is the discrete set $\{u\in I : S_{\gamma}|_u=0\}$. On $\widetilde{I}$ the fifth and fourth-order osculating bundles are spanned by $A_0,A_1,A_2,P,A_{n+1}$. This implies the existence of a smooth function $f:\widetilde{I}\to \R$ such that $$\dot{P}|_{\widetilde{I}}=fP|_{\widetilde{I}},\quad \mathrm{mod}(A_0,A_1,A_2,A_{n+1}),$$ From (\[STR1\]) and taking into account the previous identity we get $$\label{CC}\frac{d}{du}(A_0\wedge A_1\wedge A_2\wedge P\wedge A_{n+1})|_{\widetilde{I}}= f( A_0\wedge A_1\wedge A_2\wedge P\wedge A_{n+1})|_{\widetilde{I}}.$$ Let $\mathrm{Gr}_{5}(\R^{n+2})$ be the Grassmannian of the $5$-dimensional vector subspace of $\R^{n+2}$. From (\[CC\]) it follows that the map $$u\in I\to \mathrm{Span}(A_0|_u\wedge A_1|_u\wedge A_2|_u\wedge P|_u\wedge A_{n+1}|_u)\in \mathrm{Gr}_{5}(\R^{n+2})$$ is constant on $\widetilde{I}$. By continuity is constant on $I$. Then, the fourth-order osculating spaces coincide with a fixed $5$-dimensional Lorentzian subspace $\mathbb{W}_{\gamma}\subseteq \R^{2,n}$. Hence, the trajectory of $\gamma$ is contained in the $3$-dimensional Lorentzain cycle $\mathcal{E}(\mathbb{W}_{\gamma})$.

This theorem shows that if $\gamma$ is a conformal world-line then three possibilities may occur :

-   $\gamma$ is trapped in a $4$-dimensional Einstein universe and there are no 3-dimensional Lorentzian cycles containing the trajectory of $\gamma$. If this occurs, we say that $\gamma$ is a [*linearly full conformal world-line*]{}.

-   the trajectory is trapped in a $3$-dimensional Lorentzian cycle but does not lie in any $2$-dimensional Lorentzian cycle.

-   the trajectory is trapped in a $2$-dimensional Lorentzian cycle.

Linearly 