$t\in I$. For each $k=1,...,n+1$, the [*$k$-th osculating space*]{} of $\gamma$ at $\gamma(t)$ is the linear subspace $\mathcal{T}^k(\gamma)|_t\subset \R^{2,n}$ spanned by the vectors $\Gamma|_t,\Gamma'|_t,\dots ,\Gamma^k|_t$. Obviously the definition does not depend on the choice of the null lift. Putting together all the osculating spaces we get the k-th osculating sheaf $$\mathcal{T}^{k}(\gamma)=\{(t,V)\in I\times \R^{2,n} : V\in \mathcal{T}^k(\gamma)|_t\}.$$

[$\mathcal{T}^2(\gamma)|_t$ is a $3$-dimensional Lorentzian subspace of $\R^{2,n}$, for every $t\in I$.]{}

[Choose $\Gamma$ so that $\langle \Gamma',\Gamma'\rangle = -1$. Differentiating $\langle \Gamma,\Gamma'\rangle = 0$ and $\langle \Gamma',\Gamma'\rangle = -1$ we get $\langle \Gamma,\Gamma''\rangle = 1$ and $\langle \Gamma',\Gamma''\rangle = 0$. This implies that $\Gamma|_t,\Gamma'|_t$ and $\Gamma''|_t$ are linearly independent, for each $t\in I$. We put $$A_1= \frac{1}{2}(1+\langle \Gamma'',\Gamma''\rangle)\Gamma - \Gamma'',\quad A_2=\Gamma' \quad A_3=\frac{1}{2}(1-\langle \Gamma'',\Gamma''\rangle)\Gamma + \Gamma''.$$ Then, $(A_1|_t,A_2|_t,A_3|_t)$ is a basis of $\mathcal{T}^2(\gamma)|_t$ such that $$\begin{split}
 \langle A_i|_t,A_j|_t\rangle &= 0,\quad i\neq j,\\
\langle A_1|_t,A_1|_t\rangle &= \langle A_2|_t,A_2|_t\rangle=-\langle A_3|_t,A_3|_t\rangle = -1.
   \end{split}$$ This yields the result.]{}

The Lemma implies that $\mathcal{T}^k(\gamma)|_t$ is a Lorentzian subspace of $\R^{2,n}$, for every $k\ge 2$ and every $t\in I$. The Lorentzian cycle $\mathcal{E}(\mathcal{T}^k(\gamma)|_t)$ is denoted by $\mathcal{E}^k(\gamma)|_t$. We call $\mathcal{E}^k(\gamma)|_t$ the [*$k$-th osculating cycle*]{} of $\gamma$ at $\gamma(t)$. The orthogonal complement $\mathcal{N}^k(\gamma)|_t$ of $\mathcal{T}^k(\gamma)|_t$ is a space-like vector 