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

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.