and $S_{\gamma}:I\to \mathcal{N}^3(\gamma)$ be the cross section defined as in (\[SS1\]). We set $$B_3=S_{\gamma}/\sqrt{\langle S_{\gamma},S_{\gamma}\rangle}$$ and we pick the unique unit cross section $B_4:I\to \mathcal{N}^4(\gamma)$ such that $\mathrm{det}(M_0,M_1,M_2,M_3,B_4,M_5)>0$. Putting $$B_0=M_0,\quad B_1=M_1,\quad B_2=M_2,\quad B_5=M_5,$$ the map $B=(B_0,\dots, B_5)$ is a second-order frame along $\gamma$. In view of (\[LS\]) we have $$\label{FR1}\dot{B}_0=B_1,\quad \dot{B}_1=-k_1B_0-B_5,\quad \dot{B}_5=k_1B_1+B_2$$ where[^6] $k_1=h_1$. Since $D$ is a metric covariant derivative, $DB_3$ is a multiple of $B_4$ and hence $DB_3=k_3B_4$, for some non-zero function $k_3$. From (\[LS\]) we have $$\label{FR2}\dot{B}_2=B_0+k_2B_3,$$ where $k_2=\sqrt{\langle S_{\gamma},S_{\gamma}\rangle}>0$. Keeping in mind (\[EL3\]) we get $$\label{FR3}\dot{B}_3=-k_2B_2+k_3B_4.$$ Furthermore, since $B$ takes values in $\mathrm{A}^{\uparrow}_{+}(2,4)$ and using (\[FR1\])-(\[FR3\]) we obtain $$\label{FR4}\dot{B}_4=-k_3B_3.$$ Combining (\[FR1\])-(\[FR4\]) we deduce that $B$ satisfies (\[mcc\]).

Suppose that the functions $k_1,k_2$ and $k_3$ are non-constant. The first and second covariant derivatives of $S_{\gamma}$ can be written as $$DS_{\gamma}=\dot{k}_2B_3+k_2k_3B_4,\quad D^2S_{\gamma}=(\ddot{k}_2-k_2k_3^2)B_3+(2\dot{k}_2k_3+k_3\dot{k}_2)B_4.$$ Then, from variational equation $W_{\gamma}=0$ we get $$\label{EM2}
\ddot{k}_2=k_2(k_3^2+k_2^2-2k_1),\quad k_2\dot{k}_3+2k_3\dot{k}_2=0\quad \dot{k}_1=3k_2\dot{k}_2.$$ The third and the second equations in (\[EM2\]) imply the existence of two constants $c_1$ and $c_3\neq 0$ such that $$\label{EM3}k_1=\frac{3}{2}k_2^2+c_1,\quad k_3=c_3k_2^{-2}.$$ Substituting (\[EM3\]) into (\[EM2\]) we find $$\ddot{k}_2=-2k_2^{3}+c_3^2k^{-3}-4k_2-2c_1k_2.$$ Then there exist a constant $c_2$ such that $$\dot{k}_2^2+k_2^4+c_3^2k_{2}^{-2}+2c_1k_2^2+c_2=0.$$

[We call $B$ the [*canonical conformal frame*]{}. The functions $k_1,k_2,k_3$ are said the [*conformal curvatures*]{}. The constants 