Vector field on sphere I want to find the gradient vector field and flows of the function $f=x^2+2y^2+3z^2$ on the sphere $S^2$, however I've not done this in a while so would appreciate a bit of help. I'd like to see the vector field in the $(u,v)$ plane. I've parametrised the sphere with:
$\varphi_1(u,v)=\left(\pm \sqrt{1-u^2-v^2},u,v \right)$,
$\varphi_2(u,v)=\left(u,\pm \sqrt{1-u^2-v^2},v \right)$,
$\varphi_3(u,v)=\left(u,v,\pm \sqrt{1-u^2-v^2} \right)$.
So:
$\left(f \circ \varphi_1 \right)(u,v)=1+u^2+2v^2$,
$\left(f \circ \varphi_2 \right)(u,v)=2-u^2-2v^2$,
$\left(f \circ \varphi_3 \right)(u,v)=3-2u^2-v^2$
Thus:
$d(f \circ \varphi_1)=(2u,4v)$,
$d(f \circ \varphi_2)=(-2u,-4v)$
$d(f \circ \varphi_3)=(-4u,-2v)$
Now I'm stuck. This isn't giving me anything that looks sensible. I feel that my problem is in the parametrisation.
Thanks for any help.
 A: I will not solve the exercise, but will give two suggestions that might simplify your work. 
First, I suggest a different parametrization of $S^2$, called the stereographic projection (see Wikipedia): let $w=u+iv$, then $(x,y,z)=\varphi(w)=(2w,|w|^2-1)/(1+|w|^2).$ The advantage of this parametrization is that it is conformal, ie the metric induced by $S^2$ in the $u,v$ plane is the standard metric $du^2+dv^2$ multiplied by some function  $\lambda(u,v)=4/(1+|w|^2)^2$. It follows that the gradient of a function $f$ on $S^2$, expressed in the $u,v$ coordinates, is the gradient of $f(u,v)$ wrt the standard metric $du^2+dv^2$, multiplied by $1/\lambda$. Apart from being easy to calculate, this feature has the advantage that multiplying a vector field by a (non vanishing) function does not change its flow lines, just their time parametrization, so we can draw them by considering the usual gardient $(f_u, f_v)$ wrt the standard metric. 
The 2nd suggestion, before embarking on  the calculation, is to note that you can simplify the calculation by first interchanging the labeling of the $y$ and $z$ axes, so your  $f$ becomes $x^2+3y^2+2z^2$, then substract  $2(x^2+y^2+z^2)$ from your $f$. This quantity you substract is constant on $S^2$, so does not affect the gradient of $f$, but the new expression  to handle is $y^2-x^2$, which is simpler, especialy for the above parametrization (you can write it as Re$(\lambda w^2)/4$).  
