Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 118 / 3

A connection on a vector bundle is a real-linear map (complex-linear for a complex bundle)It extends to the covariant exterior derivative by $D(\eta \otimes s)=d\eta \otimes s+(-1{)}^k\eta \wedge Ds$ when $\eta$ has degree $k$. Its square is linear over smooth functions: in $D^2(fs)$ the two $df\wedge Ds$ terms have opposite signs and $d^{2}f=0$. The curvature form of a connection is therefore the endomorphism-valued two-formIn a local frame with $D=d+\theta$, it is $\Theta =d\theta +\theta \wedge \theta$.

Define the tensor product connection on decomposable sections byThe Leibniz rule for $D_1,D_2$ makes this well-defined over smooth functions and makes it a connection on a vector bundle. Applying its covariant exterior derivative again givesThe mixed terms cancel by the graded sign. Thus curvature adds on the two tensor factors:Equivalently, the mixed terms in $({\theta }_1\otimes I+I\otimes {\theta }_2)\wedge ({\theta }_1\otimes I+I\otimes {\theta }_2)$ cancel because operators on the different factors commute. This also proves the curvature of a tensor product connection formula in local frames.

Fix the convention that the Hermitian metric is conjugate-linear in its first argument. In a holomorphic local frame $e=(e_1,\dots ,e_r)$, let $H_{ij}=h(e_i,e_j)$, so $h(ev,ew)=v^{†}Hw$. Write $De=e\theta$, or equivalently $D(ev)=e(dv+\theta v)$.

Compatibility with the holomorphic vector bundle structure means $D^{0,1}={\bar{\partial }}_E$. Since the frame is holomorphic, this forces ${\theta }^{0,1}=0$. Metric compatibility meansor, in the chosen frame, $dH={\theta }^{†}H+H\theta$. Conjugate transpose here also conjugates the one-form factors. Taking the $(1,0)$ component gives $\partial H=H\theta$, and henceThis proves uniqueness and suggests existence. The proposed $\theta$ has type $(1,0)$, and its conjugate transpose obeys ${\theta }^{†}H=\bar{\partial }H$, since $H=H^{†}$. Therefore $dH={\theta }^{†}H+H\theta$, proving metric compatibility as well.

Under a change of holomorphic local frame $e^{\prime }=eG$, we have $H^{\prime }=G^{†}HG$ and $\partial G^{†}=0$. A direct computation yieldsBecause $G$ is holomorphic, $\partial G=dG$; this is exactly the transformation rule for a connection one-form. The locally defined connections therefore glue to a global connection on a vector bundle. The Chern connection exists uniquely, and its local formula for the Chern connection on a vector bundle is

The orthogonal projection ${\pi }_S$ is a smooth bundle map, since the Hermitian metric and the subbundle vary smoothly. The mapis independent of the representative, bijective on each fibre, and smooth with smooth inverse supplied by the quotient map on $S^{\perp }$. Hence it is a natural isomorphism of smooth vector bundles. It need not be holomorphic. The quotient Hermitian metric isIt is well-defined and positive definite because $S^{\perp }$ represents each quotient class uniquely.

On sections of $S$, set $D^{\prime }={\pi }_S{D}_E$. The Leibniz rule makes $D^{\prime }$ a connection on a vector bundle. For $s,t\in \Gamma (S)$, the orthogonal terms vanish in the metric pairing, soThus $D^{\prime }$ has metric compatibility. Because $S$ is a holomorphic subbundle, the $(0,1)$ part of $D_E$ preserves $S$ and restricts to ${\bar{\partial }}_S$. Therefore $(D^{\prime }{)}^{0,1}={\bar{\partial }}_S$. The uniqueness of the Chern connection in part (b) gives the projected Chern connection formulaIt follows that $A(s)=(I-{\pi }_S)D_{E}s$ is a quotient-valued one-form. The printed target $T_M\otimes Q$ is missing a dual: the cotangent bundle is the correct factor,Complex-valued forms use the complexified cotangent bundle here. In fact the $(0,1)$ part cancels, so $A$ is the second fundamental form of a holomorphic subbundle, an element of ${\mathcal{A}}^{1,0}(\mathrm{Hom}(S,Q))$. Finally the two connection Leibniz rules giveThis proves its tensoriality and all the asserted quotient and projection properties.