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Projected connection and Kato transport are well defined
Statement
Let be a finite-dimensional smooth manifold, let , and let be a map satisfying and for a fixed . Set Here a rank- complex vector bundle means a surjection that has local trivializations over open subsets of with fiber maps complex-linear and transition matrices of class . Then is such a bundle. On each open , the projected operation is complex-linear from sections to continuous -valued one-forms, satisfies for , and satisfies for sections ; these are the connection and Hermitian compatibility conditions. For every path whose derivatives extend continuously to the endpoints, the Kato equation has a unique solution. That solution is unitary and satisfies for every .
Statement references
Facts & Assumptions
Given: A finite-dimensional smooth manifold , an integer , a constant-rank orthogonal projection family , and a path .
C2 projector bundles and Kato transport: , its rank is constant, , , and the Kato equation uses .
Proof
Fix and choose an orthonormal basis of . The vectors depend on . Their Gram matrix is the identity at , so its determinant remains positive on a neighborhood of ; hence the vectors remain linearly independent there. Applying the finite Gram–Schmidt formulas uses only sums, products, division by positive norms, and square roots of positive functions, so it gives a orthonormal frame for on that neighborhood. The local bundle chart is , with inverse ; both maps are . On an overlap, the frame change is , a map into the invertible complex matrices, so the transition maps are . These charts make a rank- complex vector bundle. For the bundle is the zero bundle, and for every and the bundle is trivial.
Along , abbreviate by . Differentiating gives and . Therefore is , , and . Let , set , and recursively set . Induction gives , so converges uniformly and satisfies . The fundamental theorem of calculus gives ; since , also . If is the difference of two solutions, iteration of gives for every , whose right side tends to zero, so .
For a scalar function and a section with , the product rule gives , so satisfies the connection Leibniz rule. For two such sections , because and , ; thus preserves the Hermitian fiber metric.
Differentiation yields , hence ; a square matrix with a left inverse is invertible, so is unitary. Also , so . Multiplying by on the left gives , as required.