The function $f(x,y)=x^2+y^2$ has derivative $(2x,2y)\mathrm{\ne }0$ on $S^1=f^{-1}(1)$, so [F5] gives its usual smooth one-dimensional submanifold structure. Let $j:M\hookrightarrow {\mathbb{R}}^2\times {\mathbb{R}}^3$ be the product inclusion, put $\tilde{\alpha }=xdy-ydx$ on ${\mathbb{R}}^2\times {\mathbb{R}}^3$, and interpret the stated form as $\alpha =j^{\ast }\tilde{\alpha }$, defined by [F7] and smooth by [F8], hence a smooth cotangent section by [F4]. On the global frame, $\alpha (E_1)=1$ and $\alpha (E_0)=\alpha (E_2)=\alpha (E_3)=0$; the frame fields commute pairwise. By [F9], $d\alpha$ is a smooth two-form. Thus [F6] gives $d\alpha (E_I,E_J)=E_I(\alpha (E_J))-E_J(\alpha (E_I))-\alpha ([E_I,E_J])=0$ for every $I,J$, so $d\alpha =0$ intrinsically on $M$. Also $\alpha (E_1)=1$ and it vanishes on the other frame vectors, so the displayed frame is orthonormal for $g$; it is smooth and $E_0$ is future-directed. In the $E_2,E_3$ plane the matrix $J$ is the standard skew rotation generator, so $J^{\mathsf{T}}\eta +\eta J=0$. The displayed formula for $\mathrm{\nabla }$ is linear over smooth functions in $X$, additive in the vector field argument, and satisfies ${\mathrm{\nabla }}_X(fV)=X(f)V+f{\mathrm{\nabla }}_{X}V$ by the ordinary product rule; hence it defines a smooth affine connection with tetrad matrix $\omega$. Its curvature is $F=d\omega +\omega \wedge \omega =\frac{1}{4}Jd\alpha +\frac{1}{16}{J}^2\alpha \wedge \alpha =0$, since $d\alpha =0$ and $\alpha \wedge \alpha =0$. It is $\eta$-skew and flat, so it is a Lorentz connection under [F1].

On an angular arc, $\alpha =d\theta$. The matrix $U(t)=\exp (-\frac{1}{4}J(\theta (t)-\theta (0)))$ starts at $I_4$; termwise differentiation of the defining series, whose derivative series converges uniformly on compact parameter intervals, gives $\dot{U}(t)=-\frac{1}{4}J\dot{\theta }(t)U(t)$. By [F2] it is the transport by uniqueness, and it depends only on the endpoints on that arc. For the path $\theta (t)=2\pi t$, its endpoint is $\exp (-\frac{\pi }{2}J)$. Since $J^2=-I$ on $\mathrm{span}(E_2,E_3)$, separating the even and odd terms of the series gives $\exp (-\frac{\pi }{2}J)=\cos (\frac{\pi }{2})I-\sin (\frac{\pi }{2})J=-J$ on that plane, so it is not the identity. If the once-around loop in $M$ were thin-contractible, composing that homotopy with the smooth projection $M\to S^1$ would preserve fixed endpoints, the finite smooth-triangle subdivision, and rank at most one, and would contract the once-around circle loop, contrary to [F3]. Thus its loop class is nontrivial despite $F=0$.

For every $g\in SU(2)$, $g^{†}g=I_2$, so the defining action on ${\mathbb{C}}^2$ preserves its standard inner product; matrix multiplication gives $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$, and the matrix entries vary continuously in the Euclidean matrix topology. Thus each orbit map $g\mapsto \rho (g)v$ is continuous and $\rho$ is a strongly continuous finite-dimensional unitary representation as required by the model. The matrix $k$ is unitary and has determinant $i(-i)=1$, so $k\in SU(2)$; also $k\mathrm{\ne }{I}_2$ and $k^{-1}\mathrm{\ne }{I}_2$. By [F1], $g_{e_1}^{\mathrm{\prime }}=I_2{I}_2{k}^{-1}=k^{-1}$, $g_{e_2}^{\mathrm{\prime }}=I_2$, and $g_{e_3}^{\mathrm{\prime }}=k{I}_2{I}_2=k$. Thus individual edge matrices change.

Their later-left loop product is $g_{e_3}^{\mathrm{\prime }}{g}_{e_2}^{\mathrm{\prime }}{g}_{e_1}^{\mathrm{\prime }}=k{I}_2{k}^{-1}=I_2$, whose defining-representation trace is $2$. The original product is also $I_2$ and has trace $2$; this concrete calculation agrees with [F2].

The defining action is a group homomorphism, preserves the standard inner product because $g^{†}g=I_2$, and is strongly continuous because its matrix entries are continuous. The trivial reverse-edge actions have these properties as well. By [F1], their dual factors at $v$ are trivial one-dimensional spaces, so the displayed $W_v$ and $\iota$ are well-defined and $R_v(g)\iota =\iota (\rho (g)\otimes \rho (g))$. For $g\in SU(2)$, write the matrix of $\rho (g)$ in the chosen basis as $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$; its determinant is $\det g=1$. Expanding the tensor product gives $(g\otimes g)ϵ=(a{e}_1+c{e}_2)\otimes (b{e}_1+d{e}_2)-(b{e}_1+d{e}_2)\otimes (a{e}_1+c{e}_2)=(ad-bc)ϵ=(\det g)ϵ=ϵ$. The tensor is nonzero because $e_1\otimes e_2$ and $e_2\otimes e_1$ are distinct basis tensors.

By [F1], $P_v\iota (ϵ)={\int }_{SU(2)}{R}_v(g)\iota (ϵ)d\mu (g)$. Step 1.1 makes the integrand the constant vector $\iota (ϵ)$, and the normalized Haar probability has total mass one; hence the integral equals $\iota (ϵ)$. Applying ${\iota }^{-1}$ gives the equivalent asserted formula.