Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 17 / 5 / Solution

Use the convention that a Hermitian metric $h$ is complex-linear in its first argument and conjugate-linear in its second. It is a smoothly varying positive-definite Hermitian form on each fibre. A connection on a vector bundle is a complex-linear operator $D:\Gamma (E)\to {\Omega }^1(X,E)$ satisfying $D(as)=da\otimes s+aDs$ for smooth complex functions $a$. Metric compatibility means, for every real vector field $V$,This is the metric-compatible connection condition with the sesquilinear convention fixed.

Apply the smooth Gram-Schmidt process to a local frame to obtain an $h$-orthonormal smooth frame $s_1,\dots ,s_r$. Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write $D{s}_j={\sum }_i{A}_{ij}{s}_i$. Differentiating $h(s_i,s_j)={\delta }_{ij}$ and using compatibility givesfor every real $V$. Thus $\boxed{A(V{)}^{†}=-A(V)}$: the connection matrix is skew-Hermitian. This smooth unitary frame for a Hermitian connection is generally not holomorphic; the requested local-frame assertion requires only a smooth frame.

The holomorphic dual vector bundle $E^{\ast }$ is obtained by dualizing fibres and using transition matrices $g_{ij}^{-T}$ when $E$ has transitions $g_{ij}$. Their entries are holomorphic because matrix inversion is holomorphic on $G{L}_r(\mathbb{C})$. Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre $(E_x{)}^{\ast }$.

The conjugate vector bundle $\overline{E}$ has the same underlying real fibres but opposite scalar action: $\lambda \cdot \overline{v}=\overline{\overline{\lambda }v}$. Its smooth transition matrices are $\overline{g_{ij}}$; they need not be holomorphic on $X$. The conjugate bundle is used here as a smooth complex bundle, whereas $E^{\ast }$ is holomorphic.

Define the smooth dual tensor $H$ byIt is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of $h$ into linearity. Thus $H\in \Gamma ((E\otimes \overline{E}{)}^{\ast })$.

The conjugate connection is defined on real vector fields by ${\overline{D}}_V(\overline{s})=\overline{D_{V}s}$ and extended complex-linearly on the conjugate bundle. The tensor product connection isIts dual connection $D_0$ is uniquely characterized byThe Leibniz rule makes this a genuine connection on $(E\otimes \overline{E}{)}^{\ast }$. Evaluate it on the decomposable tensor $u=s\otimes \overline{t}$:Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every $V,s,t$:This is metric compatibility as parallelism of a Hermitian tensor.