Small-loop transport recovers curvature

Set $A={\omega }_i(p)$ and $B={\omega }_j(p)$. The integral equation $U(t)=I-{\int }_0^t\omega (\dot{\gamma })U$ gives the positive-$i$ segment expansion $S_1=I-\epsilon A+\frac{1}{2}{\epsilon }^2(A^2-{\mathrm{\partial }}_{i}A)+O({\epsilon }^3)$. On the following positive-$j$ segment, Taylor expansion at $p+\epsilon e_i$ gives $S_2=I-\epsilon B+{\epsilon }^2(\frac{1}{2}{B}^2-{\mathrm{\partial }}_{i}B-\frac{1}{2}{\mathrm{\partial }}_{j}B)+O({\epsilon }^3)$. In each case the second iterated integral contributes $\frac{1}{2}{\epsilon }^2{\omega }_k(p{)}^2$, and the first integral contributes the displayed derivative terms.

Along the negative-$i$ segment from $p+\epsilon e_i+\epsilon e_j$ to $p+\epsilon e_j$ and the negative-$j$ segment from $p+\epsilon e_j$ to $p$, the same integral expansion gives $S_3=I+\epsilon A+{\epsilon }^2(\frac{1}{2}{A}^2+\frac{1}{2}{\mathrm{\partial }}_{i}A+{\mathrm{\partial }}_{j}A)+O({\epsilon }^3)$ and $S_4=I+\epsilon B+\frac{1}{2}{\epsilon }^2(B^2+{\mathrm{\partial }}_{j}B)+O({\epsilon }^3)$. The derivative terms use starting points $p+\epsilon e_j$ and $p$; reversing the tangent changes the first-order ODE coefficient's sign.

For any fixed $q$, uniqueness in [F1] shows that adjoining a segment with identity-initial transport $S$ to an already traversed path with transport $T$ gives total matrix $ST$: $S(t)T$ solves the next segment's ODE with its required initial value. Applying this successively gives total transport $S_4{S}_3{S}_2{S}_1$; for $q=4$ this agrees with [F3] using the same component ODE of [F4]. Its linear terms cancel. The within-segment quadratic terms sum to $A^2+B^2-{\mathrm{\partial }}_{i}B+{\mathrm{\partial }}_{j}A$; the six cross-products sum to $BA-B^2-BA-AB-A^2+BA=BA-AB-A^2-B^2$. The full coefficient is therefore ${\mathrm{\partial }}_{j}A-{\mathrm{\partial }}_{i}B+BA-AB=-({\mathrm{\partial }}_i{\omega }_j-{\mathrm{\partial }}_j{\omega }_i+[A,B])=-F_{ij}(p)$ by [L1]. In the smooth tetrad specialization, evaluating $d\omega +\omega \wedge \omega$ on the commuting coordinate vectors gives exactly this expression, since $d\omega ({\mathrm{\partial }}_i,{\mathrm{\partial }}_j)={\mathrm{\partial }}_i{\omega }_j-{\mathrm{\partial }}_j{\omega }_i$ and $(\omega \wedge \omega )({\mathrm{\partial }}_i,{\mathrm{\partial }}_j)={\omega }_i{\omega }_j-{\omega }_j{\omega }_i$, in agreement with [F2]. For a uniform remainder, choose a compact neighborhood $K^{\mathrm{\prime }}$ of $K$ contained in $V$ and containing the four square sides for $0<\epsilon <{\epsilon }_0$; the $C^2$ coefficients and derivatives are bounded there. Taylor remainders in single integrals are uniformly $O({\epsilon }^3)$; replacing either coefficient in the second iterated integral by its value at $p$ also incurs $O({\epsilon }^3)$, since each coefficient differs by $O(\epsilon )$ over these length-$\epsilon$ sides. If $M$ bounds coefficient norms in the fixed submultiplicative norm, iterated-integral terms of order $m\ge 3$ are bounded by $M^m{\epsilon }^m\mathrm{/}m!$, whose sum is $O({\epsilon }^3)$. Multiplying four expansions preserves the uniform remainder order.