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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Projected connection and Kato transport are well defined

Statement

Let M be a finite-dimensional smooth manifold, let d≥1, and let P:M→End⁡(Cd) be a C2 map satisfying P(x)2=P(x)=P(x)∗ and rank⁡P(x)=r for a fixed r. Set EP={(x,v)∈M×Cd:P(x)v=v},π(x,v)=x. Here a rank-r C2 complex vector bundle means a surjection that has local trivializations over open subsets of M with fiber maps complex-linear and transition matrices of class C2. Then EP→M is such a bundle. On each open U⊆M, the projected operation ∇s=P ds is complex-linear from C1 sections to continuous EP-valued one-forms, satisfies ∇(fs)=df s+f∇s for f∈C1(U,C), and satisfies d(s∗t)=(∇s)∗t+s∗∇t for C1 sections s,t; these are the connection and Hermitian compatibility conditions. For every C2 path γ:[0,1]→M whose derivatives extend continuously to the endpoints, the Kato equation W′(t)=[Pγ′(t),Pγ(t)]W(t),W(0)=Id, has a unique C2 solution. That solution is unitary and satisfies Pγ(t)W(t)=W(t)Pγ(0) for every t∈[0,1].

Facts & Assumptions

Given: A finite-dimensional smooth manifold M, an integer d≥1, a C2 constant-rank orthogonal projection family P, and a C2 path γ:[0,1]→M.

[F1]

C2 projector bundles and Kato transport: P2=P=P∗, its rank is constant, EP,x=ran⁡P(x), ∇s=P ds, and the Kato equation uses K=[Pγ′,Pγ].

Proof

1.1givenF1constructalgebra

Fix x0∈M and choose an orthonormal basis e1,…,er of ran⁡P(x0). The vectors P(x)ej depend C2 on x. Their Gram matrix is the identity at x0, so its determinant remains positive on a neighborhood of x0; 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 C2 orthonormal frame V(x)=(v1(x),…,vr(x)) for ran⁡P(x) on that neighborhood. The local bundle chart is Φ(x,z)=(x,V(x)z), with inverse Φ−1(x,v)=(x,V(x)∗v); both maps are C2. On an overlap, the frame change is z↦Vβ(x)∗Vα(x)z, a C2 map into the invertible complex matrices, so the transition maps are C2. These charts make EP a rank-r C2 complex vector bundle. For r=0 the bundle is the zero bundle, and for r=d every P(x)=Id and the bundle is trivial.

1.2givenF1constructalgebra

Along γ, abbreviate Pγ by P. Differentiating P2=P gives P′P+PP′=P′ and PP′P=0. Therefore K=P′P−PP′ is C1, K∗=−K, and [K,P]=P′P+PP′=P′. Let MK=sup⁡t∈[0,1]∥K(t)∥, set V0(t)=Id, and recursively set Vn+1(t)=∫0tK(u)Vn(u) du. Induction gives ∥Vn(t)∥≤(MKt)n/n!, so W=∑n=0∞Vn converges uniformly and satisfies W(t)=Id+∫0tK(u)W(u) du. The fundamental theorem of calculus gives W′=KW; since K∈C1, also W∈C2. If Z is the difference of two solutions, iteration of Z(t)=∫0tK(u)Z(u) du gives ∥Z(t)∥≤sup⁡[0,1]∥Z∥(MKt)n/n! for every n, whose right side tends to zero, so Z=0.

2.1F1step 1.1algebra

For a C1 scalar function f and a C1 section s with Ps=s, the product rule gives ∇(fs)=P d(fs)=P(df s+f ds)=df s+f∇s, so ∇ satisfies the connection Leibniz rule. For two such sections s,t, d(s∗t)=(P ds)∗t+s∗(P dt) because P=P∗ and Ps=s, Pt=t; thus ∇ preserves the Hermitian fiber metric.

3.1F1step 1.2algebra∎

Differentiation yields (W∗W)′=W∗(K∗+K)W=0, hence W∗W=Id; a square matrix with a left inverse is invertible, so W is unitary. Also (W∗PW)′=W∗(P′−[K,P])W=0, so W(t)∗P(t)W(t)=P(0). Multiplying by W(t) on the left gives P(t)W(t)=W(t)P(0), as required.

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