Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 17 5 Solution Created 2026-10-03 Updated 2026-10-06
Use the convention that a Hermitian metric 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 satisfying for smooth complex functions . Metric compatibility means, for every real vector field ,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 -orthonormal smooth frame . Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write . Differentiating and using compatibility givesfor every real . Thus : 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 is obtained by dualizing fibres and using transition matrices when has transitions . Their entries are holomorphic because matrix inversion is holomorphic on . Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre .
The conjugate vector bundle has the same underlying real fibres but opposite scalar action: . Its smooth transition matrices are ; they need not be holomorphic on . The conjugate bundle is used here as a smooth complex bundle, whereas is holomorphic.
Define the smooth dual tensor byIt is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of into linearity. Thus .
The conjugate connection is defined on real vector fields by and extended complex-linearly on the conjugate bundle. The tensor product connection isIts dual connection is uniquely characterized byThe Leibniz rule makes this a genuine connection on . Evaluate it on the decomposable tensor :Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every :This is metric compatibility as parallelism of a Hermitian tensor.