How to change coordinates of a vectorfield $X$ on $\mathbb{R}P^2$

Solution 1:

The key point is that $$v_1=\frac{1}{u_1},\; v_2=\frac{u_2}{u_1}$$ From there you get $$\frac{\partial }{\partial u_1}=-\frac{1 }{u_1^2}\frac{\partial }{\partial v_1}+(-\frac{u_2 }{u_1^2})\frac{\partial }{\partial v_2},\;\operatorname {so that}\; u_1 \frac{\partial }{\partial u_1}=-\frac{1 }{u_1} \frac{\partial }{\partial v_1} -\frac{u_2}{u_1}\frac{\partial }{\partial v_2}$$ $$\frac{\partial }{\partial u_2}=0. \frac{\partial }{\partial v_1} +\frac{1 }{u_1} \frac{\partial }{\partial v_2}, \; \operatorname {so that}\; 2u_2 \frac{\partial }{\partial u_2}=2\frac{u_2 }{u_1} \frac{\partial }{\partial v_2} $$ Hence $$X= -\frac{1 }{u_1} \frac{\partial }{\partial v_1} -\frac{u_2}{u_1}\frac{\partial }{\partial v_2}+2\frac{u_2 }{u_1} \frac{\partial }{\partial v_2}= -\frac{1 }{u_1} \frac{\partial }{\partial v_1} +\frac{u_2}{u_1}\frac{\partial }{\partial v_2}$$ The desired result is thus finally $$X=-v_1\frac {\partial}{\partial v_1} + v_2\frac {\partial}{\partial v_2} . $$