$u\in I$

-   Case II : $s\wedge s_{(1)}$ vanishes identically.

[*Case I*]{}. The vectors fields $S_{\gamma}$ and $DS_{\gamma}$ are everywhere linearly independent. Then, the osculating spaces $\mathcal{T}^5(\gamma)|_u$ are six-dimensional. Using (\[STR1\]) and (\[EL3\]) we obtain $$\frac{d}{du}(A_0\wedge A_1\wedge A_2\wedge S_{\gamma}\wedge DS_{\gamma}\wedge A)=0.$$ Hence, $\mathcal{T}^5(\gamma)|_u$ coincides with a fixed $6$-dimensional Lorentzian subspace $\mathbb{W}_{\gamma}\subseteq \R^{2,n}$, for each $u\in I$. This implies that the trajectory of $\gamma$ is contained in the $4$-dimensional Lorentzian cycle $\mathcal{E}(\mathbb{W}_{\gamma})$.

[*Case II*]{}. Since $(s,s_{(1)})$ is a solution of the linear system (\[EL2\]), there are two possibilities :

-   Case II.1 : $s=s_{(1)}=0$;

-   Case II.2 : $s^t\cdot s+^ts_{(1)}\cdot s_{(1)}>0$.

[*Case II.1*]{}. If $s=\dot{s}=0$, then (\[STR1\]) implies $$\frac{d}{du}(A_0\wedge A_1\wedge A_2\wedge A_{n+1})=0.$$ Therefore, $\mathcal{T}^3(\gamma)|_u$ coincides with a fixed $4$-dimensional Lorentzian subspace $\mathbb{W}_{\gamma}\subseteq \R^{2,n}$, for every $u\in I$. Hence, the trajectory of $\gamma$ is contained in the $2$-dimensional Lorentzain cycle $\mathcal{E}(\mathbb{W}_{\gamma})$.

[*Case II.2* ]{} If $s^t\cdot s+^ts_{(1)}\cdot s_{(1)}>0$ and $s\wedge s_{(1)}=0$, then $\mathrm{Span}(S_{\gamma}|_u,DS_{\gamma}|_u)$ is a $1$-dimensional space-like subspace, for every $u\in I$. Thus $$\mathcal{P}=\{(u,V)\in I\times \R^{2,n}: V\in \mathrm{Span}(S_{\gamma}|_u,DS_{\gamma}|_u)\}$$ is a rank $1$ vector bundle. Let $P$ be a unit-length cross section of $\mathcal{P}$. The identity $s\wedge s_{(1)}=0$ implies that $S_{\gamma}$ and $DS_{\gamma}$ are both proportional to $P$. We put $\widetilde{I}=I-I_*$, where $I_*$ 