $G_m=E\otimes F^{-m}$ with the tensor-product metric. The curvature of a tensor product connection givesOn $G_m$-valued zero-forms, the Lefschetz commutator is $[L,\Lambda ]=-n\mathrm{id}$. Thus the Bochner-Kodaira-Nakano identity givesThis last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite $C$ such that $⟨R_{E}v,v⟩\ge -C\Vert v{\Vert }^2$ at every point. For a holomorphic section $s$ of $G_m$, $D^{\prime \prime }s=0$ and $D^{\prime \prime \ast }s=0$ by degree, so integrating the identity givesChoose an integer $m_0$ with $m_{0}n>C$. Then $\boxed{H^0(X,E\otimes F^{-m})=0(m\ge m_0)}$. The threshold depends on the fixed bundle $E$, as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.

No threshold can work uniformly over all holomorphic vector bundles $E$. Given a proposed threshold $M$, choose an integer $m\ge M$ and let $E=F^m$. The canonical trivialization of a line bundle tensored with its dual givesThis contradicts vanishing at that value of $m$. In the proof above, the curvature of $E=F^m$ exactly offsets the curvature contribution from $F^{-m}$, explaining why the bound cannot be independent of $E$.