I have to solve this one:
"Does $Y(x,y,z)=(xz,yz,z^2-1)$ is a smooth vector field on $S^2$?"
First of all $Y$ is smooth since I can compute all the partial derivatives. For any $p\in S^2$ we have that $T_p S^2=\{v\in\mathbb{R}^3$ such that $p\cdot v=0\}$. If we make the identification $Y=xz\frac{\partial}{\partial x}+yz\frac{\partial}{\partial y}+(z^2-1)\frac{\partial}{\partial z}\leftrightarrow (xz,yz,z^2-1)\in\mathbb{R}^3$ we have that for every $p=(x,y,z)\in S^2$
\begin{equation} (x,y,z)\cdot(xz,yz,z^2-1)=x^2z+y^2z+z(z^2-1)=z(x^2+y^2+z^2-1)=0. \end{equation}
So for every $p\in S^2$, $Y(p)\in T_pS^2$ and so $Y$ is a vector field on $S^2$.
Is that all right?
Any other idea?
Thanks