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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.