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Consider $\mathbb{R}^{n+1}$ with the metric given by $$ g(x,x) = 2x_1x_2 + \sum_{i=3}^{n+1}x_i^2 $$ and $M$ the set of $x$ such that $g(x,x)=1$. Further, take a basis $e_1,...,e_n$ for $\mathbb{R}^{n+1}$ such that $g(e_1,e_1) = -1$ let $Y$ be the vector field on $M$ given by $Y(z) = e_2 - g(z,e_2)z$.

I want to prove that $t \mapsto \alpha_z(t)$ defines the flow of $Y$ with $$\alpha_z(t) = \frac{1}{1+g(z,e_2)t} (z + \frac{t}{2} (2 + g(z,e_2)t )e_2 ).$$

So far, I have showed that $$ \dot{\alpha}_z(t) = e_2 - \frac{1}{(1+g(z,e_2)t)^2}g(z,e_2)z $$ which looks somewhat familiar. I proceeded to compute $Y(\alpha_z(t))$ but the resulting term is quite large and does not look very helpfull.

As of now, all I have is $\dot{\alpha}_z(0) = Y(\alpha_z(0))$ but I do not know how to prove it for arbitrary $t$.

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I managed to solve it if that interests someone. The key was that $g(e_2,e_2)=0$. Using this the large expression became much more concise and a few computations show $$ Y(\alpha_z(t)) = (2 - \frac{1}{\delta})e_2 - \frac{1}{\delta}g(z,e_2)z$$ where $\delta = (1 + g(z,e_2)t)^2$. Further, it also holds that $$ 2 - \frac{1}{\delta} = 1$$ thus yielding the desired assertion.

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