# Flow of vector field on semi-Riemannian manifold

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 $$t$$.

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.