We assume that $m = 0$, that is $$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$ and that moreover there exists $\lambda > 0$ such that $$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$. Show that for all $x \in \mathbb{R}$, we have $\|T^{\prime}(x)\| = \lambda$.
We assume that $m = 0$, that is
$$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$
We assume that $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$ satisfies
$$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$
and that moreover there exists $\lambda > 0$ such that
$$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$
We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$.
Show that for all $x \in \mathbb{R}$, we have $\|T^{\prime}(x)\| = \lambda$.