Path transport composes and transforms covariantly

On the constant path the coefficient $\omega (\dot{\gamma })$ is zero, so the unique solution is $U(t)=I$. If $\gamma$ is followed by $\delta$, solve on the two parameter intervals with the terminal value from the first interval as initial value for the second. The second solution is $U_{\delta }(t)U_{\gamma }(1)$, so uniqueness gives $U_{\delta \ast \gamma }(1)=U_{\delta }(1)U_{\gamma }(1)$.

Write $\eta =\mathrm{diag}(-1,1,1,1)$. Along each smooth segment the derivative is $\frac{d}{dt}(U^{\mathsf{T}}\eta U)=U^{\mathsf{T}}(-{\omega }^{\mathsf{T}}\eta -\eta \omega )U=0$ by [F4]; continuity across the finite segment endpoints gives $U(t{)}^{\mathsf{T}}\eta U(t)=\eta$ throughout. Since $\eta$ is nonsingular, each $U(t)$ is invertible.

Let $U(t)$ solve along $\gamma$. The matrix $V(s):=U(1-s)U(1{)}^{-1}$ starts at $I$ and, since $\dot{\bar{\gamma }}(s)=-\dot{\gamma }(1-s)$, satisfies $V^{\mathrm{\prime }}(s)=-\omega (\dot{\bar{\gamma }}(s))V(s)$. Hence uniqueness gives $U_{\bar{\gamma }}(1)=U(1{)}^{-1}$. By [F3], transport is independent of the representative of a thin class; [F1]–[F2] and the identities in steps 1.1–1.2 therefore give well-defined component maps on the thin path groupoid.

For $[\gamma ]:x\to y$ and $[\delta ]:y\to z$, the associated tangent arrows compose as $(E_z{U}_{\delta }{E}_y^{-1})\circ (E_y{U}_{\gamma }{E}_x^{-1})=E_z{U}_{\delta }{U}_{\gamma }{E}_x^{-1}=E_z{U}_{\delta \ast \gamma }{E}_x^{-1}$ by step 1.1. Constant paths give identity tangent maps, and reversed paths give inverse tangent maps by step 2.1. Thus the stated object and arrow assignments define a functor into $\mathcal{T}(U)$ by [F6], with arrows in the stated groupoid by [F5].

Put $\tilde{U}(t)=\mathrm{\Lambda }(\gamma (t))U(t)\mathrm{\Lambda }(\gamma (0){)}^{-1}$. Differentiating and using $U^{\mathrm{\prime }}=-\omega (\dot{\gamma })U$ gives ${\tilde{U}}^{\mathrm{\prime }}(t)=(d\mathrm{\Lambda }(\dot{\gamma }){\mathrm{\Lambda }}^{-1}-\mathrm{\Lambda }\omega (\dot{\gamma }){\mathrm{\Lambda }}^{-1})\tilde{U}(t)=-{\omega }^{\mathrm{\prime }}(\dot{\gamma })\tilde{U}(t)$ by the transformation law [F4]. Its initial value is $I$, so uniqueness identifies $\tilde{U}$ with transport in the changed frame and gives the endpoint formula. Since $E_x^{\mathrm{\prime }}=E_x\mathrm{\Lambda }(x{)}^{-1}$, the corresponding tangent map is frame independent: $E_y^{\mathrm{\prime }}{U}_{\gamma }^{\mathrm{\prime }}(1)(E_x^{\mathrm{\prime }}{)}^{-1}=E_y{U}_{\gamma }(1)E_x^{-1}\mathrm{.}$ This is a local frame covariance statement; it is a principal-bundle gauge action only when $\mathrm{\Lambda }$ has the global extension required by the cited gauge-group definition.